## Pythagorean circles

Here’s yet another Catriona Shearer problem:

But let’s start with a related picture, one with only two circles:

Again, the circles have unit radius. What are the dimensions of the rectangle?

The key to figuring this out is that if you have two lines through a point both tangent to a circle, then the line segments from the intersection point to the tangent points are congruent. We know the distance from the top right corner of the rectangle to the top tangent point on the yellow circle is 3. We don’t know the distance from the bottom right corner of the rectangle to the left tangent point on the yellow circle; call it x. Then the length of the diagonal is 3+x. By Pythagoras,

$(3+x)^2 = 4^2+(1+x)^2$

whose solution is x=2. So the height of the rectangle is 3, its width is 4, and its diagonal is 5: the diagonal and two sides make a 3:4:5 triangle, where 3:4:5 is a Pythagorean triple.

You can do the 3-circle problem the same way; here the length of the diagonal is 5+x, the equation is

$(5+x)^2 = 6^2+(1+x)^2$

whose solution is x=3/2. The rectangle then is 2.5 by 6 with diagonal 6.5. Those three numbers are in the proportion 5:12:13, another Pythagorean triple!

Coincidence? No. Of course if x is rational then the lengths of the sides and diagonal are in the proportion of a Pythagorean triple. But in general, with n circles, does x have to be rational? You might expect not. However, the solution for n circles is given by

$(2n-1+x)^2 = (2n)^2+(1+x)^2$

whose solution is

$x = \frac{n}{n-1}$

So the rectangle height:width:diagonal is

$1+\frac{n}{n-1}$ : $2n$ : $2n-1+\frac{n}{n-1}$

or

$\frac{2n-1}{n-1}$ : $\frac{2n(n-1)}{n-1}$ : $\frac{2n(n-1)+1}{n-1}$

which is the integer proportion $2n-1$ : $2n(n-1)$ : $2n(n-1)+1$, scaled by $n-1$. That’s a Pythagorean triple, and in particular it’s a triple in which the hypotenuse is one greater than the longer side. For four circles, for example, the triple is 7:24:25 and the rectangle’s dimensions are 7/3 by 8.

## Semicircles, semi mysterious

This puzzle by Catriona Shearer has been driving me nuts today:

I stared at a while, didn’t come up with anything good, and finally gave up and started looking at the solutions people were posting. They fell into three categories: Answers with no real attempt to show reasoning; algebraic solutions that were kind of opaque and didn’t really give any insight; and geometric solutions that depended on assertions I couldn’t see a justification for.

I still haven’t seen a solution that makes me particularly happy.

The geometric solutions in particular mostly looked like they assumed the left small semicircle’s base makes a 45° with the large semicircle’s base (or something equivalent). Which it does, but how do you know it does? In particular, it appeared people were assuming the square diagonal and the perpendicular bisector shown below are the same; in this diagram they’re not, because the large “semicircle” is really a semiellipse, but how do you know they do coincide when it’s a semicircle?

The clearest solution posted which I felt didn’t make any unjustified assumptions was Bilal Aydin’s. Here’s my version of it.

Let the radii of the small and large semicircles be $r$ and $R$. In the following diagram $D$ is the center of the large semicircle, $B$ is the center of the small one on the left, and $E$ is the center of the small one on the right. $C$ is the point where the base of the large semicircle is tangent to the angled semicircle, and radius $BC$ is perpendicular to that base. Since the base of the angled semicircle is a chord of the large one, its perpendicular bisector passes through $B$ and $D$.

Then $BE=2r$ and $BC = r$ so $CE = \sqrt{3}r$. Let $DE=x=R-r$. Now $(BD)^2 = r^2+(\sqrt{3}r-x)^2$. $AB=r$, so $AD^2 = r^2+r^2+(\sqrt{3}r-x)^2$. But $AD = R$. So

$r^2+r^2+(\sqrt{3}r-x)^2 = (r+x)^2$

$2r^2+3r^2-2\sqrt{3}rx+x^2 = r^2+2rx+x^2$

$4r^2= 2rx(1+\sqrt{3})$

$x = 2r/(1+\sqrt{3}) = r(\sqrt{3}-1)$

And then $R = x+r = \sqrt{3}r$. So the large semicircle is three times the area of each small one; the shaded area is $2/3$ of the whole.

That works, but it’s very mysterious. Stuff cancels on both sides for no obvious reason. And if you try to figure out $CD = \sqrt{3}r-x$ you find it’s just $r$, which means triangle $BCD$ is isosceles and the angled semicircle’s base is 45° from the large semicircle’s base. Those seem like things that should be easy to prove geometrically, and yet… how?

## Four hexagons

I had to stare at this puzzle a good long time before I got it:

I had a very hard time figuring out where to start. What finally broke it for me was getting rid of unnecessary complications. For instance, in both the white and the yellow hexagons, three of the vertices don’t enter into the problem. You can replace those hexagons with triangles.

