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

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\usepackage{ gensymb, amsmath }
 
 
==Problem 15==
 
==Problem 15==
 
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=\tfrac{2}{3}</math>, <math>\sin\varphi_B=\tfrac{3}{5}</math>, and <math>\sin\varphi_C=\tfrac{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=\tfrac{2}{3}</math>, <math>\sin\varphi_B=\tfrac{3}{5}</math>, and <math>\sin\varphi_C=\tfrac{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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By S.B.
 
By S.B.
 +
 +
===Note===
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[[File:CyclIntersect.png|400px]]
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 +
The solution uses <cmath>\varphi_A=a+c.</cmath>
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 +
We can see that this follows because <math>\varphi_A = \frac12 (2a+2c)=a+c,</math> where <math>a</math> and <math>c</math> are the central angles of opposite sides.
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____Shen Kislay Kai
  
 
==Solution 2==
 
==Solution 2==
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Therefore, without losing generality,  
 
Therefore, without losing generality,  
  
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    &\varphi_A=a+b\\
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<cmath>\varphi_A=a+b</cmath>
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    &\varphi_B=b+c\\
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<cmath>\varphi_B=b+c</cmath>
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    &\varphi_C=a+c
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<cmath>\varphi_C=a+c</cmath>
  
 
<math>(1)+(3)-(2)</math>, <math>(1)+(2)-(3)</math>, and <math>(2)+(3)-(1)</math> yields
 
<math>(1)+(3)-(2)</math>, <math>(1)+(2)-(3)</math>, and <math>(2)+(3)-(1)</math> yields
  
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    &2a=\varphi_A+\varphi_C-\varphi_B\\
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<cmath>2a=\varphi_A+\varphi_C-\varphi_B</cmath>
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    &2b=\varphi_A+\varphi_B-\varphi_C\\
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<cmath>2b=\varphi_A+\varphi_B-\varphi_C</cmath>
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    &2c=\varphi_B+\varphi_C-\varphi_A
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<cmath>2c=\varphi_B+\varphi_C-\varphi_A</cmath>
  
 
Because <math>2d=360^\circ-2a-2b-2c,</math>
 
Because <math>2d=360^\circ-2a-2b-2c,</math>
 
Therefore,
 
Therefore,
  
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    2d=360\degree-\varphi_A-\varphi_B-\varphi_C
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<cmath>2d=360^\circ-\varphi_A-\varphi_B-\varphi_C</cmath>
  
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Our area of the quadrilateral <math>K</math> then would be  
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Using the [[trigonometric identity|sum-to-product identities]], our area of the quadrilateral <math>K</math> then would be
  
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    K&=\frac{1}{2}(\sin(2a)+\sin(2b)+\sin(2c)+\sin(2d))\\
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<cmath>
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    &=\frac{1}{2}(\sin(\varphi_A+\varphi_B-\varphi_C)+\sin(\varphi_B+\varphi_C-\varphi_A)+\sin(\varphi_C+\varphi_A-\varphi_B)-\sin(\varphi_A+\varphi_B+\varphi_C))\\
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\begin{align*}
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    &=\frac{1}{2}(2\sin\varphi_B\cos(\varphi_A-\varphi_C)-2\sin\varphi_B\cos(\varphi_A+\varphi_C))\\
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K&=\frac{1}{2}(\sin(2a)+\sin(2b)+\sin(2c)+\sin(2d))\\
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    &=\frac{1}{2}\cdot2\cdot2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\
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&=\frac{1}{2}(\sin(\varphi_A+\varphi_B-\varphi_C)+\sin(\varphi_B+\varphi_C-\varphi_A)+\sin(\varphi_C+\varphi_A-\varphi_B)-\sin(\varphi_A+\varphi_B+\varphi_C))\\
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    &=2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\
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&=\frac{1}{2}(2\sin\varphi_B\cos(\varphi_A-\varphi_C)-2\sin\varphi_B\cos(\varphi_A+\varphi_C))\\
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    &=\frac{24}{35}
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&=\frac{1}{2}\cdot2\cdot2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\
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&=2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\
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&=\frac{24}{35}\\
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\end{align*}
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</cmath>
  
 
Therefore, our answer is <math>24+35=\boxed{059}</math>.
 
