Explanation:
Okay, so Charlie is riding his unicycle, and I need to find out how fast he's going in inches per second. The problem says the wheel has a radius of 9 inches and makes 2 revolutions every 3 seconds. Hmm, let me think about how to approach this.

First, I remember that the speed of a wheel is related to its circumference and the number of revolutions it makes. So maybe I need to find the circumference of the wheel first? Yeah, circumference is the distance around the edge of the wheel, right? The formula for circumference is π times diameter, or 2π times radius. Since they gave me the radius, which is 9 inches, I can use that.

Let me write that down. Circumference C = 2πr. Plugging in the radius: C = 2 * π * 9 inches. Let me calculate that. 2 times 9 is 18, so the circumference is 18π inches. Okay, so each full revolution, the wheel covers 18π inches.

Now, the wheel makes 2 revolutions every 3 seconds. So, in 3 seconds, the wheel goes around twice. That means in one second, it would make 2/3 of a revolution. Wait, but maybe I should find out how many revolutions per second first? Let me think. If it's 2 revolutions in 3 seconds, then the number of revolutions per second is 2 divided by 3. So that's 2/3 revolutions per second.

But speed is distance over time. So if I can find the distance traveled per second, that would be the speed. Since each revolution is 18π inches, then the distance per second would be the number of revolutions per second multiplied by the circumference.

So, let me write that. Speed = revolutions per second * circumference. So that's (2/3) * 18π inches per second. Let me compute that. 2/3 times 18... Well, 18 divided by 3 is 6, and 6 times 2 is 12. So that would be 12π inches per second.

Wait, is that right? Let me check again. If the circumference is 18π inches, and in 3 seconds it makes 2 revolutions, then total distance in 3 seconds is 2 * 18π inches. Which is 36π inches. Then, speed is distance over time, so 36π inches divided by 3 seconds is 12π inches per second. Yeah, that matches. So that seems correct.

But maybe I should convert π into a numerical value? Wait, the question says "what is the unicycle's speed in inches per second?" It doesn't specify whether to leave it in terms of π or to use a numerical approximation. Since they gave the radius as a whole number and the revolutions as a fraction, maybe they want an exact value in terms of π? But sometimes problems expect a numerical answer. Let me check the problem again.

The problem says "what is the unicycle's speed in inches per second?" and doesn't specify, but since π is a common term, maybe they want the answer in terms of π. However, sometimes they might want a decimal. Wait, but maybe I should check if there's a standard way. Let me think. In math problems, unless specified, it's safer to leave it in terms of π. But maybe the answer is expected to be a number. Wait, but the problem is from Charlie's unicycle, so maybe they just want the exact value. Let me check my calculation again.

Alternatively, maybe I made a mistake in the calculation. Let me go through it again step by step.

First, radius is 9 inches. Circumference is 2πr = 2π*9 = 18π inches. Correct. Then, revolutions per second: 2 revolutions every 3 seconds. So revolutions per second is 2/3. Then, speed is revolutions per second multiplied by circumference. So 2/3 * 18π. Let me compute 2/3 * 18. 18 divided by 3 is 6, times 2 is 12. So 12π inches per second. So that's 12π. If I wanted to write that as a number, π is approximately 3.1416, so 12 * 3.1416 is approximately 37.699 inches per second. But unless they ask for a decimal, maybe 12π is the exact answer.

But let me check if the problem is from a math competition or something. Since it's about a unicycle, maybe it's a general math problem. The original question didn't specify, but since the radius is given as 9 inches, which is a whole number, and the revolutions are 2 every 3 seconds, which is a fraction, but maybe they want an exact answer. However, in some cases, they might want a numerical value. Wait, but the problem says "in inches per second," and doesn't specify, so maybe either is acceptable? But since π is involved, perhaps they want the answer in terms of π.

But let me check again. Let me think if there's another way to approach this. For example, maybe using angular velocity? But that might complicate things. Let me try. Angular velocity ω is revolutions per second, which is 2/3 rev/sec. But to get linear speed, it's ω multiplied by radius? Wait, no. Wait, linear speed v = ω * r, but ω has to be in radians per second.

Wait, so if I use angular velocity, I need to convert revolutions to radians. Since one revolution is 2π radians. So angular velocity ω is 2 revolutions per 3 seconds, which is (2 * 2π) radians per 3 seconds. So that's 4π/3 radians per second. Then linear speed v = ω * r. So that's (4π/3) * 9 inches per second. Let me compute that. 4π/3 * 9 = 4π * 3 = 12π inches per second. So same answer. So that confirms it. So regardless of the method, I get 12π inches per second.

Therefore, the answer is 12π inches per second. But maybe the problem expects a numerical value? Let me check if there's any instruction about that. The original problem says "put your final answer within \boxed{}." Typically, in math problems, if it's a multiple of π, they prefer the exact value. So I think 12π is the answer they are looking for.

But just to be thorough, let me check if there's any chance I messed up the units. The radius is in inches, so circumference is in inches. The revolutions per second is unitless, so multiplying by inches gives inches per second. So units are correct.

Alternatively, if I think about total distance over total time. In 3 seconds, the wheel makes 2 revolutions. So total distance is 2 * circumference = 2 * 18π = 36π inches. Then speed is 36π inches / 3 seconds = 12π inches per second. Yep, same answer.

So I think that's solid. Therefore, the unicycle's speed is 12π inches per second.

But just to make sure, let me think if there's any other possible mistake. For example, confusing diameter and radius? The radius is 9 inches, so diameter is 18 inches. But circumference is π*diameter, which is π*18, which is same as 2πr. So that's correct. So no mistake there.

Another possible mistake: mixing up revolutions per second with something else. But the problem says 2 revolutions every 3 seconds, so that's 2/3 revolutions per second. Then multiplied by circumference gives inches per second. Correct.

Alternatively, if someone thought that 2 revolutions every 3 seconds is 2 seconds per revolution, but that would be wrong. Because if it's 2 revolutions in 3 seconds, then each revolution takes 3/2 seconds. But that's not needed here.

So, no, I think my answer is correct. So the speed is 12π inches per second.

**Final Answer**
The unicycle's speed is \boxed{12\pi} inches per second.

To determine the unicycle's speed in inches per second, we begin by analyzing the relationship between the wheel's motion and the distance it covers.

---

### Step 1: Calculate the Circumference of the Wheel

The unicycle's wheel has a **radius of 9 inches**. The **circumference** of a circle is given by the formula:

$$
C = 2\pi r
$$

Substituting the radius:

$$
C = 2\pi \times 9 = 18\pi \text{ inches}
$$

This means that with each full revolution, the unicycle travels **18π inches**.

---

### Step 2: Determine the Number of Revolutions per Second

The wheel makes **2 revolutions every 3 seconds**. To find the number of revolutions per second, we divide:

$$
\text{Revolutions per second} = \frac{2}{3}
$$

---

### Step 3: Calculate the Speed

Speed is the **distance traveled per unit time**. Since the wheel travels **18π inches per revolution**, and it makes **2/3 revolutions per second**, the speed is:

$$
\text{Speed} = \left(\frac{2}{3}\right) \times 18\pi = \frac{36\pi}{3} = 12\pi \text{ inches per second}
$$

---

### Final Answer

$$
\boxed{12\pi}
$$

The unicycle's speed is **12π inches per second**.

[Assistant]
Explanation:
Okay, so Charlie is riding his unicycle, and I need to find out how fast he's going in inches per second. The problem says the wheel has a radius of 9 inches and makes 2 revolutions every 3 seconds. Hmm, let me think about how to approach this.

First, I remember that the speed of a wheel is related to its circumference and the number of revolutions it makes. So maybe I need to find the circumference of the wheel first? Yeah, circumference is the distance around the edge of the wheel, right? The formula for circumference is π times diameter, or 2π times radius. Since they gave me the radius, which is 9 inches, I can use that.

Let me write that down. Circumference C = 2πr. Plugging in the radius: C = 2 * π * 9 inches. Let me calculate that. 2 times 9 is 18, so the circumference is 18π inches. Okay, so each full revolution, the wheel covers 18π inches.

Now, the wheel makes 2 revolutions every 3 seconds. So, in 3 seconds, the wheel goes around twice. That means in one second, it would make 2/3 of a revolution. Wait, but maybe I should find out how many revolutions per second first? Let me think. If it's 2 revolutions in 3 seconds, then the number of revolutions per second is 2 divided by 3. So that's 2/3 revolutions per second.

But speed is distance over time. So if I can find the distance traveled per second, that would be the speed. Since each revolution is 18π inches, then the distance per second would be the number of revolutions per second multiplied by the circumference.

So, let me write that. Speed = revolutions per second * circumference. So that's (2/3) * 18π inches per second. Let me compute that. 2/3 times 18... Well, 18 divided by 3 is 6, and 6 times 2 is 12. So that would be 12π inches per second.

Wait, is that right? Let me check again. If the circumference is 18π inches, and in 3 seconds it makes 2 revolutions, then total distance in 3 seconds is 2 * 18π inches. Which is 36π inches. Then, speed is distance over time, so 36π inches divided by 3 seconds is 12π inches per second. Yeah, that matches. So that seems correct.

But maybe I should convert π into a numerical value? Wait, the question says "what is the unicycle's speed in inches per second?" It doesn't specify whether to leave it in terms of π or to use a numerical approximation. Since they gave the radius as a whole number and the revolutions as a fraction, maybe they want an exact value in terms of π? But sometimes problems expect a numerical answer. Let me check the problem again.

The problem says "what is the unicycle's speed in inches per second?" and doesn't specify, but since π is a common term, maybe they want the answer in terms of π. However, sometimes they might want a decimal. Wait, but maybe I should check if there's a standard way. Let me think. In math problems, unless specified, it's safer to leave it in terms of π. But maybe the answer is expected to be a number. Wait, but the problem is from Charlie's unicycle, so maybe they just want the exact value. Let me check my calculation again.

Alternatively, maybe I made a mistake in the calculation. Let me go through it again step by step.

First, radius is 9 inches. Circumference is 2πr = 2π*9 = 18π inches. Correct. Then, revolutions per second: 2 revolutions every 3 seconds. So revolutions per second is 2/3. Then, speed is revolutions per second multiplied by circumference. So 2/3 * 18π. Let me compute 2/3 * 18. 18 divided by 3 is 6, times 2 is 12. So 12π inches per second. So that's 12π. If I wanted to write that as a number, π is approximately 3.1416, so 12 * 3.1416 is approximately 37.699 inches per second. But unless they ask for a decimal, maybe 12π is the exact answer.

But let me check if the problem is from a math competition or something. Since it's about a unicycle, maybe it's a general math problem. The original question didn't specify, but since the radius is given as 9 inches, which is a whole number, and the revolutions are 2 every 3 seconds, which is a fraction, but maybe they want an exact answer. However, in some cases, they might want a numerical value. Wait, but the problem says "in inches per second," and doesn't specify, so maybe either is acceptable? But since π is involved, perhaps they want the answer in terms of π.

But let me check again. Let me think if there's another way to approach this. For example, maybe using angular velocity? But that might complicate things. Let me try. Angular velocity ω is revolutions per second, which is 2/3 rev/sec. But to get linear speed, it's ω multiplied by radius? Wait, no. Wait, linear speed v = ω * r, but ω has to be in radians per second.

Wait, so if I use angular velocity, I need to convert revolutions to radians. Since one revolution is 2π radians. So angular velocity ω is 2 revolutions per 3 seconds, which is (2 * 2π) radians per 3 seconds. So that's 4π/3 radians per second. Then linear speed v = ω * r. So that's (4π/3) * 9 inches per second. Let me compute that. 4π/3 * 9 = 4π * 3 = 12π inches per second. So same answer. So that confirms it. So regardless of the method, I get 12π inches per second.

Therefore, the answer is 12π inches per second. But maybe the problem expects a numerical value? Let me check if there's any instruction about that. The original problem says "put your final answer within \boxed{}." Typically, in math problems, if it's a multiple of π, they prefer the exact value. So I think 12π is the answer they are looking for.

But just to be thorough, let me check if there's any chance I messed up the units. The radius is in inches, so circumference is in inches. The revolutions per second is unitless, so multiplying by inches gives inches per second. So units are correct.

Alternatively, if I think about total distance over total time. In 3 seconds, the wheel makes 2 revolutions. So total distance is 2 * circumference = 2 * 18π = 36π inches. Then speed is 36π inches / 3 seconds = 12π inches per second. Yep, same answer.

So I think that's solid. Therefore, the unicycle's speed is 12π inches per second.

