A geometry problem by Rudresh Tomar

Let the triangle be $\triangle ABC$ , with $A$ , $B$ and $C$ be the bottom left, bottom right and top vertices respectively. Let the centre of the inscribed circle be $O$ and the point it touches $AB$ be $P$ , its radius be $r$ and $AP$ be $x$ .

Since $O$ is the meeting point of the angle divisors of $\angle CAB$ and $\angle CBA$ . Let $\frac {1}{2} \angle CBA$ be $\theta$ . Then we have:

$\tan {\frac {\pi} {8} } = \dfrac {r}{x} \quad \quad \tan {\theta} = \dfrac {r} {25 - x}$

$\Rightarrow x + (25 - x) = 25 = \dfrac {r}{\tan{\frac {\pi}{8}}} + \dfrac {r}{\tan{\theta} }\quad \Rightarrow r = \dfrac {25} {\frac {1}{\tan{\frac {\pi}{8}}} + \frac {1}{\tan{\theta} }}$

Now we have

$\tan {\frac {\pi}{4} } = \dfrac {2\tan {\frac {\pi}{8} } }{1 - \tan^2 {\frac {\pi}{8}}} = 1 \quad \Rightarrow \tan^2 {\frac {\pi}{8}} +2\tan {\frac {\pi}{8}} - 1 = 0\quad \Rightarrow \tan {\frac {\pi}{8}} = \sqrt{2} - 1$

Similarly, $\quad \tan {2\theta} = 2\quad \Rightarrow \tan^2 {\theta} +\tan {\theta} - 1 = 0\quad \Rightarrow \tan {\theta} = \frac {\sqrt{5} - 1} {2}$

Therefore, $\quad r = \dfrac {25} {\frac {1}{\sqrt{2}-1} + \frac {2}{\sqrt{5}-1}} \approx \boxed {6.20}$

First find the three sides of the triangle and let the side of the triangle be side a,side b and side c. Then use the formula r^2 = (s - a) (s - b) (s - c) / s. Where "s = (a + b + c)/2" and "r = radius of the circle inscribed in the triangle". Then we get the answer 6.20

The base of the triangle with the inscribed circle is 25. It is similar to the triangle with which it shares a vertical angle by a factor of 2. Therefore the height of the triangle is 2/3 of 25, so its area is 1/2 * 25 * 2/3 * 25.

The diagonal of the square is sqrt(25^2+25^2) by the Pythagorean theorem, so the left leg of the bottom triangle is 2/3 of that. Use the Pythagorean theorem again to get the length of the other segment that cuts the square: sqrt(25^2+(25/2)^2). Again, the right leg of the triangle is 2/3 of that.

Now you need to know that the area of a triangle is equal to the semiperimeter times the radius of the incircle. Now you have everything you need:

1/2 * 25 * 2/3 * 25 = r * 1/2 * (25 + 50 * sqrt(2)/3 + 25 * sqrt(5)/3)

After a mess of scratchwork, you get:

r = 50/(3+2*sqrt(2)+sqrt(5))

So r is about 6.20.

a/4=25/4=6.25

a/4=25/4=6.25