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The butterfly theorem is notoriously tricky to prove using only "high-school geometry" but it can be proved elegantly once you think in terms of projective geometry, as explained in Ruelle's book The Mathematician's Brain or Shifman's book You Failed Your Math Test, Comrade Einstein.

Are there other good examples of simply stated theorems in Euclidean geometry that have surprising, elegant proofs using more advanced concepts? Such examples are valuable pedagogically since they illustrate the power of the advanced methods.

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Every theorem from classical geometry has short and elegant proof for "high-school students". But one can find very difficult proof which use algebraic topology or category theory. – Arseniy Akopyan Jul 11 '10 at 6:03

11 Answers 11

Somebody has this as a hobby.

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Another post where he does the same thing:… – Qiaochu Yuan Jul 11 '10 at 5:11

A nice example is Pascal's theorem for the circle:

If a hexagon is inscribed in a circle then the intersections of opposite sides are collinear.

Plücker gave an elegant proof of Pascal's theorem as a consequence of Bézout's theorem. View the unions of alternate sides of the hexagon as cubic curves

$l_{135}=0$ and $l_{246}=0$.

They meet in 9 points, 6 of which are the vertices on the circle $c=0$. But we can choose constants $a,b$ so that the cubic


passes through any point. Taking this point on the circle, the circle and the cubic have at least 7 points in common. By Bézout's theorem, the curves have a common component, necessarily the circle $c=0$, since $c$ is irreducible.

Hence $al_{135}+bl_{246}=cp$, for some polynomial $p$, which must be linear. Since $al_{135}+bl_{246}=0$ contains all 9 points common to $l_{135}=0$ and $l_{246}=0$, while $c=0$ contains only 6, the remaining 3 (intersections of opposite sides of the hexagon) must lie on the line $p=0$.

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Yes. Actually, Pascal's (and Pappus' too) theorem is partial case of associativity low for (degenerated) elliptic curve. – Fedor Petrov Jul 11 '10 at 7:35
This is one of my favorites! Cf… – Victor Protsak Jul 11 '10 at 7:44

What about the proof of the isoperimetric inequality using Fourier analysis?

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The isoperimetric inequality hardly is more "Euclidean geometry" than Fourier analysis... – darij grinberg Feb 1 '12 at 18:31
@darij grinberg: This answer may not satisfy the letter of the question, but I think it very much satisfies the spirit. The isoperimetric inequality is a classic result dating back to the fourth century, if not earlier (…), while Fourier analysis is a 19th-century method based on very modern sensibilities (function spaces, new notions of convergence and integration, etc.). – Vectornaut Feb 1 '12 at 21:01
It does not satisfy the spirit either, even though the isoperimetric inequality appeared in the fourth century. The problem is that every formalization of the isoperimetric inequality already requires most of the advanced tools needed for its proof. Otherwise, the Jordan curve theorem would be an equally good example. – darij grinberg Feb 2 '12 at 11:55

I like the following result which plays a role in Marden's Theorem:

Given any triangle, there is an inscribed ellipse which meets the midpoints of all three edges.

Proving this directly is rather difficult (in fact, I'm not sure how to), but it is very easy to do if you know anything about linear transformations.

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Linear transformations also give a nice proof of – Gjergji Zaimi Jul 11 '10 at 5:12
Gjergij, Steiner chain deserves its own entry, but the key issue is that the transformation be $\textit{conformal},$ so that it preserves circles, and not merely linear. – Victor Protsak Jul 11 '10 at 7:46
Yes, sorry, I meant fractional linear :) – Gjergji Zaimi Jul 12 '10 at 20:27
Generalization: Let $ABC$ be a triangle. Let $X$ and $X'$ be points on the line $BC$. Let $Y$ and $Y'$ be points on the line $CA$. Let $Z$ and $Z'$ be points on the line $AB$. Then, the points $X$, $X'$, $Y$, $Y'$, $Z$, $Z'$ lie on one conic (possibly degenerate) if and only if $\frac{BX}{XC}\cdot\frac{BX'}{X'C}\cdot\frac{CY}{YA}\cdot\frac{CY'}{Y'A}\cdot\fr‌​ac{AZ}{ZB}\cdot\frac{AZ'}{Z'B}=1$, where the segments are directed. This is easy to prove using Pascal and Menelaos. When two points like $X$ and $X'$ coincide, a conic passing through $X$ and $X'$ has to be understood as a conic ... – darij grinberg Feb 1 '12 at 18:36
... touching the line $BC$ at $X$, and so on, and thus you obtain as a particular case the following fact: Let $X$, $Y$, $Z$ be points on the sidelines $BC$, $CA$, $AB$ of a triangle $ABC$. Then, there exists a conic touching $BC$, $CA$, $AB$ at $X$, $Y$, $Z$ if and only if either the lines $AX$, $BY$, $CZ$ concur or the points $X$, $Y$, $Z$ are collinear (in which case it is a degenerate conic). This uses Ceva and Menelaos. Not to say this is simpler than the affine-transformation proof, but in my eyes it shows that the problem has nothing to do with midpoints of edges. – darij grinberg Feb 1 '12 at 18:38

