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Difference between revisions of "2024 SSMO Accuracy Round Problems/Problem 8"

(Created page with "==Problem== <math>ABCD</math> is a convex cyclic quadrilateral with <math>AB = 2, BC = 5, CD = 10,</math> and <math>AD = 11.</math> Let <math>W, Y, X,</math> and <math>Z</mat...")
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
We will use complex numbers. For all points <math>P,</math> let <math>p</math> denote it's complex number representation. Since <math>AB^2+AD^2 = BC^2+CD^2 = 125,</math> the diameter of the circumcircle of <math>ABCD</math> is <math>BD.</math> From <math>AB = 2, BC = 5, CD = 10,</math> and <math>AD = 11,</math> we have
 +
<cmath>\begin{align*}
 +
|a-b| &= 2\implies |a-b|^2 = 2^2\implies (a-b)\left(\frac{1}{a}-\frac{1}{b}\right) = \frac{4}{\frac{125}{4}}\implies \frac{a}{b}+\frac{b}{a} = 2-\frac{4}{\frac{125}{4}},\\
 +
|b-c| &= 2\implies |b-c|^2 = 5^2\implies (b-c)\left(\frac{1}{b}-\frac{1}{c}\right) = \frac{25}{\frac{125}{4}}\implies \frac{b}{c}+\frac{b}{c} = 2-\frac{25}{\frac{125}{4}},\\
 +
|c-d| &= 2\implies |c-d|^2 = 10^2\implies (c-d)\left(\frac{1}{c}-\frac{1}{d}\right) = \frac{100}{\frac{125}{4}}\implies \frac{c}{d}+\frac{c}{d} = 2-\frac{100}{\frac{125}{4}},\text{ and }\\
 +
|d-a| &= 2\implies |d-a|^2 = 11^2\implies (d-a)\left(\frac{1}{d}-\frac{1}{a}\right) = \frac{121}{\frac{125}{4}}\implies \frac{d}{a}+\frac{d}{a} = 2-\frac{121}{\frac{125}{4}}.
 +
\end{align*}</cmath>
 +
Now,
 +
<cmath>\begin{align*}
 +
WX^2 &= \left|\frac{a+b-c-d}{2}\right|^2 = \left(\frac{a+b-c-d}{2}\right)\left(\frac{\frac{1}{a}+\frac{1}{b}-\frac{1}{c}-\frac{1}{d}}{2}\right)\implies\\
 +
4WX^2 &= 4+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{d}+\frac{d}{c}\right)-\left(\frac{a}{c}+\frac{c}{a}\right)-\left(\frac{a}{d}+\frac{d}{a}\right)\\&-\left(\frac{b}{c}+\frac{c}{b}\right)-\left(\frac{b}{d}+\frac{d}{b}\right).
 +
\end{align*}</cmath>
 +
In the same manner,
 +
<cmath>\begin{align*}
 +
YZ^2 &= \left|\frac{-a+b+c-d}{2}\right|^2 = \left(\frac{-a+b+c-d}{2}\right)\left(\frac{-\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{d}}{2}\right)\implies\\
 +
4YZ^2 &= 4+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)-\left(\frac{a}{c}+\frac{c}{a}\right)-\left(\frac{c}{d}+\frac{d}{c}\right)\\&-\left(\frac{b}{a}+\frac{a}{b}\right)-\left(\frac{b}{d}+\frac{d}{b}\right).
 +
\end{align*}</cmath>
 +
So,
 +
<cmath>\begin{align*}
 +
4(WX^2-YZ^2) &= 2\left(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{d}+\frac{d}{c}\right)-\left(\frac{b}{c}+\frac{c}{b}\right)-\left(\frac{d}{a}+\frac{a}{d}\right)\right)\\
 +
&=2\left(\frac{4+100-25-121}{\frac{125}{4}}\right) = \frac{-336}{125}\implies\\
 +
|WX^2-YZ^2| &= \frac{84}{125}\implies 84+125 = \boxed{209}.
 +
\end{align*}</cmath>

