Point in an Isosceles Triangle

Refer to this diagram while reading my solution. alt text Let $D$ , $E$ , and $F$ be the midpoints of $\overline{BC}$ , $\overline{AB}$ , and $\overline{CA}$ , respectively. Let $H$ be the intersection of the angle bisector of $\angle B$ and $\overline{DE}$ , and let $I$ be the intersection of the angle bisector of $\angle C$ and $\overline{DF}$ . Let $G$ be the intersection of the two angle bisectors.

First, let us find the area where the point is closer to $A$ than $B$ or $C$ . If the point is closer to $A$ , it must lie on the same side of the perpendicular bisector of $\overline{AB}$ , which is $\overline{ED}$ , as $A$ . It must also lie on the same side of the perpendicular bisector of $\overline{AC}$ , which is $\overline{FD}$ , as $A$ . This area is a square, $AEDF$ .

Next, let us find the area where the point is closer to $\overline{BC}$ than $\overline{AB}$ or $\overline{AC}$ . The point must lie on the same side of the angle bisector of $\angle B$ , which is $\overline{BG}$ , as $C$ , and must lie on the same side of the angle bisector of $\angle C$ , which is $\overline{CG}$ , as $B$ . This area is a triangle, $\triangle BGC$ .

The intersection of these two areas is the successful area. This intersection is a kite, $GHDI$ . The area can be expressed as $\frac{HI\cdot GD}{2}$ .

WLOG, let $CD=1$ , so $AC=\sqrt{2}$ . Now, let us find $GD$ . By the angle bisector theorem on $\triangle ACD$ , $\dfrac{AC}{CD}=\dfrac{AG}{GD}\implies AG=GD\sqrt{2}$ Since $AG+GD=1$ , we can solve for $GD$ , which is $\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$ . To find $HI$ , we see that $HI=HD\sqrt{2}$ . By the angle bisector theorem on $\triangle EBD$ , $\dfrac{EB}{BD}=\dfrac{EH}{HD}\implies HD=EH\sqrt{2}$ Since $EH+HD=\dfrac{1}{\sqrt{2}}$ , we can solve to get $HD=GD=\sqrt{2}-1$ (this can also be found using similar triangles). Then, $HI=HD\sqrt{2}=2-\sqrt{2}$ .

Finally, the area desired is $\dfrac{GD\cdot HI}{2}=\dfrac{3\sqrt{2}-4}{2}$ and the probability is this area over the total area, which is $1$ . Converting the number into the proper form, $\dfrac{3\sqrt{2}-4}{2}=\sqrt{\dfrac{9}{2}}-2$ and the answer is $9+2+2=\boxed{13}$ .

W.L.O.G $A=(2,2)$ , $B=(0,0)$ , and $(C=4,0)$

Supposse that $P(x,y)$ closer to the point $A$ than $B$ or $C$

$PA \leq PB$ and $PA \leq PC$

From $PA \leq PB$ $(x-2)^2+(y-2)^2 \leq x^2+y^2, \rightarrow x+y \geq 2$ ,

From $PA \leq PC$

$(x-2)^2+(y-2)^2 \leq (x-4)^2+y^2, \rightarrow x-y \leq 2$ ,

From my drawing in here

Suppose that $R(x_1,y_1)$ in line $x+y=2$

Distance R to segmen AB $\geq$ Distance R to segmen BC

$(2-y_1)-y_1 \geq \sqrt{2} y_1$

$y_1 \leq 2-\sqrt{2}$ , so $x_1 \geq \sqrt{2}$

Suppose $T(x_2,y_2)$

Distance T to segmen AB $\geq$ Distance T to segmen BC

and we get $x_2 \leq 4-\sqrt{2}$

Suppose that $S(x_3,y_3)$ and we get:

Distance S to segmen AB $\geq$ Distance S to segmen BC

$x_3-y_3 \geq \sqrt{2} y_3$

$y_3=2\sqrt{2}-2$

So, probability are $\displaystyle PA=\frac{[QRST]}{[ABC]}=\frac{\frac{(2\sqrt{2}-2)(4-2\sqrt{2})}{2}}{4}=\sqrt{\frac{9}{2}}-2$

So, $a+b+c=9+2+2=13$