2008 AIME I 第 14 题

Put the center at the origin; if $T = (9\cos t, 9\sin t),$ the tangent line is $x\cos t + y\sin t = 9$

The foot is $P = A + 9(1 + \cos t)(\cos t, \sin t),$ and $BP^2$ simplifies to a quadratic in $\cos t$ — maximize it

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 $C = (\frac{9}{\cos t}, 0),$ 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 $\sin^2 t = 1 - u^2:$ $$\begin{aligned} \frac{BP^2}{81} &= (u^2 + u - 2)^2 \\ &\quad {}+ (1 + u)^2(1 - u^2) \\ &= 5 - 2u - 3u^2. \end{aligned}$$ 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 $m^2 = 81 \cdot \frac{16}{3} = 432.$