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

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

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

The question is about the angle APB. So, we need to consider the positions of A and B at some time t after they start running, and then find the angle at point P between points A and B. The goal is to find the maximum value of this angle.

Let me think about how to model this. Let's denote the position of runner A at time t as (x_A(t), 0), and runner B as (x_B(t), 0). Since they start at O, which is (0,0), and run in the same direction (let's say positive x-direction), then:

x_A(t) = v * t

x_B(t) = 3v * t

But since they are moving along the track, their positions are along the x-axis. The observer is at point P, which is at (0, h). Wait, in the Asymptote code, P is at (0,1), but maybe h is arbitrary? Or maybe we can assume h is 1 for simplicity? Or maybe it's better to keep h as a variable? Hmm.

But since the problem doesn't specify the distance from O to P, maybe the maximum angle is independent of h? Or maybe not. Wait, but the angle APB might depend on h. However, since the problem says "an observer stands at point P so that OP is perpendicular to the track," but doesn't specify where P is. Wait, but maybe the maximum angle is achieved regardless of where P is? Or maybe there's a specific position? Wait, no, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, OP is perpendicular, meaning P is somewhere off the track, but the exact position isn't given. Wait, but maybe the problem is assuming that P is fixed? Wait, the Asymptote code shows P at (0,1), but maybe that's just an example.

Wait, but the problem says "Find the maximum of angle APB." So, perhaps we need to consider all possible positions of P? Or is P fixed? Wait, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, P is fixed, but the problem doesn't specify where. Wait, but maybe the maximum angle is independent of the position of P? Or maybe there's a specific position where the maximum occurs? Wait, but the problem says "Find the maximum of angle APB," so maybe we need to find the maximum possible angle that can be achieved as the runners move, given that P is fixed at some position where OP is perpendicular. Wait, but the problem doesn't specify where P is. Wait, maybe the answer is the same regardless of where P is? Or maybe there's a specific position?

Wait, maybe the problem is similar to the classic problem where you have a point P off the track, and you want to find the maximum angle subtended by two moving points. In such cases, the maximum angle is achieved when the circle through A and B is tangent to the track. But I need to think more carefully.

Alternatively, maybe we can parametrize the positions of A and B as functions of time, then express the angle APB in terms of time, and then find its maximum. Let me try that approach.

Let me set up coordinates. Let me assume the track is the x-axis, O is at (0,0), and P is at (0, h), where h is some positive number. Let me assume that h is fixed. Then, as time t increases, runner A is at (vt, 0) and runner B is at (3vt, 0).

We need to find angle APB. So, angle at point P between points A and B. To find this angle, we can use coordinates. Let me recall that the angle between two lines can be found using the tangent formula or using vectors.

Alternatively, we can use coordinates to find vectors PA and PB, then compute the angle between them.

Let me denote coordinates:

Point P: (0, h)

Point A: (vt, 0)

Point B: (3vt, 0)

So vector PA is A - P = (vt - 0, 0 - h) = (vt, -h)

Similarly, vector PB is B - P = (3vt - 0, 0 - h) = (3vt, -h)

Then, the angle between vectors PA and PB is angle APB.

The formula for the angle θ between two vectors u and v is:

cosθ = (u · v)/(|u| |v|)

So let's compute the dot product of PA and PB:

PA · PB = (vt)(3vt) + (-h)(-h) = 3v²t² + h²

The magnitude of PA is sqrt( (vt)^2 + h² ) = sqrt(v²t² + h² )

Similarly, magnitude of PB is sqrt( (3vt)^2 + h² ) = sqrt(9v²t² + h² )

Therefore,

cosθ = [3v²t² + h²] / [ sqrt(v²t² + h² ) * sqrt(9v²t² + h² ) ]

So θ = arccos [ (3v²t² + h²) / ( sqrt(v²t² + h² ) * sqrt(9v²t² + h² )) ]

Our goal is to find the maximum value of θ. Since arccos is a decreasing function, maximizing θ is equivalent to minimizing the expression [ (3v²t² + h²) / ( sqrt(v²t² + h² ) * sqrt(9v²t² + h² )) ].

Alternatively, maybe it's easier to work with tanθ or something else. Let me think.

