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P,Q,B are collinear
MNJ2357   29
N 20 minutes ago by cj13609517288
Source: 2020 Korea National Olympiad P2
$H$ is the orthocenter of an acute triangle $ABC$, and let $M$ be the midpoint of $BC$. Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$, and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$. Prove that $P,Q,B$ are collinear.
29 replies
MNJ2357
Nov 21, 2020
cj13609517288
20 minutes ago
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Angle Relationships in Triangles
steven_zhang123   0
2 hours ago
In $\triangle ABC$, $AB > AC$. The internal angle bisector of $\angle BAC$ and the external angle bisector of $\angle BAC$ intersect the ray $BC$ at points $D$ and $E$, respectively. Given that $CE - CD = 2AC$, prove that $\angle ACB = 2\angle ABC$.
0 replies
steven_zhang123
2 hours ago
0 replies
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acute triangle and its circumcenter and orthocenter
N.T.TUAN   6
N 2 hours ago by MathLuis
Source: USA TST 2005, Problem 2
Let $A_{1}A_{2}A_{3}$ be an acute triangle, and let $O$ and $H$ be its circumcenter and orthocenter, respectively. For $1\leq i \leq 3$, points $P_{i}$ and $Q_{i}$ lie on lines $OA_{i}$ and $A_{i+1}A_{i+2}$ (where $A_{i+3}=A_{i}$), respectively, such that $OP_{i}HQ_{i}$ is a parallelogram. Prove that
\[\frac{OQ_{1}}{OP_{1}}+\frac{OQ_{2}}{OP_{2}}+\frac{OQ_{3}}{OP_{3}}\geq 3.\]
6 replies
N.T.TUAN
May 14, 2007
MathLuis
2 hours ago
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Imtersecting two regular pentagons
Miquel-point   2
N 3 hours ago by ohiorizzler1434
Source: KoMaL B. 5093
The intersection of two congruent regular pentagons is a decagon with sides of $a_1,a_2,\ldots ,a_{10}$ in this order. Prove that
\[a_1a_3+a_3a_5+a_5a_7+a_7a_9+a_9a_1=a_2a_4+a_4a_6+a_6a_8+a_8a_{10}+a_{10}a_2.\]
2 replies
1 viewing
Miquel-point
Yesterday at 6:27 PM
ohiorizzler1434
3 hours ago
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Chinese Girls Mathematical Olympiad 2017, Problem 7
Hermitianism   45
N 4 hours ago by Ilikeminecraft
Source: Chinese Girls Mathematical Olympiad 2017, Problem 7
This is a very classical problem.
Let the $ABCD$ be a cyclic quadrilateral with circumcircle $\omega_1$.Lines $AC$ and $BD$ intersect at point $E$,and lines $AD$,$BC$ intersect at point $F$.Circle $\omega_2$ is tangent to segments $EB,EC$ at points $M,N$ respectively,and intersects with circle $\omega_1$ at points $Q,R$.Lines $BC,AD$ intersect line $MN$ at $S,T$ respectively.Show that $Q,R,S,T$ are concyclic.
45 replies
Hermitianism
Aug 16, 2017
Ilikeminecraft
4 hours ago
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Easy Geometry
pokmui9909   6
N 5 hours ago by reni_wee
Source: FKMO 2025 P4
Triangle $ABC$ satisfies $\overline{CA} > \overline{AB}$. Let the incenter of triangle $ABC$ be $\omega$, which touches $BC, CA, AB$ at $D, E, F$, respectively. Let $M$ be the midpoint of $BC$. Let the circle centered at $M$ passing through $D$ intersect $DE, DF$ at $P(\neq D), Q(\neq D)$, respecively. Let line $AP$ meet $BC$ at $N$, line $BP$ meet $CA$ at $L$. Prove that the three lines $EQ, FP, NL$ are concurrent.
6 replies
pokmui9909
Mar 30, 2025
reni_wee
5 hours ago
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Old hard problem
ItzsleepyXD   3
N 5 hours ago by Funcshun840
Source: IDK
Let $ABC$ be a triangle and let $O$ be its circumcenter and $I$ its incenter.
Let $P$ be the radical center of its three mixtilinears and let $Q$ be the isogonal conjugate of $P$.
Let $G$ be the Gergonne point of the triangle $ABC$.
Prove that line $QG$ is parallel with line $OI$ .
3 replies
ItzsleepyXD
Apr 25, 2025
Funcshun840
5 hours ago
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Beautiful Angle Sum Property in Hexagon with Incenter
Raufrahim68   0
Yesterday at 6:53 PM
Hello everyone! I discovered an interesting geometric property and would like to share it with the community. I'm curious if this is a known result and whether it can be generalized.

