Question

Sides $AB$ and $AC$ in an equilateral triangle $ABC$ with side length $3$ is extended to form two rays emanating from the point $A$ as shown in the figure. A point $P$ is chosen outside the triangle $ABC$ and between the two rays such that $\angle ABP+\angle BCP={180}^{\circ }.$ If the maximum length of $CP$ is $M,$ then the value of $\frac{M^2}{2}$ is

Answer 1

Given, $\alpha +\beta ={180}^{\circ }$
In $△BCP,$
$\alpha -{60}^{\circ }+\beta +\angle P={180}^{\circ }$
$\Rightarrow \angle P={60}^{\circ }$
Using sine law in $△BCP$
$\frac{3}{\sin {60}^{\circ }}=\frac{x}{\sin (\alpha -{60}^{\circ })}\Rightarrow x=2\sqrt{3}\sin (\alpha -{60}^{\circ })x_{max}=2\sqrt{3}\equiv M\Rightarrow \frac{M^2}{2}=\frac{1}{2}\times 4\times 3=6$