# Proper Integer Heronian Triangle whose Area is 24

## Theorem

There exists exactly one proper integer Heronian triangle whose area equals $24$.

That is, the obtuse triangle whose sides are of length $4$, $13$ and $15$.

## Proof

First we show that the $\tuple {4, 13, 15}$ triangle is actually Heronian.

Heron's Formula gives us that:

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

where:

$\AA$ denotes the area of the triangle
$a$, $b$ and $c$ denote the lengths of the sides of the triangle
$s = \dfrac {a + b + c} 2$ denotes the semiperimeter of the triangle.

Hence:

 $\ds s$ $=$ $\ds \frac {4 + 13 + 15} 2$ $\ds$ $=$ $\ds 16$ $\ds \leadsto \ \$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ $=$ $\ds \sqrt {16 \paren {16 - 4} \paren {16 - 13} \paren {16 - 15} }$ $\ds$ $=$ $\ds \sqrt {16 \times 12 \times 3 \times 1}$ $\ds$ $=$ $\ds \sqrt {576}$ $\ds$ $=$ $\ds 24$

It can be constructed by taking a $9-12-15$ Pythagorean triangle and removing a $5-12-13$ Pythagorean triangle from it:

Let $\tuple {a, b, c}$ be the sides of such a triangle.

Using Heron's Formula, we have:

 $\ds 24$ $=$ $\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ $\ds \leadsto \ \$ $\ds 576$ $=$ $\ds s \paren {s - a} \paren {s - b} \paren {s - c}$

Note that:

$\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$

Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:

$x y z \paren {x + y + z} = 576$

Hence $s, x, y, z \in \N_{>0}$.

Without loss of generality suppose $x \le y \le z$.

Then:

 $\ds 576$ $=$ $\ds x y z \paren {x + y + z}$ $\ds$ $\ge$ $\ds x^3 \paren {3 x}$ $\ds$ $=$ $\ds 3 x^4$ $\ds \leadsto \ \$ $\ds x$ $\le$ $\ds \sqrt [4] {\frac {576} 3}$ $\ds$ $\approx$ $\ds 3.72$

so we need to check $1 \le x \le 3$.

Note that for fixed $x$:

 $\ds 576$ $=$ $\ds x y z \paren {x + y + z}$ $\ds$ $>$ $\ds x y^2 \paren {2 y}$ $\ds$ $=$ $\ds 2 x y^3$ $\ds \leadsto \ \$ $\ds y$ $\le$ $\ds \sqrt [3] {\frac {576} {2 x} }$

For $x = 1$:

$1 \le y \le \sqrt [3] {\frac {576} 2} \approx 6.60$

For $x = 2$:

$2 \le y \le \sqrt [3] {\frac {576} 4} \approx 5.24$

For $x = 3$:

$3 \le y \le \sqrt [3] {\frac {576} 6} \approx 4.58$

Finally, for fixed $x$, $y$:

 $\ds x y z \paren {x + y + z}$ $=$ $\ds 576$ $\ds \leadsto \ \$ $\ds z^2 + \paren {x + y} z - \frac {576} {x y}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds z$ $=$ $\ds \frac {- x - y + \sqrt {\paren {x + y}^2 + \frac {2304} {x y} } } 2$ Quadratic Formula

Since $z \in \Z$:

$\sqrt {\paren {x + y}^2 + \dfrac {2304} {x y} } = 2 z + x + y \in \Z$

We check the value of $\sqrt {\paren {x + y}^2 + \dfrac {2304} {x y} }$ for the valid values of $x, y$.

We do not need to check $y = 5$ since $5 \nmid 576$.

We have:

$\tuple {1, 1}: \sqrt {\paren {1 + 1}^2 + \dfrac {2304} {1 \times 1} } = \sqrt {2308} \notin \Z$
$\tuple {1, 2}: \sqrt {\paren {1 + 2}^2 + \dfrac {2304} {1 \times 2} } = \sqrt {1161} \notin \Z$
$\tuple {1, 3}: \sqrt {\paren {1 + 3}^2 + \dfrac {2304} {1 \times 3} } = \sqrt {784} = 28$
$\tuple {1, 4}: \sqrt {\paren {1 + 4}^2 + \dfrac {2304} {1 \times 4} } = \sqrt {601} \notin \Z$
$\tuple {1, 6}: \sqrt {\paren {1 + 6}^2 + \dfrac {2304} {1 \times 6} } = \sqrt {433} \notin \Z$
$\tuple {2, 2}: \sqrt {\paren {2 + 2}^2 + \dfrac {2304} {2 \times 2} } = \sqrt {592} \notin \Z$
$\tuple {2, 3}: \sqrt {\paren {2 + 3}^2 + \dfrac {2304} {2 \times 3} } = \sqrt {409} \notin \Z$
$\tuple {2, 4}: \sqrt {\paren {2 + 4}^2 + \dfrac {2304} {2 \times 4} } = \sqrt {324} = 18$
$\tuple {3, 3}: \sqrt {\paren {3 + 3}^2 + \dfrac {2304} {3 \times 3} } = \sqrt {292} \notin \Z$
$\tuple {3, 4}: \sqrt {\paren {3 + 4}^2 + \dfrac {2304} {3 \times 4} } = \sqrt {241} \notin \Z$

Now we calculate $z$ for $\tuple {x, y} = \tuple {1, 3}, \tuple {2, 4}$.

$\tuple {1, 3}: z = \dfrac {- 1 - 3 + 28} 2 = 12$
$\tuple {2, 4}: z = \dfrac {- 2 - 4 + 18} 2 = 6$

Using:

$a = s - x = x + y + z - x = y + z$
$b = s - y = x + z$
$c = s - z = x + y$

the possible sets of side lengths are:

$\tuple {15, 13, 4}, \tuple {10, 8, 6}$

But note that the $\tuple {6, 8, 10}$ triangle is right-angled, and hence not a proper Heronian Triangle.

Hence the result.

$\blacksquare$