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

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==Solutions (Misplaced problem?)==
 
==Solutions (Misplaced problem?)==
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Assume the incircle touches <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EA</math> at <math>P,Q,R,S,T</math> respectively. Then let <math>PB=x=BQ=RD=SD</math>, <math>ET=y=ES=CR=CQ</math>, <math>AP=AT=z</math>. So we have <math>x+y=6</math>, <math>x+z=5</math> and <math>y+z</math>=7, solve it we have <math>x=2</math>, <math>z=3</math>, <math>y=4</math>. Let the center of the incircle be <math>I</math>, by SAS we can proof triangle <math>BIQ</math> is congruent to triangle <math>DIS</math>, and triangle <math>CIR</math> is congruent to triangle <math>SIE</math>. Then we have <math>\angle AED=\angle BCD</math>, <math>\angle ABC=\angle CDE</math>. Extend <math>CD</math>, cross ray <math>AB</math> at <math>M</math>, ray <math>AE</math> at <math>N</math>, then by AAS we have triangle <math>END</math> is congruent to triangle <math>BMC</math>. Thus <math>\angle M=\angle N</math>. Let <math>EN=MC=a</math>, then <math>BM=DN=a+2</math>. So by low of cosine in triangle <math>END</math> and triangle <math>ANM</math> we can obtain <math></math>\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}<math>, solved it gives us </math>a=8<math>, which yield triangle </math>ANM<math> to be a triangle with side length 15, 15, 24, draw a height from </math>A<math> to </math>NM<math> divides it into two triangles with side lengths 9, 12, 15, so the area of triangle </math>ANM<math> is 108. Triangle </math>END<math> is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is </math>108-48=\boxed{60}$.
+
Assume the incircle touches <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, <math>EA</math> at <math>P,Q,R,S,T</math> respectively. Then let <math>PB=x=BQ=RD=SD</math>, <math>ET=y=ES=CR=CQ</math>, <math>AP=AT=z</math>. So we have <math>x+y=6</math>, <math>x+z=5</math> and <math>y+z</math>=7, solve it we have <math>x=2</math>, <math>z=3</math>, <math>y=4</math>. Let the center of the incircle be <math>I</math>, by SAS we can proof triangle <math>BIQ</math> is congruent to triangle <math>DIS</math>, and triangle <math>CIR</math> is congruent to triangle <math>SIE</math>. Then we have <math>\angle AED=\angle BCD</math>, <math>\angle ABC=\angle CDE</math>. Extend <math>CD</math>, cross ray <math>AB</math> at <math>M</math>, ray <math>AE</math> at <math>N</math>, then by AAS we have triangle <math>END</math> is congruent to triangle <math>BMC</math>. Thus <math>\angle M=\angle N</math>. Let <math>EN=MC=a</math>, then <math>BM=DN=a+2</math>. So by low of cosine in triangle <math>END</math> and triangle <math>ANM</math> we can obtain <cmath>\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}</cmath>, solved it gives us <math>a=8</math>, which yield triangle <math>ANM</math> to be a triangle with side length 15, 15, 24, draw a height from <math>A</math> to <math>NM</math> divides it into two triangles with side lengths 9, 12, 15, so the area of triangle <math>ANM</math> is 108. Triangle <math>END</math> is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is <math>108-48=\boxed{60}</math>.
  
 
-Fanyuchen20020715
 
-Fanyuchen20020715

Revision as of 09:34, 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 low 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

Video Solution

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

2020 AIME II (Problems • Answer Key • Resources)
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