If $P$ is a point inside the triangle $ABC$, such that $d(P, BC) \le \text{min} \{d(P, AB), d(P, AC)\}$, find the maximum of $d(P, BC)$

Question

$A(3, 4), B(0, 0)$ and $C(3, 0)$ are vertices of a triangle $ABC$. If $P$ is a point inside the triangle $ABC$, such that $d(P, BC) \le \text{min} \{d(P, AB), d(P, AC)\}$, find the maximum of $d(P, BC)$. ($d(P,BC)$ represents the distance of the point $P$ from the side $BC$)

Equation of $AB$ is $4x-3y=0$

Let $P$ be $(h,k)$

$d(P,BC)\le d(P,AB)\implies k\le\left|\dfrac{4h-3k}{5}\right|$. Here, do we need to consider both plus and minus sign or because of them being in the first quadrant, we could eliminate one sign easily? With plus sign, I am getting $2k\le h$. With minus sign, I am getting $2h+k\le0$

Also, $d(P,BC)\le d(P,AC)\implies k\le 3-h$

How to use these inequalities to get to the answer?

In the solution, they have simply found the incenter. Why?

Answer 1

I think this is clear if you draw a picture.

The points closer to $BC$ than to $AB$ lie below the segment $BD$, then angle bisector. Similarly, those closer to $BC$ than to $AC$ lie below the segment $EC$, so the candidate points lie in triangle $BCF$ and clearly, $F$ has the greatest distance from $BC$.