Samir's circle

Let O be the center of the circle. OP meets AB at H. Since triangle PAO is a right triangle at A, AH is the altitude then it's common knowledge that $1/AH^2=1/AP^2+1/AO^2=1/AP^2+1/R^2$

Substituting in gives $R=66\sqrt{23}/23$ The answer is 112

Let $M$ denote the centre of $\Gamma$ . The lines $PM$ and $AB$ are perpendicular. Denote the point of intersection of $PM$ and $AB$ by $S$ . By Pythagoras we obtain $PS = \sqrt{PA^2 - AS^2} = \frac12 \sqrt{23}$ .

The triangles $\Delta APM$ and $\Delta SPA$ are similar, because $\angle MAP = 90^\circ = \angle ASP \ \ \ \mbox{and} \ \ \ \angle APM = \angle SPA.$ This similarity yields the ratio $\frac{6}{\frac12 \sqrt{23}} =\frac{AP}{SP} = \frac{MA}{AS}= \frac{r}{5\frac12},$ and hence $r = \frac{66}{23}\sqrt{23}$ .

Let $O$ be the center of $\Gamma$ . Since $PA$ and $PB$ are both tangent to $\Gamma$ it results that $OA \perp PA$ and $OB \perp PB$ .

Taking into account that $OA = OB = r$ (where $r$ is the radius of $\Gamma$ ) we get that $\Delta OAP \equiv \Delta OBP$ from which $OP \perp AB$ .

Let $\alpha$ be $\angle AOP$ . In $\Delta OAP$ we have $AP = r \tan \alpha$ .

Similarly, we have $AB = 2r\sin \alpha$ which get's transformed into $AB=2AP\cos\alpha$ from the above. From here $\cos\alpha = \frac{AB}{2AP}$ .

Now, $r = \frac{AB}{2\sin\alpha}\ = \frac{AB}{2\sqrt{1 - \cos^2\alpha}} = \frac{AB}{2\sqrt{1-\frac{AB^2}{4AP^2}}} = \frac{AB * AP * \sqrt{4AP^2 - AB^2}}{4AP^2-AB^2} = \frac{66\sqrt{23}}{23}$ and the result is $66 + 23 + 23 = 112$

Let $O$ be the center of the circle $Γ$ . Since $\Delta OAB$ is an isosceles triangle with $OA = OB$ , thus $\angle OAB = \angle OBA$ . Based on the hypothesis of tangents, $\angle OAP = \angle OBP = 90^\circ$ . Thus, $\angle PAB = \angle PBA$ , and $\Delta PAB$ is also an isosceles triangle with $PA = PB$ .

Since $P$ and $O$ is equidistant from $A$ and $B$ , $OP$ is the perpendicular bisector of $AB$ and thus $OP$ is perpendicular to $AB$ . Let $H$ is the intersection of $OP$ and $AB$ , thus $H$ is the midpoint of $AB$ , and $AH = BH = \frac {11}{2}$ . We have $\angle AHO = \angle OAP = 90^\circ$ , thus $\angle HAP = \angle HOA$ .

$\Delta AHO$ and $\Delta AHP$ is right triangles, so $\cos \angle HAP = \frac {AH}{AP} = \frac {11}{12}$ and $\sin \angle HOA = \frac {AH}{AO} = \frac {11}{2R}$ . Since $\angle HAP = \angle HOA$ , we have:

$\cos^2 \angle HAP + \sin^2 \angle HOA = 1$ $\Rightarrow (\frac {11}{12})^2 + (\frac {11}{2R})^2 = 1$ $\Rightarrow R = \frac {66\sqrt{23}}{23}$

Thus, the final solution is $66 + 23 + 23 = 112$

Let $O$ denote the center of circle $\Gamma$ and let $r$ be the radius of $\Gamma.$ Because both $AP$ and $BP$ are tangent to $\Gamma,$ we know that $AP = BP = 6$ and $\angle OAP = \angle OBP = \dfrac{\pi}{2}.$ Let $\angle APB = x$ ; then because the angles of quadrilateral $APBO$ sum to $2\pi,$ we have $\angle AOB = \pi-x.$ Applying the Law of Cosines to $\triangle AOB$ and $\triangle APB,$ we get: $121 = 72 - 72 \cos x$ $121 = 2r^2 - 2r^2 \cos(\pi-x) = 2r^2[1 - \cos(\pi-x)]$

