Difference between revisions of "1991 AIME Problems/Problem 2"
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== Solution == | == Solution == | ||
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The length of the diagonal is <math>\sqrt{3^2 + 4^2} = 5</math> (a 3-4-5 [[right triangle]]). For each <math>k</math>, <math>\overline{P_kQ_k}</math> is the [[hypotenuse]] of a 3-4-5 right triangle with sides of <math>3 \cdot \frac{k}{168}, 4 \cdot \frac{k}{168}</math>. Thus, its length is <math>5 \cdot \frac{k}{168}</math>. | The length of the diagonal is <math>\sqrt{3^2 + 4^2} = 5</math> (a 3-4-5 [[right triangle]]). For each <math>k</math>, <math>\overline{P_kQ_k}</math> is the [[hypotenuse]] of a 3-4-5 right triangle with sides of <math>3 \cdot \frac{k}{168}, 4 \cdot \frac{k}{168}</math>. Thus, its length is <math>5 \cdot \frac{k}{168}</math>. | ||
− | The sum we are looking for is <math>2 \cdot \left(\sum_{k = 1}^{167} 5 \cdot \frac{k}{168}\right) + 5 = \frac{5}{84}\sum_{k=1}^{167}k + 5</math>. Using the formula for the sum of the first n numbers, we find that the solution is <math>\frac{5}{84} \cdot \frac{168 \cdot 167}{2} + 5 = 5 \cdot 167 + 5 = 840</math>. | + | The sum we are looking for is <math>2 \cdot \left(\sum_{k = 1}^{167} 5 \cdot \frac{k}{168}\right) + 5 = \frac{5}{84}\sum_{k=1}^{167}k + 5</math>. Using the formula for the sum of the first n numbers, we find that the solution is <math>\frac{5}{84} \cdot \frac{168 \cdot 167}{2} + 5 = 5 \cdot 167 + 5 = \boxed{840}</math>. |
== See also == | == See also == |
Revision as of 16:51, 23 October 2007
Problem
Rectangle has sides of length 4 and of length 3. Divide into 168 congruent segments with points , and divide into 168 congruent segments with points . For , draw the segments . Repeat this construction on the sides and , and then draw the diagonal . Find the sum of the lengths of the 335 parallel segments drawn.
Solution
The length of the diagonal is (a 3-4-5 right triangle). For each , is the hypotenuse of a 3-4-5 right triangle with sides of . Thus, its length is .
The sum we are looking for is . Using the formula for the sum of the first n numbers, we find that the solution is .
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |