Lei it \(ABCD\) be a non-cyclical, convex quadrilateral, with no parallel sides. The lines \(AB\) and \(CD\) meet in \(E\). Let it \(M \not= E\) be the intersection of circumcircles of \(ADE\) and \(BCE\). The internal angle bisectors of \(ABCD\) form an convex, cyclical quadrilateral with circumcenter \(I\). The external angle bisectors of \(ABCD\) form an convex, cyclical quadrilateral with circumcenter \(J\). Show that \(I,J,M\) are colinear.
Problem
Source: 38th Brazilian MO (2016) - Second Day, Problem 6
Tags: Brazilian Math Olympiad 2016, geometry, circumcircle, angle bisector
jaobundao
23.11.2016 19:42
Let $ W $ be the intersection of the external bisector of $ \angle A, $ $ \angle B $ and define $ X, $ $ Y, $ $ Z $ similarly. Let $ F $ be the intersection of $ AD, $ $ BC $ and let $ \Phi_M $ be the inversion with center $ M, $ power $ MA $ $ \cdot $ $ MC $ $ = $ $ MB $ $ \cdot $ $ MD $ $ = $ $ ME $ $ \cdot $ $ MF $ followed by reflection on the bisector $ \tau $ of $ \angle AMC, $ $ \angle BMD, $ $ \angle EMF. $ Since $ \measuredangle AZE $ $ = $ $ \measuredangle FYC, $ so by conformal $ \Longrightarrow $ $ \Phi_M(Z) $ lies on $ \odot (CFY). $ Similarly, we can prove $ \Phi_M(Z) $ lies on $\odot (BFW), $ so $ \Phi_M(Z) $ is the Miquel point of the complete quadrilateral $ BWYC $ and $ \Phi_M(Z) $ $ \in $ $ \odot (J). $ Similarly, we can prove $ \Phi_M(W), $ $ \Phi_M(X), $ $ \Phi_M(Y) $ lie on $ \odot (J), $ so $ J $ $ \in $ $ \tau. $ Analogously, we can prove $ I $ $ \in $ $ \tau. $
no_name1998
18.05.2017 16:27
Could someone solve this promblem by elementary mathematics, please?
DSD
20.06.2017 06:50
Use search function before posting the problem from contest !
Achillys
20.06.2017 08:56
Erkhes wrote: Your solution is wrong Erkhes wrote: Actually I think this solution is wrong What is wrong exactly? I've gone through each step of the solution several times and found no error
Erkhes
20.06.2017 13:29
Z prime lies on circumscribed circle of CFY
khina
26.08.2020 23:53
Let the internal angles of $A$ and $B$ intersect at $P$, let the external angle bisectors of $A$ and $B$ intersect at $W$, and let $P_1$ be the midpoint of $WP$. Define the points $Q$, $R$, $S$, $X$, $Y$, $Z$, $Q_1$, $R_1$, $S_1$ similarly. Finally, denote the intersection of $IJ$ with $PR$ and $QS$ as $K$ and $L$ similarly. A quick angle chase finds that $K$ and $L$ are the insimilicenter and exsimilmicenter of the two circles, so it suffices to prove $M$ lies on $KL$ (which one is which depends on your diagram). Note that $P_1$ is the midpoint of the arc $BA$ not containing $M$ on the circumcircle of $MBA$. This gives us that $BP_1A \mapsto^M CR_1D$, which gives: \begin{align*} \frac{MP_1}{MR_1} &= \frac{P_1P}{R_1R} \\ &= \frac{P_1K}{KR_1} \end{align*}Thus, $KM$ bisects $\angle{P_1MR_1}$, and similarly $LM$ bisects $\angle{Q_1MS_1}$. But a quick angle chase reveals the two angles share a bisector, so we are done.
