The incircle $\omega$ of a triangle $ABC$ centered at $I$ touches $BC$ at point $D$. Let $P$ be the projection of the orthocenter of $ABC$ to the median from $A$. Prove that the circle $AIP$ and $\omega$ cut off equal chords on $AD$.
Problem
Source: Sharygin Finals 2023 9.4
Tags: geometry, Sharygin Geometry Olympiad, Sharygin 2023
keglesnit
02.08.2023 12:18
Let $N$ and $M$ be the midpoints of $AD$ and $BC$ , the incircle touches $AC$ and $AB$ at $E$ and $F$, $L=EF\cap BC$. We will prove that $N$ is on the radical axis of $\omega$ and $(AIP)$ which will finish the problem. Claim 1. $N$, $I$, and $M$ are colinear. Proof. Let $DI$ intersect $\omega$ at $D'$. Line $AD'$ intersects $BC$ at $A_1$ which is the touch point of the $A$-excircle with $BC$. It is well-known that $D$ and $A_1$ are symmetric wrt. $M$. Hence the claim follows by a homothety centered at $D$ with factor $0.5$. Let $Q=\overline{MIN}\cap (DIL)$ and $R=AD\cap \omega\neq D$. Claim 2. $R\in(DIL)$. Proof. Since $(DR,EF)=-1$ we get $LR$ tangent to $\omega$. Hence $\angle LRI=90^{\circ}=\angle LDI$ so the claim follows. Claim 3. $Q\in(AIP)$. Proof. We will prove that $APDL$ is cyclic. Then we would be done by power of a point from $M$. Indeed, note that $(BC;DL)=-1$ so $MB^2=MD\cap ML$. But also since $P$ is the humpty point, its reflection in $M$ will lie on $(ABC)$ so $MB\cdot MC=MP\cdot MA$. I.e. $APDL$ cyclic, so we are done.
starchan
03.08.2023 11:48
here's a somewhat motivatable solution described to me by mueller.25
First, a lemma:
Lemma: Let $ABC$ be a triangle and $X$ a point in its interior. Let $AX$ meet $BC$ at point $K$. Then for any circle $\omega$ through $A$ and $X$ the quantity \[\text{Pow}(B, \omega) \cdot CK + \text{Pow}(C, \omega) \cdot BK\]is fixed.
Proof: We'll use Lin of Pop. Define the function $f(X) = \text{Pow}(X, \omega) - XK^2$. This is linear in $X$. Thus \[f(B) \cdot CK + f(B) \cdot CK = BC \cdot f(K)\]This finishes the proof of the lemma as $K$ is fixed.
Let $\omega$ be the circumcircle of $API$. Let $p_b, p_c$ be the power of $B, C$ wrt $\omega$ respectively. Using the above lemma twice for $P$ and $I$, we can obtain the following facts (relatively painlessly):
$p_b+p_c = a^2$
$bp_b+cp_c = abc$
The problem is equivalent to proving that $\text{Pow}(D, \omega) = (s-a)^2$. Define the function $f(X) = \text{Pow}(X, \omega)-XD^2$. By Lin of pop: \[(s-c)f(B) + (s-b)f(C) = af(D)\]We can compute: \[(s-c)f(B)+(s-b)f(C) = (s-c)p_b - (s-c)(s-b)^2 + (s-b)p_c - (s-b)(s-c)^2 = (s-c)p_b+(s-b)p_c - a(s-b)(s-c)\]We have $(s-c)p_b + (s-b)p_c = (s-b-c)(p_b+p_c)+(bp_b+cp_c) = (s-b-c)a^2+abc$ and then \[(s-c)f(B)+(s-b)f(C) = (s-b-c)a^2+abc-a(s-b)(s-c) = a(s-a)^2\]after some (come on it's not that much effort) algebra. Thus $f(D) = (s-a)^2$ and the problem is solved.
VicKmath7
03.08.2023 19:46
Let $AD$ meet the incircle and $(IAP)$ at $S, R$ and let $T$ be the midpoint of $RS$. If $D'$ is the antipode of $D$ in the incircle, then $IT$ is midline in $\triangle ADD'$ and it is known that $IM \parallel AD'$, so $M, I, T$ are collinear. If $MI \cap (AIP)=Q$, then we will show that $DISQ$ cyclic, which will mean that $T$ has equal powers with respect to $\omega$ and $(IAP)$, i.e. that $AR=DS$. Indeed, if the tangent at $S$ to $\omega$ meets $BC$ at $V$, then $(B, C; D, V)=-1$, so $MD \cdot MV=MB^2=MP \cdot MA=MI \cdot MQ$, so $Q$ lies on the circle with diameter $IV$, as desired.