# given 3 circles, find relation of the regions

I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.

I had no idea how to find it nor where to start

Note that this picture are DRAWN ON SCALE

any explanation of how to find the area would be very appreciated

We can say, at least, that the option (C) is correct.

Let $p,q,r,s,t,u$ be the areas of white parts :

• $p$ is the upper left area of $AB$, inside the circle whose diameter is $BC$.

• $q$ is lower of $AB$, upper of $BC$, outside the circle whose diameter is $AC$.

• $r$ is lower of $BC$, inside the circle whose diameter is $AB$.

• $s$ is lower of $BC$, inside the circle whose diameter is $AC$.

• $t$ is upper of $BC$, lower of $AC$, outside the circle whose diameter is $AB$.

• $u$ is upper of $AC$, inside the circle whose diameter is $CB$.

Then, we have four equations :

• For the circle whose diameter is $AB$ : $A_1+p=q+r+A_2\quad\tag1$

• For the circle whose diameter is $AC$ : $u+A_3=A_2+t+s\quad\tag2$

• For the circle whose diameter is $BC$ : $p+q+t+u+A_2=s+r+A_4\quad\tag3$

• For the right triangle $ABC$ : $(A_1+p)+(u+A_3)=s+r+A_4\quad\tag4$

From $(1)$, we have $$A_1=A_2+q+r-p\tag5$$ From $(2)$, we have $$A_3=A_2+t+s-u\tag6$$ From $(3)$, we have $$A_4=A_2+p+q+t+u-s-r\tag7$$ From $(4)(5)(6)(7)$, we have $$(A_2+q+r-p)+p+u+(A_2+t+s-u)=s+r+(A_2+p+q+t+u-s-r),$$ i.e. $$A_2=p+u-s-r\tag8$$ From $(8)(1)(2)(3)$, we have $$A_1=q+u-s$$ $$A_3=p+t-r$$ $$A_4=2p+2u+q+t-2s-2r.$$

From these, we can have $$A_1+A_2+A_3=2p+2u+q+t-2s-2r=A_4.$$

• it is interesting to note that the area of the triangle itself is equal to $A_4-A_2$ and equal to $A_1+A_3$ – David Quinn Jul 26 '15 at 12:16
• why triangle area is equal to $\frac 12 CircleArea_{AB} + \frac 12 CircleArea_{AC} = \frac 12 CircleArea_{ABC}$? – wuiyang Jul 26 '15 at 12:34
• @wuiyang: Sorry but, I don't get your question. The equation you wrote is correct, but the LHS in the equation is not equal to the area of the triangle. (sorry if I misunderstood your question) – mathlove Jul 26 '15 at 12:48
• @DavidQuinn: Indeed. Nice catch! – mathlove Jul 26 '15 at 12:49
• @mathlove oh, now I get it, thank you very much. – wuiyang Jul 26 '15 at 13:41

Observe that from $AB^2+AC^2=BC^2$, the sum of the area of the two smaller circles is the area of the biggest circle. Then $A_1+$white part$+A_2+A_2+A_3 = A_2+A_4+$white part, $\Rightarrow A_1+A_2+A_3 = A_4$.