1960 IMO Problems/Problem 6
Contents
Problem
Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so that one of its bases lies in the base of the cone. Let be the volume of the cone and be the volume of the cylinder.
a) Prove that ;
b) Find the smallest number for which ; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.
Solution 1
Part (a):
Let denote the radius of the cone, and let denote the radius of the cylinder and sphere. Let denote the slant height of the cone, and let denote the height of the cone.
Consider a plane that contains the axis of the cone. This plane will slice the cone and sphere into a circle inscribed in an isosceles triangle .
The area of may be computed in two different ways: From this, we deduce that .
Now, we calculate our volumes: Now, we will compute the quantity and prove that it is always greater than . Let . Clearly, can be any positive real number. Define and . We will calculate and in terms of and then compute the desired quantity .
It is easy to see that:
Now, let . Since , it follows that . We now have:
Define . It follows that:
We see that for all allowed values of . Thus, , meaning that . We have thus proved that , as desired.
Part (b):
From our earlier work in calculating the volumes and , we easily see that: Re-expressing and simplifying, we have: By the AM-GM Inequality, , meaning that . Equality holds if and only if , meaning that and .
If we check the case , we may calculate and : Indeed, we have , meaning that our minimum of can be achieved.
Thus, we have proved that the minimum value of such that is .
Now, let be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following: From the double-angle formula for tangent, This angle is easy to construct. Simply take any segment and treat it as a unit segment. Create a right triangle with legs of lengths and . This is straightforward, and the angle opposite the leg of length will be the desired angle .
It follows that we have successfully constructed the desired angle .
Solution 2
Part a):
Let and denote the radius of the sphere and the radius of the base of the cone, respectively.
Consider a plane that contains a cross section of the cone. We get a circle (the sphere) inscribed in both an isosceles triangle (the cone) and a rectangle (the cylinder). Let and be the vertices at the base of the triangle and let be the altitude through the third vertex . Let be the angle with arms and , where is the center of the circle; .
We have ; . For the two volumes we have and .
Knowing , and using the formula for a double-angle tangent, we can simplify the result:
, where
We see that the biquadratic equation has no real roots for , which means that . Therefore .
Part b):
Let . Consider the function . The coefficient in front of is negative, which means the function has an upper bound, i.e. k has a lower bound (since it is the reciprocal of ). Using the vertex formula for a quadratic function, we get , which gives us . From there .
For we have . We defined to be in the interval (0;1), so . That gives us . To find we construct a right triangle with legs equal to and using a unit length reference.
See Also
1960 IMO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 7 |