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

Revision as of 15:41, 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 $ABCDE$ .

Solution 1 (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 (Complex Bash)

Suppose that the circle intersects $\overline{AB}$ , $\overline{BC}$ , $\overline{CD}$ , $\overline{DE}$ , and $\overline{EA}$ at $P$ , $Q$ , $R$ , $S$ , and $T$ respectively. Then $AT = AP = a$ , $BP = BQ = b$ , $CQ = CR = c$ , $DR = DS = d$ , and $ES = ET = e$ . So $a + b = 5$ , $b + c = 6$ , $c + d = 6$ , $d + e = 6$ , and $e + a = 7$ . Then $2a + 2b + 2c + 2d + 2e = 30$ , so $a + b + c + d + e= 15$ . Then we can solve for each individually. $a = 3$ , $b = 2$ , $c = 4$ , $d = 2$ , and $e = 4$ . To find the radius, we notice that $4 \arctan(\frac{2}{r}) + 4 \arctan(\frac{4}{r}) + 2 \arctan (\frac{3}{r}) = 360 ^ \circ$ , or $2 \arctan(\frac{2}{r}) + 2 \arctan(\frac{4}{r}) + \arctan (\frac{3}{r}) = 180 ^ \circ$ . Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, $(r + 2i)^2 \cdot (r + 4i)^2 \cdot (r + 3i)$ is real. Expanding, we get: $(r^2 + 4ir - 4)(r^2 + 6i -9)(r + 3i)$ $(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)$ . On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0. $15ir^4 - 252ir^2 + 192i = 0$ . $5r^4 - 84r^2 + 64 = 0$ $(5r^2 - 4)(r^2 - 16) = 0$ . $r$ must equal 4, as r cannot be negative or be approximately equal to 1. Thus, the area of $ABCDE$ is $4 \cdot (a + b + c + d + e) = 4 \cdot 15 = \boxed{60}$

-nihao4112

Solution 3 (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 actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the $2.9$ into $3$ thus turning the area into $\frac{180}{3}$ 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)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 2: / Line 2: @@
 Convex pentagon <math>ABCDE</math> has side lengths <math>AB=5</math>, <math>BC=CD=DE=6</math>, and <math>EA=7</math>. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of <math>ABCDE</math>.
-==Solutions (Misplaced problem?)==
+==Solution 1 (Misplaced problem?)==
 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 law 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>.

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