2018 AIME I Problems/Problem 8

Revision as of 19:30, 7 March 2018 by Cooljoseph (talk | contribs) (Solution Diagram)

Let $ABCDEF$ be an equiangular hexagon such that $AB=6, BC=8, CD=10$, and $DE=12$. Denote $d$ the diameter of the largest circle that fits inside the hexagon. Find $d^2$.

Solutions

Solution Diagram

[asy] draw((0,0)--(12,20.78)--(24,0)--cycle); draw((1,1.73)--(2,0)); draw((9,15.59)--(15,15.59)); draw((14,0)--(19,8.66)); label("$A$",(9,15.59),NW); label("$B$",(15,15.59),NE); label("$C$",(19,8.66),NE); label("$D$",(14,0),S); label("$E$",(2,0),S); label("$F$",(1,1.73),NW); pair O; O=(11.25,7.36); dot(O); label("$O$",O,SW); draw(Circle(O,6.06)); [/asy] asymptote code for a picture - cooljoseph First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that $EF=2, FA=16$. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length $6+8+10=24$. Then, if you drew it to scale, notice that the "widest" this circle can be according to $AF, CD$ is $7\sqrt{3}$. And it will be obvious that the sides won't be inside the circle, so our answer is $\boxed{147}$.

-expiLnCalc