Comparing Areas with Circles.

Let the area of the two white regions together be $W$ . Then $B = 10^2 \pi - W = 100\pi - W$ and $R = 8^2\pi + 6^2\pi - W = 100\pi - W$ . Implying that $\boxed{B=R}$ .

Sum of Red areas before intersection is same as area of blue circle. And then after intersection equal parts are cut out from both the total red and the total blue so are net remains unchanged.

Since the answers are all equations, the size of the white region must not matter. You might as well let the circles be tangent. Then $B=100\pi$ and $R=64\pi+36\pi=100\pi$ and the solution is obvious.

This would be a much better problem had it specified there must be some overlap and the choices been

$B<R$ , $B=R$ , $B>R$

Let the white areas be X & Y. Then:

The red area is $({16}^{2} \times \pi$ ) + $({12}^{2} \times \pi$ ) $- (x+y)$ = $400\pi - (x+y)$

The blue area is $({20}^{2} \times \pi) - (x+y)$ = $400\pi - (x+y)$

The red area equals the blue area

Since we are not told the area of the overlap, the only solution that works for any amount is that the areas are equal! Obviously I did the 10^2 = 8^2 + 6^2 first before realizing this logic-based approach...

We can ignore the 'white' overlaps as that would be equally removed from both. What we then notice is that the diameters 20,16,12 gives us radii 10,8,6 which we might also recognize as a scalar multiple of a 3,4,5 triangle. Since area is a factor of r^2, the areas are equiv.

The answer choices suggest that a proportional relationship exists between B and R no matter the configuration. Therefore, we consider the configuration where none of the circles touch. Since $16^2+12^2=20^2$ , $B=R$ .

This one can be solved by looking at the Pythagorean Triple $\boxed{3^2+4^2=5^2}$ and its multiplied statement $(4\cdot3)^2+(4\cdot4)^2=(4\cdot5)^2$ , which leads to our diameter triple $\{12,16,20\}$ .

Using the area formula for the circle $\boxed{A=\pi/4\cdot d^2}$ gives us a factored version of our diameter triple in Pythagorean form.

Just knowing $\boxed{d_{red1}^2+d_{red2}^2=d_{blue}^2}$ shows us that: The FULL blue circle area $(B+White_1+White_2)$ equals the FULL combined red circle areas $(R_1+White_1)+(R_2+White_2)$ .

Combined: $\boxed{B+ White_1+ White_2 = R_1+ White_1 +R_2 +White_2}$

Finally we can subtract both white regions from both sides, leading to $\boxed{B=R_{1+2}}$ .

Well I took a less common approach: since the area of the big circle and the combined area of the two smaller circles remain the same, we can be sure that (B-R) must be a constant; however it is obvious that the value of B and R can change (the position of the circles are not determined; this is also the main reason why my method works in this question). Under this situation, the only answer with the possiblilty to apply with the change of B and R is the first answer, B=R.

(Situations like B=X*R where X is positive and doesn't equals to 1 exists, but only in special conditions; plus the values of R & B would be determined)

(Also, just being nit-picking, the question could be improved by an additional note that the two red circles do not intersect with each other.)

The only way that the ratio between R and B can be independent of the white areas (constant) is if B = R. Due to the way that the question is formulated, we can deduce this without even calculating the areas, which are of course equal.

Let's star by A(r) = πr²

We can see that the red circle is missing some fancy shape. But, the blue circlesvis missing the same shape. So, we can ignore the fancy shape

(12² + 16²)π = 20²π

In conclusion the areas are equal

The white areas are subtracted from both the blue and red areas equally, so it will not matter. And the blue and red areas are equal initially without the cut. Therefore B=R.

Area of largest circle equals sum of areas of two smaller circles. Blue area is area of largest circle minus sum of white areas. Red area is sum of areas of two smaller circles minus sum of white areas. Therefore blue area equals red area

Area of a circle = π x radius^2

10 x 10 x π = 100π

6 x 6 x π = 36π

8 x 8 x π = 64π

36 + 64 = 100

The white areas are deducted from both circles equally, so they do not matter!

The distance between circle centres is not specified. If a unique solution exists then it matters not whether the white areas exist. 100 pi = (64 +36) pi.

The "white" area removes as much red as blue area, so the ratio of removed area is 1 : 1. Now notice that we don't have any exact measurement of the white area. There is no way to calculate this. Than it must not matter for this problem. The only reasonable answer is thus B = R. The ratio R : B must be 1 : 1. Every other ratio would be changed if we change how far we would "slide the circle in" (resulting in an increase of white are).

Let's give an example to make the last statement more clear. Let's say B = 1.5R. If R = 10 than B = 15. Removing a little white area (let's remove an amount of 5) would result in R = 5 and B = 10. This means that B = 2R, clearly the ratio B : R has changed.

12,16,20 are the numbers of right angle triangle. $12^2+16^2=20^2$ so $R=B$

Note that the diagram is not very specific, so you can move the circles. Note that a change in the white region changes both the blue and red regions by the same amount, so we move both the red circles outside of the blue region. Now the area of the red region is $(12^2+16^2)\pi=400\pi$ and the area of the blue region is $20^2\pi=400\pi$ . So, $\boxed{B=R}$ .

I figured since the degree of overlap doesn't seem to matter, set it to 0 and you're home free

Kudos to all the 'engineering' solutions.

But I like the good old inclusion exclusion.

Let A be the total area

A1 be area of the top circle, A2, be the area of the middle circle, and A3 be the area of the bottom circle.

A = A1 + A2 + A3 - (A1 ∩ A2 + A2 ∩ A3) (the other terms A1∩A3 and A1∩A2∩A3 are 0 here since the top and bottom circles don't intersect) 1) Now if we observe, A2 - (A1 ∩ A2 + A2 ∩ A3) is A(Green)

So A = A1 + A3 + A(Green) = 100π + A(Green)

2) Also, A1 - (A1 ∩ A2) + A3 - (A2 ∩ A3) is A(Red)

so A = A2 + A(Red) = 100π + A(Red)

From 1) and 2) we get A(Green) = A(Red)

I think it related to the Pythagorean theorem. $12^2 + 16^2 = 20^2$

I worked it out as follows: 1)The blue area is most of a circle with total area 100pi and the red area is most of two circles whose combined area adds up to 100pi (64pi + 36pi). 2)Both the red and blue areas are reduced from 100pi by the same amount (the white area). 3)Therefore the red and blue areas are equal.

From the graph, one can get

$\begin{cases}B+X_1+X_2=100\pi\\64\pi=R_1+X_1\\36\pi=R_2+X_2\end{cases}$

Adding these three equations together, get you $\boxed{B=R}$ .