Ireland

(a) Let $I$ be the incentre of $△ABC$, 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 $△ADE$, 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 $△OIM$, we have $|OM|=|MI|\sin (\angle I)$, while from the cosine formula in $△IMJ$, $\cos (\angle I)=\frac{2r_1^2-r_2^2}{2r_1^2}=1-\frac{r_1^2}{2r_1^2}$, thus $$\sin (\angle I)=\sqrt{1-{\cos }^2(\angle I)}=\frac{r_2\sqrt{4r_1^2-r_2^2}}{2r_1^2}\text{and so}$$ $$|MN|=2|OM|=\frac{r_2\sqrt{4r_1^2-r_2^2}}{r_1}.$$

On the other hand, $△JPQ$ and $△IDE$ 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 $$|QE{|}^2=r_1^2-(r_1-r_2{)}^2=2r_1{r}_2-r_2^2.$$ $$\text{Thus }|PQ|=\frac{r_2}{r_1}|DE|=\frac{2r_2}{r_1}|QE|=\frac{2r_2}{r_1}\sqrt{2r_1{r}_2-r_2^2}.$$

From the above, $$\begin{array}{rl}|MN|=|PQ|& ⟺\sqrt{4r_1^2-r_2^2}=2\sqrt{2r_1{r}_2-r_2^2}\\ & ⟺(2r_1-r_2)(2r_1+r_2)=4r_2(2r_1-r_2)\\ & ⟺2r_1=3r_2.\end{array}$$

(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. $$|OJ|\cdot (2r_1-|OJ|)=(r_2-|OJ|)(r_2+|OJ|),\text{hence}|OJ|=\frac{r_2^2}{2r_1}.$$

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