Likewise only two vertices of each of the other two hexagons are important; you can replace them with line segments.

Once you’ve done that the solution becomes obvious, in the sense of “I only had to stare at it another five minutes before saying ‘oh duhh'”. These two line segments are congruent.

As are these two.

These two angles are congruent, since both differ from 60° by the same amount.

So these triangles are congruent.

Now put back the original hexagons.

A side of one hexagon is congruent to a diameter of the other, so it’s twice a side of the other, so the areas of the two hexagons are in a 4:1 ratio. The “pink” (or I’d call it magenta) hexagon’s area is 48.

## Identity 2

This is prompted by another Catriona Shearer puzzle. Would you say it’s at all obvious that

$2 \arctan (\sqrt{3}/7) = \arccos (23/26)$ ?

Me neither. True, though. Look at this triangle:

By the law of cosines, $3 = 13 + 13 - 2\times13\cos(\theta)$ so $\theta = \arccos (23/26)$. But if you slice that triangle in half, you find its height is $7/2$ and so $\theta = 2\arctan(\sqrt{3}/7)$.

More generally, the half angle formula for the tangent is

$\tan(\theta/2) = \frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}$

implying

$\cos\theta = \frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$.

Then if $\theta/2 = \arctan(a/b)$,

$2\arctan\left(\frac{a}{b}\right) = \arccos\left(\frac{b^2-a^2}{b^2+a^2}\right)$

and with $a=\sqrt{3}$, $b=7$ you get $2\arctan(\sqrt{3}/7) = \arccos(46/52) = \arccos(23/26)$.

## Semi similar

DE is diameter of a semicircle with center O. Triangle ADE intersects the semicircle at points B and C. Prove triangle ACB is similar to triangle ADE.

I keep feeling like there should be a simpler and more obvious proof but this is the best I’ve come up with:

$OB = OD \implies \angle OBD = \angle ODB \implies \angle DOB = \pi-2\angle OBD$

$OC = OE \implies \angle OCE = \angle OEC \implies \angle DOC = \angle DOB + \angle BOC = 2\angle OEC$

$OB = OC \implies \angle OBC = \angle OCB \implies \angle BOC = \pi - 2\angle OBC$

So $\pi-2\angle OBD + \angle BOC = 2\angle OEC$

$\implies 2\pi - 2 \angle OBD - 2 \angle OBC = 2 \angle OEC$

$\implies \angle OBD + \angle OBC + \angle OEC = \pi$

But $\angle OBD + \angle OBC + \angle CBA = \pi$

$\implies \angle OEC = \angle CBA$

and $\angle BAC = \angle EAD \implies \triangle ACB \sim \triangle ADE$ QED.

## Three rectangles

Catriona Shearer posted this problem to Twitter and it got a lot of discussion, so I thought I’d post my solution here in more detail than Twitter permits.

This design is made of three 2×1 rectangles. What fraction of it is shaded?

Label vertices and draw in a couple line segments:

Considering triangles CFJ and CKJ, angles CFJ and CKJ are right, line segment CJ is common, and CF = CK = 2, so the two triangles are congruent and FJ = JK = 1. But then angles CGE and JGK are equal, angles CEG and GKJ are right, and CE = JK, so those triangles are congruent and CG = GJ.

So JK = 1, and if KG = x, GJ = CG = 2-x, so 1²+x² = (2-x)². The solution is x = 3/4. The area of each shaded triangle is 3/8. The area of the whole pattern is three rectangles minus the shaded area: 6-3/4 = 21/5. The shaded area is 1/7 of the area of the whole pattern.

Notice the shaded triangles are 3:4:5.

Can the shaded area be dissected into pieces, seven of each of which will fill the pattern? Yes.

Here are the relative sizes of the pieces, in case you’re interested:

• Blue: 9:12:15
• Green: 12:16:20
• Red: 9:25:30:38
• Magenta: 10:16:38:40
• Shaded triangles (blue+red and green+magenta): 30:40:50

## 2^n-gons

There’s nothing at all new about this, in fact it’s ancient, but maybe it’s new to you? So here goes.

Here’s a square inscribed in a unit circle:

Or to keep down the clutter, here’s just one quadrant of that:

One side of the square along with the radii at each endpoint forms a triangle. The length of one side of the square we’ll call $s_4$ and it is of course $\sqrt 2$. That means the perimeter of the square is $p_4 = 4\sqrt 2\approx 5.656854249$.

Now consider an inscribed octagon. Again, here’s just one quadrant.

One side of the square is labeled $s_8$; what’s its length? Well, drop a perpendicular from the vertex in the middle:

And now notice the right triangle with hypotenuse 1 and sides $x_8$ and $h_8$ is just half of a quadrant of an inscribed square, which means $h_8=s_4/2$. Then $x_8^2 = 1-h_8^2 = 1-s_4^2/4$. From that you can get $y_8=1-x_8$ and from that, $s_8^2=h_8^2+y_8^2$. The perimeter of the octagon is then $p_8 = 8s_8\approx 6.122934918$.

