Semi-Circles Within a Circle

bob012345

Gold Member

2,392

1,101

• Thread starter
• May 29, 2026

• #1

Here is a puzzle from Catriona Shearer. Find the area ratio of the shaded semi-circles to the circle which they fit in. The bases of the semi-circles are parallel.

Hint: consider the extreme cases to figure what the answer might be then prove the general case.

 

anuttarasammyak

Gold Member

3,161

1,594

• May 29, 2026

• #2

NOT ELEGANT solution
The eqation of the circles
$$x^2+y^2=R^2$$
$$(x-x_1)^2+y^2=R^2-x_1^2$$
$$(x-x_2)^2+y^2=R^2-x_2^2$$
By subtraction of the two smaller circles equatins, we see they touch at ##x=x_1+x_2,y=0##. Thus
$$x_1^2=R^2-x_2^2$$
$$x_1^2+x_2^2=R^2$$
Area of the sum of small circles divided by ##\pi##
$$2R^2-x_1^2-x_2^2=R^2$$
equals the area of the large circle divided by ##\pi##. So the solution is 1/2.

 

Last edited: May 30, 2026

bob012345

Gold Member

2,392

1,101

• May 29, 2026

• #3

The extreme cases I mentioned are when the radius of the smaller semi-circle goes to zero, the larger semi-circle fills exactly half the circle and also when the radii of the semi-circles are the same length.

 

kuruman

Science Advisor

Homework Helper

Education Advisor

Insights Author

Gold Member

2025 Award

16,040

9,254

• May 30, 2026

• #4

bob012345 said:

The extreme cases I mentioned are when the radius of the smaller semi-circle goes to zero, the larger semi-circle fills exactly half the circle and also when the radii of the semi-circles are the same length.

Is this your proof? If so, don't you still have to show that the ratio of areas is independent of the radii of the semicircles?

 

bob012345

Gold Member

2,392

1,101

• May 30, 2026

• #5

kuruman said:

Is this your proof? If so, don't you still have to show that the ratio of areas is independent of the radii of the semicircles?

No, that is not my proof, just an observation.

 

bob012345

Gold Member

2,392

1,101

• May 30, 2026

• #6

Here is my proof.


$$r_1^2+(r_1+d)^2=R^2=r_2^2+(r_2-d)^2$$
Solving for ##d##$$d=\frac{r_2^2-r_1^2}{r_2+r_1}=r_2-r_1$$
Therefore $$r_1^2+r_2^2=R^2$$
And area ratio is $$\frac{\frac{1}{2}\pi r_1^2+\frac{1}{2}\pi r_2^2}{\pi R^2}=\frac{1}{2}$$

 

anuttarasammyak

Gold Member

3,161

1,594

• May 30, 2026

• #7

Another solution

F is on AC.
##\angle ACD = \pi/4##
##\angle AOD=\pi/2##
##\triangle AEO \equiv \triangle OGD##
##EO=GD##
Thus ##r_1^2+r_2^2=R^2## where ##r_1## and ##r_2## are radii of two smaller circles.

 

Last edited: May 30, 2026