A very nice example is given by the Villarceau circles: a revolution torus is cut by a bitangent hyperplane along the union of two circles. You can of course make the computation, but when you know some projective algebraic geometry you can prove it in a few words. Roughly:

  1. The revolution torus has an algebraic equation of degree four, so that it intersects any plane along a degree four curve.

  2. If the plane is bitangent, then this curve has two double points so that it must be the union of two ellipses.

  3. It is easily checked that in the complex world, the torus as well as the plane contain the circular points at infinity, so that in fact the ellipses are circles.

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Take a triangle with a circle $\Gamma_0$ tangent to two of three sides (you may also think that sides of the triangle are made out of circle arcs). Construct a chain of circles $\Gamma_1,\Gamma_2,\dots$ on such a way that $\Gamma_{n+1}\not=\Gamma_{n-1}$ is tangent to $\Gamma_n$ and two of the sides of triangle. Prove that $$\Gamma_6=\Gamma_0.$$

I do not know the proof, but I was told that it is hard to do without knowing elliptic functions.

P.S. I do not know the references --- please feel free to add it :)

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This problem from IMO Shortlist. You can find solution for high school students in Prasolov book. But this problem is not so easy in hyperbolic geometry. – Arseniy Akopyan Jul 11 '10 at 17:19
Just a small note: your description is missing the constraint that the two sides of the triangle have to change - otherwise you could just take for the $\Gamma_n$ a Pappus chain converging towards one of the triangle's vertices and tangent to the two sides incident on that vertex. A fuller description of the six-circles theorem (with a couple of references to elementary-looking proofs) is at . – Steven Stadnicki Jul 12 '10 at 20:11

The elementary proof of Morleys theorem is rather incomprehensible compared to the non elementary ones.

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There is a recent, simple and elementary proof in a recent volume of Elemente der Mathematik but I do not remember the name of its author. You start with an equilateral triangle and show in a few line that you can realise it as the Morley triangle of a triangl with arbitrary angles. – Benoît Kloeckner Jul 11 '10 at 12:56
That sounds like a proof I read 25 years ago in a book from 1960s. Maybe it got shortened? – Victor Protsak Jul 11 '10 at 13:05
Oh thanks for reminding me how Conway managed to essentially stop all graduate research for a few days by telling us about the theorem and that he has a simple proof which he won't tell us. – Willie Wong Jul 11 '10 at 13:32
LOL. Please don't cite Zeilberger's troll as an example of an elementary proof. – darij grinberg Feb 1 '12 at 18:43

The "most elementary theorem of Euclidean geometry" as proved in (p.9). The Poncelet's Porism in Gjergji Zaimi's answer is also proved there.

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(Edited after Willie's comment)

Many theorems about a triangle can be proved easily using the fact that all triangles are linearly equivalent. One example is that the segments that join each vertex to the midpoint of the opposite side intersect in a single point.

(This actually appeared on a written qualifying exam I took as a graduate student, and I did in fact use the linear algebra approach, since there was no chance I could recall the Euclidean geometric proof.)

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Deane, I am a bit confused: I don't think angle bisection is preserved under linear transformations. Did I misunderstand what you mean by "triangles being linearly equivalent"? – Willie Wong Jul 11 '10 at 14:13
Ouch. Willie, you appear to be right. It's only the theorem about segments joining the vertices to the opposite midpoint that works using this approach. – Deane Yang Jul 11 '10 at 14:41

Let's see. Most of the results of projective geometry are much harder to show by more elementary methods. I think the reason for this is that they rest more on the incidence axioms while the elementary methods play more with the metric properties.

An interesting question that arises for any of these examples is to detect why is that the case that the elementary proofs are harder.

A classical example is the constructibility of regular polyhedral with ruler and compass.

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The classification of conics would be an example, but I don't know if you count matrix reduction as a "more advanced concept".

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