Difference between revisions of "1991 AIME Problems/Problem 2"
(→Solution) |
(Adding another possible solution) |
||
(6 intermediate revisions by 3 users not shown) | |||
Line 2: | Line 2: | ||
[[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 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{ | + | 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{168-k}{168}, 4 \cdot \frac{168-k}{168}</math>. Thus, its length is <math>5 \cdot \frac{168-k}{168}</math>. Let <math>a_k=\frac{5(168-k)}{168}</math>. We want to find <math>2\sum\limits_{k=1}^{168} a_k-5</math> since we are over counting the diagonal. |
− | <math>2\sum\limits_{k=1}^{168} \frac{ | + | <math>2\sum\limits_{k=1}^{168} \frac{5(168-k)}{168}-5 |
− | =<math>2\ | + | =2\frac{(0+5)\cdot169}{2}-5 |
− | =<math> | + | =168\cdot5 |
− | =<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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | First, count the diagonal which has length <math>5</math>. For the rest of the segments, think about pairing them up so that each pair makes <math>5</math>. For example, the parallel lines closest to the diagonal would have length <math>\frac{167}{168}\cdot{5}</math> while the parallel line closest to the corner of the rectangle would have length <math>\frac{1}{168}\cdot{5}</math> by similar triangles. If you add the two lengths together, it is <math>\frac{167}{168}\cdot{5} + \frac{1}{168}\cdot{5} = 5.</math> There are <math>\frac{335-1}{2}</math> pairs of these segments, for a total of <math>5+(167)(5)=168(5)=\boxed{840}.</math> | ||
+ | ~justlearningmathog | ||
== See also == | == See also == |
Latest revision as of 11:51, 17 June 2021
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
Solution 3
First, count the diagonal which has length . For the rest of the segments, think about pairing them up so that each pair makes . For example, the parallel lines closest to the diagonal would have length while the parallel line closest to the corner of the rectangle would have length by similar triangles. If you add the two lengths together, it is There are pairs of these segments, for a total of ~justlearningmathog
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.