Tanya Khovanova's Math Blog

Tim Gowers discussed the age of consent on his blog, which I can no longer find. I will talk about his post here based on my old notes and my memory. The age of consent is a legal term to protect young people from being manipulated into agreeing to sex. Having consensual sex with people under the age of consent may be considered statutory rape or child sexual abuse.

Gowers starts with several assumptions.

• Non-triviality: There should exist an age at which a person is qualified to consent to sex and, consequently, have it.
• Simplicity: Whether or not two people are allowed to have sex with each other should depend only on their age in years.
• Monotonicity: If two people are allowed to have sex with each other today, they should be allowed to have sex with each other at all times in the future.

From these assumptions, the following theorem can be deduced.

Theorem. The only possible rule satisfying these assumptions would allow any two people to have sex with each other as long as they both reached some fixed age k.

There is a problem with this type of rule. Suppose k is 18. If two people who are slightly younger than 18 have consensual sex, they can’t both be predators. These are two children with raging hormones. There is no reason to punish anyone. Now imagine that one of the partners turns 18. Society would still consider this a Romeo-and-Juliet case and would tend not to punish such a partner. Now imagine a child younger than 18 having sex with a partner over 40. The older partner has no raging hormones, knows what they are doing, and probably knows how to manipulate little children into having sex. So, it might be desirable to have a rule that differentiates between these two cases. The rule would take into account the difference in ages while forgiving younger offenders and still punishing predators.

Consider the most common type of law to resolve this issue: Anyone older than 18 can have sex, and, in addition, a person who is not older than 20 can have sex with someone between the ages of 16 and 18. This law doesn’t satisfy monotonicity. It could be that one day the older partner is not yet 20, and the next day, oops, they have a birthday. So, as a birthday gift, they are not allowed to have sex with each other anymore.

Here is a simple idea to resolve the issue by having the law focus on the age gap instead of the age of the older partner. We can have an adjusted law: Anyone older than 18 can have sex, and, in addition, a person can have sex with someone between the ages of 16 and 18 as long as the age gap is not more than four years. This rule doesn’t satisfy the simplicity assumption above, but it is simple enough. It is close in spirit to the previous rule and satisfies monotonicity. The problem with this rule is continuity.

• Continuity: If the age gap between couple A is only slightly larger than the age gap between couple B, then couple A should not have to wait significantly longer to be allowed to have sex.

According to the adjusted rule, the couple with the age gap of four years and one day might have to wait two years longer to have sex than the couple with the age gap of four years. This seems unfair.

Tim Gowers suggests dropping the simplicity rule. We can use days rather than years. For example, the rule might be that if one person in a couple is under 18, but at least 16, and has age x, then the other partner has to be not more than age y, where for example, y − x = 4 + (x − 16)/2. So when one partner turns 16, their partner has to be not older than 20. When one partner is 16 and two months, the other cannot be older than 20 and three months. With the younger partner getting older, the allowable age gap is increasing slowly. By the time the younger partner is a day from turning 18, their partner can be almost five years older.

It might be complicated for two people to calculate if they are allowed to have sex according to this formula. But Gowers’ big idea was that apps and websites could do this easily: two people plug in their birthdays and know whether they are allowed to have sex.

Do you know what an isosceles tetrahedron is? I didn’t until recently. An isosceles tetrahedron has all of its faces congruent. Equivalently, all pairs of opposite edges are of equal length. Such tetrahedra are also called disphenoids. Here are some cute statements about them.

• If the four faces of a tetrahedron all have the same perimeter, then the tetrahedron is a disphenoid.
• If the four faces of a tetrahedron all have the same area, then the tetrahedron is a disphenoid.

Disphenoids have three nontrivial symmetries: 180-degree rotations around three lines connecting the midpoints of opposite edges.

Is it possible to have a tetrahedron with fewer symmetries than disphenoids? Yes, it is. Dan Klain just published a fascinating paper about such tetrahedra: Tetrahedra with Congruent Face Pairs. The results are so elegant and simple that I was surprised that they were new. I got curious and started to google aggressively. I found an official name for this kind of tetrahedron: phyllic disphenoid, but no theorems about them. Their name is quite unappealing: no wonder people didn’t want to study them much. Obviously, Dan wasn’t as aggressive at googling, so he didn’t find this official name and called them reversible tetrahedra. But my favorite name for them is the name Dan thought of but didn’t use in his paper: bi-isosceles tetrahedra.

Here is the picture of a bi-isosceles tetrahedron from Dan’s paper. It has two faces with sides a, b, and c, and two faces with sides a, b, and d. The edges of this tetrahedron are: a, a, b, b, c, and d. It has one nontrivial symmetry: a 180-degree rotation around the line connecting the midpoints of the unequal opposite edges c and d. The picture emphasizes this symmetry. The figure in the picture is a projection of a bi-isosceles tetrahedron onto a plane, such that the line of symmetry is projected onto the point of symmetry of the projection. The two cute statements above, about disphenoids, can be generalized to the case of bi-isosceles tetrahedra.

• If the four faces of a tetrahedron can be split into two pairs with the same perimeter, then the tetrahedron is bi-isosceles.
• If the four faces of a tetrahedron can be split into two pairs with the same area, then the tetrahedron is bi-isosceles.

The first property is trivial, while the second one is proven in Dan’s paper. Dan also shows how to calculate the volume of a bi-isosceles tetrahedron:

Here is a problem from the 1978 Kyiv Olympiad for 7 graders.

Is it possible to place seven points on a plane so that among any three of them, two will be at distance 1 from each other?

Puzzle. Two girls were born to the same mother, at the same time, on the same day, in the same month, in the same year, and yet somehow they’re not twins. Why not?

I won’t tell you the expected answer, but my students are inventive. They suggested all sorts of scenarios.

Scenario 1. There are two different fathers. I had to google this and discovered that, indeed, it is possible. This phenomenon is called heteropaternal superfecundation. It happens when two of a woman’s eggs are fertilized by sperm from two different men. Unfortunately for my students, such babies would still be called twins.

Scenario 2. The girls are born on the same date, but not on the same day. This could happen when transitioning from the Julian to Gregorian calendar. The difference in birth times could be up to two weeks. I had to google this and discovered that twins can be born months apart. The record holders have a condition called uterus didelphys, which means that the mother has two wombs. Unfortunately for my students, such babies would still be called twins.

Scenario 3. The second girl is a clone. This scenario can potentially happen in the future. Fortunately for that student, I suspect that such babies would be called clones, not twins.

I decided to invent my own scenario outside of the actual answer, and I did.

Scenario 4. Two girls are from the same surrogate mother, but they are not twins. I had to google this and discovered that this actually happened: Surrogate mother of ‘twins’ finds one is hers.

Sometimes life is more interesting than math puzzles.

In his article on Möbius strips, Martin Gardner included a cute construction.

Construction. Cut out a cross from a piece of paper. Then glue one pair of opposite ends to make a cylinder and the other pair to make a Möbius strip. Then Martin instructs, “Trisect the twisted band, then bisect the untwisted one, and open up for a big surprise!”

In my effort to reuse Möbius strips, I started making them out of zippers. So Martin’s construction had its destiny zipped. The first picture shows the construction before it is being dissected. I was quite happy with my plan to use zippers as it had its advantages over paper. For the surprise to work, the twisted band shouldn’t be cut completely. Meaning, the middle part of the Möbius strip shouldn’t be cut. I sewed my zipper monstrosity not to cause any ambiguities: there is nothing to cut, just unzip the zippers.

Little did I know that unzipping is not the issue, but zipping back up is. I tried to zip the surprise back up several times; all of them unsuccessfully, as you can see on the two other pictures. I finally had to invite the real expert — my grandson — to do it.

I used to hate crocheting. Now it’s been growing on me.

To prepare for this magic trick, take all the spades out of a deck and place them in the following order: seven, ace, queen, two, eight, three, jack, four, nine, five, king, six, and ten. Turn the assembled deck face down, so that the seven is on top. Now you are ready to do the trick.

Magic trick. Transfer the top card to the bottom of your deck and deal the new top card face-up on the table. Repeat this process until all the cards are dealt. And — abracadabra — the cards are dealt in order.

I showed this trick to my grandchildren, and they decided to reproduce it. They tried to calculate where each card goes, without too much success. Then my son showed them another trick: how to arrange the cards without calculation. He started building his arranged deck from the end of the trick with all the cards in order face-up on the table with the king on top. He took the king and put it face-down into his hand. Then he repeated the following procedure until all the cards were in his hand: He took the next card from the table and put it face-down on top of the one in his hand. Then he moved the card from the bottom of his deck to the top. And — abracadabra — the cards are arranged correctly for the trick.

