# Elementary problem of geometry involving equilateral triangles

I have the following exercise of elementary geometry: Given a triangle $ABC$, and draw (to the outside of this triangle) two equilateral triangles $ABE$ and $ACF$. Let $M,P$ be the midpoints of $BC,EF$ respectively and $H$ the projection of $A$ on $EF$. Prove that $MP=MH$.

Sorry if this problem bothers you, but I do not have any idea to do it, except using coordinates. Does someone have any idea?

• Why would it bother us ? – Jean Marie Nov 3 '16 at 15:07
• I have added a picture. I strongly advise you to provide a figure for geometrical problems. It's so easy with, for example, a software like Geogebra! – Jean Marie Nov 3 '16 at 15:26
• Could you use the law of cosines, using the fact that $AB=AE$, $AC =AF$ and $\angle EAF = \frac {4\pi} 3 - \angle BAC$? That would at least get you $EF$ in terms of $BC$. – Gabriel Burns Nov 3 '16 at 15:31
• @JeanMarie thanks for your picture! In fact I have not heard about Geogebra before... – mapping Nov 3 '16 at 15:37
• Note that by the symmetry of the problem, they must also equal the distance from $P$ to the projection of $A$ onto $BC$. – Gabriel Burns Nov 3 '16 at 15:55

Denote by $L$ and $N$ the midpoints of segments $BE$ and $CF$ respectively. Then since triangles $BAE$ and $CAF$ are equilateral, $$\angle \, ALE = \angle \, AHE = 90^{\circ}$$ $$\angle \, ANF = \angle \, AHF = 90^{\circ}$$ so quads $LAHE$ and $NAHF$ are inscribed in circles respectively, so $$\angle \, LHA = \angle \, LEA = 60^{\circ}$$ $$\angle \, NHA = \angle \, NFA = 60^{\circ}$$ and so $$\angle \, LHN = \angle \, LHA + \angle \, NHA = 120^{\circ}$$

Perform a $60^{\circ}$ counterclockwise rotation around point $A$. Then, since triangles $BAE$ and $CAF$ are equilateral, point $E$ is mapped to point $B$ and point $C$ is mapped to point $F$. Thus, triangle $EAC$ is mapped to triangle $BAF$ so they are congruent and moreover segment $EC$ is mapped to segment $BF$, which means that $EC = BF$ and the two of them intersect at angle $60^{\circ}$ (as well as $120^{\circ}$ as an external angle).

The segments $MN$ and $LP$ are mid-segments in triangles $BCF$ and $BEF$ respectively and so $MN$ and $LP$ are both parallel to $BF$ and $MN = \frac{1}{2} \, BF = LP$. Analogously, segments $PN$ and $LM$ are both parallel to $EC$ and $PN = \frac{1}{2} \, EC = LM$. But as already shown, $EC = BF$ and they intersect at angle $60^{\circ}$, the quadrilateral $LMNP$ is a rhombus with angles $$\angle\, LPN = \angle \, LMN = 120^{\circ}$$ $$\angle\, MNP = \angle \, MLP = 60^{\circ}$$ This means that triangles $MNP$ and $MLP$ are equilateral so $$MP = MN = ML$$

Furthermore, observe that $\angle \, LHN = \angle\, LPN = 120^{\circ}$ so the quad $LHPN$ is inscribed in a circle $k,$ which means that the point $H$ lies on the circle $k$ circumscribed around triangle $LPN$. However, the circle with center $N$ and radius $MP = MN = ML$ passes through the three points $L, P, N$ and so it is the circle $k$ circumscribed around triangle $LPN$ and so $H$ lies on $k$ and since $M$ is the center of circle $k$, $$MH = MP = MN = ML$$

• Very nice solution! Thanks a lot! – mapping Nov 6 '16 at 15:42
• @mapping Thank you for appreciating it and you are welcome! You can mark the solution as an accepted answer so that it is decided that this question has been answered according to your expectations. Cheers – Futurologist Nov 6 '16 at 18:07

My work is quite similar to that of @futurologist .

Fact-1 If L is the midpoint of BE, then $\angle BLA = 90^0$. This means the red dotted circle can be formed. As a consequence, $\angle LHA = \angle BEA = 60^0$. Similarly, $\angle NHA = \angle CFA = 60^0$.

Fact-2 After drawing an extra equilateral triangle BCQ outside of the line BC, we obtained the Fermat Point, X such that the blue lines AQ, BF, and CE are equal in length. In addition, they cut each other at angles of $60^0$ (i. e. $\angle FXC = 60^0$).

Fact-3 LMNP is a rhombus (because each side is equal to half of a blue line according to the midpoint theorem). In addition, $\angle PNM$, $\angle MPN$, $\angle LPM$, and $\angle PLM$ are all equal to $\angle FXC = 60^0$ (because UXVN is also a parallelogram). This means M is the center of the circle passing through L, P, and N.

Putting the result of Fact-1 in, we can say that L, H, P and N are con-cyclic. Required result follows.

Let us put everything in complex plane. Clearly $B,C\in \mathbb{R}$ and $A\in i\mathbb{R}$. Let $O$ be origin.

Instead of proving $MH=MP$ we will prove $OP=PM$ or $Re(P) ={1\over 2}M$.

Let $\varepsilon = \cos {\pi\over 3}+ i \sin {\pi\over 3}$. Then $-\varepsilon ^{-1}= \varepsilon ^2 = \varepsilon -1$ and

\begin{eqnarray} F &=& A+\varepsilon (C-A)\\ E &=& A-\varepsilon^2 (B-A) \end{eqnarray} Since $M={1\over 2}(B+C)$ and

$$P ={1\over 2}(E+F) = {1\over 2}\Big(A(2-\varepsilon +\varepsilon ^2) +\varepsilon C - \varepsilon ^2 B \Big) = {1\over 2}\Big(A +\varepsilon C - \varepsilon ^2 B \Big)$$ we have \begin{eqnarray} Re(P) &=& {1\over 2}(P+\overline{P}) \\ &=& {1\over 4}\Big(A +\varepsilon C - \varepsilon ^2 B + \overline{A +\varepsilon C - \varepsilon ^2B} \Big)\\ &=& {1\over 4}\Big(A +\varepsilon C - \varepsilon ^2 B -A +\overline{\varepsilon }C - \overline{\varepsilon }^2B \Big)\\ &=& {1\over 4}\Big((\underbrace{\varepsilon +\overline{\varepsilon }}_{=1})C - (\underbrace{\varepsilon ^2 + \overline{\varepsilon }^2}_{-1})B \Big)\\ &=& {1\over 4}(B+C)\\ &=& {1\over 2} M \end{eqnarray}