Archive for December, 2016
I’m between 6’1” and 6’4”, depending on which convenience store I’m leaving. My wife, on the other hand, doesn’t frequent convenience stores, because she can barely see over the counter, not to mention her extreme disdain for Slurpees.
The discrepancy between our heights has its pros and cons. On the plus side, I can reach the pasta pot on the top shelf in the kitchen, and she barely has to bend over to retrieve the Tupperware containers from the bottom shelf. On the down side, we both strain our necks when kissing unless she’s standing on the second step.
Moreover, our sons were measured at 33” tall at their two-year check-ups, which puts them in the 20th percentile. This is when I realized that marrying a woman only 5’2” tall makes it unlikely that either of my boys will earn a spot on the varsity basketball team.
This was a distressing thought. If my sons continued to track at the 20th percentile, they’d reach a height of only 5’7” as adults. When I mentioned this to my mother-in-law, she attempted to allay my fears. “Kids always grow up to be twice their height at age 2,” she told me. That formula predicted their adult height to be 5’6”, which did not reassure me in the least, despite the fact that I didn’t believe it for a second. (Recently, I learned that the Mayo Clinic actually considers this a reasonable formula for predicting height, although they claim that it only works for boys. For girls, they suggest doubling the height at age 18 months.)
Growing up, the shortest person in my family was my oldest sister, who is 5’10” tall. Thinking that my sons might not even reach 5’8” was a major concern for me. As a result, I began to do some research. Although I knew I could not alter the course of their physical development, I had to at least know if they stood a chance at being above average in height.
As it turns out, predicting the adult height of a child is very difficult, and not just for kids whose growth has been stunted by wrestling or gymnastics. I found no fewer than a dozen formulas in widespread use, yet none of them were very reliable. Which is to say, none of them gave the results I was looking for. I spoke with multiple pediatricians who said the best predictor of a child’s eventual height is the mean of the mother’s and father’s heights. That didn’t sit well with me, though, partially because it just seemed too easy, but mostly because it said my sons would be 5’8” tall at adulthood. Still shorter than I’d hoped, so I looked some more.
The Mayo Clinic suggests the following method to predict boys’ height:
- Add the mother’s height and the father’s height.
- Add 5 inches to that sum.
- Divide by 2.
This formula may not be superior to the “double height at age 2” formula above, but at least it felt more mathematically rigorous. Sadly, it predicted that my sons would reach a height of 5’8½“, which is better but still not what I was looking for. So, my search continued.
After several months of research, I finally stumbled on a formula that I thought I could trust. It began like many others, taking the mean of the mother’s and father’s height. But then it compensated for height differences due to gender. For boys, multiply the mean by 13/12; for girls, multiply the mean by 12/13. That is,
where m is the mother’s height and f is the father’s height. It roughly means that a boy has an extra inch added to the prediction for every foot in the mean, whereas a girl has an inch subtracted.
Finally, I had stumbled on a formula that predicted my sons would be above average, stretching the tape to just shy of 6’2” tall. This was a win-win, too — this formula predicted that they would be tall, but not quite as tall as their dad.
What’s interesting about this formula is its long-term prognostication. It implies that the human population will continue to increase in size indefinitely. Let me explain.
Google says that the average man is 5’6” tall and the average woman is 5’2” tall. This means that the average human is approximately 5’4”, or 64”, tall. Then an average man and average woman would have an average son who is 69.33” tall and an average daughter who is 59.07” tall — according to the formulas above — and the average human would be 64.20” tall. When this next generation has children, the sons would be 69.55” tall and the daughters would be 59.26” tall, and the average human height would be 64.41” tall. Do you see what’s happening? The average height is increasing slightly with each generation.
It’s not hard to see why this happens. The average of 13/12 and 12/13 is
which means that the average height will increase by 0.3% with each generation. Admittedly, that’s not much, but it portends almost certain doom! By Y3K, the average human will be almost 6’3” tall; and by Y4K, the average human will be nearly 7’3” tall.
There is some historical data to suggest that human height has been increasing for quite some time. In the early 18th century, researchers found that the average English male was 5’5” tall; today, the average English male is 5’9” tall. That’s an increase of about 0.6% per generation. Further, archaeological research shows that the average Greek woman was about 153 cm tall, and the average Greek woman today is closer to 165 cm, an increase of about 0.07% per generation.
Should we expect humans to someday stretch the tape to 8’ or more? Probably not. The paragraph above is a little lesson on how to lie with statistics. While it may be the case that the modern Greek beauty Helena Paparizou (5’7”) is taller than ancient beauty Helen of Troy (5’6”), it hasn’t necessarily been a steady increase. The average female height of a Greek woman in the 10th century was over 162 cm; just two centuries later, the average height dipped below 160 cm.
