Henry Ernest Dudeney/Puzzles and Curious Problems/203 - A Triangle Puzzle/Solution

by : $203$

 * A Triangle Puzzle

Solution
The triangle whose sides are $2701$, $2702$ and $2703$ has an area of $3 \, 161 \, 340$.

Proof
Extend the following table as you like:


 * $\begin{array} {rrrrr}

n & p_n & q_n & \text{Height} & \text{Area} \\ \hline 1 & 2 & 4 & 3 & 6 \\ 2 & 8 & 14 & 12 & 84 \\ 3 & 30 & 52 & 45 & 1170 \\ 4 & 112 & 194 & 168 & 16 \, 296 \\ 5 & 418 & 724 & 627 & 226 \, 974 \\ 6 & 1560 & 2702 & 23490 & 3 \, 161 \, 340 \\ \end {array}$

This table is governed by the recurrence relation:
 * $p_n = \begin {cases} 0 & : n = 0 \\ 2 & : n = 1 \\ 4 p_{n - 1} - p_{n - 2} \end {cases}$

and:
 * ${q_n}^2 = 3 {p_n}^2 + 4$

Each row describes a triangle $T$ whose sides are consecutive integers $q - 1$, $q$, $q + 1$, such that:


 * the height of $T$ is equal to $\dfrac {3 P} 2$
 * the area of $T$ is equal to the height multiplied by $\dfrac q 2$.

This is demonstrated in Approximations to Equilateral Triangles by Heronian Triangles, except it's not.