Fibonacci and his birds (solution)

In the previous post, we described Fibonacci’s “problem of the birds” (“the problem of the man who buys thirty birds of three kinds for 30 denari”). In English:

“A man buys 30 birds of three kinds (partridges, doves, and sparrows) for 30 denari. He buys a partridge for 3 denari, a dove for 2 denari, and 2 sparrows for 1 denaro, that is, 1 sparrow for ½ denaro. How many birds of each kind does he buy?”

The man must buy at least one of each kind of bird, or he wouldn’t be buying “birds of three kinds.” Also, he must buy less than 10 partridges, because 10 partridges (at 3 denari each) would use up all his money. Similarly, he must buy less than 15 doves. We can thus make up a table of possible solutions:

Of those 126 possible solutions, only one works out correctly in terms of cost, and that’s the answer. But that’s an unbelievably tedious way of getting the answer, and you’d be rather foolish to try to solve the problem that way. The obvious approach is to use algebra. Write p for the number of partridges bought, d for the number of doves, and s for the number of sparrows. Because the man buys 30 birds, we have the equation:

p + d + s = 30

And because the costs add up to 30 denari, we have:

3 p + 2 d + ½ s = 30

Doubling that second equation gets rid of the annoying fraction:

6 p + 4 d + s = 60

If you’ve done any high school algebra, no doubt you want to subtract the first equation from this, which will eliminate the variable s:

5 p + 3 d = 30

But now what? That gives a relationship between the variables p and d, but there doesn’t seem to be enough information to get specific values for those variables.

Fibonacci solves the problem a different way. His solution is based on a key insight – the man buys 30 birds for 30 denari, so that the birds cost, on average, 1 denaro each. Fibonacci then makes up “packages” of birds averaging 1 denaro each. There are only two ways of doing this. Package A has 1 partridge and 4 sparrows (5 birds for 5 denari), and package B has 1 dove and 2 sparrows (3 birds for 3 denari). The solution will be a combination of those two packages.

Now the man can take 1, 2, 3, 4, or 5 copies of package A, leaving 25, 20, 15, 10, or 5 birds to be made up of package B. But the birds making up package B must be multiple of 3, so that the only possible answer is 3 copies of package A and 5 copies of package B. This means that the man buys 3 partridges, 5 doves, and 3×4 + 5×2 = 22 sparrows. That’s 30 birds and 3×3 + 2×5 + ½×22 = 30 denari.

Now it turns out that, had we kept on going with the algebraic approach, we would have gotten the same answer. We had:

5 p + 3 d = 30

Given that the numbers of partridges and doves (p and d) had to be positive whole numbers, that meant that p had to be a multiple of 3, and d a multiple of 5. That could only be achieved with p = 3 and d = 5.

We can also return to the diagrammatic approach. The equation:

5 p + 3 d = 30

describes the diagonal red line in the diagram below. That line only crosses one of the possible solutions, namely the dot corresponding to 3 partridges and 5 doves.

In mathematics, there’s more than one way to skin a cat. Or, in this case, a bird.


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Fibonacci and his birds

The mathematician Leonardo of Pisa (better known as Fibonacci) is famous for his rabbits, but I was recently reminded of his “problem of the birds” or “the problem of the man who buys thirty birds of three kinds for 30 denari.” This problem appears in his influential book, the Liber Abaci.

The “problem of the birds” is expressed in terms of Italian currency of the time – 12 denari (singular: denaro) made up a soldo, and 20 soldi made up a lira. In the original Latin, the problem reads:

“Quidam emit aves 30 pro denariis 30. In quibus fuerunt perdices, columbe, et passeres: perdices vero emit denariis 3, columba denariis 2, et passeres 2 pro denario 1, scilicet passer 1 pro denariis ½. Queritur quot aves emit de unoquoque genere.”

In English, that translates to:

“A man buys 30 birds of three kinds (partridges, doves, and sparrows) for 30 denari. He buys a partridge for 3 denari, a dove for 2 denari, and 2 sparrows for 1 denaro, that is, 1 sparrow for ½ denaro. How many birds of each kind does he buy?”

How many birds of each kind does the man buy? It may help to cut out and play with the bird tokens below (click image to zoom). In a similar vein, what if the man buys birds as follows (still purchasing birds of all three kinds, and at the same price)?

  • 4 birds for 6 denari
  • between 6 and 10 birds for twice as many denari as birds
  • 8, 11, 13–14, 16–22, 24–25, or 27 birds for the same number of denari as birds
  • 8 birds for 12 denari
  • 12 birds for 18 denari
  • 16 birds for 12 denari
  • 28 birds for 21 denari
  • 6, 8–9, or 14 birds for 11 denari
  • 7–10, 12, 15, or 18 birds for 13 denari

Solution to the main problem here.


Measuring the Earth this (Southern) Christmas

In around 240 BC, Eratosthenes calculated the circumference of the Earth. The diagram above (from NOAA) shows how he did it. This Christmas, people in the Southern Hemisphere can repeat his work!

