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Call the intersection points of one (of the two) adjacent lines with the circle $(a,b)$ and $(c,d)$ (4 parameters). You know the distance between these points and the midpoints of the circle (two equations). The diameter connecting the midpoint of one of the circles and the intersection point of the adjacent point with that circle is perpendicular to the ...


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This may be cheating a bit, but why not just create two polygons that "look the same" but have a different graph? For example, we may take a triangle and insert a new vertex somewhere in one of its edges. This produces a new polygon that looks like that triangle, but is actually a rectangle with three collinear vertices.


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The curve you see is by definition a quadratic bézier curve which is always a segment of a parabola.


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well the problem was almost solved.$OM$ is perpendicular to $BC$ and $OB=OC=R$.so we can conclude that $O$ is a point on perpendicular bisector of $BC$.so $DE$ is perpendicular bisector of $BC$.Now we can easily prove that triangle $EMC$ is congruent with $EMB$. now we can says $ \angle EBC = \angle BCE$,so arc $EC$ and $BE$ are equal. and finally $\angle ...


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┌─┐ ┌┐ │ │ │└─┐ │ └─┐ │ └┐ └───┘ └───┘


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Your question really has nothing to do with “rectangles”, which @MPW has already pointed out don’t exist on a sphere. There is a way of looking at your problem that makes it an exercise in basic spherical trigonometry, and I’ll show you that. I do not make any claim that it’s the fastest way of getting your problem solved, though. You have a point, let’s ...


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Let the point at the center of the inner circle be $R$; Let the point at the center of the square be $O$; Let the point labeled at the "top" of the circle (near the number 4) be $A$; Let $r$ = radius of small circle; Let $2L$ = length of the side of the square; Let $x$ = the distance from the $O$ to $R$; We can write a few equations immediately. 1) ...


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For any triangle there is a rectangle with the same area and perimeter. proof: by Heron's formula the area of the triangle with semiperimeter $S$ and triangle sides $a,b,c$ is $$\sqrt{S(S-a)(S-b)(S-c)}\leq \sqrt{\frac{8S^4}{27}}$$ A square with semiperimeter $S$ has area $\frac{S^2}{4}\geq\sqrt{\frac{8S^4}{27}}$. We can then proceed to make the square ...


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Let $AD = DC=CB=BA=a$ and $DE =BZ=x$. a) At first note that the triangles $\triangle CDE$ and $ \triangle CBZ$ are congruent. Hence we can conclude that $CE = CZ$. As $EM =MZ$ and $CE=CZ$ it follows that $CM$ is the perpendicular bisector of segment $EZ$. b) Let $K$ the intersection point between the lines $CM$ and $AB$. The triangles $\triangle EAZ$ ...


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For the first shape, take a rectangle with sides $1,x$. For the second shape, take a right triangle with sides $4,\frac{x}{2}, \sqrt{(\frac{x}{2})^2+4}$. They both have area $x$, for every $x$. Making them have the same perimeter reduces to solving a quadratic equation; a solution is $x=\frac{3+\sqrt{33}}{2}$. Then both shapes have perimeter ...


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Just worked out a quick example, so the numbers may not be optimal: take a triangle with side lengths 2,3,4 - this has perimeter 9 and area $3\sqrt{15}/4$. It's easy enough to construct a rectangle with this data as well, by solving the equations $st = 3\sqrt{15}/4$ and $2s+2t = 9$. In fact, the sides lengths $s,t$ of the rectangle work out to be ...


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Let $A$ be a triangle with corners at the points $(0,0)$, $(2,0)$, and $(0,1)$, with area $1$ and perimeter a little larger than 5. Let $B$ be a parallellogram with corners at $(0,0)$, $(1,0)$, $(x,1)$, and $(x+1,1)$. For any $x$, the areas of $A$ and $B$ are the same, namely $1$. If $x=0$, $A$ has largest perimeter, while if $x$ is large then $B$ has ...