Well, that was so much fun, let’s do it again. Here’s a quadrant of a 16-gon:

The right triangle with hypotenuse 1 and sides $x_{16}$ and $h_{16}$ is just half of an eighth of an inscribed octagon, which means $h_{16}=s_8/2$. Then $x_{16}^2 = 1-h_{16}^2 = 1-s_8^2/4$, $y_{16}=1-x_{16}$, and $s_{16}^2=h_{16}^2+y_{16}^2$. The perimeter of the 16-gon is then $p_{16} = 16s_{16}\approx 6.242890305$.

You know the rest, right? From here you can do the 32-gon, 64-gon, 128-gon, and so on, getting $x_{2n}^2 = 1-h_{2n}^2 = 1-s_n^2/4$, $y_{2n}=1-x_{2n}$, $s_{2n}^2=h_{2n}^2+y_{2n}^2$, and perimeter $= p_{2n} = 2ns_{2n}$. For $n = 1024$, the perimeter is $p_{1024}\approx 6.283175451 = 2\times 3.1415877255$. As $n$ increases, the perimeter gets closer and closer to the circumference of the circle, so this gives a way to calculate $\pi$.

What about circumscribed polygons? If you look at the figures above, you can see the ratio of the size of a circumscribed square to an inscribed square is $1/x_8$. Likewise the ratio of the size of a circumscribed octagon to an inscribed octagon is $1/x_{16}$, and so on. So the perimeter of a circumscribed $n$-gon is $P_n = p_n/x_{2n}$. As $n$ increases, $P_n$ approaches $2\pi$ from above. And the average of $p_n$ and $P_n$ is a better approximation to $2\pi$ than either, though not by a lot.

 n h_n x_n y_n s_n p_n P_n average 4 1.41421356 5.65685425 8.00000000 6.82842712 8 0.70710678 0.70710678 0.29289322 0.76536686 6.12293492 6.62741700 6.37517596 16 0.38268343 0.92387953 0.07612047 0.39018064 6.24289030 6.36519576 6.30404303 32 0.19509032 0.98078528 0.01921472 0.19603428 6.27309698 6.30344981 6.28827340 64 0.09801714 0.99518473 0.00481527 0.09813535 6.28066231 6.28823677 6.28444954 128 0.04906767 0.99879546 0.00120454 0.04908246 6.28255450 6.28444726 6.28350088 256 0.02454123 0.99969882 0.00030118 0.02454308 6.28302760 6.28350074 6.28326417 512 0.01227154 0.99992470 0.00007530 0.01227177 6.28314588 6.28326416 6.28320502 1024 0.00613588 0.99998118 0.00001882 0.00613591 6.28317545 6.28320502 6.28319024

This is something like the way Archimedes calculated $\pi$ around 250 BC (I told you this was ancient), although he started with a hexagon rather than a square and only went up to a 96-gon. He didn’t have Google Sheets, though.

## Three squares two triangles one circle

I’ve been trying lately to post my solutions to Catriona Shearer’s geometry puzzles in a tweet with no graphics; a severe constraint. For instance, this from yesterday:

This puzzle looked daunting at first but turned out to be easier than it looked. The tweet solution may be a bit too terse to be followed easily, though.

Here is a modified diagram:

I’ve labeled five points, and the sizes of the three squares: The largest square, touching the circle at point $A$, has size $a$; the medium square, touching at $C$, has size $b$, and the smallest square, touching at $D$, has size $c$. I added a copy of the smallest square, touching the circle at $E$, and drew in the chord $DE$.

The blue triangle is right, with legs $a$ and $c$, so its area is $(ac)/2$. The blue triangle is also right, with legs $a\sqrt{2}$ and $b\sqrt{2}$; its area is $(a\sqrt{2})(b\sqrt{2})/2 = ab$.

By the intersecting chords theorem, the product of lengths $AB$ and $BC$ is equal to the product of $BD$ and $BE$. But $AB = BE = a+c$, so $BC = BD = c$; that means $b = 2c$.

So the red triangle area is $2ac$ which is four times the area of the blue triangle. That area is 5, so the area of the red triangle is 20.

Sometimes, though, you just need to have a picture:

It’s obvious the red triangles’ height is $1/2$ but the blue triangles were less obvious to me. I noted that if you drop a perpendicular from the top vertex of the bottom blue triangle to the bottom line, the resulting right triangle has legs in a ratio of $1:2$ which sum to $1$, so the height is $1/3$. Another approach another poster mentioned is to note that a diagonal of the square is divided into three equal parts by the slanting lines and so the vertical projection of one of those parts has length $1/3$.

Either way, the red and blue triangles both have base = $1/2$ so a red triangle has area $1/8$, a blue triangle has area $1/12$, and four of each add up to $5/6$. Then the octagon’s area is the square’s ($1$) minus the triangles’ ($5/6$) which equals $1/6$.