But just to make sure, let me think if there's any other possible mistake. For example, confusing diameter and radius? The radius is 9 inches, so diameter is 18 inches. But circumference is π*diameter, which is π*18, which is same as 2πr. So that's correct. So no mistake there.

Another possible mistake: mixing up revolutions per second with something else. But the problem says 2 revolutions every 3 seconds, so that's 2/3 revolutions per second. Then multiplied by circumference gives inches per second. Correct.

Alternatively, if someone thought that 2 revolutions every 3 seconds is 2 seconds per revolution, but that would be wrong. Because if it's 2 revolutions in 3 seconds, then each revolution takes 3/2 seconds. But that's not needed here.

So, no, I think my answer is correct. So the speed is 12π inches per second.

**Final Answer**
The unicycle's speed is \boxed{12\pi} inches per second.

To determine the unicycle's speed in inches per second, we begin by analyzing the relationship between the wheel's motion and the distance it covers.

---

### Step 1: Calculate the Circumference of the Wheel

The unicycle's wheel has a **radius of 9 inches**. The **circumference** of a circle is given by the formula:

$$
C = 2\pi r
$$

Substituting the radius:

$$
C = 2\pi \times 9 = 18\pi \text{ inches}
$$

This means that with each full revolution, the unicycle travels **18π inches**.

---

### Step 2: Determine the Number of Revolutions per Second

The wheel makes **2 revolutions every 3 seconds**. To find the number of revolutions per second, we divide:

$$
\text{Revolutions per second} = \frac{2}{3}
$$

---

### Step 3: Calculate the Speed

Speed is the **distance traveled per unit time**. Since the wheel travels **18π inches per revolution**, and it makes **2/3 revolutions per second**, the speed is:

$$
\text{Speed} = \left(\frac{2}{3}\right) \times 18\pi = \frac{36\pi}{3} = 12\pi \text{ inches per second}
$$

---

### Final Answer

$$
\boxed{12\pi}
$$

The unicycle's speed is **12π inches per second**.

Solution: We need to know how many inches are in one revolution of the unicycle's wheel. In other words, we need to know the circumference of the wheel. Since the radius is $9$ inches, the circumference is $18\pi$ inches. So, if there are $18\pi$ inches in one revolution, and the wheel makes $2$ revolutions every $3$ seconds, the unicycle's speed is $18\pi\cdot\frac{2}{3} = \boxed{12\pi}$ inches per second.


Explanation:
Okay, so I need to find the degree of the polynomial p(x) given by that expression. Let me first write down the problem again to make sure I have it right:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)].

And a, b, c are distinct real numbers. Alright, so the question is asking for the degree of this polynomial. Let me think about how to approach this.

First, I know that when you have a polynomial expressed as a sum of terms like this, each term is a rational function, but the problem says that p(x) is a polynomial. So maybe each term, when combined, simplifies to a polynomial? Or maybe they are designed in such a way that the denominators cancel out the denominators? Wait, but the denominators are constants, right? Because a, b, c are constants, so each denominator is just a constant. So each term is a quadratic polynomial divided by a constant. Therefore, each term is a quadratic polynomial. So if we add three quadratic polynomials together, the result could be a quadratic polynomial, but maybe there's cancellation or something?

But wait, the problem says p(x) is a polynomial. So maybe when you add them up, the denominators are actually factors that make the entire expression a polynomial. Wait, but if each term is a quadratic divided by a constant, then adding them would just be a quadratic polynomial. Unless there's some cancellation? Wait, but if they are all quadratics, adding them would give a quadratic. But maybe the coefficients of x² cancel out? Let me check.

Let me try to compute the coefficients of x² in each term. Let me take the first term: [(x - b)(x - c)] / [(a - b)(a - c)]. Let me expand the numerator:

(x - b)(x - c) = x² - (b + c)x + bc. So dividing by (a - b)(a - c), which is a constant. Similarly for the other terms.

So each term is a quadratic polynomial divided by a constant. Therefore, each term is quadratic. So when you add them up, the sum is a quadratic polynomial. But the problem says "Find the degree of the polynomial p(x)". So maybe the answer is 2? But wait, maybe there is some cancellation? Let me check with specific values.

Alternatively, maybe this is a known polynomial. Wait, the expression looks similar to Lagrange interpolation formula. Let me recall: the Lagrange interpolation polynomial for three points. If we have three points, then the Lagrange basis polynomials are of the form [(x - x_j)(x - x_k)] / [(x_i - x_j)(x_i - x_k)] for each i, j, k. So if we have three points, then the Lagrange interpolation polynomial is the sum of y_i times the basis polynomial. But in this case, the coefficients are 1 for each term. Wait, so if we have three points, and we take the sum of the basis polynomials, then that would be 1? Wait, no. Wait, if you take the sum of the Lagrange basis polynomials, then for any x, the sum would be 1? Wait, let me check.

Suppose we have three points, say, x = a, b, c. Then the Lagrange basis polynomials are L_a(x), L_b(x), L_c(x), such that L_a(a) = 1, L_a(b) = 0, etc. Then, if we take L_a(x) + L_b(x) + L_c(x), then for any x, this sum would be 1? Wait, no. Wait, if you have three points, then the sum of the Lagrange basis polynomials is 1? Let me check with specific numbers.

Let me take a simple case where a=0, b=1, c=2. Then compute L_0(x) + L_1(x) + L_2(x). Let me compute each term:

L_0(x) = [(x - 1)(x - 2)] / [(0 - 1)(0 - 2)] = [(x - 1)(x - 2)] / [(-1)(-2)] = [(x - 1)(x - 2)] / 2.

Similarly, L_1(x) = [(x - 0)(x - 2)] / [(1 - 0)(1 - 2)] = [x(x - 2)] / [1*(-1)] = -x(x - 2).

L_2(x) = [(x - 0)(x - 1)] / [(2 - 0)(2 - 1)] = [x(x - 1)] / [2*1] = [x(x - 1)] / 2.

So adding them up:

L_0(x) + L_1(x) + L_2(x) = [(x - 1)(x - 2)/2] - x(x - 2) + [x(x - 1)/2].

Let me compute each term:

First term: (x² - 3x + 2)/2.

Second term: -x² + 2x.

Third term: (x² - x)/2.

Adding them together:

First term: (x² - 3x + 2)/2.

Third term: (x² - x)/2. Adding first and third term: [x² - 3x + 2 + x² - x]/2 = [2x² - 4x + 2]/2 = x² - 2x + 1.

Then adding the second term: x² - 2x + 1 - x² + 2x = (x² - x²) + (-2x + 2x) + 1 = 0 + 0 + 1 = 1.

So indeed, the sum is 1. Therefore, in this case, the sum of the three Lagrange basis polynomials is 1, which is a constant polynomial. Therefore, the degree is 0. Wait, but in this case, the original problem's p(x) is exactly the sum of the three Lagrange basis polynomials. Therefore, p(x) is equal to 1? But that would mean that the degree is 0. But that contradicts my initial thought that it's quadratic. Wait, but in this case, when I calculated with specific numbers, the sum was 1. So maybe p(x) is the constant polynomial 1? Therefore, degree 0?

But that seems surprising. Let me check with another example. Let me take a different set of a, b, c. Let me take a=1, b=2, c=3. Then compute p(x):

First term: [(x - 2)(x - 3)] / [(1 - 2)(1 - 3)] = [(x - 2)(x - 3)] / [(-1)(-2)] = [(x - 2)(x - 3)] / 2.

Second term: [(x - 1)(x - 3)] / [(2 - 1)(2 - 3)] = [(x - 1)(x - 3)] / [(1)(-1)] = -[(x - 1)(x - 3)].

Third term: [(x - 1)(x - 2)] / [(3 - 1)(3 - 2)] = [(x - 1)(x - 2)] / [2*1] = [(x - 1)(x - 2)] / 2.

Adding them up:

First term: (x² - 5x + 6)/2.

Second term: -[x² - 4x + 3] = -x² + 4x - 3.

Third term: (x² - 3x + 2)/2.

Adding all together:

First term + third term: [x² - 5x + 6 + x² - 3x + 2]/2 = [2x² - 8x + 8]/2 = x² - 4x + 4.

Then adding the second term: x² - 4x + 4 - x² + 4x - 3 = (x² - x²) + (-4x + 4x) + (4 - 3) = 0 + 0 + 1 = 1. Again, the sum is 1. So p(x) is 1. Therefore, the degree is 0. But that seems to contradict my initial thought that it's quadratic. But according to these examples, it's a constant function. Therefore, the degree is 0. But why?

Wait, but the problem says "Find the degree of the polynomial p(x)". So maybe the answer is 0? But why does the problem give such a complicated expression? Maybe I need to check again.

Wait, but in the problem statement, the polynomial is written as the sum of three terms, each of which is a quadratic divided by a constant. But when you add them up, they cancel out to give 1. Therefore, p(x) is the constant polynomial 1, so degree 0. But then why is the problem presented as if it's non-trivial? Maybe I made a mistake here?

Wait, let me check with another example where maybe the numbers are different. Let me take a=0, b=1, c=-1. Then compute p(x):

First term: [(x - 1)(x - (-1))]/[(0 - 1)(0 - (-1))] = [(x - 1)(x + 1)] / [(-1)(1)] = (x² - 1)/(-1) = -x² + 1.

Second term: [(x - 0)(x - (-1))]/[(1 - 0)(1 - (-1))] = [x(x + 1)] / [1*2] = [x² + x]/2.

Third term: [(x - 0)(x - 1)] / [(-1 - 0)(-1 - 1)] = [x(x - 1)] / [(-1)(-2)] = [x² - x]/2.

Adding them up:

First term: -x² + 1.

Second term: (x² + x)/2.

Third term: (x² - x)/2.

Adding second and third term: [x² + x + x² - x]/2 = [2x²]/2 = x².

Then adding first term: -x² + 1 + x² = 1. Again, the sum is 1. So p(x) is 1. Therefore, regardless of a, b, c, as long as they are distinct, p(x) is 1. Therefore, the degree is 0. Therefore, the answer is 0?

But then why does the problem say "Find the degree of the polynomial p(x)"? Maybe there's a misunderstanding here. Wait, but maybe I need to check if p(x) is indeed always 1. Let me think algebraically.

Let me consider that p(x) is the sum of three Lagrange basis polynomials. Since the Lagrange interpolation polynomial for the function f(x) = 1 is the constant polynomial 1, which is the unique polynomial of degree ≤ 2 that passes through the points (a, 1), (b, 1), (c, 1). Therefore, the Lagrange interpolation polynomial for these points is 1, which is exactly p(x). Therefore, p(x) is 1, hence degree 0. Therefore, the answer is 0.

But then why does the problem present it as a polynomial? Because if you just look at the expression, each term is quadratic, but when added together, they cancel out to give a constant. Therefore, the degree is 0. Therefore, the answer is 0. But maybe I need to check if there's a mistake in my reasoning.

Alternatively, maybe the problem is not about Lagrange interpolation? Wait, let me think again. The Lagrange basis polynomials are defined as L_i(x) = product_{j ≠ i} (x - x_j)/(x_i - x_j). Therefore, in this case, the given p(x) is exactly L_a(x) + L_b(x) + L_c(x). And since the sum of the Lagrange basis polynomials for the constant function 1 is 1, then p(x) is 1. Therefore, degree 0.

But maybe the problem is designed to trick us into thinking it's quadratic, but actually it's constant. Therefore, the answer is 0. However, let me check with another approach.

Alternatively, let me try to compute p(x) algebraically. Let me denote the three terms as follows:

Term1 = [(x - b)(x - c)] / [(a - b)(a - c)]

Term2 = [(x - a)(x - c)] / [(b - a)(b - c)]

Term3 = [(x - a)(x - b)] / [(c - a)(c - b)]

Let me compute Term1 + Term2 + Term3.

First, note that denominators can be rewritten:

Denominator of Term1: (a - b)(a - c) = (a - b)(a - c)

Denominator of Term2: (b - a)(b - c) = -(a - b)(b - c)

Denominator of Term3: (c - a)(c - b) = (c - a)(c - b) = ( - (a - c)) ( - (b - c)) = (a - c)(b - c)

Wait, maybe I can write all denominators in terms of (a - b), (b - c), (c - a) or something. Alternatively, maybe I can find a common denominator? But that might be complicated. Alternatively, let me consider evaluating p(x) at specific points.

Since p(x) is a polynomial, and if I can show that p(a) = 1, p(b) = 1, p(c) = 1, then since p(x) is a polynomial of degree at most 2 (since each term is quadratic), and it takes the value 1 at three distinct points, then p(x) must be the constant polynomial 1. Therefore, degree 0.

