1960 IMO Problems/Problem 6

Revision as of 23:48, 6 July 2015 by Sujaykazi (talk | contribs) (Solution)

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 $V_1$ be the volume of the cone and $V_2$ be the volume of the cylinder.

a) Prove that $V_1 \neq V_2$;

b) Find the smallest number $k$ for which $V_1 = kV_2$; for this case, construct the angle subtended by a diamter of the base of the cone at the vertex of the cone.

Solution

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Let $R$ denote the radius of the cone, and let $r$ denote the radius of the cylinder and sphere. Let $l$ denote the slant height of the cone, and let $h$ 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 $\omega$ inscribed in an isosceles triangle $T$.

The area of $T$ may be computed in two different ways: \[[T] = \frac{1}{2} \times r \times (2l + 2R) = r(l+R)\] \[[T] = \frac{1}{2} \times 2R \times h = Rh\] From this, we deduce that $r = \frac{Rh}{l+R}$.

Now, we calculate our volumes: \[V_1 = \frac{1}{3}\pi R^2 h\] \[V_2 = \pi r^2 \times 2r = 2\pi r^3 = \frac{2R^3 h^3}{(l+R)^3}\] Now, we will compute the quantity $\frac{3}{\pi R^3} (V_1 - V_2)$ and prove that it is always greater than $0$. Let $x = \frac{h}{R}$. Clearly, $x$ can be any positive real number.

It is easy to see that: \[V_1 = \frac{\pi R^3}{3} (\sqrt{x^2+1} + 1)^3\] \[V_2 = \frac{\pi R^3}{3} 6x^2\]

Now, let $u = \sqrt{x^2+1}$. Since $x > 0$, it follows that $u > 1$. We now have: \[V_1 = \frac{\pi R^3}{3} (u + 1)^3\] \[V_2 = \frac{\pi R^3}{3} 6(u^2 - 1)\]

Define $f(u) = \frac{3}{\pi R^3} (V_1 - V_2)$. It follows that: \[f(u) = (u+1)^3 - 6(u^2 - 1)\] \[f(u) = u^3 - 3u^2 + 3u + 7\] \[f(u) = (u-1)^3 + 8 > 8 > 0\]

We see that $f(u) > 8$ for all allowed values of $u$. Thus, $V_1 - V_2 > \frac{8\pi}{3} R^3$, meaning that $V_1 \ne V_2$.

See Also

1960 IMO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 7 Followed by
Problem 7