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## Hot answers tagged euclidean-geometry

6

Let $T$ be the function in question: $$T(a,b,c,d) = a b^2(b-c) + b c^2 (c-d) + c d^2 (d-a) + d a^2 (a - b).$$ We wish to show $T(a,b,c,d)\ge 0$ if $a,b,c,d$ are the sides of a quadrilateral. (Presumably, $a$ is the side opposite $c$ and $b$ is opposite $d$, but it actually doesn't matter to the proof.) Terminology We introduce the following terminology: ...

5

Hilbert's axiomatization is second-order. However, there is a first-order axiomatization by Tarski, which is complete. (And because the theory is recursively axiomatized, it is decidable.) We cannot develop the full second-order theory of real numbers within the Tarski geometry. For instance, the "plane" $\mathbb{A}^2$, where $\mathbb{A}$ is the field of ...

4

We can assume that our square is $2\times 2$, and scale by $\frac{l}{2}$ at the end. Imagine that our square has corners $(0,0)$, $(0,2)$, $(2,2)$, $(0,2)$. Divide our square into four $1\times 1$ squares. Without loss of generality we may assume that our variable point is chosen in the northeast little square. Then the point furthest away is the origin. ...

4

There is no well defined answer. A very long pipe has area going to infinity yet can have a finite volume. For example, suppose you want $V=\pi$. The volume is $V(r,h)=\pi r^2 h$. Consider a sequence of pipes of dimension $\displaystyle r=\frac{1}{\sqrt{n}}$ and $h=n$. Now the area is $A(r,h)=2\pi r^2+2\pi rh\sim 2\pi\sqrt{n}\rightarrow \infty$ as ...

4

With the idea of the sea urchin you are on the right track. Basically, you are on the way to create a fractal surface. Here are some pictures of simple fractal surfaces (constructed similar to Koch's snowflake in 2d) taken from the Wikipedia site: You get infinite surface but finite volume and finite perimeter.

4

On one interpretation, this is the idea behind Hausdorff measure, which constructs a notion of "volume" given only a notion of "distance". The circle is the unit ball of the usual Euclidean notion of distance, so in a sense this is taking the circle as fundamental. In $\mathbb R^n$, Hausdorff measure is the same as the familiar notion of volume (up to a ...

4

(1) You found the orthogonal projections of the origin onto these lines. So we have $$\mathcal{L}_1=p_0+\mathbb{R}u \quad\mathcal{L}_2=q_0+\mathbb{R}v\quad \mbox{with}\;p_0=\pmatrix{1\\-1\\1}\;u=\pmatrix{1\\1\\0}\;q_0=\pmatrix{-1\\-1\\1}\;v=\pmatrix{2\\1\\3}$$ We are looking for a rotation, that is a direct (determinant one) isometry $f$ such that ...

3

Perhaps surprisingly, it is not. Let $\mathcal{H}$ ba a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ (that is, a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$) one of whose elements is $1$, and let $S$ be the set of all points $(x, y)$ such that the coefficient of $1$ in the expression for $x$ using $\mathcal{H}$ is $0$. Then there are ...

3

Euclid never uses the words straightedge and compass; his axioms include the idea that we can draw a circle of any known radius at any known point, and that one can extend any line indefinitely. These axioms are purely mathematical in nature, and can be given non physical interpretations (as in hyperbolic geometry or finite planes). However, they can also ...

3

The problem seems to trivial. Let $A_1,\dots, A_n\subset [0;1]^2$ be the toppings such that $\operatorname{int} A_i\not=\emptyset$ for every $i$. We can easily choose points $z_i\in\operatorname{int} A_i$ for every $i$ such that no two different points $z_i$ has equal abscissas. Therefore we can slightly inflate each point $z_i$ to a small rectangle ...

3

WLOG $a = \max(a,b,c,d)$ Let $t = \frac{1}{2} (b+c+d-a) \ge 0$ because $a$,$b$,$c$,$d$ are sides of a quadrilateral Then decreasing $(a,b,c,d)$ simultaneously by $t$ reduces the desired expression And $a-t = (b-t)+(c-t)+(d-t)$ Thus it suffices to minimize the expression when $a=b+c+d$, which reduces to: \begin{align} &(b+c+d) b^2 (b-c) + b c^2 ... 3 No, Pythagorean planes, i.e. planes satisfying Hilbert's fourteen axioms up through the Archimedian axiom), form a strictly weaker geometry than Euclidean planes. Hilbert planes correspond to constructions with straightedge and "segment transporter," which almost but can't quite do the job of a straightedge and a compass. Because of this, I get the ... 3 Perhaps you mean the inscribed angle theorem? In French, it's called arc capable, which is similar to the Spanish and Portuguese arco capaz. 3 (The diameter is2 + 8 = 10$, so the radius$OB = 5$, but that's not necessary to know.) The product of the lengths of line segments created by intersecting chords is a constant, so$AX \cdot XB = CX \cdot XD$. This means that$CX = 4,$and$AC = \sqrt{4^2 + 2^2} = 2\sqrt{5}.$2 Hint: We need to use$T$, so why not draw a circle!$O$is the center of the circle.$AH\perp BC$.$\frac{AS}{BS}=\frac{AH'}{SP}=\frac{AH'}{RS}=\frac{AH}{BC}\triangle ABF \sim \triangle AHC$2 Since the edges of the polygon have the same length as the radius of the circle, the angle subtended by an edge from the center of the circle is one angle of an equilateral triangle; that is,$60^\circ$.$\hspace{3.2cm}$Since there are$360^\circ$in a circle, so there are$6$sides to the polygon. 2 Algebraic method: Consider a general circle$x^{2}+y^{2}+2gx+2fy+c=0$. The circle passes through$(8,4)$, which gives$16g+8f+c=-80$. Also, the condition for$x^{2}+y^{2}+2mx+2ny+c=0$and$x^{2}+y^{2}+2px+2qy+d=0$to be orthogonal to each other is$2mp+2nq=c+d$. Hence, making our circle orthogonal to the given 2 circles, we get$-8f-c=0$and ... 2 A) The Fibonacci sequence is sequence of integers, not a geometrical curve. B) The golden ratio is an algebraic number$\varphi=\frac{1+\sqrt 5}2$, not a geometrical curve. D) A torus is a three-dimensional geometrical object, not a curve. Even without knowing the name of the curve$r=k\theta$, you could figure out that the answer is C. 2 A more meaningful title would be appreciated in the future. For the problem: Express the area of the triangle as the sum of the smaller triangles, i.e.:$\frac{AA'BC}{2}=\frac{MA''BC}{2} + \frac{MC''AB}{2} + \frac{MB''AC}{2}$. Then divide and get$1=\frac{MA''}{AA'} + \frac{MC''AB}{AA'BC} + \frac{MB''AC}{AA'BC}$. Then$\frac{1}{CC'}=\frac{AB}{AA'BC}$, and ... 2 Michael J. Mossinghoff, "A \$1 Problem." Amer. Math. Monthly 113 (2006), 385–402; jstor. I quote: The isodiametric problems for polygons were first studied by Karl Reinhardt, Bieberbach's first student, in 1922 [22]. He solved the area problem for odd values of $n$, showing that the regular $n$-gon is best possible. Then, in an appendix that seems to ...

