Difference between revisions of "2008 AIME I Problems/Problem 14"

Revision as of 19:33, 5 July 2020

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 $m^{2}$ .

Solution

Solution 1

$[asy] size(250); defaultpen(0.70 + fontsize(10)); import olympiad; pair O = (0,0), B = O - (9,0), A= O + (9,0), C=A+(18,0), T = 9 * expi(-1.2309594), P = foot(A,C,T); draw(Circle(O,9)); draw(B--C--T--O); draw(A--P); dot(A); dot(B); dot(C); dot(O); dot(T); dot(P); draw(rightanglemark(O,T,C,30)); draw(rightanglemark(A,P,C,30)); draw(anglemark(B,A,P,35)); draw(B--P, blue); label("$A$",A,NW); label("$B$",B,NW); label("$C$",C,NW); label("$O$",O,NW); label("$P$",P,SE); label("$T$",T,SE); label("$9$",(O+A)/2,N); label("$9$",(O+B)/2,N); label("$x-9$",(C+A)/2,N); [/asy]$

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: $\begin{align*}BP^2 &= 18^2 + \frac{9^2(x-9)^2}{x^2} + 2(18) \cdot \frac{9(x-9)}{x} \cdot \frac 9x = 405 + 729\left(\frac{2x - 27}{x^2}\right)\end{align*}$ Let $k = \frac{2x-27}{x^2} \Longrightarrow kx^2 - 2x + 27 = 0$ ; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(k)(27) \ge 0 \Longleftrightarrow k \le \frac{1}{27}$ . Thus, $BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}$ Equality holds when $x = 27$ .

Solution 1.1

Proceed as follows for Solution 1.

Once you approach the function $m=(2x-27)/x^2$ , find the maximum value by setting $dm/dx=0$ .

Simplifying $m$ to take the derivative, we have $2/x-27/x^2$ , so $dm/dx=-2/x^2+54/x^3$ . Setting $dm/dx=0$ , we have $2/x^2=54/x^3$ .

Solving, we obtain $x=27$ as the critical value. Hence, $m$ has the maximum value of $(2*27-27)/27^2=1/27$ . Since $BP^2=405+729m$ , the maximum value of $\overline {BP}$ occurs at $m=1/27$ , so $BP^2$ has a maximum value of $405+729/27=\fbox{432}$ .

Note: Please edit this solution if it feels inadequate.

Solution 2

$[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label("$B$",B,NW); label("$A$",A,NE); label("$\omega$",O,N); label("$P$",P,S); label("$T$",T,S); label("$Q$",Q,S); label("$C$",C,E); label("$\theta$",C + (-1.7,-0.2), NW); label("$9$", (B+O)/2, N); label("$9$", (O+A)/2, N); label("$9$", (O+T)/2,W); [/asy]$

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

$\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \sin\theta)^2 + 18^2\cos^2\theta\\ &= 9^2[1 + 2\sin\theta + \sin^2\theta + 4(1 - \sin^2\theta)]\\ BP^2 &= 9^2[5 + 2\sin\theta - 3\sin^2\theta]\end{align*}$

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

Solution 3 (Calculus Bash)

$[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label("$B$",B,NW); label("$A$",A,NE); label("$\omega$",O,N); label("$P$",P,S); label("$T$",T,S); label("$Q$",Q,S); label("$C$",C,E); label("$9$", (B+O)/2, N); label("$9$", (O+A)/2, N); label("$9$", (O+T)/2,W); [/asy]$

(Diagram credit goes to Solution 2)

We let $AC=x$ . From similar triangles, we have that $PC=\frac{x\sqrt{x^2+18x}}{x+9}$ . Similarly, $TP=QT=\frac{9\sqrt{x^2+18x}}{x+9}$ . Using the Pythagorean Theorem, $BQ=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2}$ . Using the Pythagorean Theorem once again, $BP=\sqrt{(x+18)^2-(\frac{(x+18)\sqrt{x^2+18x}}{x+9})^2+(\frac{18\sqrt{x^2+18x}}{x+9})^2}$ . After a large bashful simplification, $BP=\sqrt{405+\frac{1458x-6561}{x^2+18x+81}}$ . The fraction is equivalent to $729\frac{2x-9}{(x+9)^2}$ . Taking the derivative of the fraction and solving for x, we get that $x=18$ . Plugging $x=18$ back into the expression for $BP$ yields $\sqrt{432}$ , so the answer is $(\sqrt{432})^2=\boxed{432}$ .

@@ Line 26: / Line 26: @@
 where <math>\cos \angle BAP = \cos (180 - \angle TOA) = - \frac{OT}{OC} = - \frac{9}{x}</math>, so:
 <cmath>\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{2x - 27}{x^2}\right)\end{align*}</cmath>
-Let <math>m = \frac{2x-27}{x^2} \Longrightarrow mx^2 - 2x + 27 = 0</math>; this is a quadratic, and its [[discriminant]] must be nonnegative: <math>(-2)^2 - 4(m)(27) \ge 0 \Longleftrightarrow m \le \frac{1}{27}</math>. Thus,
+Let <math>k = \frac{2x-27}{x^2} \Longrightarrow kx^2 - 2x + 27 = 0</math>; this is a quadratic, and its [[discriminant]] must be nonnegative: <math>(-2)^2 - 4(k)(27) \ge 0 \Longleftrightarrow k \le \frac{1}{27}</math>. Thus,
 <cmath>BP^2 \le 405 + 729 \cdot \frac{1}{27} = \boxed{432}</cmath>
 Equality holds when <math>x = 27</math>.

2008 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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