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Here are Gordon–Webb–Wolpert polygons, they have more than just common area and perimeter, they are isospectral! Drums shaped as them would sound the same. It's a fun exercise to construct more examples using tangram pieces.


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Logarithmic spiral Approximate solution The logarithmic spiral is self-similar. As a consequence of that, the tangents make the same angle with the position vector in each point of the spiral. You can use that to draw a reasonable approximation of the spiral simply by taking unit steps in the direction of the current tangent. Perhaps it's easiest to look ...


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Counter-example without words (except for these):


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This idea is wrong, however attractive. Let's consider a polygon with $n$ unequal sides. The side lengths determine the perimeter, so any rearrangement of the sides gives the same perimeter. On the other hand any arrangement of the sides that inscribes the polygon in a circle maximizes the area contained by the polygon. The maximum area is independent of ...


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Two sides are parallel if the sum of the angles between them is a multiple of 180 degrees This is measured by starting from the ending point of the initial side to starting point of the terminal side Consider a square. We can select a side and move clockwise to the next side after encountering an angle of 90 degrees. After moving clockwise again we would ...


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Consider a two-dimensional shape of area $A$ with normal $n$. And imagine that you are looking directly at it (along $n$). You will observe it to have its true area $A$. Now imagine you rotate the shape by an angle $\theta$ about some arbitrary axis so that it now points in the direction $n'$. If you are still looking along $n$ then you will observe the ...


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This book is a great resource. See pdf page 599, actual page 567. http://www.marystarhigh.com/apps/download/7vb7ETI4n4RtLFWDnZw0xNfQRUSB1swoBHQpP7i1l9pXZS1Y.pdf/Precalculus%20Book.pdf You should go to the page before reading on and while reading the rest of the post. In it, it explains everything very coherently and breaks down the derivation into 4 ...


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Coordinatize the triangle on the unit circle via $$A = (\cos\alpha, \sin\alpha)\qquad B = (-1,0) \qquad C = (1,0)$$ Let $T$ be a point of tangency representing generically one of $D$, $E$, $F$.$$T = (\cos\theta,\sin\theta)$$ Define $T_B$ and $T_C$ as points on the tangent line, at distance $t$ from $T$; we can express these as $T \pm t\;T^\perp$, where ...


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I'll consider only your third edit. So, why is $\leq$ necessary in the C-S inequality? The theorem you are studying does not say this, but the case of equality in the C-S inequality is standard: equality holds if and only if $A$ and $B$ are linearly dependent vectors. Of course the proof cannot be based, in this particular case, on the construction of a ...


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W.l.o.g. you may assume the following coordinates: $$A=(0,0)\quad B=(a,0)\quad C=(0,1)\quad D=(b,c)$$ If you define $d:=\cos(x)$ then you can compute the cosines of multiple angles using a Chebyshev polynomial as $$\cos(5x)=T_5(d)=16d^5 - 20d^3 + 5d\;.$$ The dot product of two vectors is proportional to the cosine, so you can write the first angle ...


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If your hiker can see 1 mile in any direction, then we can define success as the probability of choosing a direction vector that will intersect a circle of radius 1 mile around the target. Now, if we let $\theta$ be the randomly chosen angle with $\theta \sim U(0,2\pi)$ (in radians), and our target is $R$ miles away from you, then the probability of finding ...


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Draw a circle of radius $r=1$ mile around landmark. From our starting point at a distance $d$ from the landmark draw a straight line L to the landmark. We want our path to intersect the circle. To do this, the direction chosen must be at an angle less than $\theta=\arcsin(r/d)$ from L (on either side). We choose a direction with a uniform distribution on ...


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Let $CZ$ be the third median of $ABC$. Consider the right triangle $AGB$ with median $GZ$. We know that, since $AGB$ is the right triangle, then $Z$ is its circumcenter and $2\times GZ=AB$. On the other hand, since $G$ is the centroid of triangle $ABC$, $2\times GZ = GC$. QED.


