Some geometry

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

Some geometry

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: St Petersburg Olympiad 2010, Grade 9, P7

The post below has been deleted. Click to close.

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

4900 posts

#1 h Sep 14, 2017, 9:43 AM • 2 Y

Y by Adventure10, Mango247

Incircle of $ABC$ tangent $AB,AC,BC$ in $C_1,B_1,A_1$ . $AA_1$ intersect incircle in $E$ . $N$ is midpoint $B_1A_1$ . $M$ is symmetric to $N$ relatively $AA_1$ . Prove that $\angle EMC= 90$

This post has been edited 1 time. Last edited by RagvaloD, Jun 22, 2018, 9:26 AM

Z K Y

The post below has been deleted. Click to close.

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

402 posts

GGPiku

#2 h Sep 14, 2017, 4:39 PM • 2 Y

Y by Adventure10, Mango247

Let $2a,2b,2c$ be the angles of the triangle.Quick angle chasing gives $\angle MA_1C=270-2a-3c$ AND $\angle EIC=270-2a-3c$ . Now $\frac{NA_1}{A_1D}=\frac{A_1C}{A_1D}=\frac{IA_1}{IC}=\frac{IE}{IC}$ . So $\triangle MA_1C~\triangle EIC$ . So $C$ is the center of the spiral similarity that sends $A_1$ to $I$ and $M$ to $E$ . Thus, $\angle EMC=\angle IA_1C$ .

Z K Y

The post below has been deleted. Click to close.

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

578 posts

#3 h May 26, 2019, 1:42 AM • 2 Y

Y by Adventure10, Mango247

Alternatively, one could observe that $EN$ and $EC$ are isogonal w.r.t. $\angle B_1EA_1$ . Therefore, we have that $\angle CEM = \angle CEA_1 + \angle MEA_1 = \angle CEA_1 + \angle NEA_1 = \angle B_1EA_1,$ and so it suffices to show that $EM = EC \cdot \cos \angle B_1EA_1.$ However, it's well-known that $EN = EC \cdot \cos \angle B_1EA_1$ , and so we're done.

$\square$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a