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Difference between revisions of "2018 AIME I Problems/Problem 15"
Line 1: Line 1:
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Basically
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Suppose our four sides lengths cut out arc lengths of <math>2a</math>, <math>2b</math>, <math>2c</math>, and <math>2d</math>, where <math>a+b+c+d=180^\circ</math>. Then, we only have to consider which arc is opposite <math>2a</math>. These are our three cases, so
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Three cases are pretty similar
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<cmath>\varphi_A=a+c</cmath>
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WLOG ABCD ACDB ADBC
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<cmath>\varphi_B=a+b</cmath>
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As the diameter is 2
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<cmath>\varphi_C=a+d</cmath>
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Let arcAB =2a arcBC=2b arcCD=2c arcDA=2d
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Our first case involves quadrilateral <math>ABCD</math> with <math>\overarc{AB}=2a</math>, <math>\overarc{BC}=2b</math>, <math>\overarc{CD}=2c</math>, and <math>\overarc{DA}=2d</math>.
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a+b+c+d=Pi
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By easy angle chasing
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Then, <math>AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)</math> and <math>BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)</math>. Therefore,
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phiA=a+c or b+d
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phiB=a+b or c+d
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<cmath>K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},</cmath>
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phiC=a+d or b+c
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so our answer is <math>24+35=\boxed{059}</math>.
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AC of case 1 = 2sin(arcABC/2)=2sin(a+b)
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BD of case 1= 2sin (arcBCD/2)=2sin(a+d)
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By S.B., LaTeX by ww4
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Sin(angleAC,BD of case 1)=sin(a+c)
 
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Area=AC*BD*sin(phiA)/2=2sin(phiA)sin(phiB)sin(phiC)=24/35
 
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ANS=24+35=059
 
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By S.B.
 

Revision as of 23:16, 9 March 2018

Suppose our four sides lengths cut out arc lengths of $2a$, $2b$, $2c$, and $2d$, where $a+b+c+d=180^\circ$. Then, we only have to consider which arc is opposite $2a$. These are our three cases, so \[\varphi_A=a+c\] \[\varphi_B=a+b\] \[\varphi_C=a+d\] Our first case involves quadrilateral $ABCD$ with $\overarc{AB}=2a$, $\overarc{BC}=2b$, $\overarc{CD}=2c$, and $\overarc{DA}=2d$.

Then, $AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)$ and $BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)$. Therefore,

\[K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},\] so our answer is $24+35=\boxed{059}$.

By S.B., LaTeX by ww4

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