Waste less time on Facebook — follow Brilliant.
×

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 ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

img

img

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

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

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...