An example of a beautiful proof that would be accessible at the high school level?

Question

The background of my question comes from an observation that what we teach in schools does not always reflect what we practice. Beauty is part of what drives mathematicians, but we rarely talk about beauty in the teaching of school mathematics.

I'm trying to collect examples of good, accessible proofs that could be used in middle school or high school. Here are two that I have come across thus far:

(1) Pick's Theorem: The area, $A,$ of a lattice polygon, with boundary points $B$ and interior points $I$ is $A = I + B/2 - 1.$

I'm actually not so interested in verifying the theorem (sometimes given as a middle school task) but in actually proving it. There are a few nice proofs floating around, like one given in "Proofs from the Book" which uses a clever application of Euler's formula. A very different, but also clever proof, which Bjorn Poonen was kind enough to show to me, uses a double counting of angle measures, around each vertex and also around the boundary. Both of these proofs involve math that doesn't go much beyond the high school level, and they feel like real mathematics.

(2) Menelaus Theorem: If a line meets the sides $BC,$ $CA,$ and $AB$ of a triangle in the points $D, E,$ and $F$ then $(AE/EC) (CD/DB) (BF/FA) = 1.$ (converse also true) See: http://www.cut-the-knot.org/Generalization/Menelaus.shtml, also for the related Ceva's Theorem.

Again, I'm not interested in the proof for verification purposes, but for a beautiful, enlightening proof. I came across such a proof by Grünbaum and Shepard in Mathematics Magazine. They use what they call the Area Principle, which compares the areas of triangles that share the same base (I would like to insert a figure here, but I do not know how. -- given triangles $ABC$ and $DBC$ and the point $P$ that lies at the intersection of $AD$ and $BC,$ $AP/PD = \operatorname{Area} (ABC) / \operatorname{Area}(DBC).$) This principle is great-- with it, you can knock out Menelaus, Ceva's, and a similar theorem involving pentagons. And it is not hard-- I think that an average high school student could follow it; and a clever student might be able to discover this principle themselves.

Anyway, I'd be grateful for any more examples like these. I'd also be interested in people's judgements about what makes these proofs beautiful (if indeed they are-- is there a difference between a beautiful proof and a clever one?) but I don't know if that kind of discussion is appropriate for this forum.

Edit: I just want to be clear that in my question I'm really asking about proofs you'd consider to be beautiful, not just ones that are neat or accessible at the high school level. (not that the distinction is always so easy to make...)

Answer 1

I like the lovely theorem in 19th century Euclidean Geometry as follows.

Let ABC be a triangle. let D,E,F be points on BC,CA,AB respectively. Then the circumcircles of AFE, BDF, CDE meet at a point.

I like this because the proof uses the property of the angles of cyclic quadrilaterals, and its converse. Also if one wants to convince students of the necessity of proof, then one should start with a result which is surprising.

It is a good thing that this situation can we worked on for more implications. Let P,Q,R be the centres of the three circles just given. Then the triangle PQR is similar to the triangle ABC.

For all these reasons I think it is a pity that some of Euclidean Geometry is not in University courses, or often school courses, in order to acquaint students with something important in our mathematical heritage. Should a student get a degree in maths without knowing why the angle in a semicircle is a right angle?

Answer 2

After two concrete answers, let me give a third, rambling answer:

A few years ago I decided to completely quit elementary geometry because it was still taking up half of my time even as I was already studying in university. Given that I have been doing it for most of my schoolyears, this should give an impression of how much there is to be done there - and all of it is comprehensible to a good school student.

I will not go into details, but this is a community wiki post ;)

First, there are so many lines in a triangle meeting at one point that one can reasonably ask whether there exist three symmetrically-defined lines not doing so. (There are, of course: e. g., the reflections of the medians in the corresponding altitudes.) This does not change the fact that each concurrence theorem is still a nontrivial result asking for a proof. Some of the basic cases can be handled with Ceva; harder results can take a dozen of pages to prove. The points where these lines concur often have several equivalent characterizations and collinearity properties (like the Euler line); this led Clark Kimberling to make an encyclopedia of such points similar to Sloane's On-Line Encyclopedia of Integer Sequences. An easy example is the concurrence of the lines $AX$, $BY$, $CZ$, where $X$, $Y$, $Z$ are the points where the incircle of triangle $ABC$ touches the sides $BC$, $CA$, $AB$. (The point where they concur is known as the Gergonne point of triangle $ABC$, known as $X_7$ in 1.) A not-so-easy example: If $O$ is the circumcenter of triangle $ABC$, then the lines connecting $A$, $B$, $C$ with the circumcenters of triangles $OBC$, $OCA$, $OAB$ concur. (This is the Kosnita point, aka $X_{54}$ in 1.) Then there are more complicated things, like: Consider the points where the excircle of triangle $ABC$ opposite to $A$ touches the extended sides $AB$ and $AC$. Let $M_a$ be the midpoint between these two points. Let $M_b$ and $M_c$ be defined similarly. Let the incircle of triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $X$, $Y$, $Z$. Then, the lines $M_aX$, $M_bY$, $M_cZ$ concur (at a point which is $X_{1122}$ in 1; this is something I have discovered back in schooltime when playing around with dynamic geometry software).

