# Heronian Triangle whose Altitude and Sides are Consecutive Integers

## Theorem

There exists exactly one Heronian triangle one of whose altitudes and its sides are all consecutive integers.

This is the Heronian triangle whose sides are $\tuple {13, 14, 15}$ and which has an altitude $12$.

## Proof

We note that a Heronian triangle whose sides are all consecutive integers is also known as a Fleenor-Heronian triangle.

From Sequence of Fleenor-Heronian Triangles, we have that the smallest such triangles are as follows:

- $\tuple {1, 2, 3}$, which has an altitude of $0$

This is the degenerate case where the Heronian triangle is a straight line.

While $0, 1, 2, 3$ is a sequence of $4$ consecutive integers, this is not technically a triangle.

- $\tuple {3, 4, 5}$ with area $6$.

It has altitudes $3$, $4$ and $\dfrac {12} 5$.

- $\tuple {13, 14, 15}$

This can be constructed by placing the $2$ Pythagorean triangles $\tuple {5, 12, 13}$ and $\tuple {9, 12, 15}$ together along their common side $12$:

Thus the altitude and sides are:

- $\tuple {12, 13, 14, 15}$

and this is the Heronian triangle we seek.

It has area $84$.

The next largest Fleenor-Heronian triangle has sides $\tuple {51, 52, 53}$.

Using Heron's Formula, its area is given by:

- $\AA = \sqrt {78 \times 25 \times 26 \times 27} = 1170$

Hence its altitudes are:

- $45 \frac {45} {51}$, $45$, $44 \frac 8 {53}$

For still larger triangles, the altitudes are never within $1$ unit of the sides:

Consider the triangle with sides $\tuple {a - 1, a, a + 1}$.

Using Heron's Formula, its area is given by:

\(\ds \AA\) | \(=\) | \(\ds \sqrt {s \paren {s - a + 1} \paren {s - a} \paren {s - a - 1} }\) | ||||||||||||

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

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

\(\ds \) | \(=\) | \(\ds \frac a 4 \sqrt {3 a^2 - 12}\) |

Its longest altitude is therefore:

\(\ds \frac {2 a} {4 \paren {a - 1} } \sqrt {3 a^2 - 12}\) | \(<\) | \(\ds \frac {a^2 \sqrt 3} {2 \paren {a - 1} }\) |

and we have:

\(\ds \frac {a^2 \sqrt 3} {2 \paren {a - 1} }\) | \(<\) | \(\ds \paren {a - 1} - 1\) | ||||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds a^2 \sqrt 3\) | \(<\) | \(\ds 2 \paren {a - 1}^2 - 2 \paren {a - 1}\) | |||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds 2 a^2 - 4 a + 2 - 2 a + 2 - \sqrt 3 a^2\) | \(>\) | \(\ds 0\) | |||||||||||

\(\ds \leadstoandfrom \ \ \) | \(\ds \paren {2 - \sqrt 3} a^2 - 6 a + 4\) | \(>\) | \(\ds 0\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds a\) | \(>\) | \(\ds \frac {6 + \sqrt {6^2 - 4 \times 4 \paren {2 - \sqrt 3} } } {2 \paren {2 - \sqrt 3} }\) | Quadratic Formula | ||||||||||

\(\ds \) | \(\approx\) | \(\ds 21.7\) |

This shows that for $a \ge 22$, all altitudes of the triangle is less than $a - 2$.

Hence there are no more examples.

$\blacksquare$

## Sources

- 1992: David Wells:
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