Eratosthenes knew that, at the summer solstice, the sun would be directly overhead at Syene (on the Tropic of Cancer) and would shine vertically down a well there. He also knew the distance to Syene.

On 21 December, the sun will be directly overhead on the Tropic of Capricorn at local noon. This table show the time of local noon on 21 December 2017, and the distance to the Tropic of Capricorn, for some Southern Hemisphere cities:

City Local Noon Distance to Tropic (km)
Adelaide 13:14 1270
Auckland 13:19 1490
Brisbane 11:46 450
Buenos Aires 12:52 1240
Darwin 12:45 1220
Hobart 13:09 2160
Johannesburg 12:06 310
Melbourne 13:18 1590
Perth 12:15 940
Santiago 13:41 1110
Sydney 12:53 1160

At exactly local noon, Eratosthenes measured the length (s) of the shadow of a tall column in his home town of Alexandria. He knew the height (h) of the column. He could then calculate the angle between the column and the sun’s rays using (in modern terms) the formula θ = arctan(s / h).

You can repeat Eratosthenes’ calculation by measuring the length of the shadow of a vertical stick (or anything else you know the height of), and using the arctan button on a calculator. Alternatively, the table below show the angles for various shadow lengths of a 1-metre stick. You could also attach a protractor to the top of the stick, run a thread from the to of the stick to the end of the shadow, and measure the angle directly.

The angle (θ) between the stick and the sun’s rays will also be the angle at the centre of the Earth (see the diagram at top). You can then calculate the circumference of the Earth using the distance to the Tropic of Capricorn and the fact that a full circle is 360° (the circumference of the Earth will be d × 360 / θ, where d is the distance to the Tropic of Capricorn).

Height (h) Shadow (s) Angle (θ)
1 0.02
1 0.03
1 0.05
1 0.07
1 0.09
1 0.11
1 0.12
1 0.14
1 0.16
1 0.18 10°
1 0.19 11°
1 0.21 12°
1 0.23 13°
1 0.25 14°
1 0.27 15°
1 0.29 16°
1 0.31 17°
1 0.32 18°
1 0.34 19°
1 0.36 20°
1 0.38 21°
1 0.4 22°
1 0.42 23°
1 0.45 24°
1 0.47 25°
1 0.49 26°
1 0.51 27°
1 0.53 28°
1 0.55 29°
1 0.58 30°
1 0.6 31°
1 0.62 32°
1 0.65 33°
1 0.67 34°
1 0.7 35°
1 0.73 36°
1 0.75 37°
1 0.78 38°
1 0.81 39°
1 0.84 40°
1 0.87 41°
1 0.9 42°
1 0.93 43°
1 0.97 44°
1 1 45°

Logic in a box!

Having recently spent some time teaching a short course on logic and critical thinking, here is the core of the course reduced down to a box of 54 cards. These include:

  • 15 logic cards (summarising basic syllogistic and propositional logic rules),
  • 19 cards illustrating logical fallacies,
  • 5 cards for testing your ability to check validity, and
  • 15 logic-puzzle cards.

If you’re interested, more details can be downloaded from the game page (see the links in the “Downloads” section). The picture below shows some of the cards:


Education in the USA

The pie chart above shows the breakdown of elementary and secondary students in the USA. Data is for 2013, from the National Center for Education Statistics (with homeschooling numbers extrapolated from 2012 data). Educational options are a hot political issue right now, with changes to education policy likely under the Trump administration. What those changes will be is unclear.

Of course, the US educational system does need some improvement. In the 2015 PISA global education survey, the US ranked equal 23th in reading, 25th in science, and only equal 39th in mathematics. Canada did much better (2nd, 7th, and 10th), and Singapore came 1st in all three categories.


Chemical Compounds: the board game!

I have previously mentioned my strong interest in science / technology / engineering / mathematics education and in networks and in board games. This has prompted me to start designing educational games, such as the World Solar Challenge game. Joining the collection is my new Chemical Compounds game, which looks like this:

The online game store (faciliated by the wonderful people at The Game Crafter) has a free download link for the rules, should anyone wish to take a look. I also have a few other educational games there.


A scale model of the Solar System

The following images of the eight major planets and Pluto are to scale, with each image 500,000 km in width (the third image also includes the Moon). For comparison, the diameter of the Sun is 1,390,000 km, so you can team these planets up with a yellow circle 2.8 times the width of each image. Click on the images (which can also be found on the Homeschool Resources section of this blog) to zoom.

    
    
    

If you want distances from the Sun to each planet to be to scale as well (like this or this), they are shown below (in millions of km and multiples of the width of each image). If the images above are printed out 5 cm or 2 inches wide (i.e. at a scale of 1 cm = 100,000 km), then the sun would be 14 cm (5.5 inches) wide, and distances from the Sun to each planet would be as per the bottom two rows of the table.

58 108 150 228 779 1,434 2,872 4,495 5,906 million km
120 220 300 460 1,560 2,870 5,700 9,000 11,800 multiples
6 11 15 23 78 140 290 450 590 metres
19 35 49 75 260 470 940 1470 1940 feet

The image below (click to zoom) shows the complete solar system, but most planets are too small to see: