Solution to May 2001 Problem

All the submitted solutions made use of the same two basic steps, but each had its own insight. To display the different approaches to the problem, for our featured solution we have combined the work of

Jeff Eggen (Regina)
Normand Laliberté (Ontario)
Leti Gimeno (Spain)
Juan Mir Pieras (Spain)
Alexander Potapenko (Russia)

Step 1. Determine the altitude from A to BC.

Approach (a).

It is possible to avoid any computation by recognizing the 13-14-15 triangle as an old friend: The altitude from A partitions triangle ABC into right triangles of side lengths 5-12-13 and 9-12-15; these share the leg of length 12, which must therefore be the height of triangle ABC. (Alternatively, Gimeno points out that by the cosine law we get
,
whence the height is 12.)
Approach (b). Use the area (Mir, and Potapenko). To apply Heron's formula for the area, we compute the semiperimeter to be s = (13 + 14 + 15)/2 = 21. Therefore

.

Since the base BC = 14, the corresponding height must be (Area)/7 = 12. Approach (c). Use the Pythagorean theorem (Eggen and Laliberté). If H is the foot of the altitude from A then

;

thus, BH = 5, so that AH = 12. Note that this approach also provides the mathematics behind the observation in (a).

Step 2. Determine the side of the square using the fact that Triangle ADG ~ triangle ABC.

(The triangles are similar because their sides are parallel.)


Let x be the side of the inscribed square: x = DE =EF = FG = DG.

Approach (a). (Eggen, Potapenko and Laliberté)



and

.

Since these ratios are equal we have  12-x/x = 12/14. Solving for x we find that the unknown side length is x = 84/13. Actually, Eggen and Laliberté used trigonometry, which leads to a similar computation. Approach (b). Use areas of the subfigures. (Gimeno). 84
= Area(ADG) + [Area(BED) + Area(GFC)] + Area(DEFG)
= (1/2)x(12 - x) + (1/2)x(14 - x) + x2
= 13x

The side of the square is therefore x = 84/13. Approach (c). (Mir)

Consider the infinite sequence of similar triangles with their inscribed squares. The original triangle ABC and its square DEFG is shrunk by a factor of k to a similar triangle ADG with its square, and so on. If the length of a side of the first square is x, then the square in ADG has side length kx, and that square determines the next triangle with its square of length k2x, and so on. Since the squares fill in the entire height we have.

   height
= x + kx + k2x + ...
= x(1 + k + k2 + ...)
= x/1 - k
To find k, note that the side of the first square comes from shrinking the base of the original triangle: x = 14k, so that k = x/14. The problem has been reduced to solving


for x. Once again we compute that x = 84/13.