Of course, concurrent lines aren't even half of the fun. There are theorems like Feuerbach's, stating that the nine-point circle touches the incircle and the excircles. This is some centuries old. Here is one found in 2000 by Floor van Lamoen: The medians of a triangle subdivide it in six little triangles (of equal area, by the way); the circumcenters of these triangles lie on one circle! Or here is another tangency property: If the incircle of triangle $ABC$ touches the side $BC$ at a point $X$, then the incircles of triangles $ABX$ and $ACX$ touch each other.

All theorems I mentioned can be proven in a synthetic way, i. e., using merely the part of elementary geometry studied in school, without coordinates or overly long computations. ("Overly long" is subjective and I am well aware of this.) This makes the field completely accessible to students. This is not to say that advanced mathematics doesn't shed some new light on it. For instance, one could try generalizing the above-mentioned Kosnita point by looking for all the points $P$ such that the lines connecting $A$, $B$, $C$ with the circumcenters of triangles $PBC$, $PCA$, $PAB$ concur. The answer turns out to be that the set of such points $P$ is a cubic curve known as the Neuberg cubic of triangle $ABC$. The sheer amount of interesting points on it allow for some neat applications of the group law on cubics to elementary geometric theorems.

I have been rather sparse with sources, as I haven't been keeping track of them for years. Some can be found on my links page, but it has not been updated since 2008 or so. Nowadays Jean-Louis Ayme's blog is the best place to find more synthetic proofs than one could ever read. There are also some good books. (Sorry for linking to my page this often; it is the one place on the internet I am most familiar with...)

Answer 3

Fermat's Little Theorem: If $p$ is prime and does not divide $a$, then $a^{p-1} \equiv 1 (\mbox{mod } p)$.

Proof: List the multiples of $a$ up to $a(p-1)$:

$$ a, a2, a3, \dots , a(p-1).$$

For any $r$ and $s$ with, $ra \equiv sa (\mbox{mod } p)$, we have $r \equiv s (\mbox{mod } p)$, so that the list above contains $p-1$ many distinct numbers.

Thus, the list above is some ordering of the list $1, 2, 3, \dots p-1$ modulo $p$. This gives us $$ a\cdot a2 \cdot a3 \cdot \cdots a(p-1) \equiv (p-1)! (\mbox{mod } p) $$

Finally we see

$$ a^{p-1} \equiv 1 (\mbox{mod } p). $$

Answer 4

Why not some elementary theorems of Euclidan geometry? As I recall, the more general and fundamental theorems were just taken as given in my schooling, but I think many of them can be given accessible and beautiful proofs. Here are some good ones:

1) The Pythagorean theorem. (many lovely proofs)

2) Parallelograms having congruent bases and heights have the same area. (Euclid's proof is pretty.)

3) Use 2 to derive that similar triangles have corresponding sides in common proportion.

4) Two distinct circles have at most 2 points of intersection.

5) Prove the formula for volume of a pyramid without using calculus.

Answer 5

I actually think that Hilbert's Third problem is one of the explainable for school guys. It's even more cool that it exists in such a famous list close to the problems that are so tempting and not yet solved. The question is: can one cut the cube in some polyhedral pieces, reglue them and get a regular tetrahedron? The answer is no and the theorem was proved by Dehn using so-called Dehn invariant. It uses some algebra and number theory but can be understood by high-school level guys. The time you need to explain this is 3-4 hours, so maybe it could be a little and nice course. See, for example, Lectures on Discrete and Polyhedral Geometry by I. Pak

Answer 6

If each brick in a tiling of a rectangle has an integer side, the rectangle does too. This has various generalizations and lots of proofs, some very accessible, like number 7 here:

https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Wagon601-617.pdf

Answer 7

The most success I have ever had teaching proofs at secondary school level is with the Peaucellier–Lipkin linkage. The proof relies on nothing more than basic geometry, namely similar triangles, but the outcome really is amazing. I found it reading Tom Körner's book called Fourier Analysis.

