1991 AIME Problems/Problem 2

Revision as of 16:51, 23 October 2007 by Azjps (talk | contribs) (Solution: +img)

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

1991 AIME-2.png

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

See also

1991 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions
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