Problem Statement:
Let
A
B
C
D
E
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ABCDEK be a convex hexagon with an incircle centered at
O
O. Prove that:

∠
A
O
B
+
∠
C
O
D
+
∠
E
O
K
=
180
∘
∠AOB+∠COD+∠EOK=180
∘
0 replies
Raufrahim68
Yesterday at 6:53 PM
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IMO 2009 P2, but in space
Miquel-point   1
N Yesterday at 6:36 PM by Miquel-point
Source: KoMaL A. 485
Let $ABCD$ be a tetrahedron with circumcenter $O$. Suppose that the points $P, Q$ and $R$ are interior points of the edges $AB, AC$ and $AD$, respectively. Let $K, L, M$ and $N$ be the centroids of the triangles $PQD$, $PRC,$ $QRB$ and $PQR$, respectively. Prove that if the plane $PQR$ is tangent to the sphere $KLMN$ then $OP=OQ=OR.$

1 reply
1 viewing
Miquel-point
Yesterday at 6:35 PM
Miquel-point
Yesterday at 6:36 PM
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Dissecting regular heptagon in similar isosceles trapezoids
Miquel-point   0
Yesterday at 6:25 PM
Source: KoMaL B. 5085
Show that a regular heptagon can be dissected into a finite number of symmetrical trapezoids, all similar to each other.

Proposed by M. Laczkovich, Budapest
0 replies
1 viewing
Miquel-point
Yesterday at 6:25 PM
0 replies
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Amazing projective stereometry
Miquel-point   0
Yesterday at 6:24 PM
Source: KoMaL B 5060
In the plane $\Sigma$, given a circle $k$ and a point $P$ in its interior, not coinciding with the center of $k$. Call a point $O$ of space, not lying on $\Sigma$, a proper projection center if there exists a plane $\Sigma'$, not passing through $O$, such that, by projecting the points of $\Sigma$ from $O$ to $\Sigma'$, the projection of $k$ is also a circle, and its center is the projection of $P$. Show that the proper projection centers lie on a circle.
0 replies
1 viewing
Miquel-point
Yesterday at 6:24 PM
0 replies
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Proving radical axis through orthocenter
azzam2912   2
N Yesterday at 6:11 PM by Miquel-point
In acute triangle $ABC$ let $D, E$ and $F$ denote the feet of the altitudes from $A, B$ and $C$, respectively. Let line $DE$ intersect circumcircle $ABC$ at points $G, H$. Similarly, let line $DF$ intersect circumcircle $ABC$ at points $I, J$. Prove that the radical axis of circles $EIJ$ and $FGH$ passes through the orthocenter of triangle $ABC$
2 replies
azzam2912
Yesterday at 12:02 PM
Miquel-point
Yesterday at 6:11 PM
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Perspectivity related to Stammler hyperbola
TelvCohl   3
N Jan 7, 2021 by TelvCohl
Source: Own
Given a $ \triangle ABC $ and a point $ P. $ Let $ \triangle P_aP_bP_c $ be the pedal triangle of $ P $ WRT $ \triangle ABC $ and let $ H_A, H_B, H_C $ be the orthocenter of $ \triangle AP_bP_c, \triangle BP_cP_a, \triangle CP_aP_b, $ respectively. Prove that $ \triangle H_AH_BH_C $ and the medial triangle of $ \triangle ABC $ are perspective if and only if $ P $ lies on the Stammler hyperbola of $ \triangle ABC. $ If that's the case, the perspector lies on the Euler line of $ \triangle ABC. $
3 replies
TelvCohl
Feb 19, 2020
TelvCohl
Jan 7, 2021
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Perspectivity related to Stammler hyperbola
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TelvCohl
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TelvCohl
#1 h  Feb 19, 2020, 5:06 PM • 8 Y
Y by amar_04, Hexagrammum16, qzc, buratinogigle, enhanced, Adventure10, Mango247, GeoKing
Given a $ \triangle ABC $ and a point $ P. $ Let $ \triangle P_aP_bP_c $ be the pedal triangle of $ P $ WRT $ \triangle ABC $ and let $ H_A, H_B, H_C $ be the orthocenter of $ \triangle AP_bP_c, \triangle BP_cP_a, \triangle CP_aP_b, $ respectively. Prove that $ \triangle H_AH_BH_C $ and the medial triangle of $ \triangle ABC $ are perspective if and only if $ P $ lies on the Stammler hyperbola of $ \triangle ABC. $ If that's the case, the perspector lies on the Euler line of $ \triangle ABC. $
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dagezjm
88 posts
dagezjm
#2 h  Mar 7, 2020, 7:50 AM • 5 Y
Y by Hexagrammum16, sameer_chahar12, amar_04, qzc, Mango247
Here is a same question I found at August 15,2019. Let $\triangle DEF$ is the pedal triangle of the orthocenter $H$ WRT $\triangle ABC$, then given a point $X$ lies on the Stammler hyperbola of $\triangle ABC$. If $ABCX\sim AEFP\sim DBFQ\sim DECR$, then $\triangle PQR$ and the median triangle of $\triangle ABC$ are perspective. And the perspector lies on the Euler line of $\triangle ABC$.(The $H_A$ in your problem is the same as the $P$ in my problem.)
And I transform this problem by a easy truth that if a point lies on Stammler hyperbola, then the cevapoint of it and the cicumcenter $O$ lies on the Euler line. So it can be expressed as the following proposition.
Given $\triangle ABC$，the circumcenter $O$ and the orthocenter $H$，$D$ is the midpoint of $BC$，$BE$ and $CF$ are two altitudes，$BO\cap AC=M$、$CO\cap AB=N$.$(BM,OP)=(CN,OQ)=-1$.Let a point X lies on $OH$，$BX\cap AC=U$，$CX\cap AB=V$，$PU\cap QV=R$，$ABCR\sim AEFY$.Prove that $AX//DY$.
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dagezjm
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dagezjm
#3 h  Mar 11, 2020, 2:44 PM • 5 Y
Y by Hexagrammum16, sameer_chahar12, ironball, qzc, AlastorMoody
Recently, one of the most important step has been solved by @jty123. Here is the complete solution of this problem.
Lemma 1.Let $BH_b,CH_c$ be two altitudes of $\triangle ABC$, then $AH_AH_bH_c\sim APBC$.
Proof of Lemma 1.$AH_A\bot P_bP_c$,so$\angle H_cAH_A=\angle PAC$, then $AH_A=P_bP_c\cot\angle BCA=AP\cos\angle BCA=AP\cdot\frac{H_bH_c}{BC}$. So $AH_AH_bH_c\sim APBC$.

