IMO Problem Shortlist

Let $k_1,k_2$ and $k_3$ be the incircles of triangles $ABM$, $MNC$, and $NDA$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_1$ which is different from $CB$ is also tangent to $k_3$.

Figure 1

To this end, let $X$ denote the point of intersection of $g$ and $h$. Then $ABCX$ and $ABCD$ are circumscribed quadrilaterals, whence $$CD-CX=(AB+CD)-(AB+CX)=(BC+AD)-(BC+AX)=AD-AX,$$ i.e. $$AX+CD=CX+AD$$ which in turn reveals that the quadrilateral $AXCD$ is also circumscribed. Thus $h$ touches indeed the circle $k_3$.

Moreover, we find that $\angle I_{3}C{I}_1=\angle I_{3}CX+\angle XC{I}_1=\frac{1}{2}(\angle DCX+\angle XCB)=\frac{1}{2}\angle DCB=\frac{1}{2}(180^{\circ }-\angle MCN)=180^{\circ }-\angle M{I}_{2}N=\angle I_3{I}_2{I}_1$, from which we conclude that $C,I_1,I_2,I_3$ are concyclic.

Let now $L_1$ and $L_3$ be the reflection points of $C$ with respect to the lines $I_2{I}_3$ and $I_1{I}_2$ respectively. Since $I_1{I}_2$ is the angle bisector of $\angle NMC$, it follows that $L_3$ lies on $g$. By analogous reasoning, $L_1$ lies on $g$.

Let $H$ be the orthocenter of $△I_1{I}_2{I}_3$. We have $\angle I_2{L}_3{I}_1=\angle I_{1}C{I}_2=\angle I_1{I}_3{I}_2=180^{\circ }-\angle I_{1}H{I}_2$, which entails that the quadrilateral $I_{2}H{I}_1{L}_3$ is cyclic. Analogously, $I_{3}H{L}_1{I}_2$ is cyclic.

Then, working with oriented angles modulo $180^{\circ }$, we have $$\angle L_{3}H{I}_2=\angle L_3{I}_1{I}_2=\angle I_2{I}_{1}C=\angle I_2{I}_{3}C=\angle L_1{I}_3{I}_2=\angle L_{1}H{I}_2,$$ whence $L_1,L_3$, and $H$ are collinear. By $L_1\ne L_3$, the claim follows.

---

Alternative solution.

We start by proving that $C,I_1,I_2$, and $I_3$ are concyclic.

Figure 2

To this end, notice first that $I_2,M,I_1$ are collinear, as are $N,I_2,I_3$ (see Figure 2). Denote by $\alpha ,\beta ,\gamma ,\delta$ the internal angles of $ABCD$. By considerations in triangle $CMN$, it follows that $\angle I_3{I}_2{I}_1=\frac{\gamma }{2}$. We will show that $\angle I_{3}C{I}_1=\frac{\gamma }{2}$, too. Denote by $I$ the incenter of $ABCD$. Clearly, $I_1\in BI$, $I_3\in DI$, $\angle I_{1}A{I}_3=\frac{\alpha }{2}$.

Using the abbreviation $[X,YZ]$ for the distance from point $X$ to the line $YZ$, we have because of $\angle BA{I}_1=\angle IA{I}_3$ and $\angle I_{1}AI=\angle I_{3}AD$ that $$\frac{[I_1,AB]}{[I_1,AI]}=\frac{[I_3,AI]}{[I_3,AD]}$$ Furthermore, consideration of the angle sums in $AIB$, $BIC$, $CID$ and $DIA$ implies $\angle AIB+\angle CID=\angle BIC+\angle DIA=180^{\circ }$, from which we see $$\frac{[I_1,AI]}{[I_3,CI]}=\frac{I_{1}I}{I_{3}I}=\frac{[I_1,CI]}{[I_3,AI]}$$ Because of $[I_1,AB]=[I_1,BC]$, $[I_3,AD]=[I_3,CD]$, multiplication yields $$\frac{[I_1,BC]}{[I_3,CI]}=\frac{[I_1,CI]}{[I_3,CD]}$$ By $\angle DCI=\angle ICB=\gamma /2$ it follows that $\angle I_{1}CB=\angle I_{3}CI$ which concludes the proof of the above statement.

Let the perpendicular from $I_1$ on $I_2{I}_3$ intersect $g$ at $Z$. Then $\angle M{I}_{1}Z=90^{\circ }-\angle I_3{I}_2{I}_1=90^{\circ }-\gamma /2=\angle MC{I}_2$. Since we have also $\angle ZM{I}_1=\angle I_{2}MC$, triangles $MZ{I}_1$ and $M{I}_{2}C$ are similar. From this one easily proves that also $M{I}_{2}Z$ and $MC{I}_1$ are similar. Because $C,I_1,I_2$, and $I_3$ are concyclic, $\angle MZ{I}_2=\angle M{I}_{1}C=\angle N{I}_{3}C$, thus $N{I}_{2}Z$ and $NC{I}_3$ are similar, hence $NC{I}_2$ and $N{I}_{3}Z$ are similar. We conclude $\angle Z{I}_3{I}_2=\angle I_{2}CN=90^{\circ }-\gamma /2$, hence $I_1{I}_2\perp Z{I}_3$. This completes the proof.