Part of my motivation for writing this series is to make a confession: some of my constructions are just a mess built of geometrically calculating a length that I've determined algebraically. Typically, my approach is to create a coordinate system, then set up a couple of equations that determine a key length for the construction, solve those equations, then use geometric operations to construct that value.
Here is an example. Theoretically, it is a spoiler for a construction in mu pack (circle tangent to two circles 12.5) but I doubt anyone will really be able to see what is going on here:
Another example is construction of the regular pentagon. I know the golden ratio figures prominently, so one approach I often use is just to build that ratio between two lengths and then impose it on the basic construction template.
Other than massively overrunning the move targets for E and L stars, the weakness of this approach is that it doesn't link with any geometric insight about the construction. In a sense, this is the power of analytic geometry: you can get results without having to find a new insight.
Over the course of the series, I'll try to work on constructions that are geometrically insightful, but won't shy away from letting you know where I've had to push through with a brute force attack.
Beta pack is the only one for which we have received all of the stars. Even so, a couple of the puzzles are worth discussing because they illustrate some interesting ideas. In particular, we like Drop a Perpendicular (2.6) and Erect a Perpendicular (2.7).
Perpendicular bisectors
One of the ideas lurking is around the perpendicular bisector, which we saw way back in alpha pack (1.2). One of the things I always found interesting about the perpendicular bisector was that it was easier to construct an object with more conditions than either constructing a perpendicular alone or finding a midpoint alone.
The other great thing about the perpendicular bisector is that it can be defined in an entirely different way: it is the locus of all points that are equidistant from our two starting points. In case that isn't clear, assume we start with two distinct points A and B. The perpendicular bisector of the segment AB is also the locus of all points C such that distance AC is equal to distance BC.
Of course, the fact that this is a line means that we only have to find two such points to construct the perpendicular bisector (which is how you solved 1.2, right?)
Drop a perpendicular
For the 2 move solution using tools, I'll let you find a solution on your own. In case you need a hint: how many combinations of 2 moves are there anyway? You could just try them all and see what you find.
Way back in my HS geometry class, I learned a construction for dropping a perpendicular that uses 4 elementary moves. For the E start, though, that's not good enough. The cool idea that breaks through here is to choose two totally arbitrary points on the line as centers of circles that we draw. For some reason, I get a kick out of the idea that arbitrary points can be helpful ("if the point we choose doesn't matter, how can it help to choose a point anyway?")
In this case, while the points on the line we choose don't matter, the circle we draw with those points as centers need to have the right radius. Click the button below if you want to see how it is done and a bit more explanation.
The key here is that the circle radii are equidistant from our target point and the new intersection point of the two circles. That means they are on the perpendicular bisector of the segment between those points.
Another observation is that the new intersection of the two circles is the reflection of our target point through the line. Kind of cool that, no matter which two center points we choose for the two circles, the new intersection point will always be the same!
Erect a perpendicular
Erecting a perpendicular also uses the idea of choosing an arbitrary point, but goes a step farther. This time, we choose any point we want that is not on the line already! Well, it also fails if we happen to choose a point that is already on the perpendicular, but that should be impossible.... In any case, just choose a point somewhere off to the side.
This construction uses Thales' Theorem. I don't know exactly why I find this result to be so cool, since the proof isn't hard. For me, it transforms a circle into a family of right triangles. Given a length for the hypotenuse, all the right triangles with that hypotenuse are living right there on the arc of the circle with that length as the diameter.
Incidentally, Thales' Theorem and the two defining properties of perpendicular bisectors will come up a lot in other Euclidea Constructions.
V stars
If you want to know where the V stars are, click below:
Angle of 30 degrees (2.3) and Double Angle (2.4) both have two solutions.
Sometime in the past two years, Sue VanHattum, introduced us to Euclid the Game. This is a nice series of classical construction puzzles (compass and straight-edge) built on top of Geogebra. We recently returned to it and saw a link to another version that we've been playing a lot recently: Euclidea.
In EtG, there is a nice discussion in the comments section. Euclidea doesn't have this feature, so I decided to write blog posts chronicling some of our struggles and, hopefully, starting a place for discussion. I'm not planning to write about every challenge or post answers to all of them, but am happy to take requests. Otherwise, I'll write about the puzzles I find challenging and/or interesting for some reason.
Alpha Pack
I think there are two things worth talking about in alpha pack: (1) the puzzle that has stumped us and (2) the location of the V stars.
That darned square (or is it a diamond?)
The last puzzle in the alpha pack is the one that has stumped us. Our target is this inscribed square:
The kicker is to construct it in 7 elementary moves!
Our thought process
We need four line moves to draw the sides of the square. That means we have only 3 moves to find the other three vertices. We can get one by drawing the diameter of the circle, so we have two moves to find the other two vertices.
We can easily find those two side vertices with three elementary moves, but are really stuck on the idea needed to get one less move.
Some spoilers
Six move square
Our approach to get the construction in 8 elementary moves serves easily to get the 6 move construction. Below, I've included the finished picture which should be enough to see the approach (it isn't very involved anyway).
Some quick puzzles created and discussed with J2 while at lunch today. For all, we made the simplifying assumptions that everyone has their birthday on the same day.
