Geometry Puzzles

Here are four puzzles about areas, in approximate order of increasing difficulty.

Mysteries of the equilateral triangle

Puzzle: Show the area of the orange circle equals the total area of the two blue regions.

In case you’re wondering, the picture above shows an equilateral triangle with a small circle inscribed in it and a big circle circumscribed around it.

The puzzle is easy if you think about it the right way. I never knew this cool fact until last month, when I read this fun free book:

• Brian McCartin, Mysteries of the equilateral triangle.

I’ve given talks on the numbers 5, 8, and 24. I’ve imagined writing a book that had chapters on all my favorite natural numbers, starting with Chapter 0. But it would be very hard to write Chapter 3, since this number has too many interesting properties! McCartin’s book covers a lot of them.

Square the lune to the tune of Claire de Lune

Puzzle: Show the crescent has the same area as the triangle.

Greek mathematicians really wanted to square the circle, by which I mean: use straightedge and compass to first draw a circle and then construct a square with the same area.

In 440 BC, Hippocrates of Chios figured out how to square the above crescent-shaped region, which lies between the circle centered at O and the smaller circle centered at D. So, this region is called the Lune of Hippocrates.

I’ve heard it said that this result gave some Greek geometers hope that the circle could be squared. I’m not sure this is true; Hippocrates himself was probably too smart to be fooled. But in any event, it would have been a false hope. Much later, around 1885, Lindemann and Weierstrass proved that squaring the circle was impossible.

Any crescent-shaped region formed by two circular arcs is called a lune. It’s widely believed that there are only 5 squarable lunes. In other words, there are only 5 shapes of lune constructible by straightedge and compass whose area equals that of a square constructible using straightedge and compass. (Obviously these lunes can come in many different sizes.)

Hippocrates discovered three squarable lunes. Two more were discovered by Martin Johan Wallenius in 1766. A proof that these are the only squarable lunes was given by Tchebatorew and Dorodnow, and summarized by the famous topologist Postnikov:

• M. M. Postnikov, The problem of squarable Lunes, translated from the Russian by Abe Shenitzer, American Mathematical Monthly, 107 (Aug.-Sep. 2000), 645–651.

However, there may be a loophole in this proof: Will Jagy claims that these Russians assumed without proof that the ratio of two angles involved in the construction of the lune is rational. With this assumption, finding a squarable lune amounts to finding a rational number u and a constructible number \sin a such that

\sin ( u a) = \sqrt{u} \sin a

Puzzle: Why should u be rational? Do you know the true state of the art here?

(My puzzles include hard questions I don’t know the answer to, and this is one.)

For a nice discussion of the 5 squarable lunes, with pictures, see this page:

The five squarable lunes, MathPages.

Twice in a blue lune

Puzzle: Show the area of this right triangle equals the total area inside the blue lunes. The outside of each lune is a semicircle. The inside of each lune is part of the circle containing the points A, B, and C.

The circle through all 3 corners of a triangle is called its circumcircle. You can construct the circumcircle using a straightedge and compass, if you want.

Again this is a famous old problem. The two blue lunes here are called the Lunes of Alhazen. This problem was later posed and solve by Leonardo da Vinci!

This puzzle is a lot of fun, so I urge you not to give up—but if you do, you can see da Vinci’s solution here.

The arbelos

Puzzle: show the area of the green shape equals the area of the circle.

The green shape is called an arbelos, which means ‘shoemaker’s knife’ in Greek, since it looks a bit like that. It shows up in Propositions 4-8 of the Book of Lemmas. This book goes back at least to Thābit ibn Qurra, a mathematician, astronomer and physician who live in Baghdad from 826 to 901 AD. Ibn Qurra said the book was written by Archimedes! But nobody is sure.

So, when you do this puzzle, you may be matching wits with Archimedes. But you’ll certainly be sharing some thoughts with Thābit ibn Qurra.

Math has been called the world’s longest conversation. Perhaps this is an exaggeration: people have been passing on stories for a long time. But Babylonians were computing the square root of two back in 1700 BC, probably using techniques that are still interesting today… so it really is remarkable how old mathematics still makes good sense, unlike the old creation myths.

For more on the arbelos try:

Arbelos, Wikipedia.


• Thomas Schoch, Arbelos references, 2013.

For the 15 propositions in the Book of Lemmas, see:

Book of Lemmas, Wikipedia.

Two semicircles is half a circle

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

Amazingly, it seems this fact was noticed only in 2011:

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

However, Andrew Jobbings is a genius when it comes to ‘recreational mathematics’. For more of his work, check out this page:

• Andrew Jobbings, Arbelos.

I learned about this puzzle from Alexander Bogomolny, who has a wonderful website full of Javascript geometry demonstrations:

• Alexander Bogomolny, A property of semicircles.

You’ll get a scary warning asking “Do you want to run this application?” Say yes. 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, and too impatient to solve the puzzle yourself, scroll down and see a proof!

However, that proof is long and confusing. With geometry I like demonstrations where after some thought you can simply see that the result is true. For this puzzle, such a demonstration was provided by Greg Egan.

