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NMTC Inter Level Problem 6

$$ABC$$ is a right angled triangle with hypotenuse $$AB$$. $$CF$$ is the altitude. The circle through $$F$$ centered at $$B$$ and another circle of the same radius centered at $$A$$ intersect on the side $$BC$$. Then the ratio $$\frac{FB}{BC}$$ is equal to

Options:

(A) $$\frac{1}{2}$$

(B) $$\frac{1}{\sqrt{2}}$$

(C) $$\frac{1}{\sqrt[3]{2}}$$

(D) $$\frac{1}{2\sqrt{2}}$$

Note by Nanayaranaraknas Vahdam
3 years, 6 months ago

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Let's call sides as $$a,b,c$$, the sides opposite to $$\angle A,\angle B,\angle C$$.

$$c \times CF = a\times b$$ (from the area)

$$CF=\dfrac{ab}{c}$$

By similarity of triangles, we have $$\triangle ACB \sim \triangle CFB$$ and thus

$$\dfrac{FB}{a}= \dfrac{a}{c} \implies FB =\dfrac{a^2}{c}$$

We have $$AD=FB=BD =\dfrac{a^2}{c}$$

In $$\triangle ACD$$ ,

$$AD^2=FB^2 = CD^2+b^2 = (a-FB)^2 +b^2$$

$$\therefore \dfrac{a^4}{c^2} =\biggl(a-\dfrac{a^2}{c}\biggr)^2+b^2$$

$$\therefore \dfrac{a^4}{c^2} =a^2-\dfrac{2a^3}{c}+\dfrac{a^4}{c^2}+b^2$$

$$\therefore 0=a^2+b^2-\dfrac{2a^3}{c}$$

$$\therefore a^2+b^2=c^2=\dfrac{2a^3}{c}$$

Thus $$c^3=2a^3 \implies \dfrac{a^3}{c^3}=\dfrac{1}{2} \implies \dfrac{a}{c} = \dfrac{1}{\sqrt[3]{2}}$$

$$\dfrac{FB}{BC} =\dfrac{\frac{a^2}{c}}{a} = \dfrac{a}{c} = \dfrac{1}{\sqrt[3]{2}}$$

- 3 years, 6 months ago