1991 AIME Problem 2

Question

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\leq k\leq167,

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.

Answer

Each segment

\overline{P_kQ_k}

is parallel to the

3

-

4

-

5

diagonal and is a fixed fraction of its length

The two side constructions give two identical arithmetic sums; remember to include

\overline{AC}

Put

A=(0,0),

B=(4,0),

and

C=(4,3).

Then

\begin{aligned}P_k&=\left(\frac{4k}{168},0\right),\\Q_k&=\left(4,3-\frac{3k}{168}\right),\end{aligned}

so the

3

-

4

-

5

ratio gives

P_kQ_k=5\left(1-\frac{k}{168}\right).

Hence one construction has total length

5\sum_{k=1}^{167}\left(1-\frac{k}{168}\right)=\frac{5\cdot167}{2}.

The construction on the other two sides has the same total, and

AC=5.

Thus the requested sum is

5(167)+5=840.

Problem 2 in other years