Radius of a Circle, Given Tangent

Let the center of the circle be O, join AB, OA, OB. Let CP=x, radius=R.

In triangle ACP, since we have $AP=19, AC=16, \angle ACP = 120 ^ \circ$ by cosine rule, $19^2=x^2+16^2-2\times 16\cos 120^\circ$ ,

Solving the quadratic equation, as x is a positive real value, we get x=5. Therefore CP=5.

By alternate segment theorem, $\angle PAC = \angle ABP$ , hence triangle ACP is similar to triangle BAP,

So we have $\frac {AB}{AP}=\frac {AC}{CP}, \frac {AB}{19}=\frac {16}{5}, AB=\frac {304}{5}$ .

$As \angle ACP = \angle BAP = 120 ^ \circ, \angle OAP = 90 ^ \circ, \angle OAB = 30 ^ \circ, OA=OB=R, \angle OAB = \angle OBA = 30 ^ \circ, \angle AOB = 120 ^ \circ$

By sine rule, we can easily obtain that $AB = \sqrt3 R$

Previously we have known that the value of AB is $\frac {304}{5}$ , therefore $R=\frac{304\sqrt3}{15}$

a+b+c = 304+3+15 = 322

First of all, we let $PC=x$ and $\angle PAC = \theta$ . So using the Cosine Rule in triangle $PAC$ , we get $x^2+256-2(x)(16)(cos 120^\circ )=361$ , ( $x-5)(x+21)=0$ . $x=5$ , since it is positive. So $PC=5$ .

Now, by Sine Rule, we get $\frac {PC}{sin \theta} = \frac {PA}{sin 120}$ , or, $sin \theta = \frac {5\sqrt 3}{38}$ .

Also $\angle CAO = \angle ACO= 90 ^\circ -\theta$ , because $\angle PAO=90^\circ$ .So $\angle AOC = 2\theta$ .

Using Sine rule in triangle $AOC$ , we get $\frac {r}{sin (90 ^\circ -\theta)}=\frac{AC}{sin (2\theta)}$ , $\frac {r}{cos \theta} = \frac {AC}{2sin\theta cos\theta}$ , $r=\frac {AC}{2sin\theta} = \frac {16}{2 (\frac {5\sqrt 3}{38})}$ , $r= \frac {304\sqrt 3}{15}$ .

So the required answer is $304+3+15=322$ .

The cosine rule: $a^2 = b^2 + c^2 -2bc cosA$

We find PC by using this rule: $16^2 + PC^2 -2\cos120^\circ\times16\times PC=19^2 \rightarrow PC=5$

The tangent which touches $\Gamma$ at C cuts AP at M $\rightarrow AM=CM \rightarrow OA\times\tan \angle MOA=R\times \tan \angle CAP =CM$

1. Using cosine rule for triangle ACP, we deduce that $\tan \angle CAP=\frac{5\sqrt{3}}{37}$ and $\cos \angle CPA =\frac{13}{19}$
2. Using cosine rule for triangle CPM, we have: $CP^2=5^2+(19-CP)^2-2\times 5\times(19-CP)\times \frac{13}{19}\rightarrow CP=\frac{304}{37}$

$\Rightarrow R=\frac{304\sqrt{3}}{15} \rightarrow a+b+c=322$

from cosine theorem in triangle APC cos(120)=(AC^2+PC^2-AP^2)/(2AC PC) we get that PC=5, then from triangles PAC similiar to PBA we have AC/AB=PC/PA=5/19, this results to AB=16 19/5=304/5. Now from sine theorem in triangle ABC we have AB/sin(60)/2R from here R=AB/(2sin60)=(304/15)*sqrt(3)

Given $$PA=19 and AC=16$$.Let AC subtend $$\angle \theta$$ to the centre O.Now $$PA^2=PC \times PB$$ (Power theorem).Applying cosine rule to $$\triangle ACP,\cos 120^\circ=\frac {16^2+PC^2-19^2}{32 \times PC}$$Hence,we get PC=5.Applying sine rule to $$\triangle ACP,\frac {19}{sin120^\circ}=\frac {5}{\sin \frac {\theta}{2}}$$. or, $$\sin \frac {\theta}{2}=\frac {5\sqrt3}{19 \times 2}.....(1)$$ Now, $$\angle OAP=90^\circ$$.(since,the tangent drawn from an external point to the circle is perp to the radius of the circle).Thus, $$\angle OAC=\angle OCA=90^\circ-\frac {\theta}{2}$$ Thus, $$\angle CAP=\frac {\theta}{2}$$ Applying cosine rule to $$\triangle OAC,\cos(90- \frac {\theta}{2})=\frac {r^2+256-r^2}{32 \times r}$$ or,$$\sin \frac {\theta}{2}=\frac {r^2+256-r^2}{32 \times r}$$ From eq. (1), $$\frac {5\sqrt3}{19 \times 2}=\frac {256}{32 \times r}$$ or, $$r=\frac {304 \times \sqrt3}{15}=\frac {a \times \sqrt b}{c}$$ $$Thus,a+b+c=322$$

