2000 USAMO Problems/Problem 5

Problem

Let $A_1A_2A_3$ be a triangle and let $\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\omega_2, \omega_3, \dots, \omega_7$ such that for $k = 2, 3, \dots, 7,$ $\omega_k$ is externally tangent to $\omega_{k - 1}$ and passes through $A_k$ and $A_{k + 1},$ where $A_{n + 3} = A_{n}$ for all $n \ge 1$ . Prove that $\omega_7 = \omega_1.$

Solution

Let the circumcenter of $\triangle ABC$ be $O$ , and let the center of $w_k$ be $O_k$ . $w_k$ and $w_{k-1}$ are externally tangent at the point $A_k$ , so $O_k, A_k, O_{k-1}$ are collinear.

$O$ is the intersection of the perpendicular bisectors of $\overline{A_1A_2}, \overline{A_2A_3}, \overline{A_3A_1}$ , and each of the centers $O_k$ lie on the perpendicular bisector of the side of the triangle that determines $w_k$ . It follows from $\triangle OA_kO_k \cong \triangle OA_{k+1}O_k$ that $\angle OA_kO_k = \angle OA_{k+1}O_k$ .

$[asy] [/asy]$

Since $O, A_k$ , and the perpendicular bisector of $\overline{A_kA_{k+1}}$ are fixed, the angle $OA_kO_k$ determines the position of $O_k$ (since $O_k$ lies on the perpendicular bisector). Let $\theta_k = m\angle OA_kO_k$ ; then, $\theta_i = \theta_j$ and $i \equiv j \pmod{3}$ together imply that $O_i \equiv O_j$ .

Now $\theta_1 = \angle OA_1O_1 = \angle OA_2O_1 = 180 - \angle OA_2O_2 = 180 - \theta_2$ (due to collinearility). Hence, we have the recursion $\theta_k = 180 - \theta_{k-1}$ , and so $\theta_k = \theta_{k-2}$ . Thus, $\theta_{1} = \theta_{7}$ .

$\theta_{1} = \theta_{7}$ implies that $O_1 \equiv O_7$ , and circles $w_1$ and $w_7$ are the same circle since they have the same center and go through the same two points.

2000 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6
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2000 USAMO Problems/Problem 5

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