Let ${I}$ be the incenter of $\triangle {ABC}$ and ${BX}$, ${CY}$ are its two angle bisectors. ${M}$ is the midpoint of arc $\overset{\frown}{BAC}$. It is known that $MXIY$ are concyclic. Prove that the area of quadrilateral $MBIC$ is equal to that of pentagon $BXIYC$. Proposed by Dominik Burek - Poland
Problem
Source: IGO 2023 Advanced P2
Tags: geometry, incenter, angle bisector
VicKmath7
18.01.2024 15:01
This one was indeed really hard.
Let the lines through $Y, X$ parallel to $BI, CI$ meet at $Z$. Then $S_{BZC}=S_{BCXIY}$, so we should show that $S_{BZC}=S_{MBIC}$, or that $\text{dist}(I, BC)+\text{dist}(Z, BC)=\text{dist}(M, BC)$. Let $IZ \cap XY=N$; then the main idea is that $2\text{dist}(N,BC)=\text{dist}(I, BC)+\text{dist}(Z, BC)$ (which follows by noticing a midline in a rectangular trapezoid); the rest is straightforward. Indeed, we are left to show that $\text{dist}(M, BC)=2\text{dist}(N,BC)$, which means that we are left to show that the reflection of $M$ with respect to $N$ lies on $BC$. Notice that if a point $W \in BC$ satisfies the requirement that $MXWY$ is a parallelogram, then $\angle MXW=180^{\circ}-\angle XMY=\angle MBC$, so $MXWB$ is cyclic. With that being said, define $W=(MXB) \cap (MYC)$; by reversing the angle chase above, $XW \parallel MY$ and $YW \parallel MX$, so $MXWY$ is a parallelogram. Finally, the condition that $MXIY$ is cyclic translates to $\angle BWM+\angle CWM=\angle IXM+\angle IYM=180^{\circ}$, so $W \in BC$, as desired.
LoloChen
18.01.2024 15:29
Really nightmare to me in IGO, it took me 2.5 hours during the test. Though what I got at last seems simple, it is really hard. Solution at the contest: Let the pedal of ${M}$ to lines $IB,IC$ be $G, H$. Obviously $\triangle MGX \sim \triangle MHY$, so $\frac{XG}{YH}=\frac{MG}{MH}=\frac{\sin \frac{\angle C}{2}}{\sin \frac{\angle B}{2}}$, so $S_{GXC}=S_{HYB}$. $$S_{BYIXC}+S_{BIC}=S_{BYC}+S_{BXC}=S_{BHC}+S_{BGC}=\frac{BC \cdot BM \cdot \sin \frac{\angle B}{2} \cos \frac{\angle C}{2}+BC \cdot CM \cdot \sin \frac{\angle C}{2} \cos \frac{\angle B}{2}}{2}=\frac{BC \cdot BM \cdot \cos \frac{\angle A}{2}}{2}=S_{MBC}=S_{MBIC}+S_{BIC}$$Done.
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starchan
18.01.2024 17:18
here's my solution to this problem. highly misplaced, definitely should've been at place 4 or higher
First we remark that circles $\odot(BXM)$ and $\odot(CYM)$ meet on $BC$ again at the same point. Indeed if $S_1, S_2$ denote the corresponding intersections, we have \[\angle BS_1M + \angle CS_2M = \angle BXM + \angle CYM = \angle IXM + \angle IYM = 180^{\circ}\]forcing $S_1 \equiv S_2$. Let $S$ denote this common intersection. Now we note that \[\angle MYS = 180^{\circ}-\angle MCS = 90^{\circ}+A/2 = 180^{\circ}-\angle YMX\]implying that $MX \parallel YS$. Similarily one obtains $MY \parallel SX$. Hence $MXSY$ is a parallelogram.
Let $f(P)$ denote the (signed) area of triangle $PBC$. We note that $f$ is linear in the location of $P$ and since $MXSY$ is a parallelogram, we must have \[f(M)+f(S) = f(X)+f(Y)\]Noting that the areas of $BYIXC$ and $MBIC$ are $f(X)+f(Y)-f(I)$ and $f(M)-f(I)$ respectively, and that $f(S) = 0$ the result follows immediately now. (observe that the signs don't affect us here since all considered point $P$ are on the same side of $BC$ as $A$)
Thus pentagon $BYIXC$ and quadrilateral $MBIC$ have the same area. $\square$
GeoKing
18.01.2024 19:37
really nice and elegant problem. I have solved it now . I will post sol later.
