Determine the area of region satisfying the given condition

Question

A triangle ABC has coordinates $A \equiv (0, 0)$, $B \equiv (1, 3)$, and $C \equiv (4, 0)$. Consider all points $P$ on or within the triangle which satisfy the relation $ d(P, A) \leq \min \{ d(P, B), d(P, C) \}$ where $d(M, N)$ denotes the distance of point $M$ from point $N$. If $R$ is the region containing all possible positions of point $P$, find the area of region $R$.

My approach

Circumcentre is the point where all vertices are equidistant. So the required region should be the area of the arc having $radius = circumradius$ and angle equal to $A$.[region between triangle and arc]

This is, however, not giving the correct answer. Pls help.

[Ans : $\frac{9}{4}$ sq. units]

Answer 1

All points left of the vertical blue line (through the middle between $A$ and $C$) are closer to $A$ than to $C$. Likewise, all points under the other blue line are closer to $A$ than to $B$. Thus $R$ is the region under/left of the blue lines. Compute the areas of the two right triangles in it.

The upper left triangle is nasty. Change of plan:

The midpoint between $A$ and $B$ is (0.5, 1.5). Compute the area of the quadrilateral up to the purple line. Height is 1.5, bottom width is 2, top width is 1.5, so the area is $1.5*(2+1.5)/2$. Subtract the upper triangle right under the purple line, which has area $1.5*0.5/2$ (the triangle height is 0.5 because its width is 1.5 and the slope of its hypotenuse is -1/3, so the change of x by 1.5 means a change of y by 1.5/3).

Result is $1.5*(2+1.5)/2 - 1.5*0.5/2 = 2.25 = 9/4$.

Answer 2

The perpendicular bisectors of the two segments $[AB]$ and $[AC]$ meet at $O=(2,1)$. The region of which we want the area $\mathcal A$ is made of the two triangles $IAO$ and $JAO$ (below and left to these two bisector lines), where $I,J$ are the respective midpoints of these two segments.

Therefore, $$\begin{align}\mathcal A&=\operatorname{area}(IAO)+\operatorname{area}(JAO)\\&=\frac12\operatorname{area}(BAO)+\frac12\operatorname{area}(CAO)\\&=\frac14\left(|x_By_O-x_Oy_B|+|x_Cy_O-x_Oy_C|\right)\\ &=\frac14\left(|1\cdot1-2\cdot3|+|4\cdot1-2\cdot0|\right)\\ &=\frac94. \end{align}$$

Answer 3

There are two possible cases:

$\color{brown}{\text{case }1)\;\min\{d(P,B),d(P,C)\}=d(P,B)}$

In this case, it results that $\,d(P,B)\leqslant d(P,C)\,,\,$ so the points $P$ are above the line $ND$ which is the axis of the segment $BC$.

Since $\,d(P,A)\leqslant\min\{d(P,B),d(P,C)\}=d(P,B)\,,\,$ the points $P$ are below the line $\,ME\,$ which is the axis of the segment $AB\,.$

Hence the points $P$ belong to the quadrilateral $AMOD$.

$\color{brown}{\text{case }2)\;\min\{d(P,B),d(P,C)\}=d(P,C)}$

In this case, it results that $\,d(P,C)\leqslant d(P,B)\,,\,$ so the points $P$ are below the line $ND$ which is the axis of the segment $BC$

Since $\,d(P,A)\leqslant\min\{d(P,B),d(P,C)\}=d(P,C)\,,\,$ the points $P$ are on the left of the line $\,LF\,$ which is the axis of the segment $\,AC\,.$

Hence the points $P$ belong to the triangle $ODL$.

$\color{brown}{\text{In any case the points }P\text{ belong to the quadrilateral }AMOL\\\text{which is the union of the right triangle }AMO\text{ and the right}}$
$\color{brown}{\text{triangle }AOL.}$

The coordinates of the midpoints of segments $AB$ and $AC$ are $\,M\!\left(\dfrac12,\dfrac32\right)\,$ and $\,L\big(2,0\big).$
It is very easy to get the coordinates of the circumcenter $O(2,1)$ by intersecting two axes, for example $ME$ and $LF$.

Now we have all we need in order to calculate the area of the quadrilateral $AMOL$ :

$\text{Area}(AMOL)=\text{Area}(AMO)+\text{Area}(AOL)=$

$=\dfrac{AM\cdot MO}2+\dfrac{AL\cdot LO}2=$

$=\dfrac{\sqrt{10}/2\cdot\sqrt{10}/2}2+\dfrac{2\cdot1}2=\dfrac{10}8+1=\dfrac54+1=\dfrac94\,.$