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2008 AIME I Problems/Problem 14

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Problem

Let $\overline{AB}$ be a diameter of circle $\omega$ . Extend $\overline{AB}$ through $A$ to $C$ . Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$ . Point $P$ is the foot of the perpendicular from $A$ to line $CT$ . Suppose $AB = 18$ , and let $m$ denote the maximum possible length of segment $BP$ . Find $m^{2}$ .

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

Solution

Solution 1

size(250); defaultpen(0.70 + fontsize(10)); import olympiad;pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * ...

Let $x = OC$ . Since $OT, AP \perp TC$ , it follows easily that $\triangle APC \sim \triangle OTC$ . Thus $\frac{AP}{OT} = \frac{CA}{CO} \Longrightarrow AP = \frac{9(x-9)}{x}$ . By the Law of Cosines on $\triangle BAP$ , $\begin{align*}BP^2 = AB^2 + AP^2 - 2 \cdot AB \cdot AP \cdot \cos \angle BAP \end{align*}$ where $\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}$ , so: $\begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{...$ Let $m = \frac{2x-27}{x^2} \Longrightarrow mx^2 - 2x + 27 = 0$ ; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \ge 0 \Longleftrightarrow m \le \frac{1}{27}$ . Thus, $BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}$ Equality holds when $x = 27$ .

Solution 2

unitsize(3mm);pair B=(0,13.5), C=(23.383,0);pair O=(7.794, 9), P=(2*7.794,0);pair T=(7.794,0), Q=(0,0);pair A=(2*7.794,4.5);d...

From the diagram, we see that $BQ = \omega T + B\omega\sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)$ , and that $QP = BA\cos\theta = 18\cos\theta$ .

$\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \sin\theta)^2 + 18^2\cos^2\theta\\&= 9^2[1 + 2\sin\theta + \sin^2\theta +...$

This is a quadratic equation, maximized when $\sin\theta = \frac { - 2}{ - 6} = \frac {1}{3}$ . Thus, $m^2 = 9^2[5 + \frac {2}{3} - \frac {1}{3}] = \boxed{432}$ .

See also

2008 AIME I (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Category: Intermediate Geometry Problems

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