Click here if you give up on this puzzle and want to see Egan’s solution. You may need to ponder it a bit, but when you’re done you should say “yes, this result is clear!

As a separate hint: the answers to this puzzle and the previous one are similar, in a very nice way!

18 Responses to Geometry Puzzles

  1. Todd Trimble says:

    (Rot 13) Sbe gur svefg chmmyr: nethr gung gur enqvhf bs gur fznyy pvepyr zhfg or unys gur enqvhf bs gur ynetr pvepyr, naq abgvpr gurer ner guerr ertvbaf pbatehrag gb gur oyhr ertvba. Sbe gur frpbaq: hfr gur snpg gung gur nern vafvqr gur fznyy frzvpvepyr vf unys bs gung bs gur ynetr frzvpvepyr.

    • To read Todd’s encrypted spoilers, shift each letter by 13. (‘s’->’f’). I hope Todd doesn’t mind me pointing this out, but if people are are needing help with the puzzles, then they might also need help with this.

      • Todd Trimble says:

        Don’t mind at all. When I tried clicking on my link to rot-13 (“rotate by 13”), it seemed to be broken (it used to be there, where you could just enter the text into a box and it would decode it for you). But the site is still there, so let me try again: rot-13.

  2. Todd Trimble says:

    (Rot 13) Sbe gur guveq chmmyr: guvf vf n qvfthvfrq irefvba bs gur Clguntberna gurberz, jurer gur nern vafvqr gur ynetr frzvpvepyr (jvgu qvnzrgre tvira ol gur ulcbgrahfr) vf rdhny gb gur fhzf bs gur nernf vafvqr gur fznyy frzvpvepyrf (jvgu qvnzrgref tvira ol gur yrtf bs gur evtug gevnatyr — abgvpr guvf vf va snpg n evtug gevnatyr!).

  3. Todd Trimble says:

    Oh, I’ve just clicked on Greg’s solution! Yes, that’s where I wanted to relocate the yellow semicircle all right — but I hadn’t seen the little folding maneuver quite so clearly as that. Very nice, Greg!

  4. Buddha Buck says:

    I had seen these when you posted them on Google+, but I must have missed the “Twice in a blue lune” one. It didn’t take me long to see the solution. Mine is basically the same as Todd’s.

    • John Baez says:

      ‘Twice in a blue lune’ is a new one, to make this article worth a look even if you’d seen the puzzles I posted on Google+. Now I’m going to post that one to Google+.

  5. Greg Egan says:

    These puzzles are great fun. Next time I visit a friend of mine who’s a painter I’ll have to show him the Lunes of Alhazen and challenge him to match da Vinci!

  6. JhM says:

    I am a bit disapointed. To my mind synthetic geometry was (is) the high water mark of mathematical reasoning. I see algebra as a cheap, inconsistent (symbols differ from one text to the other) means of arriving at a solution which is oft times only half correct (the quadratic formula, for example). You guys need to read anything and everything written by Sir Thomas Heath.

    • Todd Trimble says:

      Well, perhaps you could take this opportunity to show us how to do it right. Then again, I don’t know what exactly it is that you’re criticizing. What algebra are you referring to here?

  7. Reminds me of Sangaku; native Japanese mathematics.

    Click to access templegeometry.pdf

  8. @whut says:

    \sin ( u a) = \sqrt{u} \sin a

    I have a generic question. Has anyone encountered solutions to 2D problems that have this formulation ?

    f(x)=sin(k sin(x))

    And if so, does this formulation have a name?

    • John Baez says:

      Bessel functions are used to describe the standing waves for an oscillating disk:

      and they’re integrals of a function similar to yours, though not exactly the same:

      \displaystyle{ J_n (x) = \frac{1}{2 \pi} \int_{-\pi}^\pi e^{i(n \tau - x \sin(\tau))} \,d\tau}

      Of course that complex exponential has imaginary part

      \sin(n \tau - x \sin(\tau))

      and if we use the formula for \sin(a + b) the expression \sin(x\sin(\tau)) makes its appearance, multiplied by some other stuff.

      • @whut says:

        Thanks John, that helps a lot.

        BTW, that Bessel function also shows up in Frequency Modulation applications, as in FM radio

        \sin(\sin(x)) = J_1(1) \sin(x) + J_3(1) \sin(3x) + J_5(1) \sin(5x) + \dots

        I found the same StackExchange question and added it there:

        Those guys are kind of grumpy so I will see if it passes moderation.

        • @whut says:

          That formula parsed on StackExchange, is it the ending ellipsis ?

        • John Baez says:

          It was the ellipsis. Three periods works fine, and \dots is more classy, but you entered some sort of UNICODE ellipsis character.

        • @whut says:

          I also have a derivation of a standing wave equation different than yours that reduces to sin(sin). I think the physical significance of the behavior is that with the classical wave equation, increasing the forcing energy to the system results in an increasing amplitude of the sinusoid. But with a different formulation, increasing the amplitude results in frequency folding and so the extra energy goes into higher frequency harmonics.

  9. […] Azimuth Project (John Carlos Baez) Geometry Puzzles […]

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