geometrytriangles
I have a triangle $ABC$ with inradius $r$. Points $D$, $E$, $F$ are chosen on side $BC$, $CA$, $AB$, respectively, such that $\triangle AFE$, $\triangle BDF$, and $\triangle CED$ have same inradius $r_1$. Compute the inradius of $\triangle DEF$ in terms of $r$ and $r_1$.
I have an approach in mind by taking the various side lengths as x,y,z and so on but I know for sure it will turn out to be lengthy. Any better method please?
I guess the problems you (and maybe your friend :)) propose, are (in general) supposed to be solved in an entirely geometric way. However, I am posting my solution which follows a very straightforward path.
Let's assume $\angle DBC=\theta$. Then,
$$\frac{\sin (115^{\circ}-\theta)}{\sin \theta}=\frac{BC}{DC}=\frac{DE}{DC}=\frac{\sin 115^{\circ}}{\sin 50^{\circ}}\implies \\\cos (65^{\circ}-\theta) -\cos(165^{\circ}-\theta)=\cos(115^{\circ}-\theta)-\cos(115^{\circ}+\theta)\implies \\-\cos(165^{\circ}-\theta)=\cos(115^{\circ}-\theta)\implies \\ \cos (15^{\circ}+\theta)=\cos(115^{\circ}-\theta)\implies \\ 15^{\circ}+\theta=115^{\circ}-\theta\implies \theta=50^{\circ};$$
and the rest is trivial.
Here is a solution from one of my coworkers.
Assume $AD=DE$ and $OF=KG=r$. We will prove that $HL=r$.
Define $\alpha=\angle{DAE}$. Simple angle chasing gives the other angles shown in the diagram.
$$AT=CO+OE+EK$$ $$\frac1r(OE+CO)=\frac1r(AT-EK)$$ $$\cot{(60^0-\alpha)}+\sqrt3=\cot{\left(30^0-\frac{\alpha}{2}\right)}-\cot{\left(30^0+\frac{\alpha}{2}\right)}$$ $$\frac{\cos{\alpha+\sqrt3 \sin{\alpha}}}{\sqrt3 \cos{\alpha}-\sin{\alpha}}+\sqrt3=\frac{\sin{(30^0+\alpha/2)\cos{(30^0-\alpha/2)-\cos{(30^0+\alpha/2)}\sin{(30^0-\alpha/2)}}}}{\sin{(30^0-\alpha/2)\sin{(30^0+\alpha/2)}}}$$ $$\frac{\cos{\alpha}+\sqrt3 \sin{\alpha}}{\sqrt3\cos{\alpha}-\sin{\alpha}}+\sqrt3=\frac{2\sin{\alpha}}{\cos{\alpha}-\cos{60^0}}$$ $$2\cos^2{\alpha}-\cos{\alpha}=\sqrt3 \sin{\alpha}\cos{\alpha}-\sin^2{\alpha}$$ $$\cos^2{\alpha}-\cos{\alpha}+1=\sqrt3\sin{\alpha}\cos{\alpha}$$ $$(\cos^2{\alpha}-\cos{\alpha}+1)^2=3(1-\cos^2{\alpha})\cos^2{\alpha}$$ $$(2\cos{\alpha}-1)(2\cos^3{\alpha}-1)=0$$ Obviously, $\cos{\alpha}\ne1/2$.
Let $m=2^{1/3}$.
$$\cos{\alpha}=\frac1m=\frac{1-\tan^2{(\alpha/2)}}{1+\tan^2{(\alpha/2)}}$$ $$\tan{\alpha}=\sqrt{m^2-1}$$ $$\tan{\frac{\alpha}{2}}=\sqrt{\frac{m-1}{m+1}}$$ $$\begin{align} HL&=DE\cos{\alpha}\tan{\frac{\alpha}{2}}\\ &=(DM+ME)\cos{\alpha}\tan{\frac{\alpha}{2}}\\ &=r(\cot{\alpha}+\cot{(60^0-\alpha)})\cos\alpha\tan{\frac{\alpha}{2}}\\ &=r\left(\frac{1}{\sqrt{m^2-1}}+\frac{1+\sqrt3\sqrt{m^2-1}}{\sqrt3-\sqrt{m^2-1}}\right)\frac1m\sqrt{\frac{m-1}{m+1}}\\ &=\frac{\sqrt3mr}{(m+1)(\sqrt3-\sqrt{m^2-1})}\\ &=\frac{\sqrt3mr}{\sqrt3m+\sqrt3-(m+1)\sqrt{m^2-1}}\\ &=\frac{\sqrt3mr}{\sqrt3m+\sqrt3-\sqrt{\color{red}{(m^2-1)(m+1)^2}}}\end{align}$$
$$\color{red}{(m^2-1)(m+1)^2}=(m+2)(m^3-2)+3=0+3=3$$
$$\therefore HL=r$$
Best Answer
$$\begin{eqnarray*}r_1\left(p_{AFE}+p_{BDF}+p_{CED}\right)&=&2\left([AFE]+[BDF]+[CED]\right)\\&=&2\left([ABC]-[DEF]\right)\\&=&r\,p_{ABC}-r_{DEF}\,p_{DEF}\\&=&r_1\left(p_{ABC}+p_{DEF}\right)\end{eqnarray*}$$ leads to $$ r_{DEF}\, p_{DEF} = (r-r_1) p_{ABC} - r_1 p_{DEF}. $$ Now the interesting fact is that the perimeter of $DEF$ is fixed by $r,r_1,p_{ABC}$.
By considering the distances of $D,E,F$ from the tangency points marked with a red cross we have that $p_DEF$ equals the perimeter of $I_A I_B I_C$, with $I_A,I_B,I_C$ being the incenters of $AEF,BDF,CDE$. It follows that
$$ p_{DEF} = \frac{r-r_1}{r} p_{ABC} $$ and
$$ r_{DEF} = r- r_1.$$
Another interesting fact is that $D,E,F$ and $I_A,I_B,I_C$ belong to the same ellipse.