Let me check p(a):

p(a) = [(a - b)(a - c)] / [(a - b)(a - c)] + [(a - a)(a - c)] / [(b - a)(b - c)] + [(a - a)(a - b)] / [(c - a)(c - b)].

Simplify each term:

First term: [(a - b)(a - c)] / [(a - b)(a - c)] = 1.

Second term: [0 * (a - c)] / [...] = 0.

Third term: [0 * (a - b)] / [...] = 0.

Therefore, p(a) = 1 + 0 + 0 = 1.

Similarly, p(b) = 1, and p(c) = 1. Therefore, since p(x) is a polynomial of degree at most 2 (since each term is quadratic, and adding them gives degree at most 2), and it takes the value 1 at three distinct points, then p(x) must be the constant polynomial 1. Therefore, the degree is 0.

Therefore, the answer is 0. Therefore, the degree of p(x) is 0.

But let me check again why I thought initially that it might be quadratic. Because each term is quadratic, but when added together, they cancel out the quadratic terms. For example, in the first example, when I calculated with a=0, b=1, c=2, the quadratic terms canceled out. Similarly in the other examples. Therefore, the conclusion is that p(x) is constant. Therefore, degree 0.

But maybe I need to check the leading coefficients. Let me compute the coefficient of x² in each term.

First term: [(x - b)(x - c)] / [(a - b)(a - c)].

Expanding numerator: x² - (b + c)x + bc. Therefore, coefficient of x² is 1 / [(a - b)(a - c)].

Second term: [(x - a)(x - c)] / [(b - a)(b - c)].

Expanding numerator: x² - (a + c)x + ac. Coefficient of x² is 1 / [(b - a)(b - c)].

Third term: [(x - a)(x - b)] / [(c - a)(c - b)].

Expanding numerator: x² - (a + b)x + ab. Coefficient of x² is 1 / [(c - a)(c - b)].

Therefore, total coefficient of x² in p(x) is:

[1 / ((a - b)(a - c))] + [1 / ((b - a)(b - c))] + [1 / ((c - a)(c - b))].

Let me compute this sum. Let me denote the three terms as Term1, Term2, Term3.

Term1: 1 / [(a - b)(a - c)]

Term2: 1 / [(b - a)(b - c)] = -1 / [(a - b)(b - c)]

Term3: 1 / [(c - a)(c - b)] = 1 / [( - (a - c))(- (b - c))] = 1 / [(a - c)(b - c)].

Therefore, total coefficient:

Term1 + Term2 + Term3 = [1 / ((a - b)(a - c))] - [1 / ((a - b)(b - c))] + [1 / ((a - c)(b - c))].

Let me factor out 1 / [(a - b)(a - c)(b - c)]? Maybe not. Let me compute each term with common denominators.

Let me write all terms with denominator (a - b)(b - c)(a - c). Let me check:

Term1: [ (b - c) ] / [ (a - b)(a - c)(b - c) ) ].

Wait, Term1 is 1 / [(a - b)(a - c)] = [ (b - c) ] / [ (a - b)(a - c)(b - c) ) ].

Similarly, Term2: -1 / [(a - b)(b - c)] = - [ (a - c) ] / [ (a - b)(b - c)(a - c) ) ].

Term3: 1 / [(a - c)(b - c)] = [ (a - b) ] / [ (a - b)(b - c)(a - c) ) ].

Therefore, adding them together:

[ (b - c) - (a - c) + (a - b) ] / [ (a - b)(b - c)(a - c) ) ].

Compute numerator:

(b - c) - (a - c) + (a - b) = b - c - a + c + a - b = (b - b) + (-c + c) + (-a + a) = 0 + 0 + 0 = 0.

Therefore, the coefficient of x² is 0. Therefore, the x² terms cancel out. Therefore, p(x) has no x² term. Similarly, maybe the x term cancels out? Let me check.

But since we already know that p(x) is 1, which is a constant, then all coefficients of x², x, and the constant term must be 0 except the constant term. But since we saw that the coefficient of x² is 0, and if we check the coefficient of x, it should also be 0. Let me check that.

Alternatively, since we already know that p(x) is 1, then the coefficient of x is 0, and the constant term is 1. Therefore, the degree is 0. Therefore, the answer is 0.

But just to be thorough, let me check the coefficient of x. Let me compute the coefficient of x in each term.

First term: [(x - b)(x - c)] / [(a - b)(a - c)].

Expanding numerator: x² - (b + c)x + bc. Therefore, coefficient of x is - (b + c) / [(a - b)(a - c)].

Second term: [(x - a)(x - c)] / [(b - a)(b - c)].

Expanding numerator: x² - (a + c)x + ac. Coefficient of x is - (a + c) / [(b - a)(b - c)].

Third term: [(x - a)(x - b)] / [(c - a)(c - b)].

Expanding numerator: x² - (a + b)x + ab. Coefficient of x is - (a + b) / [(c - a)(c - b)].

Therefore, total coefficient of x in p(x) is:

- [ (b + c) / ((a - b)(a - c)) + (a + c) / ((b - a)(b - c)) + (a + b) / ((c - a)(c - b)) ].

Let me compute this expression. Let me denote this as Coeff_x.

Let me rewrite each term:

First term: - (b + c) / [(a - b)(a - c)].

Second term: - (a + c) / [(b - a)(b - c)] = (a + c) / [(a - b)(b - c)].

Third term: - (a + b) / [(c - a)(c - b)] = - (a + b) / [ ( - (a - c)) ( - (b - c)) ] = - (a + b) / [ (a - c)(b - c) ].

Therefore, Coeff_x is:

- (b + c)/[(a - b)(a - c)] + (a + c)/[(a - b)(b - c)] - (a + b)/[(a - c)(b - c)].

Let me factor out 1 / [(a - b)(a - c)(b - c)]? Let me try to combine these terms.

Let me write all terms with denominator (a - b)(a - c)(b - c):

First term: - (b + c)(b - c) / [ (a - b)(a - c)(b - c) ) ].

Second term: (a + c)(a - c) / [ (a - b)(a - c)(b - c) ) ].

Third term: - (a + b)(a - b) / [ (a - b)(a - c)(b - c) ) ].

Wait, let me check:

First term: original denominator is (a - b)(a - c). To get denominator (a - b)(a - c)(b - c), multiply numerator and denominator by (b - c). Therefore, first term becomes - (b + c)(b - c) / [ (a - b)(a - c)(b - c) ) ].

Similarly, second term: original denominator is (a - b)(b - c). Multiply numerator and denominator by (a - c): numerator becomes (a + c)(a - c), denominator becomes (a - b)(a - c)(b - c).

Third term: original denominator is (a - c)(b - c). Multiply numerator and denominator by (a - b): numerator becomes - (a + b)(a - b), denominator becomes (a - b)(a - c)(b - c).

Therefore, Coeff_x is:

[ - (b + c)(b - c) + (a + c)(a - c) - (a + b)(a - b) ] / [ (a - b)(a - c)(b - c) ) ].

Let me compute the numerator:

First part: - (b + c)(b - c) = - (b² - c²) = -b² + c².

Second part: (a + c)(a - c) = a² - c².

Third part: - (a + b)(a - b) = - (a² - b²) = -a² + b².

Adding them together:

(-b² + c²) + (a² - c²) + (-a² + b²) = (-b² + b²) + (c² - c²) + (a² - a²) = 0 + 0 + 0 = 0.

Therefore, the coefficient of x is also 0. Therefore, p(x) has no x² term, no x term, and the constant term is 1. Therefore, p(x) is the constant polynomial 1. Therefore, degree 0.

Therefore, the answer is 0. Therefore, the degree of p(x) is 0.

But just to make sure, let me think again. Suppose I didn't know about Lagrange interpolation. How else could I approach this?

Alternatively, since p(x) is a polynomial, and we can consider that p(x) - 1 is a polynomial that has roots at x = a, x = b, x = c. Because p(a) = 1, p(b) = 1, p(c) = 1. Therefore, p(x) - 1 has three roots: a, b, c. Therefore, p(x) - 1 is a polynomial of degree at most 2 (since p(x) is degree at most 2) with three roots. Therefore, p(x) - 1 must be the zero polynomial, so p(x) = 1. Therefore, degree 0. Therefore, this confirms again that the degree is 0.

Therefore, regardless of the approach, the conclusion is that p(x) is the constant polynomial 1, so degree 0. Therefore, the answer is 0.

But just to make sure, let me check with another perspective. Suppose I consider the general case. Let me assume that p(x) is a quadratic polynomial. Let me write p(x) = Ax² + Bx + C. Then, since p(a) = 1, p(b) = 1, p(c) = 1. Therefore, we have the system:

Aa² + Ba + C = 1,

Ab² + Bb + C = 1,

Ac² + Bc + C = 1.

Subtracting the first equation from the second: A(b² - a²) + B(b - a) = 0 => A(b + a) + B = 0.

Similarly, subtracting the second from the third: A(c² - b²) + B(c - b) = 0 => A(c + b) + B = 0.

Therefore, we have:

A(a + b) + B = 0,

A(b + c) + B = 0.

Subtracting these two equations: A(a + b - b - c) = 0 => A(a - c) = 0.

Since a ≠ c, then A = 0. Therefore, A = 0. Then from A(a + b) + B = 0, we get B = 0. Therefore, p(x) = C. Then from p(a) = 1, we get C = 1. Therefore, p(x) = 1. Therefore, degree 0. Therefore, this algebraic approach also confirms that p(x) is constant. Therefore, the degree is 0.

Therefore, regardless of the method, the conclusion is that the degree is 0. Therefore, the answer is 0.

But just to make sure, let me check if there's any possibility that p(x) could be a higher degree polynomial. For example, if the denominators were not constants? But no, the denominators are constants because a, b, c are constants. Therefore, each term is a quadratic polynomial divided by a constant, so each term is quadratic. Therefore, the sum is a quadratic polynomial. However, as shown, the quadratic terms cancel out, leading to a constant. Therefore, the degree is 0.

Therefore, I think the answer is 0. Therefore, the degree of p(x) is 0.

**Final Answer**
\boxed{0}

To determine the degree of the polynomial

$$
p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)},
$$

we begin by analyzing the structure of the expression.

---

### Step 1: Recognize the Form

Each term in the sum is of the form:

$$
\frac{(x - x_j)(x - x_k)}{(x_i - x_j)(x_i - x_k)},
$$

where $ x_i, x_j, x_k $ are distinct real numbers. This is the standard form of the **Lagrange basis polynomial** for interpolation at three distinct points $ a, b, c $.

Thus, the given expression is the **sum of the three Lagrange basis polynomials** corresponding to the points $ a, b, c $, each with coefficient 1. That is:

$$
p(x) = L_a(x) + L_b(x) + L_c(x),
$$

where $ L_a(x) $, $ L_b(x) $, and $ L_c(x) $ are the Lagrange basis polynomials for interpolation at $ a, b, c $, respectively.

---

### Step 2: Use the Property of Lagrange Basis Polynomials

A key property of Lagrange basis polynomials is that the **sum of all Lagrange basis polynomials** for a set of points is the **constant polynomial 1**. That is:

$$
L_a(x) + L_b(x) + L_c(x) = 1.
$$

This is because the Lagrange interpolation polynomial for the constant function $ f(x) = 1 $ is the constant polynomial 1, and the sum of the basis polynomials gives that.

Therefore, we conclude:

$$
p(x) = 1.
$$

---

### Step 3: Determine the Degree

Since $ p(x) = 1 $, it is a **constant polynomial**. The degree of a constant polynomial is **0**.

---

### Final Answer

$$
\boxed{0}
$$

[Assistant]
Explanation:
Okay, so I need to find the degree of the polynomial p(x) given by that expression. Let me first write down the problem again to make sure I have it right:

p(x) = [(x - b)(x - c)] / [(a - b)(a - c)] + [(x - a)(x - c)] / [(b - a)(b - c)] + [(x - a)(x - b)] / [(c - a)(c - b)].

And a, b, c are distinct real numbers. Alright, so the question is asking for the degree of this polynomial. Let me think about how to approach this.

First, I know that when you have a polynomial expressed as a sum of terms like this, each term is a rational function, but the problem says that p(x) is a polynomial. So maybe each term, when combined, simplifies to a polynomial? Or maybe they are designed in such a way that the denominators cancel out the denominators? Wait, but the denominators are constants, right? Because a, b, c are constants, so each denominator is just a constant. So each term is a quadratic polynomial divided by a constant. Therefore, each term is a quadratic polynomial. So if we add three quadratic polynomials together, the result could be a quadratic polynomial, but maybe there's cancellation or something?

