Inellipse Foci

The key geometrical fact behind this question is the fact that $P$ and $Q$ are isogonal conjugates, so that the ray $AQ$ is the reflection of the ray $AP$ in the angle bisector of $A$ , and similarly for the other two angles of the triangle. This result is worth proving in detail. Suppose that $ABC$ is any triangle, and that $P,Q$ are isogonal conjugate points inside the triangle. Let the feet of the three perpendiculars from $P$ to the sides of the triangle be $D,E,F$ , and let the feet of the perpendiculars from $Q$ to the three sides of the triangle be $D',E',F'$ , as indicated. Let $K$ be the intersection of $EF$ and $AQ$ . Now $AFPE$ is a cyclic quadrilateral, and hence $\angle AEF = \angle APF$ . Since $P,Q$ are isogonal conjugates, $\angle PAF = \angle KAE$ , so triangles $AFP$ and $AKE$ are similar, and hence $\angle AKE = \angle AFP = 90^\circ$ . Thus $AQ$ is orthogonal to $EF$ . Thus we deduce that $EE'QK$ is a cyclic quadrilateral, and hence the Intersecting Chords Theorem tells us that $AE\cdot AE’ = AK\cdot AQ$ . Similarly $FF'QK$ is cyclic, and hence $AF\cdot AF’ = AK\cdot AQ$ . Thus we deduce that $AE\cdot AE' = AF \cdot AF'$ , and hence $EFF'E'$ is cyclic, and its circumcircle has as centre the intersection of the perpendicular bisectors of $EE'$ and $FF'$ , namely the midpoint $M$ of $PQ$ . We can similarly show that $DD'F'F$ and $DD'EE'$ are cyclic quadrilaterals, and hence we deduce that $D,D',E,E',F,F'$ all lie on the same circle, which has centre at $M$ . This result is normally phrased by saying that the pedal circles of $P$ and $Q$ are the same. Applying the homothety $H(P,2)$ (the enlargement with centre $P$ and scale factor $2$ ), we see that $Q$ is the centre of the circumcircle of the three points $D_1,E_1.F_1$ which are the reflections of the point $P$ in the three sides of the triangle $ABC$ (this homothety maps $D,E,F$ to $D_1,E_1,F_1$ , and maps $M$ to $Q$ ).

If $R$ is the radius of the common pedal circle of $P$ and $Q$ , then the locus of the points $X$ such that $XP + XQ = 2R$ is the desired inellipse with foci $P,Q$ . Conversely, if $P$ is the focus of an inellipse, then the second focus must be equidistant from $D_1,E_1,F_1$ , and hence must be the centre of the circumcircle of $D_1E_1F_1$ . From the above it follows that the second focus must be the isogonal conjugate of $P$ . It is interesting to note that the inellipse is tangential to the common pedal triangle at the ends of its major axis.

It is a standard result that if $(u,v,w)$ are the barycentric coordinates of a point, then $(a^2u^{-1},b^2v^{-1},c^2w^{-1})$ are the barycentric coordinates of its isogonal conjugate.

On to the main problem. Set up Cartesian coordinates such that $A$ is at the origin and $B$ lies on the positive $x$ -axis. Then we have coordinates $A\;(0,0)$ , $B\;(63,0)$ , $C\;(15,20)$ and $P\;(22,16)$ . The area of the triangle $ABC$ is $630$ cm ${}^2$ . Calculating the areas of $PBC$ , $PAC$ and $PAB$ , we deduce that $P$ has absolute barycentric coordinates $(26,100,504)$ . Thus the absolute barycentric coordinates of $Q$ are $\left(\lambda\tfrac{52^2}{26},\lambda\tfrac{25^2}{100},\lambda\tfrac{63^2}{504}\right) \; = \; \left(104\lambda,\tfrac{25}{4}\lambda,\tfrac{63}{8}\lambda\right)$ where $\lambda$ is such that the three coordinates sum to $630$ . Thus we deduce that $\lambda = \tfrac{16}{3}$ , and hence that the absolute barycentric coordinates of $Q$ are $(\tfrac{1664}{3},\tfrac{100}{3},42)$ . Then $42 = \tfrac12 \times 63 \times y$ , so that $y = \tfrac43$ , while $\tfrac{100}{3} = \tfrac12(20x - 15y) = 10x - 10$ , so that $x = \tfrac{13}{3}$ . Thus the coordinates of $Q$ are $(\tfrac{13}{3},\tfrac{4}{3})$ , and hence the answer is $(13 + 4) \times 3 = \boxed{51}$ .