Just another problem...

Calculus Level pending

That guy Maths is an irresistible fellow. However much you may be frustrated with him, his very presence is tempting and you can't resist solving his problems. What will you do if Maths r e a l l y really was a human I wonder...

That apart, this is another of Maths' problems. Give it a try:

The normal at a point ( x , y ) (x,y) on a curve meets the co-ordinate axes at A , B A , B . If 1 O A + 1 O B = 1 \frac{1}{OA} + \frac{1}{OB} = 1 where O O is the origin and ( 5 , 4 ) (5 , 4) lies on the curve,

its equation can be:

( x − a ) 2 + ( y − a ) 2 = b (x-a)^2 + (y-a)^2 = b

or

( x − a ) 2 + ( y + a ) 2 = c (x-a)^2 + (y+a)^2 = c

or

( x + a ) 2 + ( y + a ) 2 = d (x+a)^2 + (y+a)^2 = d

or

( x + a ) 2 + ( y − a ) 2 = e (x+a)^2 + (y-a)^2 = e

where a , b , c , d a , b , c , d and e e are positive integers.

Find a + b + c + d + e − 4 \sqrt{ a + b + c + d + e - 4 }


The answer is 13.

Chinmay Raut by Chinmay Raut , 23, India

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1 solution

Chinmay Raut
Jan 11, 2015

x + y f ′ ( x 0 ) = x 0 + y 0 f ′ ( x 0 ) x + y f'(x_0) = x_0 + y_0 f'(x_0)

A ≡ ( x 0 + y 0 f ′ ( x 0 ) , 0 ) A \equiv (x_0 + y_0f'(x_0), 0 )

B ≡ ( 0 , y 0 + x 0 f ′ ( x 0 ) ) B \equiv (0, y_0+\frac{x_0}{f'(x_0)})

1 O A + 1 O B = 1 \frac{1}{OA} + \frac{1}{OB} = 1

⇔ 1 ∣ x 0 + y 0 f ′ ( x 0 ) ∣ + ∣ f ′ ( x 0 ) ∣ ∣ x 0 + y 0 f ′ ( x 0 ) ∣ = 1 \Leftrightarrow \frac{1}{|x_0+y_0f'(x_0)|} + \frac{|f'(x_0)|}{|x_0+y_0f'(x_0)|} = 1

⇔ 1 + ∣ f ′ ( x 0 ) ∣ = ∣ x 0 + y 0 f ′ ( x 0 ) ∣ \Leftrightarrow 1 + |f'(x_0)| = |x_0 + y_0f'(x_0)|

⇔ ± ( 1 + ∣ f ′ ( x 0 ) ∣ ) = x 0 + y 0 f ′ ( x 0 ) \Leftrightarrow \pm (1+ |f'(x_0)|) = x_0 + y_0f'(x_0)

or

⇔ ± ( 1 ± f ′ ( x 0 ) ) = x 0 + y 0 f ′ ( x 0 ) \Leftrightarrow \pm(1 \pm f'(x_0)) = x_0 + y_0f'(x_0)

Integrating, we get,

x 2 + y 2 ± 2 x ± 2 y = c x^2 + y^2 \pm 2x \pm 2y = c . . . . . . . . . . . . . . . . . . . . . . .(c ∈ \in Real constant ( y = f ( x ) ) (y=f(x)) )

( 5 , 4 ) (5 , 4) lies on y = f ( x ) y= f(x)

Hence, possible curves are,

( x − 1 ) 2 + ( y − 1 ) 2 = 25 (x-1)^2 + (y-1)^2 = 25

or

( x − 1 ) 2 + ( y + 1 ) 2 = 41 (x-1)^2 + (y+1)^2 = 41

or

( x + 1 ) 2 + ( y + 1 ) 2 = 61 (x+1)^2 + (y+1)^2 = 61

or

( x + 1 ) 2 + ( y − 1 ) 2 = 45 (x+1)^2 + (y-1)^2 = 45

∴ 1 + 25 + 41 + 61 + 45 − 4 = 13 \therefore \sqrt{1+25+41+61+45-4} = \boxed {13}

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