Equilateral Triangle in Square with the Cevians through Its Apex

Source

Equilateral Triangle in Square with the Cevians through Its Apex, source

Problem

Equilateral Triangle in Square with the Cevians through Its Apex, problem

Solution

Choose $A=-1+2i,\,$ $B=1+2i,\,$ $C=1,\,$ $D=-1.\,$ Then $P=i\sqrt{3}\,$ and we expect $M=1+i.\,$ Use the barycentric coordinates. From $P=\alpha A+\beta B+\gamma C=1\,$ and $\alpha+\beta+\gamma=1,\,$ we find $\displaystyle \alpha=\frac{1}{2},\,$ $\displaystyle \beta=\frac{\sqrt{3}-1}{2},\,$ $\displaystyle \gamma=\frac{2-\sqrt{3}}{2}.$

We have $\displaystyle \frac{B-Z}{C-Z}=-\frac{\gamma}{\beta}\,$ and, since $Z\,$ is on the line $x=1,\,$ $Z=1+2(\sqrt{3}-1)i.$

Similarly, $\displaystyle X=-\frac{2-\sqrt{3}}{\sqrt{3}}+2i\,$ and $\displaystyle Y=-\frac{1}{\sqrt{3}}+\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)i.\,$ Suffice it to show that $\displaystyle \frac{Z-M}{Z-X}\cdot\frac{Y-X}{Y-M}\,$ is real. Which is equivalent to having the following expressions real:

$\displaystyle \begin{align} &\frac{\displaystyle \sqrt{3}(2-\sqrt{3})i}{\displaystyle 2\left(\frac{1}{\sqrt{3}}-(2-\sqrt{3})i\right)}\cdot\frac{\displaystyle -\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)(1+i)}{\displaystyle \frac{1}{\sqrt{3}}(-(\sqrt{3}+1)+i)}\\ &\qquad\qquad\qquad=\frac{\displaystyle i}{\displaystyle \frac{1}{\sqrt{3}}-(2-\sqrt{3})i}\cdot\frac{1+i}{-(\sqrt{3}+1)+i}\\ &\qquad\qquad\qquad=\frac{\displaystyle -1+i}{\displaystyle \left(\frac{4-\sqrt{3}}{\sqrt{3}}\right)(-1+i)}\\ &\qquad\qquad\qquad=\frac{\displaystyle 1}{\displaystyle \frac{4-\sqrt{3}}{\sqrt{3}}} \end{align}$

which is clearly real.

Acknowledgment

I am grateful to Leo Giugiuc who supplied the solution and a link to the original post at the Peru Geometrico facebook group of the problem by Tran Quang Hung.

 

|Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62050621

Search by google: