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Sides Of A Triangle (Posted on 2003-11-23) Difficulty: 3 of 5
The sides of a triangle are in arithmetic progression and its area is 3/5th the area of an equilateral triangle with the same perimeter.

Find the ratio of the sides of the triangle.

No Solution Yet Submitted by Ravi Raja    
Rating: 4.1111 (9 votes)

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Solution | Comment 13 of 14 |
Heron's formula states that the area of any triangle is √p(p-a)(p-b)(p-c), where p is the semiperimeter, and a, b, and c are the sides. Let b = x, a = x - d, and c = x + d; then the area is √(3x/2)(x/2 + d)(x/2)(x/2 - d). Simplifying, this is x/2 * √3x^2/4 - d^2.

The area of a triangle with side length s is s^2 * (√3)/4. s = x in this case, so:

x/2 * √(3x^2/4 - d^2) = x^2 * (√3)/4 * 3/5
√(3x^2/4 - d^2) = 3x/10 * √3
3x^2/4 - d^2 = 27x^2/100
12/25 * x^2 = d^2
d = 2x/5 * √2

The ratio is then x - 2x/5 * √2 : x : x + 2x/5 * √2. Cancelling out the x's:

(5 - 2√2)/5 : 1 : (5 + 2√2)/5

(Honestly, did you guys think it would come out with rationals? The area of an equilateral triangle always involves √3 in some way.)
  Posted by Rachel Mantis on 2004-01-17 11:58:56
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