Explanation:
Okay, so I need to find the area of triangle OAB, given the vectors OA and OB. Let me recall how to do this. I think the area of a triangle formed by two vectors can be found using the cross product. The formula is something like half the magnitude of the cross product of the two vectors. Let me check if that's right.

Yes, I remember that if you have two vectors from the same point, say OA and OB, then the area of the triangle OAB is (1/2) times the magnitude of OA cross OB. So, first, I need to compute the cross product of vectors OA and OB, then find its magnitude, and then divide by 2. That should give me the area.

Alright, let me write down the vectors again. Vector OA is given as (1, 2, 3) and vector OB is (-3, -2, 1). So, OA = [1, 2, 3] and OB = [-3, -2, 1]. Now, I need to compute OA × OB.

The cross product of two vectors in three dimensions can be calculated using the determinant of a matrix with the standard unit vectors i, j, k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row. Let me set that up.

So, OA × OB = |i   j   k|
             |1   2   3|
             |-3 -2   1|

Calculating this determinant, I can expand it along the first row. So, it would be:

i * (2*1 - 3*(-2)) - j * (1*1 - 3*(-3)) + k * (1*(-2) - 2*(-3))

Let me compute each component step by step.

First, the i component: 2*1 is 2, and 3*(-2) is -6, but since it's minus 3*(-2), it becomes +6. So, 2 + 6 = 8. Therefore, the i component is 8i.

Wait, hold on. Let me make sure. The formula for the determinant is:

i*( (2)(1) - (3)(-2) ) - j*( (1)(1) - (3)(-3) ) + k*( (1)(-2) - (2)(-3) )

Yes, that's right. So for the i component: 2*1 - 3*(-2) = 2 + 6 = 8. So that's 8i.

Then the j component: it's subtracted, so - [ (1)(1) - (3)(-3) ] = - [1 + 9] = -10. Wait, hold on. Let me check again. The j component is - [ (1)(1) - (3)(-3) ]? Wait, no. Wait, the formula is:

For the cross product, the j component is subtracted, so it's - [ (OA_x * OB_z - OA_z * OB_x) ]? Wait, maybe I should recall the formula properly.

Alternatively, maybe I should use the standard formula for cross product components. Let me recall:

If vector OA = (a1, a2, a3) and vector OB = (b1, b2, b3), then OA × OB = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). Wait, is that right? Wait, no, the cross product components are:

First component: a2b3 - a3b2

Second component: a3b1 - a1b3

Third component: a1b2 - a2b1

But I need to be careful with the signs. Alternatively, using the determinant method:

i*(a2b3 - a3b2) - j*(a1b3 - a3b1) + k*(a1b2 - a2b1)

Which is equivalent to (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). So, yes, that's correct.

So, applying this to our vectors OA = (1, 2, 3) and OB = (-3, -2, 1):

First component (i): a2b3 - a3b2 = (2)(1) - (3)(-2) = 2 + 6 = 8

Second component (j): a3b1 - a1b3 = (3)(-3) - (1)(1) = -9 - 1 = -10

Third component (k): a1b2 - a2b1 = (1)(-2) - (2)(-3) = -2 + 6 = 4

Therefore, OA × OB = (8, -10, 4). Wait, but in the determinant method, the j component is subtracted, so when I calculated earlier, I had - [ (1)(1) - (3)(-3) ] which would be - [1 + 9] = -10. So that matches. So the cross product vector is (8, -10, 4). Let me confirm once again:

First component: 2*1 - 3*(-2) = 2 + 6 = 8. Correct.

Second component: 3*(-3) - 1*1 = -9 -1 = -10. Correct.

Third component: 1*(-2) - 2*(-3) = -2 + 6 = 4. Correct. So OA × OB = (8, -10, 4).

Now, the magnitude of this cross product vector is sqrt(8^2 + (-10)^2 + 4^2). Let me compute that.

First, 8 squared is 64, (-10) squared is 100, and 4 squared is 16. Adding them up: 64 + 100 + 16 = 180. Therefore, the magnitude is sqrt(180). Let me simplify sqrt(180).

180 factors into 36 * 5, since 36*5=180. Therefore sqrt(36*5) = 6*sqrt(5). So the magnitude of OA × OB is 6√5.

Therefore, the area of triangle OAB is half of that, so (1/2)*6√5 = 3√5.

Wait, let me check again. So cross product magnitude is sqrt(8² + (-10)² + 4²) = sqrt(64 + 100 + 16) = sqrt(180). Yes, sqrt(180) is indeed 6√5. Then half of that is 3√5. So the area is 3√5.

But let me just make sure I didn't make any mistakes in the cross product calculation. Let me recompute OA × OB.

OA is (1, 2, 3), OB is (-3, -2, 1). So cross product components:

i: (2)(1) - (3)(-2) = 2 + 6 = 8. Correct.

j: -( (1)(1) - (3)(-3) ) = - (1 + 9) = -10. Correct.

k: (1)(-2) - (2)(-3) = -2 + 6 = 4. Correct. So cross product is (8, -10, 4).

Yes, that seems right. Then magnitude squared is 8² + (-10)² + 4² = 64 + 100 + 16 = 180. So magnitude is sqrt(180) = 6*sqrt(5). Then area is half that, so 3*sqrt(5).

Therefore, the area of triangle OAB is 3√5.

But just to be thorough, let me think if there's another way to approach this problem. For example, using coordinates, maybe compute the lengths of OA, OB, and AB, then use Heron's formula? Although that might be more complicated, but let me try it as a check.

First, compute the lengths of OA, OB, and AB.

Length of OA: sqrt(1² + 2² + 3²) = sqrt(1 + 4 + 9) = sqrt(14).

Length of OB: sqrt((-3)^2 + (-2)^2 + 1^2) = sqrt(9 + 4 + 1) = sqrt(14). Interesting, both OA and OB have the same length, sqrt(14).

Now, length of AB. Since AB = OB - OA. Wait, vector AB is OB - OA? Wait, no. Wait, vector OA is from O to A, vector OB is from O to B. So vector AB is OB - OA. So coordinates of AB would be (-3 - 1, -2 - 2, 1 - 3) = (-4, -4, -2). Therefore, the length of AB is sqrt((-4)^2 + (-4)^2 + (-2)^2) = sqrt(16 + 16 + 4) = sqrt(36) = 6.

So sides of triangle OAB are OA = sqrt(14), OB = sqrt(14), AB = 6.

Now, using Heron's formula, the semi-perimeter s = (sqrt(14) + sqrt(14) + 6)/2 = (2*sqrt(14) + 6)/2 = sqrt(14) + 3.

Then area = sqrt( s(s - OA)(s - OB)(s - AB) )

So plugging in:

s = sqrt(14) + 3

s - OA = sqrt(14) + 3 - sqrt(14) = 3

Similarly, s - OB = 3

s - AB = sqrt(14) + 3 - 6 = sqrt(14) - 3

Therefore, area = sqrt( [sqrt(14) + 3] * 3 * 3 * [sqrt(14) - 3] )

Let me compute that. First, note that [sqrt(14) + 3][sqrt(14) - 3] is a difference of squares: (sqrt(14))² - 3² = 14 - 9 = 5.

Therefore, the expression under the square root becomes 5 * 3 * 3 = 45. Therefore, area = sqrt(45) = 3*sqrt(5).

Which matches the previous result! So that's reassuring. Therefore, the area is indeed 3√5.

So both methods give the same answer, so I think that's correct.

Alternatively, another way to think about it is that the area of the parallelogram formed by OA and OB is |OA × OB|, so the area of the triangle is half that. Which is exactly what I did first.

Therefore, I can be confident that the area of triangle OAB is 3√5.

Just to recap, steps were:

1. Find vectors OA and OB.

2. Compute their cross product.

3. Find the magnitude of the cross product.

4. Divide by 2 to get the area of the triangle.

Alternatively, using Heron's formula with the lengths of the sides also gives the same answer, which is a good check.

Therefore, the answer is 3√5.

**Final Answer**
The area of triangle $ OAB $ is \boxed{3\sqrt{5}}.

To find the area of triangle $ OAB $, we are given the position vectors of points $ A $ and $ B $ with respect to the origin $ O $:

$$
\vec{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \vec{OB} = \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix}
$$

---

### Step 1: Use the Cross Product to Find the Area

The area of triangle $ OAB $ can be calculated using the formula:

$$
\text{Area} = \frac{1}{2} \left\| \vec{OA} \times \vec{OB} \right\|
$$

We compute the cross product $ \vec{OA} \times \vec{OB} $ using the determinant method:

$$
\vec{OA} \times \vec{OB} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
-3 & -2 & 1
\end{vmatrix}
= \mathbf{i}(2 \cdot 1 - 3 \cdot (-2)) - \mathbf{j}(1 \cdot 1 - 3 \cdot (-3)) + \mathbf{k}(1 \cdot (-2) - 2 \cdot (-3))
$$

$$
= \mathbf{i}(2 + 6) - \mathbf{j}(1 + 9) + \mathbf{k}(-2 + 6)
= 8\mathbf{i} - 10\mathbf{j} + 4\mathbf{k}
$$

So,

$$
\vec{OA} \times \vec{OB} = \begin{pmatrix} 8 \\ -10 \\ 4 \end{pmatrix}
$$

---

### Step 2: Compute the Magnitude of the Cross Product

$$
\left\| \vec{OA} \times \vec{OB} \right\| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180}
$$

Simplify:

$$
\sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}
$$

---

### Step 3: Compute the Area of the Triangle

$$
\text{Area} = \frac{1}{2} \cdot 6\sqrt{5} = 3\sqrt{5}
$$

---

### Final Answer

$$
\boxed{3\sqrt{5}}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of triangle OAB, given the vectors OA and OB. Let me recall how to do this. I think the area of a triangle formed by two vectors can be found using the cross product. The formula is something like half the magnitude of the cross product of the two vectors. Let me check if that's right.

Yes, I remember that if you have two vectors from the same point, say OA and OB, then the area of the triangle OAB is (1/2) times the magnitude of OA cross OB. So, first, I need to compute the cross product of vectors OA and OB, then find its magnitude, and then divide by 2. That should give me the area.

Alright, let me write down the vectors again. Vector OA is given as (1, 2, 3) and vector OB is (-3, -2, 1). So, OA = [1, 2, 3] and OB = [-3, -2, 1]. Now, I need to compute OA × OB.

The cross product of two vectors in three dimensions can be calculated using the determinant of a matrix with the standard unit vectors i, j, k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row. Let me set that up.

So, OA × OB = |i   j   k|
             |1   2   3|
             |-3 -2   1|

Calculating this determinant, I can expand it along the first row. So, it would be:

i * (2*1 - 3*(-2)) - j * (1*1 - 3*(-3)) + k * (1*(-2) - 2*(-3))

Let me compute each component step by step.

First, the i component: 2*1 is 2, and 3*(-2) is -6, but since it's minus 3*(-2), it becomes +6. So, 2 + 6 = 8. Therefore, the i component is 8i.

Wait, hold on. Let me make sure. The formula for the determinant is:

i*( (2)(1) - (3)(-2) ) - j*( (1)(1) - (3)(-3) ) + k*( (1)(-2) - (2)(-3) )

Yes, that's right. So for the i component: 2*1 - 3*(-2) = 2 + 6 = 8. So that's 8i.

Then the j component: it's subtracted, so - [ (1)(1) - (3)(-3) ] = - [1 + 9] = -10. Wait, hold on. Let me check again. The j component is - [ (1)(1) - (3)(-3) ]? Wait, no. Wait, the formula is:

For the cross product, the j component is subtracted, so it's - [ (OA_x * OB_z - OA_z * OB_x) ]? Wait, maybe I should recall the formula properly.

Alternatively, maybe I should use the standard formula for cross product components. Let me recall:

If vector OA = (a1, a2, a3) and vector OB = (b1, b2, b3), then OA × OB = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). Wait, is that right? Wait, no, the cross product components are:

First component: a2b3 - a3b2

Second component: a3b1 - a1b3

Third component: a1b2 - a2b1

But I need to be careful with the signs. Alternatively, using the determinant method:

i*(a2b3 - a3b2) - j*(a1b3 - a3b1) + k*(a1b2 - a2b1)

Which is equivalent to (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1). So, yes, that's correct.

So, applying this to our vectors OA = (1, 2, 3) and OB = (-3, -2, 1):

First component (i): a2b3 - a3b2 = (2)(1) - (3)(-2) = 2 + 6 = 8

Second component (j): a3b1 - a1b3 = (3)(-3) - (1)(1) = -9 - 1 = -10

Third component (k): a1b2 - a2b1 = (1)(-2) - (2)(-3) = -2 + 6 = 4

Therefore, OA × OB = (8, -10, 4). Wait, but in the determinant method, the j component is subtracted, so when I calculated earlier, I had - [ (1)(1) - (3)(-3) ] which would be - [1 + 9] = -10. So that matches. So the cross product vector is (8, -10, 4). Let me confirm once again:

First component: 2*1 - 3*(-2) = 2 + 6 = 8. Correct.

