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Puzzle: show the total area of the two semicircles is half the area of the large circle.

If you get stuck, go to +Alexander Bogomolny's wonderful website:

On top you'll get an applet that lets you slide the point where the semicircles touch - no matter where it is, the semicircles have the same total area!  Click "hint" and you'll get a hint.  If you're still stuck, scroll down and see a proof.

This puzzle is a lot harder than my other recent area puzzles.   Indeed, it seems this fact was proved only in 2011!  

• Andrew K. Jobbings, Two semicircles fill half a circle, The Mathematical Gazette 95 (Nov. 2011), 538-540.

I find that amazing, since people have been thinking about this stuff for millennia!  However, Andrew Jobbings is a genius when it comes to 'recreational mathematics' - by which I mean math that's not considered 'serious', which people do just because it's fun.  (This is a curious concept, now that I think about it.) 

Check out more of his stuff here:

Stuart Presnell's profile photoMichel Plungjan's profile photoJoe Frambach's profile photoJohn Baez's profile photo
Sheez what a lot of work - cannot use Chrome, must install JAVA 7 in Fx on OSX to run... 50MB download. Disappointed in JAVA applets again
Did not help installing 151MB of stuff. Fx still does not run it
OK... Here's my opening salvo that does not quite go all the way.  What we're postulating is that the circle with radius r has double the area of the semi-circles with radius s and radius t.  Translate to some algebra...

pi * r^2 = 2 ( (pi * s^2)/2 + (pi * t^2)/2)
pi * r^2 = pi*s^2 + pi*t^2
r^2 = s^2 + t^2

which looks an awful lot like the Pythagorean theorem about right triangles.  My guess is that this is somehow key, in particular, looking at the triangles formed in the slices between the semi-circles, the horizontal line, and the outer circle.

One side of each of those triangles is the radius of the semi-circles (i.e. s and t), and we know that the other short edges (along the horizontal line) have the relationship:

(shortedgeS + shortedgeT) = r - (s + t)

But I don't know where to go from there.  Does that look like a promising start to anyone else?
+Dan Thompson - that looks good to me!  I got to the Pythagorean

r^2 = s^2 + t^2

and started looking for right triangles that made this obvious.  But I didn't find one.  It seems more work is required.
Excellent puzzle, I loved doing it.  Thanks.  
+John Baez on OSX no such luck yet. t
Too bad really. Use JAVA on the server big time. JavaScript on the clients only
+John Baez The right triangles are sort of there already, except that they're two straight edges and an arc, i.e. to the right of the yellow semi-circle and to the left of the green semi-circle.  Just slice off the arc by connecting the points where the straight lines hit the arc, and you have instant right triangles.  Those are the triangles I was talking about in my earlier comment.
I don't know if they are looking for "purely" geometrical proof. Algebraically, it is quite simple really. But, even for geometrical approach, it can be done easily. Thanks, it was fun!

Spoiler alert!

Let O (origin) be the center of big circle of radius R. Draw the line between O and the upper edge of the green circle (radius R1), where it touches the big circle. Pythagorean theorem for this triangle says that the center of the green circle, is distance DG=sqrt(R^2-R1^2) from the origin. Repeat the construction for the yellow circle, radius R2, to get DY=sqrt(R^2-R2^2). But DG+DY=R1+R2, so that
R1+R2=sqrt(R^2-R1^2)+sqrt(R^2-R2^2). The solution of this is R^2=R1^2+R2^2, and by multiplying by Pi/2 we get the required relationship for the area.
Here's a simpler proof that boils down to just the Pythagorean theorem: consider the maximal secants of the two quarter-circles in the top half of the larger circle. In Dan's notation, these have length sqrt(2)*s and sqrt(2)*t. These two secants each form 45-degree angles with the horizontal, so they are perpendicular. They form a right triangle whose hypotenuse is a secant of the larger circle. All we have to do is prove that this larger secant is also a quarter-secant, since then the Pythagorean theorem will give that 2r^2 = 2s^2 + 2t^2 which is a linear multiple of the desired relationship between areas.

This follows directly from noting that extending the smaller quarter-secants into secants forms two equal arcs and applying the basic result of geometry that the secant angle is half the sum of the two opposing arc angles (see ; this result was also used in the original linked proof).

