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Difference between revisions of "2023 SSMO Speed Round Problems/Problem 8"

(Created page with "==Problem== ==Solution== Let <math>r</math> be the radius of <math>\omega</math> and let <math>C</math> be the midpoint of <math>AB</math> and let <math>OC = x.</math> Note...")
 
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==Problem==
 
==Problem==
−
 
+
Circle <math>\omega</math> has chord <math>AB</math> of length <math>18</math>. Point <math>X</math> lies on chord <math>AB</math> such that <math>AX = 4.</math> Circle <math>\omega_1</math> with radius <math>r_1</math> and <math>\omega_2</math> with radius <math>r_2</math> lie on two different sides of <math>AB.</math> Both <math>\omega_1</math> and <math>\omega_2</math> are tangent to <math>AB</math> at <math>X</math> and <math>\omega.</math> If the sum of the maximum and minimum values of <math>r_1r_2</math> is <math>\frac{m}{n},</math> find <math>m+n</math>.
  
 
==Solution==
 
==Solution==
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</cmath>Thus,  
 
</cmath>Thus,  
 
<cmath>
 
<cmath>
−
r_1r_2 = \frac{((r^2-x^2)-5^2)^2}{4(r^2-x^2)} = \frac{784}{81},
+
r_1r_2 = \frac{((r^2-x^2)-5^2)^2}{4(r^2-x^2)} = \frac{784}{81},
 
</cmath>
 
</cmath>
 
meaning that the minimum and maximum value of <math>r_1r_2</math> are both <math>\frac{784}{81}</math> so the answer is <math>\boxed{1649}.</math>
 
meaning that the minimum and maximum value of <math>r_1r_2</math> are both <math>\frac{784}{81}</math> so the answer is <math>\boxed{1649}.</math>

Revision as of 13:22, 3 July 2023

Problem

Circle $\omega$ has chord $AB$ of length $18$. Point $X$ lies on chord $AB$ such that $AX = 4.$ Circle $\omega_1$ with radius $r_1$ and $\omega_2$ with radius $r_2$ lie on two different sides of $AB.$ Both $\omega_1$ and $\omega_2$ are tangent to $AB$ at $X$ and $\omega.$ If the sum of the maximum and minimum values of $r_1r_2$ is $\frac{m}{n},$ find $m+n$.

Solution

Let $r$ be the radius of $\omega$ and let $C$ be the midpoint of $AB$ and let $OC = x.$ Note that $r^2 - x^2 = 81$. WLOG assume that $r_2\geq r_1.$

Since $AX = 4$ and $AB = 18,$ we have $XC = \frac{AB}{2}-AX = 5.$

By the Pythagorean Theorem, we have \[(O_1X+CO)^2+(XC)^2 = (OO_1)^2\] \[(O_2X-CO)^2+(XC)^2 = (OO_2)^2.\]which is the same as \[(r_1+x)^2+25 = (r-r_1)^2 \\ (r_2-x)^2+25 =(r-r_2)^2.\] Solving for $r_1$ and $r_2,$ we have that \[r_1 = \frac{r^2-x^2-25}{2(r+x)}\] \[r_2 = \frac{r^2-x^2-25}{2(r-x)}.\]Thus, \[r_1r_2 = \frac{((r^2-x^2)-5^2)^2}{4(r^2-x^2)} = \frac{784}{81},\] meaning that the minimum and maximum value of $r_1r_2$ are both $\frac{784}{81}$ so the answer is $\boxed{1649}.$

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