Second component: 3*(-3) - 1*1 = -9 -1 = -10. Correct.

Third component: 1*(-2) - 2*(-3) = -2 + 6 = 4. Correct. So OA × OB = (8, -10, 4).

Now, the magnitude of this cross product vector is sqrt(8^2 + (-10)^2 + 4^2). Let me compute that.

First, 8 squared is 64, (-10) squared is 100, and 4 squared is 16. Adding them up: 64 + 100 + 16 = 180. Therefore, the magnitude is sqrt(180). Let me simplify sqrt(180).

180 factors into 36 * 5, since 36*5=180. Therefore sqrt(36*5) = 6*sqrt(5). So the magnitude of OA × OB is 6√5.

Therefore, the area of triangle OAB is half of that, so (1/2)*6√5 = 3√5.

Wait, let me check again. So cross product magnitude is sqrt(8² + (-10)² + 4²) = sqrt(64 + 100 + 16) = sqrt(180). Yes, sqrt(180) is indeed 6√5. Then half of that is 3√5. So the area is 3√5.

But let me just make sure I didn't make any mistakes in the cross product calculation. Let me recompute OA × OB.

OA is (1, 2, 3), OB is (-3, -2, 1). So cross product components:

i: (2)(1) - (3)(-2) = 2 + 6 = 8. Correct.

j: -( (1)(1) - (3)(-3) ) = - (1 + 9) = -10. Correct.

k: (1)(-2) - (2)(-3) = -2 + 6 = 4. Correct. So cross product is (8, -10, 4).

Yes, that seems right. Then magnitude squared is 8² + (-10)² + 4² = 64 + 100 + 16 = 180. So magnitude is sqrt(180) = 6*sqrt(5). Then area is half that, so 3*sqrt(5).

Therefore, the area of triangle OAB is 3√5.

But just to be thorough, let me think if there's another way to approach this problem. For example, using coordinates, maybe compute the lengths of OA, OB, and AB, then use Heron's formula? Although that might be more complicated, but let me try it as a check.

First, compute the lengths of OA, OB, and AB.

Length of OA: sqrt(1² + 2² + 3²) = sqrt(1 + 4 + 9) = sqrt(14).

Length of OB: sqrt((-3)^2 + (-2)^2 + 1^2) = sqrt(9 + 4 + 1) = sqrt(14). Interesting, both OA and OB have the same length, sqrt(14).

Now, length of AB. Since AB = OB - OA. Wait, vector AB is OB - OA? Wait, no. Wait, vector OA is from O to A, vector OB is from O to B. So vector AB is OB - OA. So coordinates of AB would be (-3 - 1, -2 - 2, 1 - 3) = (-4, -4, -2). Therefore, the length of AB is sqrt((-4)^2 + (-4)^2 + (-2)^2) = sqrt(16 + 16 + 4) = sqrt(36) = 6.

So sides of triangle OAB are OA = sqrt(14), OB = sqrt(14), AB = 6.

Now, using Heron's formula, the semi-perimeter s = (sqrt(14) + sqrt(14) + 6)/2 = (2*sqrt(14) + 6)/2 = sqrt(14) + 3.

Then area = sqrt( s(s - OA)(s - OB)(s - AB) )

So plugging in:

s = sqrt(14) + 3

s - OA = sqrt(14) + 3 - sqrt(14) = 3

Similarly, s - OB = 3

s - AB = sqrt(14) + 3 - 6 = sqrt(14) - 3

Therefore, area = sqrt( [sqrt(14) + 3] * 3 * 3 * [sqrt(14) - 3] )

Let me compute that. First, note that [sqrt(14) + 3][sqrt(14) - 3] is a difference of squares: (sqrt(14))² - 3² = 14 - 9 = 5.

Therefore, the expression under the square root becomes 5 * 3 * 3 = 45. Therefore, area = sqrt(45) = 3*sqrt(5).

Which matches the previous result! So that's reassuring. Therefore, the area is indeed 3√5.

So both methods give the same answer, so I think that's correct.

Alternatively, another way to think about it is that the area of the parallelogram formed by OA and OB is |OA × OB|, so the area of the triangle is half that. Which is exactly what I did first.

Therefore, I can be confident that the area of triangle OAB is 3√5.

Just to recap, steps were:

1. Find vectors OA and OB.

2. Compute their cross product.

3. Find the magnitude of the cross product.

4. Divide by 2 to get the area of the triangle.

Alternatively, using Heron's formula with the lengths of the sides also gives the same answer, which is a good check.

Therefore, the answer is 3√5.

**Final Answer**
The area of triangle $ OAB $ is \boxed{3\sqrt{5}}.

To find the area of triangle $ OAB $, we are given the position vectors of points $ A $ and $ B $ with respect to the origin $ O $:

$$
\vec{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \vec{OB} = \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix}
$$

---

### Step 1: Use the Cross Product to Find the Area

The area of triangle $ OAB $ can be calculated using the formula:

$$
\text{Area} = \frac{1}{2} \left\| \vec{OA} \times \vec{OB} \right\|
$$

We compute the cross product $ \vec{OA} \times \vec{OB} $ using the determinant method:

$$
\vec{OA} \times \vec{OB} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
-3 & -2 & 1
\end{vmatrix}
= \mathbf{i}(2 \cdot 1 - 3 \cdot (-2)) - \mathbf{j}(1 \cdot 1 - 3 \cdot (-3)) + \mathbf{k}(1 \cdot (-2) - 2 \cdot (-3))
$$

$$
= \mathbf{i}(2 + 6) - \mathbf{j}(1 + 9) + \mathbf{k}(-2 + 6)
= 8\mathbf{i} - 10\mathbf{j} + 4\mathbf{k}
$$

So,

$$
\vec{OA} \times \vec{OB} = \begin{pmatrix} 8 \\ -10 \\ 4 \end{pmatrix}
$$

---

### Step 2: Compute the Magnitude of the Cross Product

$$
\left\| \vec{OA} \times \vec{OB} \right\| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180}
$$

Simplify:

$$
\sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}
$$

---

### Step 3: Compute the Area of the Triangle

$$
\text{Area} = \frac{1}{2} \cdot 6\sqrt{5} = 3\sqrt{5}
$$

---

### Final Answer

$$
\boxed{3\sqrt{5}}
$$

Solution:
The area of the parallelogram generated by $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is given by
\[\|\overrightarrow{OA} \times \overrightarrow{OB}\| = \left\| \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix} \right\| = \left\| \begin{pmatrix} 8 \\ -10 \\ 4 \end{pmatrix} \right\| = 6 \sqrt{5}.\][asy]
unitsize(0.4 cm);

pair A, B, C, D;

A = (0,0);
B = (7,2);
C = (1,3);
D = B + C;

draw(A--B,Arrow(6));
draw(A--C,Arrow(6));
draw(B--C);
draw(B--D--C,dashed);

label("$O$", A, SW);
label("$A$", B, SE);
label("$B$", C, W);
[/asy]

Therefore, the area of triangle $OAB$ is $\boxed{3 \sqrt{5}}.$


Explanation:
Okay, so I need to simplify the fraction 3 over the square root of 27. Let me think about how to approach this. I remember that simplifying expressions with square roots often involves rationalizing the denominator, which means getting rid of the square root in the denominator. But first, maybe I should simplify the square root of 27 as much as possible?

Alright, let me recall that 27 is a multiple of 9, which is a perfect square. So, maybe I can break down the square root of 27 into the square root of 9 times the square root of 3. Let me check that. Yes, because 9 times 3 is 27. So, √27 = √(9*3). And since the square root of a product is the product of the square roots, that would be √9 * √3. And √9 is 3, right? So that simplifies to 3√3.

So, substituting back into the original expression, the denominator becomes 3√3. Therefore, the original fraction 3 over √27 simplifies to 3 over 3√3. Wait, so that would be 3 divided by 3√3. Hmm, the 3 in the numerator and the 3 in the denominator can cancel each other out. So that leaves me with 1 over √3.

But now I have a square root in the denominator again. I think the standard form is to have no radicals in the denominator. So, I need to rationalize the denominator here. How do I do that? I remember that you can multiply the numerator and the denominator by the same radical to eliminate the square root in the denominator.

In this case, the denominator is √3, so if I multiply both the numerator and the denominator by √3, that should rationalize it. Let me try that.

So, multiplying numerator and denominator by √3 gives me (1 * √3) over (√3 * √3). The numerator becomes √3, and the denominator becomes √3 squared, which is 3. So, that simplifies to √3 over 3.

Wait, so putting it all together, starting from 3/√27, simplifying √27 to 3√3, then simplifying the fraction to 1/√3, and then rationalizing to get √3/3. Is that the simplest form? Let me check if there's another way or if I made a mistake somewhere.

Alternatively, maybe I could have simplified the original expression differently. Let me try another approach. Let's take the original expression: 3 divided by √27. Instead of simplifying √27 first, maybe I can express 3 as a multiple of √27? Wait, that might not be helpful. Let me think.

Alternatively, maybe I can write both numerator and denominator in terms of exponents. Let me recall that √27 is 27^(1/2). And 27 is 3^3. So, √27 is (3^3)^(1/2) which is 3^(3/2). Therefore, the original expression is 3 divided by 3^(3/2).

Using exponent rules, dividing exponents with the same base means subtracting exponents. So, 3 is 3^1, so 3^1 divided by 3^(3/2) is 3^(1 - 3/2) = 3^(-1/2). Which is the same as 1 over 3^(1/2), which is 1 over √3. Then, rationalizing gives √3 over 3. So, same result as before.

Therefore, regardless of the method, I end up with √3 over 3. Let me check if that's correct by approximating the original expression and the simplified one numerically.

Original expression: 3 divided by √27. Let me compute √27 first. √25 is 5, √27 is a bit more than 5. Let me calculate it. 5.1 squared is 26.01, 5.2 squared is 27.04. So √27 is approximately 5.196. Therefore, 3 divided by 5.196 is approximately 0.577.

Now, √3 is approximately 1.732, so √3 divided by 3 is approximately 1.732 / 3 ≈ 0.577. So, same value. Therefore, that seems correct.

Alternatively, if I didn't rationalize, 1 over √3 is approximately 1 / 1.732 ≈ 0.577, which is the same. So, both forms are equivalent, but the rationalized form is considered more simplified.

Therefore, the simplified form of 3 over √27 is √3 over 3.

Wait, but let me check if there's a way to simplify it further. For example, can √3 over 3 be simplified more? I don't think so. √3 is already in its simplest radical form, and 3 is a rational number. So, that's as simplified as it gets.

Alternatively, maybe I can write it as (1/3)√3, but that's essentially the same thing. So, both forms are acceptable, but typically, it's written with the radical in the numerator.

Therefore, I think the answer is √3 over 3.

But let me go back through the steps again to make sure I didn't skip anything.

Original problem: 3 / √27.

Step 1: Simplify √27. Since 27 = 9 * 3, √27 = √(9*3) = √9 * √3 = 3√3.

So, substituting back, 3 / (3√3).

Step 2: Simplify numerator and denominator. 3 divided by 3 is 1, so we have 1 / √3.

Step 3: Rationalize the denominator. Multiply numerator and denominator by √3.

(1 * √3) / (√3 * √3) = √3 / 3.

Yes, that's correct.

Alternatively, if I didn't simplify √27 first, but instead rationalized the original expression. Let me try that approach.

Original expression: 3 / √27.

Multiply numerator and denominator by √27 to rationalize.

So, (3 * √27) / (√27 * √27) = (3√27) / 27.

Then, simplify numerator and denominator.

First, simplify √27. As before, √27 = 3√3. Therefore, numerator becomes 3 * 3√3 = 9√3.

Denominator is 27.

So, 9√3 / 27. Simplify by dividing numerator and denominator by 9: √3 / 3.

Same result.

Therefore, regardless of the method, the simplified form is √3 over 3.

I think that's solid. Let me check if there's any alternative interpretation of the original problem. For example, is there a possibility that the original expression was meant to be something else? Like, maybe 3 divided by sqrt(27), which is what I assumed.

Alternatively, if someone misreads it as sqrt(3)/27, but that's not the case here. The original problem is written as 3 over sqrt(27), so the fraction is 3 divided by the square root of 27.

Another check: Let me compute both the original expression and the simplified expression using exact values.

Original expression: 3 / √27.

Simplified expression: √3 / 3.

Let me square both to see if they are equal.

Original squared: (3 / √27)^2 = 9 / 27 = 1/3.

Simplified squared: (√3 / 3)^2 = 3 / 9 = 1/3.

Same result. Therefore, they are equal.

Therefore, the simplification is correct.

So, after all these steps and checks, I can confidently say that the simplified form of 3 over the square root of 27 is √3 over 3.

**Final Answer**
The simplified form is \boxed{\dfrac{\sqrt{3}}{3}}.

To simplify the expression:

$$
\frac{3}{\sqrt{27}}
$$

we follow a systematic approach, focusing on simplifying the square root in the denominator and rationalizing it if necessary.

