2022 SSMO Speed Round Problems/Problem 8

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 \begin{align*} (O_1X+CO)^2+(XC)^2 &= (OO_1)^2 \\ (O_2X-CO)^2+(XC)^2 &= (OO_2)^2. \end{align*} which is the same as \begin{align*} (r_1+x)^2+25 = (r-r_1)^2 \\ (r_2-x)^2+25 =(r-r_2)^2. \end{align*}

Solving for and we have that \begin{align*} r_1 &= \frac{r^2-x^2-25}{2(r+x)} \\

   r_2 &= \frac{r^2-x^2-25}{2(r-x)}.

\end{align*} Thus, meaning that the minimum and maximum value of are both so the answer is

\begin{center} \begin{asy}

   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, dir(145));
   dot("", b, dir(30));
   dot("", c, dir(90));
   dot("", x, dir(60));
   dot("", o, dir(45));
   dot("", o1);
   dot("", o2);

\end{asy} \end{center}