Reflect the vertices of a triangle on the opposite sides

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Reflect the vertices of a triangle on the opposite sides

G H J

Reveal topic tags

geometric transformation

geometry unsolved

L

G H BBookmark kLocked kLocked NReply

Source: Kürschák 1985, problem 3

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1059 posts

#1 h Jul 27, 2014, 11:30 AM • 2 Y

Y by Adventure10, Mango247

We reflected each vertex of a triangle on the opposite side. Prove that the area of the triangle formed by these three reflection points is smaller than the area of the initial triangle multiplied by five.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4145 posts

#2 h Jul 31, 2014, 8:03 PM • 2 Y

Y by Adventure10, Mango247

Label $\triangle ABC$ the given triangle with circumcircle $(O,R)$ and $A',B',C'$ denote the reflections of $A,B,C$ on $BC,CA,AB.$ $\triangle A'B'C'$ is homothetic to the pedal triangle $\triangle XYZ$ of its 9-point center $N$ through the homothety with center $G,$ the centroid of $\triangle ABC,$ and coefficient 4 (this has been posted before) $\Longrightarrow$ $[A'B'C']=16[XYZ].$ Hence, it suffices to prove that $16[XYZ] <5[ABC].$ By Euler's theorem for $\triangle XYZ,$ we get

$\frac{[XYZ]}{[ABC]}= \frac{|p(N,(O))|}{4R^2} \Longrightarrow 4\cdot |p(N,(O))|< 5R^2.$

Assume that $N$ lies outside $(O),$ otherwise the inequality is trivial. Then $4(ON^2-R^2)<5R^2$ $\Longrightarrow$ $ON^2 < \frac{_9}{^4}R^2.$ Since $ON=\frac{_3}{^2}OG,$ it follows that $\frac{_9}{^4}OG^2< \frac{_9}{^4}R^2$ $\Longrightarrow$ $OG <R,$ which is certainly true as $G$ is always inside $(O).$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a