Let $\triangle ABC$ a scalene triangle with incenter $I$, circumcenter $O$ and circumcircle $\Gamma$. The incircle of $\triangle ABC$ is tangent to $BC, CA$ and $AB$ at points $D, E$ and $F$, respectively. The line $AI$ meet $EF$ and $\Gamma$ at $N$ and $M\neq A$, respectively. $MD$ meet $\Gamma$ at $L\neq M$ and $IL$ meet $EF$ at $K$. The circumference of diameter $MN$ meet $\Gamma$ at $P\neq M$. Prove that $AK, PN$ and $OI$ are concurrent.
Problem
Source: 2018 Olympic Revenge, Problem 2
Tags: geometry
WizardMath
11.02.2018 22:32
From here, $L$ is the foot of $D$ on $EF$. Also, $M_A, N, P$ are collinear, by the $90^\circ$ at $P$. Now observe that $ABC \cup M_A$ is homothetic to the orthic triangle of $DEF$ along with the midpoint of $EF$, from the npc of $DEF$, so the homothety center $AK \cap PN$ is on the line joining $O$ and the npc of $DEF$. But the Euler line of $DEF$ is $OI$, so we are done.
UK2019Project
17.03.2019 05:13
Let $T$ be the midpoint of arc $BAC$. Let $H$ be the orthocenter of $\triangle DEF$. Let $U$ be the midpoint of $BC$. Fact 1: $I, K, L$ are collinear. Fact 2: $P, N, T$ are collinear. Fact 3: $H, I, O$ are collinear. Then we just want to prove that $AK, TN, HI$ are concurrent. Since $HK \parallel IA$, $KN \parallel AT$, it is enough to prove that $\triangle HKN \simeq \triangle IAT$. It is clear that $\triangle HFE \simeq \triangle IBC \implies \triangle HKN \simeq \triangle IDU.$ And $MI^2=MU \cdot MT \implies \angle TIA=\angle TUI= \angle DIU$, so $\triangle IDU \simeq \triangle IAT$ and we are done.
Arefe
01.06.2020 11:33
UK2019Project wrote: Let $T$ be the midpoint of arc $BAC$. Let $H$ be the orthocenter of $\triangle DEF$. Let $U$ be the midpoint of $BC$. Fact 1: $I, K, L$ are collinear. Fact 2: $P, N, T$ are collinear. Fact 3: $H, I, O$ are collinear. Then we just want to prove that $AK, TN, HI$ are concurrent. Since $HK \parallel IA$, $KN \parallel AT$, it is enough to prove that $\triangle HKN \simeq \triangle IAT$. It is clear that $\triangle HFE \simeq \triangle IBC \implies \triangle HKN \simeq \triangle IDU.$ And $MI^2=MU \cdot MT \implies \angle TIA=\angle TUI= \angle DIU$, so $\triangle IDU \simeq \triangle IAT$ and we are done. You can say the last part more easily : We know $KN||AT , HK||IA$ so it's enough to say $\frac{KN}{AT}=\frac{KH}{AI}$ $\frac{KH}{AI}=\frac{KH}{HE}\times\frac{HE}{r}\times\frac{r}{AI}=2sin{\frac{A}{2}}.sin{\frac{B}{2}}.sin{\frac{C}{2}}$ $\frac{KN}{AT}=\frac{r\times{sin{\frac{|B-C|}{2}}}}{2R\times{sin{\frac{|B-C|}{2}}}}=\frac{r}{2R}=\frac{r\times{sin{\frac{A}{2}}}}{BM}=\frac{IB}{BM}.sin{\frac{A}{2}}.sin{\frac{B}{2}}=2sin{\frac{A}{2}}.sin{\frac{B}{2}}.sin{\frac{C}{2}}$ So we get what we want