AIME 1991 · 第 2 题

Solution 1

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 $3 \cdot \frac{168-k}{168}, 4 \cdot \frac{168-k}{168}$. Thus, its length is $5 \cdot \frac{168-k}{168}$. Let $a_k=\frac{5(168-k)}{168}$. We want to find $2\sum\limits_{k=1}^{168} a_k-5$ since we are over counting the diagonal. $2\sum\limits_{k=1}^{168} \frac{5(168-k)}{168}-5 =2\frac{(0+5)\cdot169}{2}-5 =168\cdot5 =\boxed{840}$

Solution 2

Using the above diagram, we have that $\Delta ABC \sim \Delta P_k B Q_k$ and each one of these is a dilated 3-4-5 right triangle (This is true since $\Delta ABC$ is a 3-4-5 right triangle). Now, for all $k$, we have that $\overline{P_k Q_k}$ is the hypotenuse for the triangle $P_k B Q_k$. Therefore we want to know the sum of the lengths of all $\overline{P_k Q_k}$.This is given by the following:

$$2 \cdot(\sum_{k=1}^{168} P_kQ_k) + 5$$ $$= 2 \cdot \frac{ 0+5+10+...+835}{168} +5$$ Then by the summation formula for the sum of the terms of an arithmetic series,

$$= \frac{835 \cdot 168}{168} +5 = 835+5 = \boxed{840}$$ ~qwertysri987

Solution 3

First, count the diagonal which has length $5$. For the rest of the segments, think about pairing them up so that each pair makes $5$. For example, the parallel lines closest to the diagonal would have length $\frac{167}{168}\cdot{5}$ while the parallel line closest to the corner of the rectangle would have length $\frac{1}{168}\cdot{5}$ by similar triangles. If you add the two lengths together, it is $\frac{167}{168}\cdot{5} + \frac{1}{168}\cdot{5} = 5.$ There are $\frac{335-1}{2}$ pairs of these segments, for a total of $5+(167)(5)=168(5)=\boxed{840}.$ ~justlearningmathog