Question

In a circle of radius $42,$ two chords of length $78$ intersect at a point whose distance from the center is $18.$ The two chords divide the interior of the circle into four regions. Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form $m\pi-n\sqrt d,$ where $m,$ $n,$ and $d$ are positive integers and $d$ is not divisible by the square of any prime. Find $m+n+d.$

Answer

Answer: 378

Concepts:sector chord coordinate geometry

Difficulty rating: 2650

Small Hint:

Each chord is $9\sqrt3$ from the center, so determine the two possible line directions through the intersection point

Big Hint:

The unequal chord segments have lengths $30$ and $48,$ and their endpoints subtend $60^\circ$ at the center

Solution:

Put the center at $O=(0,0)$ and the intersection at $P=(18,0).$ A length- $78$ chord is $9\sqrt3$ from $O,$ so a line through $P$ containing such a chord makes angle $60^\circ$ or $120^\circ$ with $OP.$ Solving along either line gives segment lengths $30$ and $48.$

For either region bordered by unequal segments, the two arc endpoints subtend $60^\circ$ at $O.$ Its area is the sector minus $\triangle OAB$ plus $\triangle PAB:$ $\begin{aligned}\frac{60}{360}\pi(42)^2&-\frac12(42)^2\sin60^\circ\\&+\frac12(30)(48)\sin60^\circ\\&=294\pi-81\sqrt3.\end{aligned}$ Hence $m+n+d=378.$

1995 AIME Problem 14

Answer: 378

Difficulty rating: 2650

Small Hint:

Big Hint:

Solution:

Problem 14 in Other Years