- This puzzle was submitted on Google+ by Mustapha B.Bouzid.
One cuts out an equilateral triangle in two pieces along an altitude. One turns over one of the two pieces, and one slips it in order to obtain the figure. The surface of the equilateral triangle was 600 cm2. What is, in cm2 (and round if necessary), the surface of the covering of the two pieces, in red on the figure?
The answer is (5/24)*600 = 125cm2
Here's how I worked it out.
I start by finding the fraction of the original equilateral triangle that is now colored red. Using arbitrary units, I let the original equilateral triangle's sides be 2 units, so it's area is sqrt(3) units2.
Next, I focus my attention on the 30-60-90 triangle with the horizontal leg, viewing the other triangle as mostly a distraction. Then I label this triangle clockwise A, B, C, with C as its right angle. Sides a, b, c (in increasing order of length) are 1, sqrt(3), 2, hence area sqrt(3)/2.
I should note at this point that the topmost white triangle is an equilateral triangle (angle-chasing gives you each angle is 60 degrees) with side length 1 (because one of the sides is the short side of a triangle with sides 1, sqrt(3), 2).
The two white areas in this larger triangle (the 30-60-90 triangle I've been focusing on) are also 30-60-90 triangles. I know the long leg of the larger triangle, containing point B, is 1, because it's the short side of the 30-60-90 triangle I've been focusing on. And from the note in the previous paragraph, I know the hypotenuse of the smaller one is 1. The larger one has sides (from shortest to longest) 1/sqrt(3), 1, 2/sqrt(3), hence area 1/(2sqrt(3)). The smaller one has sides 1/2, sqrt(3)/2, 1, hence area sqrt(3)/8.
So the red area, in these units, is the original 30-60-90 triangle's area minus the area of these two smaller white triangles.
- sqrt(3)/2 - 1/(2sqrt(3)) - sqrt(3)/8 = sqrt(3)(1/2-1/6-1/8) = sqrt(3)(5/24)
Since the area of the original equilateral triangle was sqrt(3), the fraction of that area colored red is 5/24. So the answer to the question is (5/24)*600 = 125cm2
More thoughts about this puzzleEdit
Having done this work, I now see there's a nice geometrical solution. Each of the small white triangles has an area that's a simple fraction of the area of the original equilateral triangle. The topmost white triangle is an equilateral triangle (angle-chasing gives you each angle is 60 degrees) with side that is 1/2 of one of the sides of the original equilateral triangle, so the area of this top white triangle is 1/4 that of the original equilateral triangle.
The hypotenuse of each of the 30-60-90 triangles is bisected by a corner of the top white equilateral triangle. Hence the leftmost white triangle is exactly half of a triangle congruent to the topmost white equilateral triangle, hence its area is 1/8 that of the original equilateral triangle.
Now look for the isosceles (angle chasing gives us that it's isosceles) triangle composed of the leftmost white triangle and the red area. Chop it in half at its shortest altitude, and you'll see the result is congruent to the rightmost white triangle, hence this triangle can be placed exactly six times in the original equilateral triangle, so it's area is 1/6 that of the original triangle.
This gives us right away that the pink area is 1/2-1/8-1/6 = 5/24 that of the original equilateral triangle, and so the smallest bottom-most white triangle must be 1/2-1/4-5/24 = 1/24 that of the original equilateral triangle.
So now, we have the following areas, in order from largest to smallest, expressed as fractions of the original equilateral triangle:
- 1 = original equilateral triangle
- 1/2 = large 30-60-90 triangles
- 1/4 = topmost white equilateral triangle
- 5/24 = red area
- 1/6 = rightmost white 30-60-90 triangle
- 1/4 = leftmost white 30-60-90 triangle
- 1/24 = bottom-most white 30-60-90 triangle