Definition.Let $\triangle P^1P^2P^3$ is the anticevian triangle of $P$ WRT $\triangle ABC$, $\triangle X_1X_2X_3$ is the cevian triangle of $X$ WRT $\triangle ABC$, if the perspector of $\triangle P^1P^2P^3$ and $\triangle X_1X_2X_3$ is $U$, then we say that $X$ is the ceva point of $P$ and $U$, we note that $X=cevamul(P,U)$.

Lemma 2.Let the trilinear coordinates of $P$ and $U$ are $(x,y,z)$ and $(u,v,w)$, then $cevamul(P,U)=((uz + wx)(uy + vx),(vz + wy) (uy + vx),(vz + wy) (uz + wx))$.
Proof of Lemma 2.Trivial.

Corollary 1.$cevamul(P,U)=cevamul(U,P)$.
Proof of Corollary 1.Trivial by Lemma 2.

Lemma 3.Let $\triangle P^1P^2P^3$ is the anticevian triangle of $P$ WRT $\triangle ABC$, $c$ is a conic that $P^1,P^2,P^3$ lie on it.Then $cevamul(P,c)$ is the tangent line of $c$ at $P$.
Proof of Lemma 3. Given three points $X,Y,Z$ lie on $c$, $cevamul(P,X)=U,cevamul(P,Y)=V,cevamul(P,Z)=W$.Then by $A(UV,WB)=(U_1V_1,W_1B)=P^1(XY,ZP^3)=P^2(XY,ZP^3)=(U_2V_2,W_2A)=B(UV,WA)$, we got that $U,V,W$ are collinear. And by $P$ is the fixed point of $cevamul(P,\cdot)$, so $cevamul(P,c)$ is the tangent line of $c$ at $P$.

Corollary 2. Given the cicumcenter $O$ then the image of Stammler hyperbola under $cevamul(O,\cdot)$ is the Euler line.
Proof of Corollary 2. $cevamul(O,O)=O$, then given the symmedian point $K$ and the centroid $G$, we have that $cevamul(O,K)=G$, so by Lemma 3, it holds.

Lemma 4.Let $\triangle DEF$ is the intouch triangle of $\triangle ABC$, $DT$ is an altitude of $\triangle DEF$. Given the incenter $I$ and the orthocenter $H$, we have $\angle HTD=\angle DTI$.
Proof of Lemma 4. Well-known. There is a similar proposition when $\triangle DEF$ is the pedal triangle of a excenter.