2

It is not true in general if you pass a line through the centroid of a figure, then the line will cut the figure in two equal areas. The simplest example is an equilateral triangle, e.g. those with vertices $(-\frac{1}{\sqrt{3}},0), (\frac{1}{2\sqrt{3}}, \frac12 )\text{ and } (\frac{1}{2\sqrt{3}}, -\frac12 )$. The centroid is the origin and yet if you cut ...

2

Try triangulating the square using the center. The proof should follow naturally by extending your argument above about the side lengths of certain triangles being equal, then noting that the area on either side is just the sum of the three triangles. I've colored triangles that are equal in the decomposition. This should solve the problem at hand, ...

2

for any $\theta \in \mathbb{R}$ the transformation $\psi: x \rightarrow 2x^2 -1$ sends $cos \theta$ to $cos 2\theta$ . hence $\psi^2$ sends $cos \theta$ to $cos4\theta$ if $\alpha = \frac{2\pi}5$ then $cos 4\alpha = cos \alpha$ so that $cos \alpha$ is a fixed point for $\psi$ and if $c=cos \alpha$ we have $$\psi^2(c) = c$$ i.e. $$2(2c^2-1)^2-1= c$$ or ...

2

How about this for a simple proof? Let $A, B, C$ be the vertices of the triangle on a circle. Fix a side, say $AB$, and then we see that the area is maximised by having $AC = BC$ (the base is fixed, height would be maximised if $C$ is on the perpendicular bisector and at the farthest point). As we can choose any side for the base and the other two sides ...

2

You are essentially asking if there exists an orthogonal matrix $B$ such that $$B^TAB=\pmatrix{x&2a&0\\ 0&y&2b\\ 2c&0&z}\tag{1}$$ for some $x,y,z,a,b,c\in\mathbb R$. By splitting into symmetric and skew-symmetric parts, we can rewrite the equation as \begin{align*} B^T\pmatrix{1\\ &4\\ &&6}B &= ...

1

for circumcenter,it is same as incenter, the area is a hex. for centroid, it is a equilateral triangle which inscribe the circle and the position is similar to ABC. to find the area, you can find fix two points on the special points and move one point along the side which you specified. if you can find incenter area, then the other two should be easy ...

1

By $S(...)$ I mean area. Suppose $S(AOE)=x,S(OEC)=y,S(OCD)=z$. We have $AO=2OD$ and $AF/BF=4/3$ so: $x+y=2z \\ \frac{x+y+80}{z+130}=\frac{4}{3}$ Now you can find $z$ and after that you can find $\frac{x}{y}$. Good luck! How to find $x/y$: $\frac{x}{S(AOB)}=\frac{y}{S(BOC)}=\frac{OE}{OB}$

1

I'll prove the formula for a triangle, and it'll be "obvious" that the argument extends to any-dimensional space. Consider $\triangle ABC$ with incenter $O$ and inradius $r$. Recall the fundamental observation that point $O$ helps dissect the triangle into sub-triangles of equal height ($r$), so that \begin{align} |\triangle ABC| &= |\triangle OBC| + ...

1

In the proof, the relation $|\angle R_1RR_2| = \alpha + \beta - \gamma$ comes from canceling the "$+\pi/3$"s in the expression $\alpha^+ + \beta^+ - \gamma^{++}$ (where "$x^+$" abbreviates "$x+\pi/3$"). The goal here, of course, is to find the simplest expression for the angle measure. Reducing the number of "$+\pi/3$"s certainly helps, as does later ...

1

we take E on BC such that BE=BD; so EC=BC$-$BE=BC-BD=AD now we have a theorem that $\frac{AB}{BC}$=$\frac{AD}{DC}$; so in $\bigtriangleup$CED and $\bigtriangleup$CAB WE HAVE A COMMON ANGLE AND $\frac{CE}{CD}$=$\frac{AD}{CD}$=$\frac{AB}{CB}$=$\frac{CA}{CB}$ so we get $\Delta$CED~$\Delta$CAB so we have $\angle$CDE=$\angle$DCE=$\angle$ABC=2x[let] hence ...

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