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== | ||
| Line 11: | Line 11: | ||
<cmath>(O_2X-CO)^2+(XC)^2 = (OO_2)^2. | <cmath>(O_2X-CO)^2+(XC)^2 = (OO_2)^2. | ||
</cmath>which is the same as | </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> | </cmath> | ||
Solving for <math>r_1</math> and <math>r_2,</math> we have that | Solving for <math>r_1</math> and <math>r_2,</math> we have that | ||
| Line 19: | Line 19: | ||
</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}, | |
</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> | ||
Latest revision as of 13:23, 3 July 2023
Problem
Circle 
has chord 
of length 
. Point 
lies on chord such that 
Circle 
with radius 
and 
with radius 
lie on two different sides of 
Both and are tangent to at and 
If the sum of the maximum and minimum values of 
is 
find 
.
Solution
Let 
be the radius of and let 
be the midpoint of and let 
Note that 
. WLOG assume that 
Since 
and 
we have 
By the Pythagorean Theorem, we have
![\[(O_1X+CO)^2+(XC)^2 = (OO_1)^2\]](http://latex.artofproblemsolving.com/2/4/2/242482d3d4550fae1c3b944f041b3bee1dc637c1.png)
![\[(O_2X-CO)^2+(XC)^2 = (OO_2)^2.\]](http://latex.artofproblemsolving.com/6/7/1/67103950ab39fd0b22bb87343b8af996f2c62f6b.png)
which is the same as
![\[(r_1+x)^2+25 = (r-r_1)^2\]](http://latex.artofproblemsolving.com/5/5/f/55fd948dbd73994089edb65b7687b88959849580.png)
![\[(r_2-x)^2+25 =(r-r_2)^2.\]](http://latex.artofproblemsolving.com/9/1/e/91e374d7e9f95dc8da6171429a99a5ac81f1168b.png)
Solving for and 
we have that
![\[r_1 = \frac{r^2-x^2-25}{2(r+x)}\]](http://latex.artofproblemsolving.com/a/0/b/a0b87b5097f448bd9fa4615cca7ba91305e191c2.png)
![\[r_2 = \frac{r^2-x^2-25}{2(r-x)}.\]](http://latex.artofproblemsolving.com/c/3/a/c3a9ec2a13f65f71d4f0b2a52df8471f585f5bf5.png)
Thus,
![\[r_1r_2 = \frac{((r^2-x^2)-5^2)^2}{4(r^2-x^2)} = \frac{784}{81},\]](http://latex.artofproblemsolving.com/2/6/3/263f3c6988b43d7ae1129ea87bf5688c24fb2fdc.png)
meaning that the minimum and maximum value of are both 
so the answer is 
