AL bisects MS, square, midpoint, angle bisector, 45^o

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

AL bisects MS, square, midpoint, angle bisector, 45^o

G H J

Reveal topic tags

bisects segment

L

G H BBookmark kLocked kLocked NReply

Source: 2021 European Math Tournament Seniors 2 Math Battle 4.5

The post below has been deleted. Click to close.

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

30650 posts

#1 h Mar 29, 2021, 7:17 AM

Y by

In square $ABCD$ , point $M$ is the midpoint of $BC$ . Point $L$ on the side of $CD$ is such that $AM$ is bisector of $\angle BAL$ . Point $S$ on $AD$ is such that $\angle LMS = 45^o$ . Prove that the segment $MS$ is divided in half by $AL$ .

Z K Y

The post below has been deleted. Click to close.

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

332 posts

#2 h Mar 29, 2021, 8:36 AM • 1 Y

Y by Mango247

Let $A=(12,0),B=(0,0),C=(0,12),D = (12,12)$
Then $M = (6,0)$ , Equation of $AM$ is $2x+y-2 = 0$

Note that $\angle MAB = \tan^{-1}{0.5}$ . Hence the slope of line $AL$ will be $- \tan (90 - 2\tan^{-1}{0.5})=\frac{-1}{\tan (2 \tan^{-1}{0.5})} = \frac{-3}{4}$
Hence equation of $AL$ is $3x+4y = 48$

Hence we get that $L = (12,3)$

Now $\tan (\angle LMC) = 0.5$ . Hence slope of $SM = tan(\frac{\pi}{4}+\tan^{-1}(0.5)) = 3$

Hence $S$ has the coordinates $(10,12)$

Now midpoint of $SM = (8,6)$ which satisfies $3x+4y=48$ . Hence it lies on $AL$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a