Difference between revisions of "1951 AHSME Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | {{ | + | The perimeter of the first triangle is <math>3a</math>. The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing <math>3a+\frac{3a}{2}+\frac{3a}{4}+\cdots</math>. |
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| + | The starting term is <math>3a</math>, and the common ratio is <math>1/2</math>. Therefore, the sum is <math>\frac{3a}{1-\frac{1}{2}}=\boxed{\textbf{(D)}\ 6a} </math> | ||
== See Also == | == See Also == | ||
Revision as of 20:14, 10 April 2013
Problem
An equilateral triangle is drawn with a side of length 
. A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is:
Solution
The perimeter of the first triangle is 
. The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing 
.
The starting term is , and the common ratio is 
. Therefore, the sum is 
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
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| All AHSME Problems and Solutions | ||
