Let ABC be an acute triangle with orthocentre H, and let M be the midpoint of AC. The point C1 on AB is such that CC1 is an altitude of the triangle ABC. Let H1 be the reflection of H in AB. The orthogonal projections of C1 onto the lines AH1, AC and BC are P, Q and R, respectively. Let M1 be the point such that the circumcentre of triangle PQR is the midpoint of the segment MM1. Prove that M1 lies on the segment BH1.
Problem
Source: Balkan MO 2010, Problem 2
Tags: geometry, geometric transformation, reflection, circumcircle, cyclic quadrilateral, geometry proposed
04.05.2010 19:18
I'll post just my idea, because i don't have time to write full proof. *It can be easily proven that H' lies on the circumcircle of ABC. *Now, AH'BC is a cyclic quadrilateral with perpendicular diagonals. *In fact there are 8 points that lie on a same circle with center S (the midpoint of OC'): -the 4 midpoints of the sides of the quadrilateral AH'BC and -the 4 feet of the perpendiculars from C' to the sides of the quadrilateral AH'BC. *The Brahmagupta theorem may also be helpful Cheers, Steff
04.05.2010 19:19
Quote: Let ABC be an acute triangle with orthocentre H, and let M be the midpoint of AC. The point C1 on AB is such that CC1 is an altitude of the triangle ABC. Let H1 be the reflection of H in AB. The orthogonal projections of C1 onto the lines AH1, AC and BC are P, Q and R, respectively. Let M1 be the point such that the circumcentre of triangle PQR is the midpoint of the segment MM1. Prove that M1 lies on the segment BH1. Let N, M′1 and L respectively be the midpoints of CB, BH1, AH1. Then it is an obvious fact that MNM′1L is a rectangle. Therefore, it is inscribed in a circle with centered at I. Now, let P′ be the intersection of C1N with AH1. Then it is followed that ∠H1C1P= ∠NC1C= ∠NCC1≡ ∠BCH1 =∠C1AH1, which implies NC1⊥ AH1. As the result, P′ must ≡ P ⟹ P∈(I). The same argument holds for Q and R. Thus, (PQR)≡ (I) ⟹ M′1≡M1. The result of the problem is lead as follow. Our proof is completed then. ◻
09.05.2010 17:51
Does anybody know the official solution? )
23.05.2010 18:30
Here's my solution. Recall that a podar triangle corresponding to a given point P inside a triangle ABC is the triangle formed by the feet of the orthogonal projections of P onto the sides of the triangle ABC. It can also be easily proved that the projections of two isogonal points in triangle ABC are concycic, which means the podar triangles corresponding to isogonal points have the same circumcircle and its center is the midpoint of the segment determined by those two points. Returning to our problem, the isogonal of C1 with respect to the triangle formed by intersecting AH1 with BC is exactly O, the circumcenter of the triangle ABC. Therefore, the circle PQR passes through M, thus we would have to prove that the point M1 is also on the circle and MM1 is a diameter for (PQR). This is easy, since the circumcenter of (PQR) is the midpoint of OC1 which is the midpoint of MM1 (where M1 is the midpoint of BH1 ).
02.06.2010 09:50
http://mathproblems123.wordpress.com/2010/05/23/bmo-2010-geometry-problem/
24.11.2015 19:29
This was nice except that it is pretty obvious if one has done the famous USA TST 2000 kite problem. Here is a brief sketch of my idea:- Let S be the last perpendicular we can drop from C1. The it is evident that PQRS is a cyclic quad and then by our Lemma we have that the midpoints M,M1 of AC,BH1 are on this circle and that the same holds for the other midpoints. But then MM1 is a diameter and we conclude.
