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

Latest revision as of 13:23, 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 $\boxed{1649}.$

@@ Line 11: / Line 11: @@
 <cmath>(O_2X-CO)^2+(XC)^2 = (OO_2)^2.
 </cmath>which is the same as
-<cmath>(r_1+x)^2+25 = (r-r_1)^2 \\
+<cmath>(r_1+x)^2+25 = (r-r_1)^2</cmath>
-(r_2-x)^2+25 =(r-r_2)^2.
+<cmath>(r_2-x)^2+25 =(r-r_2)^2.
 </cmath>
 Solving for <math>r_1</math> and <math>r_2,</math> we have that

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

Latest revision as of 13:23, 3 July 2023

Problem

Solution