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I need to solve the following problem only by using Pythagoras Theorem and congruent triangles. Find the sides of an isosceles triangle ABC with circumradius R=25 and inradius r=12.

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From the figure

incircle

one imediately derives the equations $${\rho\over h}={a\over 2R}\ ,\qquad{\rm i.e.,}\qquad a h=600\ ,$$ $$h+\rho+d=2R\ ,\qquad{\rm i.e.,}\qquad h+d=38\ ,$$ $$2R d=a^2 \ ,\qquad{\rm i.e.,}\qquad 50d = a^2\ .$$ Eliminating $h$ and $d$ one obtains a cubic equation for $a$, two of whose solutions are natural numbers. Given $a$ and $d$ one computes $s=\sqrt{4R^2-a^2}$, and the base $b$ of the triangle is $b=2\sqrt{a^2-d^2}$.

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The side lengths will be $40$, $40$, and $48$.

The circumradius of an isosceles triangle is

$$\frac{a^2}{2\sqrt{a^2 - \frac{b^2}{4}}},$$

where two sides are of length $a$ and the third is of length $b$.

The inradius of an isoceles triangle is

$$\frac{ab - \frac{b^2}{2}}{2\sqrt{a^2 - \frac{b^2}{4}}}.$$

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