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2000 USAMO Problems/Problem 5

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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

Solution 1

Let the circumcenter of $\triangle ABC$ be $O$, and let the center of $\omega_k$ be $O_k$. $\omega_k$ and $\omega_{k-1}$ 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 $\omega_k$. It follows from $OA_k = OA_{k+1}, O_kA_k = O_kA_{k+1}, OO_k = OO_k \Longrightarrow \triangle OA_kO_k \cong \triangle OA_{k+1}O_k$ that $\angle OA_kO_k = \angle OA_{k+1}O_k$.

[asy] size(300); pathpen = linewidth(0.7); pen t = linetype("2 2"); pair A = (0,0), B=3*expi(1), C=(3.5)*expi(0); /* arbitrary points */ pair O=circumcenter(A,B,C), O1 = O + 5*( ((B+C)/2) - O ), O2 = IP(O -- O + 100*( ((A+C)/2) - O ), O1 -- O1 + 10*( C - O1 )); D(MP("A_3",A,SW)--MP("A_1",B,N)--MP("A_2",C,SE)--cycle); D(MP("O",O,NW)); D(MP("O_1",O1,E)); D(MP("O_2",O2)); D(O--B--O1--C--O--A--O2--C, linetype("4 4") + linewidth(0.7)); D(O1--O--O2,linetype("4 4") + linewidth(0.6)); D(CP(O1,C),t);D(CP(O2,C),t);D(circumcircle(A,B,C),t); [/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 $\omega_1$ and $\omega_7$ are the same circle since they have the same center and go through the same two points.

Solution 2

Using the collinearity of certain points and the fact that $A_k A_{k+1} O_k$ is isosceles, we quickly deduce that \[\angle{A_1 A_2 O_1} = 180^\circ - \angle{A_2} - (180^\circ - \angle{A_3}) + (180^\circ - \angle{A_1}) - (180^\circ - \angle{A_2}) + (180^\circ - \angle{A_3}) - (180^\circ - \angle{A_1}) + \angle{A_7 A_8 O_7}.\] From ASA Congruence we deduce that $A_1 A_2 O_1$ and $A_7 A_8 O_7$ are congruent triangles, and so $O_1 A_1 = O_7 A_7$, that is $\omega_1 = \omega_7$.

See Also

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

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