Some Math Problems Seem Impossible. That Can Be a Good Thing.

Date : June 11, 2021
Category(ies) : Classroom practices

Struggling with math problems that can’t be solved helps us better understand the ones we can.

Construct a convex octagon with four right angles.

It probably says a lot about me as a teacher that I assign problems like this. I watch as students try to arrange the right angles consecutively. When that doesn’t work, some try alternating the right angles. Failing again, they insert them randomly into the polygon. They scribble, erase and argue. The sound of productive struggle is music to a teacher’s ears.

Then they get suspicious and start asking questions. “You said four right angles. Did you really mean three?” “Are you sure you meant to say convex?” “Four right angles would basically make a rectangle. How can we get four more sides in our octagon?” I listen attentively, nodding along, acknowledging their insights.

Finally someone asks the question they’ve been tiptoeing around, the question I’ve been waiting for: “Wait, is this even possible?”

This question has the power to shift mindsets in math. Those thinking narrowly about specific conditions must now think broadly about how those conditions fit together. Those working inside the system must now take a step back and examine the system itself. It’s a question that’s been asked over and over in the history of math, by those working on problems ranging from squaring the circle to circumambulating the city of Königsberg. And it’s a question that helps us shape what mathematics is and how we understand it.

For example, finding an octagon with certain properties is a very different mathematical task than showing that no such octagon could possibly exist. In playing around with different octagons, we might just stumble upon one that has four right angles.

A non-example. This octagon does not really have four right angles.

But luck doesn’t play a role in proving that such an octagon can’t exist. It takes deep knowledge, not just of polygons, but of mathematics itself. To consider impossibility, we need to understand that just asserting that a thing exists doesn’t make it so. Mathematical definitions, properties and theorems all live in a tension born of interconnectedness. In trying to imagine our octagon with four right angles, we work inside those interconnected rules.

But to realize our octagon is impossible, we need to step back and look at the big picture. What mathematical and geometric principles might be violated by an octagon with four right angles? Here, the polygon angle sum theorem is a good place to start.

The sum of the interior angles of an n-sided polygon is given by the formula:

S = (n – 2) × 180º

This is because every n-sided polygon can be cut into (n − 2) triangles, each with total internal angles of 180º.

For an octagon, this means the interior angles add up to (8 – 2) × 180º = 6 × 180º = 1080º. Now, if four of those angles are right, each with a measure of 90º, that accounts for 4 × 90º = 360º of the angle sum, which leaves 1080º – 360º = 720º to divide up among the octagon’s remaining four angles.

That means the average measure of those four remaining angles must be:

720º4 = 180º

But the interior angles of a convex polygon must each measure less than 180º, so this is impossible. A convex octagon with four right angles cannot exist.

Proving impossibility in this way requires stepping back and seeing how different mathematical rules — like the polygon angle sum formula and the definition of a convex polygon — exist in tension. And since proofs of impossibility rely on thinking broadly across rules, there’s often more than one way to construct the proof.

Let’s take a look at our earlier remark about four right angles making a rectangle.

If an octagon had four right angles, walking around just those angles would bring us full circle: It would be as though we had walked completely around a rectangle. This insight leads us to a rule that gives us a different proof of impossibility. It is known that the sum of the exterior angles of a convex polygon is always 360º. Since an exterior angle of a right angle is also a right angle, our four right angles would take up the entire 360º of the octagon’s exterior-angle measure. This leaves nothing for the remaining four angles, again establishing that our octagon is impossible.

Proving that something is impossible is a powerful act of mathematics. It shifts our perspective from that of rule follower to that of rule enforcer. And to enforce the rules, you must first understand them. You need to know not just how to apply them, but when they don’t apply. And you also need to be on the lookout for situations where rules might conflict with one another. Our octagon exploration exposes the interplay between polygons, convexity, right angles and angle sums. And it highlights how S = (n – 2) × 180º isn’t just a formula: It’s one condition in a world of competing conditions.

Proofs of impossibility can help us better understand all areas of math. In school, probability lessons often begin with flipping lots of imaginary coins. I invite students to create an unfair coin — one that is biased toward coming up heads or tails — that has the following property: When the coin is flipped twice, the results of the two flips are more likely to be different than the same. In other words, you’re more likely to get heads and tails than to get heads and heads or tails and tails.