You can get the proof from Wikipedia, and there are some videos around if you google for them. Körner goes into the history of the problem, he believes Tchebychev was of the opinion that the problem couldn't be solved! And there is a great quote attributed to Kelvin, which I'll leave you with. When Sylvester showed a working model to him...

'[he] nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied "No! I have not had nearly enough of it - it is the most beautiful thing I have ever seen in my life."'

Answer 8

Sperner's lemma (in dimension 2 to keep it visual). The proof in Francis Su's Monthly paper, Rental harmony: Sperner's lemma in fair division is especially easy to visualize. Theris a non-empty content, you can have students ponder the role of the hypotheses. And fair division applications allow to motivate it via concrete applications.

Answer 9

I think the fundamentals of the theory of quadratic residues modulo a prime qualify.

It is easy to explain what residue classes modulo a prime $p$ are, and to formulate statements of this kind:

1. The product of two quadratic residues modulo $p$ is a quadratic residue.

2. The product of a quadratic residue and a quadratic nonresidue modulo $p$ is a quadratic nonresidue.

3. The product of two quadratic nonresidues modulo $p$ is a quadratic residue.

(Note that I am not counting $0$ as a quadratic residue, nor as a quadratic nonresidue.)

Now 1 and 2 are very easy to show. 3 is not. What do we do?

First, it is easy to see that every quadratic residue is the square of exactly $2$ distinct residues modulo $p$. Thus there are exactly $\frac{p-1}{2}$ quadratic residues modulo $p$. Hence, there are exactly $\left(p-1\right)-\frac{p-1}{2}=\frac{p-1}{2}$ quadratic nonresidues modulo $p$. Now let $a$ and $b$ be two quadratic nonresidues. If $ab$ is a quadratic nonresidue, then there are at least $\frac{p-1}{2}+1$ different residues $x$ modulo $p$ for which $ax$ is a quadratic nonresidue (namely, each of the $\frac{p-1}{2}$ quadratic residues qualifies as such $x$ (by statement 2), but the quadratic nonresidue $b$ also qualifies), which leads to at least $\frac{p-1}{2}+1$ different quadratic nonresidues (since distinct $x$'es lead to distinct $ax$'es), contradicting the fact that there are only $\frac{p-1}{2}$ quadratic nonresidues modulo $p$. Thus, $ab$ must be a quadratic residue, and 3 is proven.

This indirect argument is, I believe, understandable to high school students. The only two theorems we used are:

A. Every quadratic residue is the square of exactly $2$ distinct residues modulo $p$.

B. If $a$ is a nonzero residue modulo $p$, then distinct $x$'es lead to distinct $ax$'es.

Both of these theorems can be derived from the following well-known fact:

F. If a prime divides a product of two integers, then it divides one of these two integers.

Proof of A: Assume that $a^2 \equiv b^2 \equiv c^2 \mod p$ for three integers $a$, $b$, $c$ pairwise incongruent modulo $p$. Then, $a^2 \equiv b^2 \mod p$ rewrites as $p\mid \left(a+b\right)\left(a-b\right)$. Hence (by fact F), at least one of $a+b$ and $a-b$ is divisible by $p$. Since $a$ and $b$ are incongruent modulo $p$, this can only mean that $a+b$ is divisible by $p$. Similarly, $b+c$ and $c+a$ are divisible by $p$. But therefore $2a=\left(a+b\right)+\left(c+a\right)-\left(b+c\right)$ must also be divisible by $p$. Since $p$ cannot be $2$ (as there are no three integers pairwise incongruent modulo $2$), this yields that $a$ is divisible by $p$. Similarly, $b$ and $c$ are divisible by $p$, which contradicts with their being incongruent. This proves A.

The proof of B is much simpler. The fact F is also used in one possible proof of statement 2. (However we can also prove 2 using 1 by the same trick as we used to prove 3 using 2.)

We have thus used the fact F a lot of times, but other than that, we didn't apply anything nontrivial - not even the theorem that a nonzero polynomial over a field cannot have more roots than its degree (this fact is often used in university-level treatises of quadratic residues).

The hard part is to tell students what is interesting about quadratic residues. Maybe cryptography?

Answer 10

This is maybe ambitious, for the details are obviously not completely accessible to the high school level; but the beauty of the ideas is, and this video is really a superb example of divulgation. Smale's theorem on the eversion of the 2-sphere and Thurston's construction.

Answer 11

There are a lot of good suggestions in this feed, but here are a few problems that let you introduce modular arithmetic.