Lemma 5. Let $H,O$ be the orthocenter and circumcenter of $\triangle ABC$, then the ceva point of $O$ and the orthocenter of the tangential triangle is $H$.
Proof of Lemma 5. We only prove the situation when $\triangle ABC$ is acute. Let $AO\cap BC=O_1,AH\cap BC=H_1$, $O^1$ is the A-vertex of the anticevian triangle of $O$ WRT $\triangle ABC$, then by $(AO_1,OO^1)=-1$ and $AH_1\bot BC$, $\angle OH_1O_1=\angle O_1H_1O^1$, then use Lemma 4 in the tangential triangle we prove it.

Lemma 6. Let $R$ be a point lies on Stammler hyperbola, and $O$ is the circumcenter, $K$ is the symmedian point, $X=cevamul(O,R)$, $K^1$ is the A-vertex of the anticevian triangle of $K$ WRT $\triangle ABC$. Then let $S$ be a point lies on Jerabek hyperbola of $\triangle ABC$, and $AS//K^1R$, then $S$ is the isogonal conjugate of $X$ WRT $\triangle ABC$.
Proof of Lemma 6.Because of the isogonal conjugate of Euler line is the Jerabek hyperbola, so we only need to prove that $\angle BAX=\angle SAC$. Now we let $G,H$ be the centroid and the orthocenter of $\triangle ABC$, $H'$ is the orthocenter of the tangential triangle. then by $cevamul(O,K)=G,cevamul(O,O)=O,cevamul(O,H')=H$. Now we note $A\cdot\cap BC=\cdot_1$, and $O^1$ is the A-vertex of the anticevian triangle of $O$ WRT $\triangle ABC$.
Then we have $A(SH,KO)=K^1(RO,KH')=O^1(RO,KH')=(X_1O_1,G_1H_1)=A(X_1O_1,G_1H_1)=A(XO,GH)$. So by $H,K,O$ are the isogonal conjugate of $O,G,H$, we get that $S,X$ are isogonal conjugate.

Now we can go back to the beginning problem.
Proof of the problem. Let $cevamul(O,P)=X$, the nine-point center is $Ni$, the symmedian point is $K$, the midpoint of $BC$ is $D$, and $K^1$ is the A-vertex of the anticevian triangle of $K$ WRT $\triangle ABC$, $BE,CF$ are two altitudes of $\triangle ABC$. By Lemma 1, we have $DEH_AF\sim K^1BPC$, so $\angle H_ADNi=\angle DK^1P$, Let $Y$ be the isogonal conjugate of $X$, then by Lemma 6, $AY//K^1P$. So $\angle H_ADNi=\angle DK^1P=\angle HAY=\angle XAO$, then by $DNi//AO$, we get $DH_a//AX$. So $\triangle H_AH_BH_C$ and medial triangleof $\triangle ABC$ are perspective and the perspector is the complement of the ceva point of $O$ and $P$, so it naturally lies on Euler line.
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TelvCohl
2312 posts
TelvCohl
#6 h  Jan 7, 2021, 3:51 AM • 5 Y
Y by enhanced, ForeverHaibara, qzc, amar_04, centslordm
Property (well-known) : Given two (not homothetic) triangles $ \triangle A_1B_1C_1, \triangle A_2B_2C_2 $. Then the locus of points $ P $ such that the parallel from $ A_2, B_2, C_2 $ to $ A_1P, B_1P, C_1P, $ resp. are concurrent is a circumconic of $ \triangle A_1B_1C_1 $ union the line at infinity.

Back to the main problem :

Let $ \triangle M_aM_bM_c, \triangle H_aH_bH_c, \triangle XYZ $ be the medial triangle, orthic triangle, tangential triangle of $ \triangle ABC, $ respectively, then note that $$ AH_AH_bH_cM_a \stackrel{-}{\sim} APBCX,\ \qquad BH_BH_cH_aM_b \stackrel{-}{\sim} BPCAY,\ \qquad CH_CH_aH_bM_c \stackrel{-}{\sim} CPABZ, $$we get the isogonal conjugate of $ H_AM_a, H_BM_b, H_CM_c $ WRT $ \angle M_a, \angle M_b, \angle M_c $ is parallel to $ XP, YP, ZP, $ respectively, so from Property we know that $ \triangle H_AH_BH_C, \triangle M_aM_bM_c $ are perspective iff $ P $ lies on the Stammler hyperbola $ \mathcal{H} $ of $ \triangle ABC. $ If $ P $ lies on $ \mathcal{H}, $ then from Property again we get the isogonal conjugate of the perspector WRT $ \triangle M_aM_bM_c $ lies on the Jerabek hyperbola of $ \triangle M_aM_bM_c, $ so the perspector of $ \triangle H_AH_BH_C, \triangle M_aM_bM_c $ lies on the Euler line of $ \triangle ABC. $
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