Question

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 $m^2.$

答案

Solution:

Place the center $O$ at the origin with radius $9,$ so $A = (-9, 0)$ and $B = (9, 0).$ If the point of tangency is $T = (9\cos t, 9\sin t),$ the tangent line is $x\cos t + y\sin t = 9;$ it meets the $x$ -axis at $C = (9/\cos t, 0),$ which lies beyond $A$ exactly when $-1 \lt \cos t \lt 0.$ Writing $u = \cos t,$ the signed distance from $A$ to the line is $-9u - 9,$ so the foot of the perpendicular is $P = A + 9(1 + u)(\cos t, \sin t).$

Then $P - B = \bigl(9(u^2 + u - 2),\ 9(1 + u)\sin t\bigr),$ and using $\sin^2 t = 1 - u^2:$ $\frac{BP^2}{81} = (u^2 + u - 2)^2 + (1 + u)^2(1 - u^2) = 5 - 2u - 3u^2.$ This quadratic in $u$ is maximized at $u = -\frac{1}{3},$ which is inside $(-1, 0)$ (there $C = (-27, 0)$ ), giving $\frac{BP^2}{81} = 5 + \frac{2}{3} - \frac{1}{3} = \frac{16}{3}.$

Therefore $m^2 = 81 \cdot \frac{16}{3} = 432.$

2008 AIME I Problem 14

Solution:

Problem 14 in Other Years