1995 AIME Problems/Problem 14

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Problem

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 number. Find $m+n+d.$

Solution

Let the center of the circle be $O$ , and the two chords be $\overline{AB}, \overline{CD}$ and intersecting at $E$ , such that $AE = CE < BE = DE$ . Let $F$ be the midpoint of $\overline{AB}$ . Then $\overline{OF} \perp \overline{AB}$ .

By the Pythagorean Theorem, $OF = \sqrt{OB^2 - BF^2} = \sqrt{42^2 - 39^2} = 9\sqrt{3}$ , and $EF = \sqrt{OE^2 - OF^2} = 9$ . Then $OEF$ is a $30-60-90$ right triangle, so $\angle OEB = \angle OED = 60^{\circ}$ . Thus $\angle BEC = 60^{\circ}$ , and by the Law of Cosines,

$BC^2 = BE^2 + CE^2 - 2 \cdot BE \cdot CE \cos 60^{\circ} = 42.$

It follows that $\triangle BCO$ is an equilateral triangle, so $\angle BOC = 60^{\circ}$ . The desired area can be broken up into two regions, $\triangle BCE$ and the region bounded by $\overline{BC}$ and minor arc $\stackrel{\frown}{BC}$ . The former can be found by Heron's formula to be $[BCE] = \sqrt{60(60-48)(60-42)(60-30)} = 360\sqrt{3}$ . The latter is the difference between the area of sector $BOC$ and the equilateral $\triangle BOC$ , or $\frac{1}{6}\pi (42)^2 - \frac{42^2 \sqrt{3}}{4} = 294\pi - 441\sqrt{3}$ .

Thus, the desired area is $360\sqrt{3} + 294\pi - 441\sqrt{3} = 294\pi - 81\sqrt{3}$ , and $m+n+d = \boxed{378}$ .

1995 AIME (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1995 AIME Problems/Problem 14

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Problem

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