I think this solution is very tricky to find, so I'll give a bit of insight on what I did. I had been trying a lot of spiral similarity stuff, because setting $PQRS$ to a single point reduces the problem to a spiral + Miquel exercise (not easy, but more standard than the given problem). I was fiddling around a lot while trying to use $WXYZ$ in a spiral sim, but geometers more careful than I am should note that $BXC$ and $AQD$ are actually not similar, and so really the only points you can use for a spiral sim are $P_1$ and it's cyclic variants. Some time after I added $K$ and $L$, and everything just clicked.
note: while the above solution is short, it's very hard to find a good way to actually use spiral similarities to kill the problem, and that's where the majority of the difficulty comes from (in a similar vein to v_Enhance's 2015 G7 solution)
RodSalgDomPort
15.12.2020 04:11
For a complete solution follow the guide below: $1)$ Note that $MA.MC=MB.MD=ME.MF$ by looking at some spiral similarities with center $M$. Thus, take an inversion centered at $M$ with ratio: $\sqrt{MA.MC}$. $2)$ Then consider one of the ex-incenter $J_{DA}$ and proves that $J_{DA}'$ belongs to the circle passing through: $C,F, J_{CD}$ (simply angle chasing) $3)$ Analogously, it will belong to the circle passing through: $B,F,J_{AB}$. Then, $J_{DA}'$ will be the miquel point of quadrilateral: $BCJ_{CD}J_{AB}$ implying that $J_{DA}'$ belongs to the circle passing through all $J's$. $4)$ This implies that the image of $J_{DA}$ under the inversion lies on the circle as well $\implies$ the circle is fixed by inversion. $5)$ Reproduce all steps before (or simply say it's analagous) to state that the circle of the incenters will also stay the same by the same inversion. $6)$ As both circles are fixed, the line joining their centers is collinear with the center of the inversion which is $M$ and we're done.
Ruy
27.10.2021 23:12
Full solution in portuguese here. If you don't understand portuguese and don't like inversion (wich I think is completely unreasonable), follow the steps bellow. I believe that if you see the diagrams and math writing of my portuguese solution it will allow you to understand my step by step guide.
1. $M$ is the Miquel point of the quadrilateral;
2. Making the quadrilateral complete, we see that the bisectors at $E$ and the other point of intersection keep each one 4 important points. Prove this. (Use incenter/excenter lemma to do so);
2.1. Now you have to do some simple angle chasing at the bisector quadrilaterals (mark all angles, please);
3. As you see this, you have now to handle the circumcenters, wich was my major dificulty seeing the problem at a first glance. We can eliminate the necessity of the circumcenters drawing the radius and noticing two equal scale factor homotheties. Use law of sines and the sine of suplementary angles to find the homothety;
4. Intersecting lines, easily get the centers of this homotheties. This points will substitute the centers we found ugly;
4.1. Get the centers of the notable cyclic quadrilaterals in the figure;
5. Now you have to do some ratio chasing and be aware of the spiral similarities of the Miquel point;
6. If you follow all the above steps, I think that you reached the major dificulty. Use the spiral similarities, the ratio chasing and the internal bisector theorem to prove that $MI$ and $MJ$ are two important known bisectors. They actually are the same but proving this will give you this result.
Also see useful diagrams attached.
Attachments:
Ruy
27.10.2021 23:17
no_name1998 wrote: Could someone solve this promblem by elementary mathematics, please? I made it.
ILOVEMYFAMILY
06.02.2022 17:54
Ruy wrote: Full solution in portuguese here. If you don't understand portuguese and don't like inversion (wich I think is completely unreasonable), follow the steps bellow. I believe that if you see the diagrams and math writing of my portuguese solution it will allow you to understand my step by step guide.
1. $M$ is the Miquel point of the quadrilateral;
2. Making the quadrilateral complete, we see that the bisectors at $E$ and the other point of intersection keep each one 4 important points. Prove this. (Use incenter/excenter lemma to do so);
2.1. Now you have to do some simple angle chasing at the bisector quadrilaterals (mark all angles, please);
3. As you see this, you have now to handle the circumcenters, wich was my major dificulty seen the problem in a first look. We can eliminate the necessity of the circuncenters drawing the radius and noticing two equal scale factor homotheties. Use law of sines and the sine of suplementary angles to logically find the homothety;
4. Intersecting lines, easily get the centers of this homotheties. This points will substitute the centers we found ugly;
4.1. Get the centers of the notable cyclic quadrilaterals in the figure;
5. Now you have to do some ratio chasing and be aware of the spiral similarities of the Miquel point;
6. If you follow all the above steps, I think that you reached the major dificulty. Use the spiral similarities, the ratio chasing and the internal bisector theorem to prove that $MI$ and $MJ$ are two important known bisectors. They actually are the same but proving this will give you this result.
