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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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==Solution==
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==Solution 1==
  
 
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
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By S.B.
 
By S.B.
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 +
==Solution 2==
 +
 +
Let the four stick lengths be <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math>. WLOG, let’s say that quadrilateral <math>A</math> has sides <math>a</math> and <math>d</math> opposite each other, quadrilateral <math>B</math> has sides <math>b</math> and <math>d</math> opposite each other, and quadrilateral <math>C</math> has sides <math>c</math> and <math>d</math> opposite each other. The area of a cyclic quadrilateral can be written as <math>\frac{1}{2} d_1 d_2 \sin{\theta}</math>, where <math>d_1</math> and <math>d_2</math> are the lengths of the diagonals of the quadrilateral and <math>\theta</math> is the angle formed by the intersection of <math>d_1</math> and <math>d_2</math>. By Ptolemy's theorem <math>d_1 d_2 = ad+bc</math> for quadrilateral <math>A</math>, so, defining <math>K_A</math> as the area of <math>A</math>,
 +
<cmath>K_A = \frac{1}{2} (ad+bc)\sin{\varphi_A}</cmath>
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Similarly, for quadrilaterals <math>B</math> and <math>C</math>,
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<cmath>K_B = \frac{1}{2} (bd+ac)\sin{\varphi_B}</cmath>
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and
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<cmath>K_C = \frac{1}{2} (cd+ab)\sin{\varphi_C}</cmath>
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Multiplying the three equations and rearranging, we see that
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<cmath>K_A K_B K_C = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\sin{\varphi_A}\sin{\varphi_B}\sin{\varphi_B}</cmath>
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<cmath>K^3 = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\left(\frac{2}{3}\right) \left(\frac{3}{5}\right) \left(\frac{6}{7}\right)</cmath>
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<cmath>\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)</cmath>
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The circumradius <math>R</math> of a cyclic quadrilateral with side lengths <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> and area <math>K</math> can be computed as <math>R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}</math>.
 +
Inserting what we know,
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<cmath>1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}</cmath>
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<cmath>4K = \sqrt{\frac{70}{3}K^3}</cmath>
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<cmath>16K^2 = \frac{70}{3}K^3</cmath>
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<cmath>\frac{24}{35} = K</cmath>
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So our answer is <math>24 + 35 = \boxed{059}</math>
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~Solution by divij04
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2018|n=I|num-b=14|after=Last question}}
 
{{AIME box|year=2018|n=I|num-b=14|after=Last question}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:44, 17 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 1

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.

Solution 2

Let the four stick lengths be $a$, $b$, $c$, and $d$. WLOG, let’s say that quadrilateral $A$ has sides $a$ and $d$ opposite each other, quadrilateral $B$ has sides $b$ and $d$ opposite each other, and quadrilateral $C$ has sides $c$ and $d$ opposite each other. The area of a cyclic quadrilateral can be written as $\frac{1}{2} d_1 d_2 \sin{\theta}$, where $d_1$ and $d_2$ are the lengths of the diagonals of the quadrilateral and $\theta$ is the angle formed by the intersection of $d_1$ and $d_2$. By Ptolemy's theorem $d_1 d_2 = ad+bc$ for quadrilateral $A$, so, defining $K_A$ as the area of $A$, \[K_A = \frac{1}{2} (ad+bc)\sin{\varphi_A}\] Similarly, for quadrilaterals $B$ and $C$, \[K_B = \frac{1}{2} (bd+ac)\sin{\varphi_B}\] and \[K_C = \frac{1}{2} (cd+ab)\sin{\varphi_C}\] Multiplying the three equations and rearranging, we see that \[K_A K_B K_C = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\sin{\varphi_A}\sin{\varphi_B}\sin{\varphi_B}\] \[K^3 = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\left(\frac{2}{3}\right) \left(\frac{3}{5}\right) \left(\frac{6}{7}\right)\] \[\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)\] The circumradius $R$ of a cyclic quadrilateral with side lengths $a$, $b$, $c$, and $d$ and area $K$ can be computed as $R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}$. Inserting what we know, \[1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}\] \[4K = \sqrt{\frac{70}{3}K^3}\] \[16K^2 = \frac{70}{3}K^3\] \[\frac{24}{35} = K\] So our answer is $24 + 35 = \boxed{059}$

~Solution by divij04

See Also

2018 AIME I (Problems • Answer Key • Resources)
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