Baltic Way 1991 · Problem 16

Geometry

Let two circles $C_{1}$ and $C_{2}$ (with radii $r_{1}$ and $r_{2}$ ) touch each other externally, and let $l$ be their common tangent. A third circle $C_{3}$ (with radius $r_{3}<\min \left(r_{1}, r_{2}\right)$ ) is externally tangent to the two given circles and tangent to the line $l$ . Prove that

\frac{1}{\sqrt{r_{3}}}=\frac{1}{\sqrt{r_{1}}}+\frac{1}{\sqrt{r_{2}}}

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Review

Topics

Circles and tangency

Solutions

Solution

Let $O_{1}, O_{2}, O_{3}$ be the centres of the circles $C_{1}, C_{2}, C_{3}$ , respectively. Let $P_{1}, P_{2}, P_{3}$ be the perpendicular projections of $O_{1}, O_{2}, O_{3}$ onto the line $l$ and let $Q$ be the perpendicular projection of $O_{3}$ onto the line $P_{1} O_{1}$ (see Figure 2). Then $\left|P_{1} P_{3}\right|^{2}=\left|Q O_{3}\right|^{2}=\left|O_{1} O_{3}\right|^{2}-\left|Q O_{1}\right|^{2}=\left(r_{1}+r_{3}\right)^{2}-\left(r_{1}-r_{3}\right)^{2}=4 r_{1} r_{3}$ . Similarly we get $\left|P_{1} P_{2}\right|^{2}=4 r_{1} r_{2}$ and $\left|P_{2} P_{3}\right|^{2}=4 r_{2} r_{3}$ . Since $\left|P_{1} P_{2}\right|=\left|P_{1} P_{3}\right|+\left|P_{2} P_{3}\right|$ we have $\sqrt{r_{1} r_{2}}=$ $\sqrt{r_{1} r_{3}}+\sqrt{r_{2} r_{3}}$ , which implies the required equality.

Official solution diagram for Baltic Way 1991 Problem 16 (Figure 1).

Figure 1

Official solution diagram for Baltic Way 1991 Problem 16 (Figure 2).

Figure 2

Official solution diagram for Baltic Way 1991 Problem 16 (Figure 3).

Figure 3

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AoPS · Baltic Way 1991