Revision as of 14:34, 10 September 2025

Problem

$ABCD$ is a convex cyclic quadrilateral with $AB = 2, BC = 5, CD = 10,$ and $AD = 11.$ Let $W, Y, X,$ and $Z$ be the midpoints of sides $AB, BC, CD,$ and $DA$ respectively. If $|(WX^2-YZ^2)|$ can be expressed as $\frac{m}{n},$ for relatively prime positive integers $m$ and $n,$ find $m+n.$

Solution

We will use complex numbers. For all points $P,$ let $p$ denote it's complex number representation. Since $AB^2+AD^2 = BC^2+CD^2 = 125,$ the diameter of the circumcircle of $ABCD$ is $BD.$ From $AB = 2, BC = 5, CD = 10,$ and $AD = 11,$ we have \begin{align*} |a-b| &= 2\implies |a-b|^2 = 2^2\implies (a-b)\left(\frac{1}{a}-\frac{1}{b}\right) = \frac{4}{\frac{125}{4}}\implies \frac{a}{b}+\frac{b}{a} = 2-\frac{4}{\frac{125}{4}},\\ |b-c| &= 2\implies |b-c|^2 = 5^2\implies (b-c)\left(\frac{1}{b}-\frac{1}{c}\right) = \frac{25}{\frac{125}{4}}\implies \frac{b}{c}+\frac{b}{c} = 2-\frac{25}{\frac{125}{4}},\\ |c-d| &= 2\implies |c-d|^2 = 10^2\implies (c-d)\left(\frac{1}{c}-\frac{1}{d}\right) = \frac{100}{\frac{125}{4}}\implies \frac{c}{d}+\frac{c}{d} = 2-\frac{100}{\frac{125}{4}},\text{ and }\\ |d-a| &= 2\implies |d-a|^2 = 11^2\implies (d-a)\left(\frac{1}{d}-\frac{1}{a}\right) = \frac{121}{\frac{125}{4}}\implies \frac{d}{a}+\frac{d}{a} = 2-\frac{121}{\frac{125}{4}}. \end{align*} Now, \begin{align*} WX^2 &= \left|\frac{a+b-c-d}{2}\right|^2 = \left(\frac{a+b-c-d}{2}\right)\left(\frac{\frac{1}{a}+\frac{1}{b}-\frac{1}{c}-\frac{1}{d}}{2}\right)\implies\\ 4WX^2 &= 4+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{d}+\frac{d}{c}\right)-\left(\frac{a}{c}+\frac{c}{a}\right)-\left(\frac{a}{d}+\frac{d}{a}\right)\\&-\left(\frac{b}{c}+\frac{c}{b}\right)-\left(\frac{b}{d}+\frac{d}{b}\right). \end{align*} In the same manner, \begin{align*} YZ^2 &= \left|\frac{-a+b+c-d}{2}\right|^2 = \left(\frac{-a+b+c-d}{2}\right)\left(\frac{-\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{d}}{2}\right)\implies\\ 4YZ^2 &= 4+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)-\left(\frac{a}{c}+\frac{c}{a}\right)-\left(\frac{c}{d}+\frac{d}{c}\right)\\&-\left(\frac{b}{a}+\frac{a}{b}\right)-\left(\frac{b}{d}+\frac{d}{b}\right). \end{align*} So, \begin{align*} 4(WX^2-YZ^2) &= 2\left(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{d}+\frac{d}{c}\right)-\left(\frac{b}{c}+\frac{c}{b}\right)-\left(\frac{d}{a}+\frac{a}{d}\right)\right)\\ &=2\left(\frac{4+100-25-121}{\frac{125}{4}}\right) = \frac{-336}{125}\implies\\ |WX^2-YZ^2| &= \frac{84}{125}\implies 84+125 = \boxed{209}. \end{align*}

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