Three Circle Problem

[bob012345]

TL;DR

The problem is to determine if the intersection area of the three circles is less than or greater than one quarter the area of one circle.

Three identical circles intersect in such a way that each intersects the centers of the other two as in the figure. The problem is to determine if the shaded area is less than or greater than one quarter the area of one circle. Of course one can calculate the answer but the exercise is to show it by more subtle arguments if one can. This is adapted from a book by Martin Gardner. Enjoy!

 

[Hill]

I still need to compare $\sqrt 3$ and $\pi /2$. Not even calculation like this?

 

[bob012345]

I still need to compare $\sqrt 3$ and $\pi /2$. Not even calculation like this?

You don’t need numbers but of course you can calculate things if you want.

 

[Ibix]

Spoiler

I've divided the shaded region into an equilateral triangle of area $A_T$ and three segments of area $a$ each. Each circle has area $A_\circ$.

Observe that the triangle plus one segment is a 60° sector of a circle. Hence $$\begin{equation}\frac{A_\circ}6=A_T+a\end{equation}$$

The area of the shaded region is obviously $A_T+3a$. So the statement "the shaded area is more than a quarter circle" can be written
$$\begin{equation}\frac{A_\circ}4<A_T+3a\end{equation}$$Using (1) to eliminate $A_T$ we get$$\begin{eqnarray}\frac{A_\circ}4&<&\frac{A_\circ}6+2a\\A_\circ&<&24a\end{eqnarray}$$

I can't draw this accurately on my phone. Draw a circle and inscribe a hexagon, then draw diameters to split the hexagon into six equilateral triangles. Draw circles (all the same size as the original) so that every one of the straight lines has a 60° segment sitting on each side of it. The result looks like this:

(Adapted from Wikipedia). Notice that each green region is two back-to-back segments of area $a$ and there are twelve such regions, non-overlapping, and there is (red) space left over in the circle.

Thus it is not true that $A_\circ<24a$, and hence the shaded area is less than one quarter of the circle.

 

[Ibix]

Spoiler: Shorter solution

Start with the "flower of life" diagram:

(Adapted from Wikipedia - see previous post.) You generate this with thirteen unit circles. Draw one, then draw six with their centers evenly spaced on the perimeter of the first, then draw six with their centers on the concave corners of the outer perimeter of that figure. The central circle will look like the diagram.

Observe that one red region and the surrounding three green ones replicate the shaded region in the original post. There are six red areas and twelve green ones in the circle, but four of the shaded region requires only four red regions and twelve green ones. Thus the area of the circle is more than four times the area of the shaded region, by twice the area of a red region.

 

[Gavran]

Spoiler

$A_S$ - the shaded area from the original post
$A_T$ - the area of the triangle ABC
$A_C$ - the area of the circle
Triangles ABC, ACF, and BCD are three congruent triangles.
$A_T>3a\implies A_T+2A_T+3a>3a+2A_T+3a\implies 3(A_T+a)>2(A_T+3a)$
$3(A_T+a)>2(A_T+3a)\implies A_C/2>2A_S\implies A_S< A_C/4$

 

[A.T.]

Spoiler: Shorter solution

Start with the "flower of life" diagram:
View attachment 369369
(Adapted from Wikipedia - see previous post.) You generate this with thirteen unit circles. Draw one, then draw six with their centers evenly spaced on the perimeter of the first, then draw six with their centers on the concave corners of the outer perimeter of that figure. The central circle will look like the diagram.

Observe that one red region and the surrounding three green ones replicate the shaded region in the original post. There are six red areas and twelve green ones in the circle, but four of the shaded region requires only four red regions and twelve green ones. Thus the area of the circle is more than four times the area of the shaded region, by twice the area of a red region.

Yes, this is how you are supposed to solve that.

It's similar to solving that one: Which area is greater, the red or the blue one?

Spoiler

https://www.theguardian.com/science...e-pi-day-puzzle-that-will-spin-you-in-circles

 

[bob012345]

Here is my solution;

Spoiler

Isolate a quarter circle which includes half of the shaded shape.

It seems obvious that the blue part is larger than the yellow part but to be sure, let’s identify what’s common and drop those from each part leaving only the differences.

Then since half the original shape is less than half a quarter circle then the whole shape is less than a quarter circle.

 

[renormalize]

Then since half the original shape is less than half a quarter circle then the whole shape is less than a quarter circle.

Yes, the blue area is pretty obviously larger than the yellow. But that also means that the blue area is larger than "half a quarter circle". So I don't see the validity of your "solution".