Largest Cone In Sphere

Calculus Level 4

The volume of the largest cone that can be inscribed inside a unit sphere is of the form A π B \frac{ A \pi }{ B} for coprime positive integers A A and B B . What is A + B A + B ?


The answer is 113.

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2 solutions

Guilherme Niedu
Nov 16, 2016

Using the formula formula for circumradius of an isosceles triangle of base a a and height h h :

R = 1 8 ( a 2 h + h ) R = \frac{1}{8}(\frac{a^2}{h} + h)

Taking a section of the sphere along its central axis, we will have a triangle whose base is the cone base diameter and whose height is the cone height, which have to be circumscribed by a circle of radius 1. Thus, denoting cone radius as r r and height as h h :

1 = 1 8 ( 4 r 2 h + h ) 1 = \frac{1}{8}(\frac{4r^2}{h} + h)

r 2 = 2 h h 2 r^2 = 2h - h^2

So cone volume is:

v = π 3 r 2 h = π 3 ( 2 h 2 h 3 ) v = \frac{\pi}{3}r^2 h = \frac{\pi}{3} (2h^2 - h^3)

Differetiating v v with respect to h h and making it equal to 0 0 :

π 3 ( 4 h 3 h 2 ) = 0 \frac{\pi}{3}(4h - 3h^2) = 0

h = 4 3 h = \frac{4}{3}

Plugging this in v v :

v = π 3 ( 2 16 9 64 27 ) v = \frac{\pi}{3}(2\cdot\frac{16}{9} - \frac{64}{27})

v = π 32 81 v = \pi\cdot\frac{32}{81}

Hence, A = 32 A = 32 , B = 81 B = 81 and A + B = 113 A+B = \fbox{113}

Nice way for using circumradius to obtain the relationship between the height and the base diameter. (I set up the equations to solve for it manually)

Calvin Lin Staff - 4 years, 7 months ago
Tom Engelsman
Nov 15, 2016

Let's examine this problem via a plane section of the unit sphere (using the xy-plane and the circle x^2 + y^2 = 1):

Let the slanted edges of the cone in question be modeled as the lines y = (+/-)m*x + 1 (with the apex situated at the point (0,1). Substitution of this expression into that of our unit circle gives:

x^2 + [1 (+/-) mx]^2 = 1;

or x^2 + (1 (+/-)2mx + m^2 * x^2) = 1;

or (m^2 + 1)*x^2 (+/-)2mx = 0;

or x = 0 and x = (+/-) 2m/(m^2 + 1). Solving for the corresponding ordinate gives y = -sqrt(1 - x^2) = -sqrt[1 - (2m/(m^2+1))^2] = -[(m^2 - 1)/(m^2 + 1)]. So the endpoints of the cone's base are (x,y) = (-2m/(m^2+1), -(m^2-1)/(m^2+1)) and (2m/(m^2+1), -(m^2-1)/(m^2+1)). The radius of the cone equals 2m/(m^2 + 1) and its height equals 1 + (m^2 - 1)/(m^2 + 1) = (2m^2)/(m^2 + 1). So, the volume can now be expressed as a function of the slope m:

V(m) = (pi/3) * [2m/(m^2 + 1)]^2 * (2m^2)/(m^2 + 1) = (pi/3) * (8m^4)/(m^2 + 1)^3 (i). Differentiating (i) and setting it equal to zero produces:

dV/dm = (pi/3) * [(32m^3)(m^2 + 1)^3 - (8m^4) * 3(m^2+1)^2 * (2m)] / (m^2 + 1)^6 = 0;

or (pi/3) * [(32m^3)*(m^2 + 1) - (48m^5)] / (m^2 + 1)^4 = 0;

or 32m^5 + 32m^3 - 48m^5 = 0;

or 32m^3 - 16m^5 = 0;

or (16m^3)*(m^2 - 2) = 0;

or m = 0, (+/-)sqrt(2). Thus, the minimum volume is attained when m = 0, or V(0) = 0. The maximum volume must therefore be attained at either m = sqrt(2) or -sqrt(2), which both critical values result in V_max = (pi/3) * (8)(4)/(2+1)^3 = (32/81)*pi.

Nice way of representing the volume of the cone in 1 variable.

Calvin Lin Staff - 4 years, 7 months ago

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