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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


(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
2 years, 4 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)


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 \(c^3=2a^3 \implies \dfrac{a^3}{c^3}=\dfrac{1}{2} \implies \dfrac{a}{c} = \dfrac{1}{\sqrt[3]{2}}\)

\(\dfrac{FB}{BC} =\dfrac{\frac{a^2}{c}}{a} = \dfrac{a}{c} = \dfrac{1}{\sqrt[3]{2}}\) Aditya Raut · 2 years, 4 months ago

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