# Determine the area of region satisfying the given condition

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]

• Don't forget the reqirement "$P$ on or within the triangle" (not the disc). Commented Aug 5 at 11:49
• @AnneBauval I believe that's what I did.found the area sweeped by angle A (inside the triangle) with radius equalling circumradius. Commented Aug 5 at 11:54
• Please edit your post to include the details and result of your solution. Commented Aug 5 at 12:27
• A picture of your approach (similar to my answer's) would make it clearer. Commented Aug 5 at 12:28
• @AnneBauval , done! pls see the q now Commented Aug 5 at 12:43

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

• It does give the answer. I computed the result from your diagram. Got my mistake, thank you :-) Commented Aug 5 at 12:48
• @rohit1729 I computed it as well now, but a little differently, see the added part. Commented Aug 5 at 12:59

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}

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 $$AMOL$$ :

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

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

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