What Is This Point?

We shall work with the configuration shown in the diagram throughout the solution. I'm too lazy to check which configuration the given side lengths produce, but the solution for both configurations should be similar. I shall use the following fact without mention in my solution.

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Fact: If $S$ is the circumcenter of $\triangle PQR,$ $\angle SPQ = 90^{\circ} - \angle QPR.$

The proof of this simple fact is left to the reader.

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Image link: http://s23.postimg.org/gevqcnh2j/Untitled.png

Let $K,L,M$ be the midpoints of $AC, AB, BC$ respectively. Since $OM \perp BC$ and $OL \perp AB,$ quadrilateral $BMOL$ is cyclic with diameter $BO.$

Since $OM$ and $AD$ are both perpendicular to $BC,$ $OM \parallel AD.$ Then note that $\begin{array}{lcl} \angle OQP & = & 180^{\circ} - \angle QOM \\ & = & 180^{\circ} - (180^{\circ} - \angle ABC) \\ & = & \angle ABC. \end{array}$ Similarly, $\angle OPQ = \angle ACB.$ Therefore, triangles $\triangle OPQ$ and $\triangle ABC$ are similar, and $\angle POQ = \angle BAC.$ Similar arguments to the first paragraph show that quadrilateral $ALOK$ is also cyclic, so $\begin{array}{lcl} \angle AOQ & = & \angle AKL \\ & = & \angle ACB \\ & = & \angle QPO, \end{array}$ which implies $AO$ is tangent to the circumcircle of $\triangle OPQ.$ This gives $OA \perp SO.$

Since $DM \perp AD,$ $OS \perp OA,$ and $AD \parallel OM,$ $DM$ equals the length of the altitude from $O$ of $\triangle OPQ$ (if this is not obvious, consider the foot of perpendicular from $O$ to $PQ$ and note that that point forms a rectangle with $O,D,M$ using the aforementioned facts). Since $\triangle OPQ$ and $\triangle ABC$ are similar, $\dfrac{DM}{AD} = \dfrac{PQ}{BC}.$ Again, $OS$ and $OA$ are the circumradii of $\triangle OPQ$ and $\triangle ABC$ respectively, so $\dfrac{OS}{OA}= \dfrac{PQ}{BC}= \dfrac{DM}{AD}.$ Since $\angle ADC = \angle AOS = 90^{\circ},$ $\triangle ASO$ and $\triangle MAD$ are similar. Consequently, $\angle SAO = \angle MAD.$ Finally, $\begin{array}{lcl} \angle CAS & = & \angle CAO + \angle SAO \\ & = & 90^{\circ} - \angle ABC + \angle MAD \\ & = & \angle BAD + \angle MAD \\ & = & \angle BAM. \end{array}$

It suffices to find $\cos \angle BAM.$ Let $AM= x.$ By Stewart's theorem on $\triangle ABC$ and cevian $AM,$ we find out that $5^2 + 8^2 = 2 \left(x^2 + \left(\dfrac{7}{2}\right) ^2 \right) \implies x = \dfrac{\sqrt{129}}{2},$ and by Cosine rule on $\triangle ABM,$ $\cos \angle BAM = \dfrac{5^2+x^2- \left( \dfrac{7}{2} \right) ^2}{2 \cdot 5x} = \frac{ 3 \sqrt{129} } { 43 } ,$

so $a=3, b= 129, c= 43,$ and $a+b+c= \boxed{175}.$

My solution turns out to be a bit long and operational. As many will have noticed, both triangles are similar. $\triangle OPQ.$ is similar to $\triangle ACB$ .

Angle BAC = 60°. We take advantage of the fact that the triangles are similar and we draw a perpendicular OJ to AD.

We find the area of the triangle ABC by Heron's theorem, [ABC]= $10\sqrt 3$ .

Then AD = $\frac{20\sqrt 3}{7}$

We draw the peperndicular bisector of BC, such that it intersects it at point G; AD = $\frac{5}{7}$ , DG = $\frac{39}{14}$ , GC = $\frac{7}{2}$ . OGDJ is a rectangle, so OJ = GD = 39/14. We already have the relation of sides that have both triangles, then we draw SI perpendicular to AD. By similarity; SI = $\frac{91}{80}$ , IQ = $\frac{91\sqrt 3}{80}$ , QA= $\frac{623\sqrt 3}{240}$ .

We use composite angle cosine between SAD and CAD to find the cosine of CAS. We can obtain the sine and cosine of CAD by drawing a perpendicular to AC and using the circumradio in the triangle formed.

Sorry for the bad english, is the first time I publish a solution.