First, one can easily prove that an integer mod 9 is equal to the sum of its digits mod 9.

Second, you can prove Fermat's little theorem k^p mod p = k where p is prime.

I suppose that even (a+b) mod n = (a mod n + b mod n) mod n is kind of neat too.

You can prove that the calendar repeats itself every 28 years.

Answer 12

I love the incredibly clever proof of Sylvester’s theorem (https://en.m.wikipedia.org/wiki/Sylvester–Gallai_theorem) by Kelly. It is described nicely in the wiki page.

Answer 13

Well, I can't guarantee that I can make you happy, but atleast guess that I can. A simple problem: Determine the set of all points lying in the plane of a triangle ABC (say P), for which (PA)^2+(PB)^2+(PC)^2 is minimum. We all know that this set is the singleton set: {Centroid G of ABC}. Unfortunately, a previous knowledge of the answer makes it easy to prove the assertion, but more unfortunately, even after stating:"We will show that the centroid is the only such point", most books give a long, boring proof, involving non-trivial and non-motivating constructions and lengthy calculations. Right, I'm going to give a what-I-think beautiful proof, which is due to me! I solved it while preparing for the Indian National Mathematical Olympiad last year. Just join P with C1,A1,B1, the midpoints of AB,BC,CA,resp and form the triangle A1B1C1. Note that A1B1C1 is the image of ABC under a homothety of factor -0.5 about their common centroid. Next, applying the Apollonius' theorem on the 3 triangles APB, BPC & CPA and adding the 3 relations, note that PA^2+PB^2+PC^2 is minimum, if and only if PA1^2+PB1^2+PC1^2 is minimum. That the set above is singleton, is immediate from the extremal principle applied to 2 possible points having the same property and showing that their midpoint has the sum of squares less than them. Now, let P' be the image of P under the homothety -0.5. Then, by properties of homothety, P' and P both have minimum sum-of-squares with respect to A1B1C1. But the set is singleton, so P=P'. However, the only point that remains invariant under a homothety is the centre, which in this case, is the centroid!!:) Right, whenever one sees the sum-of-squares form one guesses to apply Apollonius' theorem and this proof doesn't even require a pre-knowledge of the answer! Please tell me if you've enjoyed it or not!

Answer 14

Well I personally liked the Euler's Theorem when I first saw, and I feel one can easily understand it at the High-School level.

There is a Youtube video with William Dunham's lecture on Euler which contains this theorem at 32:30 : https://www.youtube.com/watch?v=fEWj93XjON0

Answer 15

Quite elementary and quite beautiful: There exist two (infinitely many) irrational numbers $a,b$ such that $a^b$ is rational; and the usual proof with $\sqrt{2}^\sqrt{2}$. (Maybe starting with the proof of $\sqrt{p}$ being irrational for any prime $p>0$.)

Answer 16

I would put on such a list the proof of the Cauchy-Schwarz inequality directly from the axioms of an inner product using the discriminant criterion for the classification of the roots of a second-order polynomial with real coefficients. The latter is something all high school students either learn or already know, but it is a common target in those jokes about the parts of school math that most people find "useless"... The axioms for an inner product themselves are easy to motivate from $\mathbb{R}^2$ and $\mathbb{R}^3$.

Let me recall the proof here because it is that simple... Let $V$ be a (real) vector space with an inner product $\langle\cdot,\cdot\rangle$, and let $\vec{x},\vec{y}\in V$. We want to prove that $$\langle\vec{x},\vec{y}\rangle^2\leq\langle\vec{x},\vec{x}\rangle\langle\vec{y},\vec{y}\rangle\ ,$$ with equality iff $\vec{x}$ and $\vec{y}$ are collinear (i.e. linearly dependent). Since $\langle\vec{y},\vec{y}\rangle=0$ iff $\vec{y}=\vec{0}$ (nondegeneracy axiom) and in that case $\langle\vec{x},\vec{y}\rangle=0$ for all $\vec{x}\in V$ (due to the bilinearity axiom) the Cauchy -Schwarz inequality holds (with equality) if $\vec{y}=\vec{0}$, so there is no loss of generality if we assume $\vec{y}\neq\vec{0}$. In that case, define for $t\in\mathbb{R}$ $$P(t)=\langle\vec{x}+t\vec{y},\vec{x}+t\vec{y}\rangle=\langle\vec{y},\vec{y}\rangle t^2+2\langle\vec{x},\vec{y}\rangle t+\langle\vec{x},\vec{x}\rangle\ ,$$ where the last equality follows from the bilinearity and symmetry axioms. Now, due to the positive definiteness axiom (of which, let us recall, the nondegeneracy axiom is a consequence), we must have $P(t)\geq 0$ for all $t\in\mathbb{R}$. Recall now that the roots $t_\pm$ of a second-order polynomial $Q(t)=at^2+bt+c$ with real coefficients $a,b,c$ ($a\neq 0$) are given by the discriminant formula $$t_\pm=\frac{-b\pm\sqrt{\Delta}}{2a}\ ,$$ where the discriminant $\Delta$ of $Q(t)$ is given by $\Delta=b^2-4ac$. In the case that $Q(t)=P(t)$, one has that $\Delta=4(\langle\vec{x},\vec{y}\rangle^2-\langle\vec{x},\vec{x}\rangle\langle\vec{y},\vec{y}\rangle)$. Since $P(t)\geq 0$ for all $t$, this implies that $P(t)$ has at most one real root, which is only possible if $\Delta\leq 0$, whence the above inequality follows. The equality case happens if and only if $t=t_+=t_-$ is the only real root of $P(t)$ - for such a $t$, the nondegeneracy axiom implies that $\vec{x}=-t\vec{y}$, that is, $\vec{x}$ and $\vec{y}$ are collinear.

