54

Even more is true for this theorem. Check out this drawing from Arseniy Akopyan wonderful book of Geometry in Figures (Second, extended edition, 2017). On page 65 we find Figure 4.7.29)
In the foreword, Arseniy Akopyan writes
"It is commonly very hard to determine who the author of a certain
result is."
He nevertheless provides source for many of the ...

48

When I was in high school (in the early 1960's), Euclidean geometry was the only course in the standard curriculum that required us to write proofs. These proofs, however, were in a very rigid format, with statements on the left side of the page and a reason for every statement on the right side. So I fear that many students got an inaccurate idea of what ...

42

This question was explored here:
Lenhart, William J., and Sue H. Whitesides. "Reconfiguring closed polygonal chains in Euclidean $d$-space." Discrete & Computational Geometry 13, no. 1 (1995): 123-140; DOI: 10.1007/BF02574031, eudml.
From the Abstract: "It is shown that in three or more dimensions, reconfiguration is always possible, but that in ...

answered Aug 28 '19 at 11:32

Joseph O'Rourke

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40

There are two things you need to understand. The first is how to prove sphere packing bounds via harmonic analysis ("linear programming bounds"). My lecture notes from PCMI 2014 give an exposition of this theory, which covers the period up to, but not including, Viazovska's paper on eight dimensions. This reduces the problem to finding radial functions in ...

39

These are the so-called Regge symmetries, described by T. Regge in a 1970-ish paper. For a bit on it, with references, see the paper
Philip P. Boalch, MR 2342290 Regge and Okamoto symmetries, Comm. Math. Phys. 276 (2007), no. 1, 117--130.\

answered Jan 15 '17 at 21:13

Igor Rivin

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39

Euclidean case
Using the formula for the tan of the half solid angle that Robin Houston quotes, and expressing everything in terms of edge lengths by using the cosine law to convert the dot products, I end up with the following linear relationship:
$$\frac{1}{P_i} = \alpha_E \cot\left(\frac{\Omega_i}{2}\right)+\beta_E$$
where:
$$\alpha_E = \frac{12 V}{...

35

I try to keep my answer short.
Fact: Euclidean geometry is still taught in Iranian middle and high schools.
Observation (based on research): Most teachers do not like to teach geometry. They say, when you teach geometry, you are always faced with problems that you don't know how to solve them. But, it seems that they haven't got that problem with the rest ...

35

Arguably, the so-called "area method" of Chou, Gao and Zhang represents the state of the art in the field of machine proofs of Olympiad-style geometry problems. Their book Machine Proofs in Geometry features over 400 theorems proved by their computer program. Many of the proofs are human-readable, or nearly so.
The area method is less powerful than Tarski&...

answered Aug 5 '19 at 21:41

Timothy Chow

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30

This is not an answer to the question, but an experimental observation that suggests a sharper conjecture: it’s only written as an answer because I’d like to flesh it out a bit more than there’s room for in a comment.
The image of $e^{i\Omega}$ under the Möbius transformation $z\mapsto\frac{i-zi}{1+z}$ is $\tan(\frac\Omega2)$, so an equivalent statement of ...

25

Did you ever find any answer to this?
I find it intriguing that figuring out which shapes of holes a given solid object can pass through is widely considered to be a suitable puzzle for 2 year olds, yet we still don't know the smallest possible hole even for a regular simplex.
edit 29/04/21 — as pointed out by @mjqxxxx in the comments, there's a smaller area ...

24

Here is a partial answer: in $\mathbb R^3$ there are cycles of length $2$. To show this, it is enough to find non-trivial systems of unit vectors $x_i, y_j$ ($i,j=1,\dots,d+1$) such that $(x_i,y_j)=t$ for some $t>0$ and all $i\ne j$. If $d=3$, we can take the normalized copies of $(1,0,4),(1,0,-4),(-1,4,0),(-1,-4,0)$ and $(-2,0,-1),(-2,0,1),(2,-1,0),(2,1,...

24

I strongly recommend to read this paper of Sharygin.
(It is in Russian, but it worth to translate.)
You will see the reasons to return EG in school, you will also the reasons why it disappears.
Sharygin is my hero, he is the author of many very good math books for school students, he also wrote the best (the opinion is mine) text book in Euclidean geometry ...

23

As long as this question is open I might as well throw in my two cents. I think it is not useful to teach Euclidean geometry to high school students. Here are some reasons I can think of for people to teach Euclidean geometry to high school students and why I think they are bad reasons:
As an introduction to the notion of a proof. As I said in the comments, ...

23

Your question amounts to treating construction problems in
geometry as decision problems, and so it makes sense to me to
adopt the terminology of computability theory. This same kind of
distinction arises in computability theory, where we have the
following terminology:
A set $A$ is decidable if we can computably verify yes-or-no whether a given input $a$ ...

answered Nov 16 '14 at 21:50

Joel David Hamkins

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22

No, because there exists a projective map which preserves the parabola but does not preserve its focus (this is so for any conic and any point.)

answered Dec 31 '14 at 16:48

Fedor Petrov

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21

Euclidean geometry is still taught in American high schools, but I am strongly against it. I think it should be replaced with linear algebra.
Arguments against Euclidean geometry:
Most of what you prove in a high school Euclidean geometry class seems pretty obvious until you learn about non-Euclidean geometry. It makes students think that proofs are ...

