Circles Geometry!

First, compare the big circle to the small circles in the figure. If we say that the radius of each small circle is $r$ , than the radius of the big circle is $2r$ . So the area of the big circle, $B$ is 4 times the area of each small circle, $S$ :
$B = \pi (2r)^2\ = 4\pi r^2 = 4S \hspace{10mm} (1)$

Now, lets use the symmetry of the figure. We're given that one of the four equal yellow pieces has area $\color{#ffa700}{Y}$ . Let's also name each of the four symmetric blue pieces , $\color{#3D99F6}B$ and each red piece , $\color{#D61F06}R$ .

Now we can construct each type of circle out of the $\color{#ffa700}{yellow}, \color{#D61F06}{red},$ and $\color{#3D99F6}{blue}$ pieces:

Each small circle, $S$ is made out of 2 yellow pieces and 1 red piece, so: $\color{#ffa700}{2Y} + \color{#D61F06}R = S \hspace{10mm} (2)$

The large circle, $B$ is made out of 4 blue pieces 4 red pieces, and 4 yellow pieces, so: $\color{#3D99F6}{4B}+ \color{#D61F06}{4R}+ \color{#ffa700}{4Y} = B \hspace{10mm} (3)$

Combining all three equations together:

$\begin{aligned} \color{#3D99F6}{4B}+ \color{#D61F06}{4R}+ \color{#ffa700}{4Y} &=& B = 4S \\ \color{#3D99F6}{B}+ \color{#D61F06}{R}+ \color{#ffa700}{Y} &=& S = \color{#ffa700}{2Y} + \color{#D61F06}R \\ \color{#3D99F6}{B}+ \color{#ffa700}{Y} &=& \color{#ffa700}{2Y} \\ \color{#3D99F6}{B} &= &\color{#ffa700}{Y} \end{aligned}$ So the blue area is the same as the yellow area.

Each small circle has 1/4 the area of the big circle, so their total area is equal to that of the big circle. Thus, the area of their over lap (yellow) must be equal to the area that is in their complement of the big circle (blue). That is, for all area lost by them overlapping, that area shows up in the blue region.

I just solved it using intuition. Blue areas look like the inverse of yellow areas.

lucky guess yes yes

Such problems almost always have an equal area

The blue area can be found by realizing that it is the difference between the large circle and (the sum of the 4 small circles minus the overlap). The overlap meaning the yellow part.

In other words, we can write:

Ar(blue)=Ar(Large Circle)-Ar(4 small circles –overlap)

Ar(blue)=Ar(Large Circle)-Ar(4 small circles) + Ar(overlap)

Let us assume that the radius of the small circle is r. Then the area of the larger circle would be 2r. Ar(blue)=pi(2r)^2 – 4 (pi r^2)+ Ar(overlap)

Ar(blue)=pi 4 r^2 – pi 4 r^2)+ Ar(overlap)

Ar(blue)=Ar(overlap)

Ar(blue)= Ar(yellow)

Hence both are equal.

Just a Lucky Answer!

Assign to each point in the large circle a number that tells you how many small circles cover that point. This number is 2 in the yellow regions, 1 in the red regions, and 0 in the blue regions. The average value of this number across the whole circle must be exactly 1, because the area of the four small circles is exactly the same as the area of the big circle (because each small circle has half the diameter of the big circle, and area scales as the square of length). The only way this can happen is if the area of the blue regions is exactly the same as the area of the yellow regions.

Lets' assign one unit to the radius of each of the 4 circles. 1) The yellow area (4 lens-spaped regions) is 2(π-2) (See http://mathworld.wolfram.com/Circle-CircleIntersection.html ). 2) The radius of the big circle (circumscribing the 4 circles) is double of the radius of each of the 4 circles, so its area is π2^2 = 4π . 3) The total area of the 4 small circles is 4πr^2 = 4π . 4) The blue area is the area of the blue circle minus the area of the 4 small circles plus the yellow area, i.e. 4π - 4π + 2(π-2) = 2(π-2) , i.e. same with the yellow area.

$Blue=CircleBig-(4 \times CircleSmall-Yellow)$

$Blue-Yellow=CircleBig-4 \times CircleSmall$

$Blue-Yellow=(2 RadiusSmall)^{2} \times \pi - 4 \times RadiusSmall^{2} \times \pi$

$Blue-Yellow=4 \times RadiusSmall^{2} \times \pi - 4 \times RadiusSmall^{2} \times \pi$

$Blue-Yellow=0$

I note that the area of the blue areas is equal to the area of the large circles minus the 4 smaller circles, then adding in the yellow areas: πr² - 4π( $\frac{r}{2}$ )² + Yellow = Blue

It appears that πr² - 4π( $\frac{r}{2}$ )² = 0. thus Yellow = Blue.

Note: This is my first time making a solution guide on Brilliant, apologies if it seems confusing.

Assume that radius of small circle is 2, radius of bigger circle is therefore 4; If area of each yellow region is "a" and area of each blue region is "b" then area of the four overlapped smaller circles is 4(4 Pi) - 4a = 16 Pi - 4a

Area of the bigger circle = Pi (4)^2 - 4 b = 16 Pi - 4b

the two areas are the same; therefore 16 Pi - 4a = 16 Pi - 4b; hence a = b; QED

Use inclusion-exclusion :

The (area of the big circle) equals (the sum of the areas of the four small circles) - sum of areas of intersection (the yellow area) + area not in four small circles (the blue area).

Let $A(B),A(Y)$ denote the areas of the blue and yellow areas, respectively. If $r =$ radius of small circle, then $2r =$ radius of big circle. Then

$(2r)^2 = 4*r^2 - A(Y) + A(B)$

Simplify:

$A(Y) + 4r^2 = 4r^2 + A(B)$

Subtract $4r^2$ from each side...

$A(Y) = A(B)$

Really I'm love this type of problem ..and I want to more problems

... I just looked at the yellow, looked at the blue, saw the same amount of each color, chose equal amount of both as my answer