Simple but hard

by Lukariman, May 16, 2025, 2:47 AM

Given triangle ABC. Outside the triangle, construct rectangles ACDE and BCFG with equal areas. Let M be the midpoint of DF. Prove that CM passes through the center of the circle circumscribing triangle ABC.

geometry

Angle Relationships in Triangles

by steven_zhang123, May 14, 2025, 11:09 PM

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 $\angle ACB = 2\angle ABC$ .

angle

geometry

angle bisector

Perpendicular passes from the intersection of diagonals, \angle AEB = \angle CED

by NO_SQUARES, May 5, 2025, 5:34 PM

Inside of convex quadrilateral $ABCD$ point $E$ was chosen such that $\angle DAE = \angle CAB$ and $\angle ADE = \angle CDB$ . Prove that if perpendicular from $E$ to $AD$ passes from the intersection of diagonals of $ABCD$ , then $\angle AEB = \angle CED$ .

geometry

GMO 2024 P1

by Z4ADies, Oct 20, 2024, 10:39 PM

Let $ABC$ be an acute triangle. Define $I$ as its incenter. Let $D$ and $E$ be the incircle's tangent points to $AC$ and $AB$ , respectively. Let $M$ be the midpoint of $BC$ . Let $G$ be the intersection point of a perpendicular line passing through $M$ to $DE$ . Line $AM$ intersects the circumcircle of $\triangle ABC$ at $H$ . The circumcircle of $\triangle AGH$ intersects line $GM$ at $J$ . Prove that quadrilateral $BGCJ$ is cyclic.

Author:Ismayil Ismayilzada (Azerbaijan)

This post has been edited 1 time. Last edited by Z4ADies, Oct 21, 2024, 7:20 AM

geometry

concyclic wanted, PQ = BP, cyclic quadrilateral and 2 parallelograms related

by parmenides51, Sep 25, 2020, 4:27 AM

Let $ABCD$ be a cyclic quadrilateral in which the lines $BC$ and $AD$ meet at a point $P$ . Let $Q$ be the point of the line $BP$ , different from $B$ , such that $PQ = BP$ . We construct the parallelograms $CAQR$ and $DBCS$ . Prove that the points $C, Q, R, S$ lie on the same circle.

This post has been edited 1 time. Last edited by parmenides51, Sep 25, 2020, 4:30 AM

geometry

cyclic quadrilateral

parallelogram

Collinearity with orthocenter

by liberator, Jan 4, 2016, 9:38 PM

Let $ABC$ be an acute triangle with orthocenter $H$ , and let $W$ be a point on the side $BC$ , lying strictly between $B$ and $C$ . The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$ , respectively. Denote by $\omega_1$ is the circumcircle of $BWN$ , and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$ . Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$ , and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$ . Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand

geometry

IMO

RMM 2013 Problem 3

by dr_Civot, Mar 2, 2013, 10:43 AM

Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$ . The lines $AB$ and $CD$ meet at $P$ , the lines $AD$ and $BC$ meet at $Q$ , and the diagonals $AC$ and $BD$ meet at $R$ . Let $M$ be the midpoint of the segment $PQ$ , and let $K$ be the common point of the segment $MR$ and the circle $\omega$ . Prove that the circumcircle of the triangle $KPQ$ and $\omega$ are tangent to one another.

geometry

circumcircle

geometric transformation

homothety

reflection

Prove angles are equal

by BigSams, May 13, 2011, 2:20 AM

Let $ABC$ be an acute triangle. Let $AD$ be the altitude on $BC$ , and let $H$ be any interior point on $AD$ . Lines $BH,CH$ , when extended, intersect $AC,AB$ at $E,F$ respectively. Prove that $\angle EDH=\angle FDH$ .

Two circles, a tangent line and a parallel

by Valentin Vornicu, Oct 24, 2005, 10:15 AM

Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$ . Let $AB$ be the line tangent to these circles at $A$ and $B$ , respectively, so that $M$ lies closer to $AB$ than $N$ . Let $CD$ be the line parallel to $AB$ and passing through the point $M$ , with $C$ on $G_1$ and $D$ on $G_2$ . Lines $AC$ and $BD$ meet at $E$ ; lines $AN$ and $CD$ meet at $P$ ; lines $BN$ and $CD$ meet at $Q$ . Show that $EP = EQ$ .

Problem 5 (Second Day)

by darij grinberg, Jul 13, 2004, 2:49 PM

In a convex quadrilateral $ABCD$ , the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$ . The point $P$ lies inside $ABCD$ and satisfies $\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.$ Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$ .

Attachments:

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math_exploration

by Lukariman, May 16, 2025, 2:47 AM

by steven_zhang123, May 14, 2025, 11:09 PM

by NO_SQUARES, May 5, 2025, 5:34 PM

by Z4ADies, Oct 20, 2024, 10:39 PM

by parmenides51, Sep 25, 2020, 4:27 AM

by liberator, Jan 4, 2016, 9:38 PM

by dr_Civot, Mar 2, 2013, 10:43 AM

by BigSams, May 13, 2011, 2:20 AM

by Valentin Vornicu, Oct 24, 2005, 10:15 AM

by darij grinberg, Jul 13, 2004, 2:49 PM