I just ran across this in an old Math Horizons article — Andy Liu, vice president of the International Mathematics Tournament of the Towns, calls it “one all-time favorite geometric gem.” Given the four angles shown, compute angle CAD. “It sounds like a trivial exercise at first, and therein lies its charm.”
Liu doesn’t give the solution, but he does give a hint — I’ll put that in a spoiler box in case you want to work on the problem first.
If you fit Pascal’s triangle into a grid, then each cell displays the number of distinct paths that lead to that cell from the upper left (assuming only rightward and downward movements and no backtracking).
So, above, the cell marked 4 in the second row, fourth column, can be reached in 4 ways: 1-1-1-4, 1-1-3-4, 1-2-3-4, and 1-2-3-4.
A man who enters a public restroom has to make a complex choice quickly. He wants to choose a urinal that maximizes his chances of maintaining privacy — that is, that minimizes the chance that someone will occupy a urinal next to him. Which choice is best?
Computer scientists Evangelos Kranakis and Danny Krizanc modeled a number of strategies: lazily choosing the closest urinal that provides privacy; tacitly cooperating in the decision with other men; maximizing one’s distance from other occupants; and making the choice randomly. Happily, their findings support the general intuition:
Our main conclusion is that when faced with the decision of what urinal to choose upon entering the men’s room, in order to maximize your privacy, you should probably choose the one furthest from the door if it is available and the one next to it is unoccupied. For a vast majority of the (what we consider) natural behaviors that men choosing urinals might follow, this choice is optimal.
Related: In 1984 Donald E. Knuth noticed that the toilet paper dispensers in Stanford’s computer science department hold two rolls of tissue, both of which are available for use. Suppose there are two sorts of people in the world, those who are disposed to draw from the larger roll and those who draw from the smaller roll, and that each user takes exactly one sheet from his favored roll. What’s the expected number of sheets remaining just after one of the two rolls has been emptied? Donald E. Knuth’s Toilet Paper Problem.
(Evangelos Kranakis and Danny Krizanc, “The Urinal Problem,” in Paolo Boldi and Luisa Gargano, eds., Fun With Algorithms: 5th International Conference, Fun 2010, Iscia, Italy, June 2010: Proceedings.)
If two people want to split up amicably, the easiest solution is to divide their assets equally, with each partner getting 0.5. But suppose that one partner goes to a lawyer who charges a fee f but promises to get more, by an amount m + f, leaving his client better off by the amount m. If this happens, then the second partner will get only 0.5 – m – f. If the second partner engages their own lawyer then the split is equal again, except that now the lawyers’ fees must be paid:
This is an example of the so-called prisoner’s dilemma: Both sides would be better off if they left the lawyers out of it, but if one engages a lawyer than the other had better do so as well.
Now suppose that each partner can choose the amount of lawyer time to buy, and that they get a payoff that’s proportional to the amount they spend. If one spends x on lawyers and the other spends y, each measured as a fraction of the total assets, then the first partner should receive an amount given by:
An industrious divorcee can now use calculus to maximize this expression, varying x and keeping y constant. The optimum value of x turns out to be . If my partner spends 9%, or 0.09, of our assets on lawyers, then I should spend . Then my partner will get 0.21 of the assets, and I’ll get 0.49, and the lawyers get the rest.
Well, now what? Knowing all this, what’s our best course? If we could trust each other then we’d each pay a pittance on lawyers and get nearly 0.5 each. But I’m aware that if you pay a millionth and I pay a thousandth (still nearly a pittance), I’ll get nearly 99.9% of our assets. And simply resolving to outspend you won’t work: If you spend 0.36 then I should spend 0.24; I’ll come away with less than you, but this is the best I can do.
“Looking at the graph of , above, we (the author and reader) see that y = 0.25 gives us x = 0.25, and this gives us a sort of stability,” writes Anthony C. Robin in the Mathematical Gazette. “Neither partner can pull a fast one over the other, and it results in the assets being equally shared between us, them, and the lawyers. No doubt this is the reason why lawyers are so rich in our society!”
(Anthony C. Robin, “How Lawyers Make a Living,” Mathematical Gazette 88:512 [July 2004], 313-315.)
“Experience never misleads; what you are misled by is only your judgment, and this misleads you by anticipating results from experience of a kind that is not produced by your experiments.” — Leonardo
The area of the largest equilateral triangle that fits inside a unit square exactly equals the side of the largest square that fits inside a unit equilateral triangle.
This proof is due to John Conway. The largest square that fits inside a unit equilateral triangle is shown at the top; the side of the square is . The largest equilateral triangle that fits inside a unit square is one that’s canted at a 15° angle, shown at the bottom.
Now, the area of the shaded parallelogram is just its base times its height, or . And the parallelogram can be dissected into three pieces that fit precisely into the canted triangle. So the side of the square at the top equals the area of the triangle at the bottom.
Do animals reason? British psychologist C. Lloyd Morgan offered the case of his own dog:
Tony, the fox-terrier, already introduced to my readers, when he wants to go out into the road, puts his head under the latch of the gate, lifts it, and waits for the gate to swing open. Now an observer of the dog’s intelligent action might well suppose that he clearly perceived how the end in view was to be gained, and the most appropriate means for effecting his purpose. The following chain of ideas might be supposed to pass through the dog’s mind, not, indeed, in a clear-cut logical form, but at any rate in a rough and practically serviceable way: ‘Why does that gate remain shut? The latch holds it. I’ll lift the latch. Now it is no longer held, therefore it swings open.’ But is it necessary to assume that there were ideas involving, even in the most rudimentary way, the why and the wherefore? May not the action be quite well explained on the hypothesis that the dog acted under the sole guidance of sense-experience?
Two observers might debate that question for some time, Morgan wrote, one arguing that Tony’s feat must be based on experience and the other insisting that it showed an understanding of how the gate operates. But Morgan himself had had the benefit of seeing the trick evolve. “I was sitting at a window above the garden, and heard the dog put out of the door. I therefore watched him. He ran up and down the low wall, and put his head out between the iron bars, now here, now there, now elsewhere, anxiously gazing into the road. This he did for quite three or four minutes. At length it so happened that he put out his head beneath the latch, which, as I have said, is at a convenient height for his doing so, being about a foot above the level of the wall. The latch was thus lifted. He withdrew his head, and began to look out elsewhere, when he found that the gate was swinging open, and out he bolted.”
So Tony’s trick was the product of a fortunate accident, not abstract reasoning. Morgan proposed a general rule: “In no case is an animal activity to be interpreted in terms of higher psychological processes if it can be fairly interpreted in terms of processes which stand lower in the scale of psychological evolution and development.” This has become known as Morgan’s canon.
(From Morgan’s 1903 book An Introduction to Comparative Psychology.)
Russian science writer Yakov Perelman asks: How many times must an 8-toothed cogwheel rotate on its axis to circle around a 24-toothed cogwheel?
Meteors are more commonly seen between midnight and dawn than between dusk and midnight. Why?
In 1653, Blaise Pascal composed a triangular array in which the number in each cell is the sum of the two directly above it:
In 1915, Polish mathematician Waclaw Sierpinski described an equilateral triangle in which the central fourth is removed and the same procedure is applied to all the succeeding smaller triangles. Perplexingly, the resulting structure has zero area:
Interestingly, if the odd numbers in Pascal’s triangle are shaded, they produce an approximation to Sierpinski’s triangle:
And as this triangle grows toward infinity, it becomes Sierpinski’s triangle — an arrangement of numbers that takes the shape of a geometrical object.