Explanation:
Okay, let's tackle this problem step by step. So, we have a quadratic polynomial P(x) with real coefficients. It's given that for all real numbers x, the inequality x² - 2x + 2 ≤ P(x) ≤ 2x² - 4x + 3 holds. Additionally, we know that P(11) = 181, and we need to find P(16).

First, since P(x) is a quadratic polynomial, let's denote it as P(x) = ax² + bx + c, where a, b, c are real numbers. Our goal is to find the coefficients a, b, c such that the inequalities hold for all real x, and then use the given value P(11) = 181 to determine them. Once we have the coefficients, we can compute P(16).

But before jumping into solving for a, b, c, let's analyze the inequalities given. The polynomial P(x) is squeezed between two quadratics: the lower bound is x² - 2x + 2 and the upper bound is 2x² - 4x + 3. Since both bounds are quadratics, and P(x) is also a quadratic, it's possible that P(x) is a linear combination of these two quadratics or maybe they touch each other at certain points?

Wait, but since the inequalities must hold for all real numbers x, the difference between P(x) and the lower bound must be non-negative for all x, and similarly, the difference between the upper bound and P(x) must also be non-negative for all x. So, that suggests that P(x) - (x² - 2x + 2) ≥ 0 for all x and (2x² - 4x + 3) - P(x) ≥ 0 for all x.

Therefore, P(x) is sandwiched between these two quadratics. For the difference between two quadratics to be non-negative for all x, the resulting quadratic must be a perfect square or something that opens upwards? Wait, actually, if we have two quadratics, and their difference is non-negative for all x, then the difference must be a non-negative quadratic. Similarly for the upper bound.

But since both inequalities must hold for all real x, perhaps P(x) is equal to both bounds at some points, and maybe touches them? Let me think. If P(x) is between two quadratics for all x, and it's a quadratic itself, then perhaps P(x) is a linear combination of the two bounds?

Alternatively, maybe the two bounds are touching P(x) at their minimum or maximum points? For example, if the lower bound is x² - 2x + 2, which can be rewritten as (x - 1)^2 + 1. Similarly, the upper bound is 2x² - 4x + 3. Let me check that.

Let me compute the lower bound first: x² - 2x + 2. Completing the square: x² - 2x + 1 + 1 = (x - 1)^2 + 1. So that's a parabola opening upwards with vertex at (1, 1). The upper bound is 2x² - 4x + 3. Let me complete the square for that as well. Factor out the 2: 2(x² - 2x) + 3. Then complete the square inside the parentheses: x² - 2x + 1 - 1. So that becomes 2[(x - 1)^2 - 1] + 3 = 2(x - 1)^2 - 2 + 3 = 2(x - 1)^2 + 1. So, the upper bound is 2(x - 1)^2 + 1, which is also a parabola opening upwards with vertex at (1, 1). Interesting, both bounds have the same vertex at (1, 1), but different leading coefficients. The lower bound has a leading coefficient of 1, and the upper bound has 2.

So, since both bounds have the same vertex at (1, 1), and since they are both opening upwards, the lower bound is "wider" (since coefficient is smaller) and the upper bound is "narrower" (since coefficient is larger). Therefore, the region between them is the area between these two parabolas. Now, P(x) is a quadratic polynomial sitting between them for all real x.

Now, since both bounds are parabolas with the same vertex, perhaps P(x) is also a parabola with the same vertex? If that's the case, then maybe P(x) is a linear combination of the lower and upper bounds? Let me think.

Alternatively, since the region between the two bounds is between two parabolas with the same vertex, maybe P(x) is tangent to both bounds? Wait, but if P(x) is squeezed between them for all x, and they have the same vertex, maybe P(x) has the same vertex? Let me check.

If P(x) is between the two bounds for all x, then at the vertex x=1, the lower bound and upper bound both equal 1. Therefore, P(1) must be between 1 and 1, so P(1) = 1? Wait, because at x=1, both bounds equal (1)^2 - 2(1) + 2 = 1 - 2 + 2 = 1, and similarly the upper bound is 2(1)^2 - 4(1) + 3 = 2 - 4 + 3 = 1. Therefore, at x=1, both bounds equal 1, so P(1) must equal 1. Therefore, P(1) = 1.

So that gives us a point that P(x) must pass through: (1, 1). Also, since P(x) is a quadratic, maybe the vertex of P(x) is also at x=1? Because if P(x) is between two parabolas that both have their minimum at x=1, then if P(x) is also a parabola, its vertex could be at x=1. However, the upper bound is a narrower parabola, so if P(x) is between them, then maybe P(x) is also symmetric around x=1?

Alternatively, maybe P(x) is of the form k(x - 1)^2 + 1, where k is between 1 and 2? Because if you take k between 1 and 2, then k(x - 1)^2 + 1 would lie between the lower bound (1*(x - 1)^2 + 1) and upper bound (2*(x - 1)^2 + 1). However, that seems plausible. But let me check.

Wait, if that's the case, then P(x) = k(x - 1)^2 + 1, with 1 ≤ k ≤ 2. Then, for example, P(1) = 1, which matches. However, then we can use the condition P(11) = 181 to find k. Let me try that approach.

Let me compute P(11). If P(x) = k(x - 1)^2 + 1, then P(11) = k(10)^2 + 1 = 100k + 1. According to the problem, this is equal to 181. Therefore:

100k + 1 = 181
100k = 180
k = 1.8

Which is 9/5. Since 9/5 is between 1 and 2, that seems valid. Then, P(x) would be (9/5)(x - 1)^2 + 1. Then, to find P(16), we compute:

P(16) = (9/5)(15)^2 + 1 = (9/5)(225) + 1 = 9 * 45 + 1 = 405 + 1 = 406.

But wait, before accepting this answer, let me check if this approach is valid. The assumption was that P(x) is of the form k(x - 1)^2 + 1. But is this necessarily the case?

The problem says that for all real numbers x, the inequality holds. So if we assume that P(x) is a quadratic that lies between the two bounds, which are both of the form (x - 1)^2 + 1 and 2(x - 1)^2 + 1, then perhaps P(x) is a multiple of (x - 1)^2 + 1? But actually, the lower bound is (x - 1)^2 + 1 and upper bound is 2(x - 1)^2 + 1. However, the difference between the upper and lower bound is (2(x - 1)^2 + 1) - ((x - 1)^2 + 1) = (x - 1)^2. Therefore, the region between them is all quadratics that can be written as (x - 1)^2 + 1 + t(x - 1)^2, where t is between 0 and 1? Wait, but that would be (1 + t)(x - 1)^2 + 1. However, the upper bound is when t = 1, leading to 2(x - 1)^2 + 1 and lower bound when t = 0. However, in this case, the coefficient of (x - 1)^2 is varying between 1 and 2. Therefore, if P(x) is of the form k(x - 1)^2 + 1, with k in [1, 2], then that would fit between the two bounds.

But is this the only possibility? Suppose that P(x) is not of this form. For example, suppose that P(x) is a quadratic that is not symmetric around x=1. Then, could it still lie between the two bounds for all x?

Alternatively, perhaps we need to consider that the difference between P(x) and the lower bound is a non-negative quadratic, and the difference between the upper bound and P(x) is also a non-negative quadratic. So, let's denote:

Let me define Q(x) = P(x) - (x² - 2x + 2) ≥ 0 for all x, and R(x) = (2x² - 4x + 3) - P(x) ≥ 0 for all x.

Since Q(x) and R(x) are quadratics that are always non-negative, they must be perfect squares or have discriminant less than or equal to zero. Wait, but Q(x) is a quadratic. If Q(x) is non-negative for all x, then its discriminant must be ≤ 0, and its leading coefficient must be ≥ 0. Similarly for R(x).

Let me compute Q(x):

Q(x) = P(x) - (x² - 2x + 2) = (ax² + bx + c) - x² + 2x - 2 = (a - 1)x² + (b + 2)x + (c - 2).

Similarly, R(x) = (2x² - 4x + 3) - P(x) = 2x² - 4x + 3 - ax² - bx - c = (2 - a)x² + (-4 - b)x + (3 - c).

So, for Q(x) to be non-negative for all x, we need:

1. The leading coefficient (a - 1) ≥ 0, so a ≥ 1.

2. The discriminant of Q(x) is ≤ 0. The discriminant D_Q = (b + 2)^2 - 4*(a - 1)*(c - 2) ≤ 0.

Similarly, for R(x) to be non-negative for all x:

1. The leading coefficient (2 - a) ≥ 0 ⇒ a ≤ 2.

2. The discriminant of R(x) is ≤ 0. The discriminant D_R = (-4 - b)^2 - 4*(2 - a)*(3 - c) ≤ 0.

Therefore, combining the two conditions from the leading coefficients, we have 1 ≤ a ≤ 2.

Additionally, the discriminants D_Q and D_R must be ≤ 0.

Furthermore, since P(x) is a quadratic polynomial, and we also know that P(1) = 1. Let me check that. Earlier, we saw that since at x=1, both bounds equal 1, so P(1) must be between 1 and 1, hence P(1) = 1. Let me confirm that.

Yes, if x=1, then lower bound is 1 and upper bound is 1. Therefore, P(1) must be exactly 1. Therefore, substituting x=1 into P(x):

P(1) = a(1)^2 + b(1) + c = a + b + c = 1.

So that gives us another equation: a + b + c = 1.

So now, we have the following constraints:

1. 1 ≤ a ≤ 2.

2. For Q(x): (a - 1)x² + (b + 2)x + (c - 2) ≥ 0 for all x ⇒ discriminant D_Q ≤ 0 and a ≥ 1.

3. For R(x): (2 - a)x² + (-4 - b)x + (3 - c) ≥ 0 for all x ⇒ discriminant D_R ≤ 0 and a ≤ 2.

Additionally, a + b + c = 1.

Moreover, we have the value P(11) = 181, which gives:

a*(11)^2 + b*(11) + c = 121a + 11b + c = 181.

So, we have equations:

Equation (1): a + b + c = 1

Equation (2): 121a + 11b + c = 181

And the discriminant conditions for Q(x) and R(x).

This seems a system of equations with variables a, b, c, but also the discriminant conditions. However, maybe we can find relations between them. Let me first try to express b and c in terms of a, using equations (1) and (2).

From equation (1): c = 1 - a - b.

Substitute into equation (2):

121a + 11b + (1 - a - b) = 181
Simplify:

121a + 11b + 1 - a - b = 181
(121a - a) + (11b - b) + 1 = 181
120a + 10b + 1 = 181
Subtract 1:

120a + 10b = 180
Divide both sides by 10:

12a + b = 18
So, equation (3): b = 18 - 12a.

Now, since we have c = 1 - a - b, substitute b:

c = 1 - a - (18 - 12a) = 1 - a - 18 + 12a = (1 - 18) + (-a + 12a) = -17 + 11a.

Therefore, c = 11a - 17.

So, now we have expressions for b and c in terms of a. So:

b = 18 - 12a
c = 11a - 17.

Now, we need to ensure that the discriminants D_Q and D_R are less than or equal to zero.

First, let's compute D_Q. Recall:

Q(x) = (a - 1)x² + (b + 2)x + (c - 2).

So, substituting b and c:

First, compute the coefficients:

Leading coefficient: a - 1.

Middle term coefficient: b + 2 = (18 - 12a) + 2 = 20 - 12a.

Constant term: c - 2 = (11a - 17) - 2 = 11a - 19.

Therefore, discriminant D_Q is:

(20 - 12a)^2 - 4*(a - 1)*(11a - 19) ≤ 0.

Similarly, for R(x):

R(x) = (2 - a)x² + (-4 - b)x + (3 - c).

Compute coefficients:

Leading coefficient: 2 - a.

Middle term coefficient: -4 - b = -4 - (18 - 12a) = -4 - 18 + 12a = -22 + 12a.

Constant term: 3 - c = 3 - (11a - 17) = 3 - 11a + 17 = 20 - 11a.

Therefore, discriminant D_R is:

(-22 + 12a)^2 - 4*(2 - a)*(20 - 11a) ≤ 0.

So, we need to compute D_Q ≤ 0 and D_R ≤ 0. Let me compute D_Q first.

Compute D_Q:

(20 - 12a)^2 - 4*(a - 1)*(11a - 19) ≤ 0.

First, expand (20 - 12a)^2:

= 400 - 480a + 144a².

Then, compute 4*(a - 1)*(11a - 19):

First compute (a - 1)(11a - 19) = 11a² -19a -11a +19 = 11a² -30a +19.

Multiply by 4: 44a² -120a +76.

Therefore, D_Q = [400 - 480a + 144a²] - [44a² -120a +76] = 400 - 480a + 144a² -44a² +120a -76.

Simplify:

Combine like terms:

a²: 144a² - 44a² = 100a²

a terms: -480a + 120a = -360a

Constants: 400 - 76 = 324.

So D_Q = 100a² - 360a + 324 ≤ 0.

Let me factor this quadratic expression:

First, check if it's a perfect square. Let me compute discriminant of D_Q as a quadratic in a:

Discriminant D = (-360)^2 - 4*100*324.

Wait, but wait, we already know that D_Q is 100a² - 360a + 324. Let me check whether this is a perfect square.

Let me see:

Suppose 100a² - 360a + 324 = (10a - k)^2. Let's compute:

(10a - k)^2 = 100a² - 20k a + k².

Comparing to 100a² - 360a + 324:

So:

-20k = -360 ⇒ k = 18.

Then k² = 324. Which matches. Perfect! Therefore, D_Q = (10a - 18)^2 ≤ 0.

But a square is always ≥ 0, so (10a - 18)^2 ≤ 0 implies that (10a - 18)^2 = 0, so 10a - 18 = 0 ⇒ a = 18/10 = 9/5 = 1.8.

So, D_Q = 0 only when a = 9/5. Therefore, this gives us that a must be 9/5 for D_Q ≤ 0. Since the discriminant is a square, which is ≤ 0 only when it's zero. Therefore, this condition forces a = 9/5.

Similarly, let's check the discriminant D_R.

Recall D_R is:

(-22 + 12a)^2 - 4*(2 - a)*(20 - 11a) ≤ 0.

First, let me compute each part.

First, (-22 + 12a)^2 is the same as (12a -22)^2 = [same as (22 - 12a)^2] which is equal to:

(12a - 22)^2 = 144a² - 528a + 484.

Second, compute 4*(2 - a)*(20 - 11a):

First compute (2 - a)(20 - 11a):

Multiply out: 2*20 + 2*(-11a) - a*20 + a*11a = 40 -22a -20a + 11a² = 11a² -42a +40.

Multiply by 4: 44a² -168a +160.

Therefore, D_R is [144a² - 528a + 484] - [44a² -168a +160] = 144a² -528a +484 -44a² +168a -160.

Simplify:

a² terms: 144a² -44a² = 100a²

a terms: -528a + 168a = -360a

constants: 484 -160 = 324.

Therefore, D_R = 100a² - 360a + 324 ≤ 0.

Wait! That's the same expression as D_Q! So D_R is also equal to 100a² - 360a + 324, which we saw is (10a - 18)^2. Hence, the same condition applies. Therefore, D_R ≤ 0 implies (10a - 18)^2 ≤ 0 ⇒ a = 9/5.

Therefore, from both discriminant conditions, we must have a = 9/5. So, that's our value for a.

Therefore, a = 9/5. Then, from equation (3): b = 18 - 12a = 18 - 12*(9/5) = 18 - 108/5 = (90/5 - 108/5) = (-18/5).

Similarly, c = 11a - 17 = 11*(9/5) - 17 = 99/5 - 85/5 = 14/5.

So, now we have:

a = 9/5, b = -18/5, c = 14/5.

So, the polynomial P(x) is:

P(x) = (9/5)x² - (18/5)x + 14/5.

Alternatively, we can write this as P(x) = (9x² - 18x + 14)/5.

Let me check if this satisfies the original inequalities. For example, let's check at x = 1:

P(1) = (9 - 18 + 14)/5 = (5)/5 = 1. Which matches our earlier conclusion.

Also, check if the inequalities hold for all x. Since we derived these coefficients under the condition that the discriminant is zero, meaning that Q(x) and R(x) are perfect squares. Therefore, Q(x) = (10a - 18)^2 = 0 when a = 9/5, so Q(x) must be a perfect square?

Wait, earlier we found that D_Q = (10a - 18)^2 ≤ 0. Since a = 9/5, then 10a = 18, so D_Q = 0. Therefore, Q(x) is a perfect square. Similarly for R(x). Therefore, Q(x) is a perfect square, meaning that P(x) - (lower bound) is a perfect square, hence non-negative for all x. Similarly for R(x).

Therefore, this polynomial is indeed between the two bounds for all real x. So, our coefficients are correct.

Now, let's confirm that this polynomial satisfies P(11) = 181. Let's compute:

P(11) = (9*(11)^2 - 18*(11) + 14)/5 = (9*121 - 198 + 14)/5.

Compute numerator:

9*121: 9*120 + 9*1 = 1080 + 9 = 1089

Then, 1089 - 198 = 891

891 +14 = 905

Then 905 / 5 = 181. Perfect, matches the given condition.

Therefore, our polynomial is correct. Now, the question is to find P(16).

So, compute P(16):

P(16) = [9*(16)^2 - 18*(16) + 14]/5.

First, compute 16² = 256

Multiply by 9: 9*256. Let me compute 256*9:

Well, 250*9 = 2250, and 6*9=54, so total is 2250 + 54 = 2304.

Then, compute 18*16: 18*10 + 18*6 = 180 + 108 = 288.

So numerator is 2304 - 288 + 14.

Compute 2304 - 288: Let me subtract 288 from 2304.

2304 - 200 = 2104
2104 - 88 = 2016

Then add 14: 2016 + 14 = 2030.

Therefore, numerator is 2030, so P(16) = 2030 / 5 = 406.

So, P(16) = 406.

Therefore, the answer is \boxed{406}.

But let me just check once again. Alternatively, since we had earlier thought that P(x) = (9/5)(x - 1)^2 + 1, let me verify that this is the same as our coefficients.

Expanding (9/5)(x - 1)^2 + 1:

First, (x - 1)^2 = x² - 2x + 1

Multiply by 9/5: (9/5)x² - (18/5)x + 9/5

Add 1: (9/5)x² - (18/5)x + 9/5 + 5/5 = (9/5)x² - (18/5)x + 14/5. Yes, exactly the same as our coefficients. Therefore, this confirms that our polynomial is correct. Therefore, using this form, P(16) = (9/5)(15)^2 + 1 = (9/5)(225) + 1 = 405 + 1 = 406. Same result. So that's consistent.

Therefore, the answer is 406.

But just to be thorough, let me check if there's another possible solution. But given that both discriminants forced a = 9/5, and given that we have unique solution for a, hence unique solution for b and c. Therefore, there is only one such quadratic polynomial that satisfies the given conditions, so P(16) must be 406.

Alternatively, maybe there's another approach? Let me think.

Another way is to consider that since both the lower and upper bounds have the same vertex at x=1, and that P(x) is squeezed between them. Given that at x=1, P(1)=1, and since the bounds have the same value there, P(x) might be tangent to both bounds at x=1? Wait, but since both bounds have the same vertex, and P(x) is in between, maybe P(x) is tangent to both bounds at that point?

If that's the case, then maybe the derivatives at x=1 are equal to the derivatives of both bounds. Let me check.

Let me compute the derivatives of lower bound, upper bound, and P(x).

Lower bound: f(x) = x² - 2x + 2. f’(x) = 2x - 2. At x=1, f’(1) = 2(1) - 2 = 0.

Upper bound: g(x) = 2x² - 4x + 3. g’(x) = 4x - 4. At x=1, g’(1) = 4(1) - 4 = 0.

Therefore, both bounds have derivative zero at x=1. Similarly, if P(x) is tangent to both bounds at x=1, then P’(1) must also be zero. Let's check our polynomial's derivative.

P(x) = (9/5)x² - (18/5)x + 14/5. Then P’(x) = (18/5)x - 18/5. At x=1, P’(1) = (18/5)(1) - 18/5 = 0. Which matches. So indeed, P(x) has a critical point at x=1, same as both bounds. Therefore, this suggests that P(x) touches both bounds at x=1, meaning it's tangent to both at that point. Therefore, since both bounds have the same vertex, and P(x) also has the same vertex, and lies between them, this seems to be the only possibility. Therefore, this approach confirms that our previous result is correct.

Alternatively, if we think about the difference between P(x) and the lower bound, which is a perfect square, so Q(x) = [something]^2. Since Q(x) = (a - 1)x² + ... and we found that with a = 9/5, Q(x) is a perfect square. Let me check that.

Earlier, we had Q(x) = (a - 1)x² + (b + 2)x + (c - 2). With a = 9/5, b = -18/5, c = 14/5. Let me compute that:

a - 1 = 9/5 - 5/5 = 4/5

b + 2 = -18/5 + 10/5 = -8/5

c - 2 = 14/5 - 10/5 = 4/5

Therefore, Q(x) = (4/5)x² - (8/5)x + 4/5 = (4/5)(x² - 2x + 1) = (4/5)(x - 1)^2. Which is indeed a perfect square. Therefore, Q(x) = [ (2/√5)(x - 1) ]², so non-negative for all x. Similarly, R(x) is also a perfect square.

Similarly, R(x) = (2 - a)x² + (-4 - b)x + (3 - c). With a=9/5, 2 - a = 2 - 9/5 = 1/5.

-4 - b = -4 - (-18/5) = -4 + 18/5 = (-20/5 + 18/5) = -2/5.

3 - c = 3 - 14/5 = 15/5 - 14/5 = 1/5.

Thus, R(x) = (1/5)x² - (2/5)x + 1/5 = (1/5)(x² - 2x + 1) = (1/5)(x - 1)^2. Which is also a perfect square. Hence, R(x) = [ (1/√5)(x - 1) ]², which is non-negative for all x. Therefore, this confirms that our polynomial satisfies the given inequalities for all x.

Therefore, all conditions are satisfied, so our answer of 406 is indeed correct.

I think that's thorough enough. Initially, I thought of assuming the form of P(x) as k(x - 1)^2 + 1, which worked, but then verifying via the discriminant conditions led me to the same result. So, regardless of the approach, the answer is 406.

**Final Answer**
\boxed{406}

We are given a quadratic polynomial $ P(x) $ with real coefficients satisfying the inequality:

$$
x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3 \quad \text{for all } x \in \mathbb{R}
$$

Additionally, it is given that:

$$
P(11) = 181
$$

We are asked to find $ P(16) $.

---

### Step 1: General Form of P(x)

Let:
$$
P(x) = ax^2 + bx + c
$$

Using the given inequalities:

$$
x^2 - 2x + 2 \le ax^2 + bx + c \le 2x^2 - 4x + 3 \quad \forall x
$$

Subtracting $ x^2 - 2x + 2 $ from all sides:
$$
0 \le (a - 1)x^2 + (b + 2)x + (c - 2)
$$

Similarly, subtracting $ P(x) $ from the upper bound:
$$
0 \le (2 - a)x^2 + (-b - 4)x + (3 - c)
$$

These expressions must be **non-negative for all real** $ x $, so their **discriminants must be non-positive**, and their **leading coefficients must be non-negative**. These lead to constraints on $ a, b, c $.

---

### Step 2: Use the Fact That Both Bounds Have the Same Vertex

We are given:

$$
x^2 - 2x + 2 = (x - 1)^2 + 1
$$
$$
2x^2 - 4x + 3 = 2(x - 1)^2 + 1
$$

Both bounds have the same **vertex** at $ x = 1 $, with values:
$$
P(1) \ge 1 \text{ (from lower bound)}
$$
$$
P(1) \le 1 \text{ (from upper bound)}
$$

Therefore:
$$
P(1) = 1
$$

---

### Step 3: Express P(x) in Terms of $ (x - 1)^2 $

Since both upper and lower bounds have the same vertex at $ x = 1 $, we can assume:

$$
P(x) = k(x - 1)^2 + 1
$$

This form ensures $ P(1) = 1 $, and since the coefficient $ k $ is a constant, this gives a quadratic that lies between the two bounds depending on the value of $ k $, where $ 1 \le k \le 2 $.

---

### Step 4: Use the Given Value $ P(11) = 181 $

$$
P(x) = k(x - 1)^2 + 1
\Rightarrow P(11) = k(10)^2 + 1 = 100k + 1
$$

Set this equal to 181:
$$
100k + 1 = 181 \Rightarrow 100k = 180 \Rightarrow k = \frac{180}{100} = \frac{9}{5}
$$

Thus:
$$
P(x) = \frac{9}{5}(x - 1)^2 + 1
$$

---

### Step 5: Find $ P(16) $

$$
P(16) = \frac{9}{5}(15)^2 + 1 = \frac{9}{5} \cdot 225 + 1 = \frac{2025}{5} + 1 = 405 + 1 = 406
$$

---

### Final Answer

$$
\boxed{406}
$$

[Assistant]
Explanation:
Okay, let's tackle this problem step by step. So, we have a quadratic polynomial P(x) with real coefficients. It's given that for all real numbers x, the inequality x² - 2x + 2 ≤ P(x) ≤ 2x² - 4x + 3 holds. Additionally, we know that P(11) = 181, and we need to find P(16).

First, since P(x) is a quadratic polynomial, let's denote it as P(x) = ax² + bx + c, where a, b, c are real numbers. Our goal is to find the coefficients a, b, c such that the inequalities hold for all real x, and then use the given value P(11) = 181 to determine them. Once we have the coefficients, we can compute P(16).

But before jumping into solving for a, b, c, let's analyze the inequalities given. The polynomial P(x) is squeezed between two quadratics: the lower bound is x² - 2x + 2 and the upper bound is 2x² - 4x + 3. Since both bounds are quadratics, and P(x) is also a quadratic, it's possible that P(x) is a linear combination of these two quadratics or maybe they touch each other at certain points?

Wait, but since the inequalities must hold for all real numbers x, the difference between P(x) and the lower bound must be non-negative for all x, and similarly, the difference between the upper bound and P(x) must also be non-negative for all x. So, that suggests that P(x) - (x² - 2x + 2) ≥ 0 for all x and (2x² - 4x + 3) - P(x) ≥ 0 for all x.

Therefore, P(x) is sandwiched between these two quadratics. For the difference between two quadratics to be non-negative for all x, the resulting quadratic must be a perfect square or something that opens upwards? Wait, actually, if we have two quadratics, and their difference is non-negative for all x, then the difference must be a non-negative quadratic. Similarly for the upper bound.

But since both inequalities must hold for all real x, perhaps P(x) is equal to both bounds at some points, and maybe touches them? Let me think. If P(x) is between two quadratics for all x, and it's a quadratic itself, then perhaps P(x) is a linear combination of the two bounds?

Alternatively, maybe the two bounds are touching P(x) at their minimum or maximum points? For example, if the lower bound is x² - 2x + 2, which can be rewritten as (x - 1)^2 + 1. Similarly, the upper bound is 2x² - 4x + 3. Let me check that.

Let me compute the lower bound first: x² - 2x + 2. Completing the square: x² - 2x + 1 + 1 = (x - 1)^2 + 1. So that's a parabola opening upwards with vertex at (1, 1). The upper bound is 2x² - 4x + 3. Let me complete the square for that as well. Factor out the 2: 2(x² - 2x) + 3. Then complete the square inside the parentheses: x² - 2x + 1 - 1. So that becomes 2[(x - 1)^2 - 1] + 3 = 2(x - 1)^2 - 2 + 3 = 2(x - 1)^2 + 1. So, the upper bound is 2(x - 1)^2 + 1, which is also a parabola opening upwards with vertex at (1, 1). Interesting, both bounds have the same vertex at (1, 1), but different leading coefficients. The lower bound has a leading coefficient of 1, and the upper bound has 2.

So, since both bounds have the same vertex at (1, 1), and since they are both opening upwards, the lower bound is "wider" (since coefficient is smaller) and the upper bound is "narrower" (since coefficient is larger). Therefore, the region between them is the area between these two parabolas. Now, P(x) is a quadratic polynomial sitting between them for all real x.

Now, since both bounds are parabolas with the same vertex, perhaps P(x) is also a parabola with the same vertex? If that's the case, then maybe P(x) is a linear combination of the lower and upper bounds? Let me think.

Alternatively, since the region between the two bounds is between two parabolas with the same vertex, maybe P(x) is tangent to both bounds? Wait, but if P(x) is squeezed between them for all x, and they have the same vertex, maybe P(x) has the same vertex? Let me check.

If P(x) is between the two bounds for all x, then at the vertex x=1, the lower bound and upper bound both equal 1. Therefore, P(1) must be between 1 and 1, so P(1) = 1? Wait, because at x=1, both bounds equal (1)^2 - 2(1) + 2 = 1 - 2 + 2 = 1, and similarly the upper bound is 2(1)^2 - 4(1) + 3 = 2 - 4 + 3 = 1. Therefore, at x=1, both bounds equal 1, so P(1) must equal 1. Therefore, P(1) = 1.

So that gives us a point that P(x) must pass through: (1, 1). Also, since P(x) is a quadratic, maybe the vertex of P(x) is also at x=1? Because if P(x) is between two parabolas that both have their minimum at x=1, then if P(x) is also a parabola, its vertex could be at x=1. However, the upper bound is a narrower parabola, so if P(x) is between them, then maybe P(x) is also symmetric around x=1?

Alternatively, maybe P(x) is of the form k(x - 1)^2 + 1, where k is between 1 and 2? Because if you take k between 1 and 2, then k(x - 1)^2 + 1 would lie between the lower bound (1*(x - 1)^2 + 1) and upper bound (2*(x - 1)^2 + 1). However, that seems plausible. But let me check.

Wait, if that's the case, then P(x) = k(x - 1)^2 + 1, with 1 ≤ k ≤ 2. Then, for example, P(1) = 1, which matches. However, then we can use the condition P(11) = 181 to find k. Let me try that approach.

Let me compute P(11). If P(x) = k(x - 1)^2 + 1, then P(11) = k(10)^2 + 1 = 100k + 1. According to the problem, this is equal to 181. Therefore:

100k + 1 = 181
100k = 180
k = 1.8

Which is 9/5. Since 9/5 is between 1 and 2, that seems valid. Then, P(x) would be (9/5)(x - 1)^2 + 1. Then, to find P(16), we compute:

P(16) = (9/5)(15)^2 + 1 = (9/5)(225) + 1 = 9 * 45 + 1 = 405 + 1 = 406.

But wait, before accepting this answer, let me check if this approach is valid. The assumption was that P(x) is of the form k(x - 1)^2 + 1. But is this necessarily the case?

The problem says that for all real numbers x, the inequality holds. So if we assume that P(x) is a quadratic that lies between the two bounds, which are both of the form (x - 1)^2 + 1 and 2(x - 1)^2 + 1, then perhaps P(x) is a multiple of (x - 1)^2 + 1? But actually, the lower bound is (x - 1)^2 + 1 and upper bound is 2(x - 1)^2 + 1. However, the difference between the upper and lower bound is (2(x - 1)^2 + 1) - ((x - 1)^2 + 1) = (x - 1)^2. Therefore, the region between them is all quadratics that can be written as (x - 1)^2 + 1 + t(x - 1)^2, where t is between 0 and 1? Wait, but that would be (1 + t)(x - 1)^2 + 1. However, the upper bound is when t = 1, leading to 2(x - 1)^2 + 1 and lower bound when t = 0. However, in this case, the coefficient of (x - 1)^2 is varying between 1 and 2. Therefore, if P(x) is of the form k(x - 1)^2 + 1, with k in [1, 2], then that would fit between the two bounds.

But is this the only possibility? Suppose that P(x) is not of this form. For example, suppose that P(x) is a quadratic that is not symmetric around x=1. Then, could it still lie between the two bounds for all x?

Alternatively, perhaps we need to consider that the difference between P(x) and the lower bound is a non-negative quadratic, and the difference between the upper bound and P(x) is also a non-negative quadratic. So, let's denote:

Let me define Q(x) = P(x) - (x² - 2x + 2) ≥ 0 for all x, and R(x) = (2x² - 4x + 3) - P(x) ≥ 0 for all x.

Since Q(x) and R(x) are quadratics that are always non-negative, they must be perfect squares or have discriminant less than or equal to zero. Wait, but Q(x) is a quadratic. If Q(x) is non-negative for all x, then its discriminant must be ≤ 0, and its leading coefficient must be ≥ 0. Similarly for R(x).

Let me compute Q(x):

Q(x) = P(x) - (x² - 2x + 2) = (ax² + bx + c) - x² + 2x - 2 = (a - 1)x² + (b + 2)x + (c - 2).

Similarly, R(x) = (2x² - 4x + 3) - P(x) = 2x² - 4x + 3 - ax² - bx - c = (2 - a)x² + (-4 - b)x + (3 - c).

So, for Q(x) to be non-negative for all x, we need:

1. The leading coefficient (a - 1) ≥ 0, so a ≥ 1.

2. The discriminant of Q(x) is ≤ 0. The discriminant D_Q = (b + 2)^2 - 4*(a - 1)*(c - 2) ≤ 0.

Similarly, for R(x) to be non-negative for all x:

1. The leading coefficient (2 - a) ≥ 0 ⇒ a ≤ 2.

2. The discriminant of R(x) is ≤ 0. The discriminant D_R = (-4 - b)^2 - 4*(2 - a)*(3 - c) ≤ 0.

Therefore, combining the two conditions from the leading coefficients, we have 1 ≤ a ≤ 2.

Additionally, the discriminants D_Q and D_R must be ≤ 0.

Furthermore, since P(x) is a quadratic polynomial, and we also know that P(1) = 1. Let me check that. Earlier, we saw that since at x=1, both bounds equal 1, so P(1) must be between 1 and 1, hence P(1) = 1. Let me confirm that.

Yes, if x=1, then lower bound is 1 and upper bound is 1. Therefore, P(1) must be exactly 1. Therefore, substituting x=1 into P(x):

P(1) = a(1)^2 + b(1) + c = a + b + c = 1.

So that gives us another equation: a + b + c = 1.

So now, we have the following constraints:

1. 1 ≤ a ≤ 2.

2. For Q(x): (a - 1)x² + (b + 2)x + (c - 2) ≥ 0 for all x ⇒ discriminant D_Q ≤ 0 and a ≥ 1.

3. For R(x): (2 - a)x² + (-4 - b)x + (3 - c) ≥ 0 for all x ⇒ discriminant D_R ≤ 0 and a ≤ 2.

Additionally, a + b + c = 1.

Moreover, we have the value P(11) = 181, which gives:

a*(11)^2 + b*(11) + c = 121a + 11b + c = 181.

So, we have equations:

Equation (1): a + b + c = 1

Equation (2): 121a + 11b + c = 181

And the discriminant conditions for Q(x) and R(x).

This seems a system of equations with variables a, b, c, but also the discriminant conditions. However, maybe we can find relations between them. Let me first try to express b and c in terms of a, using equations (1) and (2).

From equation (1): c = 1 - a - b.

Substitute into equation (2):

121a + 11b + (1 - a - b) = 181
Simplify:

121a + 11b + 1 - a - b = 181
(121a - a) + (11b - b) + 1 = 181
120a + 10b + 1 = 181
Subtract 1:

120a + 10b = 180
Divide both sides by 10:

12a + b = 18
So, equation (3): b = 18 - 12a.

Now, since we have c = 1 - a - b, substitute b:

c = 1 - a - (18 - 12a) = 1 - a - 18 + 12a = (1 - 18) + (-a + 12a) = -17 + 11a.

Therefore, c = 11a - 17.

So, now we have expressions for b and c in terms of a. So:

b = 18 - 12a
c = 11a - 17.

Now, we need to ensure that the discriminants D_Q and D_R are less than or equal to zero.

First, let's compute D_Q. Recall:

Q(x) = (a - 1)x² + (b + 2)x + (c - 2).

So, substituting b and c:

First, compute the coefficients:

Leading coefficient: a - 1.

Middle term coefficient: b + 2 = (18 - 12a) + 2 = 20 - 12a.

Constant term: c - 2 = (11a - 17) - 2 = 11a - 19.

Therefore, discriminant D_Q is:

(20 - 12a)^2 - 4*(a - 1)*(11a - 19) ≤ 0.

Similarly, for R(x):

R(x) = (2 - a)x² + (-4 - b)x + (3 - c).

Compute coefficients:

Leading coefficient: 2 - a.

Middle term coefficient: -4 - b = -4 - (18 - 12a) = -4 - 18 + 12a = -22 + 12a.

Constant term: 3 - c = 3 - (11a - 17) = 3 - 11a + 17 = 20 - 11a.

Therefore, discriminant D_R is:

(-22 + 12a)^2 - 4*(2 - a)*(20 - 11a) ≤ 0.

So, we need to compute D_Q ≤ 0 and D_R ≤ 0. Let me compute D_Q first.

Compute D_Q:

(20 - 12a)^2 - 4*(a - 1)*(11a - 19) ≤ 0.

First, expand (20 - 12a)^2:

= 400 - 480a + 144a².

Then, compute 4*(a - 1)*(11a - 19):

First compute (a - 1)(11a - 19) = 11a² -19a -11a +19 = 11a² -30a +19.

Multiply by 4: 44a² -120a +76.

Therefore, D_Q = [400 - 480a + 144a²] - [44a² -120a +76] = 400 - 480a + 144a² -44a² +120a -76.

Simplify:

Combine like terms:

a²: 144a² - 44a² = 100a²

a terms: -480a + 120a = -360a

Constants: 400 - 76 = 324.

So D_Q = 100a² - 360a + 324 ≤ 0.

Let me factor this quadratic expression:

First, check if it's a perfect square. Let me compute discriminant of D_Q as a quadratic in a:

Discriminant D = (-360)^2 - 4*100*324.

Wait, but wait, we already know that D_Q is 100a² - 360a + 324. Let me check whether this is a perfect square.

Let me see:

Suppose 100a² - 360a + 324 = (10a - k)^2. Let's compute:

(10a - k)^2 = 100a² - 20k a + k².

Comparing to 100a² - 360a + 324:

So:

-20k = -360 ⇒ k = 18.

Then k² = 324. Which matches. Perfect! Therefore, D_Q = (10a - 18)^2 ≤ 0.

But a square is always ≥ 0, so (10a - 18)^2 ≤ 0 implies that (10a - 18)^2 = 0, so 10a - 18 = 0 ⇒ a = 18/10 = 9/5 = 1.8.

So, D_Q = 0 only when a = 9/5. Therefore, this gives us that a must be 9/5 for D_Q ≤ 0. Since the discriminant is a square, which is ≤ 0 only when it's zero. Therefore, this condition forces a = 9/5.

Similarly, let's check the discriminant D_R.

Recall D_R is:

(-22 + 12a)^2 - 4*(2 - a)*(20 - 11a) ≤ 0.

First, let me compute each part.

First, (-22 + 12a)^2 is the same as (12a -22)^2 = [same as (22 - 12a)^2] which is equal to:

(12a - 22)^2 = 144a² - 528a + 484.

Second, compute 4*(2 - a)*(20 - 11a):

First compute (2 - a)(20 - 11a):

Multiply out: 2*20 + 2*(-11a) - a*20 + a*11a = 40 -22a -20a + 11a² = 11a² -42a +40.

Multiply by 4: 44a² -168a +160.

Therefore, D_R is [144a² - 528a + 484] - [44a² -168a +160] = 144a² -528a +484 -44a² +168a -160.

Simplify:

a² terms: 144a² -44a² = 100a²

a terms: -528a + 168a = -360a

constants: 484 -160 = 324.

Therefore, D_R = 100a² - 360a + 324 ≤ 0.

Wait! That's the same expression as D_Q! So D_R is also equal to 100a² - 360a + 324, which we saw is (10a - 18)^2. Hence, the same condition applies. Therefore, D_R ≤ 0 implies (10a - 18)^2 ≤ 0 ⇒ a = 9/5.

Therefore, from both discriminant conditions, we must have a = 9/5. So, that's our value for a.

Therefore, a = 9/5. Then, from equation (3): b = 18 - 12a = 18 - 12*(9/5) = 18 - 108/5 = (90/5 - 108/5) = (-18/5).

Similarly, c = 11a - 17 = 11*(9/5) - 17 = 99/5 - 85/5 = 14/5.

So, now we have:

a = 9/5, b = -18/5, c = 14/5.

So, the polynomial P(x) is:

P(x) = (9/5)x² - (18/5)x + 14/5.

Alternatively, we can write this as P(x) = (9x² - 18x + 14)/5.

Let me check if this satisfies the original inequalities. For example, let's check at x = 1:

P(1) = (9 - 18 + 14)/5 = (5)/5 = 1. Which matches our earlier conclusion.

Also, check if the inequalities hold for all x. Since we derived these coefficients under the condition that the discriminant is zero, meaning that Q(x) and R(x) are perfect squares. Therefore, Q(x) = (10a - 18)^2 = 0 when a = 9/5, so Q(x) must be a perfect square?

Wait, earlier we found that D_Q = (10a - 18)^2 ≤ 0. Since a = 9/5, then 10a = 18, so D_Q = 0. Therefore, Q(x) is a perfect square. Similarly for R(x). Therefore, Q(x) is a perfect square, meaning that P(x) - (lower bound) is a perfect square, hence non-negative for all x. Similarly for R(x).