01.08.2016 20:27
My Solution. let L be the midpoint of BH1. since H1 is the reflection of H in AB we have, ∡BH1C=∡H1HB=90o−∡HBA=∡BAC therefore the quadrilateral ACBH1 is Cyclic. and, ∡H1C1L=90o−12∡HBH1=H1HB, hence, LC1 and BH are parallel, and thus LC1 is perpendicular to AC and we know that C1Q is perpendicular to AC. hence, the points L,C1,Q are collinear. this mean that Q lie on the Circle (Ω) (with diametre ML). Now, we have, →LP⋅→MP=(→LC1+→C1P)(→MA+→AP)=→C1P⋅→MA−→LC1⋅→PA=sin∡H1AC⋅(MA⋅C1P−LC1⋅PA)=X and we have, LC1C1P=12H1BH1C1⋅sinβ=121cos∡BACsinβ=12ACAC1⋅sin(90o−∡H1AB)=MAPA. Or, X=0. Therefore P lie on the Circle (Ω). analogously, we prove that R lie on the Circle (Ω) (By Symmetric!), and we deduce that L=M1, and the result will be Clear.
Attachments:

23.04.2017 11:24
Ithoughtitwouldbebothinterestingandusefultopostthegradingscheme
Attachments:
BMO_2010_Grading_scheme.pdf (250kb)
22.05.2017 16:21
It's quite trivial that AH′BC is a cyclic quadrilateral. Let J,K,L be the midpoint of segments AH′,H′B,BC. JKLM is the varignon quadrilateral of AH′BC. So JKLM is a parallelogram. Also, as AB⊥CH′, JKLM is a rectangle. Let us now drop perpendicular from J,K,L,M to the opposite sides are J′,K′,L′,M′. All these perpendiculars go through C′. Hence we get that J,K,L,M,J′,K′,L′,M′ is cyclic and we are done with the problem.
11.02.2019 12:16
Let AH′ meet BC at B′. See that PQR is the pedal triangle of C1 WRT AB′C. Now, observe that O, the circumcenter of ABC is the isogonal conjugate of C1 WRT AB′C. Let M′ be the mid point of BH1. Now, since AH1BC is a cyclic quadrilateral with perpendicular diagonals, R−C1−M′. So, OM∥CM′. Also, C1M′=H1B2=HB2=OM. So, OMM′C1 is a parallelogram, which means MM′,OC1 share the same midpoint. Also, the circumcenter of PQR is the midpoint of OC1. So, M1=M′ which implies M1∈¯H1B. ◼.
11.02.2019 15:48
Special case of shortlist 2011 G3
28.01.2022 07:14
By chasing angles: H1∈(ABC) Notice that: (PQR) is eight point circle. ⟹M∈(PQR) M1 it is diametrically opposite to M w.r.t (PQR). Let M′ be the midpoint of BH1 by eight point circle MM′ is diameter of (PQR) i.e M′ it is diametrically opposite to M w.r,t (PQR) ⟹M′=M1.◼
13.12.2022 16:10
Note O,O1 be circumcenter of (ABC) and (PQR) ,respectively. Just an easy angle-chasing yields H1∈(ABC) Hence (PQR) is eight point circle ⟹O1 is midpoint of (C1O) Let us continue with easy complex bash.Set (ABC) as unit circle.Then h=a+b+c∧ˉh=ab+bc+caabch1=(a−b)ˉh+ˉab−aˉbˉa−ˉb=−abc∧¯h1=−cabm1=2o1−m=c1−a+c2=b(c−a)2c∧¯m1=a−c2abIt remains to show that B,M1,H1 are collinear ⟺|b1b1b(c−a)2ca−c2ab1−abc−cab1|=0⟺|111c−a2ca−c2a1−ac−ca1|=0⟺0=0 so we are done
21.12.2024 01:55
Really easy with eight-point circle properties: Obviously AH1BC is cyclic and notice that (PQR) is its eight-point circle, so M∈(PQR) which means that M1 is diametrically opposite to M, but this implies that M1 is the midpoint of H1B (because of the eight-point circle) and we're done.