Basics/assumptions
Our friends were discussing their current ages: Jin 9, Jate 7, Panelia 6, Sophia 5, and Jane 4. The first two are boys, the latter three girls. All have their birthday on the same day of the year and are having their birthday today!
Add them up
The boys wondered: will it or has it ever happened that the sum of our ages is the same as the sum of the ages of the girls?
Doubles
A little twist on the previous question: when will the sum of the girls ages be twice that of the boys?
For those that know algebra: can you make sense of the solution?
Halves
The previous puzzle was a bit strange. What if we go the other way: when will (or were) the sum of the ages of the boys twice that of the girls?
Murky waters with products
What is the current product of the boys ages? The product of the girls ages?
Have those products ever been equal? We argued this based on the intermediate value theorem.
When were those products equal? We numerically approximated to the closest half year.
Some other families
J2 asked me to include these, which we'd discussed during a dinner last week.
Bill is five years younger than his sister. In seven years, Bill will be 2/3 his sister's age. How old are they now?
John is ten years older than his brother Joe. In six years, John will be twice as old as Joe. How old is Joe now?
What are your favorites?
If you have any of these types of puzzles, please let us know in the comments!
Order of operations is a major pet peeve of mine for two reasons:
(1) Some people love to use it in "gotcha" challenges to make other people feel bad about their math abilities. Here are some examples: facebook meme, a similar one, and this:
(2) For some students, it stands as a clear example of the idea that math is a set of arbitrary rules they have to accept and/or memorize (It isn't!)
I think there is a very different way of approaching this issue which is much more mathematically rich and fun.
1a. Multiple answers
The first and most obvious is to treat these memes as games and see how many answers you can justify by making the order of calculation explicit (use parentheses). Implicitly, this is the idea behind games like 24 or the traditional New Year's challenge (use the digits of the new year to make all values from 1 to 100).
2. Talk about history
Some launching questions: Where did the order of operations come from? Is it universally agreed?
This article from Tara Haelle does a good job of talking about this perspective and gives some further references.
3. Ask students for their own thinking
What do they think the order of operations should be? Why?
A tantalizing question: should there be an agreed (implicit) order of operations at all?
4. Talk about redundancy and error flagging (and error correcting) codes
Jordan Ellenberg's How Not to Be Wrong has a great discussion about redundancy in language and related code concepts. One key point is that we always face a trade-off between brevity and transmission errors. In other words, we can write short messages where every character carries critical, independent information, or we can use a system in which our messages are longer, but carry duplication and internal references that make our meaning more robustly clear. (compare the previous two sentences!)
Relying on a convention, like the order of operations, means that we can use fewer symbols to convey a mathematical expression. The great danger comes if the author and the reader don't share the same conventions! A less obvious danger is if a symbol gets garbled in the transmission, it may be hard to identify the error or even see that there was an error.
These are my notes working through problems posted by Mike Lawler on his blog. You'll have to go there to see the problem statements.
The intended value of this write-up is to show examples of the actual problem solving thought process someone has followed, not just a polished solution.
Problem 19
Since the circle has area 156 π the radius squared is 156, which is 4 * 3* 13. That seems like a strange number, I'm curious to see where it will make calculations come out nicely, later in the problem.
We're told OA has length 4 sqrt(3). Squaring that only gives 48, so A is well within the circle. That means triangle ABC has vertex pointing toward O along the perpendicular bisector of side BC.
This lets me set up a picture of a right triangle with legs length $x$ (half the side of the equilateral triangle), $x$ sqrt(3) + 4 sqrt(3) (the altitude of the equilateral triangle plus OA) and hypotenuse r (2 sqrt(3*13)).
Applying the pythagorean theorem and simplifying along the way:
Visually factoring gets me $x$ is either -9 or 3. 3 is much more reasonable for half the length of the side of a triangle, so my answer is 6.
Note: If this weren't a contest problem, I might try to think about the -9 root a little more carefully.
Equiangular hexagon
I had already solved this problem from a recent tweet of Mikes, so I can't fully recreate my thought process. Here were some of the highlights:
wonder why 70%
draw a picture: fail to notice that triangle ACE is equilateral
Split the hexagon into two trapezoids by line CF
Calculate the area of those two trapezoids
Calculate the length from C the intersection with AE and segment CF.
Calculate the area of ACE based on the two subtriangles split by CF.
Obtain the quadratic equation for r based on the formulae for the two areas and the given 70% parameter.
School Competition
Let's call Andrea's rank $m$ for median. Since she is the unique median, there must be $2m - 1$ total contestants. We also know:
$m \leq 36$ because Andrea scored higher than Beth who was ranked 37th
$64 \leq 2m -1 $ because Carla ranked 64th, so there were at least that many competitors
$3 \mid (2m - 1)$ since every school sent three competitors
The first two inequalities tell us that $33 \leq m \leq 36$. Because $2m -1$ is a multiple of three, $m$ can't be a multiple of 3, so it has to be 34 or 35. Calculating mod 3, we can quickly check both and see that $m$ has to be 35.
That means the competition had 69 competitors from 23 schools.
Using multiple choice
After putting together these notes, I saw a commenter on Mike's blog use "solution by multiple choice." When I was doing timed tests/contests, I would make use of the options as part of my strategy. However, I don't do that now, since I'm much more interested in exploring the mathematics of the problems than getting the final answer.
