Surprising Length Dependence In Equilateral Triangle by Miguel Ochoa Sanchez

Source

Surprising Length Dependence In Equilateral Triangle by Miguel Ochoa Sanchez, source

Problem

Surprising Length Dependence In Equilateral Triangle by Miguel Ochoa Sanchez, problem

Solution 1

Surprising Length Dependence In Equilateral Triangle by Miguel Ochoa Sanchez, Solution 1

Surprising Length Dependence In Equilateral Triangle by Miguel Ochoa Sanchez, Solution 1

Solution 2

Let the actual distance trilinear coordinates of point $P$ be $(m,n,l)$ and $(m,n,p)$ with respect to the triangles $AEF$ and $ABC$, respectively.

From the equation of the circumcircle of an equilateral triangle in trilinear coordinates (for $AEF$), we have

$mn+nl+lm=0.$

In any equilateral triangle, the trilinear coordinates $(x,y,z)$ satisfy $a(x+y+z)=2S,\,$ where $a$ is the side and $S$ is the area. Noting that $\displaystyle\frac{S}{a}$ for $ABC$ is twice that for $AEF$.

$(m+n+p)=2(m+n+l).$

Eliminating $l$ from the two equations,

$\displaystyle p=\frac{m^2+n^2}{m+n}.$

Solution 3

Let $Q'\,$ be the intersection of $MN\,$ and $QP.\,$ Then the quadrilaterals $ENPQ'\,$ and $FMQ'P\,$ are cyclic; $\angle EPF=120^{\circ}=\angle MPN.\,$ For every point $X\,$ on the arc $EPN\,$ of $(ENPQ'),\,$ the second intersection $Y\,$ of $XQ'\,$ with $(FMQ'P)\,$ has the property that $XPY=120^{\circ},\,$ implying that $MN\,$ passes through $Q'\,$ for an assumption to contrary leads to a contradiction.

Surprising Length Dependence In Equilateral Triangle by Miguel Ochoa Sanchez, Solution 2

It follows that $PQ'\,$ is the bisector of angle at $P\,$ in $\Delta MNP,\,$ from which we conclude that $\displaystyle PQ'=\frac{mn}{m+n}.$

Assuming, WLOG, $AB=4\,$ we have $QQ'=\sqrt{3}\,$ and, according to Viviani's theorem, $m+n+PQ=2\sqrt{3}.\,$ Since also $\displaystyle \sqrt{3}-\frac{mn}{m+n}=PQ,\,$ we have

$\displaystyle (m+n+PQ)+2\left(\sqrt{3}-\frac{mn}{m+n}\right)=2\sqrt{3}+2\cdot PQ,$

from which $\displaystyle PQ=m+n-2\frac{mn}{m+n}=\frac{m^2+n^2}{m+n}.$

Acknowledgment

This is a problem by Miguel Ochoa Sanchez that was posted at the Peru Geometrico facebook group. Solution 1 is by Thanos Kalogerakis; Solution 2 is by Amit Itagi.

 

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]