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 $\overline{AB} = 18$, 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 $k=(2x-27)/x^2$, find the maximum value by setting $dk/dx=0$.

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

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

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("$$O$$",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 $BQ = OT + BO \sin\theta = 9 + 9\sin\theta = 9(1 + \sin\theta)$, 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 $(\sqrt{432})^2=\boxed{432}$.

Solution 4

$[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)

Let $AC=x$. The only constraint on $x$ is that it must be greater than $0$. Using similar triangles, we can deduce that $PA=\frac{9x}{x+9}$. Now, apply law of cosines on $\triangle PAB$. $$BP^2=\left(\frac{9x^2}{x+9}\right)^2+18^2-2(18)\left(\frac{9x}{x+9}\right)\cos(\angle PAB).$$ We can see that $\cos(\angle PAB)=\cos(180^{\circ}-\angle PAC)=\cos(\angle PAC -90^{\circ})=-\sin(\angle PCA)$. We can find $-\sin(\angle PCA)=-\frac{9}{x+9}$. Plugging this into our equation, we get: $$BP^2=\left(\frac{9x^2}{x+9}\right)^2+18^2-2(18)\left(\frac{9x}{x+9}\right)\left(-\frac{9}{x+9}\right).$$ Eventually, $$BP^2 = 81\left(\frac{x^2+36x}{(x+9)^2}+4\right).$$ We want to maximize $\frac{x^2+36x}{(x+9)^2}$. There are many ways to maximize this expression, discussed here: https://artofproblemsolving.com/community/c4h2292700_maximization. The maximum result of that expression is $\frac{4}{3}$. Finally, evaluating $BP^2$ for this value $81\left(\frac{4}{3}+4\right) = \boxed{432}$.

~superagh