Maximizing the Area of a Triangle with Integer Side Lengths and a Given Perimeter

In mathematics, particularly in the realm of geometry, a common problem involves finding the maximum area of a triangle given specific constraints. One such problem is determining the maximum possible area of a triangle with integer side lengths and a perimeter of 7. This problem requires an understanding of the triangle inequality and Heron's formula.

Problem Statement

Given a triangle with integer side lengths and a perimeter of 7, find the maximum possible area.

Mathematical Setup

Let's denote the side lengths of the triangle as a, b, and c. According to the problem, the perimeter of the triangle is 7, so we have:

[ a b c 7 ]

The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. Therefore, we have the following conditions:

[ a b > c ] [ a c > b ] [ b c > a ]

Given that c is the longest side, we can rewrite the first inequality as:

[ a b > 7 - a - b implies 2a 2b > 7 implies a b > 3.5 ]

Since a, b, and c are integers, we get:

[ a b geq 4 ]

Possible Integer Combinations

Given the constraint that the perimeter is 7, we need to find all possible combinations of integer side lengths that satisfy both the perimeter and the triangle inequality. The possible integer combinations for a, b, and c that sum to 7 are:

[ (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 2, 3) ]

Checking the Combinations

Now, we will check which of these combinations satisfy the triangle inequality:

(1, 1, 5)

[ 1 1 2 This combination fails the triangle inequality.

(1, 2, 4)

[ 1 2 3 This combination also fails the triangle inequality.

(1, 3, 3)

[ 1 3 4 geq 3 ] [ 3 3 6 geq 1 ]

This combination satisfies the triangle inequality.

(2, 2, 3)

[ 2 2 4 geq 3 ] [ 2 3 5 geq 2 ] [ 2 3 5 geq 2 ]

This combination satisfies the triangle inequality.

Calculating the Area

Now we will calculate the area for the valid triangles using Heron's formula. Heron's formula states that the area of a triangle with sides a, b, and c is given by:

[ A sqrt{s(s-a)(s-b)(s-c)} ]

where:

[ s frac{a b c}{2} ]

(1, 3, 3) Triangle

[ s frac{1 3 3}{2} 3.5 ] [ A sqrt{3.5(3.5-1)(3.5-3)(3.5-3)} sqrt{3.5 times 2.5 times 0.5 times 0.5} ] [ A sqrt{2.1875} approx 1.479 ]

(2, 2, 3) Triangle

[ s frac{2 2 3}{2} 3.5 ] [ A sqrt{3.5(3.5-2)(3.5-2)(3.5-3)} sqrt{3.5 times 1.5 times 1.5 times 0.5} ] [ A sqrt{3.9375} approx 1.984 ]

Conclusion

The maximum possible area of a triangle with integer side lengths and a perimeter of 7 is obtained from the triangle with sides (2, 2, 3), which gives an area of approximately 1.984 square units.

Interesting Insights

While the above solution finds the optimal triangle with integer side lengths, it's interesting to note that if all sides were equal (making it an equilateral triangle), the area would be higher. The equilateral triangle with side length 10, for example, has:

[ text{Area} frac{1}{2} a b sin C frac{1}{2} times 10 times 10 times sin 60^circ 50 times sin 60^circ 50 times frac{sqrt{3}}{2} 25 sqrt{3} approx 43.3 ]

Keeping the perimeter at 30 and trying all possible side length combinations, none of them will have an area as high as the equilateral triangle. This highlights the elegance of equilateral triangles in maximizing the area for a given perimeter.