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IMO 2014 Problem 4
ipaper   169
N 37 minutes ago by YaoAOPS
Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$.

Proposed by Giorgi Arabidze, Georgia.
169 replies
ipaper
Jul 9, 2014
YaoAOPS
37 minutes ago
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Tangents forms triangle with two times less area
NO_SQUARES   1
N an hour ago by Luis González
Source: Kvant 2025 no. 2 M2831
Let $DEF$ be triangle, inscribed in parabola. Tangents in points $D,E,F$ forms triangle $ABC$. Prove that $S_{DEF}=2S_{ABC}$. ($S_T$ is area of triangle $T$).
From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
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NO_SQUARES
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Luis González
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Symmetric orthotransversal
buratinogigle   5
N Jun 3, 2015 by jayme
Source: Own
Prove that orthotransversal of incenter wrt reference triangle and wrt incentral triangle are symmetric through such this incenter.
5 replies
buratinogigle
May 18, 2015
jayme
Jun 3, 2015
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Symmetric orthotransversal
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buratinogigle
2344 posts
buratinogigle
#1 h  May 18, 2015, 6:55 PM • 2 Y
Y by Adventure10, Mango247
Prove that orthotransversal of incenter wrt reference triangle and wrt incentral triangle are symmetric through such this incenter.
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Luis González
4148 posts
Luis González
#2 h  May 18, 2015, 9:38 PM • 3 Y
Y by buratinogigle, Adventure10, Mango247
$I$ is incenter of $\triangle ABC$ and $AI,BI,CI$ cut $BC,CA,AB$ at $D,E,F.$ Perpendicular to $AI$ at $I$ cuts $BC,EF$ at $X,X^*,$ respectively. Points $\{Y,Y^*\}$ and $\{Z,Z^*\}$ are defined cyclically. $\overline{XYZ}$ and $\overline{X^*Y^*Z^*}$ are the orthotransversals of $I$ WRT $\triangle ABC$ and $\triangle DEF.$

If $XX^*$ cuts $AC,AB$ at $U,V,$ then by Desargues involution theorem for $BFEC,$ it follows that $X \mapsto X^*,$ $U \mapsto V$ is an involution where $I$ is double and $U,V$ are symmetric WRT $I$ $\Longrightarrow$ it coincides with the central symmetry WRT $I$ $\Longrightarrow$ $X,X^*$ are symmetric WRT $I$ and similarly $Y,Y^*$ and $Z,Z^*$ are symmetric WRT $I$ $\Longrightarrow$ $\overline{XYZ}$ and $\overline{X^*Y^*Z^*}$ are symmetryc WRT $I.$
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TelvCohl
2312 posts
TelvCohl
#3 h  May 18, 2015, 11:44 PM • 3 Y
Y by buratinogigle, enhanced, Adventure10
My solution :

Let $ I $ be the incenter of $ \triangle ABC $ and $ \triangle DEF $ be its cevian triangle WRT $ \triangle ABC $ .
Let the orthotransversal of $ I $ WRT $ \triangle ABC $ cuts $ BC, CA, AB $ at $ X, Y, Z $, respectively .
Let the orthotransversal of $ I $ WRT $ \triangle DEF $ cuts $ EF, FD, DE $ at $ X^*, Y^*, Z^* $, respectively .
Let $ T $ be the intersection of $ BC $ with A-external bisctor of $ \triangle ABC $ ( it's well-known $ T \in EF $ ) .

From $ T(A,I;X^*,X)=-1 $ and $ XX^* \parallel TA $ ( both $ \perp AI $ ) $ \Longrightarrow XI=X^*I $ .
Similarly, we can prove $ YI=Y^*I $ and $ ZI=Z^*I \Longrightarrow \overline{XYZ} $ and $ \overline{X^*Y^*Z^*} $ are symmetry WRT $ I $ .

Q.E.D
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buratinogigle
2344 posts
buratinogigle
#4 h  May 19, 2015, 1:20 AM • 2 Y
Y by Adventure10, Mango247
Thank you so much, actually I have found solution the same as Telv base on idea of Luis but Telv posted. I only note that. Then line passes through I cuts $BC, EF$ at $G,H$ then $AG,AH$ are isogonal conjugate wrt $\angle BAC$.
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tranquanghuy7198
253 posts
tranquanghuy7198
#5 h  Jun 2, 2015, 4:49 PM • 3 Y
Y by buratinogigle, Adventure10, Mango247
Mr.Buratino wrote:
Prove that orthotransversal of incenter wrt reference triangle and wrt incentral triangle are symmetric through such this incenter (post #1).
My solution:

Lemma 1. Let $X, Y$ be 2 points on the side $BC$ of $\triangle{ABC}$. We have: $\odot{(AXY)}$ is tangent to $\odot{(ABC)} \Longleftrightarrow AX, AY$ are isogonal in $\angle{BAC}$
Proof. Simple angle chasing.

Lemma 2. Given $\triangle{ABC}, E = R_{AC}(B), F = R_{AB}(C)$. The line passing through $A$ which is tangent to $\odot{(AEF)}$ intersects $BC, EF$ at $X, Y$. Prove that: $AX = AY$

Back to our main problem:
Consider $\triangle{ABC}$ and the bisectors $AD, BE, CF$ concur at $I$
The orthotransversal of $I$ WRT $\triangle{ABC}$ intersects $BC, CA, AB$ at $X, Y, Z$
The orthotransversal of $I$ WRT $\triangle{DEF}$ intersects $EF, FD, DE$ at $M, N, P$
In order to prove $\overline{X, Y, Z} = S_I(\overline{M, N, P})$, we only need to indicate that $IX = IM$
$K = R_{IF}(E), L = R_{IE}(F)$ $\Rightarrow K, L\in{BC}$
Notice that $\angle{CIK} = \angle{CIE} = \angle{BIF} = \angle{BIL}$
$\Rightarrow (IKL), (IBC)$ are tangent to each other at $I$ (lemma 1)
$\Rightarrow (IKL)$ is tangent to $\overline{X, I, M}$
Now apply the lemma 2 for $\triangle{IEF}$ we receive: $IX = IM$ and our proof is completed!
Q.E.D
This post has been edited 1 time. Last edited by tranquanghuy7198, Jun 2, 2015, 4:50 PM
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jayme
9782 posts
jayme
#6 h  Jun 3, 2015, 8:11 AM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

a proof based on anharmonic pencils can be very nice...

Sincerely
Jean-Louis
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