37 of 100: Carved Triangles

All 3 areas in triangle A are congruent
Triangle B is divided into 4 congruent triangles and all purple area are congruent with the area of $\frac{1}{4}$ of red areas
So the sum of area in 2 triangles is equal.

Notice that $\alpha=\frac{\pi}{6}$ . Thus $\sin \alpha= \frac{r}{x+r}= \frac{R}{x+3r}=\frac{1}{2}$ Thus $x=r$ and $R=2r$ . Hence the area of big circle equals the area of four small circles combined, that is $\pi R^2=\pi (2r)^2=4 \pi r^2$ Consequently areas of shaded regions are equal.

I would quibble with the wording in the problem statement, or perhaps more succinctly with the wording in the clarification, which was I believe was intended to make solutions more accessible ...

BTW, the wording of the clarification was revised, adding the comment that the four inner circles are congruent, while I was writing this commentary ... Yay! The reason for the revision, which I am not responsible for, is made clear by the following commentary.

@Daniel Willard ... The figures given in the problem imply but do not assert that in the second (i.e. purple) equilateral triangle the inner circles are congruent. If you assume this fact ( i.e. the four inner circles are congruent ) then you get the suggested correct answer. However, according to the wording in the clarification, you can have arrangements of inner circles which will fall between the two extremes outlined below. In Figure 1 below, the middle circle is the same size as the red triangle circle given in the problem. In this case the blue shaded area is obvious less than the red shaded area of the problem because of the addition of three more circles subtracting away area. On the other extreme, Figure 2 below, you can shrink the middle circle to its smallest size while growing the other three circles to their maximum size. In this case the yellow shaded area is also less than the red shaded area of the problem. In fact, the only time when the shaded areas of the two situations are equal is in the special case when the four inner circles are congruent! Otherwise, the four circle figure will have less shaded area than the one circle figure. Try this: in Figure 2 below, derive the exact radius of the three large circles relative to the side length of the equilateral triangle ( approximately 18.3% ). For a super challenge, derive the exact radius of the tiny inner circle in Figure 2 below.

Let each side of the overall triangle be x. If we drop a perpendicular from the center of the circle to each side ,say the base, it will bisect at the middle. The length of the perpendicular will be the radius of the incircle, r. Now if we join the center to the leftmost vertex , we get a 30-60-90 trng. Now to find r, 2*(r/x) = tan(30) = 1/sqrt(3)

r = x/(2*sqrt(3))

Therefore, area of the incircle = pi(r)^2 = pi x^2 / (2 sqrt(3))^2 = pi*x^2 / 12...(1)

Now, we would like to establish the area of each of the smaller incircles. If we join the midpoints of the 2 segments of the large trng (other than the base), we get a triangle which is similar to the large trng and the length of the base say y ((by similarity) is x/2. Again ,this is a 30-60-90 trng and by similar arguments , the radius of each small incircle = r2 = y/(2 sqrt(3)))) = x / (4 sqrt(3)) . So the area of the smaller incircles =pi x^2/(4 sqrt(3))^2 = pi x^2 / 48. Since there are 4 of these, the combined area of the 4 smaller circles= 4 pi x^2 / 48 = pi x^2/12...(2) which is the same as (1)

Therefore A(large incircle) = A(sum of areas of 4 smaller incircles).

Therefore A(Trng) - A(large incircle) = A(Trng) - A(sum of areas of 4 smaller incircles)

Sum of red regions = sum of purple regions

Do we assume the four circles in the purple triangle are equal?

Draw an equilateral triangle inside the large circle that touches the points where the circle hits the triangle (upside down versus the big red triangle. Do the same for each of the small circles. The ratio of the size of each triangle to its circle is constant- you don't have to calculate it. So, if the four small triangles are bigger than the one big triangle, the small circles are bigger, and the remaining area is smaller. But if you fill the purple triangle with small triangles you will see that four small triangles fit in the big triangle you made. So the small triangles are the same area as the big triangle, the circles are the same area, and the remaining area is identical.

First, let us set the side lengths of each of the equilateral triangles equal to $1$ . Setting any other side length will give us the third choice in the end, so it doesn't matter which side length we choose to use.

Let us first find the area of triangle A's incircle.

We know that the inradius of any equilateral triangle is $\frac{s\sqrt{3}}{6}$ , where $s$ is the side length of the triangle. Since $s=1$ , our inradius is $\frac{\sqrt{3}}{6}$ . Using the circle area formula $\pi*r^2$ where $r=\frac{\sqrt{3}}{6}$ , we know that the area of triangle A's incircle is $\frac{\pi}{12}$ .

Now, we find the area of triangle B's circles.

Since the center circle in equilateral triangle B is tangent to the three other circles, which in turn are tangent to the side lengths of the equilateral triangle, we can connect the midpoints of each side, giving us line segments that are tangent to the circles. This creates four equilateral triangles of equal area with four incircles of equal area respectively.

Since the side length of each of the smaller equilateral triangles is $\frac{1}{2}$ , the inradius is $\frac{\frac{\sqrt{3}}{2}}{6}=\frac{\sqrt{3}}{12}$ . Using the circle area formula $\pi*r^2$ , where $r=\frac{\sqrt{3}}{6}$ , we know that the area of each of the smaller equilateral triangles' incircles is $\frac{\pi}{48}$ . Multiplying this by $4$ for the total area of the smaller incircles in equilateral triangle B gives us $\frac{\pi}{12}$ , which is the same result along with the answer we received for triangle A's incircle. Since the areas of each of the white regions in equilateral triangles A and B are the same, the red region in triangle A has the same area as the purple region in triangle B.