---

### Step 1: Simplify the Square Root in the Denominator

We begin by simplifying the square root of 27. Since 27 is not a perfect square, we factor it into a product of a perfect square and another number:

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$

So, the original expression becomes:

$$
\frac{3}{\sqrt{27}} = \frac{3}{3\sqrt{3}}
$$

---

### Step 2: Simplify the Fraction

Now, we simplify the fraction by canceling the common factor of 3 in the numerator and denominator:

$$
\frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}
$$

---

### Step 3: Rationalize the Denominator

To express the fraction in a standard form with no radicals in the denominator, we rationalize the denominator. Multiply both the numerator and the denominator by $\sqrt{3}$:

$$
\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}
$$

---

### Final Answer

The simplified form of the expression is:

$$
\boxed{\dfrac{\sqrt{3}}{3}}
$$

[Assistant]
Explanation:
Okay, so I need to simplify the fraction 3 over the square root of 27. Let me think about how to approach this. I remember that simplifying expressions with square roots often involves rationalizing the denominator, which means getting rid of the square root in the denominator. But first, maybe I should simplify the square root of 27 as much as possible?

Alright, let me recall that 27 is a multiple of 9, which is a perfect square. So, maybe I can break down the square root of 27 into the square root of 9 times the square root of 3. Let me check that. Yes, because 9 times 3 is 27. So, √27 = √(9*3). And since the square root of a product is the product of the square roots, that would be √9 * √3. And √9 is 3, right? So that simplifies to 3√3.

So, substituting back into the original expression, the denominator becomes 3√3. Therefore, the original fraction 3 over √27 simplifies to 3 over 3√3. Wait, so that would be 3 divided by 3√3. Hmm, the 3 in the numerator and the 3 in the denominator can cancel each other out. So that leaves me with 1 over √3.

But now I have a square root in the denominator again. I think the standard form is to have no radicals in the denominator. So, I need to rationalize the denominator here. How do I do that? I remember that you can multiply the numerator and the denominator by the same radical to eliminate the square root in the denominator.

In this case, the denominator is √3, so if I multiply both the numerator and the denominator by √3, that should rationalize it. Let me try that.

So, multiplying numerator and denominator by √3 gives me (1 * √3) over (√3 * √3). The numerator becomes √3, and the denominator becomes √3 squared, which is 3. So, that simplifies to √3 over 3.

Wait, so putting it all together, starting from 3/√27, simplifying √27 to 3√3, then simplifying the fraction to 1/√3, and then rationalizing to get √3/3. Is that the simplest form? Let me check if there's another way or if I made a mistake somewhere.

Alternatively, maybe I could have simplified the original expression differently. Let me try another approach. Let's take the original expression: 3 divided by √27. Instead of simplifying √27 first, maybe I can express 3 as a multiple of √27? Wait, that might not be helpful. Let me think.

Alternatively, maybe I can write both numerator and denominator in terms of exponents. Let me recall that √27 is 27^(1/2). And 27 is 3^3. So, √27 is (3^3)^(1/2) which is 3^(3/2). Therefore, the original expression is 3 divided by 3^(3/2).

Using exponent rules, dividing exponents with the same base means subtracting exponents. So, 3 is 3^1, so 3^1 divided by 3^(3/2) is 3^(1 - 3/2) = 3^(-1/2). Which is the same as 1 over 3^(1/2), which is 1 over √3. Then, rationalizing gives √3 over 3. So, same result as before.

Therefore, regardless of the method, I end up with √3 over 3. Let me check if that's correct by approximating the original expression and the simplified one numerically.

Original expression: 3 divided by √27. Let me compute √27 first. √25 is 5, √27 is a bit more than 5. Let me calculate it. 5.1 squared is 26.01, 5.2 squared is 27.04. So √27 is approximately 5.196. Therefore, 3 divided by 5.196 is approximately 0.577.

Now, √3 is approximately 1.732, so √3 divided by 3 is approximately 1.732 / 3 ≈ 0.577. So, same value. Therefore, that seems correct.

Alternatively, if I didn't rationalize, 1 over √3 is approximately 1 / 1.732 ≈ 0.577, which is the same. So, both forms are equivalent, but the rationalized form is considered more simplified.

Therefore, the simplified form of 3 over √27 is √3 over 3.

Wait, but let me check if there's a way to simplify it further. For example, can √3 over 3 be simplified more? I don't think so. √3 is already in its simplest radical form, and 3 is a rational number. So, that's as simplified as it gets.

Alternatively, maybe I can write it as (1/3)√3, but that's essentially the same thing. So, both forms are acceptable, but typically, it's written with the radical in the numerator.

Therefore, I think the answer is √3 over 3.

But let me go back through the steps again to make sure I didn't skip anything.

Original problem: 3 / √27.

Step 1: Simplify √27. Since 27 = 9 * 3, √27 = √(9*3) = √9 * √3 = 3√3.

So, substituting back, 3 / (3√3).

Step 2: Simplify numerator and denominator. 3 divided by 3 is 1, so we have 1 / √3.

Step 3: Rationalize the denominator. Multiply numerator and denominator by √3.

(1 * √3) / (√3 * √3) = √3 / 3.

Yes, that's correct.

Alternatively, if I didn't simplify √27 first, but instead rationalized the original expression. Let me try that approach.

Original expression: 3 / √27.

Multiply numerator and denominator by √27 to rationalize.

So, (3 * √27) / (√27 * √27) = (3√27) / 27.

Then, simplify numerator and denominator.

First, simplify √27. As before, √27 = 3√3. Therefore, numerator becomes 3 * 3√3 = 9√3.

Denominator is 27.

So, 9√3 / 27. Simplify by dividing numerator and denominator by 9: √3 / 3.

Same result.

Therefore, regardless of the method, the simplified form is √3 over 3.

I think that's solid. Let me check if there's any alternative interpretation of the original problem. For example, is there a possibility that the original expression was meant to be something else? Like, maybe 3 divided by sqrt(27), which is what I assumed.

Alternatively, if someone misreads it as sqrt(3)/27, but that's not the case here. The original problem is written as 3 over sqrt(27), so the fraction is 3 divided by the square root of 27.

Another check: Let me compute both the original expression and the simplified expression using exact values.

Original expression: 3 / √27.

Simplified expression: √3 / 3.

Let me square both to see if they are equal.

Original squared: (3 / √27)^2 = 9 / 27 = 1/3.

Simplified squared: (√3 / 3)^2 = 3 / 9 = 1/3.

Same result. Therefore, they are equal.

Therefore, the simplification is correct.

So, after all these steps and checks, I can confidently say that the simplified form of 3 over the square root of 27 is √3 over 3.

**Final Answer**
The simplified form is \boxed{\dfrac{\sqrt{3}}{3}}.

To simplify the expression:

$$
\frac{3}{\sqrt{27}}
$$

we follow a systematic approach, focusing on simplifying the square root in the denominator and rationalizing it if necessary.

---

### Step 1: Simplify the Square Root in the Denominator

We begin by simplifying the square root of 27. Since 27 is not a perfect square, we factor it into a product of a perfect square and another number:

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$

So, the original expression becomes:

$$
\frac{3}{\sqrt{27}} = \frac{3}{3\sqrt{3}}
$$

---

### Step 2: Simplify the Fraction

Now, we simplify the fraction by canceling the common factor of 3 in the numerator and denominator:

$$
\frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}
$$

---

### Step 3: Rationalize the Denominator

To express the fraction in a standard form with no radicals in the denominator, we rationalize the denominator. Multiply both the numerator and the denominator by $\sqrt{3}$:

$$
\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}
$$

---

### Final Answer

The simplified form of the expression is:

$$
\boxed{\dfrac{\sqrt{3}}{3}}
$$

Solution:
We have:

$\frac{3}{\sqrt{27}}=\frac{3\sqrt{3}}{\sqrt{81}}=\frac{3\sqrt{3}}{9}=\boxed{\frac{\sqrt{3}}{3}}$.


Explanation:
Okay, so there's this problem about two runners, A and B, starting from the same point O on a linear track. They run in the same direction, and B is three times as fast as A. There's an observer at point P, where OP is perpendicular to the track. We need to find the maximum angle APB, in degrees.

First, let me visualize the setup. The track is a straight line, and OP is perpendicular to it. So, if I imagine the track as the x-axis, then point O is at the origin (0,0), and point P is somewhere along the y-axis, say at (0, h) for some height h. The Asymptote code shows P at (0,1), but maybe the actual coordinates aren't important, just the relative positions.

Runners A and B start at O and run in the same direction. Let me assume they run along the positive x-axis. Since B is three times as fast as A, if A's speed is v, then B's speed is 3v.

The question is about the angle APB. So, we need to consider the positions of A and B at some time t, then find the angle between PA and PB. The goal is to find the maximum value of this angle as the runners move along the track.

Let me think about how to model this. Let's set up coordinates. Let me define the track as the x-axis. Let O be at (0,0), and P at (0, h). Let me assume h is some positive number. Since the problem doesn't specify the distance from O to P, maybe it cancels out or we can assume a specific value for simplicity. Maybe h = 1? The Asymptote code shows P at (0,1), so maybe that's a hint. Let me go with that. So P is at (0,1).

Let me denote the position of runner A at time t as (x(t), 0), and runner B as (y(t), 0). Since they start at O, which is (0,0), and run in the same direction. Let me assume A's speed is v, so x(t) = v*t. Similarly, B's speed is 3v, so y(t) = 3v*t.

But maybe to make things simpler, I can set v = 1, so that A moves at speed 1, and B at speed 3. Then x(t) = t, y(t) = 3t.

So, positions of A and B at time t are (t, 0) and (3t, 0), respectively.

Now, we need to find angle APB. So, points A, P, B. Let me recall that angle APB is the angle at point P between points A and B. So, if I can find vectors PA and PB, then the angle between them can be found using the dot product formula.

Alternatively, maybe using coordinates, we can compute the angle. Let me try that.

First, coordinates:

Point P is (0,1).

Point A is (t, 0).

Point B is (3t, 0).

So, vector PA is A - P = (t - 0, 0 - 1) = (t, -1).

Similarly, vector PB is B - P = (3t - 0, 0 - 1) = (3t, -1).

Then, the angle between vectors PA and PB can be found using the dot product formula:

cos(theta) = (PA · PB) / (|PA| |PB|)

Compute PA · PB: (t)(3t) + (-1)(-1) = 3t² + 1

Compute |PA|: sqrt(t² + (-1)^2) = sqrt(t² + 1)

Similarly, |PB|: sqrt((3t)^2 + (-1)^2) = sqrt(9t² + 1)

Therefore,

cos(theta) = (3t² + 1) / [sqrt(t² + 1) * sqrt(9t² + 1)]

Our goal is to find the maximum angle theta, which corresponds to minimizing cos(theta), since cosine is a decreasing function in [0, π].

Therefore, we need to find the value of t that minimizes the expression (3t² + 1)/[sqrt(t² + 1) * sqrt(9t² + 1)].

Alternatively, since dealing with square roots can be messy, maybe square the expression to make it easier. Let me denote f(t) = [ (3t² + 1)^2 ] / [ (t² + 1)(9t² + 1) ]

Then, minimizing f(t) will correspond to minimizing the original expression, hence maximizing theta.

So, let's compute f(t):

f(t) = (9t^4 + 6t² + 1) / [ (t² + 1)(9t² + 1) ]

Let me compute the denominator:

(t² + 1)(9t² + 1) = 9t^4 + t² + 9t² + 1 = 9t^4 + 10t² + 1

So, f(t) = (9t^4 + 6t² + 1) / (9t^4 + 10t² + 1)

So, we need to minimize f(t). Let me denote u = t², since t² is non-negative. Let u ≥ 0.

Then, f(u) = (9u² + 6u + 1) / (9u² + 10u + 1)

So, we need to find the minimum of f(u) for u ≥ 0.

Let me compute derivative of f(u) with respect to u, set it to zero.

Let me denote numerator as N = 9u² + 6u + 1, denominator as D = 9u² + 10u + 1.

Then, f(u) = N/D, so derivative f’(u) = (N’ D - N D’) / D².

Compute N’ = 18u + 6

Compute D’ = 18u + 10

Therefore,

f’(u) = [ (18u + 6)(9u² + 10u + 1) - (9u² + 6u + 1)(18u + 10) ] / (D²)

Let me compute the numerator:

Let me expand both products.

First term: (18u + 6)(9u² + 10u + 1)

Let me compute this:

Multiply 18u by each term:

18u * 9u² = 162u³

18u * 10u = 180u²

18u * 1 = 18u

Then, 6 * 9u² = 54u²

6 * 10u = 60u

6 * 1 = 6

So adding all together:

162u³ + 180u² + 18u + 54u² + 60u + 6

Combine like terms:

162u³ + (180u² + 54u²) + (18u + 60u) + 6

Which is:

162u³ + 234u² + 78u + 6

Second term: (9u² + 6u + 1)(18u + 10)

Let me compute this:

Multiply 9u² by each term:

9u² * 18u = 162u³

9u² * 10 = 90u²

Multiply 6u by each term:

6u * 18u = 108u²

6u * 10 = 60u

Multiply 1 by each term:

1 * 18u = 18u

1 * 10 = 10

Adding all together:

162u³ + 90u² + 108u² + 60u + 18u + 10

Combine like terms:

162u³ + (90u² + 108u²) + (60u + 18u) + 10

Which is:

162u³ + 198u² + 78u + 10

Therefore, the numerator of f’(u) is:

[First term] - [Second term] = [162u³ + 234u² + 78u + 6] - [162u³ + 198u² + 78u + 10]

Compute term by term:

162u³ - 162u³ = 0

234u² - 198u² = 36u²

78u - 78u = 0

6 - 10 = -4

Therefore, numerator is 36u² - 4.

Therefore, f’(u) = (36u² - 4)/D²

Set derivative equal to zero:

36u² - 4 = 0 => 36u² = 4 => u² = 4/36 = 1/9 => u = 1/3 or u = -1/3. But since u ≥ 0, only u = 1/3 is critical point.

Therefore, the function f(u) has critical point at u = 1/3. Now, we need to check if this is a minimum or maximum.

Since f’(u) changes sign from negative to positive or vice versa. Let me check the sign of f’(u) around u = 1/3.

For u < 1/3, say u = 0: numerator is 36*(0) - 4 = -4 < 0, so f’(u) < 0.

For u > 1/3, say u = 1: numerator is 36*(1) - 4 = 32 > 0, so f’(u) > 0.

Therefore, the function f(u) decreases until u = 1/3, then increases. Therefore, u = 1/3 is the point of minimum. Therefore, the minimum of f(u) occurs at u = 1/3, which corresponds to t² = 1/3, so t = 1/√3.

Therefore, the minimum value of f(u) is f(1/3):

Compute f(1/3) = [9*(1/3)^2 + 6*(1/3) + 1] / [9*(1/3)^2 + 10*(1/3) + 1]

Compute numerator:

9*(1/9) + 6*(1/3) + 1 = 1 + 2 + 1 = 4

Denominator:

9*(1/9) + 10*(1/3) + 1 = 1 + 10/3 + 1 = 2 + 10/3 = 16/3

Therefore, f(1/3) = 4 / (16/3) = 4 * 3 / 16 = 12 / 16 = 3/4.

Therefore, the minimum value of f(u) is 3/4, which means the minimum value of cos(theta) is 3/4. Therefore, the maximum angle theta is arccos(3/4).

Wait, but the question says "Find the maximum of angle APB". So, arccos(3/4) is the angle when cos(theta) is minimized. Therefore, theta_max = arccos(3/4). But we need to convert this to degrees.

But maybe there's a way to find the exact value? Or maybe it's a standard angle? Let me check.

Alternatively, maybe I made a mistake here. Let me verify.

Wait, arccos(3/4) is approximately... Well, cos(41.4 degrees) is approximately 0.75. Let me check with calculator:

cos(41.4°) ≈ cos(41.4) ≈ 0.75. Yes, because cos(60°) is 0.5, cos(30°) ≈ 0.866. So 0.75 is between 30 and 60, closer to 41 degrees. But the problem says to find the maximum angle, and maybe it's a nice angle? Wait, but maybe there's a different approach where we can find an exact value?

Alternatively, maybe I made a mistake in the calculation. Let me check again.

Wait, let me check the calculation of f(1/3):

Numerator: 9u² + 6u + 1 with u = 1/3:

9*(1/3)^2 = 9*(1/9) = 1

6*(1/3) = 2

So total numerator: 1 + 2 + 1 = 4. Correct.

Denominator: 9u² + 10u + 1:

Same 9*(1/3)^2 = 1, 10*(1/3) = 10/3, so total denominator: 1 + 10/3 + 1 = 2 + 10/3 = 16/3. Correct. So 4 divided by 16/3 is 3/4. Correct. So cos(theta) = 3/4. Therefore theta = arccos(3/4).

But the problem says "Find the maximum of angle APB, in degrees." So unless there's a way to express arccos(3/4) in exact degrees, which I don't think so, but maybe there's a different approach where we can find it?

Alternatively, maybe I made a wrong assumption in setting up the coordinates? Let me check.

Wait, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, OP is perpendicular, so P is somewhere off the track. But in my coordinate system, I assumed P is at (0,1). But maybe the position of P is arbitrary? Wait, but the problem doesn't specify where P is, just that OP is perpendicular. However, since the track is linear, and OP is perpendicular, then P is at some fixed distance from O. However, in the Asymptote code, they have P at (0,1), but maybe the actual position of P is arbitrary? Wait, but if we don't know where P is, how can we find the maximum angle? Wait, but maybe the maximum angle is independent of the position of P? Or maybe the problem assumes that P is fixed? Wait, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, maybe P is fixed? But since the track is linear, and OP is perpendicular, then P is at a fixed distance from O. However, the problem doesn't specify the distance. Wait, but in the Asymptote code, they have P at (0,1), but maybe that's just for illustration. However, in the problem statement, since they don't specify the distance, maybe the maximum angle is independent of the distance? Wait, but in my calculation, I assumed P is at (0,1). But if I take P at (0, h), would the result be the same?

Let me check. Let me redo the calculation with P at (0, h). Let me see if the angle depends on h or not.

So, let me redo the problem with P at (0, h). Then, coordinates:

Point P: (0, h)

Point A: (t, 0)

Point B: (3t, 0)

Vectors PA = (t - 0, 0 - h) = (t, -h)

Vector PB = (3t - 0, 0 - h) = (3t, -h)

Dot product PA · PB = t*(3t) + (-h)*(-h) = 3t² + h²

|PA| = sqrt(t² + h²)

|PB| = sqrt(9t² + h²)

Therefore, cos(theta) = (3t² + h²)/[sqrt(t² + h²) * sqrt(9t² + h²)]

So, same as before, except with h instead of 1. Then, if we do the same substitution u = t², then:

Let me define f(u) = [3u + h²]/[sqrt(u + h²) * sqrt(9u + h²)] ?

Wait, no, wait. Wait, the numerator is 3t² + h², which is 3u + h². The denominator is sqrt(u + h²) * sqrt(9u + h²). Therefore, if we square it, we get [ (3u + h²)^2 ] / [ (u + h²)(9u + h²) ]

So, same as before, except with h² instead of 1. So, if we do the same steps as before, we can see that the critical point occurs at u = 1/3, regardless of h? Wait, let me check.

Let me do the same steps. Let me set u = t², then f(u) = [ (3u + h²)^2 ] / [ (u + h²)(9u + h²) ]

Let me compute the derivative. Let me denote N = (3u + h²)^2, D = (u + h²)(9u + h²). Then f(u) = N/D. Then f’(u) = [N’ D - N D’]/D².

Compute N’ = 2*(3u + h²)*3 = 6*(3u + h²)

Compute D = (u + h²)(9u + h²) = 9u² + 10u h² + h^4

Wait, no, expanding (u + h²)(9u + h²) = 9u² + u h² + 9u h² + h^4 = 9u² + 10u h² + h^4. Wait, but earlier when h was 1, we had 9u² + 10u + 1. So, yes, similar.

But D’ = derivative of D with respect to u is 18u + 10 h².

Wait, let me compute D’:

D = (u + h²)(9u + h²). So, derivative is (1)(9u + h²) + (u + h²)(9) = 9u + h² + 9u + 9h² = 18u + 10 h². Correct.

Therefore, f’(u) = [6*(3u + h²)*(9u² + 10u h² + h^4) - (3u + h²)^2*(18u + 10 h²)] / D²

Wait, this seems complicated. However, note that in the previous case when h = 1, we found that the numerator simplifies to 36u² - 4. But here, with general h, maybe there's a similar simplification?

Alternatively, maybe we can factor out terms. Let me factor out (3u + h²):

Numerator: 6*(3u + h²)*D - (3u + h²)^2 * D’ = (3u + h²)[6D - (3u + h²) D’]

So, let me compute 6D - (3u + h²) D’:

First, D = (u + h²)(9u + h²)

D’ = 18u + 10 h²

So,

6D - (3u + h²) D’ = 6(u + h²)(9u + h²) - (3u + h²)(18u + 10 h²)

Let me factor out terms. Let me denote A = u + h², B = 9u + h², C = 3u + h², D’ = 18u + 10 h².

But maybe expanding:

First term: 6(u + h²)(9u + h²) = 6[9u² + u h² + 9u h² + h^4] = 6[9u² + 10u h² + h^4] = 54u² + 60u h² + 6 h^4

Second term: (3u + h²)(18u + 10 h²) = 3u*18u + 3u*10 h² + h²*18u + h²*10 h² = 54u² + 30u h² + 18u h² + 10 h^4 = 54u² + 48u h² + 10 h^4

Therefore, 6D - (3u + h²) D’ = [54u² + 60u h² + 6 h^4] - [54u² + 48u h² + 10 h^4] = (54u² - 54u²) + (60u h² - 48u h²) + (6 h^4 - 10 h^4) = 12u h² - 4 h^4 = 4 h² (3u - h²)

Therefore, numerator of f’(u) is (3u + h²) * 4 h² (3u - h²) = 4 h² (3u + h²)(3u - h²) = 4 h² (9u² - h^4)

Wait, no:

Wait, (3u + h²)(3u - h²) = 9u² - h^4. Therefore, numerator is 4 h² (9u² - h^4). Therefore, f’(u) = [4 h² (9u² - h^4)] / D²

Set derivative equal to zero: 9u² - h^4 = 0 => 9u² = h^4 => u² = h^4 / 9 => u = h² / 3 (since u ≥ 0)

Therefore, critical point at u = h² / 3. Therefore, regardless of h, the critical point is at u = h² / 3. Therefore, t² = h² / 3 => t = h / √3.

Therefore, the minimum of f(u) occurs at u = h² / 3. Therefore, the value of cos(theta) at this point is:

Let me compute f(u) at u = h² / 3:

Numerator: (3u + h²)^2 = [3*(h² / 3) + h²]^2 = [h² + h²]^2 = (2 h²)^2 = 4 h^4

Denominator: (u + h²)(9u + h²) = [ (h² / 3) + h² ] * [ 9*(h² / 3) + h² ] = [ (4 h² / 3) ] * [ 3 h² + h² ] = (4 h² / 3) * 4 h² = (16 h^4)/3

Therefore, f(u) = 4 h^4 / (16 h^4 / 3) ) = 4 / (16 / 3) ) = 3/4.

So, regardless of h, the minimum value of cos(theta) is 3/4. Therefore, theta_max = arccos(3/4). Therefore, the maximum angle APB is arccos(3/4) radians, which we need to convert to degrees.

But the problem says "Find the maximum of angle APB, in degrees." So, unless there's a way to express arccos(3/4) in exact degrees, which I don't think so, but maybe there's a different approach where we can find it?

Alternatively, maybe I made a mistake in assuming that the maximum occurs at this critical point? Let me check with specific values.

For example, let me take t approaching 0. When t approaches 0, points A and B are both near O. Then, angle APB approaches the angle between PO and PA, and PO and PB. Since A and B are both near O, the angle APB approaches the angle between PO and PO, which is zero? Wait, no. Wait, when t approaches 0, points A and B are both near O. So, PA is from P to A, which is near O, and PB is from P to B, which is also near O. So, the angle between PA and PB would approach the angle between two lines from P to O, which is zero? Wait, but O is the point where both A and B start. Wait, but when t approaches 0, A and B are both approaching O, so PA and PB are approaching PO. Therefore, angle APB approaches zero.

When t approaches infinity, what happens? Let me see. As t becomes very large, points A and B are far away on the track. The vectors PA and PB are approximately (t, -h) and (3t, -h). The angle between them can be approximated by the angle between (1, -h/t) and (3, -h/t). As t approaches infinity, h/t approaches zero, so vectors approach (1, 0) and (3, 0). The angle between them approaches zero. Therefore, as t approaches infinity, angle APB approaches zero. Therefore, the angle APB has a maximum somewhere in between, which we found at t = h / √3, giving angle arccos(3/4).

Therefore, the maximum angle is arccos(3/4). Now, converting this to degrees.

But the problem says "Find the maximum of angle APB, in degrees." So, unless there's a specific value, but maybe it's a standard angle? Let me check.

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

Alternatively, maybe using calculus, we can find the exact value. But arccos(3/4) is approximately 41.4 degrees. However, the problem might expect an exact value. Wait, but maybe there's a way to find it using some geometric properties?

Alternatively, maybe using the Law of Sines or Law of Cosines on triangle APB?

Let me try that. Let me consider triangle APB. Let me find the lengths of PA, PB, and AB, then use the Law of Cosines to find angle APB.

