Algebra II Question? Please!! I need help.?

Question

Ok. Here is the problem.

In a large equilateral triangle is a inscribed smaller one, its vertices at trisection points of sides of the larger. Is the perimeter of the larger triangle is 9, what is the perimeter of the smaller equilateral triangle?

ANY help would be appreciated. I also have to explain my answer so if you could include that it would be great!!

Answer 1

If you draw this out, you'll get the larger triangle divided up into the interior equilateral triangle, with side length x, say, and three congruent triangles with side lengths 2, 1, and x, with an angle of pi/3 radians (60 degrees) opposite the side of length x (we know this angle because it's one of the angles of the large equilateral triangle). Note that for a given large triangle, there are two choices of which trisection points to use to construct the inner triangle, but it doesn't matter which set you use.

If you know the cosine rule, this gives you
x^2 = 1^2 + 2^2 - 2.1.2.cos pi/3
and hence x = sqrt(3) (noting that cos pi/3 = 1/2) and the perimeter of the smaller equilateral triangle is 3 sqrt(3).

Otherwise, I'd suggest the following procedure:
- Let the base of the larger triangle have end points (0, 0) and (3, 0). Show that the apex has coordinates (3/2, 3sqrt(3)/2).
- Find the coordinates of the two trisection points defining one edge of the inner equilateral triangle. For instance, depending on which trisection points you use, one edge of the smaller triangle might have endpoints (2, 0) and (2, sqrt(3)). Note that the trisection points of a line going from (x1, y1) to (x2, y2) are
((1/3)x1 + (2/3) x2, (1/3) y1 + (2/3) y2)
and
((2/3)x1 + (1/3) x2, (2/3) y1 + (1/3) y2)
- Determine the length between these points to be sqrt(3) and hence find the perimeter to be 3 sqrt(3). Alternatively, show that the triangle is a right-angled triangle and use the known angle of pi/3 to find the side length and hence perimeter.