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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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