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Difference between revisions of "2020 AIME II Problems/Problem 13"

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==Solution 2 (Guess)==
 
==Solution 2 (Guess)==
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The area of a regular pentagon with side lengths <math>s</math> is <math>\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}</math>. <math>5-2\sqrt{5}</math> is slightly greater than <math>\frac{1}{2}</math> given that <math>2\sqrt{5}</math> is slightly less than <math>\frac{9}{2}</math>. <math>4\sqrt{5-2\sqrt{5}}</math> is then slightly greater than <math>2\sqrt{2}</math>. We will approximate that to be <math>2.9</math>. The area is now roughly <math>\frac{180}{2.9}</math>, but because the pentagon is not regular we can say the area is <math>\frac{180}{3}</math> which is <math>60</math> and since <math>60</math> is a multiple of the semiperimeter <math>15</math>, we can safely say that the answer is most likely <math>60</math>.
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This pentagon is very close to a regular pentagon with side lengths <math>6</math>. The area of a regular pentagon with side lengths <math>s</math> is <math>\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}</math>. <math>5-2\sqrt{5}</math> is slightly greater than <math>\frac{1}{2}</math> given that <math>2\sqrt{5}</math> is slightly less than <math>\frac{9}{2}</math>. <math>4\sqrt{5-2\sqrt{5}}</math> is then slightly greater than <math>2\sqrt{2}</math>. We will approximate that to be <math>2.9</math>. The area is now roughly <math>\frac{180}{2.9}</math>, but because the pentagon is not regular we can say the area is <math>\frac{180}{3}</math> which is <math>60</math> and since <math>60</math> is a multiple of the semiperimeter <math>15</math>, we can safely say that the answer is most likely <math>60</math>.
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:53, 8 June 2020

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 law of cosine in triangle $END$ and triangle $ANM$ we can obtain \[\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}\], 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

Solution 2 (Guess)

This pentagon is very close to a regular pentagon with side lengths $6$. The area of a regular pentagon with side lengths $s$ is $\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}$. $5-2\sqrt{5}$ is slightly greater than $\frac{1}{2}$ given that $2\sqrt{5}$ is slightly less than $\frac{9}{2}$. $4\sqrt{5-2\sqrt{5}}$ is then slightly greater than $2\sqrt{2}$. We will approximate that to be $2.9$. The area is now roughly $\frac{180}{2.9}$, but because the pentagon is not regular we can say the area is $\frac{180}{3}$ which is $60$ and since $60$ is a multiple of the semiperimeter $15$, we can safely say that the answer is most likely $60$.

Video Solution

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

2020 AIME II (Problems • Answer Key • Resources)
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