2021 JMC 10 Problems/Problem 21

Problem

Two identical circles $\omega_{a}$ and $\omega_{b}$ with radius $1$ have centers that are $\tfrac{4}{3}$ units apart. Two externally tangent circles $\omega_1$ and $\omega_2$ of radius $r_1$ and $r_2$ respectively are each internally tangent to both $\omega_a$ and $\omega_b$ . If $r_1 + r_2 = \tfrac{1}{2}$ , what is $r_1r_2$ ?

$\textbf{(A) }\dfrac{1}{21}\qquad\textbf{(B) }\dfrac{1}{14}\qquad\textbf{(C) }\dfrac{5}{63}\qquad\textbf{(D) }\dfrac{2}{21}\qquad\textbf{(E) }\dfrac{1}{7}$

Solution

Let $A$ and $B$ be the centers of $\omega_a$ and $\omega_b$ respectively. Let $R=1$ be the radius of $\omega_a$ and $\omega_b$ and $D=\tfrac{4}{3}$ be the distance between $A$ and $B$ . Note that the centers of $\omega_1$ and $\omega_2$ , say $O_1$ and $O_2$ respectively, lie on a line that is both perpendicular to $AB$ and equidistant from $A$ and $B$ .

Because $\triangle AO_1 O_2 \cong \triangle BO_1 O_2$ , we have that $\tfrac{D}{2}$ is the length of the $A$ -altitude of $AO_1 O_2$ . We have $AO_1 = R-r_1$ , $AO_2 = R-r_2$ , and $O_1 O_2 =r_1 +r_2$ , so $\triangle AO_1 O_2$ 's perimeter is $2R$ . Thus, by Heron's Formula $[\triangle{AO_1 O_2}]=\sqrt{Rr_1 r_2 (R-r_1 -r_2)} =\tfrac{1}{2} \cdot \tfrac{D}{2} \cdot (r_1 + r_2)$ . Substituting known values, we have $\sqrt{1 \cdot r_1 r_2 (1-\tfrac{1}{2})} =\frac{\frac{4}{3}}{4} \cdot \frac{1}{2},$ whence $r_1 \cdot r_2= \tfrac{1}{18}$ .

$[asy] size(5cm); draw(circle((0,0),1/3),purple); draw(circle((0,-1/2),1/6),purple); draw(circle((-2/3,0),1)); draw(circle((2/3,0),1)); dot((0,0)); dot((0,-1/2)); dot((-2/3,0)); dot((2/3,0)); label("$A$", (-2/3,0), SW); label("$B$", (2/3,0), SE); draw((-2/3,0)--(2/3,0)); draw((-2/3,0)--(0,-1/2)--(2/3,0)); draw((0,-1/2)--(0,0), dotted); [/asy]$

Remark: In this specific case, $\triangle{AO_1 O_2}$ is actually a right triangle with lengths in the ratio $3:4:5$ , which is why the diagram has one of the centers lying on $AB$ .

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2021 JMC 10 Problems/Problem 21

Problem

Solution