# Show that $MNPQ$ is a square

Let $$ABCD$$ a quadrilateral s.t. $$AC=BD$$ and $$m (\angle AOD)=30°$$ where $$O=AC\cap BD$$.

Let $$\triangle ABM, \triangle DCN, \triangle ADN, \triangle CBQ$$ equilateral triangles with $$Int (\triangle ABM)\cap Int (ABCD)=\emptyset$$, $$Int (\triangle DCP)\cap Int (ABCD)=\emptyset$$, $$Int (\triangle ADN)\cap Int (ABCD) \neq \emptyset$$, $$Int (\triangle CBQ)\cap Int (ABCD)\neq\emptyset$$.

Show that $$MNPQ$$ is a square.

I have no idea how to start.

• Just drawn a picture and it didn't look like a square to me. – Quang Hoang Jan 10 at 21:02
• Edit: I think I see the problem here. All equilateral triangles should be drawn in the opposite direction and the statement would be true. I still don't know how to approach though... – Edward H. Jan 10 at 21:23
• Ok so here's an idea assuming that the equilateral triangles are meant to be drawn oppositely: Rotate ΔABC by 60° around B and you'll get ΔMBQ. Do this 4 times and use AC=BD, and you'll find that MNPQ is a rhombus. Moreover if you keep track of all the angles of rotation and use ∠AOD = 30°, you'll find that the angles of the rhombus are all right, and hence it is a square. – Edward H. Jan 10 at 21:49
• As given it's clearly false. $m\angle AMB =60$ and $m\angle MBA = 60$ and $m\angle QBC = 60$ so $Q$ is in the interior of $\angle AMB$. As is $N$ so $m\angle NMQ < m\angle AMB = 60 < 90$. Can't be a square. But if all the intersections are empty so the equi triangles pop out. – fleablood Jan 10 at 22:29

Let $$R^{\alpha}_O$$ be a rotation in the plain by an angle $$\alpha$$ around a point $$O$$.

Easy to see that to rotate a vector by an angle $$\alpha$$ it's the same to rotate this vector around his tail.

Now, by using the beautiful Daniel Mathias's picture we obtain: $$R^{90^{\circ}}\left(\vec{NM}\right)=R^{30^{\circ}}\left(R^{60^{\circ}}\left(\vec{NA}+\vec{AM}\right)\right)=R^{30^{\circ}}\left(\vec{DA}+\vec{AB}\right)=$$ $$=R^{30^{\circ}}\left(\vec{DB}\right)=R^{60^{\circ}}\left(\vec{AC}\right)=R^{60^{\circ}}\left(\vec{AB}+\vec{BC}\right)=\vec{MB}+\vec{BQ}=\vec{MQ},$$ which says $$NM\perp MQ$$ and $$NM=MQ.$$

Also, $$R^{90^{\circ}}\left(\vec{QP}\right)=R^{30^{\circ}}\left(R^{60^{\circ}}\left(\vec{QC}+\vec{CP}\right)\right)=R^{30^{\circ}}\left(\vec{BC}+\vec{CD}\right)=$$ $$=R^{30^{\circ}}\left(\vec{BD}\right)=R^{60^{\circ}}\left(\vec{CA}\right)=R^{60^{\circ}}\left(\vec{CB}+\vec{BA}\right)=\vec{QB}+\vec{BM}=\vec{QM},$$ which says $$QM\perp PQ$$ and $$QM=PQ.$$

Can you end it now?

• There is a synthetic proof? I try to prove it with congruences of triangles. – Problemsolving Jan 11 at 5:44
• I like your solution +1, do you like mine? – greedoid Jan 14 at 20:42
• Do you have some more problems like this solved vith rotation? – greedoid 2 days ago
• @greedoid Yes, of course. See here: math.stackexchange.com/questions/3079152 – Michael Rozenberg 2 days ago
• Yes I saw that one, it is nice. That is all you have? Do you have some pdf file with problems like this? – greedoid 2 days ago

I believe this is the desired result. Note that the intersection/non-intersection is opposite from the description.

Knowing complex numbers, this one is easy to solve. We need just this lemma:

Lemma: If $$|\vec{XY}| =|\vec{ZT}|$$ and $$\angle (\vec{XY},\vec{ZT}) = \alpha$$ then $$ZT = \varepsilon\cdot XY$$ where $$\varepsilon = \cos \alpha + i \sin \alpha$$

So it is enought to prove $$MN = i\cdot MQ\;\;\; (*)$$. We have $$DB=\varepsilon AC \;\;\;\;\;\;\;\;{\rm where}\;\;\;\;\varepsilon = \cos {\pi \over 6} + i \sin {\pi \over 6}$$ and if $$\delta = \cos {\pi \over 3} + i \sin {\pi \over 3}$$ then $$MA = \delta MB \implies M = {A-\delta B\over 1-\delta}$$ $$QC = \delta QB \implies Q = {C-\delta B\over 1-\delta}$$ $$NA = \delta ND \implies N = {A-\delta D\over 1-\delta}$$

thus $$MN ={A-\delta D\over 1-\delta}-{A-\delta B\over 1-\delta} = {\delta\over 1-\delta}DB$$ $$= {\delta\over 1-\delta}\varepsilon \cdot AC=i{AC\over 1-\delta} = iMQ$$

and we are done.