$P$ and $Q$ are two points on a circle of center $C$ and radius $a$, the angle $\widehat{PCQ}$ being $2\theta$, then the radius of the circle inscribed in the triangle $CPQ$ is maximum when

  • $\sin\theta=\frac{\sqrt3-1}{2\sqrt2}$
  • $\sin\theta=\frac{\sqrt5-1}{2}$
  • $\sin\theta=\frac{\sqrt5+1}{2}$
  • $\sin\theta=\frac{\sqrt5-1}{4}$

$CPQ$ is a isosceles triangle with vertex angle $2\theta$ and two equal sides $a$.
I have learnt a formula from Wolfram for inradius of isosceles triangle is given by $$r=\frac{a(\sqrt{a^2+4h^2}-a)}{4h}$$ Here $h$ is the base of the isosceles triangle and $a$ is the congruent side. So $\sin\theta = \frac{h}{2a}$
Making $r$ in terms of $\sin\theta$, we get $$r=\frac{a(\sqrt{1+16\sin^2\theta}-1)}{8\sin\theta}$$ When i differentiated this equation for $r$ and put it equal to zero, I get $\sin\theta=0$.
Is this method not correct? Is there some better method possible?


The inradius is $$r=\frac{2A}{p}= \frac{\sin(2\theta) a^2}{2a+2a\sin\theta}= \frac a 2 \cdot \frac{\sin2\theta}{1+\sin\theta}$$ where $A$ is the area of the triangle, $p$ its perimeter, $a$ the radius of the first circle.

If you derive it, you get $$\begin{align}r'(\theta) &= \frac a 2 \cdot \frac{2\cos 2\theta\,(1+\sin\theta)- \sin 2\theta \cos\theta}{(1+\sin\theta)^2} \\ &= a \cdot\frac{\left(2\cos^2\theta-1\right)(1+\sin\theta) - \sin\theta\cos^2\theta}{(1+\sin\theta)^2} \\ &= a \cdot \frac{2\cos^2\theta+\sin\theta\cos^2\theta-\sin\theta-1}{(1+\sin\theta)^2} \\ &= a \cdot \frac{\cos^2\theta+\sin\theta\cos^2\theta-\sin\theta-\sin^2\theta}{(1+\sin\theta)^2} \\ &= a \cdot \frac{\left(1-\sin^2\theta\right)(1+\sin\theta)-\sin\theta(1+\sin\theta)}{(1+\sin\theta)^2} = a \cdot \left(\frac 1{1+\sin\theta} -\sin\theta \right) \end{align}$$ Then you solve the inequality $r'(\theta) \le 0$, that is $$\sin^2 \theta+\sin \theta - 1 \ge 0$$ And you can thus find when $r$ is maximum. Note that $0 \le \theta \le \frac\pi 2$.

  • $\begingroup$ How do you find the $$r'(\theta)= \frac{1}{\sin\theta +1}-\sin\theta$$?@Timon $\endgroup$ – user1557 Apr 8 '16 at 16:01
  • $\begingroup$ I have added some steps. ;) $\endgroup$ – Τίμων Apr 8 '16 at 19:58

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