A triangle inside an equilateral triangle

Question

$\triangle ABC$ is an equilateral triangle of side $1$. $ \triangle BDC$ is an isosceles triangle with $D$ opposite to $A$ and $DB = DC$ and $\angle BDC = 120^\circ$. If points $M$ and $N$ are on $AB$ and $AC$, respectively, such that $\angle MDN = 60^\circ$, find the perimeter of $\triangle AMN$.

My answer comes out to be $1$ by taking the points $M$ and $N$ in such a way that the quadrilateral $AMDN$ becomes a rhombus. This is all I could do and please help in out extending the result of the perimeter (if correct) to all points $M$ and $N$ in general.

Question is from the Facebook page created by my teacher

Answer 1

I used the same idea as you, let put $M$ and $N$ in such a way that $AMDN$ is rhombus. Let $E$ be midpoint of $BC$, then $DC=\frac{CE}{\cos 30°}=\frac{1}{\sqrt{3}}$. Notice that $m\angle{NCD}=90°$ then $DN=\frac{DC}{\cos 30°}=\frac{2}{3}$. But $DN=MN=AM=AN$ so perimeter of $\triangle{AMN}=3\cdot \frac{2}{3}=2$

Answer 2

Like Vasya and Qurultay in the comments, I too get 2.

To find the length of $BD$, let $K$ be a perpendicular bisector of $BC$, which also bisects $\angle BDC$, thus we have a right triangle $\triangle BKD$. Thus: $$\sin 60^\circ=\frac{0.5}{BD}\Rightarrow BD=\frac{0.5}{\sin 60^\circ}$$

Note that $\angle MBD $ is a right angle${}^{\color{red}{2}}$. We know that $\angle MDB$ is $30^\circ {}^{\color{red}{1}}$. To find $MD$: $$\cos30^\circ=\frac{BD}{MD}\Rightarrow MD=\frac{BD}{\cos30^\circ}=\frac{0.5}{\cos30^\circ\sin60^\circ}=\frac{0.5}{0.75}=\frac23$$

From this information, we can deduce that $MB=\frac13$ (you can work this out yourself) and thus $AM=\frac23$. We must note that $\triangle MDN$ is isosceles${}^{\color{red}{3}}$, and so is $\triangle AMN$ and thus, knowing $AM$, the perimiter of $\triangle AMN$ is shown as: $$\mathrm{P}_{\triangle AMN}=3\cdot AM=3\cdot\frac23=2$$

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${}^{\color{red}{1}}$If $KD$ bisects $BDC$, then $\angle KDB=60^\circ$ and $\angle KMD=30^\circ$ which leaves $\angle MDB=30^\circ$

${}^{\color{red}{2}}$ see comments.

${}^{\color{red}{3}}$ Since $MD=AM=\frac23$ and $\angle MAN=\angle MDN=60^\circ$, though I argue there could be a more rigorous proof than this.