Difference between revisions of "2001 AMC 10 Problems/Problem 21"
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Now we create a second equation. We must choose some point <math>(x,y)</math> on the line <math>y=\frac{12}{5}x</math> such that <math>y=10-2x</math>, which implies that the cylinder’s diameter, <math>10-2x</math>, must be equal to its height, <math>y</math>. Solving yields <math>x=\frac{25}{11}</math>, and the radius is thus <math>\frac{10-2x}{2}=\frac{\frac{60}{11}}{2}=\boxed{\textbf{(B)}\ \frac{30}{11}}</math>. | Now we create a second equation. We must choose some point <math>(x,y)</math> on the line <math>y=\frac{12}{5}x</math> such that <math>y=10-2x</math>, which implies that the cylinder’s diameter, <math>10-2x</math>, must be equal to its height, <math>y</math>. Solving yields <math>x=\frac{25}{11}</math>, and the radius is thus <math>\frac{10-2x}{2}=\frac{\frac{60}{11}}{2}=\boxed{\textbf{(B)}\ \frac{30}{11}}</math>. | ||
− | ==Solution 5 ( | + | ==Solution 5 (Without similar triangles)== |
Like in Solution 2, we draw a diagram. | Like in Solution 2, we draw a diagram. | ||
Line 84: | Line 84: | ||
draw((30/11,60/11)--(30/11,0)); | draw((30/11,60/11)--(30/11,0)); | ||
draw((0,0)--(0,12)); | draw((0,0)--(0,12)); | ||
− | label("$ | + | label("$2x$",(0,30/11),E); |
− | label("$ | + | label("$2x$",(0,60/11),S); |
− | + | label("$H$",(0,0),S); | |
− | label("$ | ||
label("$A$",(0,12),N); | label("$A$",(0,12),N); | ||
label("$B$",(-5,0),SW); | label("$B$",(-5,0),SW); | ||
Line 93: | Line 92: | ||
label("$D$",(-30/11,60/11),W); | label("$D$",(-30/11,60/11),W); | ||
label("$E$",(30/11,60/11),E); | label("$E$",(30/11,60/11),E); | ||
+ | label("$F$",(30/11,0),S); | ||
+ | label("$G$",(-30/11,0),S); | ||
</asy> | </asy> | ||
− | + | It is known that <math>\overline{AH}</math> has length <math>12</math> and <math>\overline{BC}</math> has length <math>10</math>, so triangle <math>\triangle ABC</math> has area <math>60</math>. Also, let <math>x</math> be equal to the radius of the cylinder. | |
+ | |||
+ | |||
+ | Triangles <math>\triangle DBG</math> and <math>\triangle ECF</math> can be combined into one triangle with base <math>10-2x</math> and height <math>2x</math>. The area of this new triangle is <math>\frac{2x(10-2x)}{2} = 2x(5-x)</math>. | ||
+ | |||
+ | Triangle <math>\triangle ADE</math> has base <math>2x</math> and height <math>12-2x</math>, so its area is<math>\frac{2x(12-2x)}{2} = 2x(6-x)</math>. | ||
+ | |||
+ | Finally, square <math>DEFG</math> has area <math>4x^2</math>. | ||
+ | |||
+ | |||
+ | Now we can construct an equation to find <math>x</math>: | ||
+ | |||
+ | <cmath>2x(5-x) + 2x(6-x) + 4x^2 = 60</cmath> | ||
+ | <cmath>\Rightarrow 2x(5-x+6-x+2x) = 60</cmath> | ||
+ | <cmath>\Rightarrow 2x(11) = 60</cmath> | ||
+ | <cmath>\Rightarrow 22x = 60</cmath> | ||
+ | <cmath>\Rightarrow x = \frac{60}{22} = \boxed{(B) \frac{30}{11}}</cmath> | ||
+ | |||
+ | ~Dreamer1297 | ||
==Trivia== | ==Trivia== |
Latest revision as of 23:14, 10 February 2024
Contents
Problem
A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter and altitude
, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
Solution 1 (video solution)
Solution 2
Let the diameter of the cylinder be . Examining the cross section of the cone and cylinder, we find two similar triangles. Hence,
which we solve to find
. Our answer is
.
Solution 3 (Very similar to solution 2 but explained more)
We are asked to find the radius of the cylinder, or so we can look for similarity. We know that
and
, thus we have similarity between
and
by
similarity.
Therefore, we can create an equation to find the length of the desired side. We know that:
Plugging in yields:
Cross multiplying and simplifying gives:
Since the problem asks us to find the radius of the cylinder, we are done and the radius of the cylinder is .
~etvat
Solution 4 (graphical)
Assume that a point on a given diameter of the cone is the point on a two-dimensional representation of the cone as shown in Solution 2. The top point of the cone is thus
and the line that goes through both points is
.
Now we create a second equation. We must choose some point on the line
such that
, which implies that the cylinder’s diameter,
, must be equal to its height,
. Solving yields
, and the radius is thus
.
Solution 5 (Without similar triangles)
Like in Solution 2, we draw a diagram.
It is known that has length
and
has length
, so triangle
has area
. Also, let
be equal to the radius of the cylinder.
Triangles and
can be combined into one triangle with base
and height
. The area of this new triangle is
.
Triangle has base
and height
, so its area is
.
Finally, square has area
.
Now we can construct an equation to find :
~Dreamer1297
Trivia
This problem appeared in AoPS's Introduction to Geometry as a challenge problem.
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.