euclidean-geometrygeometrytrianglestrigonometry
If $DEF$ is the orthic triangle of $\triangle ABC$, then prove that
$$\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R} $$
where $r$ and $R$ are the inradius and the circumradius of $\triangle ABC$.
My attempt is very simple , I put the side length of orthic triangle in terms of $\cos$ and the side length of $\triangle ABC$ but I can't get the required 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.
In the standard notation we have: $$a+b>c,$$ $$c^2>a^2+b^2$$ and $$\frac{\sin\beta}{b}=\frac{\sin\alpha}{a}>\frac{\sin2\beta}{a}=\frac{2\sin\beta\cos\beta}{a},$$ which gives $$\frac{a}{b}>\frac{a^2+c^2-b^2}{ac}$$ or $$a^2(c-b)>b(c^2-b^2)$$ or $$a^2>bc+b^2.$$ If $b=1$ we obtain $$c<a+1$$ or $$c-a<1,$$ which is impossible.
Thus, $b\geq2$.
Now, $$a<c<\frac{a^2-b^2}{b},$$ which gives $$a+1\leq c\leq\frac{a^2}{b}-b-1$$ and $$a^2-ab-b^2-2b\geq0,$$ which gives $$a\geq\frac{b+\sqrt{5b^2+8b}}{2}\geq4$$ and since $$c^2>4^2+2^2,$$ we obtain $$c\geq5$$ and $$a+b+c\geq11.$$ Can you end it now?
Best Answer
$$\sin{2A}+\sin{2B}+\sin2C=2\sin{(A+B)}\cos{(A-B)}+2\sin{C}\cos{C}$$$$=2\sin{C}(\cos{(A-B)}-\cos{(A+B)})=4\sin{A}\sin{B}\sin{C}$$
$$\text{Now, }\frac{EF}{\sin A}=\frac{AE}{\sin C}=\frac{c\cos A}{\sin C}\implies EF=2R\sin A\cos A=R\sin{2A}\text{ .}$$
$$\therefore \text{Perimeter of }\triangle DEF=4R\sin{A}\sin{B}\sin{C}=\frac{2\Delta}{R}=\frac{r\cdot 2s}{R}$$$$\implies\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R}$$
$\blacksquare$
Explanation: $\Delta=\frac{1}{2}ab\sin C=\frac{1}{2}(2R\sin A)(2R \sin B)\sin C=2R^2\sin{A}\sin{B}\sin{C}$