Therefore, this polynomial is indeed between the two bounds for all real x. So, our coefficients are correct.

Now, let's confirm that this polynomial satisfies P(11) = 181. Let's compute:

P(11) = (9*(11)^2 - 18*(11) + 14)/5 = (9*121 - 198 + 14)/5.

Compute numerator:

9*121: 9*120 + 9*1 = 1080 + 9 = 1089

Then, 1089 - 198 = 891

891 +14 = 905

Then 905 / 5 = 181. Perfect, matches the given condition.

Therefore, our polynomial is correct. Now, the question is to find P(16).

So, compute P(16):

P(16) = [9*(16)^2 - 18*(16) + 14]/5.

First, compute 16² = 256

Multiply by 9: 9*256. Let me compute 256*9:

Well, 250*9 = 2250, and 6*9=54, so total is 2250 + 54 = 2304.

Then, compute 18*16: 18*10 + 18*6 = 180 + 108 = 288.

So numerator is 2304 - 288 + 14.

Compute 2304 - 288: Let me subtract 288 from 2304.

2304 - 200 = 2104
2104 - 88 = 2016

Then add 14: 2016 + 14 = 2030.

Therefore, numerator is 2030, so P(16) = 2030 / 5 = 406.

So, P(16) = 406.

Therefore, the answer is \boxed{406}.

But let me just check once again. Alternatively, since we had earlier thought that P(x) = (9/5)(x - 1)^2 + 1, let me verify that this is the same as our coefficients.

Expanding (9/5)(x - 1)^2 + 1:

First, (x - 1)^2 = x² - 2x + 1

Multiply by 9/5: (9/5)x² - (18/5)x + 9/5

Add 1: (9/5)x² - (18/5)x + 9/5 + 5/5 = (9/5)x² - (18/5)x + 14/5. Yes, exactly the same as our coefficients. Therefore, this confirms that our polynomial is correct. Therefore, using this form, P(16) = (9/5)(15)^2 + 1 = (9/5)(225) + 1 = 405 + 1 = 406. Same result. So that's consistent.

Therefore, the answer is 406.

But just to be thorough, let me check if there's another possible solution. But given that both discriminants forced a = 9/5, and given that we have unique solution for a, hence unique solution for b and c. Therefore, there is only one such quadratic polynomial that satisfies the given conditions, so P(16) must be 406.

Alternatively, maybe there's another approach? Let me think.

Another way is to consider that since both the lower and upper bounds have the same vertex at x=1, and that P(x) is squeezed between them. Given that at x=1, P(1)=1, and since the bounds have the same value there, P(x) might be tangent to both bounds at x=1? Wait, but since both bounds have the same vertex, and P(x) is in between, maybe P(x) is tangent to both bounds at that point?

If that's the case, then maybe the derivatives at x=1 are equal to the derivatives of both bounds. Let me check.

Let me compute the derivatives of lower bound, upper bound, and P(x).

Lower bound: f(x) = x² - 2x + 2. f’(x) = 2x - 2. At x=1, f’(1) = 2(1) - 2 = 0.

Upper bound: g(x) = 2x² - 4x + 3. g’(x) = 4x - 4. At x=1, g’(1) = 4(1) - 4 = 0.

Therefore, both bounds have derivative zero at x=1. Similarly, if P(x) is tangent to both bounds at x=1, then P’(1) must also be zero. Let's check our polynomial's derivative.

P(x) = (9/5)x² - (18/5)x + 14/5. Then P’(x) = (18/5)x - 18/5. At x=1, P’(1) = (18/5)(1) - 18/5 = 0. Which matches. So indeed, P(x) has a critical point at x=1, same as both bounds. Therefore, this suggests that P(x) touches both bounds at x=1, meaning it's tangent to both at that point. Therefore, since both bounds have the same vertex, and P(x) also has the same vertex, and lies between them, this seems to be the only possibility. Therefore, this approach confirms that our previous result is correct.

Alternatively, if we think about the difference between P(x) and the lower bound, which is a perfect square, so Q(x) = [something]^2. Since Q(x) = (a - 1)x² + ... and we found that with a = 9/5, Q(x) is a perfect square. Let me check that.

Earlier, we had Q(x) = (a - 1)x² + (b + 2)x + (c - 2). With a = 9/5, b = -18/5, c = 14/5. Let me compute that:

a - 1 = 9/5 - 5/5 = 4/5

b + 2 = -18/5 + 10/5 = -8/5

c - 2 = 14/5 - 10/5 = 4/5

Therefore, Q(x) = (4/5)x² - (8/5)x + 4/5 = (4/5)(x² - 2x + 1) = (4/5)(x - 1)^2. Which is indeed a perfect square. Therefore, Q(x) = [ (2/√5)(x - 1) ]², so non-negative for all x. Similarly, R(x) is also a perfect square.

Similarly, R(x) = (2 - a)x² + (-4 - b)x + (3 - c). With a=9/5, 2 - a = 2 - 9/5 = 1/5.

-4 - b = -4 - (-18/5) = -4 + 18/5 = (-20/5 + 18/5) = -2/5.

3 - c = 3 - 14/5 = 15/5 - 14/5 = 1/5.

Thus, R(x) = (1/5)x² - (2/5)x + 1/5 = (1/5)(x² - 2x + 1) = (1/5)(x - 1)^2. Which is also a perfect square. Hence, R(x) = [ (1/√5)(x - 1) ]², which is non-negative for all x. Therefore, this confirms that our polynomial satisfies the given inequalities for all x.

Therefore, all conditions are satisfied, so our answer of 406 is indeed correct.

I think that's thorough enough. Initially, I thought of assuming the form of P(x) as k(x - 1)^2 + 1, which worked, but then verifying via the discriminant conditions led me to the same result. So, regardless of the approach, the answer is 406.

**Final Answer**
\boxed{406}

We are given a quadratic polynomial $ P(x) $ with real coefficients satisfying the inequality:

$$
x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3 \quad \text{for all } x \in \mathbb{R}
$$

Additionally, it is given that:

$$
P(11) = 181
$$

We are asked to find $ P(16) $.

---

### Step 1: General Form of P(x)

Let:
$$
P(x) = ax^2 + bx + c
$$

Using the given inequalities:

$$
x^2 - 2x + 2 \le ax^2 + bx + c \le 2x^2 - 4x + 3 \quad \forall x
$$

Subtracting $ x^2 - 2x + 2 $ from all sides:
$$
0 \le (a - 1)x^2 + (b + 2)x + (c - 2)
$$

Similarly, subtracting $ P(x) $ from the upper bound:
$$
0 \le (2 - a)x^2 + (-b - 4)x + (3 - c)
$$

These expressions must be **non-negative for all real** $ x $, so their **discriminants must be non-positive**, and their **leading coefficients must be non-negative**. These lead to constraints on $ a, b, c $.

---

### Step 2: Use the Fact That Both Bounds Have the Same Vertex

We are given:

$$
x^2 - 2x + 2 = (x - 1)^2 + 1
$$
$$
2x^2 - 4x + 3 = 2(x - 1)^2 + 1
$$

Both bounds have the same **vertex** at $ x = 1 $, with values:
$$
P(1) \ge 1 \text{ (from lower bound)}
$$
$$
P(1) \le 1 \text{ (from upper bound)}
$$

Therefore:
$$
P(1) = 1
$$

---

### Step 3: Express P(x) in Terms of $ (x - 1)^2 $

Since both upper and lower bounds have the same vertex at $ x = 1 $, we can assume:

$$
P(x) = k(x - 1)^2 + 1
$$

This form ensures $ P(1) = 1 $, and since the coefficient $ k $ is a constant, this gives a quadratic that lies between the two bounds depending on the value of $ k $, where $ 1 \le k \le 2 $.

---

### Step 4: Use the Given Value $ P(11) = 181 $

$$
P(x) = k(x - 1)^2 + 1
\Rightarrow P(11) = k(10)^2 + 1 = 100k + 1
$$

Set this equal to 181:
$$
100k + 1 = 181 \Rightarrow 100k = 180 \Rightarrow k = \frac{180}{100} = \frac{9}{5}
$$

Thus:
$$
P(x) = \frac{9}{5}(x - 1)^2 + 1
$$

---

### Step 5: Find $ P(16) $

$$
P(16) = \frac{9}{5}(15)^2 + 1 = \frac{9}{5} \cdot 225 + 1 = \frac{2025}{5} + 1 = 405 + 1 = 406
$$

---

### Final Answer

$$
\boxed{406}
$$

Solution: Rewriting the given quadratics in vertex form, we have \[1 + (x-1)^2 \le P(x) \le 1 + 2(x-1)^2.\]Both of those quadratics have vertex at $(1, 1)$; considering the shape of the graph of a quadratic, we see that $P$ must also have its vertex at $(1,1)$. Therefore, \[P(x) = 1 + k(x-1)^2\]for some constant $k$. Setting $x = 11$, we have $181 = 1 +100k$, so $k = \tfrac{9}{5}$. Then \[P(16) = 1 + \tfrac{9}{5} \cdot 15^2 = \boxed{406}.\]


Explanation:
Okay, so I need to solve this equation: (3)5² - 4(5 - a)² ÷ 3 = 63. Hmm, let me break this down step by step. First, I should probably make sure I understand the equation correctly. Let me rewrite it to make sure I have the order of operations right.

The equation is: 3 times 5 squared minus 4 times (5 minus a) squared divided by 3 equals 63. Wait, the original problem is written as (3)5² - 4(5 - a)² ÷ 3 = 63. Let me check if there are any parentheses that might change the order.

So, according to the original problem: (3)5² - 4(5 - a)² ÷ 3 = 63. Let me parse this. The first term is (3) times 5 squared. Then subtract 4 times (5 - a) squared divided by 3. So the division is only on the 4(5 - a)² part, right? Because there's no parentheses around the entire 4(5 - a)² ÷ 3. So that would mean the equation is:

3 * 5² - [4(5 - a)² / 3] = 63.

Okay, that seems right. Let me confirm with order of operations: multiplication and division have the same precedence and are evaluated left to right. So first, compute 3 * 5². Then compute 4*(5 - a)^2, then divide by 3, and subtract that from the first term.

Alright, so let me compute each part step by step.

First, compute 3 * 5².

5 squared is 25. Then 3 times 25 is 75. So the first term is 75.

Then, the second term is 4*(5 - a)² divided by 3. Let me write that as [4*(5 - a)^2]/3.

So the equation becomes: 75 - [4*(5 - a)^2]/3 = 63.

Okay, now I need to solve for a. Let me rearrange the equation to isolate the term with a.

First, subtract 75 from both sides:

- [4*(5 - a)^2]/3 = 63 - 75.

Calculating 63 - 75: 63 - 75 is -12.

So, - [4*(5 - a)^2]/3 = -12.

Hmm, maybe I can multiply both sides by -1 to make it simpler.

Multiplying both sides by -1: [4*(5 - a)^2]/3 = 12.

Okay, so [4*(5 - a)^2]/3 equals 12. Let me solve for (5 - a)^2.

First, multiply both sides by 3 to eliminate the denominator.

4*(5 - a)^2 = 12*3.

12*3 is 36.

So, 4*(5 - a)^2 = 36.

Now, divide both sides by 4:

(5 - a)^2 = 36 / 4.

36 divided by 4 is 9.

Therefore, (5 - a)^2 = 9.

Now, to solve for a, take the square root of both sides. Remembering that square roots can be positive or negative.

So, 5 - a = ±√9.

√9 is 3, so:

5 - a = 3 or 5 - a = -3.

Now, solving each equation for a:

First equation: 5 - a = 3.

Subtract 5 from both sides: -a = 3 - 5.

- a = -2.

Multiply both sides by -1: a = 2.

Second equation: 5 - a = -3.

Subtract 5 from both sides: -a = -3 -5.

Which is -a = -8.

Multiply both sides by -1: a = 8.

Therefore, the solutions are a = 2 and a = 8.

The question asks for the sum of the values of a. So, 2 + 8 = 10.

Wait, let me check if I did everything correctly. Let me verify my steps again.

Original equation: 3*5² - [4*(5 - a)²]/3 = 63.

Calculations: 3*25 = 75. So 75 - [4*(5 - a)^2]/3 = 63.

Subtract 75: - [4*(5 - a)^2]/3 = -12. Multiply by -1: [4*(5 - a)^2]/3 = 12. Multiply by 3: 4*(5 - a)^2 = 36. Divide by 4: (5 - a)^2 = 9. Square roots: 5 - a = ±3. Then solving: a = 5 ∓ 3. So a = 5 - 3 = 2, and a = 5 + 3 = 8.

Yes, that seems correct. So the solutions are 2 and 8. Sum is 10.

But just to be thorough, let me plug them back into the original equation to check.

First, check a = 2.

Original equation: 3*5² - [4*(5 - 2)^2]/3 = ?

Compute 3*25 = 75.

Then, 5 - 2 = 3. 3 squared is 9. Multiply by 4: 36. Divide by 3: 12.

So 75 - 12 = 63. Which is equal to the right-hand side. Correct.

Now check a = 8.

Original equation: 3*5² - [4*(5 - 8)^2]/3.

Compute 3*25 = 75.

5 - 8 = -3. Squared is 9. Multiply by 4: 36. Divide by 3: 12.

75 - 12 = 63. Correct again.

So both solutions work.

Therefore, the sum of the values of a is 2 + 8 = 10.

Hence, the answer is 10.

Wait, but hold on. Let me check if there was any step where I might have made an error in algebra.

Starting from (5 - a)^2 = 9. When taking square roots, we have 5 - a = ±3. So solving for a:

Case 1: 5 - a = 3 → a = 5 - 3 = 2. Correct.

Case 2: 5 - a = -3 → a = 5 + 3 = 8. Correct.

