1991 AIME Problems/Problem 14

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Problem

A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$ , has length $31$ . Find the sum of the lengths of the three diagonals that can be drawn from $A_{}^{}$ .

Solution

Let $x=AC=BF$ , $y=AD=BE$ , and $z=AE=BD$ .

Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$ , and Ptolemy on $ACDE$ gives $x\cdot z+81^2=y^2$ . Subtracting these equations give $y^2-81y-112\cdot 81=0$ , and from this $y=144$ . Ptolemy on $ADEF$ gives $81y+81^2=z^2$ , and from this $z=135$ . Finally, plugging back into the first equation gives $x=105$ , so $x+y+z=105+144+135=\boxed{384}$ .

1991 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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1991 AIME Problems/Problem 14

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Problem

Solution

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