But wait, the problem says p(x) is a polynomial. So maybe when you add them up, the denominators are actually factors that make the entire expression a polynomial. Wait, but if each term is a quadratic divided by a constant, then adding them would just be a quadratic polynomial. Unless there's some cancellation? Wait, but if they are all quadratics, adding them would give a quadratic. But maybe the coefficients of x² cancel out? Let me check.

Let me try to compute the coefficients of x² in each term. Let me take the first term: [(x - b)(x - c)] / [(a - b)(a - c)]. Let me expand the numerator:

(x - b)(x - c) = x² - (b + c)x + bc. So dividing by (a - b)(a - c), which is a constant. Similarly for the other terms.

So each term is a quadratic polynomial divided by a constant. Therefore, each term is quadratic. So when you add them up, the sum is a quadratic polynomial. But the problem says "Find the degree of the polynomial p(x)". So maybe the answer is 2? But wait, maybe there is some cancellation? Let me check with specific values.

Alternatively, maybe this is a known polynomial. Wait, the expression looks similar to Lagrange interpolation formula. Let me recall: the Lagrange interpolation polynomial for three points. If we have three points, then the Lagrange basis polynomials are of the form [(x - x_j)(x - x_k)] / [(x_i - x_j)(x_i - x_k)] for each i, j, k. So if we have three points, then the Lagrange interpolation polynomial is the sum of y_i times the basis polynomial. But in this case, the coefficients are 1 for each term. Wait, so if we have three points, and we take the sum of the basis polynomials, then that would be 1? Wait, no. Wait, if you take the sum of the Lagrange basis polynomials, then for any x, the sum would be 1? Wait, let me check.

Suppose we have three points, say, x = a, b, c. Then the Lagrange basis polynomials are L_a(x), L_b(x), L_c(x), such that L_a(a) = 1, L_a(b) = 0, etc. Then, if we take L_a(x) + L_b(x) + L_c(x), then for any x, this sum would be 1? Wait, no. Wait, if you have three points, then the sum of the Lagrange basis polynomials is 1? Let me check with specific numbers.

Let me take a simple case where a=0, b=1, c=2. Then compute L_0(x) + L_1(x) + L_2(x). Let me compute each term:

L_0(x) = [(x - 1)(x - 2)] / [(0 - 1)(0 - 2)] = [(x - 1)(x - 2)] / [(-1)(-2)] = [(x - 1)(x - 2)] / 2.

Similarly, L_1(x) = [(x - 0)(x - 2)] / [(1 - 0)(1 - 2)] = [x(x - 2)] / [1*(-1)] = -x(x - 2).

L_2(x) = [(x - 0)(x - 1)] / [(2 - 0)(2 - 1)] = [x(x - 1)] / [2*1] = [x(x - 1)] / 2.

So adding them up:

L_0(x) + L_1(x) + L_2(x) = [(x - 1)(x - 2)/2] - x(x - 2) + [x(x - 1)/2].

Let me compute each term:

First term: (x² - 3x + 2)/2.

Second term: -x² + 2x.

Third term: (x² - x)/2.

Adding them together:

First term: (x² - 3x + 2)/2.

Third term: (x² - x)/2. Adding first and third term: [x² - 3x + 2 + x² - x]/2 = [2x² - 4x + 2]/2 = x² - 2x + 1.

Then adding the second term: x² - 2x + 1 - x² + 2x = (x² - x²) + (-2x + 2x) + 1 = 0 + 0 + 1 = 1.

So indeed, the sum is 1. Therefore, in this case, the sum of the three Lagrange basis polynomials is 1, which is a constant polynomial. Therefore, the degree is 0. Wait, but in this case, the original problem's p(x) is exactly the sum of the three Lagrange basis polynomials. Therefore, p(x) is equal to 1? But that would mean that the degree is 0. But that contradicts my initial thought that it's quadratic. Wait, but in this case, when I calculated with specific numbers, the sum was 1. So maybe p(x) is the constant polynomial 1? Therefore, degree 0?

But that seems surprising. Let me check with another example. Let me take a different set of a, b, c. Let me take a=1, b=2, c=3. Then compute p(x):

First term: [(x - 2)(x - 3)] / [(1 - 2)(1 - 3)] = [(x - 2)(x - 3)] / [(-1)(-2)] = [(x - 2)(x - 3)] / 2.

Second term: [(x - 1)(x - 3)] / [(2 - 1)(2 - 3)] = [(x - 1)(x - 3)] / [(1)(-1)] = -[(x - 1)(x - 3)].

Third term: [(x - 1)(x - 2)] / [(3 - 1)(3 - 2)] = [(x - 1)(x - 2)] / [2*1] = [(x - 1)(x - 2)] / 2.

Adding them up:

First term: (x² - 5x + 6)/2.

Second term: -[x² - 4x + 3] = -x² + 4x - 3.

Third term: (x² - 3x + 2)/2.

Adding all together:

First term + third term: [x² - 5x + 6 + x² - 3x + 2]/2 = [2x² - 8x + 8]/2 = x² - 4x + 4.

Then adding the second term: x² - 4x + 4 - x² + 4x - 3 = (x² - x²) + (-4x + 4x) + (4 - 3) = 0 + 0 + 1 = 1. Again, the sum is 1. So p(x) is 1. Therefore, the degree is 0. But that seems to contradict my initial thought that it's quadratic. But according to these examples, it's a constant function. Therefore, the degree is 0. But why?

Wait, but the problem says "Find the degree of the polynomial p(x)". So maybe the answer is 0? But why does the problem give such a complicated expression? Maybe I need to check again.

Wait, but in the problem statement, the polynomial is written as the sum of three terms, each of which is a quadratic divided by a constant. But when you add them up, they cancel out to give 1. Therefore, p(x) is the constant polynomial 1, so degree 0. But then why is the problem presented as if it's non-trivial? Maybe I made a mistake here?

Wait, let me check with another example where maybe the numbers are different. Let me take a=0, b=1, c=-1. Then compute p(x):

First term: [(x - 1)(x - (-1))]/[(0 - 1)(0 - (-1))] = [(x - 1)(x + 1)] / [(-1)(1)] = (x² - 1)/(-1) = -x² + 1.

Second term: [(x - 0)(x - (-1))]/[(1 - 0)(1 - (-1))] = [x(x + 1)] / [1*2] = [x² + x]/2.

Third term: [(x - 0)(x - 1)] / [(-1 - 0)(-1 - 1)] = [x(x - 1)] / [(-1)(-2)] = [x² - x]/2.

Adding them up:

First term: -x² + 1.

Second term: (x² + x)/2.

Third term: (x² - x)/2.

Adding second and third term: [x² + x + x² - x]/2 = [2x²]/2 = x².

Then adding first term: -x² + 1 + x² = 1. Again, the sum is 1. So p(x) is 1. Therefore, regardless of a, b, c, as long as they are distinct, p(x) is 1. Therefore, the degree is 0. Therefore, the answer is 0?

But then why does the problem say "Find the degree of the polynomial p(x)"? Maybe there's a misunderstanding here. Wait, but maybe I need to check if p(x) is indeed always 1. Let me think algebraically.

Let me consider that p(x) is the sum of three Lagrange basis polynomials. Since the Lagrange interpolation polynomial for the function f(x) = 1 is the constant polynomial 1, which is the unique polynomial of degree ≤ 2 that passes through the points (a, 1), (b, 1), (c, 1). Therefore, the Lagrange interpolation polynomial for these points is 1, which is exactly p(x). Therefore, p(x) is 1, hence degree 0. Therefore, the answer is 0.

But then why does the problem present it as a polynomial? Because if you just look at the expression, each term is quadratic, but when added together, they cancel out to give a constant. Therefore, the degree is 0. Therefore, the answer is 0. But maybe I need to check if there's a mistake in my reasoning.

Alternatively, maybe the problem is not about Lagrange interpolation? Wait, let me think again. The Lagrange basis polynomials are defined as L_i(x) = product_{j ≠ i} (x - x_j)/(x_i - x_j). Therefore, in this case, the given p(x) is exactly L_a(x) + L_b(x) + L_c(x). And since the sum of the Lagrange basis polynomials for the constant function 1 is 1, then p(x) is 1. Therefore, degree 0.

But maybe the problem is designed to trick us into thinking it's quadratic, but actually it's constant. Therefore, the answer is 0. However, let me check with another approach.

Alternatively, let me try to compute p(x) algebraically. Let me denote the three terms as follows:

Term1 = [(x - b)(x - c)] / [(a - b)(a - c)]

Term2 = [(x - a)(x - c)] / [(b - a)(b - c)]

Term3 = [(x - a)(x - b)] / [(c - a)(c - b)]

Let me compute Term1 + Term2 + Term3.

First, note that denominators can be rewritten:

Denominator of Term1: (a - b)(a - c) = (a - b)(a - c)

Denominator of Term2: (b - a)(b - c) = -(a - b)(b - c)

Denominator of Term3: (c - a)(c - b) = (c - a)(c - b) = ( - (a - c)) ( - (b - c)) = (a - c)(b - c)

Wait, maybe I can write all denominators in terms of (a - b), (b - c), (c - a) or something. Alternatively, maybe I can find a common denominator? But that might be complicated. Alternatively, let me consider evaluating p(x) at specific points.

Since p(x) is a polynomial, and if I can show that p(a) = 1, p(b) = 1, p(c) = 1, then since p(x) is a polynomial of degree at most 2 (since each term is quadratic), and it takes the value 1 at three distinct points, then p(x) must be the constant polynomial 1. Therefore, degree 0.

Let me check p(a):

p(a) = [(a - b)(a - c)] / [(a - b)(a - c)] + [(a - a)(a - c)] / [(b - a)(b - c)] + [(a - a)(a - b)] / [(c - a)(c - b)].

Simplify each term:

First term: [(a - b)(a - c)] / [(a - b)(a - c)] = 1.

Second term: [0 * (a - c)] / [...] = 0.

Third term: [0 * (a - b)] / [...] = 0.

Therefore, p(a) = 1 + 0 + 0 = 1.

Similarly, p(b) = 1, and p(c) = 1. Therefore, since p(x) is a polynomial of degree at most 2 (since each term is quadratic, and adding them gives degree at most 2), and it takes the value 1 at three distinct points, then p(x) must be the constant polynomial 1. Therefore, the degree is 0.

Therefore, the answer is 0. Therefore, the degree of p(x) is 0.

But let me check again why I thought initially that it might be quadratic. Because each term is quadratic, but when added together, they cancel out the quadratic terms. For example, in the first example, when I calculated with a=0, b=1, c=2, the quadratic terms canceled out. Similarly in the other examples. Therefore, the conclusion is that p(x) is constant. Therefore, degree 0.

But maybe I need to check the leading coefficients. Let me compute the coefficient of x² in each term.

First term: [(x - b)(x - c)] / [(a - b)(a - c)].

Expanding numerator: x² - (b + c)x + bc. Therefore, coefficient of x² is 1 / [(a - b)(a - c)].

Second term: [(x - a)(x - c)] / [(b - a)(b - c)].

Expanding numerator: x² - (a + c)x + ac. Coefficient of x² is 1 / [(b - a)(b - c)].

Third term: [(x - a)(x - b)] / [(c - a)(c - b)].

Expanding numerator: x² - (a + b)x + ab. Coefficient of x² is 1 / [(c - a)(c - b)].

Therefore, total coefficient of x² in p(x) is:

[1 / ((a - b)(a - c))] + [1 / ((b - a)(b - c))] + [1 / ((c - a)(c - b))].

Let me compute this sum. Let me denote the three terms as Term1, Term2, Term3.

Term1: 1 / [(a - b)(a - c)]

Term2: 1 / [(b - a)(b - c)] = -1 / [(a - b)(b - c)]

Term3: 1 / [(c - a)(c - b)] = 1 / [( - (a - c))(- (b - c))] = 1 / [(a - c)(b - c)].

Therefore, total coefficient:

Term1 + Term2 + Term3 = [1 / ((a - b)(a - c))] - [1 / ((a - b)(b - c))] + [1 / ((a - c)(b - c))].

Let me factor out 1 / [(a - b)(a - c)(b - c)]? Maybe not. Let me compute each term with common denominators.

Let me write all terms with denominator (a - b)(b - c)(a - c). Let me check:

Term1: [ (b - c) ] / [ (a - b)(a - c)(b - c) ) ].

Wait, Term1 is 1 / [(a - b)(a - c)] = [ (b - c) ] / [ (a - b)(a - c)(b - c) ) ].

Similarly, Term2: -1 / [(a - b)(b - c)] = - [ (a - c) ] / [ (a - b)(b - c)(a - c) ) ].

Term3: 1 / [(a - c)(b - c)] = [ (a - b) ] / [ (a - b)(b - c)(a - c) ) ].

