# Henry Ernest Dudeney/Puzzles and Curious Problems/162 - Find the Triangle/Solution

*Puzzles and Curious Problems* by Henry Ernest Dudeney: $162$

- Find the Triangle

*The sides and height of a triangle are four consecutive whole numbers.**What is the area of the triangle?*

## Solution

*The sides of the triangle are $13$, $14$, and $15$, making $14$ the base, the height $12$, and the area $84$.*

## Proof

As Dudeney put it:

*There is an infinite number of rational triangles composed of three consecutive integers, like $3$, $4$ and $5$, and $13$, $14$, and $15$,**but there is no other case in which the height will comply with our conditions.*

Rational triangles whose sides are three consecutive integers with integer area are given by the recurrence relation:

- $U_n = \begin {cases} 4 & : n = 1 \\ 14 & : n = 2 \\ 4 U_{n - 1} - U_{n - 2} & : n > 2 \end {cases}$

where the lengths of the sides of triangle $n$ are $U_n - 1$, $U_n$ and $U_n + 1$.

This needs considerable tedious hard slog to complete it.In particular: Prove this recurrence relationTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

This can be presented as follows:

Let lengths of the sides of a triangle with integer area be $2 x - 1$, $2 x$ and $2 x + 1$.

Then $3 \paren {x^2 - 1}$ is a square number.

This is demonstrated in Approximations to Equilateral Triangles by Heronian Triangles.

Hence the rational triangles with integer area have the sides whose lengths are:

- $\begin{array} {rrr} 3 & 4 & 5 \\ 13 & 14 & 15 \\ 51 & 52 & 53 \\ 193 & 194 & 195 \\ 723 & 724 & 725 \end {array}$

From Heronian Triangle whose Altitude and Sides are Consecutive Integers, the only one fitting the condition is the $13$, $14$, $15$ one.

$\Box$

It remains to exclude the possibility that we have not overlooked the possibility of a triangle whose sides are not consecutive integers.

There are two such cases:

- the side lengths are in the pattern $n$, $n + 1$, $n + 3$ with height $n + 2$;
- the side lengths are in the pattern $n$, $n + 2$, $n + 3$ with height $n + 1$.

For the first case, we have:

- $s = \dfrac {3 n + 4} 2$

Thus the area of this triangle is:

\(\ds \) | \(\) | \(\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }\) | Heron's Formula | |||||||||||

\(\ds \) | \(=\) | \(\ds \sqrt {\paren {\frac {3 n + 4} 2} \paren {\frac {n + 4} 2} \paren {\frac {n + 2} 2} \paren {\frac {n - 2} 2} }\) |

As the lengths of the sides and the height $n + 2$ are all integers, the area must be half of an integer.

Since $3 n + 4, n + 4, n + 2, n - 2$ are all of the same parity, if they are all odd, the area cannot be half of an integer.

Hence they are all even, and so is $n$.

Writing $n = 2 x$:

\(\ds \sqrt {\paren {\frac {3 n + 4} 2} \paren {\frac {n + 4} 2} \paren {\frac {n + 2} 2} \paren {\frac {n - 2} 2} }\) | \(=\) | \(\ds \sqrt {\paren {3 x + 2} \paren {x + 2} \paren {x + 1} \paren {x - 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\paren {side} \paren {x + 2} } 2\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {3 x + 2} \paren {x + 2} \paren {x + 1} \paren {x - 1}\) | \(=\) | \(\ds \paren {\frac {\paren {side} \paren {2 x + 2} } 2}^2\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {x + 1}^2 \paren {side}^2\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {3 x + 2} \paren {x + 2} \paren {x - 1}\) | \(=\) | \(\ds \paren {x + 1} \paren {side}^2\) |

so $\paren {x + 1}$ must be a factor of the left hand side.

Note that, by GCD with Remainder:

- $\gcd \set {3 x + 2, x + 1} = \gcd \set {x, x + 1} = 1$
- $\gcd \set {x + 2, x + 1} = 1$

By Euclid's Lemma, we must have $\paren {x + 1} \mid \paren {x - 1}$.

Since $x - 1 < x + 1$, by Absolute Value of Integer is not less than Divisors, $x + 1 \nmid x - 1$.

This is a contradiction.

Therefore no such rational triangle exist.

For the second case, we have:

- $s = \dfrac {3 n + 5} 2$

Thus the area of this triangle is:

\(\ds \) | \(\) | \(\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }\) | Heron's Formula | |||||||||||

\(\ds \) | \(=\) | \(\ds \sqrt {\paren {\frac {3 n + 5} 2} \paren {\frac {n + 5} 2} \paren {\frac {n + 1} 2} \paren {\frac {n - 1} 2} }\) |

As the lengths of the sides and the height $n + 1$ are all integers, the area must be half of an integer.

Since $3 n + 5, n + 5, n + 1, n - 1$ are all of the same parity, if they are all odd, the area cannot be half of an integer.

Hence they are all even, and thus $n$ must be odd.

Writing $n = 2 x + 1$:

\(\ds \sqrt {\paren {\frac {3 n + 5} 2} \paren {\frac {n + 5} 2} \paren {\frac {n + 1} 2} \paren {\frac {n - 1} 2} }\) | \(=\) | \(\ds \sqrt {\paren {3 x + 4} \paren {x + 3} \paren {x + 1} \paren {x} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\paren {side} \paren {n + 1} } 2\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {3 x + 4} \paren {x + 3} \paren {x + 1} \paren {x}\) | \(=\) | \(\ds \paren {\frac {\paren {side} \paren {2 x + 2} } 2}^2\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {x + 1}^2 \paren {side}^2\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {3 x + 4} \paren {x + 3} \paren {x}\) | \(=\) | \(\ds \paren {x + 1} \paren {side}^2\) |

so $\paren {x + 1}$ must be a factor of the left hand side.

Note that, by GCD with Remainder:

- $\gcd \set {3 x + 4, x + 1} = \gcd \set {1, x + 1} = 1$
- $\gcd \set {x, x + 1} = 1$

By Euclid's Lemma, we must have $\paren {x + 1} \mid \paren {x + 3}$.

For $x > 1$, $2 \paren {x + 1} > x + 3$.

Hence we must have $x = 1$, which leads to the $3 - 5 - 6$ triangle.

- $\sqrt {\paren {3 + 4} \paren {1 + 3} \paren {1 + 1} \paren {1} } = \sqrt {56}$

which is not an integer, and the height is not an integer either.

Therefore no such rational triangle exist.

$\blacksquare$

## Sources

- 1932: Henry Ernest Dudeney:
*Puzzles and Curious Problems*... (previous) ... (next): Solutions: $162$. -- Find the Triangle - 1968: Henry Ernest Dudeney:
*536 Puzzles & Curious Problems*... (previous) ... (next): Answers: $230$. Find the Triangle