Difference between revisions of "1960 IMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
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+ | 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}{\pi R^3} (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. | ||
+ | |||
+ | It is easy to see that: | ||
+ | <cmath>V_1 = \frac{\pi R^3}{3} (\sqrt{x^2+1} + 1)^3</cmath> | ||
+ | <cmath>V_2 = \frac{\pi R^3}{3} 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>V_1 = \frac{\pi R^3}{3} (u + 1)^3</cmath> | ||
+ | <cmath>V_2 = \frac{\pi R^3}{3} 6(u^2 - 1)</cmath> | ||
+ | |||
+ | Define <math>f(u) = \frac{3}{\pi R^3} (V_1 - V_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}{3} R^3</math>, meaning that <math>V_1 \ne V_2</math>. | ||
==See Also== | ==See Also== | ||
{{IMO7 box|year=1960|num-b=5|num-a=7}} | {{IMO7 box|year=1960|num-b=5|num-a=7}} |
Revision as of 00:48, 7 July 2015
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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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.
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
.
See Also
1960 IMO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 7 |