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

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

#Geometry

Easy Math Editor

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Comments

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 - 6 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$