Alternatively, since angle APB is formed by lines PA and PB, maybe we can use coordinates to find tanθ. Let me try that.

Alternatively, since we have coordinates, maybe we can compute the slopes of PA and PB, then compute the angle between them.

But maybe using coordinates is more complicated. Let me think again.

Alternatively, since we have coordinates, maybe we can use coordinates to find tanθ. Let me recall that if we have two lines with slopes m1 and m2, then tanθ = |(m2 - m1)/(1 + m1m2)|. But since angle APB is the angle between PA and PB, which are two lines from P to A and P to B.

So, let me compute the slopes of PA and PB.

Slope of PA: (0 - h)/(vt - 0) = -h / (vt)

Slope of PB: (0 - h)/(3vt - 0) = -h / (3vt)

So, m1 = -h/(vt), m2 = -h/(3vt)

Then tanθ = |(m2 - m1)/(1 + m1m2)|

Let me compute that:

Numerator: m2 - m1 = [ -h/(3vt) - (-h/(vt)) ] = -h/(3vt) + h/(vt) = ( -1/3 + 1 ) h/(vt) = (2/3) h/(vt)

Denominator: 1 + m1m2 = 1 + [ (-h/(vt)) * (-h/(3vt)) ] = 1 + [ h² / (3v²t²) ]

Therefore tanθ = | (2/3 h/(vt)) / (1 + h²/(3v²t²)) | = (2/3 h/(vt)) / (1 + h²/(3v²t²))

Simplify denominator:

Multiply numerator and denominator by 3v²t²:

Numerator becomes: (2/3 h/(vt)) * 3v²t² = 2 h v t

Denominator becomes: 3v²t² + h²

Therefore tanθ = (2 h v t) / (3v²t² + h² )

So tanθ = (2 h v t) / (3v²t² + h² )

So now, we have tanθ expressed in terms of t. Our goal is to find the maximum θ, which corresponds to maximizing tanθ, since tanθ is increasing in [0, π/2).

Therefore, we can find the maximum of tanθ as a function of t, then take arctangent to find θ.

So let me denote f(t) = (2 h v t) / (3v²t² + h² )

We need to find the maximum of f(t). Let me set u = t, so f(u) = (2 h v u) / (3v²u² + h² )

To find maximum, take derivative with respect to u, set to zero.

Let me compute f'(u):

Using quotient rule:

f'(u) = [ (2 h v)(3v²u² + h² ) - (2 h v u)(6v²u) ] / (3v²u² + h² )²

Simplify numerator:

First term: 2 h v (3v²u² + h² )

Second term: - 2 h v u * 6v²u = -12 h v³ u²

So numerator:

2 h v * 3v²u² + 2 h v * h² - 12 h v³ u²

= 6 h v³ u² + 2 h² v - 12 h v³ u²

= (6 h v³ u² - 12 h v³ u² ) + 2 h² v

= (-6 h v³ u² ) + 2 h² v

Set numerator equal to zero:

-6 h v³ u² + 2 h² v = 0

Divide both sides by 2 h v (assuming h ≠ 0, v ≠ 0):

-3 v² u² + h = 0

So:

-3 v² u² + h = 0

=> 3 v² u² = h

=> u² = h / (3 v² )

=> u = sqrt( h / (3 v² )) = (sqrt(h)/v ) / sqrt(3 )

Therefore, critical point at t = sqrt( h ) / ( v sqrt(3) )

Now, check if this is a maximum. Since f(t) approaches 0 as t approaches 0 and as t approaches infinity, this critical point must be a maximum.

Therefore, maximum tanθ is f(t) evaluated at t = sqrt( h ) / ( v sqrt(3) )

Let me compute that:

First, compute numerator: 2 h v t = 2 h v * [ sqrt(h) / (v sqrt(3)) ] = 2 h v * sqrt(h) / (v sqrt(3)) ) = 2 h sqrt(h) / sqrt(3) = 2 h^(3/2)/sqrt(3)

Denominator: 3v² t² + h²

Compute t²: h / (3 v² )

So 3v² t² = 3v² * [ h / (3 v² ) ] = h

Therefore denominator is h + h² ?

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Wait denominator is 3v² t² + h² = h + h² ?