21

Similar issues come up in studying gerrymandering (drawing political districts with partisan objectives), where it's useful to have a measure of how "irregular" a region is.
You can read about various classical irregularity measures in this political science paper: Measuring the Compactness of Legislative Districts (Young, Legislative Studies Quarterly, ...

dg.differential-geometry mg.metric-geometry riemannian-geometry euclidean-geometry conformal-geometry

21

To address the scissors congruence question at the end of the post: the Regge symmetries produce tetrahedra which are scissors congruent. This is proved in Section 6 (Theorem 9 and Corollary 10) of
J. Roberts. Classical $6j$-symbols and the tetrahedron. Geom. & Top. 3 (1999), pp. 21-66. (link to paper on arXiv)
The argument is indirect, proving that ...

answered Jan 16 '17 at 19:57

Matthias Wendt

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20

It suffices to consider the case when $\Omega$ is a circumcircle, so let it be.
At first, the points $A_b, A_c, B_c, B_a, C_a, C_b$ lie on a conic if and only if
$$
\frac{AB_a\cdot AB_c}{AC_a\cdot AC_b}\cdot \frac{BC_a\cdot BC_b}{BA_b\cdot BA_c}\cdot
\frac{CA_b\cdot CA_c}{CB_a\cdot CB_c}=1\quad\quad\quad\quad(\heartsuit)
$$
(by Pascal theorem, they lie on a ...

19

A classic problem in this category is Alhazen's billiard problem. I reproduce a quote from 100 Great Problems of Elementary Mathematics. The problem could not be solved using compass and ruler because its solution requires taking a cube root (see references at MathWorld).

19

Your quote about Cartan thinking of $B_n$ and $D_n$ as 'projective groups..." is actually Cartan describing the lowest dimensional homogeneous space of these groups (except, of course, for a few exceptional cases such as $D_2$, which is not simple, and therefore should be left out of the description).
If you go just a little bit further in Cartan's 1894 ...

answered Jun 18 '20 at 9:12

Robert Bryant

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18

Edit: a preprint concerning this problem can now be found on the arXiv: http://arxiv.org/abs/1407.0683
Let me give an exhaustive answer. Croft closes his paper with a list of the unsolved cases:
Here $\kappa$ denotes the maximal edge length of the inner polyhedron. The ratio of the volumes can be easily computed when $\kappa$ is known.
Let's fill in the ...

18

With the caveats mentioned by Andreas, I think Euclidean Geometry makes excellent sense as a high-school course. (My high school experience was not dissimilar to Andreas's -- still the two-column format, but I also had a teacher who understood mathematics beyond what was in the textbook.)
The basic point of agreement (between those Bourbakistes and those ...

18

Let me give a geometric interpretation for the case of tetrahedra of volume zero. The statement becomes as follows:
Given four positive numbers $a,b,c,d$ that satisfy the quadrangle inequalities, denote
$$
s=\frac{a+b+c+d}2, \quad a'=s-a \text{ etc.}
$$
Take a quadrilateral with side lengths $a,b,c,d$ (in this cyclic order). If its diagonals have lengths $x,...

17

Chromatic Number of the Plane or Hadwiger–Nelson problem asks for the minimum number of colors required to color the plane such that no two points at distance $1$ from each other have the same color.
It is only known that
$$5 ≤ \chi ≤ 7.$$

17

Have a look at Hartshorne's Geometry: Euclid and Beyond. He uses Hilbert's axioms for geometry and discusses (section 11) the following "circle-circle intersection axiom (E)":
Given two circles $\Gamma,\Delta$, if $\Delta$ contains at least one point inside $\Gamma$, and $\Delta$ contains at least one point outside $\Gamma$, then $\Gamma$ and $\Delta$ ...

17

Here goes the elementary proof of the claim by Robert Houston that the quadraples $(P_1^{-1},P_2^{-1},P_3^{-1},P_4^{-1})$ and $(\cot \frac{\Omega_1}2,\cot \frac{\Omega_2}2,\cot \frac{\Omega_3}2,\cot \frac{\Omega_4}2)$ are affinely equivalent. In the spherical case the first quadraple should be replaced to $\{\cot\frac{P_i}2\}$, in the hyperbolic case to ...

answered Aug 27 '19 at 21:53

Fedor Petrov

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16

W. Wernick has tabulated 139 triangle construction problems using a
list of sixteen points associated with the triangle. In each case three points are given and the goal is to construct a triangle for which the three 'special' points listed are the points given. There are still some open problems in Wernick's list. Notation:
$A$, $B$, $C$ Three vertices,
...

16

Let O be the point of intersection of the angle bisectors of the initial triangle ABC (the incenter). Then the length of OB' is equal to the length of OC, |OC'|=|OA| and |OA'|=|OB|. Now the statement of the problem will follow from the following fact: given point O and three length x,y,z, consider all triangles EDF that satisfy |OE|=x, |OD|=y, |OF|=z; among ...

16

Your guess is correct. If the elements outside the diagonal have absolute values less than $1/(n-1)$, the matrix has 'diagonal dominance', thus it is nonsingular.
To make the answer self-contained, I give a proof. If $x=(x_1,\dots,x_n)^t$ satisfies $Ax=0$, take $k$ such that $|x_k|$ is maximal and look at $\sum a_{ki}x_i$. The summand $a_{kk}x_k$ has ...

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