While the formulas above may serve to make reasonable predictions now, they may not work so well in the future. So if you’re 5’4” tall, don’t be bummed if your great-great-great-great-great-great-grandchild still isn’t able to dunk.
Alex and Eli are now 9 years old, and both of them are 4’3” tall. I strongly suspect that neither of them will reach 6’0”, and that’s just fine. My wife may not be tall, but she is kind, smart, funny, and compassionate, traits that she has generously shared with her children. As a result, my sons are well above average in ways that actually matter.
Through the Academic and Creative Endeavors (ACE) program at their school, my sons participate in the Math Olympiad for Elementary and Middle School (MOEMS). While passing the door to the ACE room yesterday, I noticed a sign with the names of those who scored a perfect 5 out of 5 on the most recent contest — and my sons’ names were conspicuously absent. Last night at dinner, I asked Alex and Eli what happened, and they told me about the problem that they both missed. (What? Like we’re the only family in America that discusses math problems at the dinner table.) Here’s how they explained it to me:
Nine 1-cm by 1-cm squares are arranged to form a 3 × 3 square, as shown below. The 3 × 3 square is divided into two pieces by cutting along gridlines only. What is the greatest total (combined) perimeter for the two shapes?
The answer to this problem appears below. Pause here if you’d like to solve it before reaching the spoiler.
[Ed. Note: I didn’t see the actual exam, so the presentation of the problem above is based entirely on my sons’ description. Apologies to MOEMS for any substantive differences.]
This problem epitomizes what I love about math competitions.
- The answer to the problem is not obvious. This is the case with many competition problems, unlike the majority of problems that appear in a traditional textbook.
- The solution does not rely on rote mechanics. Again, this differentiates it from a standard textbook problem or — shudder! — from the problems that often populate the databases of many skill-based online programs.
- Students have to get messy. That is, they’ll need to try something, see what happens, then decide if they can improve the result.
- Students have to convince themselves when to stop. Or more precisely, they’ll need to convince themselves that they’ve found the correct answer. For instance, let’s say a student divides the square into a 3 × 2 and a 3 × 1 rectangle. The combined perimeter is 18 units. Is that good enough, or can you do better? This is different from, say, a typical algebra problem, for which students are taught how to check their answer.
The problem also epitomizes what many people hate about math competitions.
- There’s a time limit. Students have 26 minutes to solve 5 problems. Which means that if students spend more than 5 minutes on this one, they may not have time to finish the other four. (There was a student of John Benson who, when asked about his goal for an upcoming math competition, replied, “I hope to solve half the problems during the competition and all of them by the end of the week.” That’s the way mathematicians work.)
- It’s naked math. Sorry, nothing real-world about this one. (But maybe that’s okay, because real may not be better.)
- The problem is presented as a neat little bundle. This is rarely how mathematics actually works. True problems often don’t present themselves all at once; it’s through investigation and research that the constraints become known and the nuances are revealed.
All that said, I believe that the pros far outweigh the cons. Benjamin Franklin Finkel said, “Many dormant minds have been aroused into activity through the mastery of a single problem.” I don’t remember the last time a mind was aroused by the solution to x + 7 = -3, but I’ve witnessed awakenings when students solve problems like the one above.
And here’s the tragedy in all of this: Many teachers believe that only kids who participate in math competitions can handle — or appreciate — math competition questions. No! Quite the opposite, in fact. Students who have tuned out have done so because they’ve never been challenged and, worse, have never felt the thrill of solving a problem on their own.
What I really love about the problem, though, is it made me think about other questions that could be asked:
- How many ways are there to divide a 3 × 3 square into two pieces that will yield the maximum total perimeter?
- What is the maximum total perimeter if a 3 × 3 square is divided into three pieces?
- What is the maximum total perimeter if a 4 × 4 square is divided into two pieces? …a 5 × 5 square? …a 6 × 6 square?
- The answer to the problem about the 4 × 4 square appears below. Pause here if you’d like to solve it before reaching the spoiler.
- (wait for it) What is the maximum total perimeter if an n × n square is divided into two pieces?
It was that last question that really got the blood pumping.
Here’s a solution for how to divide a 3 × 3 square to yield the greatest total perimeter:
And here’s a solution for how to divide a 4 × 4 square to yield the greatest total perimeter:
What’s the solution for an n × n square? That’s left as an exercise for the reader.