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

Revision as of 12:56, 15 February 2019

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.

@@ Line 15: / Line 15: @@
 so our answer is <math>24+35=\boxed{059}</math>.
 By S.B.
-LaTeX by willwin4sure
 ==See Also==
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Difference between revisions of "2018 AIME I Problems/Problem 15"

Revision as of 12:56, 15 February 2019

Problem 15

Solution

See Also