Perspective Triangles

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
$a_1a_3+a_3a_5+a_5a_7+a_7a_9+a_9a_1=a_2a_4+a_4a_6+a_6a_8+a_8a_{10}+a_{10}a_2.$

1 reply

Miquel-point

3 hours ago

Edward_Tur

35 minutes ago

P,Q,B are collinear

MNJ2357 28

N an hour ago by Ilikeminecraft

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.

28 replies

MNJ2357

Nov 21, 2020

Ilikeminecraft

an hour ago

Chinese Girls Mathematical Olympiad 2017, Problem 7

Hermitianism 45

N an hour 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

an hour ago

D1031 : A general result on polynomial 1

Dattier 1

N an hour ago by Dattier

Source: les dattes à Dattier

Let $P(x,y) \in \mathbb Q(x,y)$ with $\forall (a,b) \in \mathbb Z^2, P(a,b) \in \mathbb Z$ .

Is it true that $P(x,y) \in \mathbb Q[x,y]$ ?

1 reply

Dattier

4 hours ago

Dattier

an hour ago

Asymmetric FE

sman96 18

N an hour ago by jasperE3

Source: BdMO 2025 Higher Secondary P8

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(xf(y)-y) + f(xy-x) + f(x+y) = 2xy$ for all $x, y \in \mathbb{R}$ .

18 replies

sman96

Feb 8, 2025

jasperE3

an hour ago

Easy Geometry

pokmui9909 6

N an hour 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

an hour ago

Old hard problem

ItzsleepyXD 3

N 2 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

2 hours ago

\frac{2^{n!}-1}{2^n-1} be a square

AlperenINAN 10

N 2 hours ago by Nuran2010

Source: Turkey JBMO TST 2024 P5

Find all positive integer values of $n$ such that the value of the
$\frac{2^{n!}-1}{2^n-1}$ is a square of an integer.

10 replies

AlperenINAN

May 13, 2024

Nuran2010

2 hours ago

Beautiful Angle Sum Property in Hexagon with Incenter

Raufrahim68 0

3 hours ago

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
K
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

3 hours ago

0 replies

Anything real in this system must be integer

Assassino9931 7

N 3 hours ago by Leman_Nabiyeva

Source: Al-Khwarizmi International Junior Olympiad 2025 P1

Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
$\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}$ and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia

7 replies

Assassino9931

May 9, 2025

Leman_Nabiyeva

3 hours ago

CIIM 2011 First day problem 3

Ozc 2

N 3 hours ago by pi_quadrat_sechstel

Source: CIIM 2011

Let $f(x)$ be a rational function with complex coefficients whose denominator does not have multiple roots. Let $u_0, u_1,... , u_n$ be the complex roots of $f$ and $w_1, w_2,..., w_m$ be the roots of $f'$ . Suppose that $u_0$ is a simple root of $f$ . Prove that
$\sum_{k=1}^m \frac{1}{w_k - u_0} = 2\sum_{k = 1}^n\frac{1}{u_k - u_0}.$

2 replies

Ozc

Oct 3, 2014

pi_quadrat_sechstel

3 hours ago

IMO 2009 P2, but in space

Miquel-point 1

N 3 hours ago 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

Miquel-point

3 hours ago

Miquel-point

3 hours ago

Perspective Triangles

MNJ2357 2

N Jan 26, 2020 by Pathological

Source: Korea Winter Program Practice Test 2 P4

$\triangle ABC$ and $\triangle A_1B_1C_1$ are perspective triangles. $(ABB_1)$ and $(ACC_1)$ meet at $A_2 (\neq A)$ . Define $B_2,C_2$ analogously. Prove that $AA_2, BB_2,CC_2$ are concurrent.

2 replies

MNJ2357

Jan 23, 2020

Pathological

Jan 26, 2020

Perspective Triangles

G H J

Reveal topic tags

geometry

G H BBookmark kLocked kLocked NReply

Source: Korea Winter Program Practice Test 2 P4

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MNJ2357

644 posts

MNJ2357

#1 h Jan 23, 2020, 3:33 PM • 2 Y

Y by stroller, Adventure10

$\triangle ABC$ and $\triangle A_1B_1C_1$ are perspective triangles. $(ABB_1)$ and $(ACC_1)$ meet at $A_2 (\neq A)$ . Define $B_2,C_2$ analogously. Prove that $AA_2, BB_2,CC_2$ are concurrent.

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stroller

894 posts

stroller

#2 h Jan 23, 2020, 6:03 PM • 3 Y

Y by MNJ2357, Adventure10, Mango247

mOvInG pOiNtS

Lemma: The map $AO \mapsto (BC_1C)$ , $A_1 \mapsto B_2$ is projective.

Proof: After inversion around $B$ , $A_1 \in (ABA_1)$ and $B_2$ is the intersection of line $AA_1$ with line $CC_1$ . Thus the map becomes projective after inversion. Since inversion preserves cross ratios, the map itself is projective, as desired.

Now we induct on $n$ , the number of points among $A_2, B_2, C_2$ that are equal to $O$ . The base case $n = 1$ is true and this follows from the fact that the radical axes of $(AA_1B), (AA_1C),(BOC)$ concur.

Suppose the statement holds for $n-1$ and we prove it for $n (\le 3)$ . WLOG $A_1 \ne O$ . The maps $A_1 \mapsto BB_2, CC_2$ are both projective, hence it suffices to verify the statement for three positions of $A_1$ . We may take $A_1 = O$ (and have the result from induction hypothesis), $\infty$ (in which case the lines concur at $A$ ) and the point on $AO$ such that $AA_1B_1B$ is cyclic.

why am I such a troll

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Pathological

578 posts

Pathological

#4 h Jan 26, 2020, 11:29 PM • 3 Y

Y by stroller, Pluto1708, Adventure10

Easy Trig

Let $O_{ab}$ denote the circumcenter of $\triangle ABB_1$ , and define $O_{ba}, O_{bc}, O_{cb}, O_{ca}, O_{ac}$ analogously.

For a point $P$ and line $\ell$ , let $f(P, \ell)$ denote the line through $P$ perpendicular to $\ell.$

Note that $AA_2$ is the line $f(A, O_{ab}O{ac}).$ Also, $AP$ is the line $f(A, O_{ba}O_{ca}).$ Since it's obvious that $f(A, O_{ba}O_{ca}), f(B, O_{ab}O_{cb}), f(C, O_{ac}O_{bc})$ are concurrent (at $P$ ), we get from sine Ceva in $\triangle ABC$ that $O_{ab} \cdot O_{bc} \cdot O_{ca} = O_{ba} \cdot O_{cb} \cdot O_{ac}$ .

As this condition is clearly symmetric in $a, b, c$ , it follows that $f(A, O_{ab}O_{ac}), f(B, O_{ba}O_{bc}), f(C, O_{ac}O_{bc})$ are concurrent as well.

$\square$

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