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# Triangle Perimeters

Alignments to Content Standards: G-GPE.B.7

Below is a picture of two triangles with vertices on coordinate grid points:

1. What are the perimeters of $\triangle ABC$ and $\triangle PQR$?
2. What is the smallest perimeter possible for a triangle with vertices on grid points and with whole number side lengths? Explain.

## IM Commentary

The purpose of this task is to apply the Pythagorean theorem to calculate distances and areas. While the first part of the task is routine, part (b) requires creative initiative in order to argue that all cases have been treated. The general flavor of this task is similar to http://www.illustrativemathematics.org/illustrations/1684.  The teacher may wish to give students a picture of a (5,12,13) right triangle where the leg of length 5 is the hypotenuse of a (3,4,5) right triangle to give them a hint for part (b).

There is a very challenging follow-up question which the teacher may pose if desired: what is the smallest area possible for a triangle with vertices on grid points and with whole number side lengths? This is a difficult problem because it requires finding (or bounding) the area of a triangle in terms of its side lengths. It can be done, for example, via Heron's Formula, http://en.wikipedia.org/wiki/Heron%27s_formula although this is an advanced result. This question does provide some motivation for studying Heron's Formula as a special project.

## Solution

1. To find the perimeter of $\triangle ABC$ we use the Pythagorean Theorem which tells us that $$|AB|^2 = |AC|^2 + |BC|^2.$$ Since $|AC| = 5$ and $|BC| = 12$ we find that $|AB| = \sqrt{169} = 13.$ The perimeter of $\triangle ABC$ is then $5 + 12 + 13 = 30$ units.  For the perimeter of $\triangle PQR$ we find similarly that $|PR| = 3$, $|QR| = 7$, and $|PR| = \sqrt{58}$.  So the perimeter of $\triangle PQR$ is $10 + \sqrt{58}$ units.
2. Part (a) produces a triangle whose perimeter is 30 units with three whole number side lengths. There are right triangles with shorter integer side lengths such as (6,8,10) and (3,4,5).  The (3,4,5) triangle has perimeter 12 units. To see whether or not there can be a triangle with perimeter less than 12 units we can begin by examining which small whole number length sides are possible with vertices on coordinate grid points. Lengths of 1, 2, 3, and 4 units are all possible but only if the segment is vertical or horizontal. To see why, note that any line segment which is not vertical or horizontal is the hypotenuse of a triangle with one vertical side and one horizontal side whose vertices all have integer coordinates. This would mean that the given segment length $c$ must satisfy $$c^2 = a^2 + b^2$$ where $a$ and $b$ are positive whole numbers. By trial and error we can check that $c = 5$ is the smallest positive integer for which we can solve this equation.

If $\triangle ABC$ has vertices with integer coordinates then we just saw that any side which is not horizontal or vertical must have length at least 5 units. If exactly one side is not vertical or horizontal this means that 12 units is the smallest possible perimeter, achieved for the (3,4,5) triangle studied above. If two sides are not vertical then the only possibility with perimeter smaller than 12 units would be a (5,5,1) isosceles triangle with the side of length 1 either vertical or horizontal. This is not possible, nor is a (5,5,2) triangle with the side of length 2 units being horizontal or vertical. Therefore the smallest possible perimeter for these triangles is 12 units and this occurs only for the (3,4,5) right triangle with one horizontal and one vertical side.