2022 AIME II Problems/Problem 11

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

2022 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2022 AIME II Problems/Problem 11

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