How to prove that the angle between two sides of that triangle is less than 60 degree?

Question

The product of two sides of triangle is equal to 8*(R*r) where R is circumradius of this triangle, and r is inradius of this triangle.

How to prove that the angle between two sides of that triangle is less than 60 degree?

Answer 1

$$ab=8Rr$$

$$absin(\gamma)=8Rrsin(\gamma)$$ Then left hand side is equal to $2A$ where $A$ is the area of triangle and we know that $2A=(a+b+c)r$

$$(a+b+c)r=8rRsin(\gamma)$$ $$(a+b+c)=8Rsin(\gamma) $$ and remember

$${a\over sin(\alpha)}={b\over sin(\beta)}={c\over sin(\gamma)}=2R$$

$${(a+b+c)\over 2R}=sin(\alpha)+sin(\beta)+sin(\gamma)=4sin(\gamma) $$

$$sin(\alpha)+sin(\beta)=3sin(\gamma) $$ Nottice that if $60^{0}\leq\gamma$ then rigth hand side is bigger than $2$ which is impossible since $sin(\alpha)+sin(\beta)\leq2$ we are done.

Answer 2

Suppose $ab = 8Rr$. We have $abc = 4RA$ where $A$ is the area of the triangle, hence $2rc = A$. The area of the triangle is also given by $rs$ where $2s=a+b+c$. It follows that $2rc = rs$ thus $4c=a+b+c$ or $3c=a+b$. It follows that $$ \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{9a^2+9b^2-(a+b)^2}{18ab} = \frac{8a^2+8b^2-2ab}{18ab} \geq \frac{14}{18}, $$ (using the fact that $a^2+b^2 \geq 2ab$) implying that $C \leq \arccos(\frac{14}{18}) = 38.9^\circ$.