Therefore, adding them together:

[ (b - c) - (a - c) + (a - b) ] / [ (a - b)(b - c)(a - c) ) ].

Compute numerator:

(b - c) - (a - c) + (a - b) = b - c - a + c + a - b = (b - b) + (-c + c) + (-a + a) = 0 + 0 + 0 = 0.

Therefore, the coefficient of x² is 0. Therefore, the x² terms cancel out. Therefore, p(x) has no x² term. Similarly, maybe the x term cancels out? Let me check.

But since we already know that p(x) is 1, which is a constant, then all coefficients of x², x, and the constant term must be 0 except the constant term. But since we saw that the coefficient of x² is 0, and if we check the coefficient of x, it should also be 0. Let me check that.

Alternatively, since we already know that p(x) is 1, then the coefficient of x is 0, and the constant term is 1. Therefore, the degree is 0. Therefore, the answer is 0.

But just to be thorough, let me check the coefficient of x. Let me compute the coefficient of x in each term.

First term: [(x - b)(x - c)] / [(a - b)(a - c)].

Expanding numerator: x² - (b + c)x + bc. Therefore, coefficient of x is - (b + c) / [(a - b)(a - c)].

Second term: [(x - a)(x - c)] / [(b - a)(b - c)].

Expanding numerator: x² - (a + c)x + ac. Coefficient of x is - (a + c) / [(b - a)(b - c)].

Third term: [(x - a)(x - b)] / [(c - a)(c - b)].

Expanding numerator: x² - (a + b)x + ab. Coefficient of x is - (a + b) / [(c - a)(c - b)].

Therefore, total coefficient of x in p(x) is:

- [ (b + c) / ((a - b)(a - c)) + (a + c) / ((b - a)(b - c)) + (a + b) / ((c - a)(c - b)) ].

Let me compute this expression. Let me denote this as Coeff_x.

Let me rewrite each term:

First term: - (b + c) / [(a - b)(a - c)].

Second term: - (a + c) / [(b - a)(b - c)] = (a + c) / [(a - b)(b - c)].

Third term: - (a + b) / [(c - a)(c - b)] = - (a + b) / [ ( - (a - c)) ( - (b - c)) ] = - (a + b) / [ (a - c)(b - c) ].

Therefore, Coeff_x is:

- (b + c)/[(a - b)(a - c)] + (a + c)/[(a - b)(b - c)] - (a + b)/[(a - c)(b - c)].

Let me factor out 1 / [(a - b)(a - c)(b - c)]? Let me try to combine these terms.

Let me write all terms with denominator (a - b)(a - c)(b - c):

First term: - (b + c)(b - c) / [ (a - b)(a - c)(b - c) ) ].

Second term: (a + c)(a - c) / [ (a - b)(a - c)(b - c) ) ].

Third term: - (a + b)(a - b) / [ (a - b)(a - c)(b - c) ) ].

Wait, let me check:

First term: original denominator is (a - b)(a - c). To get denominator (a - b)(a - c)(b - c), multiply numerator and denominator by (b - c). Therefore, first term becomes - (b + c)(b - c) / [ (a - b)(a - c)(b - c) ) ].

Similarly, second term: original denominator is (a - b)(b - c). Multiply numerator and denominator by (a - c): numerator becomes (a + c)(a - c), denominator becomes (a - b)(a - c)(b - c).

Third term: original denominator is (a - c)(b - c). Multiply numerator and denominator by (a - b): numerator becomes - (a + b)(a - b), denominator becomes (a - b)(a - c)(b - c).

Therefore, Coeff_x is:

[ - (b + c)(b - c) + (a + c)(a - c) - (a + b)(a - b) ] / [ (a - b)(a - c)(b - c) ) ].

Let me compute the numerator:

First part: - (b + c)(b - c) = - (b² - c²) = -b² + c².

Second part: (a + c)(a - c) = a² - c².

Third part: - (a + b)(a - b) = - (a² - b²) = -a² + b².

Adding them together:

(-b² + c²) + (a² - c²) + (-a² + b²) = (-b² + b²) + (c² - c²) + (a² - a²) = 0 + 0 + 0 = 0.

Therefore, the coefficient of x is also 0. Therefore, p(x) has no x² term, no x term, and the constant term is 1. Therefore, p(x) is the constant polynomial 1. Therefore, degree 0.

Therefore, the answer is 0. Therefore, the degree of p(x) is 0.

But just to make sure, let me think again. Suppose I didn't know about Lagrange interpolation. How else could I approach this?

Alternatively, since p(x) is a polynomial, and we can consider that p(x) - 1 is a polynomial that has roots at x = a, x = b, x = c. Because p(a) = 1, p(b) = 1, p(c) = 1. Therefore, p(x) - 1 has three roots: a, b, c. Therefore, p(x) - 1 is a polynomial of degree at most 2 (since p(x) is degree at most 2) with three roots. Therefore, p(x) - 1 must be the zero polynomial, so p(x) = 1. Therefore, degree 0. Therefore, this confirms again that the degree is 0.

Therefore, regardless of the approach, the conclusion is that p(x) is the constant polynomial 1, so degree 0. Therefore, the answer is 0.

But just to make sure, let me check with another perspective. Suppose I consider the general case. Let me assume that p(x) is a quadratic polynomial. Let me write p(x) = Ax² + Bx + C. Then, since p(a) = 1, p(b) = 1, p(c) = 1. Therefore, we have the system:

Aa² + Ba + C = 1,

Ab² + Bb + C = 1,

Ac² + Bc + C = 1.

Subtracting the first equation from the second: A(b² - a²) + B(b - a) = 0 => A(b + a) + B = 0.

Similarly, subtracting the second from the third: A(c² - b²) + B(c - b) = 0 => A(c + b) + B = 0.

Therefore, we have:

A(a + b) + B = 0,

A(b + c) + B = 0.

Subtracting these two equations: A(a + b - b - c) = 0 => A(a - c) = 0.

Since a ≠ c, then A = 0. Therefore, A = 0. Then from A(a + b) + B = 0, we get B = 0. Therefore, p(x) = C. Then from p(a) = 1, we get C = 1. Therefore, p(x) = 1. Therefore, degree 0. Therefore, this algebraic approach also confirms that p(x) is constant. Therefore, the degree is 0.

Therefore, regardless of the method, the conclusion is that the degree is 0. Therefore, the answer is 0.

But just to make sure, let me check if there's any possibility that p(x) could be a higher degree polynomial. For example, if the denominators were not constants? But no, the denominators are constants because a, b, c are constants. Therefore, each term is a quadratic polynomial divided by a constant, so each term is quadratic. Therefore, the sum is a quadratic polynomial. However, as shown, the quadratic terms cancel out, leading to a constant. Therefore, the degree is 0.

Therefore, I think the answer is 0. Therefore, the degree of p(x) is 0.

**Final Answer**
\boxed{0}

To determine the degree of the polynomial

$$
p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)},
$$

we begin by analyzing the structure of the expression.

---

### Step 1: Recognize the Form

Each term in the sum is of the form:

$$
\frac{(x - x_j)(x - x_k)}{(x_i - x_j)(x_i - x_k)},
$$

where $ x_i, x_j, x_k $ are distinct real numbers. This is the standard form of the **Lagrange basis polynomial** for interpolation at three distinct points $ a, b, c $.

Thus, the given expression is the **sum of the three Lagrange basis polynomials** corresponding to the points $ a, b, c $, each with coefficient 1. That is:

$$
p(x) = L_a(x) + L_b(x) + L_c(x),
$$

where $ L_a(x) $, $ L_b(x) $, and $ L_c(x) $ are the Lagrange basis polynomials for interpolation at $ a, b, c $, respectively.

---

### Step 2: Use the Property of Lagrange Basis Polynomials

A key property of Lagrange basis polynomials is that the **sum of all Lagrange basis polynomials** for a set of points is the **constant polynomial 1**. That is:

$$
L_a(x) + L_b(x) + L_c(x) = 1.
$$

This is because the Lagrange interpolation polynomial for the constant function $ f(x) = 1 $ is the constant polynomial 1, and the sum of the basis polynomials gives that.

Therefore, we conclude:

$$
p(x) = 1.
$$

---

### Step 3: Determine the Degree

Since $ p(x) = 1 $, it is a **constant polynomial**. The degree of a constant polynomial is **0**.

---

### Final Answer

$$
\boxed{0}
$$

Solution:
Note that $p(x)$ has degree at most 2.  Also, $p(a) = p(b) = p(c) = 1.$  Thus, the polynomials $p(x)$ and 1 agree at three different values, so by the Identity Theorem, they are the same polynomial.  Hence, the degree of $p(x)$ (which is the constant polynomial 1) is $\boxed{0}.$

You can check manually that
\[p(x) = \frac{(x - b)(x - c)}{(a - b)(a - c)} + \frac{(x - a)(x - c)}{(b - a)(b - c)} + \frac{(x - a)(x - b)}{(c - a)(c - b)}\]simplifies to 1.


Explanation:
Okay, so I need to find cos(angle RPS) given that sin(angle RPQ) is 7/25. Let me try to visualize the diagram first. There's a point P at the origin, Q at (2,0), and S at (-2,0). R is some point above the x-axis, created by rotating the point (1.5, 0) by an angle whose sine is 7/25. The diagram shows lines Q to S, P to R, and probably other connections.

First, let me recall that angle RPQ is the angle at point P between points R and Q. So, angle RPQ is formed by the lines PR and PQ. Similarly, angle RPS is the angle at point P between points R and S. So, angle RPS is formed by lines PR and PS. Since S is on the opposite side of P from Q, maybe these angles are supplementary or something? Wait, but I need to figure out the relationship between them.

Given that sin(angle RPQ) = 7/25, I need to find cos(angle RPS). Let me think about how these angles relate. Since points Q and S are on the x-axis, symmetric with respect to P? Wait, P is at (0,0), Q is at (2,0), and S is at (-2,0). So PQ is from P(0,0) to Q(2,0), and PS is from P(0,0) to S(-2,0). So PQ and PS are in opposite directions along the x-axis. Therefore, angle between PQ and PS is 180 degrees. But angle RPQ is between PR and PQ, and angle RPS is between PR and PS. So if I imagine PR as a line from P to R, then angle RPQ is the angle between PR and PQ (which is along the positive x-axis), and angle RPS is the angle between PR and PS (which is along the negative x-axis). Therefore, angle RPS is supplementary to angle RPQ? Wait, not necessarily. Let me think.

If I consider angle between PR and PQ (which is angle RPQ) and angle between PR and PS (which is angle RPS), since PQ and PS are in opposite directions, then angle RPS would be equal to 180 degrees minus angle RPQ? Wait, if you have a line PR making an angle θ with PQ (which is along positive x-axis), then the angle between PR and PS (which is along negative x-axis) would be 180° - θ. Therefore, angle RPS = 180° - angle RPQ. Therefore, cos(angle RPS) = cos(180° - angle RPQ) = -cos(angle RPQ). Because cosine of supplementary angles is negative cosine of the original angle. So if I can find cos(angle RPQ), then take its negative, that would be cos(angle RPS).

But wait, the problem gives me sin(angle RPQ) = 7/25. So if I can find cos(angle RPQ), then I can get cos(angle RPS). Let me confirm this relationship again. Let me draw a mental picture. Let me suppose angle RPQ is θ, so sin θ = 7/25. Then angle RPS is 180° - θ, so cos(angle RPS) = cos(180° - θ) = -cos θ. Therefore, if I can find cos θ, then multiply by -1 to get the answer.

So, first step: find cos θ where θ is angle RPQ, given sin θ = 7/25. Since sin²θ + cos²θ = 1, so cos θ = ±√(1 - (7/25)²). But we need to figure out the sign. Since angle RPQ is an angle between PR and PQ. Since R is above the x-axis (as per the Asymptote code, R is rotated from (1.5, 0) by some angle, so it's above the x-axis), so angle RPQ is between PR (going up to R) and PQ (along positive x-axis). Therefore, θ is between 0 and 180 degrees, but since R is above the x-axis, θ is between 0 and 180, but more specifically, since sin θ is positive, θ is between 0 and 180. However, depending on where R is, θ could be acute or obtuse. But since sin θ = 7/25, which is positive, but we need to figure out if θ is acute or obtuse. However, since R is constructed by rotating (1.5, 0) by an angle whose sine is 7/25. Wait, the Asymptote code says R = rotate(aSin(7/25))*(1.5,0). So that means that R is obtained by rotating the point (1.5, 0) by an angle whose sine is 7/25. So the angle of rotation is arcsin(7/25). Therefore, that angle is between 0 and 90 degrees, since arcsin(7/25) is in the first quadrant. Therefore, the angle between PR and the x-axis is arcsin(7/25), so angle RPQ is arcsin(7/25), which is acute. Therefore, θ is acute, so cos θ is positive. Therefore, cos θ = sqrt(1 - (7/25)^2). Let me compute that.

