2020 AIME II Problems/Problem 13

Problem

Convex pentagon $ABCDE$ has side lengths $AB=5$ , $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ .

Solutions (Misplaced problem?)

Assume the incircle touches $AB$ , $BC$ , $CD$ , $DE$ , $EA$ at $P,Q,R,S,T$ respectively. Then let $PB=x=BQ=RD=SD$ , $ET=y=ES=CR=CQ$ , $AP=AT=z$ . So we have $x+y=6$ , $x+z=5$ and $y+z$ =7, solve it we have $x=2$ , $z=3$ , $y=4$ . Let the center of the incircle be $I$ , by SAS we can proof triangle $BIQ$ is congruent to triangle $DIS$ , and triangle $CIR$ is congruent to triangle $SIE$ . Then we have $\angle AED=\angle BCD$ , $\angle ABC=\angle CDE$ . Extend $CD$ , cross ray $AB$ at $M$ , ray $AE$ at $N$ , then by AAS we have triangle $END$ is congruent to triangle $BMC$ . Thus $\angle M=\angle N$ . Let $EN=MC=a$ , then $BM=DN=a+2$ . So by low of cosine in triangle $END$ and triangle $ANM$ we can obtain $$ (Error compiling LaTeX. Unknown error_msg)\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)} $, solved it gives us$ a=8 $, which yield triangle$ ANM $to be a triangle with side length 15, 15, 24, draw a height from$ A $to$ NM $divides it into two triangles with side lengths 9, 12, 15, so the area of triangle$ ANM $is 108. Triangle$ END $is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is$ 108-48=\boxed{60}$.

-Fanyuchen20020715

Video Solution

https://youtu.be/bz5N-jI2e0U?t=327

2020 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2020 AIME II Problems/Problem 13

Problem

Solutions (Misplaced problem?)

Video Solution