1991 AIME Problems/Problem 14

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 $A_{}^{}$ .

File:1991 AIME-14.png

Solution

File:1991 AIME-14a.png

Let $x=AC$ , $y=AD$ , and $z=AE$ . Ptolemy's Theorem on $ABCD$ gives $81y+31\cdot 81=xz$ , and Ptolemy on $ACDE4 gives$ 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=384$.

1991 AIME (Problems • Answer Key • 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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