Let $ABC$ be a triangle with bisectors $BE$ and $CF$ meet at $I$. Let $D$ be the projection of $I$ on the $BC$. Let M and $N$ be the orthocenters of triangles $AIF$ and $AIE$, respectively. Lines $EM$ and $FN$ meet at $P.$ Let $X$ be the midpoint of $BC$. Let $Y$ be the point lying on the line $AD$ such that $XY \perp IP$. Prove that line $AI$ bisects the segment $XY$. Proposed by Tran Quang Hung - Vietnam
Problem
Source: 2023 IGO Adcanced P4
Tags: geometry, bisects segment
starchan
18.01.2024 17:19
very nice problem, I had fun with this during the test
Let $\omega$ denote the incircle and let the projections of $A$ onto lines $BI, CI$ be denoted by $U, V$ respectively. Let $B_1, C_1$ denote the touch points of $\omega$ on sides $AC, AB$ respectively. We note that $M = AV \cap IC_1$ and that $N = AU \cap IB_1$. From Pascal on cyclic hexagon $B_1VIUC_1A$ we obtain that points $B_1C_1\cap UV, E, M$ are collinear. In particular, $EM$ passes through $B_1C_1 \cap UV$. Similarily, $FN$ passes through $B_1C_1 \cap UV$; forcing $P = B_1C_1 \cap UV$.
Now let $K$ denote the orthocentre of $BIC$. It is well known that the polars of $U, V$ wrt $\omega$ are the perpendiculars from $C, B$ onto lines $IB, IC$ respectively. It therefore follows that the pole of $UV$ wrt $\omega$ is $K$. Hence, the polar of $P = B_1C_1 \cap UV$ is the line $AK$. As a result, $IP \perp AK$, which implies that $XY \parallel AK$.
Next, let $R = B_1C_1 \cap BC$. Note that $(RD;BC) = -1$. As $D$ is the foot of perpendicular in $KBC$, $I$ the orthocentre and $KX$ the median, it follows that $IR \perp KX$. But $R$ is also the pole of line $AD$ wrt $\omega$, so $IR \perp AD$. This implies that $KX \parallel AD \equiv AY$. It follows that $AKXY$ is a parallelogram.
Let $Z$ be the point such that $KBZC$ is a parallelogram. It's well known that $IBCZ$ is cyclic, and thus \[\angle CIZ = \angle CBZ = \angle KCB = 90^{\circ}-\angle IBC = 90^{\circ}-B/2\]As a result $\angle AIC + \angle CIZ = 180^{\circ}$, implying that $A, I, Z$ are collinear. Therefore, if $AI$ meets $XY$ again at a point $M$, we obtain that \[\frac{MX}{XY} = \frac{MX}{AK} = \frac{ZX}{ZK} = \frac{1}{2}\]or that $M$ is the midpoint of $XY$.
Hence, line $AI$ bisects segment $XY$. $\square$
VicKmath7
18.01.2024 18:42
Amazing problem indeed. Solved with geogebra after the contest (basically I didn't really have the chance to try it seriously due to the impossible P2).
Let the line through $A$ perpendicular to $IP$ meet $BC$ at $K$ and let $L=AI \cap BC$. If $G$ is the midpoint of $XY$, projecting $(XY;G\infty)=-1$ through $A$ onto $BC$ transforms the problem to showing that $(KL;DX)=-1$. Let $R$ be a point on the $A$-altitude such that $AD'DR$ is a parallelogram, where $D'$ is the antipode of $D$ in the incircle and let $T$ be the midpoint of $AR$. If we show that $I,K,R$ are collinear, we will be done upon projecting $(AR;T\infty)=-1$ through $I$ onto $BC$ as $I,T,X$ are collinear ($IX \parallel AD'$ is well-known).
Redefine $IR \cap BC=K$; we will show that $AK \perp IP$. Let the incircle touch $AC,AB$ at $B_1, C_1$. Then there is a homothety at $P$ sending $\triangle FC_1M$ to $\triangle NB_1E$, so $P \in B_1C_1$. Hence, $P$ lies on the polar of $A$ with respect the incircle, so La Hire implies that $A$ lies on the polar of $P$. Hence, it's sufficient to show that $AK$ is the polar of $P$, or that $K$ lies on the polar of $P$. Again by La Hire, we want $P$ to lie on the polar of $K$, or that $PD \perp IR$.
