# Equilateral Triangle: An Interesting Result

Here's an interesting problem, and result, that I wish to share with the math community here at Math SE. The above problem has two methods.

1. Pure geometry. A bit of angle chasing and standard results from circles help us arrive at the desired result - ∆MNP is equilateral. I'll put up a picture of the angle chasing part here (I hope someone edits it, and puts up a picture using GeoGebra or some similar software - I'm sorry I'm not good at editing)

I joined BP and CN for angle chasing purposes.

(Note that if M is the midpoint of arc BC, then the figure so formed is a star, with 8 equilateral triangles) 1. This can be solved beautifully using complex numbers. The vertices of the ∆ can be assumed to be 1,W,W2 on a unit circle centered at origin.

We need to prove that the new equilateral triangle is essentially a rotation of the original one, about an axis passing through center of its circumcircle perpendicular to its plane.

I haven't posted the solution, hope you fellow Math SE members try the problem and post your solutions and ideas in the answers section.

More methods (apart from geometry and complex numbers) are welcome. I'd like to know more about why this result is interesting in itself, and what other deductions can be made from it.

P.S. Please use LaTeX wherever necessary and edit this article. Thanks a lot!

• +1 for a nicely posed question. the new equilateral triangle is essentially a rotation of the original one For a different geometric hint, consider the diameter perpendicular to $AM\,$, which is orthogonal to and bisects each of $AM, BN, CP\,$ (why?). Then $\triangle MNP$ is simply the reflection of $\triangle ABC$ across this diameter. – dxiv Dec 29 '17 at 5:35
• Why don't you learn basic MathJax yourself instead of asking people to do it for you? – user21820 Dec 29 '17 at 6:46
• Can you state the result in the title, instead of claiming that it is "interesting"? (Because personally I don't find many results about triangles to be interesting at all.) – Asaf Karagila Dec 29 '17 at 10:12

## 4 Answers

A geometric proof for part a.

From the OP:

We need to prove that the new equilateral triangle is essentially a rotation of the original one, about an axis passing through center of its circumcircle perpendicular to its plane.

I think it's easier to think of it as a reflection. In the diagram below

• $GH$ is the diameter of the circumcircle that is perpendicular to $AM$.
• It's given that $BN$ and $CP$ are parallel to $AM$.
• Hence, points $M,N,P$ are reflections of points $A,B,C$ respectively in the line $GH$.
• Hence, $\triangle MNP$ is a reflection of $\triangle ABC$ through the line $GH$. Edit, to add part b.

By way of the reflective symmetry about $GH$ and the rotational symmetry of the equilateral triangles we can note that $AP=BN=CM$, and that $AN=BM=CP$. From this we can draw a system of parallel lines to show that $AM=BN+CP$. • Brilliant! Do you not think it would work out backwards, too? I mean, whether it now follows that $AN$, $BP$ and $CM$ are all parallel? Because this will give a straightforward proof for (b): let $Q$ be the intersection of $BP$ and $AM$, then $BQAN$ and $QPCM$ are parallelograms, so $AM=AQ+QM=BN+CP$. – user491874 Dec 29 '17 at 11:54
• @user8734617 Yes, it works exactly like that. I was just getting my diagram together when you commented... – nickgard Dec 29 '17 at 12:05

Geometric hint: cyclic trapezoids $APCM, AMBN$ must be isosceles, so arcs $\overparen{AP}=\overparen{CM}$ and $\overparen{BM}=\overparen{AN}\,$. Since $\,\overparen{AB}=\overparen{BC}\,$ the latter implies $\overparen{BN}=\overparen{AB}-\overparen{AN}=\overparen{BC}-\overparen{BM}=\overparen{CM}\,$, so in the end $\overparen{AP}=\overparen{CM}=\overparen{BN}\,$ and therefore $\triangle PNM$ is $\triangle ABC$ rotated by $\overparen{AP}$.

As for point b, writing Ptolemy's theorem twice:

• $APCM\,$: $\;AC^2=AP^2 + AM \cdot CP$

• $AMBN\,$: $\;AB^2=AN^2 + AM \cdot BN$

Subtracting the above and using that $AC=AB, AP=BN, CP=AN\,$:

$$0 = BN^2-CP^2+AM\cdot(CP-BN) = (BN-CP)\cdot(BN+CP-AM)$$

It follows that $\,BN+CP-AM=0 \iff AM = BN+CP\,$.  (Rigorously, the case $\,BN=CP\,$ would need to be considered separately, or could be derived by continuity).

Let me add an answer for (b), assuming (a) is proven...

Draw a line through $P$ parallel to $CM$, and let it intersect $AM$ at $Q$. $PQMC$ is a parallelogram, so $CP$=$QM$.

On the other hand, it is easy to see that $\triangle APQ$ is equilateral, as its angles are $60^\circ$ ($\angle A$ is peripheral over arc $\overparen {PM}$ and $\angle Q=\angle CMA$ is peripheral over arc $\overparen{AC}$). Thus, $AQ$=$AP$=$BN$.

Finally, $AM=AQ+QM=BN+CP$ as claimed.

This can be solved beautifully using complex numbers. The vertices of the ∆ can be assumed to be 1,W,W2 on a unit circle centered at origin.

Here is an attempt to prove it using complex numbers.

I am trying to prove it backwards.Let C,A,B be $1$,$\omega$,$\omega^2$ and M be $e^{i\theta}$.

If $\Delta$MPN is an equilateral triangle,then AM||PC which implies

AM=$e^{i\theta}-\omega$

OP=OM$\times \omega$

PC=$e^{i\theta}\omega-1$

We have to prove AM=$\lambda$PC,for some scalar $\lambda$

@schrodinger_16 ,can you help me to proceed further?