Summer Problem Solving Marathon Solution #23

Let D be the vertex of the rhombus on side AB of the triangle, E be the vertex of the rhombus on side AC, and F be the vertex of the rhombus on side BC.

Opposite sides of a rhombus are parallel, so DF is parallel to AE, and hence also to AC. So the triangles BDF and BAC are similar. Thus their sides are in the same ratio. Let s be the side length of the rhombus. Then BD = 12-s. The ratio of BD to BA is the same as the ratio of DF to AC. So we have:

(12 - s)/12 = s/6

Multiplying both sides by 12, we have:

12 - s = 2s
12 = 3s
s = 4

This gives us the side length of the rhombus, but not yet the area. To find the area of the rhombus, we view it as two congruent triangles -- ADE and FDE. To find the area of ADE, we use the area formula:

Area = (ab sin C)/2

In this case we have the two sides, so we just need the sine of the included angle. That angle is angle BAC.

We now use the law of cosines applied to triangle ABC. Recall that the law of cosines says:

c² = a² + b² - 2ab cos C

We use the particular instance in which c is side BC, a is side AB, b is side AC, and C is angle BAC (which we will call angle A). Then we have:

64 = 144 + 36 - 2(12)(6)cos A
64 = 180 - 144cos A
144cos A = 114
cos A = 114/144

However, we want sin A. So we use the fact that cos² A + sin² A = 1, so sin A = (1 - cos² A)^1/2.

So sin A = (1 - (114/144)²)^1/2 = ((144 - 114)²/144²)^1/2 = ((144 + 114)(144-114))^1/22/144 = (258 x 30)^1/2/144 = (6/144)√215 = (1/24)√215.

The area of triangle ADE is then (1/2)(4)(4)(1/24)√215 = (1/3)√215. The area of the rhombus is twice this, or (2/3)√215.