problem figure

Look at triangle $APC$ ,

$∠ACP = 120^0, AP=19, AC=16$ , using sinus rule,

$\frac{19}{\sin (120^0)} = \frac{16}{\sin (∠APC)} \Rightarrow \sin (∠APC) = \frac{8\sqrt{3}}{19}$

note that, $∠PAC =180 ^0 - (∠APC + ∠ACP)$ , then

$\sin (∠PAC)=\sin (180 ^0 - (∠APC + ∠ACP))= \sin (∠APC + ∠ACP)\\ =\sin (∠APC )\cos (∠ACP) + \cos (∠APC)\sin (∠ACP) \\ = \frac{8\sqrt{3}}{19}\cdot \frac{-1}{2} + \cos (\sin^{-1}(\frac{8\sqrt{3}}{19})) \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{38}$

Now, use the fact that $PA$ tangent to the circle, then $PA \perp AO$ , hence $∠CAO =90^0 - ∠PAC$ . so, we have

$\sin (∠CAO) = cos (∠PAC) =\frac{37}{38}$

Now, look the triangle $ACO$ ,

$OA=OC=r$ , $r$ is the radius of the circle, then $∠CAO = ∠ACO$ and $∠AOC =180^0 -(∠CAO+∠ACO)$ , hence

$\sin (∠AOC) = \sin (180^0 - 2∠CAO) =\sin (2∠CAO)\\ =2\sin (∠CAO) \cos (∠CAO) =2 \cdot \frac{37}{38} \cdot \frac{\sqrt{38^2 - 37^2}}{38} =\frac{185\sqrt{3}}{722}$

again using sinus rule,

$\frac{r}{(\frac{37}{38})} = \frac{16}{(\frac{185\sqrt{3}}{722})}$

so $r =\frac{304\sqrt{3}}{15}$ and the solution of the problem $304+3+15=322$

Applying the cosine rule to triangle $ACP$ , we have $19^2 =PA^2 = PC^2 + 16^2 - 2 \times 16 \times PC \cos 120^\circ$ $\Rightarrow (PC+21)(PC-5) = 0$ . Since $PC > 0$ we have $PC = 5$ .

Applying the sine rule on triangle $PAC$ , we have $\frac {5}{\sin PAC} = \frac {19} {\sin 120^\circ} \Rightarrow \sin PAC = \frac {5 \sqrt{3} } { 38}$ . By the Alternate Segment Theorem, $\angle PAC = \angle CBA$ . By the extended sine rule , we have $R = \frac {AC}{2 \sin ABC} = \frac {16} { 2 \sin PAC} = \frac {8 \cdot 38 \sqrt{3} } { 15} = \frac {304 \sqrt{3} } {15}$ . Thus $a + b + c = 304 + 3+15 = 322$ .

Note: Because $\angle PCA \geq 90^\circ$ , this tells us that $B, C, P$ lie in that order.

It is evident that C is closer to P than B because <ACP>90.

Since <ACP=120, by the cosine rule, (AC^2+CP^2-AP^2)/(2 AC CP)=cos(120)=-1/2. Thus (CP^2-105)/(32CP)=-1/2, and CP^2-105+16CP=0 --> (CP+21)(CP-5)=0. Since CP is positive, CP=5.

By the Power of a Point Theorem, BP CP=AP^2=19^2=361. Thus, BP=361/5, so BC=361/5-5=67.2. <ACB=60, so by the cosine rule, (AC^2+BC^2-AB^2)/(2 AC BC)=cos(60)=1/2. Thus AC^2+BC^2-AB^2=AC BC --> 256+67.2^2-AB^2=16*67.2. Solving we get AB=60.8.

By the sine rule, AB/sin(ACB)=2R --> 60.8/sin(60)=2R --> 121.6/sqrt3=2R, so R=60.8/sqrt3=304/(5sqrt3)=(304sqrt3)/15.

The a+b+c=304+3+15=322.

P.S. A bit too easy for 230 pts? Could have been exchanged with a 180 pt. one.

First we make the circle and mark the points P, A, B and C.

The lenght AP and AC are given. Using the power of a point: AP² = AC*AB So, we found AB.

We can say that <BCA equal 60°, because <BCA = 180° <ACP. Using the cossine law in the circumscribed triangle ABC, we found AB.

As the radius of a circle circumscribed in a triangle is given by R = AB AC BC/S, where S is the area of the triangle.

Using Heron formula to find S, we find R = (304√3)/15.

Triangle ACP, using sine rule, we find that angle P = 46.8, angle A = 13.2, PC = 5.

Triangle ACP and BAP are similar (angle PAC = angle ABP) so AC:AB = CP:AP => AB = 16x19/5

Since angle ACB = 60, angle AOB = 120 and AB = sqrt3 R Thus R = 304sqrt3/15