Question on #3D

Geometry Level 4

A plane passes through a fixed point ( 37 , 41 , 43 ) (37,41,43) . The l o c u s locus of the foot of perpendicular to it from from the origin is the sphere a x 2 + b y 2 + c z 2 − d x − e y − f z + g = 0 ax^2+by^2+cz^2-dx-ey-fz+g=0

Find a + b + c + d + e + f + g \large{a+b+c+d+e+f+g}

Note: a , b , c , d , e , f , g a,b,c,d,e,f,g are co prime integers .


The answer is 124.

Tanishq Varshney by Tanishq Varshney , 23, India

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1 solution

Ujjwal Rane
Apr 1, 2015

This is just a 3 D extension of a 2 D property - inscribed angle in a semicircle is a right angle! So the origin and (37,41,43) are just end points of a diameter of the sphere!

Thus the center must be ( 37 2 , 41 2 , 43 2 , (\frac{37}{2},\frac{41}{2},\frac{43}{2}, ) and radius will be half the distance of (37,41,43) from the origin.

Put this in the equation ( x − 37 2 ) 2 + ( y − 41 2 ) 2 + ( z − 43 2 ) 2 = ( 37 2 ) 2 + ( 41 2 ) 2 + ( 43 2 ) 2 (x-\frac{37}{2})^2+(y-\frac{41}{2})^2+(z-\frac{43}{2})^2=(\frac{37}{2})^2+(\frac{41}{2})^2+(\frac{43}{2})^2

Giving the equation: x 2 + y 2 + z 2 − 37 x − 41 y − 43 z = 0 x^2 + y^2 + z^2 - 37x - 41 y - 43 z = 0

(a, b, c, d, e, f, g) = (1, 1, 1, 37, 41, 43, 0) and the sum a + b + c + d + e + f + g = 124

@Niranjan Khanderia

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Would be Very helpful!If someone elaborate that a little more :)

arifuzzaman arif - 5 years, 5 months ago

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Imagine two fixed points A and B in a plane.

We will draw a family of lines passing through B and then drop perpendiculars on these lines from A. So you will have a family of right angle triangles with AB as the hypotenuse.

What will be the locus of the third vertex C where the right angle is formed? Try sketching it and you will see it is a semicircle with AB as the diameter. Now there is no reason why C cannot lie on the other side of AB and hence we get a full circle!

The above problem is just an extension of this in 3D

Ujjwal Rane - 5 years, 4 months ago

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