Manipulating the first equation gives $\cos x = -\dfrac{49}{72}.$ Then we use the formula $\cos(\pi-x) = -\cos x$ on the second equation: $121 = 2r^2(1 + \cos x) = 2r^2 \left(1-\dfrac{49}{72}\right) = \dfrac{23}{36}r^2$ From this, we find that $r^2 = \dfrac{121 \cdot 36}{23} = \dfrac{66^2}{23},$ so $r = \dfrac{66}{\sqrt{23}} = \dfrac{66\sqrt {23}}{23}.$ The answer is $66 + 23 + 23 = \boxed{112}.$

Here we have a quadrilateral APBC (C = center, A and B being on the circle, and P on the outside). To start, we know $\angle A = \angle B = 90 ^ \circ$ (due to $\overline{AP}$ and $\overline{BP}$ being tangent to the circle), and $\overline{AC}$ and $\overline{BC}$ are radii of the circle; we will label them as length R. Furthermore, it is stated that line $\overline{AP} = 6$ and due to the symmetry, $\overline{BP} = 6$ . Also stated is that $\overline{AB} = 11$ .

We can divide the quadrilateral into 4 right triangles by drawing in diagonals. We will call the point where the diagonals meet, X. $\bigtriangleup APX \cong \bigtriangleup BPX$ and $\bigtriangleup CXA\cong \bigtriangleup CXB$ . Now let us examine one triangle: $\bigtriangleup APX$ . Set $\angle PAX\ = x ^ \circ$ and $\angle APX\ = y ^ \circ$ . Now because $\angle PAC\ = 90 ^ \circ$ , and $\angle XAC$ is complementary to $\angle PAX$ , $\angle XAC = y ^ \circ$ . It follows that $\angle ACX = x ^ \circ$ and $\bigtriangleup APX \sim \bigtriangleup CXA$ .

Examining $\bigtriangleup APX$ again, we know $\overline{AP} = 6$ and $\overline{AX} = \frac{11}{2}$ , so by the Pythagorean Theorem, $\overline{PX} = \frac{\sqrt{23}}{2}$ .

Now we can use this relation: $\frac{6}{\frac{\sqrt{23}}{2}} = \frac{R}{\frac{11}{2}}$ . Solving for R, we get $R = \frac{66}{\sqrt{23}}$ . Multiplying both the numerator and denominator by $\sqrt{23}$ , we get $R = \frac{66\sqrt{23}}{23}$ . Thus $a = 66$ , $b = 23$ , $c = 23$ , and our final answer $= 66 + 23 + 23 = 112$ ,

Let $O$ be the center of the circle, and consider the triangle $OAP$ . Since $OA$ is a radius of the circle and $PA$ is a tangent, the angle $OAP$ is a right angle and the triangle is a right triangle. Letting $z=|OP|$ be its hypotenuse, we obtain

$z^2 = r^2 + 6^2$

where $r = |OA|$ is the radius of the circle.

Now, observe that $AB$ is perpendicular to $OP$ , so half the distance between $A$ and $B)$ is a height of the right triangle (the height dropped from $A$ to the hypotenuse). We can now calculate the area $S$ of the triangle $OAP$ in two ways:

$S = \frac{1}{2} \times r \times 6 = \frac{1}{2} \times z \times \frac{11}{2}$

since both of these expressions are "half of base times height", with different choices of base and height. This reads

$3r = \frac{11}{4} z$

which, when squared, is

$9 r^2 = \frac{121}{16} z^2$

or

$144 r^2 = 121 z^2$ .

Finally, since $z^2 = r^2 + 36$ , this can be rearranged into

$23 r^2 = 121 \times 36 = (11\times 6)^2$

so

$r^2 = \frac{66^2}{23}$

or

$r = \frac{66}{\sqrt{23}} = \frac{66 \sqrt{23}}{23}$ .

The last step is needed since we're asked to have the square root at the enumerator rather than the denominator (which is a standard approach to writing radical expressions).

Thus, the required answer is $66+23+23=112$ .