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.

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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).

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.

nh_nx_ny_ns_np_nP_naverage
41.414213565.656854258.000000006.82842712
80.707106780.707106780.292893220.765366866.122934926.627417006.37517596
160.382683430.923879530.076120470.390180646.242890306.365195766.30404303
320.195090320.980785280.019214720.196034286.273096986.303449816.28827340
640.098017140.995184730.004815270.098135356.280662316.288236776.28444954
1280.049067670.998795460.001204540.049082466.282554506.284447266.28350088
2560.024541230.999698820.000301180.024543086.283027606.283500746.28326417
5120.012271540.999924700.000075300.012271776.283145886.283264166.28320502
10240.006135880.999981180.000018820.006135916.283175456.283205026.28319024

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.