First, compute (7/25)^2: 49/625. Then 1 - 49/625 = (625 - 49)/625 = 576/625. Therefore, sqrt(576/625) = 24/25. Therefore, cos θ = 24/25. Therefore, cos(angle RPS) = -cos θ = -24/25. Therefore, the answer is -24/25.

Wait, but let me make sure that angle RPS is indeed 180° - θ. Let me think again. If angle RPQ is θ, which is the angle between PR and PQ (positive x-axis). Then angle between PR and PS (negative x-axis) would be angle RPS. Since PQ is along positive x-axis, PS is along negative x-axis, so the angle between PR and PS is 180° - θ. Therefore, yes, angle RPS = 180° - θ. Therefore, cos(angle RPS) = cos(180° - θ) = -cos θ. Therefore, since cos θ is 24/25, then cos(angle RPS) is -24/25.

But let me check if there's another way to approach this problem, maybe using coordinates. Let me try that.

The Asymptote code defines points: SS is (-2,0), P is (0,0), Q is (2,0), and R is obtained by rotating the point (1.5, 0) by an angle whose sine is 7/25. So, rotating (1.5, 0) by angle α where sin α = 7/25. Let me find coordinates of R.

If we rotate a point (x, y) by angle α, the new coordinates are (x cos α - y sin α, x sin α + y cos α). Since the original point is (1.5, 0), so x = 1.5, y = 0. Therefore, after rotation, coordinates of R are (1.5 cos α - 0, 1.5 sin α + 0) = (1.5 cos α, 1.5 sin α). Therefore, R is at (1.5 cos α, 1.5 sin α).

Given that sin α = 7/25, so we can find cos α as 24/25, as before. Therefore, coordinates of R are (1.5*(24/25), 1.5*(7/25)). Let me compute that.

First, 1.5 is 3/2. Therefore, x-coordinate: (3/2)*(24/25) = (72)/50 = 36/25. Y-coordinate: (3/2)*(7/25) = 21/50. Therefore, R is at (36/25, 21/50).

Now, angle RPS is the angle at P between points R and S. Since S is at (-2, 0), and P is at (0,0). So vectors PR and PS. Vector PR is from P to R: (36/25, 21/50). Vector PS is from P to S: (-2, 0).

To find angle between vectors PR and PS, which is angle RPS. The formula for cosine of the angle between two vectors u and v is (u ⋅ v)/(|u| |v|).

So let me compute the dot product of PR and PS.

PR ⋅ PS = (36/25)*(-2) + (21/50)*0 = -72/25 + 0 = -72/25.

Now, |PR| is the magnitude of vector PR. Let's compute that.

|PR| = sqrt( (36/25)^2 + (21/50)^2 ). Let me compute that.

First, square of 36/25 is (1296)/625. Square of 21/50 is (441)/2500.

To add them, convert to common denominator. 625 is 25*25, 2500 is 100*25. So common denominator is 2500.

So 1296/625 = (1296 * 4)/2500 = 5184/2500.

Then 441/2500 remains as is.

Total is (5184 + 441)/2500 = 5625/2500.

sqrt(5625/2500) = sqrt(5625)/sqrt(2500) = 75/50 = 3/2.

Wait, that's interesting. The magnitude of PR is 3/2. Which makes sense because the original point was (1.5, 0) and we rotated it, so the distance from P to R should be the same as the distance from P to (1.5, 0), which is 1.5. So 3/2. So that checks out.

Now, |PS| is the magnitude of vector PS, which is from (0,0) to (-2, 0), so that's just 2.

Therefore, cos(angle RPS) = (PR ⋅ PS)/(|PR| |PS|) = (-72/25)/( (3/2)*2 ) = (-72/25)/(3) = (-72/25)*(1/3) = -24/25.

So that's the same result as before. Therefore, cos(angle RPS) is -24/25.

Therefore, the answer is \boxed{-\dfrac{24}{25}}.

But let me just check once more. Since angle RPS is between PR and PS. Since PR is going up to R, and PS is going to the left to S. So the angle between them is obtuse? Because if angle RPQ is acute, then angle RPS is obtuse, so cosine should be negative, which matches our result.

Alternatively, if I think of coordinates, vector PR is (36/25, 21/50) and vector PS is (-2, 0). The dot product is negative, which means the angle between them is obtuse, so cosine is negative. So that makes sense.

Therefore, I think this is correct.

Another way to think about it: since angle between PR and PQ is θ with sin θ = 7/25, then coordinates of R can be considered as (PR cos θ, PR sin θ). Since PR is length 3/2, as we saw. Wait, earlier we saw that |PR| is 3/2. So coordinates of R would be ( (3/2) cos θ, (3/2) sin θ ). Which matches with our previous calculation, since cos θ = 24/25, sin θ = 7/25. Therefore, coordinates are ( (3/2)(24/25), (3/2)(7/25) ) = (36/25, 21/50).

So, if I want to find angle between PR and PS, which is angle RPS. Since PS is along negative x-axis, so direction of PS is (-1, 0). So the angle between PR and (-1, 0).

Alternatively, we can think of angle between PR and the negative x-axis. Which would be 180° - θ, as before. Therefore, cos(angle RPS) = cos(180° - θ) = -cos θ = -24/25.

So all methods lead to the same answer. Therefore, I think the answer is definitely -24/25.

Just to recap, since the problem gave sin(angle RPQ) = 7/25, we found cos(angle RPQ) = 24/25, then realized angle RPS is supplementary, so cosine is negative of that. Then confirmed with coordinates and dot product. All steps check out.

Therefore, I'm confident the answer is -24/25.

**Final Answer**
\boxed{-\dfrac{24}{25}}

To solve the problem, we are given that:

$$
\sin(\angle RPQ) = \frac{7}{25}
$$

and we are asked to find:

$$
\cos(\angle RPS)
$$

---

### Step 1: Understand the Geometry

From the diagram and the Asymptote code, we can deduce the following:

- Point $ P $ is at the origin $(0, 0)$.
- Point $ Q $ is at $(2, 0)$, and point $ S $ is at $(-2, 0)$, so they are symmetric about $ P $ along the x-axis.
- Point $ R $ is obtained by rotating the point $(1.5, 0)$ by an angle $ \alpha $, where $ \sin(\alpha) = \frac{7}{25} $.

This means that the angle $ \angle RPQ $ is the angle between the line $ PR $ and the positive x-axis (i.e., the direction of $ PQ $), and $ \angle RPS $ is the angle between $ PR $ and the negative x-axis (i.e., the direction of $ PS $).

---

### Step 2: Use Trigonometric Identities

Let $ \theta = \angle RPQ $. Then:

$$
\sin(\theta) = \frac{7}{25}
$$

We can use the Pythagorean identity to find $ \cos(\theta) $:

$$
\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}
$$

Since $ \theta $ is the angle between $ PR $ and the positive x-axis, and $ R $ is above the x-axis, $ \theta $ is acute, so $ \cos(\theta) > 0 $.

---

### Step 3: Relate $ \angle RPS $ to $ \angle RPQ $

Since $ \angle RPS $ is the angle between $ PR $ and the negative x-axis, and $ \angle RPQ $ is the angle between $ PR $ and the positive x-axis, the two angles are supplementary:

$$
\angle RPS = 180^\circ - \angle RPQ
$$

Using the identity for cosine of supplementary angles:

$$
\cos(180^\circ - \theta) = -\cos(\theta)
$$

So:

$$
\cos(\angle RPS) = -\cos(\angle RPQ) = -\frac{24}{25}
$$

---

### Step 4: Confirm with Vector Approach (Optional)

We can also confirm this using vector geometry. The coordinates of $ R $, obtained by rotating $ (1.5, 0) $ by angle $ \alpha $, are:

$$
R = (1.5 \cos \alpha, 1.5 \sin \alpha)
$$

Given $ \sin \alpha = \frac{7}{25} $, we find $ \cos \alpha = \frac{24}{25} $, so:

$$
R = \left( \frac{3}{2} \cdot \frac{24}{25}, \frac{3}{2} \cdot \frac{7}{25} \right) = \left( \frac{36}{25}, \frac{21}{50} \right)
$$

Vector $ \vec{PR} = \left( \frac{36}{25}, \frac{21}{50} \right) $, and vector $ \vec{PS} = (-2, 0) $.

The dot product is:

$$
\vec{PR} \cdot \vec{PS} = \left( \frac{36}{25} \cdot (-2) \right) + \left( \frac{21}{50} \cdot 0 \right) = -\frac{72}{25}
$$

The magnitudes are:

$$
|\vec{PR}| = \sqrt{\left( \frac{36}{25} \right)^2 + \left( \frac{21}{50} \right)^2} = \frac{3}{2}, \quad |\vec{PS}| = 2
$$

So:

$$
\cos(\angle RPS) = \frac{\vec{PR} \cdot \vec{PS}}{|\vec{PR}| \cdot |\vec{PS}|} = \frac{-\frac{72}{25}}{\frac{3}{2} \cdot 2} = \frac{-\frac{72}{25}}{3} = -\frac{24}{25}
$$

---

### Final Answer

$$
\boxed{-\dfrac{24}{25}}
$$

[Assistant]
Explanation:
Okay, so I need to find cos(angle RPS) given that sin(angle RPQ) is 7/25. Let me try to visualize the diagram first. There's a point P at the origin, Q at (2,0), and S at (-2,0). R is some point above the x-axis, created by rotating the point (1.5, 0) by an angle whose sine is 7/25. The diagram shows lines Q to S, P to R, and probably other connections.

First, let me recall that angle RPQ is the angle at point P between points R and Q. So, angle RPQ is formed by the lines PR and PQ. Similarly, angle RPS is the angle at point P between points R and S. So, angle RPS is formed by lines PR and PS. Since S is on the opposite side of P from Q, maybe these angles are supplementary or something? Wait, but I need to figure out the relationship between them.

Given that sin(angle RPQ) = 7/25, I need to find cos(angle RPS). Let me think about how these angles relate. Since points Q and S are on the x-axis, symmetric with respect to P? Wait, P is at (0,0), Q is at (2,0), and S is at (-2,0). So PQ is from P(0,0) to Q(2,0), and PS is from P(0,0) to S(-2,0). So PQ and PS are in opposite directions along the x-axis. Therefore, angle between PQ and PS is 180 degrees. But angle RPQ is between PR and PQ, and angle RPS is between PR and PS. So if I imagine PR as a line from P to R, then angle RPQ is the angle between PR and PQ (which is along the positive x-axis), and angle RPS is the angle between PR and PS (which is along the negative x-axis). Therefore, angle RPS is supplementary to angle RPQ? Wait, not necessarily. Let me think.

If I consider angle between PR and PQ (which is angle RPQ) and angle between PR and PS (which is angle RPS), since PQ and PS are in opposite directions, then angle RPS would be equal to 180 degrees minus angle RPQ? Wait, if you have a line PR making an angle θ with PQ (which is along positive x-axis), then the angle between PR and PS (which is along negative x-axis) would be 180° - θ. Therefore, angle RPS = 180° - angle RPQ. Therefore, cos(angle RPS) = cos(180° - angle RPQ) = -cos(angle RPQ). Because cosine of supplementary angles is negative cosine of the original angle. So if I can find cos(angle RPQ), then take its negative, that would be cos(angle RPS).

But wait, the problem gives me sin(angle RPQ) = 7/25. So if I can find cos(angle RPQ), then I can get cos(angle RPS). Let me confirm this relationship again. Let me draw a mental picture. Let me suppose angle RPQ is θ, so sin θ = 7/25. Then angle RPS is 180° - θ, so cos(angle RPS) = cos(180° - θ) = -cos θ. Therefore, if I can find cos θ, then multiply by -1 to get the answer.

