2022 AIME I Problems/Problem 11

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solution 1

Let the circle tangent to $BC,AD,AB$ at $P,Q,M$ separately, denote that $\angle{ABC}=\angle{D}=\alpha$

Using POP, it is very clear that $PC=20,AQ=AM=6$ , let $BM=BP=x,QD=14+x$ , using LOC in $\triangle{ABP}$ , $x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2$ , similarly, use LOC in $\triangle{DQC}$ , getting that $(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2$ . We use the second equation to minus the first equation, getting that $28x+196-(2x+12)*14*\cos\alpha=364$ , we can get $\cos\alpha=\frac{2x-12}{2x+12}$ .

Now applying LOC in $\triangle{ADC}$ , getting $(6+x)^2+400-2(6+x)*20*\frac{2x-12}{2x+12}=(3+9+16)^2$ , solving this equation to get $x=\frac{9}{2}$ , then $\cos\alpha=-\frac{1}{7}$ , $\sin\alpha=\frac{4\sqrt{3}}{7}$ , the area is $\frac{21}{2}*\frac{49}{2}*\frac{4\sqrt{3}}{7}=147\sqrt{3}$ leads to $\boxed{150}$

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2022 AIME I Problems/Problem 11

solution 1