A much simpler puzzle is this: Given that the sum of the areas is constant, how do you determine this constant?
+Harald Hanche-Olsen
Since it is constant for any two values of R1 and R2, take, e.g., R2=0, then R1=R, so the constant is 0.5*Pi*R^2.
A bit off topic: When I read +Anton Geraschenko 's comment above (shortly before I wrote my own comment), the comment he was referring to was not visible to me, nor was the one before that. I was wondering if Anton was replying in public to a private comment? But I guess it was only some G+ weirdness going on. I wish it wouldn't mess up the conversation like that.
I haven't looked at the comments above because I don't want to read any spoilers, so apologies if, as seems likely, I'm repeating things that have already been said. First, I notice that the constancy of the area in question is equivalent to the constancy of the distance between the top corners of the two semicircles. This follows from Pythagoras's theorem applied to the triangle with vertices those two corners and the point where the two semicircles touch.

Now let's consider an arbitrary point on the horizontal diameter, and two lines going out from it at 45 degrees above the horizontal. We want to prove that the distance between the two points where those lines cut the circle is constant. But if we move the point by delta to the right, then the two points in question move by the same amount clockwise round the circle, at least to first order (and in fact exactly, but we only need to prove it to first order). I haven't fully checked this claim (except in the sense that it must be true if the result is true), but I feel it should be a simple consequence of the fact that the tangent lines are at right angles. I can't do this in my head any more -- time for a piece of paper.
Hmm ... it doesn't seem obvious that those tangent lines do meet at right angles. Indeed, if they do, then the distance between the points is trivially constant. Oops.
Ah, but is that enough? If they are at right angles anywhere, then the derivative of their distance is zero there, so they are at right angles everywhere! 

The only question that leaves me (apart from a slight feeling that I ought to make that more rigorous) is whether there is a nice calculus-free proof.
+Timothy Gowers
If the green and yellow semi-circles were of the same size, the polygon connecting four points that touch the big circle would be a square with diagonal 2*R (by symmetry, this diagonal passes through the center of the big circle). Thus the radius of such semicircles is R/sqrt(2). These two equal semicircles form a circle of this radius, with area (1/2)*Pi*R^2, which is half the area of the big circle.
What a terrible classification of such problems as "non-serious" as if there is any seriousness in math in the rest of it.  Though I know that you have no intention of having that definition per se, I'm pretty sure that such snob attitude exists due to my experience with the definition applied math.
proved it but, can't see the proof in the article. How can I see it ?
I assumed that centre of the bigger circle is at distance x from the centre of bigger of semicircles then eliminated x from two distance relations.
+Rahul Kashyap wrote: "can't see the proof in the article. How can I see it?"

Scroll down!  There's a big blank space in the article to keep people from accidentally seeing the answer.   +Alexander Bogomolny's proof avoids use of analytic geometry.
+John Baez  I'm sorry, stupid of me. It's an elegant proof ! A good construction does half the work.
+Nenad Svrakic Yes. I was taking that observation as read, so the main task is to prove that the area is independent of the sizes of the semicircles.
If the radius of the big circle is 1 you want to show that the sum of the radii of the green and yellow circles equals 1 as well. Thales theorem and cosinus and tangens might do that, I suppose.
+mathias voellinger sum of the radii of the green and yellow circles equals 1
That wouldn't do it because if R1+R2=1, then R1^2+R2^2<1.
 (The task is to show that R1^2+R2^2=1)
Look at the green G and yellow Y parts of the diameter; exchanging them (G+Y->Y+G), we get a point that is at the equal distance R from all four corners of half-circles (R^2=G^2+Y^2), so it is the center of the big circle and its radius is R
Using the labeling of the solution diagram to avoid drawing one myself.
The only thing I'd mention different is I won't draw lines AD & BC, but extend DO to meet AM and show the triangle so formed is similar to ODN. From this one gets BMO & DNO are congruent. These triangles now gives the Pythagorean relationship.
What is surprising (at least I couldn't do it) is go from MO + NO = a + b to the result by using Pythagoras' to write MO & NO in terms of their respective semicircle radius and the circle radius. Quite surprised this fails, because its such a big guide to start the proof, before you realize it's not needed at all. I would have thought this would work without drawing a decent diagram. Very hard question though for such easy method of solution. I needed people here to emphasize elementary and geometrical, otherwise I'd be stuck.
+Anton Geraschenko's drawing was key for me although my solution, assuming it's right, involved too much algebra to be elegant.
I think I've finally found an elegant solution thanks to Greg Egan.  The margin of this comment is too small to contain it.  But I will post it someday!  What I wanted was a solution where you can just look at a picture and see the truth.
+John Baez, yes, this is what I strove for and often failed to achieve in my short math career.
I'm still trying to find a solution that I really like. On a long car journey yesterday I thought I had some ingredients, but I haven't had time with pen and paper.
The right triangle you're looking for is between the top of the green semicircle (P), the centre of the big circle (O), and the midpoint of the green semicircle's diameter (M).  Obviously the radius of the big circle is OP, and the radius of the green semicircle is MP.  What's not obvious is that the radius of the yellow semicircle is OM.  

