Question

Difficulty: Very hardTriangles: Properties, Perimeter, and Area

A triangle has integer side lengths aa, bb, and cc such that its perimeter is 1818. Which of the following could be the area of the triangle? Select all such values.

  1. 1212Answer
  2. 666\sqrt{6}Answer
  3. 939\sqrt{3}Answer
  4. D
    4154\sqrt{15}
  5. E
    858\sqrt{5}

Answer

The possible areas of the triangle are 1212, 666\sqrt{6}, and 939\sqrt{3}.
To form a valid triangle with integer sides and a perimeter of 18, each side length must be a positive integer strictly less than 9 by the triangle inequality theorem. Evaluating Heron's formula A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)} with s=9s=9 for all valid side combinations (a,b,c)(a,b,c) produces the set of areas {37,63,315,12,65,66,93}\{3\sqrt{7}, 6\sqrt{3}, 3\sqrt{15}, 12, 6\sqrt{5}, 6\sqrt{6}, 9\sqrt{3}\}. Among the options given, the values 12 (from side lengths 8, 5, 5), 666\sqrt{6} (from side lengths 7, 6, 5), and 939\sqrt{3} (from side lengths 6, 6, 6) are correct.

Step-by-Step Solution

1
Determine the structural bounds on side lengths using the Triangle Inequality Theorem.
For any side length xx, x<Perimeter2=9x < \frac{\text{Perimeter}}{2} = 9. Thus, a,b,c{1,2,3,4,5,6,7,8}a, b, c \in \{1, 2, 3, 4, 5, 6, 7, 8\}.
The sum of any two side lengths must be strictly greater than the third side length.
2
List all unique integer side length combinations (a,b,c)(a, b, c) where abca \ge b \ge c and a+b+c=18a + b + c = 18.
The valid triples are (8,8,2)(8, 8, 2), (8,7,3)(8, 7, 3), (8,6,4)(8, 6, 4), (8,5,5)(8, 5, 5), (7,7,4)(7, 7, 4), (7,6,5)(7, 6, 5), and (6,6,6)(6, 6, 6).
Exhaustive enumeration of integer partitions of 18 satisfying a8a \le 8.
3
Calculate the semi-perimeter ss of the triangle.
s=182=9s = \frac{18}{2} = 9.
Required parameter for Heron's formula A=s(sa)(sb)(sc)A = \sqrt{s(s-a)(s-b)(s-c)}.
4
Compute the area for candidate triples using Heron's formula.
For (8,5,5)(8, 5, 5): A=9(1)(4)(4)=12A = \sqrt{9(1)(4)(4)} = 12.
For (7,6,5)(7, 6, 5): A=9(2)(3)(4)=66A = \sqrt{9(2)(3)(4)} = 6\sqrt{6}.
For (6,6,6)(6, 6, 6): A=9(3)(3)(3)=93A = \sqrt{9(3)(3)(3)} = 9\sqrt{3}.
Direct evaluation of geometric area for the valid triangles.

Key Concept

Triangle Inequality Theorem and Area calculation via Heron's Formula
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