So, first step: find cos θ where θ is angle RPQ, given sin θ = 7/25. Since sin²θ + cos²θ = 1, so cos θ = ±√(1 - (7/25)²). But we need to figure out the sign. Since angle RPQ is an angle between PR and PQ. Since R is above the x-axis (as per the Asymptote code, R is rotated from (1.5, 0) by some angle, so it's above the x-axis), so angle RPQ is between PR (going up to R) and PQ (along positive x-axis). Therefore, θ is between 0 and 180 degrees, but since R is above the x-axis, θ is between 0 and 180, but more specifically, since sin θ is positive, θ is between 0 and 180. However, depending on where R is, θ could be acute or obtuse. But since sin θ = 7/25, which is positive, but we need to figure out if θ is acute or obtuse. However, since R is constructed by rotating (1.5, 0) by an angle whose sine is 7/25. Wait, the Asymptote code says R = rotate(aSin(7/25))*(1.5,0). So that means that R is obtained by rotating the point (1.5, 0) by an angle whose sine is 7/25. So the angle of rotation is arcsin(7/25). Therefore, that angle is between 0 and 90 degrees, since arcsin(7/25) is in the first quadrant. Therefore, the angle between PR and the x-axis is arcsin(7/25), so angle RPQ is arcsin(7/25), which is acute. Therefore, θ is acute, so cos θ is positive. Therefore, cos θ = sqrt(1 - (7/25)^2). Let me compute that.

First, compute (7/25)^2: 49/625. Then 1 - 49/625 = (625 - 49)/625 = 576/625. Therefore, sqrt(576/625) = 24/25. Therefore, cos θ = 24/25. Therefore, cos(angle RPS) = -cos θ = -24/25. Therefore, the answer is -24/25.

Wait, but let me make sure that angle RPS is indeed 180° - θ. Let me think again. If angle RPQ is θ, which is the angle between PR and PQ (positive x-axis). Then angle between PR and PS (negative x-axis) would be angle RPS. Since PQ is along positive x-axis, PS is along negative x-axis, so the angle between PR and PS is 180° - θ. Therefore, yes, angle RPS = 180° - θ. Therefore, cos(angle RPS) = cos(180° - θ) = -cos θ. Therefore, since cos θ is 24/25, then cos(angle RPS) is -24/25.

But let me check if there's another way to approach this problem, maybe using coordinates. Let me try that.

The Asymptote code defines points: SS is (-2,0), P is (0,0), Q is (2,0), and R is obtained by rotating the point (1.5, 0) by an angle whose sine is 7/25. So, rotating (1.5, 0) by angle α where sin α = 7/25. Let me find coordinates of R.

If we rotate a point (x, y) by angle α, the new coordinates are (x cos α - y sin α, x sin α + y cos α). Since the original point is (1.5, 0), so x = 1.5, y = 0. Therefore, after rotation, coordinates of R are (1.5 cos α - 0, 1.5 sin α + 0) = (1.5 cos α, 1.5 sin α). Therefore, R is at (1.5 cos α, 1.5 sin α).

Given that sin α = 7/25, so we can find cos α as 24/25, as before. Therefore, coordinates of R are (1.5*(24/25), 1.5*(7/25)). Let me compute that.

First, 1.5 is 3/2. Therefore, x-coordinate: (3/2)*(24/25) = (72)/50 = 36/25. Y-coordinate: (3/2)*(7/25) = 21/50. Therefore, R is at (36/25, 21/50).

Now, angle RPS is the angle at P between points R and S. Since S is at (-2, 0), and P is at (0,0). So vectors PR and PS. Vector PR is from P to R: (36/25, 21/50). Vector PS is from P to S: (-2, 0).

To find angle between vectors PR and PS, which is angle RPS. The formula for cosine of the angle between two vectors u and v is (u ⋅ v)/(|u| |v|).

So let me compute the dot product of PR and PS.

PR ⋅ PS = (36/25)*(-2) + (21/50)*0 = -72/25 + 0 = -72/25.

Now, |PR| is the magnitude of vector PR. Let's compute that.

|PR| = sqrt( (36/25)^2 + (21/50)^2 ). Let me compute that.

First, square of 36/25 is (1296)/625. Square of 21/50 is (441)/2500.

To add them, convert to common denominator. 625 is 25*25, 2500 is 100*25. So common denominator is 2500.

So 1296/625 = (1296 * 4)/2500 = 5184/2500.

Then 441/2500 remains as is.

Total is (5184 + 441)/2500 = 5625/2500.

sqrt(5625/2500) = sqrt(5625)/sqrt(2500) = 75/50 = 3/2.

Wait, that's interesting. The magnitude of PR is 3/2. Which makes sense because the original point was (1.5, 0) and we rotated it, so the distance from P to R should be the same as the distance from P to (1.5, 0), which is 1.5. So 3/2. So that checks out.

Now, |PS| is the magnitude of vector PS, which is from (0,0) to (-2, 0), so that's just 2.

Therefore, cos(angle RPS) = (PR ⋅ PS)/(|PR| |PS|) = (-72/25)/( (3/2)*2 ) = (-72/25)/(3) = (-72/25)*(1/3) = -24/25.

So that's the same result as before. Therefore, cos(angle RPS) is -24/25.

Therefore, the answer is \boxed{-\dfrac{24}{25}}.

But let me just check once more. Since angle RPS is between PR and PS. Since PR is going up to R, and PS is going to the left to S. So the angle between them is obtuse? Because if angle RPQ is acute, then angle RPS is obtuse, so cosine should be negative, which matches our result.

Alternatively, if I think of coordinates, vector PR is (36/25, 21/50) and vector PS is (-2, 0). The dot product is negative, which means the angle between them is obtuse, so cosine is negative. So that makes sense.

Therefore, I think this is correct.

Another way to think about it: since angle between PR and PQ is θ with sin θ = 7/25, then coordinates of R can be considered as (PR cos θ, PR sin θ). Since PR is length 3/2, as we saw. Wait, earlier we saw that |PR| is 3/2. So coordinates of R would be ( (3/2) cos θ, (3/2) sin θ ). Which matches with our previous calculation, since cos θ = 24/25, sin θ = 7/25. Therefore, coordinates are ( (3/2)(24/25), (3/2)(7/25) ) = (36/25, 21/50).

So, if I want to find angle between PR and PS, which is angle RPS. Since PS is along negative x-axis, so direction of PS is (-1, 0). So the angle between PR and (-1, 0).

Alternatively, we can think of angle between PR and the negative x-axis. Which would be 180° - θ, as before. Therefore, cos(angle RPS) = cos(180° - θ) = -cos θ = -24/25.

So all methods lead to the same answer. Therefore, I think the answer is definitely -24/25.

Just to recap, since the problem gave sin(angle RPQ) = 7/25, we found cos(angle RPQ) = 24/25, then realized angle RPS is supplementary, so cosine is negative of that. Then confirmed with coordinates and dot product. All steps check out.

Therefore, I'm confident the answer is -24/25.

**Final Answer**
\boxed{-\dfrac{24}{25}}

To solve the problem, we are given that:

$$
\sin(\angle RPQ) = \frac{7}{25}
$$

and we are asked to find:

$$
\cos(\angle RPS)
$$

---

### Step 1: Understand the Geometry

From the diagram and the Asymptote code, we can deduce the following:

- Point $ P $ is at the origin $(0, 0)$.
- Point $ Q $ is at $(2, 0)$, and point $ S $ is at $(-2, 0)$, so they are symmetric about $ P $ along the x-axis.
- Point $ R $ is obtained by rotating the point $(1.5, 0)$ by an angle $ \alpha $, where $ \sin(\alpha) = \frac{7}{25} $.

This means that the angle $ \angle RPQ $ is the angle between the line $ PR $ and the positive x-axis (i.e., the direction of $ PQ $), and $ \angle RPS $ is the angle between $ PR $ and the negative x-axis (i.e., the direction of $ PS $).

---

### Step 2: Use Trigonometric Identities

Let $ \theta = \angle RPQ $. Then:

$$
\sin(\theta) = \frac{7}{25}
$$

We can use the Pythagorean identity to find $ \cos(\theta) $:

$$
\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}
$$

Since $ \theta $ is the angle between $ PR $ and the positive x-axis, and $ R $ is above the x-axis, $ \theta $ is acute, so $ \cos(\theta) > 0 $.

---

### Step 3: Relate $ \angle RPS $ to $ \angle RPQ $

Since $ \angle RPS $ is the angle between $ PR $ and the negative x-axis, and $ \angle RPQ $ is the angle between $ PR $ and the positive x-axis, the two angles are supplementary:

$$
\angle RPS = 180^\circ - \angle RPQ
$$

Using the identity for cosine of supplementary angles:

$$
\cos(180^\circ - \theta) = -\cos(\theta)
$$

So:

$$
\cos(\angle RPS) = -\cos(\angle RPQ) = -\frac{24}{25}
$$

---

### Step 4: Confirm with Vector Approach (Optional)

We can also confirm this using vector geometry. The coordinates of $ R $, obtained by rotating $ (1.5, 0) $ by angle $ \alpha $, are:

$$
R = (1.5 \cos \alpha, 1.5 \sin \alpha)
$$

Given $ \sin \alpha = \frac{7}{25} $, we find $ \cos \alpha = \frac{24}{25} $, so:

$$
R = \left( \frac{3}{2} \cdot \frac{24}{25}, \frac{3}{2} \cdot \frac{7}{25} \right) = \left( \frac{36}{25}, \frac{21}{50} \right)
$$

Vector $ \vec{PR} = \left( \frac{36}{25}, \frac{21}{50} \right) $, and vector $ \vec{PS} = (-2, 0) $.

The dot product is:

$$
\vec{PR} \cdot \vec{PS} = \left( \frac{36}{25} \cdot (-2) \right) + \left( \frac{21}{50} \cdot 0 \right) = -\frac{72}{25}
$$

The magnitudes are:

$$
|\vec{PR}| = \sqrt{\left( \frac{36}{25} \right)^2 + \left( \frac{21}{50} \right)^2} = \frac{3}{2}, \quad |\vec{PS}| = 2
$$

So:

$$
\cos(\angle RPS) = \frac{\vec{PR} \cdot \vec{PS}}{|\vec{PR}| \cdot |\vec{PS}|} = \frac{-\frac{72}{25}}{\frac{3}{2} \cdot 2} = \frac{-\frac{72}{25}}{3} = -\frac{24}{25}
$$

---

### Final Answer

$$
\boxed{-\dfrac{24}{25}}
$$

Solution:
For any angle $x$, we have $\cos(180^\circ - x)=-\cos x$, so $\cos \angle RPS = \cos(180^\circ - \angle RPQ) =- \cos\angle RPQ$.

Since $\sin^2 \angle RPQ + \cos^2 \angle RPQ = 1$, we have $\cos^2\angle RPQ = 1 - \left(\frac{7}{25}\right)^2 = \frac{576}{625}$.  Since $\angle RPQ$ is acute, we have $\cos\angle RPQ = \frac{24}{25}$, which gives us $\cos\angle RPS = -\cos\angle RPQ = \boxed{-\frac{24}{25}}$.


Explanation:
Okay, so there's this robot that moves at a constant speed. Let me try to figure out how long the hallway is. The problem says that the robot takes 2.5 hours to travel 1 kilometer. Then, moving at the same speed, it takes 90 seconds to travel the length of a hallway. We need to find out how many meters long that hallway is.

First, I need to find the robot's speed. Since speed is distance divided by time, right? So maybe I can calculate the speed using the first piece of information: 1 kilometer in 2.5 hours. Let me write that down.

Speed = Distance / Time. So, the distance here is 1 kilometer, and the time is 2.5 hours. But wait, the time is in hours and the distance is in kilometers. Maybe I should convert the time into seconds or the distance into meters? Hmm, but maybe I can just keep it in kilometers per hour for now and then convert later? Let me think.

Alternatively, since the second part of the problem gives time in seconds, maybe I should convert the robot's speed into meters per second? That might make it easier to calculate the hallway length in meters. Let me try that approach.

First, let's convert the robot's speed from kilometers per hour to meters per second.

So, 1 kilometer is 1000 meters. 2.5 hours is how many seconds? Well, 1 hour is 60 minutes, and 1 minute is 60 seconds. So 2.5 hours is 2.5 * 60 * 60 seconds. Let me calculate that.

First, 2.5 hours * 60 minutes/hour = 150 minutes. Then, 150 minutes * 60 seconds/minute = 9000 seconds. So, 2.5 hours is 9000 seconds.

So, the robot travels 1000 meters in 9000 seconds. Therefore, its speed is 1000 meters / 9000 seconds. Let me compute that.

Dividing numerator and denominator by 1000, that's 1/9 meters per second. Wait, 1000 divided by 9000 is 1/9? Let me check. 9000 divided by 1000 is 9, so yes, 1000/9000 = 1/9. So the speed is 1/9 m/s.

Alternatively, maybe I can check that again. If it takes 2.5 hours to go 1 km, then in one hour it would go 1 / 2.5 km. Which is 0.4 km per hour. Then converting that to meters per second: 0.4 km/h is 400 meters per hour. Then divide by 3600 seconds to get meters per second. 400 / 3600 = 1/9 m/s. Yep, same result. So that's correct.

So the robot's speed is 1/9 meters per second.

