I am sorry to report that December has now fallen across the land.
It is December on the beaches. It is December in the streets. It is very specifically and intensely December outside my bedroom window, where 3-foot death icicles dangle, ready to fall and impale me as I step out into the frigid December of it all.
(Yes, we’re loving Minnesota, thanks for asking!)
Where can we find warmth in this frigid time? Yes, yes, the Southern hemisphere, but more usefully: in the life-giving geometry problems of Catriona Shearer, Puzzle Magician.
Here are three of her favorites from November. Come, friend; heat yourself by this fire.
#1: Sci-Fi Stained Glass
Why do I dub this one “stained glass”? Because it has the delicate and symmetric beauty of a stained glass window, obviously.
But why “sci-fi”? That takes more explanation.
On Twitter, acclaimed sci-fi author Greg Egan chimed in with a crucial point: as stated, the problem is under-defined, allowing for an infinite family of solutions. Thus, Catriona added another condition: the whole diagram has bilateral symmetry.
To see why this is necessary, check out the nifty animation that Greg created. I, for one, am eager to watch the sci-fi film where all stained glass windows rotate like this.
#2: Kiddle’s Riddle
(Note to my fellow U.S. readers: please forgive the Britishism “trapezium.” Actually, scratch that: don’t forgive it, embrace it! Trapezium is an adorable word.)
Catriona explained the genesis of this one:
Alison Kiddle, who coordinates my local MathsJam, asked me a question at last month’s meeting: could I draw a trapezium so that when I drew in the diagonals the four areas each had a different area? Could I draw one where 2 areas were equal, or 3, or all 4?
I was pretty confident that I could, but [spoiler] after a fair bit of scribbling and failing to convince Alison (who is very good at asking questions!) I realised I shouldn’t have been quite so hasty.
Naturally I’d brought my felt tips to the pub, so by the time I left I’d made this puzzle from my discoveries.
#3: The House of the Square Sun
I found this one the trickiest of the three. Indeed, Catriona says that these three are “in ascending order of difficulty (in my mind, at least).” She also explains, tantalizingly:
Unless I missed it, I don’t think anyone posted the solution I originally came up with. It’s unusual for me to feel like my method is unique!
As always, feel free to post your solutions in the comments (which means you should watch out for spoilers if you don’t want them)!
All three seemed pretty easy to me, though I made the mistake on the first one of assuming that the squares in the center had the diagonals parallel to the sides of the main square (without that assumption or some similar one the problem does not have a unique solution). Making the area of the side squares 1, the red square has area 2, side length sqrt(2), which makes the middle side rectangles have diagonal sqrt(2), so they must be squares also, and the overall side length is 3, making 6/9 of the figure colored.
For the second one, label the top a and the bottom b. The blue+purple triangle has area bh/2 so the purple triangle has 2/3 (bh/2) = bh/3. Each blue triangle has area bh/6. The height of the purple triangle is 2/3 h, so the height of the white one is 1/3 h, and the area of the white triangle is ah/6. But white+blue triangle has area ah/2 = ah/6+bh/6, so area of white triangle is half the area of the blue. That gives us proportions 1:2:4 for white:blue:purple areas, and 8/9 of the figure is shaded.
The last one has a lot of red herrings. If we make x be the height of the red rectangle, then the red area is 12x, the square has area 24 x and the side of the square is sqrt(24 x). The right triangle within the red rectangle has two sides known, so the third side is sqrt(24x-x^2). The right triangle above the red rectangle is similar, so the vertical side is sqrt(24x-x^2) (sqrt(24x-x^2)/x) = 24-x. Thus the height of the green rectangle is 24. I got that answer for a couple of special cases first: the square parallel to the axes and the diagonals of the square parallel to the axes, so I’m pretty sure it is correct.
For the third one, extract the right triangle angle on the left side of the green rectangle. The line from the red square forms an altitude for this right triangle, dividing it into two similar triangles to the whole.
Let’s call the side of the square x, the height of the red rectangle y and the height of the green rectangle z. All of these can be placed in the right triangle we extracted earlier; the longest leg is x, the hypothenuse is z and the longest line part of the hypothenuse that gets divided by the altitude is y.
By similarity, we can state that x/y = z/x, thus z = x^2/y
The red rectangle (12*y) is half of the square (x^2), so x^2 = 24y. Plug that into the equation we got earlier and we get z = 24y/y = 24
Here’s the easy-bake response though;
Looking more carefully at the third one, even if we mentally rotate the square clockwise so it rests on its side, the rectangles would still form naturally, so there’s no issue with dividing by zero or anything like that.
So just plonk the square on its side. Now 12x = x^2/2 thus x = 24.
Links to the original tweets?
Ack, good point! Will add pronto.
Viva New Zealand!
Can non-mathematicians reply? I don’t quite see the puzzle in No. 1 (and the others are too hard for me). Is it assuming too much to start with the observation that one little square is 1/9 of the whole and that 6 times 1/9 equals 6/9 and that’s that?