## Power chords

Here’s our next Catriona Shearer puzzle.
8.
Power Chords

What’s the area of the circle?
Spoiler!

“I never learnt the intersecting chords theorem at school, so I love anything where I get to use it!” — CS

Continue reading “Power chords”

## A couple more

Another two of those Catriona Shearer puzzles.

5.
Shear Beauty

The area of the bottom left square is 5. What’s the area of the blue triangle?

6.
All Men are Created Equilateral

Spoilers!

## Semicircle Turducken

Here’s another of those Catriona Shearer puzzles that had me stumped for a while.

7.
Semicircle Turducken

This looks ridiculous. There are no lengths, areas, or angles given, but you’re supposed to figure out an angle?

Spoiler!

## The Garden of Clocks

Last week Math With Bad Drawings published a bunch of geometry puzzles by Catriona Shearer. I’m about ready to talk about the first one!

1.
The Garden of Clocks

Spoilers!

## Easy foorp

That proof can also be done sort of backwards and is arguably easier that way. Instead of using three triangles of known dimensions and proving they can be assembled into a rectangle, start with the rectangle and find the dimensions of the triangles.

Take a right triangle with sides $a$$b$, and $c$. Draw perpendiculars to the hypotenuse at each end and a parallel line through the third vertex to make a rectangle $BCDE$.

Angles $CDA$ and $EAD$ are congruent (alternate interior angles). So are $ADE$ and $ACD$ since both are complementary with $ADC$. So triangles $ADE$ and $ACD$ are similar; the hypotenuses are in the ratio $b/c$, so the vertical leg $DE = a(b/c)$ and the horizontal leg $AE = b(b/c)$. Likewise we find $BC = b(a/c)$ (as it should be) and $AB = a(a/c)$. But then $BE = CD$ implies $a(a/c)+b(b/c) = c$ or, multiplying though by $c$$a^2+b^2=c^c$.

(Right away you know this proof fails in non Euclidean geometry, because in the first steps it relies on there being one and only one line parallel to $CD$ through $A$.)

## Easy proof

Sheesh, how have I lived this long and not seen this proof of the Pythagorean Theorem?

Here’s a right triangle, with legs $a$ and $b$ and hypotenuse $c$.Here’s the same triangle scaled up by a factor of $a$, so the legs are $a^2$ and $ab$ and the hypotenuse is $ac$.And here it is twice more, scaled up by factors of $b$… and $c$.Of course these all are similar triangles.

Now, notice the hypotenuse of the first scaled-up triangle is the same length as one leg of the third, so we can draw them juxtaposed like this:Capital letters label the vertices.

Since the triangles are similar, angles $ADC$ and $ACB$ are congruent, and since they’re right triangles, angles $ADC$ and $ACD$ add to a right angle; therefore angle $BCD$ is a right angle.

Similarly the hypotenuse of the second scaled triangle is congruent to the other leg of the third, so they can be juxtaposed as well:and as before, angle $CDE$ is a right angle. Also, angles $DAE$ and $BAC$ sum to a right angle so angle $BAE$ is two right angles, i.e.$B$, $A$, and $E$ are colinear. Therefore $BCDE$ is a rectangle, and $BE = CD$. But $BE = a^2+b^2$ and $CD = c^2$. So $a^2+b^2=c^2$. QED.

What I like about this proof is how easy it is to grasp visually. Most proofs of the Pythagorean Theorem deal with areas, specifically persuading you the areas of two squares sum to the same as the area of a third square, but summing areas isn’t easy to visualize. Here there are no areas, just lengths, and we all can see immediately that $a^2+b^2$ is just $BE$.

(I saw this proof on, um, some blog recently. And failed to make note of exactly where. Sorry.)

## Counting uniforms

Euclid’s “proof” that there are exactly five regular (convex) polyhedra is well known and pretty simple. Observe that there have to be at least three faces meeting at each vertex, and the angles of those faces have to add up to less than 360°. For instance, three hexagons surrounding a vertex add to 360° and they lie flat; they don’t make a corner. Three heptagons don’t even fit in 360°. So a vertex of a regular polyhedron can be three, four, or five triangles, three squares, or three pentagons — that’s all. And there are indeed regular polyhedrons of each of those five types: tetrahedron, octahedron, icosahedron, cube, dodecahedron. And there are no others.

“Proof” is in quotes, because Euclid omitted one thing: He didn’t prove there can’t be more than one polyhedron for any vertex type. There can’t be, and that’s pretty clear to anyone who’s ever cut out triangles and squares and pentagons and tried putting them together, but the rigorous proof of it didn’t come until, I think, around the 17th century.

What about the uniform (convex) polyhedra, though? Recall these are polyhedra with regular but not congruent (convex) polygons as faces, and congruent vertices. Actually the vertices have to be transitive, which is a stronger requirement, but for now we’ll just talk about congruence. I wrote before about how you can derive the Archimedean solids — the uniform polyhedra other than the prisms and antiprisms — from the regular polyhedra by truncation, rectification, and alternation, but how do we know we got them all? Let’s try counting them similarly to the regular polyhedra by looking for valid vertex types.

