Let $R$ and $r$ be circumradius and inradius respectively of a triangle with semiperimeter $s$. Prove that it is right if $$s=2R+r$$
It is not hard to find $$s = {a\over \sin \alpha} + (s-a)\tan{\alpha \over 2}$$
Labeling $x=\tan{\alpha\over 2}$ we have $$(2s-a)x^2-2sx +a=0 $$ which has solution $x=1$ and $x= \displaystyle{a\over b+c}$. If $x=1$ we are done else $${a\over b+c} = \tan{\alpha\over 2} = {r\over s-a}\hspace{2cm} (1)$$
We can do the same for other two angles and if some is right then we are done, else we have also
$${b\over c+a} = \tan{\beta\over 2} = {r\over s-b}\hspace{2cm} (2)$$ and $${c\over a+b} = \tan{\gamma\over 2} = {r\over s-c}\hspace{2cm} (3)$$
Multiplying all equations (1),(2) and (3) we have
$$s(s-a)(s-b)(s-c)abc = r^3s(a+b)(b+c)(c+a)$$
$$S^2abc = r^3s(a+b)(b+c)(c+a)$$
$$sabc = r(a+b)(b+c)(c+a)$$
$$sbc = (a+b)(s-a)(c+a)$$
$$(a+b+c)bc = (a(a+b+c)+bc)(b+c-a)$$
$$ abc+(b+c)bc = a(b^2+c^2+2bc-a^2)+ bc(b+c)-abc$$
$$ 0 = a(b^2+c^2-a^2)$$ and we are done.
Is there more nice synthetic solution without trigonometry?
Best Answer
Let $A$ be the area of the triangle. Note that $$ \begin{aligned} \cos {\alpha} = {} & \frac{b^2 + c^2 - a^2}{2bc} \\ = {} & \frac{(b + c)^2 - a^2}{2bc} - 1 \\ = {} & \frac{(b + c - a)(b + c + a)}{2bc} - 1 \\ = {} & \frac{(2s - 2a)(2s)}{2bc} - 1 \\ = {} & \frac{2s(s - a)}{bc} - 1 \\ = {} & \frac{2sr(s - a)}{rbc} - 1 \\ = {} & \frac{2A(s - a)}{rbc} - 1 \\ = {} & \frac{2Aa(s - a)}{rabc} - 1 \\ = {} & \frac{2Aa(s - a)}{4ARr} - 1 \\ = {} & \frac{a(s - a)}{2Rr} - 1 \\ = {} & \frac{2R (s - 2R \sin {\alpha}) \sin {\alpha} }{2Rr} - 1 \\ = {} & \frac{(s - 2R \sin {\alpha}) \sin {\alpha} }{r} - 1. \end{aligned} $$ Hence $$ r(1 + \cos {\alpha}) = (s - 2R \sin {\alpha}) \sin {\alpha}, $$ which is $$ r \cdot 2\cos {\alpha \over 2} \cos {\alpha \over 2} = (s - 2R \sin {\alpha}) \cdot 2\cos {\alpha \over 2} \sin {\alpha \over 2}, $$ which is $$ r \cdot 2\cos {\alpha \over 2} \sin {\alpha \over 2} = (s - 2R \sin {\alpha}) \cdot 2\sin {\alpha \over 2} \sin {\alpha \over 2}, $$ which is $$ r \sin {\alpha} = (s - 2R \sin {\alpha}) (1 - \cos {\alpha}), $$ which is $$ \sin {\alpha} = \frac{s (1 - \cos {\alpha})}{(2R + r) - 2R\cos {\alpha}}. $$ Hence $$ r(1 + \cos {\alpha}) = \left( s - 2R \frac{s (1 - \cos {\alpha})}{(2R + r) - 2R\cos {\alpha}} \right) \frac{s (1 - \cos {\alpha})}{(2R + r) - 2R\cos {\alpha}}, $$ which is $$ 4R^2 \cos^3 {\alpha} - 4R (R + r) \cos^2 {\alpha} + (s^2 + r^2 - 4R^2) \cos {\alpha} - (s^2 - (2R + r)^2) = 0. $$
Hence $\cos {\alpha}$, $\cos {\beta}$, $\cos {\gamma}$ are the roots of the eqaution $$ 4R^2 x^3 - 4R (R + r) x^2 + (s^2 + r^2 - 4R^2)x - (s^2 - (2R + r)^2) = 0. $$ Hence $$ \cos {\alpha} \cos {\beta} \cos {\gamma} = \frac{s^2 - (2R + r)^2}{4R^2} = 0. $$ Hence one of $\cos {\alpha}$, $\cos {\beta}$, $\cos {\gamma}$ must be zero.