Next time, I should ask my grandkids to show this trick with the whole deck.

Each time I teach my students about the Möbius strip, I bring paper, scissors, and tape to class. The students make Möbius strips, and then I ask them to cut the strips in half along the midline and predict the result.

Then we move to advanced strips. To glue the Möbius strip, you need one turn of the paper. An interesting experiment is to glue strips with two, three, or more turns. In this case, too, it is fun to cut them along the midline and predict the shape of the resulting thingy. My class ends with a big paper mess.

As you might know, I am passionate about recycling. So I have always wanted to buy Möbius strips that can be cut in half and then put back together. I didn’t find them, so I made them myself from zippers. Now I can unzip them along the midline and zip them back together. I hope to have less mess in my future classes. We’ll see.

I recently posted my new favorite probability puzzle from the fall 2019 issue of Emissary, submitted by Peter Winkler.

Puzzle. Alice and Bob each have a biased coin that flips heads with probability 51% and 100 dollars to play with. The buzzer rings, and Alice and Bob begin flipping their coins once per minute and betting one dollar on each flip against the house. The bet is at even odds: each round, each of them either loses or wins a dollar. Alice always bets on heads, and poor Bob, always on tails. As it happens, however, both eventually go broke. Who is more likely to have gone broke first?
Follow-up question: They play the same game as above, but this time Alice and Bob are flipping the same coin (biased 51% toward heads). Again, assuming both eventually go broke, who is more likely to have lost their money first?

One might assume that Bob obviously got a bad deal. He will lose his money very fast. So the answer to both questions must be that Bob loses his money first. If you know me, you should realize that I wouldn’t have given you this puzzle if the answer was that intuitive. I love counter-intuitive puzzles.

Bob has a disadvantage, so he is guaranteed to lose eventually. Alice has an advantage, so she might lose, or she might not. If she goes broke, that means there should be more tails in the flips than if she doesn’t go broke. How does this influence the second scenario, in which they use the same coin? As we are expecting a lot of tails in the flips, the situation should be better for Bob as compared to the first scenario. Given that I posed this puzzle, one might decide that in the follow-up question Bob doesn’t go broke first. Is it a tie?

But, if Bob is the first to go broke in the first scenario, why would I include the first part of the puzzle at all? If you know me, you should wonder what makes the first question interesting. Or maybe, you know Peter Winkler, in which case you should also wonder the same thing.

I gave you enough meta information to guess the amazing counter-intuitive answers to these two questions: in the first scenario, they tie; and in the second scenario, Alice goes broke first with higher probability.

To prove that they tie in the first scenario, let me first define a switch on a sequence of coin flips. A switch changes each head into tails and vice versa. The switch creates a one-to-one correspondence between the sequences of flips where Bob goes broke with the sequences where Alice goes broke. Suppose p is the probability that Alice goes broke with a particular sequence a, and q is the probability that Bob goes broke with a `switched’ sequence b. Then p = rq, where r = (49/51)²⁰⁰. The reason is that sequence a has exactly 200 more tails than sequence b. The same ratio r works for any pair of switched sequences. Bob is guaranteed to go broke eventually. But to calculate the conditional probability of Alice going broke with sequence a, given that she does go broke, we need to divide the probability of the sequence a occurring by the probability that Alice goes broke. The resulting probability distribution on sequences of length n has to be the same as for Bob because it has to sum to one. If Alice goes broke, then the probability that she goes broke after n flips is the same as for Bob. That means they tie.

To answer the second question, we use the switch again. The switch creates a one-to-one correspondence between sequences of flips where Alice goes broke first and Bob second, and vice versa. The sequence for which Alice loses second has 200 more tails than the corresponding sequence where Bob loses second. Thus, in each pair of two switched sequences, the probability of the sequence where Alice loses second is equal to r times the probability of the sequence where Bob loses second. Thus, the probability of Alice winning is r times the probability of Bob winning, and r is a very small number.

Here are my newly-made Borromean rings: two identical sets of them. They are an example of three linked rings, with any two of them not linked. The top set is positioned the way the Borromean rings are usually presented. You can see that any two rings are not linked by mentally ignoring the third one. For example, the red ring is on top of the green one, the green one is on top of the blue one, and the blue one is on top of the red one. They have a non-transitive ordering.

The bottom set of rings is arranged for a lazier thinker. Obviously, the blue and green rings are not linked.