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.

$\blacksquare$