Can we limit which kinds of polygons are used in a uniform polyhedron? No. Because now instead of requiring at least three of them to fit around a vertex, all we have to require is that one of them plus at least two other polygons fit around a vertex. And since the interior angle of any (convex) polygon is less than 180°, you can always use one with three triangles or two squares.

So what other limits do we have? Well, first, the maximum number of faces at a vertex is five, and if you have a five-face vertex and one face is a square or larger, then the rest have to be triangles. So the possible uniform 5-vertices, and their angle sums, are

3-3-3-3-4 [330°]
3-3-3-3-5 [348°]

Let’s think about the 3-vertices. If all three faces are the same then it’s a regular vertex, not uniform, so either two are the same and one is different or all three are different. The vertex is either a-b-b  (or the equivalent b-b-a or b-a-b) or a-b-c where different letters correspond to different values.

Suppose it’s a-b-c. Consider an a face: at each vertex around the face, the adjacent face to the left must be b and the adjacent face to the right must be c, or vice versa. So if one adjacent face is b the next is c, then the next is b, and so on: they alternate all the way around between b and c. But if a is odd then this doesn’t work: the last adjacent face will be b and you’ll have an a-b-b vertex instead of a-b-c. So if a, b, and are all different, then a must be even. But by the same reasoning, b and c must be even too.

If we have a-b-b then all the faces surrounding a are b: there’s no alternation, so a can be odd or even. But in that case consider a b face. An adjacent a face must be followed by a b, which must be followed by an a, and so on: again there’s an alternation around the b face, which doesn’t work if b is odd. Therefore b must be even.

All right. So we can only have a-b-b where b is even, or a-b-c where ab, and c are all even. In other words, we have to have either two equal even faces, or three unequal even faces.

Taking that and the requirement that the angle sum be < 360⁰ into account, the only possible uniform 3-vertices are:

3-6-6 [300°]
3-8-8 [330°]
3-10-10 [348°]
4-4-x for x = 3 or x > 4 [360°-360°/x]
4-6-6 [330°]
4-6-8 [345°]
4-6-10 [354°]
5-6-6 [348°]

Now the 4-vertices. Here either at least one face is unique or there are two of one kind and two of another. Also at least one face must be a triangle to fit in 360⁰.

If we have two each of two kinds of faces it’s either a-a-b-b or a-b-a-b. Without loss of generality, a = 3But then in the first case the faces around a triangle face alternate between 3 and b, and that fails since 3 is odd. In the second case the a faces are surrounded by b faces and vice versa, so there’s no problem: 3-b-3-b is viable. That (and the angle sum requirement) gives us:

3-4-3-4 [300°]
3-5-3-5 [336°]

Otherwise we can have: a-b-c-da-b-c-c, a-b-b-c, a-b-c-b, or a-b-b-b, where different letters are unequal.

In the first case, an a face must have b and d faces alternating around it, so a must be even. So must bc, and d by the same logic. But we know one face must be a triangle, so that’s ruled out.

Similarly, in a-b-c-c, a and b must be even. So must c, because if you think about it, adjacent faces around a c face must alternate between c and either a or b. Again, though, we know there’s a triangle, so this is ruled out. So is the equivalent a-b-b-c.

In a-b-c-bb must be even to allow alternation between a and c. That gives:

3-4-5-4 [348°]

And a-b-b-b gives:

3-3-3-x for x > 3 [360°-360°/x]
3-4-4-4 [330°]

So those are our allowed vertices, and sure enough, there’s a uniform polyhedron corresponding to each:

 3-6-6 Truncated tetrahedron 3-8-8 Truncated cube 3-10-10 Truncated dodecahedron 4-4-x Prisms 4-6-6 Truncated octahedron 4-6-8 Truncated cuboctahedron 4-6-10 Truncated icosidodecahedron 5-6-6 Truncated icosahedron 3-3-3-x Antiprisms 3-4-3-4 Cuboctahedron 3-4-4-4 Rhombicuboctahedron 3-4-5-4 Rhombicosidodecahedron 3-5-3-5 Icosidodecahedron 3-3-3-3-4 Snub cube 3-3-3-3-5 Snub dodecahedron

Now, as with the regular polyhedra, we’ve only shown how many valid uniform vertices there are; we haven’t proved there’s only one polyhedron for each vertex. In fact we can’t prove it, because it’s not true. There are two polyhedra with 3-4-4-4 vertices: the rhombicuboctahedron and the pseudorhombicuboctahedron. The latter’s not counted as a uniform polyhedron because while its vertices are congruent, they aren’t transitive.

So the proof runs a little short: you need to show that, aside from the pseudorhombicuboctahedron, there are no other polyhedra, or at least no vertex transitive ones, with any of these vertices. (There are no others.) But at least we know what vertices are allowed.