To see this, label the middle and bottom of the yellow semicircle's diameter N and R respectively, and the point where the two semicircles touch A.

We just have to show that angles MPO and RON are equal, which follows from these three observations:

(i) Angle MPR is 45 degrees.
(ii) Triangle POR is isoceles, so angle OPR = angle ORP.  
(iii) Angle OAR is 135 degrees, so RON plus ORP equals 45 degrees.  

Thus MPO = MPR - OPR = 45 - ORP = RON.
I've finally had a chance to draw the diagram I couldn't quite visualize in enough detail while driving, and it works! I'm afraid my technical skills are not up to reproducing it online, so instead I'll give a wordy description and try to make it as clear as I can (but not too long).

Lemma. Draw a semicircle with the straight side horizontal and the semicircle above it. Let the two ends of the straight side be A and B, and let the top point be C. Let the midpoint of A and B be D. Let P be any point on the circle between C and B. Then the angle APC is 45 degrees.

Proof. It's a well known theorem in geometry that the angle is constant. When P=B, it is clearly 45 degrees.

Note also that the angle APB is 90 degrees, by the same well known theorem in geometry. So the angle you get by going from B to P and then directly away from C is also 45 degrees.

What is the relevance of this semicircle? Well, suppose we take another semicircle, also with a horizontal base, and take two points on it V and W such that the angle VOW (where O is the centre of the circle) is 90 degrees. Then the points V, O and W lie in a unique circle, and that circle cuts the base of the semicircle in two points, one of which is O. Let Q be the other point. I want to prove that the angle VQO is 45 degrees, as is the angle you get if you go from W to Q and then directly away from O. But VW is a diameter of the circle that goes through V, O and W (since the angle VOW is 90 degrees), so if we turn the diagram round and consider the semicircle that has VW as its base and goes through O and Q, then the lemma is telling us precisely what we want.

This is sort of doing things backwards: I'm showing that if the angle VOW is 90 degrees, then V and W will be the top points of two semicircles as in John's diagram. It is then not hard to prove uniqueness: that is, if 90 degrees works, then no other angle can work.

That may sound slightly complicated but if you draw the diagrams I describe, you'll see that it's an almost instant deduction from the lemma, which itself is an almost instant consequence of the well known geometrical theorem about the angle made by three points in a circle not depending on the middle point. 

I forgot to say at the beginning of all this that I still need the Pythagoras step that converts the problem from one about areas to one about the length of the arc that joins the top points of the two semicircles. Maybe there's a nice proof that sees it done via areas rather than lengths. Now to look at +Stuart Presnell's diagram to see whether it's the same idea. 
Looking at +Stuart Presnell's diagram, I have the impression that his argument is in a similar spirit but not quite identical. 
I'm now quite convinced the approach in my comment above won't work (as it was written).
Unfortunately I can't watch the Java demonstration on the linked page, and I don't have access to the Andrew Jobbings paper either.  Could someone who does have access perhaps post a screenshot (or a re-construction) of the relevant diagram so we can see how our own solutions compare?
+Joe Frambach - Very nice!  Esthetically speaking, that horizontal gray bar looks a bit fat.  I guess that's to make it easy to notice it and put the cursor on it?  (If I could program worth a darn I'd just fiddle around and try something else.)
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