Yes, that's correct. So, no mistake there.

Alternatively, sometimes when you square both sides, you can introduce extraneous solutions, but in this case, since we're dealing with square roots and solving linear equations, both solutions are valid.

So, yes, 2 and 8 are both valid solutions.

Therefore, the sum is 2 + 8 = 10.

I think that's all. No steps seem to be wrong.

**Final Answer**
The sum of the values of $ a $ is \boxed{10}.

We are given the equation:

$$
(3)5^2 - \frac{4(5 - a)^2}{3} = 63
$$

---

### **Step 1: Simplify the First Term**

Compute the first term:

$$
3 \times 5^2 = 3 \times 25 = 75
$$

So now the equation becomes:

$$
75 - \frac{4(5 - a)^2}{3} = 63
$$

---

### **Step 2: Isolate the Second Term**

Subtract 75 from both sides:

$$
- \frac{4(5 - a)^2}{3} = 63 - 75 = -12
$$

Multiply both sides by -1:

$$
\frac{4(5 - a)^2}{3} = 12
$$

---

### **Step 3: Eliminate the Denominator**

Multiply both sides by 3:

$$
4(5 - a)^2 = 36
$$

Now divide both sides by 4:

$$
(5 - a)^2 = 9
$$

---

### **Step 4: Solve the Equation**

Take the square root of both sides:

$$
5 - a = \pm 3
$$

Solve each case:

- If $5 - a = 3$, then $a = 5 - 3 = 2$
- If $5 - a = -3$, then $a = 5 + 3 = 8$

So the two solutions for $a$ are:

$$
a = 2 \quad \text{and} \quad a = 8
$$

---

### **Step 5: Find the Sum of the Solutions**

To answer the question, we sum the valid solutions:

$$
2 + 8 = \boxed{10}
$$

---

### **Final Answer**

$$
\boxed{10}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: (3)5² - 4(5 - a)² ÷ 3 = 63. Hmm, let me break this down step by step. First, I should probably make sure I understand the equation correctly. Let me rewrite it to make sure I have the order of operations right.

The equation is: 3 times 5 squared minus 4 times (5 minus a) squared divided by 3 equals 63. Wait, the original problem is written as (3)5² - 4(5 - a)² ÷ 3 = 63. Let me check if there are any parentheses that might change the order.

So, according to the original problem: (3)5² - 4(5 - a)² ÷ 3 = 63. Let me parse this. The first term is (3) times 5 squared. Then subtract 4 times (5 - a) squared divided by 3. So the division is only on the 4(5 - a)² part, right? Because there's no parentheses around the entire 4(5 - a)² ÷ 3. So that would mean the equation is:

3 * 5² - [4(5 - a)² / 3] = 63.

Okay, that seems right. Let me confirm with order of operations: multiplication and division have the same precedence and are evaluated left to right. So first, compute 3 * 5². Then compute 4*(5 - a)^2, then divide by 3, and subtract that from the first term.

Alright, so let me compute each part step by step.

First, compute 3 * 5².

5 squared is 25. Then 3 times 25 is 75. So the first term is 75.

Then, the second term is 4*(5 - a)² divided by 3. Let me write that as [4*(5 - a)^2]/3.

So the equation becomes: 75 - [4*(5 - a)^2]/3 = 63.

Okay, now I need to solve for a. Let me rearrange the equation to isolate the term with a.

First, subtract 75 from both sides:

- [4*(5 - a)^2]/3 = 63 - 75.

Calculating 63 - 75: 63 - 75 is -12.

So, - [4*(5 - a)^2]/3 = -12.

Hmm, maybe I can multiply both sides by -1 to make it simpler.

Multiplying both sides by -1: [4*(5 - a)^2]/3 = 12.

Okay, so [4*(5 - a)^2]/3 equals 12. Let me solve for (5 - a)^2.

First, multiply both sides by 3 to eliminate the denominator.

4*(5 - a)^2 = 12*3.

12*3 is 36.

So, 4*(5 - a)^2 = 36.

Now, divide both sides by 4:

(5 - a)^2 = 36 / 4.

36 divided by 4 is 9.

Therefore, (5 - a)^2 = 9.

Now, to solve for a, take the square root of both sides. Remembering that square roots can be positive or negative.

So, 5 - a = ±√9.

√9 is 3, so:

5 - a = 3 or 5 - a = -3.

Now, solving each equation for a:

First equation: 5 - a = 3.

Subtract 5 from both sides: -a = 3 - 5.

- a = -2.

Multiply both sides by -1: a = 2.

Second equation: 5 - a = -3.

Subtract 5 from both sides: -a = -3 -5.

Which is -a = -8.

Multiply both sides by -1: a = 8.

Therefore, the solutions are a = 2 and a = 8.

The question asks for the sum of the values of a. So, 2 + 8 = 10.

Wait, let me check if I did everything correctly. Let me verify my steps again.

Original equation: 3*5² - [4*(5 - a)²]/3 = 63.

Calculations: 3*25 = 75. So 75 - [4*(5 - a)^2]/3 = 63.

Subtract 75: - [4*(5 - a)^2]/3 = -12. Multiply by -1: [4*(5 - a)^2]/3 = 12. Multiply by 3: 4*(5 - a)^2 = 36. Divide by 4: (5 - a)^2 = 9. Square roots: 5 - a = ±3. Then solving: a = 5 ∓ 3. So a = 5 - 3 = 2, and a = 5 + 3 = 8.

Yes, that seems correct. So the solutions are 2 and 8. Sum is 10.

But just to be thorough, let me plug them back into the original equation to check.

First, check a = 2.

Original equation: 3*5² - [4*(5 - 2)^2]/3 = ?

Compute 3*25 = 75.

Then, 5 - 2 = 3. 3 squared is 9. Multiply by 4: 36. Divide by 3: 12.

So 75 - 12 = 63. Which is equal to the right-hand side. Correct.

Now check a = 8.

Original equation: 3*5² - [4*(5 - 8)^2]/3.

Compute 3*25 = 75.

5 - 8 = -3. Squared is 9. Multiply by 4: 36. Divide by 3: 12.

75 - 12 = 63. Correct again.

So both solutions work.

Therefore, the sum of the values of a is 2 + 8 = 10.

Hence, the answer is 10.

Wait, but hold on. Let me check if there was any step where I might have made an error in algebra.

Starting from (5 - a)^2 = 9. When taking square roots, we have 5 - a = ±3. So solving for a:

Case 1: 5 - a = 3 → a = 5 - 3 = 2. Correct.

Case 2: 5 - a = -3 → a = 5 + 3 = 8. Correct.

Yes, that's correct. So, no mistake there.

Alternatively, sometimes when you square both sides, you can introduce extraneous solutions, but in this case, since we're dealing with square roots and solving linear equations, both solutions are valid.

So, yes, 2 and 8 are both valid solutions.

Therefore, the sum is 2 + 8 = 10.

I think that's all. No steps seem to be wrong.

**Final Answer**
The sum of the values of $ a $ is \boxed{10}.

We are given the equation:

$$
(3)5^2 - \frac{4(5 - a)^2}{3} = 63
$$

---

### **Step 1: Simplify the First Term**

Compute the first term:

$$
3 \times 5^2 = 3 \times 25 = 75
$$

So now the equation becomes:

$$
75 - \frac{4(5 - a)^2}{3} = 63
$$

---

### **Step 2: Isolate the Second Term**

Subtract 75 from both sides:

$$
- \frac{4(5 - a)^2}{3} = 63 - 75 = -12
$$

Multiply both sides by -1:

$$
\frac{4(5 - a)^2}{3} = 12
$$

---

### **Step 3: Eliminate the Denominator**

Multiply both sides by 3:

$$
4(5 - a)^2 = 36
$$

Now divide both sides by 4:

$$
(5 - a)^2 = 9
$$

---

### **Step 4: Solve the Equation**

Take the square root of both sides:

$$
5 - a = \pm 3
$$

Solve each case:

- If $5 - a = 3$, then $a = 5 - 3 = 2$
- If $5 - a = -3$, then $a = 5 + 3 = 8$

So the two solutions for $a$ are:

$$
a = 2 \quad \text{and} \quad a = 8
$$

---

### **Step 5: Find the Sum of the Solutions**

To answer the question, we sum the valid solutions:

$$
2 + 8 = \boxed{10}
$$

---

### **Final Answer**

$$
\boxed{10}
$$

Solution:
First we multiply all terms in the equation by 3 to avoid fractions and then solve for $a$. \begin{align*}
9\cdot5^2-4(5-a)^2&=3\cdot63\quad\Rightarrow\\
-4(5-a)^2&=9\cdot21-9\cdot25\quad\Rightarrow\\
&=9(-4)\quad\Rightarrow\\
(5-a)^2&=9
\end{align*} Therefore, \begin{align*}
5-a=3\quad\text{ OR }\quad 5-a=-3\quad\Rightarrow\\
2=a \quad\text{ OR }\quad 8=a.
\end{align*} The sum of the values of $a$ is $2+8=\boxed{10}$.


Explanation:
Okay, let me try to figure out this vector problem. So, the question is about finding a vector v such that when you take the cross product of the vector (1, 2, -5) with v, you get (90, 30, 30). And among all possible such vectors v, I need to find the one with the smallest magnitude.

First, let me recall what the cross product is. The cross product of two vectors in three-dimensional space results in a vector that is perpendicular to both of the original vectors. The cross product also has some properties like anti-commutativity, and the magnitude relates to the area of the parallelogram formed by the vectors. But maybe I need to approach this algebraically.

Let me denote the vector v as (a, b, c), where a, b, c are real numbers. Then, according to the problem, the cross product of (1, 2, -5) and (a, b, c) should equal (90, 30, 30).

So let me write that out. The cross product of vector u = (u₁, u₂, u₃) and vector v = (v₁, v₂, v₃) is given by the determinant:

| i   j   k  |
| u₁ u₂ u₃ |
| v₁ v₂ v₃ |

Which equals (u₂v₃ - u₃v₂)i - (u₁v₃ - u₃v₁)j + (u₁v₂ - u₂v₁)k.

So applying that to my specific case where u = (1, 2, -5) and v = (a, b, c), let me compute each component.

First component (i-direction): u₂v₃ - u₃v₂ = 2*c - (-5)*b = 2c + 5b.

Second component (j-direction): -(u₁v₃ - u₃v₁) = -(1*c - (-5)*a) = -(c + 5a) = -c - 5a.

Third component (k-direction): u₁v₂ - u₂v₁ = 1*b - 2*a = b - 2a.

So, putting it all together, the cross product is:

(2c + 5b, -c - 5a, b - 2a) = (90, 30, 30).

Therefore, we can set up the following system of equations:

1. 2c + 5b = 90
2. -c - 5a = 30
3. b - 2a = 30

Now, we need to solve this system for a, b, c. Let me write them again:

Equation 1: 5b + 2c = 90

Equation 2: -5a - c = 30

Equation 3: -2a + b = 30

So, three equations with three variables. Let me try to solve them step by step.

First, let me look at Equation 3: -2a + b = 30. Maybe I can express b in terms of a. Let me rearrange it:

b = 2a + 30

Okay, so b is expressed as 2a + 30. Now, let me substitute this into Equation 1 and Equation 2.

Starting with Equation 2: -5a - c = 30. Let me solve for c here.

-5a - c = 30 ⇒ -c = 30 + 5a ⇒ c = -30 -5a

So, c is equal to -30 -5a.

Now, let me substitute both b and c into Equation 1:

Equation 1: 5b + 2c = 90

Substituting b = 2a + 30 and c = -30 -5a:

5*(2a + 30) + 2*(-30 -5a) = 90

Let me compute this:

First term: 5*(2a + 30) = 10a + 150

Second term: 2*(-30 -5a) = -60 -10a

Adding both terms: 10a + 150 -60 -10a = (10a -10a) + (150 -60) = 0 + 90 = 90

Wait, that equals 90, which is exactly the right-hand side of Equation 1. So, this equation is satisfied for any a? That means that Equation 1 doesn't give any new information once Equations 2 and 3 are satisfied. Therefore, the system is underdetermined; there are infinitely many solutions depending on the value of a.

Therefore, the solutions are parameterized by a. So, we can write:

a is any real number.

Then, b = 2a + 30

c = -5a - 30

Therefore, the vector v is (a, 2a + 30, -5a - 30).

So, the vector v can be written as:

v = (a, 2a + 30, -5a - 30) where a ∈ ℝ.

Now, the problem asks for the vector v with the smallest magnitude. So, we need to find the value of a that minimizes the magnitude of v.

The magnitude squared of v is a² + (2a + 30)² + (-5a - 30)². Since the square root is a monotonic function, minimizing the magnitude squared will also minimize the magnitude. So, let's compute that.

Let me denote |v|² = a² + (2a + 30)^2 + (-5a -30)^2.

Let me compute each term:

First term: a²

Second term: (2a + 30)^2 = 4a² + 120a + 900

Third term: (-5a -30)^2 = (5a + 30)^2 = 25a² + 300a + 900

Therefore, adding all three terms:

a² + 4a² + 120a + 900 + 25a² + 300a + 900

Combine like terms:

a² + 4a² + 25a² = 30a²

120a + 300a = 420a

900 + 900 = 1800

So, |v|² = 30a² + 420a + 1800

Now, to find the minimum of this quadratic function in terms of a. Since the coefficient of a² is positive (30), the parabola opens upwards, so the minimum is at the vertex.

The vertex of a quadratic function ax² + bx + c is at x = -b/(2a). Wait, here the coefficient of a² is 30, coefficient of a is 420. So, the value of a (the parameter) that minimizes |v|² is:

a = -420/(2*30) = -420/60 = -7

Therefore, the minimum occurs when a = -7.

Now, let me compute the corresponding b and c.