GeoKing
18.01.2024 20:45
Let $D$ be midpoint of $XY$ , $N$ be midpoint of $BC$. $S_{BYIXC}=S_{MBIC} \iff S_{BYC}+S_{BXC}=S_{MBC} \iff S_{BDC}=\frac{S_{MBC}}{2} \iff D$ lies on $M$ midline of $\Delta BMC \iff DM=DN$. Let $IM$ meet $(BIC)$ again at $T$, $(BICT)$ and $(XIYM)$ meet again at $J$. Observe that $BNCT \sim XDYM$ and $J$ is their spiral center.Observe that $\frac{DN}{MT}=\frac{DJ}{MJ} \implies DN= \frac {MT \cdot DJ}{MJ}$ and $\frac{DM}{NT}=\frac{DJ}{NJ} \implies DM=\frac{NT \cdot DJ}{NJ}$. So $DM=DN \iff \frac{NT}{NJ}=\frac{MT}{MJ} \iff \frac{NT}{MT}=\frac{NJ}{MJ}$ which is true since $J,T$ lie on $(BIC)$ i.e. an appolonian circle wrt segment $NM$. So we are done
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Strudan_Borisov
19.01.2024 17:43
Here is the solution I submitted on the contest. First of all, we'll transform the area condition. We'll denote with $dist(L,l)$ the distance from $L$ to the line $l$. \begin{align*} S_{BCXIY}&=S_{MBIC}\iff \\ S_{BIC}+S_{BCXIY}&=S_{MBIC}+S_{BIC}\iff\\ S_{BCX}+S_{BYC}&=S_{BMC}\iff\\ dist(X,BC)+dist(Y,BC)&=dist(M,BC). \end{align*}For the cyclic quadrilateral $MXIY$ we'll apply the cotangent technique. \[\triangle MYC\colon \cot\angle MYC = \frac{\frac{CY}{MC}-\cos{\frac{\beta}{2}}}{\sin{\frac{\beta}{2}}}.\]\[\triangle MXB\colon\cot MXB=\frac{\frac{BX}{MB}-\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}}.\]We have that $\angle MYC+\angle MXB=180^{\circ}$, so $\cot\angle MYC=-\cot\angle MXB$. Now \[\frac{\frac{CY}{MC}-\cos{\frac{\beta}{2}}}{\sin{\frac{\beta}{2}}}=\frac{\frac{BX}{MB}-\cos{\frac{\gamma}{2}}}{\sin{\frac{\gamma}{2}}}\iff\]\[\frac{\displaystyle\frac{dist(Y,BC)/\sin{\frac{\gamma}{2}}}{dist(M,BC)/\cos{\frac{\alpha}{2}}}-\cos{\frac{\beta}{2}}}{\displaystyle\sin{\frac{\beta}{2}}} = \frac{\displaystyle\cos{\frac{\gamma}{2}}-\displaystyle\frac{dist(X,BC)/\sin{\frac{\beta}{2}}}{dist(M,BC)/\cos{\frac{\alpha}{2}}}}{\displaystyle\sin{\frac{\gamma}{2}}}\iff\]\[\cot{\frac{\beta}{2}}+\cot{\frac{\gamma}{2}}=\frac{dist(X,BC)+dist(Y,BC)}{dist(M,BC)}.\frac{\cos{\frac{\alpha}{2}}}{\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}}.\]The left hand side equals $\displaystyle\frac{\sin\left(\frac{\beta+\gamma}{2}\right)}{\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}}=\frac{\cos{\frac{\alpha}{2}}}{\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}}\implies \frac{dist(X,BC)+dist(Y,BC)}{dist(M,BC)}=1,$ which finishes the solution.
X.Allaberdiyev
19.01.2024 20:07
What's up to IGO advanced? Problems are so ugly, problem 2 is very complicated(spend 4 hours on it during the contest and didn't solve). It will be very good if they found beautiful GEOMETRY problem(not related with area or other stuff). Problem 3 was combinatorial, so I had to give my all time to problem 2 which is very ugly problem (since I know that I can't solve problem 4).
