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Difference between revisions of "2022 AIME I Problems/Problem 11"

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==solution 1==
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Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math>
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Using POP, it is very clear that <math>PC=20,AQ=AM=6</math>, let <math>BM=BP=x,QD=14+x</math>, using LOC in <math>\triangle{ABP}</math>,<math>x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2</math>, similarly, use LOC in <math>\triangle{DQC}</math>, getting that <math>(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2</math>. We use the second equation to minus the first equation, getting that <math>28x+196-(2x+12)*14*\cos\alpha=364</math>, we can get <math>\cos\alpha=\frac{2x-12}{2x+12}</math>.
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Now applying LOC in <math>\triangle{ADC}</math>, getting <math>(6+x)^2+400-2(6+x)*20*\frac{2x-12}{2x+12}=(3+9+16)^2</math>, solving this equation to get <math>x=\frac{9}{2}</math>, then <math>\cos\alpha=-\frac{1}{7}</math>, <math>\sin\alpha=\frac{4\sqrt{3}}{7}</math>, the area is <math>\frac{21}{2}*\frac{49}{2}*\frac{4\sqrt{3}}{7}=147\sqrt{3}</math> leads to <math>\boxed{150}</math>

Revision as of 19:07, 17 February 2022

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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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