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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 8"

(Created page with "==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> C...")
 
Line 8: Line 8:
  
 
By the Pythagorean Theorem, we have
 
By the Pythagorean Theorem, we have
−
\begin{align*}
+
<cmath>(O_1X+CO)^2+(XC)^2 = (OO_1)^2</cmath>
−
(O_1X+CO)^2+(XC)^2 &= (OO_1)^2 \\
+
<cmath>(O_2X-CO)^2+(XC)^2 = (OO_2)^2.</cmath>
−
(O_2X-CO)^2+(XC)^2 &= (OO_2)^2.
 
−
\end{align*}
 
 
which is the same as
 
which is the same as
−
\begin{align*}
+
<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.
+
\
−
\end{align*}
 
−
 
 
 
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
−
\begin{align*}
+
<cmath>r_1 = \frac{r^2-x^2-25}{2(r+x)}</cmath>
−
r_1 &= \frac{r^2-x^2-25}{2(r+x)} \\
+
<cmath>r_2 = \frac{r^2-x^2-25}{2(r-x)}.</cmath>
−
    r_2 &= \frac{r^2-x^2-25}{2(r-x)}.
 
−
\end{align*}
 
 
Thus,  
 
Thus,  
 
<cmath>
 
<cmath>
Line 29: Line 23:
 
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>
  
−
\begin{center}
+
<center>
−
\begin{asy}
+
<asy>
 
     size(7cm);
 
     size(7cm);
 
     point a, b, c, x, o, t, o1, o2;
 
     point a, b, c, x, o, t, o1, o2;
Line 57: Line 51:
 
     filldraw(circle(o2, abs(o2-x)), opacity(0.2)+palered, lightred);
 
     filldraw(circle(o2, abs(o2-x)), opacity(0.2)+palered, lightred);
  
−
     dot("<math>A</math>", a, dir(145));
+
     dot("$A$", a, dir(145));
−
     dot("<math>B</math>", b, dir(30));
+
     dot("$B$", b, dir(30));
−
     dot("<math>C</math>", c, dir(90));
+
     dot("$C$", c, dir(90));
−
     dot("<math>X</math>", x, dir(60));
+
     dot("$X$", x, dir(60));
−
     dot("<math>O</math>", o, dir(45));
+
     dot("$O$", o, dir(45));
−
     dot("<math>O_1</math>", o1);
+
     dot("$O_1$", o1);
−
     dot("<math>O_2</math>", o2);
+
     dot("$O_2$", o2);
−
\end{asy}
+
</asy>
−
\end{center}
+
</center>

Revision as of 14:03, 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 $\boxed{1649}.$ 

    size(7cm);
    point a, b, c, x, o, t, o1, o2;
    a = (0,0);
    b = (18,0);
    c = (9,0);
    x = (4,0);
    o = (9, -3);

    circle cir = circle(o, abs(a-o));
    t = intersectionpoints(cir, line(x,o))[1];

    point[] o1o2 = intersectionpoints(ellipse(x, o, (x+t)/2), line(x, x+(0,1)));

    o1 = o1o2[0];
    o2 = o1o2[1];

    draw(o1--o2, red);
    draw(a--b, blue);
    draw(c--o, blue);
    
    filldraw(cir, opacity(0.2)+lightcyan, blue);
    // draw(ellipse(x, o, (x+t)/2));

    filldraw(circle(o1, abs(o1-x)), opacity(0.2)+palered, lightred);
    filldraw(circle(o2, abs(o2-x)), opacity(0.2)+palered, lightred);

    dot("$A$", a, dir(145));
    dot("$B$", b, dir(30));
    dot("$C$", c, dir(90));
    dot("$X$", x, dir(60));
    dot("$O$", o, dir(45));
    dot("$O_1$", o1);
    dot("$O_2$", o2);
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