If $P$,$Q$,$R$ are midpoints of $BX$,$BZ$,$AC$, prove $\triangle PQR$ is equilateral.

Question

Equilateral triangles $BCX$, $CAY$ and $ABZ$ are constructed externally on the sides of $\triangle ABC$. If $P$, $Q$, $R$ are the midpoints of $BX$, $BZ$ and $AC$, prove that $\triangle PQR$ is equilateral.

My try -

I couldn't have made any progress in this question after spending an hour ... kindly provide me a hint to start.....

Sources - CTPCM

Answer 1

Algebraic proof:

Solution using complex numbers. Let $a,b,c\in \Bbb C$ be the affixes for the points $A,B,C$, and we will use the convention, that lower case letters denote the affix point of the corresponding upper case geometric point. We may and do assume $b=0$. Let $u=e^{i\pi/3}=\cos 60^\circ+i\sin 60^\circ=\frac 12(1+i\sqrt 3)$ be this primitive root of unity of order six, so the multilicaiton with $u$ is implementing geometrically the $60^\circ$-rotation centered in $B$. One computes easily $u^2=e^{2i\pi/3}=\frac 12(-1+i\sqrt 3)=u-1$, i.e. $u^2-u+1=0$, et caetera. Then: $$ \begin{aligned} x &= ce^{-i\pi/3}\ ,& p &=\frac 12(b+x)=\frac 12x=\frac 12ce^{-i\pi/3}\ ,\\ z &= ae^{+i\pi/3}\ ,& q &=\frac 12(b+z)=\frac 12z=\frac 12ae^{+i\pi/3}\ ,\\ && r&=\frac 12(a+c)\ ,\\ &&|p-q|&=\frac 12|ce^{-i\pi/3}-ae^{+i\pi/3}|=\frac 12|ae^{2i\pi/3}-c|=\frac 12|au^2-c|\ ,\\ &&|r-q|&=\frac 12|a(1-e^{i\pi/3})+c|=\frac 12|a(1-u)+c|=\frac 12|-au^2+c|\ ,\\ &&|r-p|&=\frac 12|a+(1-e^{-i\pi/3})c|=\frac 12|au^2+(u^2-u)c|=\frac 12|au^2-c|\ ,\\ \end{aligned} $$ and we get $|p-q|=|q-r|=|r-p|$, so $\Delta PQR$ equilateral. Job done.

$\square$

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Bonus 1: The proposition remains valid if the two triangles are constructed "inside" (not "outside"). Proof: Exchange $a\leftrightarrow c$.

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Bonus 2: The center of $\Delta PQR$ is located in $$ \frac 13(p+q+r) =\frac 16(a(1+u)+c(1-u))= \frac {\sqrt 3}2\cdot \frac 13(b+av+c\bar v)\ . $$ Above, the value of $(1+u)$ was calculate like $$(1+u)=1+\frac 12(1+i\sqrt 3) =\frac 12(3+i\sqrt 3) =\sqrt 3\cdot\frac 12(\sqrt 3+i)=\sqrt 3\cdot v\ ,$$ where $v=\cos 30^\circ+i\sin 30^\circ$ is a primitive root of order $12$ of unity.

This point is the following one.

• The point $av$ corresponds to a $+30^\circ$-rotation centered in $B$ applied on $A$. After multiplication with $\sqrt 3/2$, this gives the mid point of $AZ$.
• The point $c\bar v$ corresponds to a $-30^\circ$-rotation centered in $B$ applied on $C$. After multiplication with $\sqrt 3/2$, this gives the mid point of $CX$.
• Consider the above two points, together with $B$, they determine a triangle, take its centroid, let us denote it by $O$. Then $O$ is the center of $\Delta PQR$.
• The center $O$ can be alternatively constructed as follows. Let $X'$ be the (in)center of $\Delta BCX$, and let $Z'$ be the (in)center of $\Delta ABZ$. Then $O$ is the mid point of $X'Z'$.

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Bonus 3: The synthetic solution can now be easily found, knowing that we have to do something with $au^2$ (and with c\bar u^2). (See the formulas for $|p-q|$, $|r-q|$, $|r-p|$.) We construct the point corresponding to $au^2$, this is $A'$, the reflection of $A$ w.r.t. the segment $BZ$ (or w.r.t. the point $Q$), and also construct the point corresponding to $c\bar u^2$, this is $C'$, the reflection of $C$ w.r.t. the segment $BX$ (or w.r.t. the point $P$.)

Synthetic proof: A $60^\circ$-rotation around $B$ shows $\Delta ZBX\equiv \Delta A'BC$, which gives $ZX=A'C$, and passing to the corresponding parallel mid line in appropiate two triangles, $$ QP=QR\ . $$ Similarly $PQ=PR$.

$\square$

Yes, this proof is shorter, explaining how to get it makes the story longer.

Answer 2

You ask for a hint.

Let $A=(0,0),\space B=(a,0),\space C=(b,c)$ (Where $a,b,c$ are not the sides of your triangle $\triangle ABC$ at starting point).

By property of equilateral triangles you have

(1) Point $X$ is located in the $BC$ side mediatrix at a distance of $\dfrac{\sqrt3}{2}BC$.

(2) Similarly Point $Z$ is located in the $AB$ side mediatrix at a distance of $\dfrac{\sqrt3}{2}AB$.

(3) Find the midpoints of $BX,\space BZ$ and $AC$.

(4) Verify the corresponding distances are equal.