2001 Pan African MO Problems/Problem 6

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Let $S_1$ be a semicircle with centre $O$ and diameter $AB$.A circle $C_1$ with centre $P$ is drawn, tangent to $S_1$, and tangent to $AB$ at $O$. A semicircle $S_2$ is drawn, with centre $Q$ on $AB$, tangent to $S_1$ and to $C_1$. A circle $C_2$ with centre $R$ is drawn, internally tangent to $S_1$ and externally tangent to $S_2$ and $C_1$. Prove that $OPRQ$ is a rectangle.

Solution

We use Cartesian coordinates, setting $O = (0,0), B = (1,0)$, and assume without loss of generality that $Q$ is closer to $B$ than to $A$ and that the $y$-coordinate of $P$ is positive. Then we compute $P = (0, \frac{1}{2})$ (because that is the only choice that allows for tangency to both $S_1$ and the line $AB$). Letting $r_Q$ be the radius of $S_2$, we find that $Q = (1 - r_Q, 0)$. However, since $C_1$ and $S_2$ are tangent, we know that $PQ = \frac{1}{2} + r_Q$; using the Pythagorean theorem, we then solve $(\frac{1}{2})^2 + (1 - r_Q)^2 = (\frac{1}{2} + r_Q)^2$, which solves to $r_Q = \frac{1}{3}$, so that $Q = (\frac{2}{3}, 0)$.

Finally, we know that $R$ must now be uniquely determined. If $OPRQ$ was a rectangle, then $R$ would have to be located at $(\frac{2}{3}, \frac{1}{2})$, so we only need check that a circle centered there is tangent to $S_1, S_2, C_1$. Letting $r_R$ denote the radius of $C_2$, we know from tangencies that we must have $PR = \frac{1}{2} + r_R, QR = \frac{1}{3} + r_R, OR = 1 - r_R$. These equations are all satisfied for our desired choice of $R$ and the value $r_R = \frac{1}{6}$, so we conclude that $OPRQ$ is rectangular and we are done.

See also

2001 Pan African MO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
Last Problem
All Pan African MO Problems and Solutions
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