Show that $MNPQ$ is a square

Question

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.

Answer 1

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?

Answer 2

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

Answer 3

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.