Indija TS

(a) Write $BC=a$, $CA=b$ and $AB=c$. Let $\angle PTQ=2x$, $\angle KTL=2y$ and $\angle MTN=2z$. Since $TP=TQ$, we get $\angle TQP=\angle TPQ=90^{\circ }-x$. Thus $\angle 30^{\circ }+x$. Similarly, $\angle BNP=30^{\circ }+z$. It follows that $z+x=120^{\circ }-B$. Likewise $x+y=120^{\circ }-C$ and $y+z=120^{\circ }-A$. Solving these, we get $x=A-30^{\circ }$, $y=B-30^{\circ }$ and $z=C-30^{\circ }$. As each angle $A,B,C$ is greater than $30^{\circ }$, we see that $x,y,z$ are all positive. Further, $\angle BPN=30^{\circ }+x=A$ and $\angle BNP=30^{\circ }+z=C$. So the triangle $PBN$ is similar to $ABC$. So are $QKC$ and $ALM$. If we take $BN=ka$, $NP=kb$ and $PB=kc$, then $QC=k{b}^2/c$ (Note $PN=KQ$). Thus $$\begin{array}{rl}a=BP+PQ+QC=kc+2kb\sin x+\frac{k{b}^2}{c}& =\frac{k}{c}(c^2+b^2+2bc\sin (A-30^{\circ }))\\ & =\frac{k}{c}(b^2+c^2-2bc\cos (A+60^{\circ }))\\ & =\frac{k}{c}{f}^2,\end{array}$$ where $f^2=(a^2+b^2+c^2+4\sqrt{3}F)/2$. This shows that $k=ac/f^2$; $PN=kb=abc/f^2$, the radius of the circle as desired.

(b) Using cosine rule, $$\begin{array}{rl}B{T}^2& =B{P}^2+P{T}^2-2BP\cdot PT\cdot \cos \angle BPT\\ & =k^2(c^2+b^2-2bc\cos (90^{\circ }+x))\\ & =k^2(b^2+c^2-2bc\cos (A+60^{\circ }))\\ & =k^2{f}^2.\end{array}$$ So $BT=kf=ac/f$. This shows that $BT\cdot AC=abc/f$. The symmetry of right side shows that $$BT\cdot AC=AT\cdot BC=CT\cdot AB.$$