GeoKing
19.01.2024 20:51
X.Allaberdiyev wrote: What's up to IGO advanced? Problems are so ugly, problem 2 is very complicated(spend 4 hours on it during the contest and didn't solve). It will be very good if they found beautiful GEOMETRY problem(not related with area or other stuff). Problem 3 was combinatorial, so I had to give my all time to problem 2 which is very ugly problem (since I know that I can't solve problem 4). the problem is very beautiful and elegant tho it might be hard for P2
SerdarBozdag
20.01.2024 12:42
It is enough to prove that $d(X,BC) + d(Y,BC) = d(M, BC)$. Definitions. $S$ is the midpoint of $EF$, $K$ is the midpoint of $BC$, $AI \cap ABC = N$, $BI \cap (ABC) = U$, $CI \cap (ABC) =V$. $G = XY \cap BC$, $I_A$ is the $A$-excenter. $L= (GSK) \cap (IXY)$. $W$ is the antipode of $G$ in $(GSK)$, it satisfies $SW \perp XY$ and $W \in MN$. $MI \cap (ABC)=E$, $MI \cap (IBI_AC) =D$. Claim. $LYX \sim DBC$. Proof. It is enough to prove that $LSY \sim DKB$ which symmetrically implies $LSX \sim DKC$ because these together give the desired similarity. Note that $ID$ is symmedian in $BIC$ and $BDC$ because $BM$ and $CM$ are tangent to $(IBI_AC)$. $$ \angle LSY = \angle LKG \overset{BIK \sim DIC}= \angle ICD = \angle BKD, \angle LYS = \angle LIX = \angle BIK \overset{BIK \sim DIC}= \angle DIC = \angle DBC$$$\square$ $\angle YIL = \angle YXL = \angle DCB = \angle BID = \angle KIC \implies L \in IK$. Note that $IM$ is median in $IUV$ because $IVMU$ is parallelogram. Because $IK$ is median in $IBC$, $IL$ is symmedian in $VIU$. Combining it with $M \in (IXY)$ gives $SM = SL$ and because $SM$ and $SL$ symmetric wrt $SW$ we get $\angle LWS = \angle SWM = \angle SWK \implies SL = SM = SK$. Thus $$d(M, BC) = 2\cdot d(S,BC) = d(E,BC) + d(F,BC)$$$\square$
MathLuis
03.10.2024 07:06
Orthology on top. Let the projections from $M$ to $IB,IC$ be $D,E$ respectively, and let $N$ the midpoint of $DE$ as well as $M'$ the midpoint of $BC$ and $I'$ a point such that $BICI'$ is a parallelogram. Claim 1: $S_{BXIYC}=S_{BEIDC}$ Proof: Notice it's enough to prove that $S_{BYE}=S_{CXD}$ which happens when $\frac{d(B, IC)}{d(C, IB}=\frac{XD}{YE}$ but notice that we have from LoS and Ratio Lemma and the fact that $IM$ is symedian in $\triangle BIC$ that: \[ \frac{d(B, IC)}{d(C, IB)}=\frac{BI}{IC}=\frac{\sin(\angle M'IC)}{\sin(\angle M'IB)}=\frac{\sin(\angle MID)}{\sin(\angle MIE)}=\frac{MD}{ME}=\frac{XD}{YE} \]Last one being true as $\triangle MDX \sim \triangle MEY$ by the angles thus the claim is proven. The finish: Add $S_{BIC}$ on each side of the desired equality, then we have that all we need is $d(D, BC)+d(E, BC)=d(M, BC)$ or just $2d(N, BC)=d(M, BC)$ (from trapezoid lenghts). Now we prove that $\triangle DME, \triangle BIC$ are orthologic, since $BI' \perp ME$ and $CI' \perp MD$ all we need is that $II' \perp DE$ but notice from symedian we have that $IM, IM'$ are isogonal on $\angle DIE$ meaning that since $IM$ is a diameter in $(MEID)$ we have that $IM''$ must be the altitude so $II' \perp DE$ as desired, so by orthology we have that perpendiculars from $D,E,M$ to $CI,IB, BC$ meet at one single point $M_1$, note that $M_1$ has to be orthocenter of $\triangle DIE$ therefore $M,N,M_1$ are collinear by reflecting, but notice that $M_1$ lies on the perpendicular bisector of $BC$ and so does $M'$ and also $I,M',M_1$ are colinear from being perpendicular to $DE$ meaning that $M'=M_1$ so in fact $N$ is midpoint of $MM'$ so the distance claim is clearly true thus we are done .