Question

Triangle $ABC$ has side lengths $AB = 9,$ $BC = 5\sqrt{3},$ and $AC = 12.$ Points $A = P_0, P_1, P_2, \ldots, P_{2450} = B$ are on segment $\overline{AB}$ with $P_k$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, \ldots, 2449,$ and points $A = Q_0, Q_1, Q_2, \ldots, Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_k$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, \ldots, 2449.$ Furthermore, each segment $\overline{P_kQ_k},$ $k = 1, 2, \ldots, 2449,$ is parallel to $\overline{BC}.$ The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_kQ_k},$ $k = 1, 2, \ldots, 2450,$ that have rational length.

Answer

Answer: 20

Concepts:similarity area ratio perfect square

Difficulty rating: 2650

Solution:

Since the $2450$ regions have equal areas, triangle $AP_kQ_k$ (the union of the first $k$ regions) has area $\frac{k}{2450}$ of triangle $ABC.$ Each triangle $AP_kQ_k$ is similar to $ABC,$ and lengths scale as the square root of areas, so $\begin{aligned} P_kQ_k &= 5\sqrt{3}\,\sqrt{\frac{k}{2450}} \\ &= 5\sqrt{3} \cdot \frac{\sqrt{k}}{35\sqrt{2}} \\ &= \frac{\sqrt{6k}}{14}. \end{aligned}$

This is rational exactly when $6k$ is a perfect square, which happens exactly when $k = 6j^2$ for a positive integer $j.$ The condition $6j^2 \le 2450$ gives $j^2 \le 408,$ so $j = 1, 2, \ldots, 20.$ There are $20$ such segments.

2018 AIME II Problem 7

Answer: 20

Difficulty rating: 2650

Solution:

Problem 7 in Other Years