Also see useful diagrams attached. How MI be the angle bisector of angle AMC. Could u explain more clearly for me ,please?
NiltonCesar
10.08.2022 03:54
Is there any elementar solution?
NiltonCesar
07.09.2022 18:21
jaobundao wrote: Let $ W $ be the intersection of the external bisector of $ \angle A, $ $ \angle B $ and define $ X, $ $ Y, $ $ Z $ similarly. Let $ F $ be the intersection of $ AD, $ $ BC $ and let $ \Phi_M $ be the inversion with center $ M, $ power $ MA $ $ \cdot $ $ MC $ $ = $ $ MB $ $ \cdot $ $ MD $ $ = $ $ ME $ $ \cdot $ $ MF $ followed by reflection on the bisector $ \tau $ of $ \angle AMC, $ $ \angle BMD, $ $ \angle EMF. $ Since $ \measuredangle AZE $ $ = $ $ \measuredangle FYC, $ so by conformal $ \Longrightarrow $ $ \Phi_M(Z) $ lies on $ \odot (CFY). $ Similarly, we can prove $ \Phi_M(Z) $ lies on $\odot (BFW), $ so $ \Phi_M(Z) $ is the Miquel point of the complete quadrilateral $ BWYC $ and $ \Phi_M(Z) $ $ \in $ $ \odot (J). $ Similarly, we can prove $ \Phi_M(W), $ $ \Phi_M(X), $ $ \Phi_M(Y) $ lie on $ \odot (J), $ so $ J $ $ \in $ $ \tau. $ Analogously, we can prove $ I $ $ \in $ $ \tau. $ by "conformal" do you mean "conformal map"?
Ruy
30.09.2022 18:01
ILOVEMYFAMILY wrote: Ruy wrote: Full solution in portuguese here. If you don't understand portuguese and don't like inversion (wich I think is completely unreasonable), follow the steps bellow. I believe that if you see the diagrams and math writing of my portuguese solution it will allow you to understand my step by step guide.
1. $M$ is the Miquel point of the quadrilateral;
2. Making the quadrilateral complete, we see that the bisectors at $E$ and the other point of intersection keep each one 4 important points. Prove this. (Use incenter/excenter lemma to do so);
2.1. Now you have to do some simple angle chasing at the bisector quadrilaterals (mark all angles, please);
3. As you see this, you have now to handle the circumcenters, wich was my major dificulty seen the problem in a first look. We can eliminate the necessity of the circuncenters drawing the radius and noticing two equal scale factor homotheties. Use law of sines and the sine of suplementary angles to logically find the homothety;
4. Intersecting lines, easily get the centers of this homotheties. This points will substitute the centers we found ugly;
4.1. Get the centers of the notable cyclic quadrilaterals in the figure;
5. Now you have to do some ratio chasing and be aware of the spiral similarities of the Miquel point;
6. If you follow all the above steps, I think that you reached the major dificulty. Use the spiral similarities, the ratio chasing and the internal bisector theorem to prove that $MI$ and $MJ$ are two important known bisectors. They actually are the same but proving this will give you this result.
Also see useful diagrams attached. How MI be the angle bisector of angle AMC. Could u explain more clearly for me ,please? $G,H$ are intersections of $IJ$ with $EC_2$, $FC_2$. $\angle DAK=\angle KDA=\angle CBL =\angle LCB=\frac{\angle CEB}{2}$ and the similarity spiral sending $\overline{BC}$ to $\overline{AD}$ also send $L$ to $K$. This SS is centered at $M$ and it's ratio is $$\frac{\overline{AK}}{\overline{BL}}=\frac{\overline{MK}}{\overline{ML}}=\frac{\overline{KG}}{\overline{GL}}$$(This ratio is probably not so clear, but you can ask for more details). The internal bisector theorem implies $MG$ bisects $\angle KML$. Analogously, $MH$ bisects $\angle OMN$.