Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2023 SSMO Speed Round Problems/Problem 8

Revision as of 14:21, 3 July 2023 by Pinkpig (talk | contribs) (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...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

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 $\boxed{1649}.$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_SSMO_Speed_Round_Problems/Problem_8&oldid=195186"