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Difference between revisions of "2018 AIME I Problems/Problem 15"
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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>.
 
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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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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Then, by Law of Sines, <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,
  
 
<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>
 
<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>
 
so our answer is <math>24+35=\boxed{059}</math>.
 
so our answer is <math>24+35=\boxed{059}</math>.
  
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By S.B., LaTeX by ww4
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By S.B.  
 +
LaTeX by willwin4sure

Revision as of 09:01, 10 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, by Law of Sines, $AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)$ 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 willwin4sure

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