2019 IMO SL #G2

Let $ABC$ be an acute-angled triangle and let $D, E$, and $F$ be the feet of altitudes from $A, B$, and $C$ to sides $BC, CA$, and $AB$, respectively. Denote by $\omega_B$ and $\omega_C$ the incircles of triangles $BDF$ and $CDE$, and let these circles be tangent to segments $DF$ and $DE$ at $M$ and $N$, respectively. Let line $MN$ meet circles $\omega_B$ and $\omega_C$ again at $P \ne M$ and $Q \ne N$, respectively. Prove that $MP = NQ$.

Denote $I_1,I_2$ as the centers of $\omega_b,\omega_c$, respectively, and denote $N',M'$ as the feet from $I_1,I_2$ to $BC$, respectively.

First off, note $$\angle NM'Q = \tfrac12 \widehat{QN} = \angle QNE = \angle MND.$$

Next, notice that $$\angle NQM'=\tfrac12 \widehat{M'N} = \angle DNM'=90-A/2.$$

We now apply the Law of Sines to $\triangle NQM'$ to obtain $$\qquad \frac{NQ}{\sin \angle NM'Q} = \frac{M'N}{\sin \angle NQM'} \implies \frac{NQ}{\sin \angle MND} = \frac{M'N}{\sin (90-A/2)}$$

$$\implies NQ = \frac{M'N\sin \angle MND}{\sin (90-A/2)}=\frac{M'N}{ND} \cdot \frac{ND\sin \angle MND}{\sin (90-A/2)}.$$

Similarly, we see that $MP = \frac{MN'}{MD} \cdot \frac{MD\sin \angle NMD}{\sin(90-A/2)}.$

However, since $\triangle DMN' \sim \triangle DNM'$, we have $$\frac{M'N}{ND}=\frac{MN'}{MD},$$ and by Law of Sines on $\triangle MND$, we get $$ND\sin \angle MND=MD\sin \angle NMD,$$ we conclude that $MP=NQ$. $\square$

1. Anonymous2/04/2022

wao nice

2. Anonymous2/05/2022

lovely sol