Radius of a circle

Geometry Level 3

In the figure above, circle $O$ is tangent to the hypotenuse $BC$ of right isosceles $\triangle ABC.$ $AB$ and $AC$ are extended and are tangent to circle $O$ at $E$ and $F$ , respectively. If the area of $\triangle ABC$ is $50$ , find the length of the radius of circle $O$ . If your answer is of the form $a\sqrt{b}+10$ where $a$ and $b$ are positive integers with $b$ square-free, give your answer as $a+b$ .

The answer is 7.

1 solution

A Former Brilliant Member
Jun 14, 2017

From the formula of area of a triangle: $A=\dfrac{1}{2}(AB)(AC)$ but $AC=AB$ so we have

$50=\dfrac{1}{2}(AB)^2$ $\implies$ $100=(AB)^2$ $\implies$ $AB=\sqrt{100}=10$

By pythagorean theorem : $BC=\sqrt{10^2+10^2}=\sqrt{200}=10\sqrt{2}$

Since $\angle OAF=45^\circ, \triangle ADC$ is an isosceles right $\triangle$ . So $AD=DC=5\sqrt{2}$

Since $\triangle ADC \sim \triangle AFO$ , $\dfrac{DC}{OF}=\dfrac{AC}{OA}$

Substituting, we have

$\dfrac{5\sqrt{2}}{R}=\dfrac{10}{R+5\sqrt{2}}$ $\implies$ $5\sqrt{2}R+25(2)=10R$ $\implies$ $R=\dfrac{50}{10-5\sqrt{2}}$

After rationalizing the denominator, we get $R=$ $\boxed{5\sqrt{2}+10}$ .

Finally, $a+b=2+5=$ $\boxed{7}$

Radius of a circle

The answer is 7.

1 solution

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