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

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

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

The question is about the angle APB. So, we need to consider the positions of A and B at some time t after they start running, and then find the angle at point P between points A and B. The goal is to find the maximum value of this angle.

Let me think about how to model this. Let's denote the position of runner A at time t as (x_A(t), 0), and runner B as (x_B(t), 0). Since they start at O, which is (0,0), and run in the same direction (let's say positive x-direction), then:

x_A(t) = v * t

x_B(t) = 3v * t

But since they are moving along the track, their positions are along the x-axis. The observer is at point P, which is at (0, h). Wait, in the Asymptote code, P is at (0,1), but maybe h is arbitrary? Or maybe we can assume h is 1 for simplicity? Or maybe it's better to keep h as a variable? Hmm.

But since the problem doesn't specify the distance from O to P, maybe the maximum angle is independent of h? Or maybe not. Wait, but the angle APB might depend on h. However, since the problem says "an observer stands at point P so that OP is perpendicular to the track," but doesn't specify where P is. Wait, but maybe the maximum angle is achieved regardless of where P is? Or maybe there's a specific position? Wait, no, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, OP is perpendicular, meaning P is somewhere off the track, but the exact position isn't given. Wait, but maybe the problem is assuming that P is fixed? Wait, the Asymptote code shows P at (0,1), but maybe that's just an example.

Wait, but the problem says "Find the maximum of angle APB." So, perhaps we need to consider all possible positions of P? Or is P fixed? Wait, the problem says "an observer stands at point P so that OP is perpendicular to the track." So, P is fixed, but the problem doesn't specify where. Wait, but maybe the maximum angle is independent of the position of P? Or maybe there's a specific position where the maximum occurs? Wait, but the problem says "Find the maximum of angle APB," so maybe we need to find the maximum possible angle that can be achieved as the runners move, given that P is fixed at some position where OP is perpendicular. Wait, but the problem doesn't specify where P is. Wait, maybe the answer is the same regardless of where P is? Or maybe there's a specific position?

Wait, maybe the problem is similar to the classic problem where you have a point P off the track, and you want to find the maximum angle subtended by two moving points. In such cases, the maximum angle is achieved when the circle through A and B is tangent to the track. But I need to think more carefully.

Alternatively, maybe we can parametrize the positions of A and B as functions of time, then express the angle APB in terms of time, and then find its maximum. Let me try that approach.

Let me set up coordinates. Let me assume the track is the x-axis, O is at (0,0), and P is at (0, h), where h is some positive number. Let me assume that h is fixed. Then, as time t increases, runner A is at (vt, 0) and runner B is at (3vt, 0).

We need to find angle APB. So, angle at point P between points A and B. To find this angle, we can use coordinates. Let me recall that the angle between two lines can be found using the tangent formula or using vectors.

Alternatively, we can use coordinates to find vectors PA and PB, then compute the angle between them.

Let me denote coordinates:

Point P: (0, h)

Point A: (vt, 0)

Point B: (3vt, 0)

So vector PA is A - P = (vt - 0, 0 - h) = (vt, -h)

Similarly, vector PB is B - P = (3vt - 0, 0 - h) = (3vt, -h)

Then, the angle between vectors PA and PB is angle APB.

The formula for the angle θ between two vectors u and v is:

cosθ = (u · v)/(|u| |v|)

So let's compute the dot product of PA and PB:

PA · PB = (vt)(3vt) + (-h)(-h) = 3v²t² + h²

The magnitude of PA is sqrt( (vt)^2 + h² ) = sqrt(v²t² + h² )

Similarly, magnitude of PB is sqrt( (3vt)^2 + h² ) = sqrt(9v²t² + h² )

Therefore,

cosθ = [3v²t² + h²] / [ sqrt(v²t² + h² ) * sqrt(9v²t² + h² ) ]

So θ = arccos [ (3v²t² + h²) / ( sqrt(v²t² + h² ) * sqrt(9v²t² + h² )) ]

Our goal is to find the maximum value of θ. Since arccos is a decreasing function, maximizing θ is equivalent to minimizing the expression [ (3v²t² + h²) / ( sqrt(v²t² + h² ) * sqrt(9v²t² + h² )) ].