From earlier:

b = 2a + 30. When a = -7:

b = 2*(-7) + 30 = -14 + 30 = 16

c = -5a -30. When a = -7:

c = -5*(-7) -30 = 35 - 30 = 5

Therefore, the vector v is (-7, 16, 5). Let me check if this satisfies the original cross product equation.

Let me compute the cross product of (1, 2, -5) and (-7, 16, 5):

Using the cross product formula:

i component: (2*5 - (-5)*16) = 10 + 80 = 90

j component: - (1*5 - (-5)*(-7)) = - (5 - 35) = - (-30) = 30

k component: (1*16 - 2*(-7)) = 16 + 14 = 30

So, the cross product is (90, 30, 30), which matches the given vector. Great, so this checks out.

Now, to confirm that this is indeed the vector with the smallest magnitude, let me compute its magnitude.

Compute |v|² when a = -7:

Earlier, we had |v|² = 30*(-7)^2 + 420*(-7) + 1800. Let me compute that:

First term: 30*49 = 1470

Second term: 420*(-7) = -2940

Third term: 1800

Total: 1470 - 2940 + 1800 = (1470 + 1800) - 2940 = 3270 - 2940 = 330

Therefore, |v|² = 330 ⇒ |v| = sqrt(330). Let me see if there's a way to verify this is indeed the minimum. Alternatively, since we derived it through calculus (or rather, through the vertex formula), it should be the minimum.

Alternatively, we can think of this as an optimization problem with a quadratic function, which is convex, so there's only one minimum. Therefore, the value we found is indeed the minimum.

Alternatively, maybe there's a more geometric interpretation? Let me think.

The cross product equation u × v = w implies that v lies in a plane perpendicular to u × w? Wait, not sure. Wait, the set of vectors v satisfying u × v = w is a line, since we have three equations but one is dependent, leading to a one-dimensional solution space. Therefore, the set of possible v's is a line in 3D space. The vector with the smallest magnitude on a line is the one that is closest to the origin, which is the projection of the origin onto the line. Therefore, another way to approach this problem is to find the projection of the zero vector onto the line defined by the solutions of u × v = w. However, since we already solved it algebraically, and found a specific vector, maybe it's redundant, but just to make sure.

Alternatively, maybe we can write v as a particular solution plus a multiple of a direction vector. Wait, in the system of equations, the solutions are parameterized by a. So, the general solution is v = (-7, 16, 5) + a*(some direction vector). Wait, but actually, the general solution is expressed in terms of a. Let me check.

Wait, the general solution is (a, 2a + 30, -5a -30). Which can be rewritten as:

v = a*(1, 2, -5) + (0, 30, -30). Wait, let me check:

Let me factor out a:

a*(1, 2, -5) + (0, 30, -30). Wait:

If you take a*(1, 2, -5) + (0, 30, -30) = (a, 2a, -5a) + (0, 30, -30) = (a, 2a + 30, -5a -30). Yes! So, the general solution is a particular solution plus a scalar multiple of the vector u = (1, 2, -5).

But why is that? Because the cross product equation u × v = w is linear in v, so the solution set is an affine space. The homogeneous equation u × v = 0 has solutions v parallel to u, since cross product is zero iff vectors are parallel. Therefore, the general solution is a particular solution plus any vector parallel to u.

Therefore, the solutions form a line in the direction of u. Therefore, the vector v with minimal magnitude is the particular solution (when a = -7) and then we can think of adjusting that by adding multiples of u. Wait, but according to our earlier expression, the general solution is (a, 2a + 30, -5a -30) which is equivalent to (a, 2a + 30, -5a -30). Alternatively, as I noted, this is the particular solution plus a scalar multiple of u. Wait, but if we set a particular solution, say, when a = -7, we get (-7, 16, 5), and then adding any multiple of u would give the general solution. Wait, but the general solution is given by varying a. However, since u is (1,2,-5), and our solution is parameterized by a, then the direction vector is (1, 2, -5). Therefore, the line is all vectors of the form v = (-7, 16, 5) + t*(1, 2, -5).

Therefore, the minimal magnitude vector is the closest point on this line to the origin. Which, as I thought earlier, can be found by projection.

But since we already found it by calculus, maybe that's not necessary. However, maybe I can cross-verify.

Let me consider the vector v = (-7, 16, 5). Let me check if this is the projection. Alternatively, if I think of the line as parametrized by t: v(t) = (-7, 16, 5) + t*(1, 2, -5). Wait, but actually, in our previous parameterization, the general solution is v(a) = (a, 2a + 30, -5a -30). So, if we take a = -7, we get (-7, 16, 5). If we take a different a, say, a = -7 + t, then v(a) = (-7 + t, 2*(-7 + t) + 30, -5*(-7 + t) -30). Let me compute that:

First component: -7 + t

Second component: 2*(-7) + 2t + 30 = -14 + 2t + 30 = 16 + 2t

Third component: 35 - 5t -30 = 5 - 5t

So, v(a) = (-7 + t, 16 + 2t, 5 - 5t). Which is the same as (-7, 16, 5) + t*(1, 2, -5). Therefore, the line is parametrized by t. Therefore, we can think of it as the point (-7, 16, 5) plus t times the direction vector u = (1, 2, -5).

So, to find the vector on this line closest to the origin, we can use the projection formula. The closest point on the line to the origin is the projection of the origin onto the line.

But since the line is v(t) = p + t*u, where p is (-7, 16, 5) and u is (1, 2, -5). Wait, actually, the line is p + t*u, where p is a particular point on the line and u is the direction vector. However, the projection formula requires knowing the parametric form.

Alternatively, since we have already solved for the minimum using calculus, and arrived at a = -7, perhaps that is sufficient. However, to confirm, let me try using the projection formula.

The formula for the closest point on the line to the origin is given by:

Let me denote the line as L: v = p + t*u, where p is (-7, 16, 5) and u is (1, 2, -5). Wait, but actually, p is not necessarily on the line. Wait, no, p is a point on the line. Wait, no, actually in our parametrization, the line is v(t) = (-7, 16, 5) + t*(1, 2, -5). Therefore, the parametric equations are:

x = -7 + t

y = 16 + 2t

z = 5 - 5t

So, to find the closest point to the origin, we can minimize the square of the distance from the origin:

D^2 = x² + y² + z² = (-7 + t)^2 + (16 + 2t)^2 + (5 - 5t)^2

Which is exactly the same expression we had earlier, with a = t. Wait, earlier we had |v|² = 30a² + 420a + 1800. Let me check if this is the same.

Wait, expanding D²:

(-7 + t)^2 = t² -14t + 49

(16 + 2t)^2 = 4t² + 64t + 256

(5 - 5t)^2 = 25t² - 50t + 25

Adding them up:

t² -14t + 49 + 4t² + 64t + 256 + 25t² -50t +25

Combine terms:

t² + 4t² +25t² = 30t²

-14t +64t -50t = 0t

49 + 256 +25 = 330

Therefore, D² = 30t² + 0t + 330. Which is the same as 30t² + 330. Wait, but earlier when we had |v|² = 30a² +420a + 1800. Wait, this seems conflicting. What's happening here?

Wait, no! Wait, earlier, I had set v(a) as (a, 2a + 30, -5a - 30). Then, |v|² = 30a² +420a + 1800. But in this parametrization, we have D² = 30t² + 330. So which is correct?

Wait, I think there is a confusion here. Let me check.

Earlier, when I did the substitution, we found that the general solution is (a, 2a + 30, -5a -30). So, for this parametrization, the square of the magnitude is |v|² = 30a² + 420a + 1800. However, in this new parametrization with t, we have v(t) = (-7 + t, 16 + 2t, 5 -5t). So, if we compute |v(t)|², we get D² = (-7 + t)^2 + (16 + 2t)^2 + (5 -5t)^2 = 30t² + 330. But if we compute |v(t)|² using the previous parametrization, let me check for t = a + 7? Wait, because in the previous parametrization, a was arbitrary, and in this parametrization, t is a different parameter? Wait, actually, when we set a = -7 + t, which would mean that t = a +7. So, substituting a = t -7 into the previous expression for |v|²:

Original |v|² when a is general: 30a² + 420a + 1800. If a = t -7, then:

|v|² = 30(t -7)^2 + 420(t -7) + 1800

Expand that:

First term: 30(t² -14t +49) = 30t² -420t + 1470

Second term: 420t - 2940

Third term: +1800

Adding all together:

30t² -420t + 1470 + 420t - 2940 + 1800

Simplify:

-420t + 420t cancels out.

1470 -2940 + 1800 = 1470 + 1800 -2940 = 3270 - 2940 = 330.

Thus, |v|² = 30t² + 330. Which matches the other calculation. Therefore, in terms of the parameter t, the magnitude squared is 30t² + 330. Therefore, the minimal value is when t = 0, which gives |v|² = 330. So, the minimal magnitude is sqrt(330) when t = 0. Therefore, the vector is v(0) = (-7 +0, 16 +0, 5 -0) = (-7, 16, 5). Which is consistent with our previous result. So, this confirms that the minimal magnitude occurs at t = 0, corresponding to a = -7.

Therefore, this is the vector with the smallest magnitude.

Alternatively, since the minimal distance squared is 30t² + 330, which is minimized when t = 0, which is exactly the point we found earlier. So, this gives me confidence that (-7, 16, 5) is indeed the correct answer.

Let me just quickly check if there is any possible mistake here. For example, did I compute the cross product correctly? Let me verify once again with the found vector.

Given u = (1, 2, -5) and v = (-7, 16, 5). Compute u × v:

|i   j   k|
|1   2   -5|
|-7 16   5|

Calculating determinant:

i*(2*5 - (-5)*16) - j*(1*5 - (-5)*(-7)) + k*(1*16 - 2*(-7))

Compute each component:

i*(10 + 80) - j*(5 - 35) + k*(16 + 14)

Which is:

i*(90) - j*(-30) + k*(30)

Therefore, (90, 30, 30), which matches the given vector. So, cross product is correct.

Therefore, the answer is vector (-7, 16, 5). So, in boxed form, it should be \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

But let me just check if there's another way this could be approached, maybe using vector projections or something else?

Another approach: Since u × v = w, then v is such that it's related to w and u. Also, note that u × v is perpendicular to both u and v. Therefore, w is perpendicular to u. Let me check if that's true here.

Given w = (90, 30, 30). Let me check if w · u = 0?

Compute the dot product:

u · w = (1)(90) + (2)(30) + (-5)(30) = 90 + 60 - 150 = 0.

Yes! The dot product is zero, which makes sense because the cross product u × v is perpendicular to both u and v. Therefore, w must be perpendicular to u. Which it is, so that checks out.

So, another thought: since u × v = w, then v is in the plane spanned by u and w? Wait, no. Wait, the cross product u × v = w implies that w is perpendicular to u and v, so v is in the plane perpendicular to w? Wait, maybe not. Alternatively, since u × v is in the direction perpendicular to both u and v.

Alternatively, since we have u × v = w, then v can be expressed as (something). But perhaps more importantly, since we have a particular solution and the general solution is that particular solution plus a multiple of u.

But given that we already found the answer using two methods (solving the system and minimizing the magnitude), and cross-checked, I think we are safe.

Alternatively, think of this as solving for v in the equation u × v = w. This is a linear equation in v, and the solution is a particular solution plus the null space of the cross product operator. The null space consists of vectors v such that u × v = 0, which are the vectors parallel to u. Therefore, the solutions form an affine space, which is a line.

So, to find the vector v with minimal magnitude on this line, we can take the particular solution and then subtract the component of the particular solution in the direction of u. Wait, but how?

Alternatively, if we consider that the minimal vector is the orthogonal projection of some vector onto the line? Wait, but the line is already in the space. Wait, since the line is defined by the solutions, and we want the point on the line closest to the origin, which is the same as the projection of the origin onto the line. However, since the origin is not on the line, the closest point is found by the projection.

But since we already calculated that the minimal magnitude occurs at a = -7, and the vector is (-7, 16, 5), which is the same as the particular solution when we set a = -7, and since the line is parametrized as that vector plus multiples of u, and since the projection gives us that point, I think that's consistent.

Another way to think about it: Let me recall that for a line defined by a point p and direction vector u, the distance from the origin to the line is |p × u| / |u|. But here, we want the closest point on the line to the origin, which is the projection. However, since we have a particular solution, maybe we can use that.

But in our case, the line is defined by all vectors v such that u × v = w. If we take the particular solution v0 = (-7, 16, 5), and then the direction vector is u = (1, 2, -5). Then, the distance from the origin to the line is |v0 × u| / |u|. However, since v0 is a specific solution, but we can check if that gives us something.

Wait, if we compute v0 × u, that would be (-7, 16, 5) × (1, 2, -5). Let me compute that:

|i   j   k|
|-7 16   5|
|1   2   -5|

Which is i*(16*(-5) - 5*2) - j*(-7*(-5) - 5*1) + k*(-7*2 - 16*1)

Calculating each component:

i*(-80 -10) - j*(35 -5) + k*(-14 -16)

Simplify:

i*(-90) - j*(30) + k*(-30)

Therefore, (-90, -30, -30). The magnitude squared of this vector is (-90)^2 + (-30)^2 + (-30)^2 = 8100 + 900 + 900 = 9900. The magnitude is sqrt(9900) = 30*sqrt(11). The magnitude of u is sqrt(1^2 + 2^2 + (-5)^2) = sqrt(1 + 4 + 25) = sqrt(30). Therefore, the distance from the origin to the line would be |v0 × u| / |u| = 30*sqrt(11)/sqrt(30) = 30/sqrt(30) * sqrt(11) = sqrt(30)*sqrt(11) = sqrt(330). Which is exactly the magnitude of the vector we found earlier! Therefore, the distance from the origin to the line is sqrt(330), and this occurs at the point v = (-7, 16, 5). Therefore, this also confirms that this vector is the closest one.

So, this gives me even more confidence that (-7, 16, 5) is indeed the correct answer.

Therefore, after checking multiple ways, I believe the vector with the smallest magnitude is (-7, 16, 5).