Now, the problem says that it takes 90 seconds to travel the length of the hallway. So, if I can find the distance, which is the length of the hallway, using the formula: distance = speed * time.

So, distance = (1/9 m/s) * 90 seconds. Let me compute that.

Multiplying 1/9 by 90: 90 divided by 9 is 10. So, 10 meters? Wait, that seems short for a hallway, but maybe it's possible. Let me check my steps again to make sure I didn't make a mistake.

First, converting 2.5 hours to seconds: 2.5 * 60 * 60. 2.5 * 60 is 150, 150 * 60 is 9000. Correct. Then, 1 km is 1000 meters. So speed is 1000 meters / 9000 seconds = 1/9 m/s. Correct. Then, time taken for hallway is 90 seconds. So distance is speed * time = (1/9) * 90 = 10 meters.

Hmm. So according to this, the hallway is 10 meters long. But maybe I made a mistake in unit conversion? Let me check again.

Alternatively, maybe I should approach it using ratios? Since the robot's speed is constant, the time taken is proportional to the distance.

So, if 1 km takes 2.5 hours, then how long does it take to travel x kilometers? Well, but the time given is 90 seconds. Let me convert 2.5 hours to seconds to make the units match.

As before, 2.5 hours is 9000 seconds. So, 1 km takes 9000 seconds. Therefore, the time per meter is 9000 seconds / 1000 meters = 9 seconds per meter. Therefore, if it takes 90 seconds to travel the hallway, then the length is 90 seconds / 9 seconds per meter = 10 meters.

Same answer. So that seems to confirm it.

Alternatively, maybe I can think in terms of proportions.

If 1 km (which is 1000 meters) takes 2.5 hours, then how many meters does it travel in 90 seconds?

First, let's convert 2.5 hours to seconds: 2.5 * 3600 = 9000 seconds.

So, the robot travels 1000 meters in 9000 seconds. Therefore, the number of meters it travels in 1 second is 1000 / 9000 = 1/9 meters per second. Then, in 90 seconds, it travels 1/9 * 90 = 10 meters.

Same result again.

So, all methods lead to 10 meters. Maybe the hallway is indeed 10 meters long.

But just to make sure, let me check if there's another way someone might approach this problem and perhaps make a mistake.

For example, someone might think that since 2.5 hours is 2 hours and 30 minutes, and then try to convert 90 seconds into hours? Let me see.

90 seconds is 90/3600 hours, which is 1/40 hours. Then, if the robot's speed is 1 km per 2.5 hours, then in 1/40 hours, it would travel (1 km / 2.5 hours) * (1/40 hours) = (1 / 2.5) * (1 / 40) km.

Calculating that: 1 divided by 2.5 is 0.4, then 0.4 divided by 40 is 0.01 km. Then, converting 0.01 km to meters is 0.01 * 1000 = 10 meters.

Same answer again.

So regardless of the method, the answer is 10 meters.

Therefore, I think the hallway is 10 meters long.

But just to make sure, let me check if I messed up any unit conversions.

Original data: 2.5 hours for 1 km.

Convert 2.5 hours to seconds: 2.5 * 60 * 60.

Yes, 2.5 * 60 is 150 minutes, 150 * 60 is 9000 seconds.

So 1 km is 1000 meters.

So speed is 1000 meters / 9000 seconds = 1/9 m/s.

Then, time for hallway is 90 seconds.

Distance = speed * time = (1/9) * 90 = 10 meters.

Yes.

Alternatively, if I wanted to use km and hours, but then convert the time to hours.

So, 90 seconds is 90 / 3600 hours = 0.025 hours.

Then, since the robot's speed is 1 km / 2.5 hours, then distance = speed * time = (1 / 2.5) * 0.025 km.

Calculating that: 1 divided by 2.5 is 0.4, multiplied by 0.025 is 0.01 km.

Convert to meters: 0.01 * 1000 = 10 meters.

Same answer.

So, all methods confirm that the hallway is 10 meters long.

Therefore, I think the answer is 10 meters.

**Final Answer**
The hallway is \boxed{10} meters long.

To determine the length of the hallway, we begin by analyzing the robot's constant speed based on the given information:

---

### **Step 1: Determine the Robot's Speed**

The robot travels **1 kilometer (1000 meters)** in **2.5 hours**. To find its speed in **meters per second**, we first convert the time from hours to seconds:

$$
2.5 \text{ hours} = 2.5 \times 60 \times 60 = 9000 \text{ seconds}
$$

Now, calculate the speed:

$$
\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1000 \text{ meters}}{9000 \text{ seconds}} = \frac{1}{9} \text{ meters per second}
$$

---

### **Step 2: Use the Speed to Find the Hallway Length**

We are told that the robot takes **90 seconds** to travel the length of the hallway. Using the formula:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

Substitute the known values:

$$
\text{Distance} = \frac{1}{9} \text{ m/s} \times 90 \text{ s} = 10 \text{ meters}
$$

---

### **Conclusion**

After verifying through multiple approaches (unit conversions, proportionality, and direct calculation), we consistently find that the hallway is:

$$
\boxed{10}
$$

meters long.

[Assistant]
Explanation:
Okay, so there's this robot that moves at a constant speed. Let me try to figure out how long the hallway is. The problem says that the robot takes 2.5 hours to travel 1 kilometer. Then, moving at the same speed, it takes 90 seconds to travel the length of a hallway. We need to find out how many meters long that hallway is.

First, I need to find the robot's speed. Since speed is distance divided by time, right? So maybe I can calculate the speed using the first piece of information: 1 kilometer in 2.5 hours. Let me write that down.

Speed = Distance / Time. So, the distance here is 1 kilometer, and the time is 2.5 hours. But wait, the time is in hours and the distance is in kilometers. Maybe I should convert the time into seconds or the distance into meters? Hmm, but maybe I can just keep it in kilometers per hour for now and then convert later? Let me think.

Alternatively, since the second part of the problem gives time in seconds, maybe I should convert the robot's speed into meters per second? That might make it easier to calculate the hallway length in meters. Let me try that approach.

First, let's convert the robot's speed from kilometers per hour to meters per second.

So, 1 kilometer is 1000 meters. 2.5 hours is how many seconds? Well, 1 hour is 60 minutes, and 1 minute is 60 seconds. So 2.5 hours is 2.5 * 60 * 60 seconds. Let me calculate that.

First, 2.5 hours * 60 minutes/hour = 150 minutes. Then, 150 minutes * 60 seconds/minute = 9000 seconds. So, 2.5 hours is 9000 seconds.

So, the robot travels 1000 meters in 9000 seconds. Therefore, its speed is 1000 meters / 9000 seconds. Let me compute that.

Dividing numerator and denominator by 1000, that's 1/9 meters per second. Wait, 1000 divided by 9000 is 1/9? Let me check. 9000 divided by 1000 is 9, so yes, 1000/9000 = 1/9. So the speed is 1/9 m/s.

Alternatively, maybe I can check that again. If it takes 2.5 hours to go 1 km, then in one hour it would go 1 / 2.5 km. Which is 0.4 km per hour. Then converting that to meters per second: 0.4 km/h is 400 meters per hour. Then divide by 3600 seconds to get meters per second. 400 / 3600 = 1/9 m/s. Yep, same result. So that's correct.

So the robot's speed is 1/9 meters per second.

Now, the problem says that it takes 90 seconds to travel the length of the hallway. So, if I can find the distance, which is the length of the hallway, using the formula: distance = speed * time.

So, distance = (1/9 m/s) * 90 seconds. Let me compute that.

Multiplying 1/9 by 90: 90 divided by 9 is 10. So, 10 meters? Wait, that seems short for a hallway, but maybe it's possible. Let me check my steps again to make sure I didn't make a mistake.

First, converting 2.5 hours to seconds: 2.5 * 60 * 60. 2.5 * 60 is 150, 150 * 60 is 9000. Correct. Then, 1 km is 1000 meters. So speed is 1000 meters / 9000 seconds = 1/9 m/s. Correct. Then, time taken for hallway is 90 seconds. So distance is speed * time = (1/9) * 90 = 10 meters.

Hmm. So according to this, the hallway is 10 meters long. But maybe I made a mistake in unit conversion? Let me check again.

Alternatively, maybe I should approach it using ratios? Since the robot's speed is constant, the time taken is proportional to the distance.

So, if 1 km takes 2.5 hours, then how long does it take to travel x kilometers? Well, but the time given is 90 seconds. Let me convert 2.5 hours to seconds to make the units match.

As before, 2.5 hours is 9000 seconds. So, 1 km takes 9000 seconds. Therefore, the time per meter is 9000 seconds / 1000 meters = 9 seconds per meter. Therefore, if it takes 90 seconds to travel the hallway, then the length is 90 seconds / 9 seconds per meter = 10 meters.

Same answer. So that seems to confirm it.

Alternatively, maybe I can think in terms of proportions.

If 1 km (which is 1000 meters) takes 2.5 hours, then how many meters does it travel in 90 seconds?

First, let's convert 2.5 hours to seconds: 2.5 * 3600 = 9000 seconds.

So, the robot travels 1000 meters in 9000 seconds. Therefore, the number of meters it travels in 1 second is 1000 / 9000 = 1/9 meters per second. Then, in 90 seconds, it travels 1/9 * 90 = 10 meters.

Same result again.

So, all methods lead to 10 meters. Maybe the hallway is indeed 10 meters long.

But just to make sure, let me check if there's another way someone might approach this problem and perhaps make a mistake.

For example, someone might think that since 2.5 hours is 2 hours and 30 minutes, and then try to convert 90 seconds into hours? Let me see.

90 seconds is 90/3600 hours, which is 1/40 hours. Then, if the robot's speed is 1 km per 2.5 hours, then in 1/40 hours, it would travel (1 km / 2.5 hours) * (1/40 hours) = (1 / 2.5) * (1 / 40) km.

Calculating that: 1 divided by 2.5 is 0.4, then 0.4 divided by 40 is 0.01 km. Then, converting 0.01 km to meters is 0.01 * 1000 = 10 meters.

Same answer again.

So regardless of the method, the answer is 10 meters.

Therefore, I think the hallway is 10 meters long.

But just to make sure, let me check if I messed up any unit conversions.

Original data: 2.5 hours for 1 km.

Convert 2.5 hours to seconds: 2.5 * 60 * 60.

Yes, 2.5 * 60 is 150 minutes, 150 * 60 is 9000 seconds.

So 1 km is 1000 meters.

So speed is 1000 meters / 9000 seconds = 1/9 m/s.

Then, time for hallway is 90 seconds.

Distance = speed * time = (1/9) * 90 = 10 meters.

Yes.

Alternatively, if I wanted to use km and hours, but then convert the time to hours.

So, 90 seconds is 90 / 3600 hours = 0.025 hours.

Then, since the robot's speed is 1 km / 2.5 hours, then distance = speed * time = (1 / 2.5) * 0.025 km.

Calculating that: 1 divided by 2.5 is 0.4, multiplied by 0.025 is 0.01 km.

Convert to meters: 0.01 * 1000 = 10 meters.

Same answer.

So, all methods confirm that the hallway is 10 meters long.

Therefore, I think the answer is 10 meters.

**Final Answer**
The hallway is \boxed{10} meters long.

To determine the length of the hallway, we begin by analyzing the robot's constant speed based on the given information:

---

### **Step 1: Determine the Robot's Speed**

The robot travels **1 kilometer (1000 meters)** in **2.5 hours**. To find its speed in **meters per second**, we first convert the time from hours to seconds:

$$
2.5 \text{ hours} = 2.5 \times 60 \times 60 = 9000 \text{ seconds}
$$

Now, calculate the speed:

$$
\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{1000 \text{ meters}}{9000 \text{ seconds}} = \frac{1}{9} \text{ meters per second}
$$

---

### **Step 2: Use the Speed to Find the Hallway Length**

We are told that the robot takes **90 seconds** to travel the length of the hallway. Using the formula:

$$
\text{Distance} = \text{Speed} \times \text{Time}
$$

Substitute the known values:

$$
\text{Distance} = \frac{1}{9} \text{ m/s} \times 90 \text{ s} = 10 \text{ meters}
$$

---

### **Conclusion**

After verifying through multiple approaches (unit conversions, proportionality, and direct calculation), we consistently find that the hallway is:

$$
\boxed{10}
$$

meters long.

Solution: We see that 2.5 hours is the same as $2.5\cdot 60 = 150$ minutes, or $150\cdot 60 = 9000$ seconds.  This is 100 times longer than the robot was traveling the hallway, meaning the hallway is $\frac{1}{100}$ kilometers, or $\frac{1000}{100} = \boxed{10}$ meters long.