Circle is inscribed in isosceles triangle with area $2$. Find angles of triangle for which radius of circle is maximal.

I have $\displaystyle r=\frac{4}{\frac{4}{\sqrt{\sin x}}+\sqrt{\frac{8}{\sin x}-\frac{8\cos x}{\sin x}}}$ but I think this function doesn't lead to any.

  • $\begingroup$ What does "Circle is inscribed in isosceles triangle with Square 2" mean? $\endgroup$ – Umberto P. Feb 4 '15 at 17:42
  • $\begingroup$ What is x? can you give a picture? $\endgroup$ – Narasimham Feb 4 '15 at 17:45
  • $\begingroup$ x is angle between equal sides of triangle. Sorry for my bad English. $\endgroup$ – Sinister Feb 4 '15 at 17:48

Here is a quick picture:

enter image description here

We can consider just half the isosceles triangle, as shown. Let the base be $b$ and height be $h$. The length of the base of half the triangle is $b/2$. The radius of the incircle is $b/2 \tan \theta/2$. The area of the whole isosceles triangle is $hb = b^2/4\tan\theta = 2$.

Using the double-angle formula for tangent, $$ hb = \frac{b^2}{4}\frac{2\tan\theta/2}{1-\tan^2 \theta/2} = \frac{b^2}{4}\frac{2x}{1-x^2} = 2 $$ where $x = \tan\frac{\theta}{2}$. The above equation is a constraint on the base $b$ for a given $x$.

The radius is then $$ r = b/2 \tan \theta/2 = \sqrt{\frac{1-x^2}{x}}x $$ Maximizing this, we set the derivative equal to zero, giving $x = \frac{1}{\sqrt 3}$, so $\theta/2 = \frac{\pi}{6}$, and so the triangle is equilateral.

  • $\begingroup$ Did you draw that in Paint? $\endgroup$ – Dylan Feb 4 '15 at 18:21
  • $\begingroup$ Yep. It's quick and easy. $\endgroup$ – Victor Liu Feb 4 '15 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.