Two line segments $OA$ and $OB$ are inclined to each other at $120^\circ$ at point $O$. $
\quad P$ is a fixed point on the internal angle bisector of $\lt AOB$. $\quad OA = OB = 20cm, OP = 10cm$. Two points are chosen on the line segments $OA$ and $OB$ such that the two points and $P$ make an equilateral triangle. The two points maybe in the same line segment or in different line segments. If the number of such non – congruent equilateral triangles that can be made is $k$, then find the value of $[{200\over k}]$
My Work :-
According to me only $2$ non-congruent triangles can be formed;
$\ast$ One which would have vertices as $P, O$ and midpoint of $OA$
$\ast$ Second which would have vertices as $P$ and one vertex would be in $OA$ and another on $OB$
We also see another equilateral $\Delta$ which would have vertices as $P,O$ and midpoint of $OB$ however that triangle would be congruent to the first one; hence not considered.
Thus answer should be $100$. 
Best Answer
Let $C$ and $D$ be the midpoints of $AO$ and $OB$ respectively. Draw $PC$ and $PD$. Drop perpendiculars from point $P$ on $OC$ and $OB$ meeting them at point $E$ and $F$ respectively.
Notice that, if one vertex of any such equilateral triangle is between $C$ and $E$, then the other one(vertex $P$ is fixed ) has to be between $O$ and $F$. Every such triangle will have a congruent triangle which will have one of its vertices between $E$ and $O$ and the other one between $F$ and $D$. Also notice that there can't be two congruent triangles with both of them having one of their vertices between $C$ and $E$ and the other one between $O$ and $F$. Hence only those triangles are to be counted. Choose a random point $Q$ on segment $CE$. Now choose a point $R$ on segment $OF$ such that $OR=10-OQ$. Draw $PQ$ and $PR$.
Observe that, in $\triangle PCQ$ and $\triangle POR$,
$\left(i\right)$ $CQ=10-OQ=OR$
$\left(ii\right)$ $\angle PCQ=60^{\circ}=\angle POR$
$\left(iii\right)$ $PC=PO$ [ Since $\triangle PCO$ is equilateral ]
Hence, $\triangle PCQ\cong \triangle POR$ by $S-S-A$ criterion of congruence.
Thus, $\angle QPR=\angle QPO+\angle OPR=\angle QPO+\angle CPQ=\angle CPO=60^{\circ}$ and $PQ=PR$; Which indicates that $\triangle PQR$ is equilateral.
This is true for every point $Q$ on $CE$. Since there are an infinite number of such points on $CE$, $k\to\infty$.
Therefore, $\boxed {\lfloor\left(\frac{200}{k}\right)\rfloor \to 0}$