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2022 SSMO Speed Round Problems/Problem 8

Revision as of 14:03, 3 July 2023 by Pinkpig (talk | contribs)

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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