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Difference between revisions of "2018 AIME I Problems/Problem 15"

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David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, <math>A,\text{ }B,\text{ }C</math>, which can each be inscribed in a circle with radius <math>1</math>. Let <math>\varphi_A</math> denote the measure of the acute angle made by the diagonals of quadrilateral <math>A</math>, and define <math>\varphi_B</math> and <math>\varphi_C</math> similarly. Suppose that <math>\sin\varphi_A=\frac{2}{3}</math>, <math>\sin\varphi_B=\frac{3}{5}</math>, and <math>\sin\varphi_C=\frac{6}{7}</math>. All three quadrilaterals have the same area <math>K</math>, which can be written in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, <math>A,\text{ }B,\text{ }C</math>, which can each be inscribed in a circle with radius <math>1</math>. Let <math>\varphi_A</math> denote the measure of the acute angle made by the diagonals of quadrilateral <math>A</math>, and define <math>\varphi_B</math> and <math>\varphi_C</math> similarly. Suppose that <math>\sin\varphi_A=\frac{2}{3}</math>, <math>\sin\varphi_B=\frac{3}{5}</math>, and <math>\sin\varphi_C=\frac{6}{7}</math>. All three quadrilaterals have the same area <math>K</math>, which can be written in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
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[[2018 AIME I Problems/Problem 15 | Solution]]
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==Solution==
  
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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
 
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
 
<cmath>\varphi_A=a+c</cmath>
 
<cmath>\varphi_A=a+c</cmath>

Revision as of 15:42, 6 June 2018

Problem 15

David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$, which can each be inscribed in a circle with radius $1$. Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$, and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\frac{2}{3}$, $\sin\varphi_B=\frac{3}{5}$, and $\sin\varphi_C=\frac{6}{7}$. All three quadrilaterals have the same area $K$, which can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

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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