Therefore, our answer is <math>24+35=\boxed{059}</math>.
  
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~Solution by Eric Zou
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~Solution by eric-z
  
 
==Solution 3==
 
==Solution 3==

Latest revision as of 00:46, 20 November 2024

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 $\sin\varphi_A=\tfrac{2}{3}$, $\sin\varphi_B=\tfrac{3}{5}$, and $\sin\varphi_C=\tfrac{6}{7}$. 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 $24+35=\boxed{059}$.

Note that the conditions of the problem are satisfied when the lengths of the four sticks are about $0.32, 0.91, 1.06, 1.82$.

By S.B.

Note

CyclIntersect.png

The solution uses \[\varphi_A=a+c.\]

We can see that this follows because $\varphi_A = \frac12 (2a+2c)=a+c,$ where $a$ and $c$ are the central angles of opposite sides. ____Shen Kislay Kai

Solution 2

Suppose the four side lengths of the quadrilateral cut out arc lengths of $2a$, $2b$, $2c$, and $2d$. $a+b+c+d=180^\circ$. Therefore, without losing generality,

\[\varphi_A=a+b\] \[\varphi_B=b+c\] \[\varphi_C=a+c\]

$(1)+(3)-(2)$, $(1)+(2)-(3)$, and $(2)+(3)-(1)$ yields

\[2a=\varphi_A+\varphi_C-\varphi_B\] \[2b=\varphi_A+\varphi_B-\varphi_C\] \[2c=\varphi_B+\varphi_C-\varphi_A\]

Because $2d=360^\circ-2a-2b-2c,$ Therefore,

\[2d=360^\circ-\varphi_A-\varphi_B-\varphi_C\]

Using the sum-to-product identities, our area of the quadrilateral $K$ then would be

\begin{align*} K&=\frac{1}{2}(\sin(2a)+\sin(2b)+\sin(2c)+\sin(2d))\\ &=\frac{1}{2}(\sin(\varphi_A+\varphi_B-\varphi_C)+\sin(\varphi_B+\varphi_C-\varphi_A)+\sin(\varphi_C+\varphi_A-\varphi_B)-\sin(\varphi_A+\varphi_B+\varphi_C))\\ &=\frac{1}{2}(2\sin\varphi_B\cos(\varphi_A-\varphi_C)-2\sin\varphi_B\cos(\varphi_A+\varphi_C))\\ &=\frac{1}{2}\cdot2\cdot2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ &=2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ &=\frac{24}{35}\\ \end{align*}

Therefore, our answer is $24+35=\boxed{059}$.

~Solution by eric-z

Solution 3

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 convex 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, \[1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}\quad \Rightarrow \quad 16K^2 = \frac{70}{3}K^3\quad \Rightarrow \quad \frac{24}{35} = K\] So our answer is $24 + 35 = \boxed{059}$.

~Solution by divij04

Solution 4 (No words)

2018 AIME I 15.png

vladimir.shelomovskii@gmail.com, vvsss

Solution 5

Let the sides of the quadrilaterals be $a,b,c,$ and $d$ in some order such that $A$ has $a$ opposite of $c$, $B$ has $a$ opposite of $b$, and $C$ has $a$ opposite of $d$. Then, let the diagonals of $A$ be $e$ and $f$. Similarly to solution $2$, we get that $\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K$, but this is also equal to $2\cdot\tfrac{eab+ecd}{4(1)}=2\cdot\tfrac{fad+fbc}{4(1)}$ using the area formula for a triangle using the circumradius and the sides, so $\tfrac{e(ab+cd)}{2}=\tfrac{3}{5}(ab+cd)$ and $\tfrac{f(ad+bc)}{2}=\tfrac{6}{7}(ad+bc)$. Solving for $e$ and $f$, we get that $e=\tfrac{6}{5}$ and $f=\tfrac{12}{7}$, but $K=\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot{}ef$, similarly to solution $2$, so $K=\tfrac{24}{35}$ and the answer is $24+35=\boxed{059}$.

Video Solution by MOP 2024

https://youtu.be/oPG4MHzpvcc

~r00tsOfUnity

See Also

2018 AIME I (Problems • Answer Key • Resources)
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