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

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.

@@ Line 1: / Line 1: @@
 ==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 <math>V_1</math> be the volume of the cone and <math>V_2</math> be the volume of the cylinder.
-==Solution==
+a) Prove that <math>V_1 \neq V_2</math>;
-{{solution}}
+b) Find the smallest number <math>k</math> for which <math>V_1 = kV_2</math>; 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 <math>R</math> denote the radius of the cone, and let <math>r</math> denote the radius of the cylinder and sphere. Let <math>l</math> denote the slant height of the cone, and let <math>h</math> 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 <math>\omega</math> inscribed in an isosceles triangle <math>T</math>.
+The area of <math>T</math> may be computed in two different ways:
+<cmath>[T] = \frac{1}{2} \times r \times (2l + 2R) = r(l+R)</cmath>
+<cmath>[T] = \frac{1}{2} \times 2R \times h = Rh</cmath>
+From this, we deduce that <math>r = \frac{Rh}{l+R}</math>.
+Now, we calculate our volumes:
+<cmath>V_1 = \frac{1}{3}\pi R^2 h</cmath>
+<cmath>V_2 = \pi r^2 \times 2r = 2\pi r^3 = \frac{2R^3 h^3}{(l+R)^3}</cmath>
+Now, we will compute the quantity <math>\frac{3(l+R)^3}{\pi R^5 h} (V_1 - V_2)</math> and prove that it is always greater than <math>0</math>.
+Let <math>x = \frac{h}{R}</math>. Clearly, <math>x</math> can be any positive real number. Define <math>W_1 = \frac{3(l+R)^3}{\pi R^5 h} V_1</math> and <math>W_2 = \frac{3(l+R)^3}{\pi R^5 h} V_2</math>. We will calculate <math>W_1</math> and <math>W_2</math> in terms of <math>x</math> and then compute the desired quantity <math>W_1 - W_2</math>.
+It is easy to see that:
+<cmath>W_1 = (\sqrt{x^2+1} + 1)^3</cmath>
+<cmath>W_2 = 6x^2</cmath>
+Now, let <math>u = \sqrt{x^2+1}</math>. Since <math>x > 0</math>, it follows that <math>u > 1</math>. We now have:
+<cmath>W_1 = (u + 1)^3</cmath>
+<cmath>W_2 = 6(u^2 - 1)</cmath>
+Define <math>f(u) = W_1 - W_2</math>. It follows that:
+<cmath>f(u) = (u+1)^3 - 6(u^2 - 1)</cmath>
+<cmath>f(u) = u^3 - 3u^2 + 3u + 7</cmath>
+<cmath>f(u) = (u-1)^3 + 8 > 8 > 0</cmath>
+We see that <math>f(u) > 8</math> for all allowed values of <math>u</math>. Thus, <math>V_1 - V_2 > \frac{8\pi R^5 h}{3(l+R)^3}</math>, meaning that <math>V_1 > V_2</math>. We have thus proved that <math>V_1 \ne V_2</math>, as desired.
+Part (b):
+From our earlier work in calculating the volumes <math>V_1</math> and <math>V_2</math>, we easily see that:
+<cmath>\frac{V_1}{V_2} = \frac{(u+1)^3}{6(u^2 - 1)}</cmath>
+Re-expressing and simplifying, we have:
+<cmath>\frac{6V_1}{V_2} = \frac{(u+1)^2}{u-1} = (u-1) + 4 + \frac{4}{u-1}</cmath>
+By the AM-GM Inequality, <math>(u-1) + \frac{4}{u-1} \ge 4</math>, meaning that <math>\frac{V_1}{V_2} \ge \frac{4}{3}</math>.
+Equality holds if and only if <math>u-1 = \frac{4}{u-1}</math>, meaning that <math>u=3</math> and <math>x = 2\sqrt{2}</math>.
+If we check the case <math>x = \frac{h}{R} = 2\sqrt{2}</math>, we may calculate <math>V_1</math> and <math>V_2</math>:
+<cmath>V_1 = \frac{1}{3} \pi R^2 h = \frac{2\sqrt{2}}{3}\pi R^3</cmath>
+<cmath>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</cmath>
+Indeed, we have <math>\frac{V_1}{V_2} = \frac{4}{3}</math>, meaning that our minimum of <math>k = \frac{4}{3}</math> can be achieved.
