Difference between revisions of "2001 AMC 10 Problems/Problem 21"
Dreamer1297 (talk | contribs) (→Solution 5 (Similar to Solution 2)) |
Dreamer1297 (talk | contribs) (→Solution 5 (Similar to Solution 2)) |
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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 97: | Line 96: | ||
</asy> | </asy> | ||
− | Let's say you don't know similar triangles but still want to solve this problem. You can combine triangles | + | 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. |
+ | |||
+ | |||
+ | Let's say you don't know similar triangles but still want to solve this problem. You can combine triangles <math>\triangle DBG</math> and triangle <math>\triangle ECF</math> into one triangle with base <math>10-x</math> and height <math>x</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== |
Revision as of 23:10, 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.
Let's say you don't know similar triangles but still want to solve this problem. You can combine triangles and triangle
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.