Strangely enough, nobody posted a solution to this one until now, so let me do: Problem. Let ABC be an isosceles triangle with AB = AC, and let T be the midpoint of its base BC. Let P be a point on the extension of the segment BC beyound B. Let $r_{1}$ be the radius of the incircle of triangle APB, and let $r_{2}$ be the radius of the excircle of triangle APC opposite to the vertex P. Prove that $r_{1}+r_{2}=AT$. Solution. Let the incircle of triangle APB have the center $O_{1}$ and touch the side PB of this triangle at $Q_{1}$. Then, since $r_{1}$ is the radius of this incircle, we have $O_{1}Q_{1}=r_{1}$ and $O_{1}Q_{1}\perp PB$. Let the excircle of triangle APC opposite to the vertex P have the center $O_{2}$ and touch the extended side PC of this triangle at $Q_{2}$. Then, since $r_{2}$ is the radius of this excircle, we have $O_{2}Q_{2}=r_{2}$ and $O_{2}Q_{2}\perp PC$. The relations $O_{1}Q_{1}\perp PB$ and $O_{2}Q_{2}\perp PC$ rewrite as $O_{1}Q_{1}\perp BC$ and $O_{2}Q_{2}\perp BC$, so that $O_{1}Q_{1}\parallel O_{2}Q_{2}$. Thus, the quadrilateral $O_{1}O_{2}Q_{2}Q_{1}$ is a trapezoid with its bases $O_{1}Q_{1}$ and $O_{2}Q_{2}$ being perpendicular to the line BC. Hence, if M and N are the midpoints of the legs $O_{1}O_{2}$ and $Q_{1}Q_{2}$ of this trapezoid, then the line MN is parallel to its bases $O_{1}Q_{1}$ and $O_{2}Q_{2}$, and thus perpendicular to the line BC, and we have $MN=\frac{O_{1}Q_{1}+O_{2}Q_{2}}{2}$. In other words, $MN=\frac{r_{1}+r_{2}}{2}$, so that $r_{1}+r_{2}=2\cdot MN$. Being the incenter of triangle APB, the point $O_{1}$ lies on the angle bisector of the angle APB. Being the excenter of triangle APC opposite to the vertex P, the point $O_{2}$ lies on the angle bisector of the angle APC. Actually, these two angle bisectors are one and the same line - the angle bisector of the angle formed by the rays PA and PB = PC. Since the points $O_{1}$ and $O_{2}$ both lie on this angle bisector, the midpoint M of the segment $O_{1}O_{2}$ must also lie on this angle bisector; hence, < APM = < TPM. Now, we can describe our situation as follows: The points A, B, C are chosen on the sides of an angle with the vertex P such that the point B lies between the points P and C, and the point A lies on the other side of the angle. The point $O_{1}$ is the incenter of triangle APB, and the point $O_{2}$ is the excenter of triangle APC opposite to the vertex P. Then, according to the Remark at the end of http://www.mathlinks.ro/Forum/viewtopic.php?t=30389 post #2, since AB = AC, we have $O_{1}A=O_{2}A$. Thus, the triangle $O_{1}AO_{2}$ is isosceles, and the midpoint M of its base $O_{1}O_{2}$ must be the foot of its altitude from its apex A; in other words, $AM\perp O_{1}O_{2}$, so that < AMP = 90°. On the other hand, the triangle ABC is isosceles with AB = AC; hence, the midpoint T of its base BC is the foot of its altitude from its apex A; thus, $AT\perp BC$, so that < ATP = 90°. From < AMP = 90° and < ATP = 90°, it follows that the points M and T lie on the circle with diameter AP. Now, since < APM = < TPM, the chords AM and TM have equal chordal angles in this circle. Hence, these chords must have equal length: AM = TM. Thus, the triangle AMT is isosceles. Hence, the midpoint W of the base AT of this triangle is the foot of its altitude from its apex M; in other words, $MW\perp AT$. Now, the quadrilateral MNTW has three right angles: < MNT = 90° (since $MN\perp BC$), < NTW = 90° (since $AT\perp BC$) and < TWM = 90° (since $MW\perp AT$). Thus, it is a rectangle, so that MN = WT. But since the point W is the midpoint of the segment AT, we have $2\cdot WT=AT$. Hence, $r_{1}+r_{2}=2\cdot MN=2\cdot WT=AT$, and the problem is solved. Note that the problem is equivalent to the one discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=18930 . Darij

here is my solution: Every time an isosceles triangle is mentioned without a shape limitation, it becomes our international responsibility to degenerate it to a segment by reducing the base under it to zero. Once this is done here, we see that the circles have orthogonal internal and external tangents, and with this acknowlegement comes the realization that the sum of the radii in question stops being questionable and becomes fixed. Thank you. M.T. @ Darij: T is for Tolouse-or-not-to-loose-a-traction.

Ι have a trig solution. Let $k=\widehat{APC}$ From the triangle $ADP$ we have that $PA=\frac{AD}{sin2k}$. Also $PD=\frac{ADcos2k}{sin2k}$ So after some algebra we have that $PA+PD=AD\frac{ 1}{tgk}$ So $r_{1}+r_{2}=(PA+PD)cotk=tgk\cdot AD\cdot \frac{1}{tgk}=AD=ct$

Coming at Darij's solution, with his notations, to see that $M$ lies on the perpendicular bisector of $AT$, there is an easier way: $APTM$ is cyclic, $PM$ the angle bisector of $\angle APT\implies AM=MT$, q.e.d. Best regards, sunken rock

See PP6 from here.