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 $A_{}^{}$ .

Solution

Let x=AC, y=AD, and z=AE. 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= $384$ .

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
All AIME Problems and Solutions

Page

Toolbox

Search

1991 AIME Problems/Problem 14

Problem

Solution

See also