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Difference between revisions of "2022 AIME II Problems/Problem 11"
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==Solution==
 
==Solution==
  
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[[Image:2022AIME2-Q11.png|thumb|center|500px|2022 AIME II Q11(Hand-draw picture)]]
  
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According to the problem, we have <math>AB=AB'=2</math>, <math>DC=DC'=3</math>, <math>MB=MB'</math>, <math>MC=MC'</math>, and <math>B'C'=7-2-3=2</math>
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Because <math>M</math> is the midpoint of <math>BC</math>, we have <math>BM=MC</math>, so: <cmath>MB=MB'=MC'=MC.</cmath>
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Then, we can see that <math>\bigtriangleup{MB'C'}</math> is an isosceles triangle with <math>MB'=MC'</math>
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Therefore, we could start our angle chasing: <math>\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}</math>.
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This is when we found that points <math>M</math>, <math>C</math>, <math>D</math>, and <math>B'</math> are on a circle. Thus, <math>\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}</math>. This is the time we found that <math>\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}</math>.
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Thus, <math>\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (BM')^2=AB' \cdot C'D = 6</math>
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Point <math>H</math> is the midpoint of <math>B'C'</math>, and <math>MH \perp AD</math>. <math>B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}</math>.
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The area of this quadrilateral is the sum of areas of triangles: <cmath>S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}</cmath>
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<cmath>=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2</cmath>
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<cmath>=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH</cmath>
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<cmath>=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}</cmath>
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Finally, the square of the area is <math>(6\sqrt{5})^2=\boxed{180}</math>
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~DSAERF-CALMIT
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|num-b=10|num-a=12}}
 
{{AIME box|year=2022|n=II|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:27, 18 February 2022

Problem

Let $ABCD$ be a convex quadrilateral with $AB=2$, $AD=7$, and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}$. Find the square of the area of $ABCD$.

Solution

2022 AIME II Q11(Hand-draw picture)


According to the problem, we have $AB=AB'=2$, $DC=DC'=3$, $MB=MB'$, $MC=MC'$, and $B'C'=7-2-3=2$

Because $M$ is the midpoint of $BC$, we have $BM=MC$, so: \[MB=MB'=MC'=MC.\]

Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$

Therefore, we could start our angle chasing: $\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}$.

This is when we found that points $M$, $C$, $D$, and $B'$ are on a circle. Thus, $\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}$. This is the time we found that $\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}$.

Thus, $\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (BM')^2=AB' \cdot C'D = 6$

Point $H$ is the midpoint of $B'C'$, and $MH \perp AD$. $B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}$.

The area of this quadrilateral is the sum of areas of triangles: \[S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}\] \[=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2\] \[=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH\] \[=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}\]

Finally, the square of the area is $(6\sqrt{5})^2=\boxed{180}$

~DSAERF-CALMIT

See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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