The Hat Problem

Let the point of tangency of $AB$ to the circle be $X$ , and the point of tangency of $AC$ to the same circle be $Y$ . Construct the following segments: $OX$ , $OY$ , $OA$ . Note that $OX=OY=\text{radius of the circle}$ , and we will denote this by $r$ .

This construction forms two triangles: $\Delta AOB$ with base $AB$ and altitude $OX$ , and $\Delta AOC$ with base $AC$ and altitude $OY$ (Why $OX$ and $OY$ are altitudes of the respective triangles?). The areas of these triangles sum up to the area of $\Delta ABC$ , i.e. $[AOB]+[AOC]=[ABC]$ The area of $\Delta ABC$ can be found by Heron's formula. With $s=\frac{AB+AC+BC}{2}=\frac{13+14+15}{2}=21$ , we have $[ABC]=\sqrt{s(s-AB)(s-AC)(s-BC)}=\sqrt{21(21-13)(21-14)(21-15)}=84$ On the other hand, the areas of $\Delta AOB$ and $\Delta AOC$ can be found by the basic formula $A=\frac{\text{base}\times\text{altitude}}{2}$ . Thus $[AOB]+[AOC]=[ABC]$ becomes $\frac{(AB)(OX)}{2}+\frac{(AC)(OY)}{2}=84$ $\frac{(13)(r)}{2}+\frac{(14)(r)}{2}=84$ $\frac{27r}{2}=84$ $r=\frac{56}{9}$ Therefore, the area of circle $O$ is $A=\pi r^2=\pi \left(\frac{56}{9}\right)^2=\frac{3136\pi}{81}$ So $a+b=\boxed{3217}$ .

Here's a beautiful solution: Reflect triangle ABC over line BC to get a kite with an inscribed circle. The total area of the kite is twice the area of ABC. Using right triangles or Heron's formula (done thoroughly in other solutions), ABC has area 84, so the kite has area 168. Another way to calculate this area is to draw line segments from the center of the circle to the 4 vertices of the kite, creating 4 triangles with height of the radius. Since we know the length of the sides, this comes out to be: Area = (1/2) * r * (perimeter) = 27r. Equating this to 168 we get r = 56/9. Squaring this yields 3136/81, so our answer is 3136 + 81 = 3217.

$\overline{AO}$ is the angle bisector of $\angle BAC$ (Proof: triangles formed by $A$ , $O$ , and the points of tangency are congruent by SSS, so the angles must also be congruent), so by the angle bisector theorem, $\frac{CO}{CB}=\frac{14}{27}$ .

Let $D$ be the foot of the altitude from $B$ onto $\overline {AC}$ . We can see that $BD=12$ because this splits $\triangle ABC$ into a 5-12-13 triangle and a 9-12-15 triangle (This can be proved by constructing the 5-12-13 triangle and then constructing the 9-12-15 triangle on it. This resulting triangle is congruent to $\triangle ABC$ by SSS.)

Drop a perpendicular from $O$ to $\overline {AC}$ and let the foot of this perpendicular be $E$ . $\triangle BDC \sim \triangle OEC$ by AA, so $\frac{OE}{BD}=\frac{OE}{12}=\frac{CO}{CB}=\frac{13}{27}$ . Therefore, $OE=\frac{12 \times 14}{27}=\frac{56}{9}$ .

The area of the circle is $\pi \times (\frac{56}{9})^2 = \frac{3136\pi}{81}$ . $3136+81=\boxed{3217}$ .

By cosine law

$14^2 = 13^2 + 15^2 - (2)(13)(15) \cos B$

$\cos B = \frac{33}{65}$

$\cos C = \frac{3}{5}$

Therefore

$\sin B = \frac{\sqrt{65^2 - 33^2}}{65}=\frac{56}{65}$

$\sin C = \frac{4}{5}$

Solving for the radius R. From center O draw a line perpendicular to the AB and AC.

$\sin B = \frac{R}{OB} = \frac{56}{65}$

$OB = \frac{65R}{56}$

$\sin C = \frac{R}{OC} = \frac{4}{5}$

$OC = \frac{5R}{4}$

$OB + OC = 15$

$\frac{65R}{56} + \frac{5R}{4} = 15$

$R = \frac{56}{9}$

Solving for the area of the circle O:

$A_{circle} = (\frac{56}{9})^2 \times π$

$A_{circle} = \frac{3136}{81} \times π$

$a + b = 3136 + 81 = \boxed{3217}$

Here's the best way to approach such problems- Let R be the radius of the circle. We know the area of ABC from Heron's formula which comes out to be 84. Draw line AO & perpendiculars from O to AB & AC. Now ar ABC=ar ABO+ar ACO $\Rightarrow \frac{14R+13R}{2}=84$ $\Rightarrow R=\frac{56}{9}\Rightarrow R^2=\frac{3136}{81}$ Hence our answer is $3136+81=\boxed{3217}$ Hope that helped!

Let us consider the point of tangency of AB and AC be L and M respectively.....so OL = OM = r (radius of the circle)

OL perpendicular to AB (using properties of circle) OM perpendicular to AC(using properties of circle)sin

BY USING HERONS FORMULA area of triangle ABC = 84 sq unit

1/2 * AB * BC * sinB = 84
1/2 * 13 * 15 * sinB = 84 sinB = 56/65 ..............(1)

now again... 1/2 * AC * BC * sinC = 84 1/2 * 14 * 15 * sinC = 84 sinC = 4/5 ....................(2)

In triangle OLB ... OB = r / sinB OB = r /( 56/65) [ from (1)] OB = 65r /56

In triangle OLC .. OC = r / sinC OC = r / (4/5) [from (2)] OC = 5r/4

NOw, OB + OC = BC 65r/56 + 5r/4 = 15 [ BC =15 (given)]
r = 56/9

AREA OF CIRCLE = (56/9)^2 * PIE = 3136 / 81 * PIE = a / b * PIE a + b = 3136 + 81 = 3217 .ans