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Difference between revisions of "1991 AIME Problems/Problem 2"

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== Problem ==
 
== Problem ==
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[[Rectangle]] <math>ABCD_{}^{}</math> has [[side]]s <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.
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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.
  
 
== 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>.
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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 sum we are looking for is <math>2 \cdot (\displaystyle\sum_{k = 1}^{167} 5 \cdot \frac{k}{168}) + 5 = \frac{5}{84} \displaystyle\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 (\displaystyle\sum_{k = 1}^{167} 5 \cdot \frac{k}{168}) + 5 = \frac{5}{84} \displaystyle\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>.  

Revision as of 10:34, 6 July 2007

Problem

Rectangle $ABCD_{}^{}$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$. For $1_{}^{} \le k \le 167$, draw the segments $\overline {P_kQ_k}$. Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$, and then draw the diagonal $\overline {AC}$. Find the sum of the lengths of the 335 parallel segments drawn.

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

The length of the diagonal is $\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 right triangle). For each $k$, $\overline{P_kQ_k}$ is the hypotenuse of a 3-4-5 right triangle with sides of $3 \cdot \frac{k}{168}, 4 \cdot \frac{k}{168}$. Thus, its length is $5 \cdot \frac{k}{168}$.

The sum we are looking for is $2 \cdot (\displaystyle\sum_{k = 1}^{167} 5 \cdot \frac{k}{168}) + 5 = \frac{5}{84} \displaystyle\sum_{k=1}^{167}k + 5$. Using the formula for the sum of the first n numbers, we find that the solution is $\frac{5}{84} \cdot \frac{168 \cdot 167}{2} + 5 = 5 \cdot 167 + 5 = 840$.

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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