50 of 100: Lumière des Lunes

Consider the following more general figure: Lemma: Area of red region is equal to area of yellow region.

Proof: Area of red crescents = area of right triangle + $\frac{1}{2}$ area of small circle + $\frac{1}{2}$ area of medium circle - $\frac{1}{2}$ area of large circle = area of right triangle + $\frac{1}{2} \frac{\pi}{4} ( \underbrace{a^2+b^2-c^2}_{=0} )$ = area of right triangle.

Thus we have: purple area = square = $(5/\sqrt{2})^2=12.5$ .

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Alternatively:

One purple area= $\frac{1}{4}$ square + $\frac{1}{2}$ small circle - $\frac{1}{4}$ large circle

Total purple area = square + $2$ small circle - large circle = $25/2+ 2 \pi (25/8) - \pi (5/2)^2=12.5$

NOTE THANKS TO COMMENTS FROM OTHERS I HAVE CORRECTED MY SOLUTION:

This problem can be solved using composite areas.

In order to do this we first need the length of a side of the square which can either be found using a 45-45-90 special right triangle or pythagorean theorem. Here is the work using pythagorean theorem.

$a^{2}+b^{2}=c^{2}$

$s^{2}+s^{2}=5^{2}$

$2s^{2}=5^{2}$

$s^{2}=\frac{5^{2}}{2}$

$s=\sqrt{\frac{5^{2}}{2}}$

$s=\frac{5}{\sqrt{2}}$

Then we need to find the areas of the following shapes

1.The squares area is

$A=(\frac{5}{\sqrt{2}})^{2}$

$A=\frac{25}{2}$

$A=12.5$

1. The large central circle

$A=\pi(\frac{5}{2})^{2}$

$A=\frac{25\pi}{4}$

1. The area of the four semicircles

$A=4\frac{\pi(\frac{5}{2\sqrt{2}})^{2}}{2}$

$A=2\pi(\frac{5}{2\sqrt{2}})^{2}$

$A=2\pi(\frac{25}{8})$

$A=\frac{25\pi}{4}$

The area of the purple regions is then found by finding the sum area of the 4 semicircles and the square and subtracting the large circle

$\frac{25\pi}{4}$ + $12.5$ - $\frac{25\pi}{4}$ = $12.5$

Here's a slightly altered version of the diagram: I started by seeing this as a square with four semi-circles around it. We can work out the area of the whole thing by finding the area of 2 circles and adding it to the area of the square.

Then we need to subtract the bigger circle (shown here in yellow), to leave the area we want:

Rather than use 5 as the diagonal of the square, I called half of that diagonal $x$

Putting in the other diagonal of the square gave me four right-angled triangles. The hypotenuse of these is useful, because it is the side of the square and also the diameter of the semi-circles.

The area of the square is $2x^2$

The radius of each semi-circle is $\frac{\sqrt{2}}{2}x$ and this gives the total area of the four semicircles as $\pi x^2$

So, the area of the whole thing is $2x^2 + \pi x^2$ and we need to subtract the area of the big yellow circle, which is $\pi x^2$

This leads to the wonderful result that the area of the four lunes is the same as the area of the square, namely $2x^2$

In this particular case $x=2.5$ , and this gives the area of the purple lunes as 12.5 square units.

Lune of Hippocrates. Yellow area = blue area. In the given problem, the purple area = the area of the square = 12.5

If you take one-quarter of the figure, you have a triangle (A), a segment (B), and a crescent (C). The area of the triangle is easy. Now (A+B) is one-fourth of the area of the large circle and (B+C) is one-half of the area of the small circle. Both of these numbers are easily determined. So you can now write an equation for the area of the crescent as C = (B+C) - ((A+B) - A). The answer, oddly enough, is A. So 4 x C = 4 x A, which happens to be the area of the square in the middle of the figure.

One additional point........the answer does not involve pi even though the perimeter of the crescent is made up of two intersecting circles. In other words: "No pi for you!" As far as I know, this is a unique characteristic of this type of crescent. Does anyone know whether there are other "circular" shapes that share this oddity.

In the above diagram, label the areas as shown. By Pythagoras the sidelength of the square is $\frac{5√2}{2}$ (with the denominator rationalised). Therefore that is also the diameter of all of the 4 small circles, making their radius $\frac{5√2}{4}$ . As the area of a semicircle is $\frac{1}{2}$ of πr², the area a + e = $\frac{25π}{16}$ .

Therefore, the area a + b + c + d + e + f + g + h = $\frac{25π}{16}$ x 4 = $\frac{25π}{4}$ as this includes 4 lots of a + e.

The area of the square (labelled i) is $\frac{5√2}{2}$ x $\frac{5√2}{2}$ = $\frac{50}{4}$ = 12.5

The pale blue circle has diameter 5 (given) so the area is π x 2.5² = $\frac{25π}{4}$ . In the notation of areas on the diagram: e + f + g + h + i = $\frac{25π}{4}$

This gives us 3 equations: a + b + c + d + e + f + g + h = $\frac{25π}{4}$

e + f + g + h + i = $\frac{25π}{4}$

i = 12.5

Equating the first two gives us that a + b + c + d + e + f + g + h = e + f + g + h + i, or a + b + c + d = i. However, since i = 12.5, a + b + c + d = 12.5 also and from the diagram, the purple areas are labelled a, b, c & d.

Hence, the total area of the regions shaded purple is 12.5

$4π(\frac{5√2}{4}$ )² $-$ $π$ ( $\frac{5}{2}$ )² $+$ ( $\frac{5√2}{2}$ )² $-$ $2π$ ( $\frac{5√2}{4}$ )² $= 12.5$

Area of the 4 small circles - Area of the big circle - Area of the small curved cross left The long, stupid way to do it, by me

Consider the blue region above the square, then its area A is:

A = pi(5/2)^2 – 25/2 = 25pi/4 – 25/2

Since the side square a satisfies:

a^2 = 25/2

Then the pink area A’ is:

A’ = (2pi)(a/2)^2 – A = 25pi/4 – A = 25/2 = 12.5.