BC//XY wanted, altitudes, midpoint (2021 All-Siberian Qualifying II 10.4)

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

BC//XY wanted, altitudes, midpoint (2021 All-Siberian Qualifying II 10.4)

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

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

30629 posts

#1 h May 19, 2021, 6:04 PM • 1 Y

Y by Mango247

Let $AH,BP$ and $CT$ be the altitudes, and $M$ be the midpoint of the side $BC$ in an acute-angled triangle $ABC$ . Line $PM$ intersects the extension of side $AB$ beyond the vertex $B$ at point $Y$ , and line $TH$ intersects the extension of the side $AC$ beyond the vertex $C$ at the point $X$ . Prove that the lines $BC$ and $XY$ are parallel.

Z K Y

The post below has been deleted. Click to close.

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

8945 posts

#2 h May 21, 2021, 7:47 PM

Y by

Let $A(0,a),B(-b,0),C(c,0)$ .
Then $M(\frac{c-b}{2},0)$ .
$H(0,0)$ .

Equation of $AB\ :\ y=\frac{a}{b}(x+b)$ , equation of $AC\ :\ y=-\frac{a}{c}(x-c)$ .

Points $P(\frac{c(a^{2}-bc)}{a^{2}+c^{2}},\frac{ac(b+c)}{a^{2}+c^{2}})$ and $T(\frac{b(bc-a^{2})}{a^{2}+b^{2}},\frac{ab(b+c)}{a^{2}+b^{2}})$ .

Points $Y(\frac{b(bc-a^{2})}{a^{2}-2bc-c^{2}},\frac{ac(b+c)}{c^{2}+2bc-a^{2}})$ and $X(\frac{c(a^{2}-bc)}{a^{2}-2bc-c^{2}},\frac{ac(b+c)}{c^{2}+2bc-a^{2}})$ .

Because $y_{Y}=y_{X}$ , is $XY\ //\ BC$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a