Problem 1965

(a) Let $I$ be the incentre of $\triangle ABC$, and let $J$ be the point of intersection of $AI$ and $C_1$. Then, the line $AJ$ is the angle bisector of $\angle DAE$ and the triangles $ADI$ and $AEI$ are congruent. In particular, $\angle EIA = \angle AID$. We have $\angle EIJ = 2\angle EDJ$ as both are subtended by the same arc. Also, $\angle JID = 2\angle JDA$ (chord tangent angle). Thus $\angle JDA = \angle EDJ$ and $DJ$ is the angle bisector of $\angle EDA$. So $J$ is the incentre of $\triangle ADE$, and it lies on $C_1$.

(b) First Solution: Let $r_1$ and $r_2$ denote the radiuses of the circles $C_1$ and $C_2$. Let $O$ be the midpoint of the segment $MN$ and $R$ the midpoint of the segment $DE$. $PQ \parallel MN$ as they are both perpendicular on $AI$. Thus $MNPQ$ is an isosceles trapezium with bases $MN$ and $PQ$. It is a rectangle if and only if the lengths of these segments are equal. We calculate them in terms of $r_1$ and $r_2$.

In $\triangle OIM$, we have $|OM| = |MI|\sin(\angle I)$, while from the cosine formula in $\triangle IMJ$, $\cos(\angle I) = \frac{2r_1^2 - r_2^2}{2r_1^2} = 1 - \frac{r_1^2}{2r_1^2}$, thus
$$\sin(\angle I) = \sqrt{1 - \cos^2(\angle I)} = \frac{r_2\sqrt{4r_1^2 - r_2^2}}{2r_1^2} \quad \text{and so}$$
$$|MN| = 2|OM| = \frac{r_2\sqrt{4r_1^2 - r_2^2}}{r_1}.$$

On the other hand, $\triangle JPQ$ and $\triangle IDE$ are similar as they have parallel sides. This implies $|PQ|/|DE| = r_2/r_1$ and so $|PQ| = \frac{r_2}{r_1}|DE|$. We also have $|DE|/2 = |ER| = |QE|$ as both $ER$ and $EQ$ are tangent to $C_2$. From the trapezium $JQEI$ with right angles at $Q$ and $E$ we get
$$|QE|^2 = r_1^2 - (r_1 - r_2)^2 = 2r_1r_2 - r_2^2.$$
$$\text{Thus } |PQ| = \frac{r_2}{r_1}|DE| = \frac{2r_2}{r_1}|QE| = \frac{2r_2}{r_1}\sqrt{2r_1r_2 - r_2^2}.$$

From the above,
$$\begin{align*} |MN| = |PQ| &\iff \sqrt{4r_1^2 - r_2^2} = 2\sqrt{2r_1r_2 - r_2^2} \\ &\iff (2r_1 - r_2)(2r_1 + r_2) = 4r_2(2r_1 - r_2) \\ &\iff 2r_1 = 3r_2. \end{align*}$$

(b) Second Solution: Let $r_1$ and $r_2$ denote the radiuses of the circles $C_1$ and $C_2$. Let $O$ be the midpoint of the segment $MN$, $F$ the midpoint of the segment $DE$ and $R$ the midpoint of $PQ$. $PQ \parallel MN$ as they are both perpendicular on $AI$. Thus $MNPQ$ is an isosceles trapezium with bases $MN$ and $PQ$. It is a rectangle if and only if the lengths of these segments are equal. Using Pythagoras, this is easily seen to be equivalent to $|OJ| = |RJ|$.

The line through $M$ and $N$ is the radical axis of the two circles $C_1$ and $C_2$. The point $O$ has therefore the same power with respect to these circles, i.e.
$$|OJ| \cdot (2r_1 - |OJ|) = (r_2 - |OJ|)(r_2 + |OJ|), \quad \text{hence} \quad |OJ| = \frac{r_2^2}{2r_1}.$$

As $QR \parallel EF$ and $QJ \parallel EI$, the triangles $RJQ$ and $FIE$ are similar and so
$$\frac{|RJ|}{r_2} = \frac{|FI|}{r_1} = \frac{r_1 - r_2}{r_1}, \quad \text{hence} \quad |RJ| = \frac{r_2}{r_1}(r_1 - r_2). \quad \text{Finally we see}$$
$$|OJ| = |RJ| \iff \frac{r_2^2}{2r_1} = \frac{r_2}{r_1}(r_1 - r_2) \iff r_2 = 2(r_1 - r_2) \iff 3r_2 = 2r_1.$$