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Difference between revisions of "1960 IMO Problems/Problem 6"

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Latest revision as of 17:49, 29 January 2018

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 1

Part (a):

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 $\frac{3(l+R)^3}{\pi R^5 h} (V_1 - V_2)$ and prove that it is always greater than $0$. Let $x = \frac{h}{R}$. Clearly, $x$ can be any positive real number. Define $W_1 = \frac{3(l+R)^3}{\pi R^5 h} V_1$ and $W_2 = \frac{3(l+R)^3}{\pi R^5 h} V_2$. We will calculate $W_1$ and $W_2$ in terms of $x$ and then compute the desired quantity $W_1 - W_2$.

It is easy to see that: \[W_1 = (\sqrt{x^2+1} + 1)^3\] \[W_2 = 6x^2\]

Now, let $u = \sqrt{x^2+1}$. Since $x > 0$, it follows that $u > 1$. We now have: \[W_1 = (u + 1)^3\] \[W_2 = 6(u^2 - 1)\]

Define $f(u) = W_1 - W_2$. 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, $V_1 - V_2 > \frac{8\pi R^5 h}{3(l+R)^3}$, meaning that $V_1 > V_2$. We have thus proved that $V_1 \ne V_2$, as desired.

Part (b):

From our earlier work in calculating the volumes $V_1$ and $V_2$, we easily see that: \[\frac{V_1}{V_2} = \frac{(u+1)^3}{6(u^2 - 1)}\] Re-expressing and simplifying, we have: \[\frac{6V_1}{V_2} = \frac{(u+1)^2}{u-1} = (u-1) + 4 + \frac{4}{u-1}\] By the AM-GM Inequality, $(u-1) + \frac{4}{u-1} \ge 4$, meaning that $\frac{V_1}{V_2} \ge \frac{4}{3}$. Equality holds if and only if $u-1 = \frac{4}{u-1}$, meaning that $u=3$ and $x = 2\sqrt{2}$.

If we check the case $x = \frac{h}{R} = 2\sqrt{2}$, we may calculate $V_1$ and $V_2$: \[V_1 = \frac{1}{3} \pi R^2 h = \frac{2\sqrt{2}}{3}\pi R^3\] \[V_2 = 2\pi r^3 = 2\pi \left(\frac{R\times 2\sqrt{2} R}{4R}\right)^3 = \frac{\pi}{\sqrt{2}} R^3\] Indeed, we have $\frac{V_1}{V_2} = \frac{4}{3}$, meaning that our minimum of $k = \frac{4}{3}$ can be achieved.

Thus, we have proved that the minimum value of $k$ such that $V_1 = kV_2$ is $\frac{4}{3}$.

Now, let $\theta$ be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following: \[\tan\left(\frac{\theta}{2}\right) = \frac{1}{x} = \frac{\sqrt{2}}{4}\] From the double-angle formula for tangent, \[\theta = \arctan\left(\frac{4\sqrt{2}}{7}\right)\] 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 $4\sqrt{2}$ and $7$. This is straightforward, and the angle opposite the leg of length $4\sqrt{2}$ will be the desired angle $\theta$.

It follows that we have successfully constructed the desired angle $\theta$.

Solution 2

Part a):

Let $r$ and $a$ 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 $A$ and $A_1$ be the vertices at the base of the triangle and let $H_1 H_2 = h$ be the altitude through the third vertex $H_1$. Let $\theta$ be the angle with arms $AO$ and $AA_1$, where $O$ is the center of the circle; $0 < \theta < \frac{\pi}{4}$ .

We have $\tan \theta = \frac{r}{a}$ ; $0 < \tan \theta < 1$. For the two volumes we have $V_2 = \pi r^2 \times 2r = 2\pi r^3$ and $V_1 = \frac{\pi a^2 \times H_1 H_2}{3} = \frac{\pi a^3 \tan (\angle A_1 A H_1) }{3}$ .

Knowing $\angle A_1 A H_1 = 2 \times \angle A_1 A O = 2 \theta$, and using the formula for a double-angle tangent, we can simplify the result:


$V_1 = \frac{\pi a^3 \tan (\angle A_1 A H_1)}{3}$


$V_1 = \frac{2 \pi r^3 \tan \theta}{3 \tan^3 \theta (1 - \tan^2 \theta)}$


$V_1 = k V_2$, where $k = \frac{1}{3\tan^2 \theta (1 - \tan^2 \theta)}$


We see that the biquadratic equation $3 \tan^4 \theta - 3 \tan^2 \theta + 1 = 0$ has no real roots for $\tan \theta$, which means that $k \neq 1$. Therefore $V_1 \neq V_2$ .

Part b):

Let $\tan^2 \theta = x; 0 < x < 1$. Consider the function $f(x) = 3x - 3x^2$. The coefficient in front of $x^2$ is negative, which means the function has an upper bound, i.e. k has a lower bound (since it is the reciprocal of $f(x)$). Using the vertex formula for a quadratic function, we get $x = -\frac{b}{2a} = \frac{1}{2}$ , which gives us $f(x) = y_{max} = \frac{3}{4}$ . From there $k_{min} = \frac{4}{3}$ .

For $\tan^2 \theta$ we have $\tan^2 \theta = \frac{1}{2}; \tan \theta = +/- \frac{\sqrt{2}}{4}$ . We defined $\tan \theta$ to be in the interval (0;1), so $\tan \theta = \frac{\sqrt{2}}{4}$ . That gives us $\tan \angle A_1 A H_1 = \frac{4 \sqrt{2}}{7}$ . To find $2\theta$ we construct a right triangle with legs equal to $7$ and $4\sqrt{2}$ 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
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