Difference between revisions of "1991 AIME Problems/Problem 2"
(→Solution) |
(→Solution) |
||
Line 17: | Line 17: | ||
</asy></center><!-- asy replacing Image:1991 AIME-2.png by azjps --> | </asy></center><!-- asy replacing Image:1991 AIME-2.png by azjps --> | ||
− | 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 <math>3-4-5</math> 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>. Let <math>a_k=\frac{5k}{168}</math>. We want to find 2<math>\sum\limits_{k=1}^{168} a_k</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 <math>3-4-5</math> 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>. Let <math>a_k=\frac{5k}{168}</math>. We want to find 2<math>\sum\limits_{k=1}^{168} a_k-5</math> since we are over counting the diagonal. |
<math>2\sum\limits_{k=1}^{168} \frac{5k}{168}-5 | <math>2\sum\limits_{k=1}^{168} \frac{5k}{168}-5 | ||
=2\sum\limits_{k=0}^{168} \frac{5k}{168}-5 | =2\sum\limits_{k=0}^{168} \frac{5k}{168}-5 |
Revision as of 00:49, 3 January 2016
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 right triangle with sides of . Thus, its length is . Let . We want to find 2 since we are over counting the diagonal.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.