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

G
Topic
First Poster
Last Poster
Collinear points
tenplusten   2
N 3 hours ago by Blackbeam999
Let $A,B,C$ be three collinear points and $D,E,F$ three other collinear points. Let $G,H,I$ be the intersection of the lines $BE,CF$ $AD,CF$ and $AD,CE$,respectively. If $AI=HD$ and $CH=GF$.Prove that $BI=GE$



I hope you will use Pappus theorem in your solutions.
2 replies
tenplusten
Jun 20, 2016
Blackbeam999
3 hours ago
Simple Geometry
AbdulWaheed   0
3 hours ago
Source: EGMO
Try to avoid Directed angles
Let ABC be an acute triangle inscribed in circle $\Omega$. Let $X$ be the midpoint of the arc $\overarc{BC}$ not containing $A$ and define $Y, Z$ similarly. Show that the orthocenter of $XYZ$ is the incenter $I$ of $ABC$.
0 replies
AbdulWaheed
3 hours ago
0 replies
IMO Shortlist 2014 G6
hajimbrak   30
N Today at 12:01 AM by awesomeming327.
Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$ . Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$ , respectively. We call the pair $(E, F )$ $\textit{interesting}$, if the quadrilateral $KSAT$ is cyclic.
Suppose that the pairs $(E_1 , F_1 )$ and $(E_2 , F_2 )$ are interesting. Prove that $\displaystyle\frac{E_1 E_2}{AB}=\frac{F_1 F_2}{AC}$
Proposed by Ali Zamani, Iran
30 replies
hajimbrak
Jul 11, 2015
awesomeming327.
Today at 12:01 AM
60^o angle wanted, equilateral on a square
parmenides51   4
N Yesterday at 11:49 PM by MathIQ.
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 2
A square $ABCD$ is given. Over the side $BC$ draw an equilateral triangle $BCS$ on the outside. The midpoint of the segment $AS$ is $N$ and the midpoint of the side $CD$ is $H$. Prove that $\angle NHC = 60^o$.
.
(Karl Czakler)
4 replies
parmenides51
Dec 18, 2020
MathIQ.
Yesterday at 11:49 PM
Serbian selection contest for the IMO 2025 - P2
OgnjenTesic   9
N Yesterday at 11:47 PM by hectorleo123
Source: Serbian selection contest for the IMO 2025
Let $ABC$ be an acute triangle. Let $A'$ be the reflection of point $A$ over the line $BC$. Let $O$ and $H$ be the circumcenter and the orthocenter of triangle $ABC$, respectively, and let $E$ be the midpoint of segment $OH$. Let $D$ and $L$ be the points where the reflection of line $AA'$ with respect to line $OA'$ intersects the circumcircle of triangle $ABC$, where point $D$ lies on the arc $BC$ not containing $A$. If \( M \) is a point on the line \( BC \) such that \( OM \perp AD \), prove that \( \angle MAD = \angle EAL \).

Proposed by Strahinja Gvozdić
9 replies
OgnjenTesic
Yesterday at 4:02 PM
hectorleo123
Yesterday at 11:47 PM
collinear wanted, regular hexagon
parmenides51   3
N Yesterday at 11:34 PM by MathIQ.
Source: 2023 Austrian Mathematical Olympiad , Junior Regional Competition , Problem 2
Let $ABCDEF$ be a regular hexagon with sidelength s. The points $P$ and $Q$ are on the diagonals $BD$ and $DF$, respectively, such that $BP = DQ = s$. Prove that the three points $C$, $P$ and $Q$ are on a line.

(Walther Janous)
3 replies
parmenides51
Mar 26, 2024
MathIQ.
Yesterday at 11:34 PM
Geometric inequality with angles
Amir Hossein   7
N Yesterday at 11:06 PM by MathIQ.
Let $p, q$, and $r$ be the angles of a triangle, and let $a = \sin2p, b = \sin2q$, and $c = \sin2r$. If $s = \frac{(a + b + c)}2$, show that
\[s(s - a)(s - b)(s -c) \geq 0.\]
When does equality hold?
7 replies
Amir Hossein
Sep 1, 2010
MathIQ.
Yesterday at 11:06 PM
IMO 2014 Problem 3
v_Enhance   103
N Yesterday at 10:59 PM by Mysteriouxxx
Source: 0
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
103 replies
v_Enhance
Jul 8, 2014
Mysteriouxxx
Yesterday at 10:59 PM
three discs of radius 1 cannot cover entirely a square surface of side 2
parmenides51   1
N Yesterday at 8:17 PM by Blast_S1
Source: 2014 Romania NMO VIII p4
Prove that three discs of radius $1$ cannot cover entirely a square surface of side $2$, but they can cover more than $99.75\%$ of it.
1 reply
parmenides51
Aug 15, 2024
Blast_S1
Yesterday at 8:17 PM
2025 Caucasus MO Juniors P6
BR1F1SZ   2
N Yesterday at 7:38 PM by IEatProblemsForBreakfast
Source: Caucasus MO
A point $P$ is chosen inside a convex quadrilateral $ABCD$. Could it happen that$$PA = AB, \quad PB = BC, \quad PC = CD \quad \text{and} \quad PD = DA?$$
2 replies
BR1F1SZ
Mar 26, 2025
IEatProblemsForBreakfast
Yesterday at 7:38 PM
Radius of circle tangent to two equal circles and a common line
rilarfer   1
N Apr 19, 2025 by Lankou
Source: ASJTNic 2005
Two circles of radius 2 are tangent to each other and to a straight line. A third circle is placed so that it is tangent to both of the other circles and also tangent to the same straight line.

What is the radius of the third circle?

IMAGE
1 reply
rilarfer
Apr 19, 2025
Lankou
Apr 19, 2025
Radius of circle tangent to two equal circles and a common line
G H J
G H BBookmark kLocked kLocked NReply
Source: ASJTNic 2005
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
rilarfer
26 posts
#1
Y by
Two circles of radius 2 are tangent to each other and to a straight line. A third circle is placed so that it is tangent to both of the other circles and also tangent to the same straight line.

What is the radius of the third circle?

[asy]
size(150);
draw((-1,0)--(3,0)); // ground line
draw(circle((0,1),1)); // left big circle
draw(circle((2,1),1)); // right big circle
draw(circle((1,0.25),0.25)); // small circle in between
[/asy]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Lankou
1405 posts
#2 • 1 Y
Y by teomihai
We have the relation $(2+r)^2=(2-r)^2+2^2$; $r=\frac{1}{2}$
This post has been edited 1 time. Last edited by Lankou, Apr 19, 2025, 7:08 PM
Z K Y
N Quick Reply
G
H
=
a