The above proof is completely general (works also for complex vector spaces and in infinite dimensions) and the actual result lies at the basis of Euclidean geometry, for one can get from it all the properties of the Euclidean distance and it also guarantees that the definition of the (Euclidean) angle between vectors makes sense through the cosine rule $$\angle(\vec{x},\vec{y})=\arccos\left(\frac{\langle\vec{x},\vec{y}\rangle}{\sqrt{\langle\vec{x},\vec{x}\rangle\langle\vec{y},\vec{y}\rangle}}\right)\ ,$$ so one can also use the latter to tell the students: "Do you see? The discriminant formula plays a role whenever you are doing any kind of Euclidean geometry..." In other words, it displays a deep connection about two seemingly completely different areas of mathematics (geometry and algebra).

Answer 17

Figuring out the Lagrange interpolation polynomial was a pretty awesome moment for me as a high school nerd.

I was amazed a while later that you can simulate a Turing machine with just two counters, but that takes a bit of technical stuff to explain what a Turing machine is.

$x+1/x\ge 2$ if $x > 0$. Proof: $(\sqrt x-\sqrt{1/x})^2$ must be >=0, so expanding, $(x + {1\over x} - 2) \ge 0$. Not very deep, but kind of an aha moment in seeing reasoning appear from nowhere and immediately look obvious, getting rid of a calculus problem.

Proof of the triangle inequality in R**n, using Schwarz's inequality. Again, maybe the proof isn't beautiful in itself, but it was eye-opening in connecting geometry to analysis.

Answer 18

I would also like to add a different proof of Fermats little theorem ($p|(a^p-a)$ for prime $p$) to the list.

Suppose you have a colors of pearls and you want to produce pearl-chains of length $p$.

First you put $p$ pearls on a string. There are $a^p$ possibilities. Next you discard the mono-colored ones, they are boring. This leads to $a^p-a$ choices.

Next you put a knot into the string to turn it to a circular chain.

Now you have each type of chain multiple times, since cyclic permutations give you the same chain.

Finally you have to convince yourself that each type of chain arises in exactly $p$ ways using that $p$ is prime. Thus there are $(a^p-a)/p$ such different chains and hence that number has to be an integer.

Answer 19

Pick up any of the three volumes in Roger Nelsen's Proofs Without Words series and turn to any page. Many of the summation identities mentioned in this thread are included in these books.

Answer 20

If we're going for "beautiful" rather than just "neat", I'd vote for the Eckmann-Hilton argument. Although it's reasonably abstract for a high-schooler, it should still be quite accessible, and has a lot of "beautiful" symmetry, especially if you look at the nice circle of proofiness.

Answer 21

The following post from the "Everything Seminar" blog would make an excellent lesson in my opinion (link is below). It starts from a simple, but clever "hats puzzle" and then presents an infinite version of the puzzle which is solved (quite amazingly and beautifully) using the axiom of choice. It exemplifies the gap between what we expect to happen and what actually happens which is encountered in mathematics from time to time. Also, you don't really need to introduce any complicated concept, only describe what an equivalence relation is. The mere definition of an equivalence relation is beautiful mathematics in my opinion and the way in which such a simple concept can be used to solve a difficult (impossible?!) problem as above shows how interesting and intriguing mathematics can be.

The relevant post is here: http://cornellmath.wordpress.com/2007/09/13/the-axiom-of-choice-is-wrong/