Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
Sequence of projections is convergent
Filipjack   0
17 minutes ago
Source: Romanian National Olympiad 1997 - Grade 10 - Problem 3
A point $A_0$ and two lines $d_1$ and $d_2$ are given in the space. For each nonnegative integer $n$ we denote by $B_n$ the projection of $A_n$ on $d_2,$ and by $A_{n+1}$ the projection of $B_n$ on $d_1.$ Prove that there exist two segments $[A'A''] \subset d_1$ and $[B'B''] \subset d_2$ of length $0.001$ and a nonnegative integer $N$ such that $A_n \in [A'A'']$ and $B_n \in [B'B'']$ for any $n \ge N.$
0 replies
Filipjack
17 minutes ago
0 replies
Right-angled triangle if circumcentre is on circle
liberator   76
N an hour ago by numbertheory97
Source: IMO 2013 Problem 3
Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.

Proposed by Alexander A. Polyansky, Russia
76 replies
liberator
Jan 4, 2016
numbertheory97
an hour ago
APMO 2016: Great triangle
shinichiman   26
N 2 hours ago by ray66
Source: APMO 2016, problem 1
We say that a triangle $ABC$ is great if the following holds: for any point $D$ on the side $BC$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $AB$ and $AC$, respectively, then the reflection of $D$ in the line $PQ$ lies on the circumcircle of the triangle $ABC$. Prove that triangle $ABC$ is great if and only if $\angle A = 90^{\circ}$ and $AB = AC$.

Senior Problems Committee of the Australian Mathematical Olympiad Committee
26 replies
shinichiman
May 16, 2016
ray66
2 hours ago
IMO ShortList 2001, geometry problem 2
orl   48
N 2 hours ago by legogubbe
Source: IMO ShortList 2001, geometry problem 2
Consider an acute-angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$.
48 replies
orl
Sep 30, 2004
legogubbe
2 hours ago
Projective Electrostatistics
drago.7437   0
2 hours ago
Source: Own
Given two charges of any magnitude , a third charge collinear with them , exists such that it is in equillibirum , Prove that if a fourth charge in the same line exists such that it is in equillibrium , then the 3rd charge and the fourth charge are harmonic conjugates with respect to the two fixed charges . , For example if two +q charges are fixed then if in their midpoint placed a charge -q , it is in equillibrium , also if the same charge -q is placed at infinity the system is again in equillibrium , and the midpoint and the point at infinity are harmonic conjugates .
0 replies
drago.7437
2 hours ago
0 replies
Find an angle
socrates   3
N 2 hours ago by Nari_Tom
Source: Baltic Way 2016, Problem 18
Let $ABCD$ be a parallelogram such that $\angle BAD = 60^{\circ}.$ Let $K$ and $L$ be the midpoints of $BC$ and $CD,$ respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD.$
3 replies
socrates
Nov 5, 2016
Nari_Tom
2 hours ago
Two Orthocenters and an Invariant Point
Mathdreams   1
N 3 hours ago by RANDOM__USER
Source: 2025 Nepal Mock TST Day 1 Problem 3
Let $\triangle{ABC}$ be a triangle, and let $P$ be an arbitrary point on line $AO$, where $O$ is the circumcenter of $\triangle{ABC}$. Define $H_1$ and $H_2$ as the orthocenters of triangles $\triangle{APB}$ and $\triangle{APC}$. Prove that $H_1H_2$ passes through a fixed point which is independent of the choice of $P$.

(Kritesh Dhakal, Nepal)
1 reply
Mathdreams
4 hours ago
RANDOM__USER
3 hours ago
Geometry
youochange   2
N 3 hours ago by youochange
m:}
Let $\triangle ABC$ be a triangle inscribed in a circle, where the tangents to the circle at points $B$ and $C$ intersect at the point $P$. Let $M$ be a point on the arc $AC$ (not containing $B$) such that $M \neq A$ and $M \neq C$. Let the lines $BC$ and $AM$ intersect at point $K$. Let $P'$ be the reflection of $P$ with respect to the line $AM$. The lines $AP'$ and $PM$ intersect at point $Q$, and $PM$ intersects the circumcircle of $\triangle ABC$ again at point $N$.

Prove that the point $Q$ lies on the circumcircle of $\triangle ANK$.
2 replies
youochange
6 hours ago
youochange
3 hours ago
Beautiful problem
luutrongphuc   20
N 3 hours ago by r7di048hd3wwd3o3w58q
Let triangle $ABC$ be circumscribed about circle $(I)$, and let $H$ be the orthocenter of $\triangle ABC$. The circle $(I)$ touches line $BC$ at $D$. The tangent to the circle $(BHC)$ at $H$ meets $BC$ at $S$. Let $J$ be the midpoint of $HI$, and let the line $DJ$ meet $(I)$ again at $X$. The tangent to $(I)$ parallel to $BC$ meets the line $AX$ at $T$. Prove that $ST$ is tangent to $(I)$.
20 replies
luutrongphuc
Apr 4, 2025
r7di048hd3wwd3o3w58q
3 hours ago
Common tangent to diameter circles
Stuttgarden   1
N 3 hours ago by jrpartty
Source: Spain MO 2025 P2
The cyclic quadrilateral $ABCD$, inscribed in the circle $\Gamma$, satisfies $AB=BC$ and $CD=DA$, and $E$ is the intersection point of the diagonals $AC$ and $BD$. The circle with center $A$ and radius $AE$ intersects $\Gamma$ in two points $F$ and $G$. Prove that the line $FG$ is tangent to the circles with diameters $BE$ and $DE$.
1 reply
Stuttgarden
Mar 31, 2025
jrpartty
3 hours ago
Ratios in a right triangle
PNT   1
N 4 hours ago by Mathzeus1024
Source: Own.
Let $ABC$ be a right triangle in $A$ with $AB<AC$. Let $M$ be the midpoint of $AB$ and $D$ a point on $AC$ such that $DC=DB$. Let $X=(BDC)\cap MD$.
Compute in terms of $AB,BC$ and $AC$ the ratio $\frac{BX}{DX}$.
1 reply
PNT
Jun 9, 2023
Mathzeus1024
4 hours ago
parallel
phuong   2
N Sep 17, 2020 by jayme
Source: unsolved geometry
Let ABC be a triangle and its incircle touches BC,CA,AB at D,E, F, reps. AD cuts (I) again at G. H lies on line EF such that $GH \perp AD$. Prove that $AH \parallel BC$.
2 replies
phuong
Sep 26, 2016
jayme
Sep 17, 2020
parallel
G H J
G H BBookmark kLocked kLocked NReply
Source: unsolved geometry
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
phuong
210 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let ABC be a triangle and its incircle touches BC,CA,AB at D,E, F, reps. AD cuts (I) again at G. H lies on line EF such that $GH \perp AD$. Prove that $AH \parallel BC$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jayme
9775 posts
#2 • 1 Y
Y by Adventure10
Dear Mathlinkers,
have a look at
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=111468

Sincerely
Jean-Louis
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jayme
9775 posts
#3
Y by
Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/01.%201.%20Parallele%20a%20un%20cote%20du%20triangle%20passant%20par%20un%20sommet%20du%20triangle.pdf

p. 6-7.

Sincerely
Jean-Louis
Z K Y
N Quick Reply
G
H
=
a