Difference between revisions of "2016 AMC 10B Problems/Problem 10"
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Note: In general, the weight would be directly proportional to the volume. However, we are told they have equal height, so we essentially treat the problem as 2D. | Note: In general, the weight would be directly proportional to the volume. However, we are told they have equal height, so we essentially treat the problem as 2D. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/zUzgzT1kFyo | ||
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| + | ~Education, the Study of Everything | ||
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==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
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==Solution 4 (Intuition)== | ==Solution 4 (Intuition)== | ||
| − | We ignore the height in this solution because it's constant with both triangles; thus, it has no effect on the difference in their respective weights. Now here's the intuition: notice how the <math>3</math> by <math>3</math> triangle is simply scaled by a factor of <math>\frac{5}{3}</math> on each length to form the <math>5</math> by <math>5</math> triangle, which means that intuitively on one length the weight is scaled by that factor, and we note that we have 2-dimensional system since we ignored the height; therefore, we proceed by doing <math>(\frac{5}{3})^2 \cdot | + | We ignore the height in this solution because it's constant with both triangles; thus, it has no effect on the difference in their respective weights. Now here's the intuition: notice how the <math>3</math> by <math>3</math> triangle is simply scaled by a factor of <math>\frac{5}{3}</math> on each length to form the <math>5</math> by <math>5</math> triangle, which means that intuitively on one length the weight is scaled by that factor, and we note that we have 2-dimensional system since we ignored the height; therefore, we proceed by doing <math>(\frac{5}{3})^2 \cdot 12</math> to get our desired result as <math>\boxed{\textbf{(D)}\ 33.3}</math>. |
~triggod | ~triggod | ||
| + | ~edited by someone else | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2016|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:45, 2 July 2023
Contents
Problem
A thin piece of wood of uniform density in the shape of an equilateral triangle with side length 
inches weighs 
ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of 
inches. Which of the following is closest to the weight, in ounces, of the second piece?
Solution 1
We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
.
We can then solve the equation to get 
which is closest to 
Solution 2
Also recall that the area of an equilateral triangle is 
so we can give a ratio as follows:
Cross multiplying and simplifying, we get 
Which is 
- Solution by
Solution 3
Note that the ratio of the two triangle's weights is equal to the ratio of their areas, as the height is the same. The ratio of their areas is equal to the square of the ratio of their sides. So if 
denotes the weight of the second triangle, we have ![\[\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}\]](http://latex.artofproblemsolving.com/f/f/a/ffa8e6a726de50505f6c51db032d2c9aef0d51a5.png)
Solving gives us 
so the answer is .
Note: In general, the weight would be directly proportional to the volume. However, we are told they have equal height, so we essentially treat the problem as 2D.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
Solution 4 (Intuition)
We ignore the height in this solution because it's constant with both triangles; thus, it has no effect on the difference in their respective weights. Now here's the intuition: notice how the by triangle is simply scaled by a factor of 
on each length to form the by triangle, which means that intuitively on one length the weight is scaled by that factor, and we note that we have 2-dimensional system since we ignored the height; therefore, we proceed by doing 
to get our desired result as .
~triggod ~edited by someone else
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
