Difference between revisions of "1991 AIME Problems/Problem 2"
(→Solution) |
(→Solution) |
||
Line 21: | Line 21: | ||
=<math>2\sum\limits_{k=0}^{168} \frac{5k}{168}-5</math> | =<math>2\sum\limits_{k=0}^{168} \frac{5k}{168}-5</math> | ||
=<math>2\frac{(0+5)\cdot169}{2}-5</math> | =<math>2\frac{(0+5)\cdot169}{2}-5</math> | ||
− | =<math>168 | + | =<math>168\cdot5</math> |
=<math>\boxed{840}</math> | =<math>\boxed{840}</math> | ||
Revision as of 00:48, 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 2-5 since we are over counting . = = = =
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.