Alternatively, maybe it's easier to work with tanθ or something else. Let me think.

Alternatively, since angle APB is formed by lines PA and PB, maybe we can use coordinates to find tanθ. Let me try that.

Alternatively, since we have coordinates, maybe we can compute the slopes of PA and PB, then compute the angle between them.

But maybe using coordinates is more complicated. Let me think again.

Alternatively, since we have coordinates, maybe we can use coordinates to find tanθ. Let me recall that if we have two lines with slopes m1 and m2, then tanθ = |(m2 - m1)/(1 + m1m2)|. But since angle APB is the angle between PA and PB, which are two lines from P to A and P to B.

So, let me compute the slopes of PA and PB.

Slope of PA: (0 - h)/(vt - 0) = -h / (vt)

Slope of PB: (0 - h)/(3vt - 0) = -h / (3vt)

So, m1 = -h/(vt), m2 = -h/(3vt)

Then tanθ = |(m2 - m1)/(1 + m1m2)|

Let me compute that:

Numerator: m2 - m1 = [ -h/(3vt) - (-h/(vt)) ] = -h/(3vt) + h/(vt) = ( -1/3 + 1 ) h/(vt) = (2/3) h/(vt)

Denominator: 1 + m1m2 = 1 + [ (-h/(vt)) * (-h/(3vt)) ] = 1 + [ h² / (3v²t²) ]

Therefore tanθ = | (2/3 h/(vt)) / (1 + h²/(3v²t²)) | = (2/3 h/(vt)) / (1 + h²/(3v²t²))

Simplify denominator:

Multiply numerator and denominator by 3v²t²:

Numerator becomes: (2/3 h/(vt)) * 3v²t² = 2 h v t

Denominator becomes: 3v²t² + h²

Therefore tanθ = (2 h v t) / (3v²t² + h² )

So tanθ = (2 h v t) / (3v²t² + h² )

So now, we have tanθ expressed in terms of t. Our goal is to find the maximum θ, which corresponds to maximizing tanθ, since tanθ is increasing in [0, π/2).

Therefore, we can find the maximum of tanθ as a function of t, then take arctangent to find θ.

So let me denote f(t) = (2 h v t) / (3v²t² + h² )

We need to find the maximum of f(t). Let me set u = t, so f(u) = (2 h v u) / (3v²u² + h² )

To find maximum, take derivative with respect to u, set to zero.

Let me compute f'(u):

Using quotient rule:

f'(u) = [ (2 h v)(3v²u² + h² ) - (2 h v u)(6v²u) ] / (3v²u² + h² )²

Simplify numerator:

First term: 2 h v (3v²u² + h² )

Second term: - 2 h v u * 6v²u = -12 h v³ u²

So numerator:

2 h v * 3v²u² + 2 h v * h² - 12 h v³ u²

= 6 h v³ u² + 2 h² v - 12 h v³ u²

= (6 h v³ u² - 12 h v³ u² ) + 2 h² v

= (-6 h v³ u² ) + 2 h² v

Set numerator equal to zero:

-6 h v³ u² + 2 h² v = 0

Divide both sides by 2 h v (assuming h ≠ 0, v ≠ 0):

-3 v² u² + h = 0

So:

-3 v² u² + h = 0

=> 3 v² u² = h

=> u² = h / (3 v² )

=> u = sqrt( h / (3 v² )) = (sqrt(h)/v ) / sqrt(3 )

Therefore, critical point at t = sqrt( h ) / ( v sqrt(3) )

Now, check if this is a maximum. Since f(t) approaches 0 as t approaches 0 and as t approaches infinity, this critical point must be a maximum.

Therefore, maximum tanθ is f(t) evaluated at t = sqrt( h ) / ( v sqrt(3) )

Let me compute that:

First, compute numerator: 2 h v t = 2 h v * [ sqrt(h) / (v sqrt(3)) ] = 2 h v * sqrt(h) / (v sqrt(3)) ) = 2 h sqrt(h) / sqrt(3) = 2 h^(3/2)/sqrt(3)

Denominator: 3v² t² + h²

Compute t²: h / (3 v² )

So 3v² t² = 3v² * [ h / (3 v² ) ] = h

Therefore denominator is h + h² ?

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