Semi-Circles Within a Circle

[bob012345]

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]

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.

 

[bob012345]

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]

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]

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]

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]

Another more elegant solution than post#2.

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.