To proceed, we will need a further characterisation of $P$. We claim that $P$ lies on the $A$-midline of $\triangle ABC$. Let $AM \cap CI=G, AN\cap BI=H$ and let $AG, AH$ meet $BC$ at $R,L$. Then by angle-chasing $ABRI, ACLI$ are cyclic, so $IA=IR=IL$ and thus $GH$ is the $A$-midline of $\triangle ABC$. Now by simple angle-chasing again the circles $(GMH),(B_1C_1M),(NB_1E)$ concur, so the lines $ME, B_1C_1, GH$ also concur at $P$ by radical axis, so we have proven the claim.
Now, let $PD \cap (I)=Z$ and let $U,V$ be the feet of the altitudes from $B,C$ in $\triangle BIC$ and let them meet at the orthocenter $W$ of $\triangle BIC$. By Sharygin 2019 Finals 9.7 we know that $P$ lies on the radical axis of the circle with diameter $BC$ and the incircle, so $PD \cdot PZ=PB_1 \cdot PC_1=PU \cdot PV$, so $Z \in(UDXV)$. Let $O$ be the center of this circle; then it suffices to show that $I, O, R$ are collinear, as $IO$ is the perpendicular bisector of $PZ$. Let $S$ be the midpoint of the minor arc $BC$ and let $S'$ be its reflection in $BC$. It is easy to see that $ARSS'$ and $WISS'$ are parallelograms, so $R, I, S'$ and $I, O, S'$ are collinear, as desired.
Eeightqx
19.01.2024 16:18
A great beautiful and elegant diagram!
Assume that the incircle touches $AC,AB$ at $B_0,C_0$, $H$ is the orthocenter of $\triangle BIC$.
Claim 1 $P\in B_0C_0$
Has been shown above.
Claim 2 $AH\parallel XY$
Has been shown above.
Assume that $ID\cap B_0C_0=S$.
Claim 3 $A,S,X$ are collinear.
Construct a line that passes $S$ and perpendicular to $ID$, call it $\ell$. If $\ell \cap AB=S_1,\ell \cap AC=S_2$, it's trival that $S_1, C_0, I, S$ are concyclic $\Longrightarrow$ $\angle SS_1I=\angle SC_0I$. Similarly, $\angle SS_2I=\angle SB_0I$$\Longrightarrow$$\angle SS_1I=\angle SS_2I$$\Longrightarrow$$IS_1=IS_2$. Combine the fact that $S_1S_2\perp ID$, we get $S$ is the midpoint of segment $S_1S_2$$\Longrightarrow$$A,S,X$ are collinear.
Claim 4 $AI$ bisects segment $XY$
By the Iran Lemma we have $B_0C_0,CI, BH$ are concurrent. Denote the point by $Q$. Similarly we have $R$.
Consider the complete quadrilateral $HQBICR$, we can easily see that $(H,I;S,D)=-1$, which shows that $AI$ bisects segment $XY$.
axolotlx7
24.02.2024 11:19
Nice problem A solution without introducing the orthocenter of $\Delta BIC$: Let $DE_0F_0$ be the intouch triangle, $Q = AN \cap BE$, $R = AM \cap CF$. Then by Iran Lemma, $Q \in DE_0$ and $R \in DF_0$, and also $QR$ is the $A$-midline. Noting that $AE_0IF_0$ is a kite, we have $(AF_0 \cap E_0I) - (AE_0 \cap F_0I) \parallel E_0F_0 \parallel FM \parallel EN$, so by Desargues on $\Delta EE_0N$ and $\Delta MF_0F$ we have $E_0F_0 \cap EM \cap FN = P$. Note that a homothety at $P$ sends $(EE_0NQ)$ to $(FF_0MR)$, and $E_0, F_0$ are antihomologous points. But$(E_0F_0RQ)$ is the circle with diameter $AI$, so $Q$ and $R$ are also antihomologous points. In particular, $P$ lies on $QR$, the $A$-midline. Let $AI$ meet $BC$ at $G$, the external $\angle A$-bisector meet $BC$ at $G'$, $E_0F_0$ meet $BC$ at $D'$, $S$ be the point on $BC$ such that $AS$ is the polar of $P$ w.r.t. to the incircle, and $S' = AP \cap BC$. Since $AG' \parallel E_0F_0$ and $P$ is the midpoint of $AS'$, by intercept theorem we have that $D'$ is the midpoint of $G'S'$. Since $(SS';BC)=(DD';BC)=(GG';BC)=(X \infty_{BC}; BC) = -1$, by inverting about the circle with diameter $BC$ we get that $(SG;DX)=(S'G',D\infty_{BC})=-1$. Also, $AS \perp IP \perp XY$, so projecting from $A$ onto $XY$ we have $(\infty_{AS},(AI \cap XY);YX)=(SG;DX)=-1$, which proves the desired result.