Finding triangle of maximal area given two vertices and condition on internal bi-sector

Question

$A=(2,5), B=(5,11)$ and a point $P$ moves such that internal bi-sector of $\angle APB$ passes through $(4,9).$ The maximum area of $ \triangle APB\;$ is __?

My attempt:

I found that $(4,9)$ lies on the line segment of $AB.$ Then I thought that due to the internal bi-sector condition it must have something to do with ellipse. Because, the two triangles formed by connecting focii of ellipse of perimeter are a congruent. So, if $P$ is a point on perimeter and $Q= (4,9),$ then $\angle APQ= \angle BPQ.$

I initially thought the optimum would happen when $ \angle ABP = \angle BAP =45^{\circ},$ but seems like that leads to a situation where the constraint is not obeyed.

So, how exactly do I get the optimum area while holding the constraint true?

Answer 1

Hint. We know that area of any triangle is half the product of its base and height. As the base is fixed, we just need to maximize the height so as to maximize the area. Further, by angle bisector theorem, we know the ratio $$\frac{PA}{PB}=\text{ratio in which $(4,9)$ divides $AB$}=k$$Thus, locus of point $P$ is the Apollonius Circle wrt to the ratio $k$. Thus, maximum height of $P$ from $AB$ will be the radius of Apollonius Circle.

Answer 2

Since the point $D=(4,9)$ belongs to $AB$, it is a foot of the bisector of $\angle ACB$, and a known expression for it is

\begin{align} D&=\frac{aA+bB}{a+b} \tag{1}\label{1} . \end{align}

With known coordinates we have

\begin{align} (4,9)&= \left(\frac{2a+5b}{a+b},\, \frac{5a+11b}{a+b} \right) \tag{2}\label{2} \end{align}

or a system

\begin{align} \frac{2a+5b}{a+b}&=4 \tag{3}\label{3} ,\\ \frac{5a+11b}{a+b}&=9 \tag{4}\label{4} \end{align}

with solution $b=2a$.

Expression for the area of $\triangle ABC$ squared is

\begin{align} S_2(a,b,c)&= \tfrac1{16}\,(4a^2b^2-(a^2+b^2-c^2)^2) \tag{5}\label{5} , \end{align}

so for $b=2a$ we must have \begin{align} S_2(a,2a,c)&= -\tfrac9{16}\,(a^2)^2+\tfrac58\,c^2\,a^2-\tfrac1{16}\,c^4 \tag{6}\label{6} . \end{align}

It's easy to find that

\begin{align} \max_a S_2(a,2a,c)&= S_2(a,2a,c)\Big|_{a=5} =225 , \end{align} hence the maximal area of such triangle is $15$.