Triangles with Integer Area and Integer Sides in Arithmetical Sequence

Theorem

The triangles with the following sides in arithmetic sequence have integer areas:

$3, 4, 5$

$13, 14, 15$

$15, 28, 41$

$15, 26, 37$

Their areas are:

$6, 84, 126, 156$

Proof

From Heron's Formula, the area $A$ of $\triangle ABC$ is given by:

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

where $s = \dfrac{a + b + c} 2$ is the semiperimeter of $\triangle ABC$.

For $3, 4, 5$:

$\ds s$

$=$

$\ds \frac {3 + 4 + 5} 2$

$\ds $

$=$

$\ds 6$

$\ds \leadsto \ \ $

$\ds A$

$=$

$\ds \sqrt {6 \paren {6 - 3} \paren {6 - 4} \paren {6 - 5} }$

$\ds $

$=$

$\ds \sqrt {6 \times 3 \times 2 \times 1}$

$\ds $

$=$

$\ds \sqrt {6 \times 6}$

$\ds $

$=$

$\ds 6$

For $13, 14, 15$:

$\ds s$

$=$

$\ds \frac {13 + 14 + 15} 2$

$\ds $

$=$

$\ds 21$

$\ds \leadsto \ \ $

$\ds A$

$=$

$\ds \sqrt {21 \paren {21 - 13} \paren {21 - 14} \paren {21 - 15} }$

$\ds $

$=$

$\ds \sqrt {21 \times 8 \times 7 \times 6}$

$\ds $

$=$

$\ds \sqrt {\paren {3 \times 7} \times 2^3 \times 7 \times \paren {2 \times 3} }$

$\ds $

$=$

$\ds 3 \times 7 \times 2^2$

$\ds $

$=$

$\ds 84$

For $15, 28, 41$:

$\ds s$

$=$

$\ds \frac {15 + 28 + 41} 2$

$\ds $

$=$

$\ds 42$

$\ds \leadsto \ \ $

$\ds A$

$=$

$\ds \sqrt {42 \paren {42 - 15} \paren {42 - 28} \paren {42 - 41} }$

$\ds $

$=$

$\ds \sqrt {42 \times 27 \times 14 \times 1}$

$\ds $

$=$

$\ds \sqrt {\paren {2 \times 3 \times 7} \times 3^3 \times \paren {2 \times 7} }$

$\ds $

$=$

$\ds 2 \times 3^2 \times 7$

$\ds $

$=$

$\ds 126$

For $15, 26, 37$:

$\ds s$

$=$

$\ds \frac {15 + 26 + 37} 2$

$\ds $

$=$

$\ds 39$

$\ds \leadsto \ \ $

$\ds A$

$=$

$\ds \sqrt {39 \paren {39 - 15} \paren {39 - 26} \paren {39 - 37} }$

$\ds $

$=$

$\ds \sqrt {39 \times 24 \times 13 \times 2}$

$\ds $

$=$

$\ds \sqrt {\paren {3 \times 13} \times \paren {2^3 \times 3} \times 13 \times 2}$

$\ds $

$=$

$\ds 2^2 \times 3 \times 13$

$\ds $

$=$

$\ds 156$

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