Difference between revisions of "1991 AIME Problems/Problem 2"
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[[Rectangle]] <math>ABCD_{}^{}</math> has [[edge | sides]] <math>\overline {AB}</math> of [[length]] 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</math> into 168 [[congruent]] [[segment]]s with [[point]]s <math>A_{}^{}=P_0, P_1, \ldots, P_{168}=B</math>, and divide <math>\overline {CB}</math> into 168 congruent segments with points <math>C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B</math>. For <math>1_{}^{} \le k \le 167</math>, draw the segments <math>\overline {P_kQ_k}</math>. Repeat this [[construction]] on the sides <math>\overline {AD}</math> and <math>\overline {CD}</math>, and then draw the [[diagonal]] <math>\overline {AC}</math>. Find the sum of the lengths of the 335 [[parallel]] segments drawn. | [[Rectangle]] <math>ABCD_{}^{}</math> has [[edge | sides]] <math>\overline {AB}</math> of [[length]] 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</math> into 168 [[congruent]] [[segment]]s with [[point]]s <math>A_{}^{}=P_0, P_1, \ldots, P_{168}=B</math>, and divide <math>\overline {CB}</math> into 168 congruent segments with points <math>C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B</math>. For <math>1_{}^{} \le k \le 167</math>, draw the segments <math>\overline {P_kQ_k}</math>. Repeat this [[construction]] on the sides <math>\overline {AD}</math> and <math>\overline {CD}</math>, and then draw the [[diagonal]] <math>\overline {AC}</math>. Find the sum of the lengths of the 335 [[parallel]] segments drawn. | ||
− | == Solution == | + | == Solution 1 == |
<center><asy> | <center><asy> | ||
real r = 0.35; size(220); | real r = 0.35; size(220); | ||
Line 22: | Line 22: | ||
=168\cdot5 | =168\cdot5 | ||
=\boxed{840}</math> | =\boxed{840}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Using the above diagram, we have that <math> \Delta ABC \sim \Delta P_k B Q_k </math> and each one of these is a dilated 3-4-5 right triangle (This is true since <math>\Delta ABC </math> is a 3-4-5 right triangle). Now, for all <math>k</math>, we have that <math>\overline{P_k Q_k}</math> is the hypotenuse for the triangle <math>P_k B Q_k</math>. Therefore we want to know the sum of the lengths of all <math>\overline{P_k Q_k}</math>.This is given by the following: | ||
+ | <cmath> 2 \cdot(\sum_{k=1}^{168} P_kQ_k) + 5 </cmath> | ||
+ | |||
+ | <cmath> = 2 \cdot \frac{ 0+5+10+...+835}{168} +5</cmath> | ||
+ | Then by the summation formula for the sum of the terms of an arithmetic series, | ||
+ | <cmath> = \frac{835 \cdot 168}{168} +5 = 835+5 = \boxed{840}</cmath> | ||
+ | |||
+ | ~qwertysri987 | ||
== See also == | == See also == |
Revision as of 11:51, 19 July 2019
Contents
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 1
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 since we are over counting the diagonal.
Solution 2
Using the above diagram, we have that and each one of these is a dilated 3-4-5 right triangle (This is true since is a 3-4-5 right triangle). Now, for all , we have that is the hypotenuse for the triangle . Therefore we want to know the sum of the lengths of all .This is given by the following:
Then by the summation formula for the sum of the terms of an arithmetic series,
~qwertysri987
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 |
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