PA = sqrt(t² + h²)

PB = sqrt(9t² + h²)

AB = distance between A and B = 3t - t = 2t

So, in triangle APB, sides are PA, PB, AB. Let me denote angle at P as theta. Then, by Law of Cosines:

AB² = PA² + PB² - 2 PA PB cos(theta)

Therefore,

(2t)^2 = (t² + h²) + (9t² + h²) - 2 sqrt(t² + h²) sqrt(9t² + h²) cos(theta)

Simplify:

4t² = t² + h² + 9t² + h² - 2 sqrt(...) cos(theta)

Which is:

4t² = 10t² + 2h² - 2 sqrt(...) cos(theta)

Rearranged:

2 sqrt(...) cos(theta) = 10t² + 2h² - 4t² = 6t² + 2h²

Therefore,

cos(theta) = (6t² + 2h²) / [2 sqrt(t² + h²) sqrt(9t² + h²)] = (3t² + h²)/[sqrt(t² + h²) sqrt(9t² + h²)]

Which is the same expression as before. Therefore, same result. Therefore, we can't get a better expression here.

Therefore, the maximum angle is arccos(3/4). But the problem says "Find the maximum of angle APB, in degrees." So, unless there's a specific value, but maybe the answer is 90 degrees? Wait, but arccos(3/4) is less than 90 degrees, since 3/4 is positive. Wait, arccos(3/4) is acute. Wait, but maybe I made a mistake in the direction of the angle? Wait, angle APB is the angle at P between A and B. So, if the maximum angle is arccos(3/4), which is approximately 41.4 degrees. But maybe there's a different approach where we can find it?

Alternatively, maybe using calculus, we can find the exact value? Wait, but arccos(3/4) is not a standard angle. However, maybe the problem expects an exact value in terms of inverse trigonometric functions? But the problem says "in degrees," so likely expects a numerical value. However, since it's a competition problem, maybe there's a trick to find it?

Wait, let me think again. Maybe there's a geometric interpretation where the maximum angle occurs when the circle through A and B is tangent to the line OP? Wait, that might be a method used in some optimization problems.

Let me recall that for a fixed point P, the locus of points A and B moving along the track... Wait, but A and B are moving with different speeds. Wait, but maybe for the angle APB to be maximum, the circle passing through A and B with P on the circle? Wait, the maximum angle occurs when the circle through A and B is tangent to the line OP?

Alternatively, the maximum angle occurs when the circle passing through A and B is tangent to the line OP. Because for a fixed point P, the angle APB is maximized when the circle through A and B is tangent to the line OP.

Let me try to think.

If we fix P, and consider points A and B moving along the track, then the angle APB is maximized when the circle passing through A and B is tangent to the line OP. Because if the circle is tangent, then the angle is maximized.

Alternatively, this is similar to the problem where you have a fixed point and a line, and you want to find the maximum angle subtended by a point moving along the line.

But in this case, we have two points moving along the line with different speeds.

Alternatively, maybe we can parametrize the positions of A and B such that the angle APB is maximized.

But since we already did calculus and found that the maximum angle is arccos(3/4), which is approximately 41.4 degrees, but maybe the problem expects an exact value? Wait, but maybe there's a miscalculation here. Let me check again.

Wait, let me check with specific numbers. Let me take h = 1, t = 1/√3 ≈ 0.577. Then, coordinates:

Point A: (1/√3, 0)

Point B: (3/√3, 0) = (√3, 0)

Point P: (0, 1)

Compute vectors PA and PB:

PA = (1/√3 - 0, 0 - 1) = (1/√3, -1)

PB = (√3 - 0, 0 - 1) = (√3, -1)

Compute the angle between PA and PB.

Dot product PA · PB = (1/√3)(√3) + (-1)(-1) = 1 + 1 = 2

|PA| = sqrt( (1/√3)^2 + (-1)^2 ) = sqrt(1/3 + 1) = sqrt(4/3) = 2/√3

|PB| = sqrt( (√3)^2 + (-1)^2 ) = sqrt(3 + 1) = 2

Therefore, cos(theta) = 2 / [ (2/√3)(2) ] = 2 / (4 / √3) ) = (2 * √3)/4 = √3 / 2 ≈ 0.866. Wait, but earlier I thought it was 3/4. Wait, this contradicts previous result. What's going on here?

Wait, hold on! Earlier, I thought that cos(theta) was 3/4, but here, with h = 1, t = 1/√3, I get cos(theta) = √3 / 2 ≈ 0.866, which is 30 degrees. Wait, but this contradicts previous calculation where I had f(u) = 3/4. What is wrong here?

Wait, let me check my previous steps. Earlier, I had:

cos(theta) = (3t² + h²)/[sqrt(t² + h²) * sqrt(9t² + h²)]

With h = 1, t = 1/√3:

Numerator: 3*(1/3) + 1 = 1 + 1 = 2

Denominator: sqrt(1/3 + 1) * sqrt(9*(1/3) + 1) = sqrt(4/3) * sqrt(4) = (2/√3) * 2 = 4 / √3

Therefore, cos(theta) = 2 / (4 / √3) ) = (2 * √3)/4 = √3 / 2 ≈ 0.866, which is 30 degrees. But earlier, I thought that f(u) was 3/4. Wait, but in the previous calculation, when I did f(u) = [ (3u + h²)^2 ] / [ (u + h²)(9u + h²) ] with u = h² / 3, I had:

Numerator: (3u + h²)^2 = [3*(h² / 3) + h²]^2 = (h² + h²)^2 = (2h²)^2 = 4h^4

Denominator: (u + h²)(9u + h²) = [ (h² / 3) + h² ] * [ 9*(h² / 3) + h² ] = (4h² / 3) * (4h²) = 16h^4 / 3

Therefore, f(u) = 4h^4 / (16h^4 / 3) ) = 3/4. But this was the square of cos(theta). Wait! Wait, earlier I said that f(u) = [ (3t² + h²)^2 ] / [ (t² + h²)(9t² + h²) ] which is [cos(theta)]². Therefore, cos(theta) = sqrt(f(u))?

Wait, no! Wait, earlier, I had:

cos(theta) = (3t² + h²)/[sqrt(t² + h²) * sqrt(9t² + h²)]

Therefore, [cos(theta)]² = (3t² + h²)^2 / [ (t² + h²)(9t² + h²) ] = f(u). Therefore, when I calculated f(u) at u = h² / 3, I found f(u) = 3/4, so [cos(theta)]² = 3/4, so cos(theta) = sqrt(3)/2 or -sqrt(3)/2. Since theta is between 0 and 180 degrees, cos(theta) is positive, so cos(theta) = sqrt(3)/2, which implies theta = 30 degrees.

Wait, so earlier I had a mistake in thinking that cos(theta) was 3/4, but actually, [cos(theta)]² = 3/4, so cos(theta) = sqrt(3)/2, which is 30 degrees. Therefore, theta_max = 30 degrees.

Wait, so where was the mistake? Earlier, when I thought that f(u) was cos(theta), but actually f(u) is [cos(theta)]². Therefore, when I found f(u) = 3/4, that means cos(theta) = sqrt(3)/2, so theta = 30 degrees.

Therefore, the maximum angle is 30 degrees.

But why did I think earlier that cos(theta) was 3/4? Because I confused f(u) with cos(theta). So, that was the mistake.

So, correcting that, the maximum angle is arccos(sqrt(3)/2) = 30 degrees.

But let me verify with the specific example. When t = 1/√3, h = 1, we found that cos(theta) = sqrt(3)/2, which is 30 degrees. Therefore, the maximum angle is 30 degrees.

But why did the previous calculation with f(u) = 3/4? Because f(u) was [cos(theta)]², so cos(theta) = sqrt(3)/2.

Therefore, the maximum angle is 30 degrees.

But why did I get confused earlier? Because I thought that f(u) was cos(theta), but it's actually [cos(theta)]².

Therefore, the answer is 30 degrees.

But let me check with another example. Let me take t = 0. Then, points A and B are both at O. Then, angle APB is undefined, since A and B coincide. But as t approaches 0, angle approaches zero.

If I take t = 1, h = 1:

PA = sqrt(1 + 1) = sqrt(2)

PB = sqrt(9 + 1) = sqrt(10)

Dot product PA · PB = 3*1 + 1 = 4

cos(theta) = 4 / (sqrt(2)*sqrt(10)) = 4 / sqrt(20) = 4 / (2*sqrt(5)) = 2 / sqrt(5) ≈ 0.894, which is about 26.565 degrees. Which is less than 30 degrees.

If I take t = 1/√3 ≈ 0.577, then angle is 30 degrees.

If I take t approaching infinity, angle approaches zero.

Therefore, the maximum angle is indeed 30 degrees.

Therefore, the answer is 30 degrees.

But why did I get confused earlier? Because I miscalculated the relationship between f(u) and cos(theta).

So, the key steps are:

- Set up coordinates with P at (0, h), A at (t, 0), B at (3t, 0).

- Compute vectors PA and PB.

- Use dot product to find cos(theta).

- Find that [cos(theta)]² = (3t² + h²)^2 / [(t² + h²)(9t² + h²)].

- Minimize this expression, which leads to critical point at t² = h² / 3.

- At that point, [cos(theta)]² = 3/4, so cos(theta) = sqrt(3)/2, leading to theta = 30 degrees.

Therefore, the maximum angle is 30 degrees.

Therefore, the answer is 30 degrees.

But let me check with another approach. Let me consider the circle passing through A and B and tangent to OP.

If we consider the circle passing through A and B and tangent to OP at point P, then the angle APB would be 90 degrees? Wait, no. Wait, if a circle is tangent to OP at P, then the angle subtended by AB at P would be related to the circle.

Alternatively, maybe using the property that the maximum angle occurs when the circle through A and B is tangent to OP.

Let me try this approach.

Let me assume that there is a circle passing through A and B and tangent to OP at P. Then, the angle APB would be equal to the angle subtended by AB at P, which for a circle tangent at P would be... Wait, if the circle is tangent to OP at P, then P lies on the circle, and OP is tangent to the circle at P. Therefore, the radius at P is perpendicular to OP. But OP is perpendicular to the track, so the radius at P is along OP. Therefore, the center of the circle lies along OP.

Let me denote the center of the circle as C, lying on OP. Let me assume coordinates again. Let me take P at (0, h), O at (0, 0). The center C is somewhere along the y-axis, say at (0, k).

Since the circle passes through A and B, and is tangent to OP at P.

Since the circle is tangent to OP at P, the radius at P is perpendicular to OP. Since OP is vertical (from (0,0) to (0, h)), the tangent line OP is vertical, so the radius at P must be horizontal. But the center is on OP, which is vertical. Therefore, the radius from C to P must be horizontal. Therefore, the center C must be horizontally aligned with P. But since C is on OP (the y-axis), the only way for the radius CP to be horizontal is if C is at (something, y) such that the line from C to P is horizontal. But since P is at (0, h), and C is on the y-axis, the line from C to P is vertical. Therefore, this is only possible if the radius is zero, which is not possible. Therefore, there is no such circle tangent to OP at P with center on OP. Therefore, this approach might not work.

Alternatively, maybe the circle is tangent to the track? Not sure.

Alternatively, maybe using calculus was the correct approach, leading to 30 degrees.

Alternatively, let me think of the problem in terms of parametric equations.

Let me consider the angle APB as a function of t, and we found that it reaches maximum at t = h / √3, giving angle 30 degrees.

Alternatively, since the problem didn't specify the position of P, but in the Asymptote code, P is at (0,1), but we saw that the result is independent of h, so the answer is always 30 degrees regardless of where P is.

Therefore, the maximum angle is 30 degrees.

But let me check with another example. Let me take h = 2, t = 2 / √3.

Point A: (2/√3, 0)

Point B: (3*(2/√3), 0) = (2√3, 0)

Point P: (0, 2)

Vectors PA = (2/√3, -2)

PB = (2√3, -2)

Dot product PA · PB = (2/√3)(2√3) + (-2)(-2) = 4 + 4 = 8

|PA| = sqrt( (4/3) + 4 ) = sqrt( (4/3 + 12/3) ) = sqrt(16/3) = 4 / √3

|PB| = sqrt( (12) + 4 ) = sqrt(16) = 4

Therefore, cos(theta) = 8 / ( (4 / √3) * 4 ) = 8 / (16 / √3) ) = (8 * √3)/16 = √3 / 2, so theta = 30 degrees.

Yes, same result. Therefore, regardless of h, the maximum angle is 30 degrees.

Therefore, the answer is 30 degrees.

But why did I initially think it was arccos(3/4)? Because I confused [cos(theta)]² with cos(theta). Therefore, the correct maximum angle is 30 degrees.

Therefore, the answer is 30 degrees.

**Final Answer**
\boxed{30}

To determine the **maximum angle** $\angle APB$, we analyze the motion of two runners, $A$ and $B$, starting from the same point $O$ on a linear track, running in the same direction, with $B$ running **three times as fast** as $A$. An observer stands at point $P$, where $\overline{OP}$ is **perpendicular** to the track.

---

### **Coordinate Setup**

Let’s place the track along the x-axis, with point $O$ at the origin $(0, 0)$. Since $\overline{OP}$ is perpendicular to the track, point $P$ lies along the y-axis. Let’s assume $P = (0, h)$, where $h > 0$ is the height of the observer above the track.

Let the speed of runner $A$ be $v$, so the position of $A$ at time $t$ is:
$$
A = (vt, 0)
$$
Since $B$ runs three times as fast, its position is:
$$
B = (3vt, 0)
$$

---

### **Vectors and Angle Calculation**

We are interested in the angle $\angle APB$, which is the angle between vectors $\vec{PA}$ and $\vec{PB}$.

- $\vec{PA} = A - P = (vt, -h)$
- $\vec{PB} = B - P = (3vt, -h)$

The cosine of the angle $\theta = \angle APB$ is given by the dot product formula:
$$
\cos(\theta) = \frac{\vec{PA} \cdot \vec{PB}}{|\vec{PA}| \cdot |\vec{PB}|}
$$

Compute the dot product:
$$
\vec{PA} \cdot \vec{PB} = (vt)(3vt) + (-h)(-h) = 3v^2t^2 + h^2
$$

Compute the magnitudes:
$$
|\vec{PA}| = \sqrt{(vt)^2 + h^2} = \sqrt{v^2t^2 + h^2}
$$
$$
|\vec{PB}| = \sqrt{(3vt)^2 + h^2} = \sqrt{9v^2t^2 + h^2}
$$

So,
$$
\cos(\theta) = \frac{3v^2t^2 + h^2}{\sqrt{v^2t^2 + h^2} \cdot \sqrt{9v^2t^2 + h^2}}
$$

To simplify, let’s set $v = 1$ (since the ratio of speeds is what matters), so:
$$
\cos(\theta) = \frac{3t^2 + h^2}{\sqrt{t^2 + h^2} \cdot \sqrt{9t^2 + h^2}}
$$

---

### **Maximizing the Angle**

To **maximize** $\theta$, we need to **minimize** $\cos(\theta)$. Let’s define:
$$
f(t) = \frac{3t^2 + h^2}{\sqrt{t^2 + h^2} \cdot \sqrt{9t^2 + h^2}}
$$

Let’s simplify by setting $u = t^2$, so $u \geq 0$. Then:
$$
f(u) = \frac{3u + h^2}{\sqrt{u + h^2} \cdot \sqrt{9u + h^2}}
$$

We now minimize $f(u)$. After computing the derivative and setting it to zero, we find the critical point occurs at:
$$
u = \frac{h^2}{3}
$$

Substituting back, we find:
$$
\cos(\theta) = \frac{3 \cdot \frac{h^2}{3} + h^2}{\sqrt{\frac{h^2}{3} + h^2} \cdot \sqrt{9 \cdot \frac{h^2}{3} + h^2}} = \frac{2h^2}{\sqrt{\frac{4h^2}{3}} \cdot \sqrt{4h^2}} = \frac{2h^2}{\frac{2h}{\sqrt{3}} \cdot 2h} = \frac{2h^2}{\frac{4h^2}{\sqrt{3}}} = \frac{\sqrt{3}}{2}
$$

Thus:
$$
\cos(\theta) = \frac{\sqrt{3}}{2} \Rightarrow \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ
$$

---

### **Conclusion**

The **maximum angle** $\angle APB$ is achieved when the runners are at positions such that the cosine of the angle is minimized, which occurs at a specific time $t = \frac{h}{\sqrt{3}}$. This results in the angle being:

$$
\boxed{30}
$$

degrees.

[Assistant]
Explanation:
Okay, so there's this problem about two runners, A and B, starting from the same point O on a linear track. They run in the same direction, and B is three times as fast as A. There's an observer at point P, where OP is perpendicular to the track. We need to find the maximum angle APB, in degrees.

First, let me visualize the setup. The track is a straight line, and OP is perpendicular to it. So, if I imagine the track as the x-axis, then point O is at the origin (0,0), and point P is somewhere along the y-axis, say at (0, h) for some height h. The Asymptote code shows P at (0,1), but maybe the actual coordinates aren't important, just the relative positions.

Runners A and B start at O and run in the same direction. Let me assume they run along the positive x-axis. Since B is three times as fast as A, if A's speed is v, then B's speed is 3v.

The question is about the angle APB. So, we need to consider the positions of A and B at some time t, then find the angle between PA and PB. The goal is to find the maximum value of this angle as the runners move along the track.

Let me think about how to model this. Let's set up coordinates. Let me define the track as the x-axis. Let O be at (0,0), and P at (0, h). Let me assume h is some positive number. Since the problem doesn't specify the distance from O to P, maybe it cancels out or we can assume a specific value for simplicity. Maybe h = 1? The Asymptote code shows P at (0,1), so maybe that's a hint. Let me go with that. So P is at (0,1).

Let me denote the position of runner A at time t as (x(t), 0), and runner B as (y(t), 0). Since they start at O, which is (0,0), and run in the same direction. Let me assume A's speed is v, so x(t) = v*t. Similarly, B's speed is 3v, so y(t) = 3v*t.

But maybe to make things simpler, I can set v = 1, so that A moves at speed 1, and B at speed 3. Then x(t) = t, y(t) = 3t.

So, positions of A and B at time t are (t, 0) and (3t, 0), respectively.

Now, we need to find angle APB. So, points A, P, B. Let me recall that angle APB is the angle at point P between points A and B. So, if I can find vectors PA and PB, then the angle between them can be found using the dot product formula.

Alternatively, maybe using coordinates, we can compute the angle. Let me try that.

First, coordinates:

Point P is (0,1).

Point A is (t, 0).

Point B is (3t, 0).

So, vector PA is A - P = (t - 0, 0 - 1) = (t, -1).

Similarly, vector PB is B - P = (3t - 0, 0 - 1) = (3t, -1).

Then, the angle between vectors PA and PB can be found using the dot product formula:

cos(theta) = (PA · PB) / (|PA| |PB|)

Compute PA · PB: (t)(3t) + (-1)(-1) = 3t² + 1

Compute |PA|: sqrt(t² + (-1)^2) = sqrt(t² + 1)

Similarly, |PB|: sqrt((3t)^2 + (-1)^2) = sqrt(9t² + 1)

Therefore,

cos(theta) = (3t² + 1) / [sqrt(t² + 1) * sqrt(9t² + 1)]

Our goal is to find the maximum angle theta, which corresponds to minimizing cos(theta), since cosine is a decreasing function in [0, π].

Therefore, we need to find the value of t that minimizes the expression (3t² + 1)/[sqrt(t² + 1) * sqrt(9t² + 1)].

Alternatively, since dealing with square roots can be messy, maybe square the expression to make it easier. Let me denote f(t) = [ (3t² + 1)^2 ] / [ (t² + 1)(9t² + 1) ]

Then, minimizing f(t) will correspond to minimizing the original expression, hence maximizing theta.

So, let's compute f(t):

f(t) = (9t^4 + 6t² + 1) / [ (t² + 1)(9t² + 1) ]

Let me compute the denominator:

(t² + 1)(9t² + 1) = 9t^4 + t² + 9t² + 1 = 9t^4 + 10t² + 1

So, f(t) = (9t^4 + 6t² + 1) / (9t^4 + 10t² + 1)

So, we need to minimize f(t). Let me denote u = t², since t² is non-negative. Let u ≥ 0.

Then, f(u) = (9u² + 6u + 1) / (9u² + 10u + 1)

So, we need to find the minimum of f(u) for u ≥ 0.

Let me compute derivative of f(u) with respect to u, set it to zero.

Let me denote numerator as N = 9u² + 6u + 1, denominator as D = 9u² + 10u + 1.

Then, f(u) = N/D, so derivative f’(u) = (N’ D - N D’) / D².

Compute N’ = 18u + 6

Compute D’ = 18u + 10

Therefore,

f’(u) = [ (18u + 6)(9u² + 10u + 1) - (9u² + 6u + 1)(18u + 10) ] / (D²)

Let me compute the numerator:

Let me expand both products.

First term: (18u + 6)(9u² + 10u + 1)

Let me compute this:

Multiply 18u by each term:

18u * 9u² = 162u³

18u * 10u = 180u²

18u * 1 = 18u

Then, 6 * 9u² = 54u²

6 * 10u = 60u

6 * 1 = 6

So adding all together:

162u³ + 180u² + 18u + 54u² + 60u + 6

Combine like terms:

162u³ + (180u² + 54u²) + (18u + 60u) + 6

Which is:

162u³ + 234u² + 78u + 6

Second term: (9u² + 6u + 1)(18u + 10)

Let me compute this:

Multiply 9u² by each term:

9u² * 18u = 162u³

9u² * 10 = 90u²

Multiply 6u by each term:

6u * 18u = 108u²

6u * 10 = 60u

Multiply 1 by each term:

1 * 18u = 18u

1 * 10 = 10

Adding all together:

162u³ + 90u² + 108u² + 60u + 18u + 10

Combine like terms:

162u³ + (90u² + 108u²) + (60u + 18u) + 10

Which is:

162u³ + 198u² + 78u + 10

Therefore, the numerator of f’(u) is:

[First term] - [Second term] = [162u³ + 234u² + 78u + 6] - [162u³ + 198u² + 78u + 10]

Compute term by term:

162u³ - 162u³ = 0

234u² - 198u² = 36u²

78u - 78u = 0

6 - 10 = -4

Therefore, numerator is 36u² - 4.

Therefore, f’(u) = (36u² - 4)/D²

Set derivative equal to zero:

36u² - 4 = 0 => 36u² = 4 => u² = 4/36 = 1/9 => u = 1/3 or u = -1/3. But since u ≥ 0, only u = 1/3 is critical point.

Therefore, the function f(u) has critical point at u = 1/3. Now, we need to check if this is a minimum or maximum.

Since f’(u) changes sign from negative to positive or vice versa. Let me check the sign of f’(u) around u = 1/3.

For u < 1/3, say u = 0: numerator is 36*(0) - 4 = -4 < 0, so f’(u) < 0.

For u > 1/3, say u = 1: numerator is 36*(1) - 4 = 32 > 0, so f’(u) > 0.

Therefore, the function f(u) decreases until u = 1/3, then increases. Therefore, u = 1/3 is the point of minimum. Therefore, the minimum of f(u) occurs at u = 1/3, which corresponds to t² = 1/3, so t = 1/√3.

Therefore, the minimum value of f(u) is f(1/3):

Compute f(1/3) = [9*(1/3)^2 + 6*(1/3) + 1] / [9*(1/3)^2 + 10*(1/3) + 1]

Compute numerator:

9*(1/9) + 6*(1/3) + 1 = 1 + 2 + 1 = 4

Denominator:

9*(1/9) + 10*(1/3) + 1 = 1 + 10/3 + 1 = 2 + 10/3 = 16/3

Therefore, f(1/3) = 4 / (16/3) = 4 * 3 / 16 = 12 / 16 = 3/4.

Therefore, the minimum value of f(u) is 3/4, which means the minimum value of cos(theta) is 3/4. Therefore, the maximum angle theta is arccos(3/4).

Wait, but the question says "Find the maximum of angle APB". So, arccos(3/4) is the angle when cos(theta) is minimized. Therefore, theta_max = arccos(3/4). But we need to convert this to degrees.

But maybe there's a way to find the exact value? Or maybe it's a standard angle? Let me check.

Alternatively, maybe I made a mistake here. Let me verify.

Wait, arccos(3/4) is approximately... Well, cos(41.4 degrees) is approximately 0.75. Let me check with calculator:

cos(41.4°) ≈ cos(41.4) ≈ 0.75. Yes, because cos(60°) is 0.5, cos(30°) ≈ 0.866. So 0.75 is between 30 and 60, closer to 41 degrees. But the problem says to find the maximum angle, and maybe it's a nice angle? Wait, but maybe there's a different approach where we can find an exact value?

Alternatively, maybe I made a mistake in the calculation. Let me check again.

Wait, let me check the calculation of f(1/3):

Numerator: 9u² + 6u + 1 with u = 1/3:

9*(1/3)^2 = 9*(1/9) = 1

6*(1/3) = 2

So total numerator: 1 + 2 + 1 = 4. Correct.

Denominator: 9u² + 10u + 1:

Same 9*(1/3)^2 = 1, 10*(1/3) = 10/3, so total denominator: 1 + 10/3 + 1 = 2 + 10/3 = 16/3. Correct. So 4 divided by 16/3 is 3/4. Correct. So cos(theta) = 3/4. Therefore theta = arccos(3/4).

But the problem says "Find the maximum of angle APB, in degrees." So unless there's a way to express arccos(3/4) in exact degrees, which I don't think so, but maybe there's a different approach where we can find it?

Alternatively, maybe I made a wrong assumption in setting up the coordinates? Let me check.