+Thus, we have proved that the minimum value of <math>k</math> such that <math>V_1 = kV_2</math> is <math>\frac{4}{3}</math>.
+Now, let <math>\theta</math> be the angle subtended by a diameter of the base of the cone at the vertex of the cone. We have the following:
+<cmath>\tan\left(\frac{\theta}{2}\right) = \frac{1}{x} = \frac{\sqrt{2}}{4}</cmath>
+From the double-angle formula for tangent,
+<cmath>\theta = \arctan\left(\frac{4\sqrt{2}}{7}\right)</cmath>
+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 <math>4\sqrt{2}</math> and <math>7</math>. This is straightforward, and the angle opposite the leg of length <math>4\sqrt{2}</math> will be the desired angle <math>\theta</math>.
+It follows that we have successfully constructed the desired angle <math>\theta</math>.
+==Solution 2==
+Part a):
+Let <math>r</math> and <math>a</math> 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 <math>A</math> and <math>A_1</math> be the vertices at the base of the triangle and let <math>H_1 H_2 = h</math> be the altitude through the third vertex <math>H_1</math>. Let <math>\theta</math> be the angle with arms <math>AO</math> and <math>AA_1</math>, where <math>O</math> is the center of the circle; <math>0 < \theta < \frac{\pi}{4}</math> .
+We have <math>\tan \theta = \frac{r}{a}</math> ; <math>0 < \tan \theta < 1</math>. For the two volumes we have <math>V_2 = \pi r^2 \times 2r = 2\pi r^3</math> and <math>V_1 = \frac{\pi a^2 \times H_1 H_2}{3} = \frac{\pi a^3 \tan (\angle A_1 A H_1) }{3}</math> .
+Knowing <math>\angle A_1 A H_1 = 2 \times \angle A_1 A O = 2 \theta</math>, and using the formula for a double-angle tangent, we can simplify the result:
+<math>V_1 = \frac{\pi a^3 \tan (\angle A_1 A H_1)}{3}</math>
+<math>V_1 = \frac{2 \pi r^3 \tan \theta}{3 \tan^3 \theta (1 - \tan^2 \theta)}</math>
+<math>V_1 = k V_2</math>, where <math>k = \frac{1}{3\tan^2 \theta (1 - \tan^2 \theta)}</math>
+We see that the biquadratic equation <math>3 \tan^4 \theta - 3 \tan^2 \theta + 1 = 0</math> has no real roots for <math>\tan \theta</math>, which means that <math>k \neq 1</math>. Therefore <math>V_1 \neq V_2</math> .
+Part b):
+Let <math>\tan^2 \theta = x; 0 < x < 1</math>. Consider the function <math>f(x) = 3x - 3x^2</math>. The coefficient in front of <math>x^2</math> is negative, which means the function has an upper bound, i.e. k has a lower bound (since it is the reciprocal of <math>f(x)</math>). Using the vertex formula for a quadratic function, we get <math>x = -\frac{b}{2a} = \frac{1}{2}</math> , which gives us <math>f(x) = y_{max} = \frac{3}{4}</math> . From there <math>k_{min} = \frac{4}{3}</math> .
+For <math>\tan^2 \theta</math> we have <math>\tan^2 \theta = \frac{1}{2};   \tan \theta = +/- \frac{\sqrt{2}}{4}</math> . We defined <math>\tan \theta</math> to be in the interval (0;1), so <math>\tan \theta = \frac{\sqrt{2}}{4}</math> . That gives us <math>\tan \angle A_1 A H_1 = \frac{4 \sqrt{2}}{7}</math> . To find <math>2\theta</math> we construct a right triangle with legs equal to <math>7</math> and <math>4\sqrt{2}</math> using a unit length reference.
 ==See Also==
-{{IMO box|year=1960|num-b=5|after=Final Question}}
+{{IMO7 box|year=1960|num-b=5|num-a=7}}
+[[Category:Olympiad Geometry Problems]]
+[[Category:3D Geometry Problems]]
+[[Category:Geometric Construction Problems]]

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

Latest revision as of 17:49, 29 January 2018

Contents

Problem

Solution 1

Solution 2

See Also