Difference between revisions of "1991 AIME Problems/Problem 2"
m |
(solution) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Rectangle <math>ABCD_{}^{}</math> has | + | [[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. |
== Solution == | == Solution == | ||
− | {{solution}} | + | {{image}} |
+ | 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>. | ||
== See also == | == See also == | ||
{{AIME box|year=1991|num-b=1|num-a=3}} | {{AIME box|year=1991|num-b=1|num-a=3}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 18:08, 11 March 2007
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
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 (a 3-4-5 right triangle). For each
,
is the hypotenuse of a 3-4-5 right triangle with sides of
. Thus, its length is
.
The sum we are looking for is . Using the formula for the sum of the first n numbers, we find that the solution is
.
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 |