Wait, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, OP is perpendicular, so P is somewhere off the track. But in my coordinate system, I assumed P is at (0,1). But maybe the position of P is arbitrary? Wait, but the problem doesn't specify where P is, just that OP is perpendicular. However, since the track is linear, and OP is perpendicular, then P is at some fixed distance from O. However, in the Asymptote code, they have P at (0,1), but maybe the actual position of P is arbitrary? Wait, but if we don't know where P is, how can we find the maximum angle? Wait, but maybe the maximum angle is independent of the position of P? Or maybe the problem assumes that P is fixed? Wait, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, maybe P is fixed? But since the track is linear, and OP is perpendicular, then P is at a fixed distance from O. However, the problem doesn't specify the distance. Wait, but in the Asymptote code, they have P at (0,1), but maybe that's just for illustration. However, in the problem statement, since they don't specify the distance, maybe the maximum angle is independent of the distance? Wait, but in my calculation, I assumed P is at (0,1). But if I take P at (0, h), would the result be the same?

Let me check. Let me redo the calculation with P at (0, h). Let me see if the angle depends on h or not.

So, let me redo the problem with P at (0, h). Then, coordinates:

Point P: (0, h)

Point A: (t, 0)

Point B: (3t, 0)

Vectors PA = (t - 0, 0 - h) = (t, -h)

Vector PB = (3t - 0, 0 - h) = (3t, -h)

Dot product PA · PB = t*(3t) + (-h)*(-h) = 3t² + h²

|PA| = sqrt(t² + h²)

|PB| = sqrt(9t² + h²)

Therefore, cos(theta) = (3t² + h²)/[sqrt(t² + h²) * sqrt(9t² + h²)]

So, same as before, except with h instead of 1. Then, if we do the same substitution u = t², then:

Let me define f(u) = [3u + h²]/[sqrt(u + h²) * sqrt(9u + h²)] ?

Wait, no, wait. Wait, the numerator is 3t² + h², which is 3u + h². The denominator is sqrt(u + h²) * sqrt(9u + h²). Therefore, if we square it, we get [ (3u + h²)^2 ] / [ (u + h²)(9u + h²) ]

So, same as before, except with h² instead of 1. So, if we do the same steps as before, we can see that the critical point occurs at u = 1/3, regardless of h? Wait, let me check.

Let me do the same steps. Let me set u = t², then f(u) = [ (3u + h²)^2 ] / [ (u + h²)(9u + h²) ]

Let me compute the derivative. Let me denote N = (3u + h²)^2, D = (u + h²)(9u + h²). Then f(u) = N/D. Then f’(u) = [N’ D - N D’]/D².

Compute N’ = 2*(3u + h²)*3 = 6*(3u + h²)

Compute D = (u + h²)(9u + h²) = 9u² + 10u h² + h^4

Wait, no, expanding (u + h²)(9u + h²) = 9u² + u h² + 9u h² + h^4 = 9u² + 10u h² + h^4. Wait, but earlier when h was 1, we had 9u² + 10u + 1. So, yes, similar.

But D’ = derivative of D with respect to u is 18u + 10 h².

Wait, let me compute D’:

D = (u + h²)(9u + h²). So, derivative is (1)(9u + h²) + (u + h²)(9) = 9u + h² + 9u + 9h² = 18u + 10 h². Correct.

Therefore, f’(u) = [6*(3u + h²)*(9u² + 10u h² + h^4) - (3u + h²)^2*(18u + 10 h²)] / D²

Wait, this seems complicated. However, note that in the previous case when h = 1, we found that the numerator simplifies to 36u² - 4. But here, with general h, maybe there's a similar simplification?

Alternatively, maybe we can factor out terms. Let me factor out (3u + h²):

Numerator: 6*(3u + h²)*D - (3u + h²)^2 * D’ = (3u + h²)[6D - (3u + h²) D’]

So, let me compute 6D - (3u + h²) D’:

First, D = (u + h²)(9u + h²)

D’ = 18u + 10 h²

So,

6D - (3u + h²) D’ = 6(u + h²)(9u + h²) - (3u + h²)(18u + 10 h²)

Let me factor out terms. Let me denote A = u + h², B = 9u + h², C = 3u + h², D’ = 18u + 10 h².

But maybe expanding:

First term: 6(u + h²)(9u + h²) = 6[9u² + u h² + 9u h² + h^4] = 6[9u² + 10u h² + h^4] = 54u² + 60u h² + 6 h^4

Second term: (3u + h²)(18u + 10 h²) = 3u*18u + 3u*10 h² + h²*18u + h²*10 h² = 54u² + 30u h² + 18u h² + 10 h^4 = 54u² + 48u h² + 10 h^4

Therefore, 6D - (3u + h²) D’ = [54u² + 60u h² + 6 h^4] - [54u² + 48u h² + 10 h^4] = (54u² - 54u²) + (60u h² - 48u h²) + (6 h^4 - 10 h^4) = 12u h² - 4 h^4 = 4 h² (3u - h²)

Therefore, numerator of f’(u) is (3u + h²) * 4 h² (3u - h²) = 4 h² (3u + h²)(3u - h²) = 4 h² (9u² - h^4)

Wait, no:

Wait, (3u + h²)(3u - h²) = 9u² - h^4. Therefore, numerator is 4 h² (9u² - h^4). Therefore, f’(u) = [4 h² (9u² - h^4)] / D²

Set derivative equal to zero: 9u² - h^4 = 0 => 9u² = h^4 => u² = h^4 / 9 => u = h² / 3 (since u ≥ 0)

Therefore, critical point at u = h² / 3. Therefore, regardless of h, the critical point is at u = h² / 3. Therefore, t² = h² / 3 => t = h / √3.

Therefore, the minimum of f(u) occurs at u = h² / 3. Therefore, the value of cos(theta) at this point is:

Let me compute f(u) at u = h² / 3:

Numerator: (3u + h²)^2 = [3*(h² / 3) + h²]^2 = [h² + h²]^2 = (2 h²)^2 = 4 h^4

Denominator: (u + h²)(9u + h²) = [ (h² / 3) + h² ] * [ 9*(h² / 3) + h² ] = [ (4 h² / 3) ] * [ 3 h² + h² ] = (4 h² / 3) * 4 h² = (16 h^4)/3

Therefore, f(u) = 4 h^4 / (16 h^4 / 3) ) = 4 / (16 / 3) ) = 3/4.

So, regardless of h, the minimum value of cos(theta) is 3/4. Therefore, theta_max = arccos(3/4). Therefore, the maximum angle APB is arccos(3/4) radians, which we need to convert to degrees.

But the problem says "Find the maximum of angle APB, in degrees." So, unless there's a way to express arccos(3/4) in exact degrees, which I don't think so, but maybe there's a different approach where we can find it?

Alternatively, maybe I made a mistake in assuming that the maximum occurs at this critical point? Let me check with specific values.

For example, let me take t approaching 0. When t approaches 0, points A and B are both near O. Then, angle APB approaches the angle between PO and PA, and PO and PB. Since A and B are both near O, the angle APB approaches the angle between PO and PO, which is zero? Wait, no. Wait, when t approaches 0, points A and B are both near O. So, PA is from P to A, which is near O, and PB is from P to B, which is also near O. So, the angle between PA and PB would approach the angle between two lines from P to O, which is zero? Wait, but O is the point where both A and B start. Wait, but when t approaches 0, A and B are both approaching O, so PA and PB are approaching PO. Therefore, angle APB approaches zero.

When t approaches infinity, what happens? Let me see. As t becomes very large, points A and B are far away on the track. The vectors PA and PB are approximately (t, -h) and (3t, -h). The angle between them can be approximated by the angle between (1, -h/t) and (3, -h/t). As t approaches infinity, h/t approaches zero, so vectors approach (1, 0) and (3, 0). The angle between them approaches zero. Therefore, as t approaches infinity, angle APB approaches zero. Therefore, the angle APB has a maximum somewhere in between, which we found at t = h / √3, giving angle arccos(3/4).

Therefore, the maximum angle is arccos(3/4). Now, converting this to degrees.

But the problem says "Find the maximum of angle APB, in degrees." So, unless there's a specific value, but maybe it's a standard angle? Let me check.

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

Alternatively, maybe using calculus, we can find the exact value. But arccos(3/4) is approximately 41.4 degrees. However, the problem might expect an exact value. Wait, but maybe there's a way to find it using some geometric properties?

Alternatively, maybe using the Law of Sines or Law of Cosines on triangle APB?

Let me try that. Let me consider triangle APB. Let me find the lengths of PA, PB, and AB, then use the Law of Cosines to find angle APB.

PA = sqrt(t² + h²)

PB = sqrt(9t² + h²)

AB = distance between A and B = 3t - t = 2t

So, in triangle APB, sides are PA, PB, AB. Let me denote angle at P as theta. Then, by Law of Cosines:

AB² = PA² + PB² - 2 PA PB cos(theta)

Therefore,

(2t)^2 = (t² + h²) + (9t² + h²) - 2 sqrt(t² + h²) sqrt(9t² + h²) cos(theta)

Simplify:

4t² = t² + h² + 9t² + h² - 2 sqrt(...) cos(theta)

Which is:

4t² = 10t² + 2h² - 2 sqrt(...) cos(theta)

Rearranged:

2 sqrt(...) cos(theta) = 10t² + 2h² - 4t² = 6t² + 2h²

Therefore,

cos(theta) = (6t² + 2h²) / [2 sqrt(t² + h²) sqrt(9t² + h²)] = (3t² + h²)/[sqrt(t² + h²) sqrt(9t² + h²)]

Which is the same expression as before. Therefore, same result. Therefore, we can't get a better expression here.

Therefore, the maximum angle is arccos(3/4). But the problem says "Find the maximum of angle APB, in degrees." So, unless there's a specific value, but maybe the answer is 90 degrees? Wait, but arccos(3/4) is less than 90 degrees, since 3/4 is positive. Wait, arccos(3/4) is acute. Wait, but maybe I made a mistake in the direction of the angle? Wait, angle APB is the angle at P between A and B. So, if the maximum angle is arccos(3/4), which is approximately 41.4 degrees. But maybe there's a different approach where we can find it?

Alternatively, maybe using calculus, we can find the exact value? Wait, but arccos(3/4) is not a standard angle. However, maybe the problem expects an exact value in terms of inverse trigonometric functions? But the problem says "in degrees," so likely expects a numerical value. However, since it's a competition problem, maybe there's a trick to find it?

Wait, let me think again. Maybe there's a geometric interpretation where the maximum angle occurs when the circle through A and B is tangent to the line OP? Wait, that might be a method used in some optimization problems.

Let me recall that for a fixed point P, the locus of points A and B moving along the track... Wait, but A and B are moving with different speeds. Wait, but maybe for the angle APB to be maximum, the circle passing through A and B with P on the circle? Wait, the maximum angle occurs when the circle through A and B is tangent to the line OP?

Alternatively, the maximum angle occurs when the circle passing through A and B is tangent to the line OP. Because for a fixed point P, the angle APB is maximized when the circle through A and B is tangent to the line OP.

Let me try to think.

If we fix P, and consider points A and B moving along the track, then the angle APB is maximized when the circle passing through A and B is tangent to the line OP. Because if the circle is tangent, then the angle is maximized.

Alternatively, this is similar to the problem where you have a fixed point and a line, and you want to find the maximum angle subtended by a point moving along the line.

But in this case, we have two points moving along the line with different speeds.

Alternatively, maybe we can parametrize the positions of A and B such that the angle APB is maximized.

But since we already did calculus and found that the maximum angle is arccos(3/4), which is approximately 41.4 degrees, but maybe the problem expects an exact value? Wait, but maybe there's a miscalculation here. Let me check again.

Wait, let me check with specific numbers. Let me take h = 1, t = 1/√3 ≈ 0.577. Then, coordinates:

Point A: (1/√3, 0)

Point B: (3/√3, 0) = (√3, 0)

Point P: (0, 1)

Compute vectors PA and PB:

PA = (1/√3 - 0, 0 - 1) = (1/√3, -1)

PB = (√3 - 0, 0 - 1) = (√3, -1)

Compute the angle between PA and PB.

Dot product PA · PB = (1/√3)(√3) + (-1)(-1) = 1 + 1 = 2

|PA| = sqrt( (1/√3)^2 + (-1)^2 ) = sqrt(1/3 + 1) = sqrt(4/3) = 2/√3

|PB| = sqrt( (√3)^2 + (-1)^2 ) = sqrt(3 + 1) = 2

Therefore, cos(theta) = 2 / [ (2/√3)(2) ] = 2 / (4 / √3) ) = (2 * √3)/4 = √3 / 2 ≈ 0.866. Wait, but earlier I thought it was 3/4. Wait, this contradicts previous result. What's going on here?

Wait, hold on! Earlier, I thought that cos(theta) was 3/4, but here, with h = 1, t = 1/√3, I get cos(theta) = √3 / 2 ≈ 0.866, which is 30 degrees. Wait, but this contradicts previous calculation where I had f(u) = 3/4. What is wrong here?