**Final Answer**
The vector $\mathbf{v}$ with the smallest magnitude is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

To solve the problem, we are given two vectors:

$$
\mathbf{u} = \begin{pmatrix}1 \\ 2 \\ -5\end{pmatrix}, \quad \mathbf{w} = \begin{pmatrix}90 \\ 30 \\ 30\end{pmatrix}
$$

and we are told that there exists a vector $\mathbf{v} = \begin{pmatrix}a \\ b \\ c\end{pmatrix}$ such that:

$$
\mathbf{u} \times \mathbf{v} = \mathbf{w}
$$

Our goal is to find the vector $\mathbf{v}$ with the **smallest magnitude** satisfying this condition.

---

### Step 1: Use the Cross Product Formula

The cross product of two 3D vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (a, b, c)$ is given by:

$$
\mathbf{u} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & -5 \\
a & b & c
\end{vmatrix}
= \begin{pmatrix}
2c + 5b \\
- (c + 5a) \\
b - 2a
\end{pmatrix}
$$

Setting this equal to $\mathbf{w} = \begin{pmatrix}90 \\ 30 \\ 30\end{pmatrix}$, we obtain the following system of equations:

$$
\begin{cases}
2c + 5b = 90 \\
- c - 5a = 30 \\
b - 2a = 30
\end{cases}
$$

---

### Step 2: Solve the System

From the third equation:

$$
b = 2a + 30
$$

Substituting into the first equation:

$$
2c + 5(2a + 30) = 90 \Rightarrow 2c + 10a + 150 = 90 \Rightarrow 2c = -10a - 60 \Rightarrow c = -5a - 30
$$

From the second equation:

$$
- c - 5a = 30 \Rightarrow -(-5a - 30) - 5a = 30 \Rightarrow 5a + 30 - 5a = 30 \Rightarrow 30 = 30
$$

So this equation is **consistent and doesn't provide new information**.

Thus, the general solution is:

$$
\mathbf{v} = \begin{pmatrix}a \\ 2a + 30 \\ -5a - 30\end{pmatrix}
$$

---

### Step 3: Minimize the Magnitude

Let's compute the squared magnitude of $\mathbf{v}$:

$$
|\mathbf{v}|^2 = a^2 + (2a + 30)^2 + (-5a - 30)^2
$$

First, expand each term:

- $a^2$
- $(2a + 30)^2 = 4a^2 + 120a + 900$
- $(-5a - 30)^2 = 25a^2 + 300a + 900$

Add them:

$$
|\mathbf{v}|^2 = a^2 + 4a^2 + 120a + 900 + 25a^2 + 300a + 900 = 30a^2 + 420a + 1800
$$

This is a quadratic function in $a$, and the minimum occurs at the vertex:

$$
a = -\frac{420}{2 \cdot 30} = -7
$$

Substitute $a = -7$ into the general solution:

$$
b = 2(-7) + 30 = 16, \quad c = -5(-7) - 30 = 5
$$

So the vector $\mathbf{v}$ is:

$$
\mathbf{v} = \begin{pmatrix}-7 \\ 16 \\ 5\end{pmatrix}
$$

---

### Step 4: Verify the Cross Product

Compute $\mathbf{u} \times \mathbf{v}$:

$$
\mathbf{u} = \begin{pmatrix}1 \\ 2 \\ -5\end{pmatrix}, \quad \mathbf{v} = \begin{pmatrix}-7 \\ 16 \\ 5\end{pmatrix}
$$

$$
\mathbf{u} \times \mathbf{v} = \begin{pmatrix}
2(5) - (-5)(16) \\
- (1 \cdot 5 - (-5)(-7)) \\
1 \cdot 16 - 2 \cdot (-7)
\end{pmatrix}
= \begin{pmatrix}
10 + 80 \\
- (5 - 35) \\
16 + 14
\end{pmatrix}
= \begin{pmatrix}90 \\ 30 \\ 30\end{pmatrix}
$$

This matches the given $\mathbf{w}$, confirming correctness.

---

### Final Answer

$$
\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this vector problem. So, the question is about finding a vector v such that when you take the cross product of the vector (1, 2, -5) with v, you get (90, 30, 30). And among all possible such vectors v, I need to find the one with the smallest magnitude.

First, let me recall what the cross product is. The cross product of two vectors in three-dimensional space results in a vector that is perpendicular to both of the original vectors. The cross product also has some properties like anti-commutativity, and the magnitude relates to the area of the parallelogram formed by the vectors. But maybe I need to approach this algebraically.

Let me denote the vector v as (a, b, c), where a, b, c are real numbers. Then, according to the problem, the cross product of (1, 2, -5) and (a, b, c) should equal (90, 30, 30).

So let me write that out. The cross product of vector u = (u₁, u₂, u₃) and vector v = (v₁, v₂, v₃) is given by the determinant:

| i   j   k  |
| u₁ u₂ u₃ |
| v₁ v₂ v₃ |

Which equals (u₂v₃ - u₃v₂)i - (u₁v₃ - u₃v₁)j + (u₁v₂ - u₂v₁)k.

So applying that to my specific case where u = (1, 2, -5) and v = (a, b, c), let me compute each component.

First component (i-direction): u₂v₃ - u₃v₂ = 2*c - (-5)*b = 2c + 5b.

Second component (j-direction): -(u₁v₃ - u₃v₁) = -(1*c - (-5)*a) = -(c + 5a) = -c - 5a.

Third component (k-direction): u₁v₂ - u₂v₁ = 1*b - 2*a = b - 2a.

So, putting it all together, the cross product is:

(2c + 5b, -c - 5a, b - 2a) = (90, 30, 30).

Therefore, we can set up the following system of equations:

1. 2c + 5b = 90
2. -c - 5a = 30
3. b - 2a = 30

Now, we need to solve this system for a, b, c. Let me write them again:

Equation 1: 5b + 2c = 90

Equation 2: -5a - c = 30

Equation 3: -2a + b = 30

So, three equations with three variables. Let me try to solve them step by step.

First, let me look at Equation 3: -2a + b = 30. Maybe I can express b in terms of a. Let me rearrange it:

b = 2a + 30

Okay, so b is expressed as 2a + 30. Now, let me substitute this into Equation 1 and Equation 2.

Starting with Equation 2: -5a - c = 30. Let me solve for c here.

-5a - c = 30 ⇒ -c = 30 + 5a ⇒ c = -30 -5a

So, c is equal to -30 -5a.

Now, let me substitute both b and c into Equation 1:

Equation 1: 5b + 2c = 90

Substituting b = 2a + 30 and c = -30 -5a:

5*(2a + 30) + 2*(-30 -5a) = 90

Let me compute this:

First term: 5*(2a + 30) = 10a + 150

Second term: 2*(-30 -5a) = -60 -10a

Adding both terms: 10a + 150 -60 -10a = (10a -10a) + (150 -60) = 0 + 90 = 90

Wait, that equals 90, which is exactly the right-hand side of Equation 1. So, this equation is satisfied for any a? That means that Equation 1 doesn't give any new information once Equations 2 and 3 are satisfied. Therefore, the system is underdetermined; there are infinitely many solutions depending on the value of a.

Therefore, the solutions are parameterized by a. So, we can write:

a is any real number.

Then, b = 2a + 30

c = -5a - 30

Therefore, the vector v is (a, 2a + 30, -5a - 30).

So, the vector v can be written as:

v = (a, 2a + 30, -5a - 30) where a ∈ ℝ.

Now, the problem asks for the vector v with the smallest magnitude. So, we need to find the value of a that minimizes the magnitude of v.

The magnitude squared of v is a² + (2a + 30)² + (-5a - 30)². Since the square root is a monotonic function, minimizing the magnitude squared will also minimize the magnitude. So, let's compute that.

Let me denote |v|² = a² + (2a + 30)^2 + (-5a -30)^2.

Let me compute each term:

First term: a²

Second term: (2a + 30)^2 = 4a² + 120a + 900

Third term: (-5a -30)^2 = (5a + 30)^2 = 25a² + 300a + 900

Therefore, adding all three terms:

a² + 4a² + 120a + 900 + 25a² + 300a + 900

Combine like terms:

a² + 4a² + 25a² = 30a²

120a + 300a = 420a

900 + 900 = 1800

So, |v|² = 30a² + 420a + 1800

Now, to find the minimum of this quadratic function in terms of a. Since the coefficient of a² is positive (30), the parabola opens upwards, so the minimum is at the vertex.

The vertex of a quadratic function ax² + bx + c is at x = -b/(2a). Wait, here the coefficient of a² is 30, coefficient of a is 420. So, the value of a (the parameter) that minimizes |v|² is:

a = -420/(2*30) = -420/60 = -7

Therefore, the minimum occurs when a = -7.

Now, let me compute the corresponding b and c.

From earlier:

b = 2a + 30. When a = -7:

b = 2*(-7) + 30 = -14 + 30 = 16

c = -5a -30. When a = -7:

c = -5*(-7) -30 = 35 - 30 = 5

Therefore, the vector v is (-7, 16, 5). Let me check if this satisfies the original cross product equation.

Let me compute the cross product of (1, 2, -5) and (-7, 16, 5):

Using the cross product formula:

i component: (2*5 - (-5)*16) = 10 + 80 = 90

j component: - (1*5 - (-5)*(-7)) = - (5 - 35) = - (-30) = 30

k component: (1*16 - 2*(-7)) = 16 + 14 = 30

So, the cross product is (90, 30, 30), which matches the given vector. Great, so this checks out.

Now, to confirm that this is indeed the vector with the smallest magnitude, let me compute its magnitude.

Compute |v|² when a = -7:

Earlier, we had |v|² = 30*(-7)^2 + 420*(-7) + 1800. Let me compute that:

First term: 30*49 = 1470

Second term: 420*(-7) = -2940

Third term: 1800

Total: 1470 - 2940 + 1800 = (1470 + 1800) - 2940 = 3270 - 2940 = 330

Therefore, |v|² = 330 ⇒ |v| = sqrt(330). Let me see if there's a way to verify this is indeed the minimum. Alternatively, since we derived it through calculus (or rather, through the vertex formula), it should be the minimum.

Alternatively, we can think of this as an optimization problem with a quadratic function, which is convex, so there's only one minimum. Therefore, the value we found is indeed the minimum.

Alternatively, maybe there's a more geometric interpretation? Let me think.

The cross product equation u × v = w implies that v lies in a plane perpendicular to u × w? Wait, not sure. Wait, the set of vectors v satisfying u × v = w is a line, since we have three equations but one is dependent, leading to a one-dimensional solution space. Therefore, the set of possible v's is a line in 3D space. The vector with the smallest magnitude on a line is the one that is closest to the origin, which is the projection of the origin onto the line. Therefore, another way to approach this problem is to find the projection of the zero vector onto the line defined by the solutions of u × v = w. However, since we already solved it algebraically, and found a specific vector, maybe it's redundant, but just to make sure.

Alternatively, maybe we can write v as a particular solution plus a multiple of a direction vector. Wait, in the system of equations, the solutions are parameterized by a. So, the general solution is v = (-7, 16, 5) + a*(some direction vector). Wait, but actually, the general solution is expressed in terms of a. Let me check.

Wait, the general solution is (a, 2a + 30, -5a -30). Which can be rewritten as:

v = a*(1, 2, -5) + (0, 30, -30). Wait, let me check:

Let me factor out a:

a*(1, 2, -5) + (0, 30, -30). Wait:

If you take a*(1, 2, -5) + (0, 30, -30) = (a, 2a, -5a) + (0, 30, -30) = (a, 2a + 30, -5a -30). Yes! So, the general solution is a particular solution plus a scalar multiple of the vector u = (1, 2, -5).

But why is that? Because the cross product equation u × v = w is linear in v, so the solution set is an affine space. The homogeneous equation u × v = 0 has solutions v parallel to u, since cross product is zero iff vectors are parallel. Therefore, the general solution is a particular solution plus any vector parallel to u.

Therefore, the solutions form a line in the direction of u. Therefore, the vector v with minimal magnitude is the particular solution (when a = -7) and then we can think of adjusting that by adding multiples of u. Wait, but according to our earlier expression, the general solution is (a, 2a + 30, -5a -30) which is equivalent to (a, 2a + 30, -5a -30). Alternatively, as I noted, this is the particular solution plus a scalar multiple of u. Wait, but if we set a particular solution, say, when a = -7, we get (-7, 16, 5), and then adding any multiple of u would give the general solution. Wait, but the general solution is given by varying a. However, since u is (1,2,-5), and our solution is parameterized by a, then the direction vector is (1, 2, -5). Therefore, the line is all vectors of the form v = (-7, 16, 5) + t*(1, 2, -5).

Therefore, the minimal magnitude vector is the closest point on this line to the origin. Which, as I thought earlier, can be found by projection.

But since we already found it by calculus, maybe that's not necessary. However, maybe I can cross-verify.

Let me consider the vector v = (-7, 16, 5). Let me check if this is the projection. Alternatively, if I think of the line as parametrized by t: v(t) = (-7, 16, 5) + t*(1, 2, -5). Wait, but actually, in our previous parameterization, the general solution is v(a) = (a, 2a + 30, -5a -30). So, if we take a = -7, we get (-7, 16, 5). If we take a different a, say, a = -7 + t, then v(a) = (-7 + t, 2*(-7 + t) + 30, -5*(-7 + t) -30). Let me compute that:

First component: -7 + t

Second component: 2*(-7) + 2t + 30 = -14 + 2t + 30 = 16 + 2t

Third component: 35 - 5t -30 = 5 - 5t

So, v(a) = (-7 + t, 16 + 2t, 5 - 5t). Which is the same as (-7, 16, 5) + t*(1, 2, -5). Therefore, the line is parametrized by t. Therefore, we can think of it as the point (-7, 16, 5) plus t times the direction vector u = (1, 2, -5).

