Take the 2-minute tour ×

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 Grunbaum 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 = Area (ABC)/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...)

share|improve this question
Surely you know the books ''Math! Encounters with Undergraduates'', ''Math Talks for Undergraduates'' and ''The beauty of doing Mathematics'' by Serge Lang. They were written with your same intention. –  Giuseppe Tortorella Sep 8 '11 at 11:15
I remember Lang giving a talk at Caltech supposedly aimed at undergraduates, but far more advanced. Lang called on a postdoc in the front row, apparently thinking he was a student, and the postdoc couldn't answer. At one point, he wrote out a complicated formula, and challenged Prof. Ramakrishnan, "Do you teach your students this?" "No." "So you see, Caltech is not better than anywhere [sic] else." After a bit more, he went back to the formula and corrected a sign. Ramakrishnan called out, "That, we teach." –  Douglas Zare Sep 14 '11 at 0:38
Even the notion of proof may not be accessible at high school level –  Fernando Muro Sep 15 '11 at 20:33
In the US, most high school mathematics classes are not based on proofs. Theorems are introduced without reference to proof. Only in a few limited situations are students asked to prove anything, such as in some geometry classes and some calculus classes. Instead, high school math classes concentrate on introducing objects, their properties, and how to manipulate those objects. Finding neat accessible proofs to show them is reasonable, but this is very different from finding neat accessible math to show them. E.g., I think the Chaos Game is accessible, but few students can handle the proofs. –  Douglas Zare Sep 17 '11 at 0:39
As far as I remember, the thing I liked the most in high school maths (age of 14) was the so called Ruffini's rule: $(x-a)$ divides a polynomial $P(x)$ if and only if $P(a)=0$. It looked to me so incredibly easy and so full of consequences. I hope they still learn it with a proof. –  Pietro Majer Sep 18 '11 at 20:12

46 Answers 46

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.

share|improve this answer

This has been mentioned before but here is a video and motivation of Gauss proof

Gauss proof for the sum of first 100 natural numbers

There is more videos in that site that include proofs

here is the list of instructional videos from isallaboutmath.com

especially this one

Proof (stereographic projection)

this should be accessible with the most basic geometric knowledge.

share|improve this answer

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...)

share|improve this answer

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.

share|improve this answer

Using Euler's formula ($F-E+V=2$, as mentioned earlier), it can be proved that the graphs $K_5$ and $K_{3,3}$ are not planar. I think these proof and also the proof of Euler's formula are simple enough to be understood by an interested high school student.

share|improve this answer

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

share|improve this answer

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!

share|improve this answer

Here is one that I like and used it for different purposes, e.g. introduction to proofs, algebraic thinking, beauty, and so one. Shuffle a deck of cards. Divide it into two halves. Magic: The number of the red cards in one of the halves is exactly equal to the number of black cards in the other half. It has a simple algebraic proof. However, The bigger magic is yet to come.

Theorem: Vertically opposite angles are equal.

Bigger Magic: The two theorems/proofs are essentially the same!

share|improve this answer

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

share|improve this answer

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:


share|improve this answer

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.

share|improve this answer
There's a bizarre application of Sperner's lemma that's hard for high-school students (but I did give a talk to a group of them about it): one can't dissect a square into an odd number of non-overlapping triangles, all of the same area. –  paul Monsky Sep 9 '11 at 4:53

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?

share|improve this answer

The proof that there exist irrational number which irrational power is rational. Consider $\sqrt{2}^\sqrt{2}$ and $\sqrt{2}$.

share|improve this answer

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.

share|improve this answer
Do you really think middle school or high school students will appreciate proving that a monoid is commutative, or the applications to higher homotopy theory? Perhaps I went to the wrong middle school. –  Douglas Zare Sep 8 '11 at 20:44
Depends on the high school students. The assertion is certainly a nice first-round contest-level problem. And it is probably more interesting than many things done in school mathematics (then again, everything is...). –  darij grinberg Sep 8 '11 at 23:29
How compelling do you think this symbol manipulation would be for a wide audience? Say, those who would not understand what you are asking if you ask them to show that a left identity equals a right identity. –  Douglas Zare Sep 9 '11 at 0:26

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.

share|improve this answer

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/

share|improve this answer
I disagree with, "You don't need to introduce any complicated concept." Do you think the students are already familiar with the Axiom of Choice? They aren't even comfortable with infinite sets. Presenting counterintuitive results which they can't otherwise verify and which are not connected to anything they have seen may be damaging to students new to mathematics. They may well reject mathematics (and at least the AOC) as not sensible, and they may be right unless you are extremely careful in the presentation, much more so than the blog you link, which was not aimed at high school students. –  Douglas Zare Sep 16 '11 at 23:25
The positive integers are not the only infinite set involved in this example. If it were, you wouldn't need to use the Axiom of Choice. You are using the Axiom of Choice on equivalence classes of subsets of the natural numbers. I really think this example would be damaging. Normally, we start proving things we think are obvious, then prove things we believe, then use proofs to establish the truths of things which are uncertain. These students might or might not have hit the first stage and you want to jump to proving paradoxes with questionable axioms. Students will reject math. –  Douglas Zare Sep 17 '11 at 0:18
Also, I disagree that the Axiom of Choice is intuitive. I think it's only intuitive if you overlook what it is really saying, that there exist oracles for situations where we don't have oracles. (The reals can be well-ordered? Ok, tell me how to compare them, or the least element of this set.) You are asking students to accept the step of memorizing the output of an oracle they don't have, but which somehow "exists." This would have turned me off from mathematics. There are much better things to show middle school and high school students than paradoxes which follow from questionable axioms. –  Douglas Zare Sep 17 '11 at 0:26
I agree with Douglas Zare that this topic would not be appropriate for the intended audience. But I disagree with him abut the axiom of choice "really saying, that there exist oracles for situations where we don't have oracles." The axiom of choice is about sets, not oracles (unless you take "oracle" to mean simply set), and it certainly isn't about our "having" anything. After all, we don't even "have" all the natural numbers. It seems to me that part of the problem with the hats puzzle is that the solution pretends that sets (obtained by AC) can be used as oracles. –  Andreas Blass Sep 17 '11 at 15:55

Your Answer