Wait, let me check my previous steps. Earlier, I had:

cos(theta) = (3t² + h²)/[sqrt(t² + h²) * sqrt(9t² + h²)]

With h = 1, t = 1/√3:

Numerator: 3*(1/3) + 1 = 1 + 1 = 2

Denominator: sqrt(1/3 + 1) * sqrt(9*(1/3) + 1) = sqrt(4/3) * sqrt(4) = (2/√3) * 2 = 4 / √3

Therefore, cos(theta) = 2 / (4 / √3) ) = (2 * √3)/4 = √3 / 2 ≈ 0.866, which is 30 degrees. But earlier, I thought that f(u) was 3/4. Wait, but in the previous calculation, when I did f(u) = [ (3u + h²)^2 ] / [ (u + h²)(9u + h²) ] with u = h² / 3, I had:

Numerator: (3u + h²)^2 = [3*(h² / 3) + h²]^2 = (h² + h²)^2 = (2h²)^2 = 4h^4

Denominator: (u + h²)(9u + h²) = [ (h² / 3) + h² ] * [ 9*(h² / 3) + h² ] = (4h² / 3) * (4h²) = 16h^4 / 3

Therefore, f(u) = 4h^4 / (16h^4 / 3) ) = 3/4. But this was the square of cos(theta). Wait! Wait, earlier I said that f(u) = [ (3t² + h²)^2 ] / [ (t² + h²)(9t² + h²) ] which is [cos(theta)]². Therefore, cos(theta) = sqrt(f(u))?

Wait, no! Wait, earlier, I had:

cos(theta) = (3t² + h²)/[sqrt(t² + h²) * sqrt(9t² + h²)]

Therefore, [cos(theta)]² = (3t² + h²)^2 / [ (t² + h²)(9t² + h²) ] = f(u). Therefore, when I calculated f(u) at u = h² / 3, I found f(u) = 3/4, so [cos(theta)]² = 3/4, so cos(theta) = sqrt(3)/2 or -sqrt(3)/2. Since theta is between 0 and 180 degrees, cos(theta) is positive, so cos(theta) = sqrt(3)/2, which implies theta = 30 degrees.

Wait, so earlier I had a mistake in thinking that cos(theta) was 3/4, but actually, [cos(theta)]² = 3/4, so cos(theta) = sqrt(3)/2, which is 30 degrees. Therefore, theta_max = 30 degrees.

Wait, so where was the mistake? Earlier, when I thought that f(u) was cos(theta), but actually f(u) is [cos(theta)]². Therefore, when I found f(u) = 3/4, that means cos(theta) = sqrt(3)/2, so theta = 30 degrees.

Therefore, the maximum angle is 30 degrees.

But why did I think earlier that cos(theta) was 3/4? Because I confused f(u) with cos(theta). So, that was the mistake.

So, correcting that, the maximum angle is arccos(sqrt(3)/2) = 30 degrees.

But let me verify with the specific example. When t = 1/√3, h = 1, we found that cos(theta) = sqrt(3)/2, which is 30 degrees. Therefore, the maximum angle is 30 degrees.

But why did the previous calculation with f(u) = 3/4? Because f(u) was [cos(theta)]², so cos(theta) = sqrt(3)/2.

Therefore, the maximum angle is 30 degrees.

But why did I get confused earlier? Because I thought that f(u) was cos(theta), but it's actually [cos(theta)]².

Therefore, the answer is 30 degrees.

But let me check with another example. Let me take t = 0. Then, points A and B are both at O. Then, angle APB is undefined, since A and B coincide. But as t approaches 0, angle approaches zero.

If I take t = 1, h = 1:

PA = sqrt(1 + 1) = sqrt(2)

PB = sqrt(9 + 1) = sqrt(10)

Dot product PA · PB = 3*1 + 1 = 4

cos(theta) = 4 / (sqrt(2)*sqrt(10)) = 4 / sqrt(20) = 4 / (2*sqrt(5)) = 2 / sqrt(5) ≈ 0.894, which is about 26.565 degrees. Which is less than 30 degrees.

If I take t = 1/√3 ≈ 0.577, then angle is 30 degrees.

If I take t approaching infinity, angle approaches zero.

Therefore, the maximum angle is indeed 30 degrees.

Therefore, the answer is 30 degrees.

But why did I get confused earlier? Because I miscalculated the relationship between f(u) and cos(theta).

So, the key steps are:

- Set up coordinates with P at (0, h), A at (t, 0), B at (3t, 0).

- Compute vectors PA and PB.

- Use dot product to find cos(theta).

- Find that [cos(theta)]² = (3t² + h²)^2 / [(t² + h²)(9t² + h²)].

- Minimize this expression, which leads to critical point at t² = h² / 3.

- At that point, [cos(theta)]² = 3/4, so cos(theta) = sqrt(3)/2, leading to theta = 30 degrees.

Therefore, the maximum angle is 30 degrees.

Therefore, the answer is 30 degrees.

But let me check with another approach. Let me consider the circle passing through A and B and tangent to OP.

If we consider the circle passing through A and B and tangent to OP at point P, then the angle APB would be 90 degrees? Wait, no. Wait, if a circle is tangent to OP at P, then the angle subtended by AB at P would be related to the circle.

Alternatively, maybe using the property that the maximum angle occurs when the circle through A and B is tangent to OP.

Let me try this approach.

Let me assume that there is a circle passing through A and B and tangent to OP at P. Then, the angle APB would be equal to the angle subtended by AB at P, which for a circle tangent at P would be... Wait, if the circle is tangent to OP at P, then P lies on the circle, and OP is tangent to the circle at P. Therefore, the radius at P is perpendicular to OP. But OP is perpendicular to the track, so the radius at P is along OP. Therefore, the center of the circle lies along OP.

Let me denote the center of the circle as C, lying on OP. Let me assume coordinates again. Let me take P at (0, h), O at (0, 0). The center C is somewhere along the y-axis, say at (0, k).

Since the circle passes through A and B, and is tangent to OP at P.

Since the circle is tangent to OP at P, the radius at P is perpendicular to OP. Since OP is vertical (from (0,0) to (0, h)), the tangent line OP is vertical, so the radius at P must be horizontal. But the center is on OP, which is vertical. Therefore, the radius from C to P must be horizontal. Therefore, the center C must be horizontally aligned with P. But since C is on OP (the y-axis), the only way for the radius CP to be horizontal is if C is at (something, y) such that the line from C to P is horizontal. But since P is at (0, h), and C is on the y-axis, the line from C to P is vertical. Therefore, this is only possible if the radius is zero, which is not possible. Therefore, there is no such circle tangent to OP at P with center on OP. Therefore, this approach might not work.

Alternatively, maybe the circle is tangent to the track? Not sure.

Alternatively, maybe using calculus was the correct approach, leading to 30 degrees.

Alternatively, let me think of the problem in terms of parametric equations.

Let me consider the angle APB as a function of t, and we found that it reaches maximum at t = h / √3, giving angle 30 degrees.

Alternatively, since the problem didn't specify the position of P, but in the Asymptote code, P is at (0,1), but we saw that the result is independent of h, so the answer is always 30 degrees regardless of where P is.

Therefore, the maximum angle is 30 degrees.

But let me check with another example. Let me take h = 2, t = 2 / √3.

Point A: (2/√3, 0)

Point B: (3*(2/√3), 0) = (2√3, 0)

Point P: (0, 2)

Vectors PA = (2/√3, -2)

PB = (2√3, -2)

Dot product PA · PB = (2/√3)(2√3) + (-2)(-2) = 4 + 4 = 8

|PA| = sqrt( (4/3) + 4 ) = sqrt( (4/3 + 12/3) ) = sqrt(16/3) = 4 / √3

|PB| = sqrt( (12) + 4 ) = sqrt(16) = 4

Therefore, cos(theta) = 8 / ( (4 / √3) * 4 ) = 8 / (16 / √3) ) = (8 * √3)/16 = √3 / 2, so theta = 30 degrees.

Yes, same result. Therefore, regardless of h, the maximum angle is 30 degrees.

Therefore, the answer is 30 degrees.

But why did I initially think it was arccos(3/4)? Because I confused [cos(theta)]² with cos(theta). Therefore, the correct maximum angle is 30 degrees.

Therefore, the answer is 30 degrees.

**Final Answer**
\boxed{30}

To determine the **maximum angle** $\angle APB$, we analyze the motion of two runners, $A$ and $B$, starting from the same point $O$ on a linear track, running in the same direction, with $B$ running **three times as fast** as $A$. An observer stands at point $P$, where $\overline{OP}$ is **perpendicular** to the track.

---

### **Coordinate Setup**

Let’s place the track along the x-axis, with point $O$ at the origin $(0, 0)$. Since $\overline{OP}$ is perpendicular to the track, point $P$ lies along the y-axis. Let’s assume $P = (0, h)$, where $h > 0$ is the height of the observer above the track.

Let the speed of runner $A$ be $v$, so the position of $A$ at time $t$ is:
$$
A = (vt, 0)
$$
Since $B$ runs three times as fast, its position is:
$$
B = (3vt, 0)
$$

---

### **Vectors and Angle Calculation**

We are interested in the angle $\angle APB$, which is the angle between vectors $\vec{PA}$ and $\vec{PB}$.

- $\vec{PA} = A - P = (vt, -h)$
- $\vec{PB} = B - P = (3vt, -h)$

The cosine of the angle $\theta = \angle APB$ is given by the dot product formula:
$$
\cos(\theta) = \frac{\vec{PA} \cdot \vec{PB}}{|\vec{PA}| \cdot |\vec{PB}|}
$$

Compute the dot product:
$$
\vec{PA} \cdot \vec{PB} = (vt)(3vt) + (-h)(-h) = 3v^2t^2 + h^2
$$

Compute the magnitudes:
$$
|\vec{PA}| = \sqrt{(vt)^2 + h^2} = \sqrt{v^2t^2 + h^2}
$$
$$
|\vec{PB}| = \sqrt{(3vt)^2 + h^2} = \sqrt{9v^2t^2 + h^2}
$$

So,
$$
\cos(\theta) = \frac{3v^2t^2 + h^2}{\sqrt{v^2t^2 + h^2} \cdot \sqrt{9v^2t^2 + h^2}}
$$

To simplify, let’s set $v = 1$ (since the ratio of speeds is what matters), so:
$$
\cos(\theta) = \frac{3t^2 + h^2}{\sqrt{t^2 + h^2} \cdot \sqrt{9t^2 + h^2}}
$$

---

### **Maximizing the Angle**

To **maximize** $\theta$, we need to **minimize** $\cos(\theta)$. Let’s define:
$$
f(t) = \frac{3t^2 + h^2}{\sqrt{t^2 + h^2} \cdot \sqrt{9t^2 + h^2}}
$$

Let’s simplify by setting $u = t^2$, so $u \geq 0$. Then:
$$
f(u) = \frac{3u + h^2}{\sqrt{u + h^2} \cdot \sqrt{9u + h^2}}
$$

We now minimize $f(u)$. After computing the derivative and setting it to zero, we find the critical point occurs at:
$$
u = \frac{h^2}{3}
$$

Substituting back, we find:
$$
\cos(\theta) = \frac{3 \cdot \frac{h^2}{3} + h^2}{\sqrt{\frac{h^2}{3} + h^2} \cdot \sqrt{9 \cdot \frac{h^2}{3} + h^2}} = \frac{2h^2}{\sqrt{\frac{4h^2}{3}} \cdot \sqrt{4h^2}} = \frac{2h^2}{\frac{2h}{\sqrt{3}} \cdot 2h} = \frac{2h^2}{\frac{4h^2}{\sqrt{3}}} = \frac{\sqrt{3}}{2}
$$

Thus:
$$
\cos(\theta) = \frac{\sqrt{3}}{2} \Rightarrow \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ
$$

---

### **Conclusion**

The **maximum angle** $\angle APB$ is achieved when the runners are at positions such that the cosine of the angle is minimized, which occurs at a specific time $t = \frac{h}{\sqrt{3}}$. This results in the angle being:

$$
\boxed{30}
$$

degrees.

Solution:
Without loss of generality, assume that $OP = 1.$  Let $OA = x$ and $OB = 3x.$  Let $\alpha = \angle OPA$ and $\beta = \angle OPB,$ so $\tan \alpha = x$ and $\tan \beta = 3x,$ so from the angle subtraction formula,
\begin{align*}
\tan \angle APB &= \tan (\angle OPB - \angle OPA) \\
&= \tan (\beta - \alpha) \\
&= \frac{\tan \beta - \tan \alpha}{1 + \tan \alpha \tan \beta} \\
&= \frac{2x}{1 + 3x^2}.
\end{align*}We want to maximize this expression.  Maximizing this expression is equivalent to minimizing $\frac{1 + 3x^2}{2x}.$  By AM-GM,
\[\frac{1 + 3x^2}{2x} \ge \frac{2 \sqrt{1 \cdot 3x^2}}{2x} = \frac{2x \sqrt{3}}{2x} = \sqrt{3},\]so
\[\tan \angle APB \le \frac{1}{\sqrt{3}},\]which means $\angle APB \le 30^\circ.$  Equality occurs when $x = \frac{1}{\sqrt{3}},$ so the maximum of $\angle APB$ is $\boxed{30^\circ}.$