After some tinkering and a little productive frustration, the students arrive at an interesting hypothesis: Different outcomes are never more likely than the same outcome. Some algebra illuminates this and hints at an underlying symmetry.

Let’s say that the coin is biased toward heads. We’ll call the probability of flipping heads 12 + k, where 0 < k ≤ 12. The fact that k > 0 guarantees that heads is more likely than tails, which has probability 12 – k, since the two probabilities must add up to 1.

If we flip the coin twice, the probability of getting two heads or two tails will be

(12+k)2+(12−k)2.

Here we’re adding the probability of getting two heads (on the left) with the probability of getting two tails (on the right). Using algebra we can simplify the probability of getting the same result on both flips:

(12+k)2+(12−k)2 = 14 + k + _k_² + 14 – k + k_² = 12 + 2_k².

Since k > 0, we know that 12 + 2_k_² > 12, which means it’s more likely than not that the outcomes of the flips will be the same. In fact, we see that even if k = 0 (when the coin is fair), the probability of the same outcomes is exactly 12, making the probability of different flips also 12. The same outcome will never be less likely than different outcomes.

As with the octagon problem, we see competing mathematical tensions at work: Altering the likelihood of getting one side of the coin changes the likelihood of getting the other, and this interconnectedness governs what’s possible in terms of the two-flip outcomes. We expose those tensions by trying to do the impossible.

We can expose these tensions in every area of math. Try to find six consecutive integers that sum to 342, and some perseverance will lead to a better understanding of parity. (The fact that consecutive integers alternate between even and odd affects what their sums can be.) The search for a cubic polynomial with integer coefficients that has three non-real roots will teach you about the importance of complex conjugates — pairs of complex numbers whose product and sum are always real. And if you try to inscribe a non-square rhombus in a circle, you’ll walk away having discovered an important property of cyclic quadrilaterals — that the opposite angles in a quadrilateral whose vertices lie on a circle must sum to 180 degrees.

Confronting the impossible invites us to explore the boundaries of our mathematical worlds. The impossible itself is already a kind of generalization, so it’s only natural to keep generalizing: An octagon can’t have four right angles, but what about a decagon? What about a convex polygon with n > 4 sides? These kinds of questions push against the boundaries of our mathematical worlds and deepen our understanding of them.

If we push further, the impossible can even inspire the creation of new mathematical worlds. To prove the impossibility of squaring the circle — a problem that’s at least 2,000 years old — we needed the modern theory of transcendental numbers that cannot be roots of integer polynomials. To solve the bridges of Königsberg problem, Euler turned islands and bridges into vertices and edges, bringing to life the rich fields of graph theory and network theory, with their many applications. Taking the square root of −1 led to an entirely new system of arithmetic. And the logician Kurt Gödel changed the landscape of math forever when he proved that it’s impossible to prove that everything that is true is true.

So the next time you’re stuck on a math problem, ask yourself: “Is this possible?” Struggling with impossibility could give you a better understanding of what actually is possible. You might even create some new math along the way.

Exercises

1. Find the area of the triangle whose side lengths are 46, 85 and 38.

2. Let f (x) = 2_x_³ + _bx_² + cx + d. Find integers b, c and d such that f (14) = 0.

3. Find a perfect square, all of whose digits are in the set {2, 3, 7, 8}.

Answers

Answer 1:

This triangle doesn’t exist. Its side lengths fail to satisfy the triangle inequality theorem, which states that the sum of any two sides must be greater than the third. This can be seen geometrically if we start with two vertices of the triangle at opposite ends of a segment of length 85, and then construct circles of radius 38 and 46 around them. These circles don’t intersect, making it impossible to locate the third vertex of
the triangle.
It’s fun to use something like Heron’s formula to compute the area of this non-triangle.
Interesting questions will follow from this!

Answer 2:

There are different ways to establish the impossibility of this polynomial. For example, these conditions violate the rational root theorem, which says that any rational roots of a polynomial must be a ratio of a factor of the constant term divided by a factor of

the leading coefficient.

Answer 3:

A curious fact about perfect squares shows us that this task is impossible. The units digit of a perfect square can only be 0, 1, 4, 5, 6 or 9. This can be shown by simply squaring every possible units digit and observing the possible results. Since no perfect

square can end in 2, 3, 7 or 8, no perfect square exists with only those digits.

Author: Patrick Honner, Contributing Columnist and teacher

Source: https://www.quantamagazine.org/tag/quantized-academy/