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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
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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
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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
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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 $\angle NHC = 60^o$.
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(Karl Czakler)
4 replies
parmenides51
Dec 18, 2020
MathIQ.
Yesterday at 11:49 PM
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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
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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
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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
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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
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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
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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
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Is the geometric function injective?
Project_Donkey_into_M4   1
N Apr 20, 2025 by Funcshun840
Source: Mock RMO TDP and Kayak 2018, P3
A non-degenerate triangle $\Delta ABC$ is given in the plane, let $S$ be the set of points which lie strictly inside it. Also let $\mathfrak{C}$ be the set of circles in the plane. For a point $P \in S$, let $A_P, B_P, C_P$ be the reflection of $P$ in sides $\overline{BC}, \overline{CA}, \overline{AB}$ respectively. Define a function $\omega: S \rightarrow \mathfrak{C}$ such that $\omega(P)$ is the circumcircle of $A_PB_PC_P$. Is $\omega$ injective?

Note: The function $\omega$ is called injective if for any $P, Q \in S$, $\omega(P) = \omega(Q) \Leftrightarrow P = Q$
1 reply
Project_Donkey_into_M4
Apr 20, 2025
Funcshun840
Apr 20, 2025
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Is the geometric function injective?
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Source: Mock RMO TDP and Kayak 2018, P3
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Project_Donkey_into_M4
163 posts
Project_Donkey_into_M4
#1 h  Apr 20, 2025, 6:23 PM
Y by
A non-degenerate triangle $\Delta ABC$ is given in the plane, let $S$ be the set of points which lie strictly inside it. Also let $\mathfrak{C}$ be the set of circles in the plane. For a point $P \in S$, let $A_P, B_P, C_P$ be the reflection of $P$ in sides $\overline{BC}, \overline{CA}, \overline{AB}$ respectively. Define a function $\omega: S \rightarrow \mathfrak{C}$ such that $\omega(P)$ is the circumcircle of $A_PB_PC_P$. Is $\omega$ injective?

Note: The function $\omega$ is called injective if for any $P, Q \in S$, $\omega(P) = \omega(Q) \Leftrightarrow P = Q$
This post has been edited 1 time. Last edited by Project_Donkey_into_M4, Apr 20, 2025, 6:24 PM
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Funcshun840
28 posts
Funcshun840
#2 h  Apr 20, 2025, 9:04 PM
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I’m confused, doesn’t the injectivity of the function follow from the fact that the center of $\omega(P)$ is the isogonal conjugate of $P$?
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