So, to find the vector on this line closest to the origin, we can use the projection formula. The closest point on the line to the origin is the projection of the origin onto the line.

But since the line is v(t) = p + t*u, where p is (-7, 16, 5) and u is (1, 2, -5). Wait, actually, the line is p + t*u, where p is a particular point on the line and u is the direction vector. However, the projection formula requires knowing the parametric form.

Alternatively, since we have already solved for the minimum using calculus, and arrived at a = -7, perhaps that is sufficient. However, to confirm, let me try using the projection formula.

The formula for the closest point on the line to the origin is given by:

Let me denote the line as L: v = p + t*u, where p is (-7, 16, 5) and u is (1, 2, -5). Wait, but actually, p is not necessarily on the line. Wait, no, p is a point on the line. Wait, no, actually in our parametrization, the line is v(t) = (-7, 16, 5) + t*(1, 2, -5). Therefore, the parametric equations are:

x = -7 + t

y = 16 + 2t

z = 5 - 5t

So, to find the closest point to the origin, we can minimize the square of the distance from the origin:

D^2 = x² + y² + z² = (-7 + t)^2 + (16 + 2t)^2 + (5 - 5t)^2

Which is exactly the same expression we had earlier, with a = t. Wait, earlier we had |v|² = 30a² + 420a + 1800. Let me check if this is the same.

Wait, expanding D²:

(-7 + t)^2 = t² -14t + 49

(16 + 2t)^2 = 4t² + 64t + 256

(5 - 5t)^2 = 25t² - 50t + 25

Adding them up:

t² -14t + 49 + 4t² + 64t + 256 + 25t² -50t +25

Combine terms:

t² + 4t² +25t² = 30t²

-14t +64t -50t = 0t

49 + 256 +25 = 330

Therefore, D² = 30t² + 0t + 330. Which is the same as 30t² + 330. Wait, but earlier when we had |v|² = 30a² +420a + 1800. Wait, this seems conflicting. What's happening here?

Wait, no! Wait, earlier, I had set v(a) as (a, 2a + 30, -5a - 30). Then, |v|² = 30a² +420a + 1800. But in this parametrization, we have D² = 30t² + 330. So which is correct?

Wait, I think there is a confusion here. Let me check.

Earlier, when I did the substitution, we found that the general solution is (a, 2a + 30, -5a -30). So, for this parametrization, the square of the magnitude is |v|² = 30a² + 420a + 1800. However, in this new parametrization with t, we have v(t) = (-7 + t, 16 + 2t, 5 -5t). So, if we compute |v(t)|², we get D² = (-7 + t)^2 + (16 + 2t)^2 + (5 -5t)^2 = 30t² + 330. But if we compute |v(t)|² using the previous parametrization, let me check for t = a + 7? Wait, because in the previous parametrization, a was arbitrary, and in this parametrization, t is a different parameter? Wait, actually, when we set a = -7 + t, which would mean that t = a +7. So, substituting a = t -7 into the previous expression for |v|²:

Original |v|² when a is general: 30a² + 420a + 1800. If a = t -7, then:

|v|² = 30(t -7)^2 + 420(t -7) + 1800

Expand that:

First term: 30(t² -14t +49) = 30t² -420t + 1470

Second term: 420t - 2940

Third term: +1800

Adding all together:

30t² -420t + 1470 + 420t - 2940 + 1800

Simplify:

-420t + 420t cancels out.

1470 -2940 + 1800 = 1470 + 1800 -2940 = 3270 - 2940 = 330.

Thus, |v|² = 30t² + 330. Which matches the other calculation. Therefore, in terms of the parameter t, the magnitude squared is 30t² + 330. Therefore, the minimal value is when t = 0, which gives |v|² = 330. So, the minimal magnitude is sqrt(330) when t = 0. Therefore, the vector is v(0) = (-7 +0, 16 +0, 5 -0) = (-7, 16, 5). Which is consistent with our previous result. So, this confirms that the minimal magnitude occurs at t = 0, corresponding to a = -7.

Therefore, this is the vector with the smallest magnitude.

Alternatively, since the minimal distance squared is 30t² + 330, which is minimized when t = 0, which is exactly the point we found earlier. So, this gives me confidence that (-7, 16, 5) is indeed the correct answer.

Let me just quickly check if there is any possible mistake here. For example, did I compute the cross product correctly? Let me verify once again with the found vector.

Given u = (1, 2, -5) and v = (-7, 16, 5). Compute u × v:

|i   j   k|
|1   2   -5|
|-7 16   5|

Calculating determinant:

i*(2*5 - (-5)*16) - j*(1*5 - (-5)*(-7)) + k*(1*16 - 2*(-7))

Compute each component:

i*(10 + 80) - j*(5 - 35) + k*(16 + 14)

Which is:

i*(90) - j*(-30) + k*(30)

Therefore, (90, 30, 30), which matches the given vector. So, cross product is correct.

Therefore, the answer is vector (-7, 16, 5). So, in boxed form, it should be \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

But let me just check if there's another way this could be approached, maybe using vector projections or something else?

Another approach: Since u × v = w, then v is such that it's related to w and u. Also, note that u × v is perpendicular to both u and v. Therefore, w is perpendicular to u. Let me check if that's true here.

Given w = (90, 30, 30). Let me check if w · u = 0?

Compute the dot product:

u · w = (1)(90) + (2)(30) + (-5)(30) = 90 + 60 - 150 = 0.

Yes! The dot product is zero, which makes sense because the cross product u × v is perpendicular to both u and v. Therefore, w must be perpendicular to u. Which it is, so that checks out.

So, another thought: since u × v = w, then v is in the plane spanned by u and w? Wait, no. Wait, the cross product u × v = w implies that w is perpendicular to u and v, so v is in the plane perpendicular to w? Wait, maybe not. Alternatively, since u × v is in the direction perpendicular to both u and v.

Alternatively, since we have u × v = w, then v can be expressed as (something). But perhaps more importantly, since we have a particular solution and the general solution is that particular solution plus a multiple of u.

But given that we already found the answer using two methods (solving the system and minimizing the magnitude), and cross-checked, I think we are safe.

Alternatively, think of this as solving for v in the equation u × v = w. This is a linear equation in v, and the solution is a particular solution plus the null space of the cross product operator. The null space consists of vectors v such that u × v = 0, which are the vectors parallel to u. Therefore, the solutions form an affine space, which is a line.

So, to find the vector v with minimal magnitude on this line, we can take the particular solution and then subtract the component of the particular solution in the direction of u. Wait, but how?

Alternatively, if we consider that the minimal vector is the orthogonal projection of some vector onto the line? Wait, but the line is already in the space. Wait, since the line is defined by the solutions, and we want the point on the line closest to the origin, which is the same as the projection of the origin onto the line. However, since the origin is not on the line, the closest point is found by the projection.

But since we already calculated that the minimal magnitude occurs at a = -7, and the vector is (-7, 16, 5), which is the same as the particular solution when we set a = -7, and since the line is parametrized as that vector plus multiples of u, and since the projection gives us that point, I think that's consistent.

Another way to think about it: Let me recall that for a line defined by a point p and direction vector u, the distance from the origin to the line is |p × u| / |u|. But here, we want the closest point on the line to the origin, which is the projection. However, since we have a particular solution, maybe we can use that.

But in our case, the line is defined by all vectors v such that u × v = w. If we take the particular solution v0 = (-7, 16, 5), and then the direction vector is u = (1, 2, -5). Then, the distance from the origin to the line is |v0 × u| / |u|. However, since v0 is a specific solution, but we can check if that gives us something.

Wait, if we compute v0 × u, that would be (-7, 16, 5) × (1, 2, -5). Let me compute that:

|i   j   k|
|-7 16   5|
|1   2   -5|

Which is i*(16*(-5) - 5*2) - j*(-7*(-5) - 5*1) + k*(-7*2 - 16*1)

Calculating each component:

i*(-80 -10) - j*(35 -5) + k*(-14 -16)

Simplify:

i*(-90) - j*(30) + k*(-30)

Therefore, (-90, -30, -30). The magnitude squared of this vector is (-90)^2 + (-30)^2 + (-30)^2 = 8100 + 900 + 900 = 9900. The magnitude is sqrt(9900) = 30*sqrt(11). The magnitude of u is sqrt(1^2 + 2^2 + (-5)^2) = sqrt(1 + 4 + 25) = sqrt(30). Therefore, the distance from the origin to the line would be |v0 × u| / |u| = 30*sqrt(11)/sqrt(30) = 30/sqrt(30) * sqrt(11) = sqrt(30)*sqrt(11) = sqrt(330). Which is exactly the magnitude of the vector we found earlier! Therefore, the distance from the origin to the line is sqrt(330), and this occurs at the point v = (-7, 16, 5). Therefore, this also confirms that this vector is the closest one.

So, this gives me even more confidence that (-7, 16, 5) is indeed the correct answer.

Therefore, after checking multiple ways, I believe the vector with the smallest magnitude is (-7, 16, 5).

**Final Answer**
The vector $\mathbf{v}$ with the smallest magnitude is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

To solve the problem, we are given two vectors:

$$
\mathbf{u} = \begin{pmatrix}1 \\ 2 \\ -5\end{pmatrix}, \quad \mathbf{w} = \begin{pmatrix}90 \\ 30 \\ 30\end{pmatrix}
$$

and we are told that there exists a vector $\mathbf{v} = \begin{pmatrix}a \\ b \\ c\end{pmatrix}$ such that:

$$
\mathbf{u} \times \mathbf{v} = \mathbf{w}
$$

Our goal is to find the vector $\mathbf{v}$ with the **smallest magnitude** satisfying this condition.

---

### Step 1: Use the Cross Product Formula

The cross product of two 3D vectors $\mathbf{u} = (u_1, u_2, u_3)$ and $\mathbf{v} = (a, b, c)$ is given by:

$$
\mathbf{u} \times \mathbf{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & -5 \\
a & b & c
\end{vmatrix}
= \begin{pmatrix}
2c + 5b \\
- (c + 5a) \\
b - 2a
\end{pmatrix}
$$

Setting this equal to $\mathbf{w} = \begin{pmatrix}90 \\ 30 \\ 30\end{pmatrix}$, we obtain the following system of equations:

$$
\begin{cases}
2c + 5b = 90 \\
- c - 5a = 30 \\
b - 2a = 30
\end{cases}
$$

---

### Step 2: Solve the System

From the third equation:

$$
b = 2a + 30
$$

Substituting into the first equation:

$$
2c + 5(2a + 30) = 90 \Rightarrow 2c + 10a + 150 = 90 \Rightarrow 2c = -10a - 60 \Rightarrow c = -5a - 30
$$

From the second equation:

$$
- c - 5a = 30 \Rightarrow -(-5a - 30) - 5a = 30 \Rightarrow 5a + 30 - 5a = 30 \Rightarrow 30 = 30
$$

So this equation is **consistent and doesn't provide new information**.

Thus, the general solution is:

$$
\mathbf{v} = \begin{pmatrix}a \\ 2a + 30 \\ -5a - 30\end{pmatrix}
$$

---

### Step 3: Minimize the Magnitude

Let's compute the squared magnitude of $\mathbf{v}$:

$$
|\mathbf{v}|^2 = a^2 + (2a + 30)^2 + (-5a - 30)^2
$$

First, expand each term:

- $a^2$
- $(2a + 30)^2 = 4a^2 + 120a + 900$
- $(-5a - 30)^2 = 25a^2 + 300a + 900$

Add them:

$$
|\mathbf{v}|^2 = a^2 + 4a^2 + 120a + 900 + 25a^2 + 300a + 900 = 30a^2 + 420a + 1800
$$

This is a quadratic function in $a$, and the minimum occurs at the vertex:

$$
a = -\frac{420}{2 \cdot 30} = -7
$$

Substitute $a = -7$ into the general solution:

$$
b = 2(-7) + 30 = 16, \quad c = -5(-7) - 30 = 5
$$

So the vector $\mathbf{v}$ is:

$$
\mathbf{v} = \begin{pmatrix}-7 \\ 16 \\ 5\end{pmatrix}
$$

---

### Step 4: Verify the Cross Product

Compute $\mathbf{u} \times \mathbf{v}$:

$$
\mathbf{u} = \begin{pmatrix}1 \\ 2 \\ -5\end{pmatrix}, \quad \mathbf{v} = \begin{pmatrix}-7 \\ 16 \\ 5\end{pmatrix}
$$

$$
\mathbf{u} \times \mathbf{v} = \begin{pmatrix}
2(5) - (-5)(16) \\
- (1 \cdot 5 - (-5)(-7)) \\
1 \cdot 16 - 2 \cdot (-7)
\end{pmatrix}
= \begin{pmatrix}
10 + 80 \\
- (5 - 35) \\
16 + 14
\end{pmatrix}
= \begin{pmatrix}90 \\ 30 \\ 30\end{pmatrix}
$$

This matches the given $\mathbf{w}$, confirming correctness.

---

### Final Answer

$$
\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}
$$

Solution:
Let $\mathbf{v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}.$  Then
\[\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix},\]so
\[\begin{pmatrix} 5b + 2c \\ -5a - c \\ -2a + b \end{pmatrix} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}.\]Comparing the components, we get
\begin{align*}
5b + 2c &= 90, \\
-5a - c &= 30, \\
-2a + b &= 30.
\end{align*}From the second equation, $c = -5a - 30.$  From the third equation, $b = 2a + 30.$  We want to minimize the magnitude of $\mathbf{v},$ which is equivalent to minimizing
\[a^2 + b^2 + c^2 = a^2 + (2a + 30)^2 + (-5a - 30)^2 = 30a^2 + 420a + 1800.\]Completing the square, we get $30 (a + 7)^2 + 330,$ so the magnitude is minimized when $a = -7.$  Then $b = 16$ and $c = 5,$ so the vector $\mathbf{v}$ we seek is $\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.$