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They've made this just a little too verbose, and they've ignored special cases along the way, and that's why you're getting focused on that. Just hit the high points: If both $A$ and $B$ are $0$, then $|A\cdot B| \le \|A\|\|B\|$ is trivial to show because both sides are $0$. So, without loss of generality, assume $A \ne 0$ (rename if necessary.) Then $$ ...


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These inequalities hold because 1) $\lVert A \rVert^2$ equals $c^2$ plus something positive (top of page 27) and 2) $c^2\lVert B \rVert^2$ equals $\lVert A \rVert^2$ plus something positive (bottom of page 27). They are not assumptions, but follow from the subadditivity, positivity and homogeneity of the norm.


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A sufficient condition could be the following for the case of a polyline bounded on all 4 sides: For every pair of vertical sides with common length $y$, let $y_1, y_2,..., y_n$ be the lengths of $n$ sides parallel to the original pair. Then the square cover number is equal to the number of horizontal sides if the horizontal distance between the $y$ ...


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Replace Math.log(b * theta) by Math.exp(b * theta) and things should agree with the formulas from the article you referenced.


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The synthetic viewpoint is more general! This is a benefit since stating things abstractly in terms of primitive notions is probably easier to grasp rather than delving immediately into coordinates. Compare: Two lines intersect in at most one point The system $\{ax+by+c=0;dx+ey+f=0\}$ has at most one solution. The first comes with less baggage than the ...


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Your 'extreme height of a triangle' is called an arc's sagitta. See http://en.wikipedia.org/wiki/Sagitta_(geometry)


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Hint: let $\theta$ be the angle $AOB$. Then $R\theta = L_{AB}=84$ and $2R\sin(\theta/2)=AB=74$. Height of triangle $ACB$ is $R-R\cos(\theta/2)$.


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The equation $(x-a)^2 + (y-b)^2 = r^2$ describes a circle on the x/y plane; of radius $r$ and centre $(a, b)$. It's solutions include all the combinations of $x$ and $y$ that make up a two dimensional circle, and no other points. Now for 3d space we have a z-axis too, For which values of $z$ is the equation true? Since $z$ is not mention in the equation, ...


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Another way to think about this is the equations of the axes themselves. For instance, take the z-axis. All points on it have x=0 and y=0. Clearly it is a line with any z-allowed. You could write the equation of the axis as $x^2+y^2=0$, because in the reals the only solution to this equation is $x=0$ and $y=0$. This shows you that the z-axis is like an ...


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From JimmyK4542's formulas it follows that $a={2\over3}\sqrt{2s^2-m_a^2}\>$, where $s:=\sqrt{m_b^2+m_c^2}$. From this one derives the following construction of ${3\over2}a$: Construct $s$ as hypotenuse of a right triangle with legs $m_b$, $m_c$; then draw a square with side $s$ and find $\sqrt{2}\>s$ as length of the diagonal $d$. Draw a Thales ...


1

I'd call this a self-similar logarithmic spiral which is the image of a square under an iterated similarity transformation. There is a similarity transformation which maps the corners of the outermost square to those of the second-to-largest one, and iterating that transformation will yield an infinite sequence of squares converging to the center. The ...


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First off, note that the the term “rectangle” is not well defined if you actually consider the spherical case. There are no spherical quadrilaterals with geodesic edges and four right angles, although small quadrilaterals can come close. I guess that you have a bounding box in mind which comes from the extrema of the latitudes and longitudes of some data ...


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There is way easier solution mentioned here without the need for a solver.


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I commented on the previous post as well! There is a good reason you couldn't find an example by hand: the smallest example of what's called a perfect squared square is a $112\times 112$ (link). There is much more research here.


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This is the implicit equation of a cylinder: a point $(x,y,z)$ lies on the cylinder if it satisfies the equation. The important thing here is in fact that $z$ does not occur in the equation. Which means that if some point $(x,y,z)$ lies on the cylinder, then another point $(x,y,z')$ which differs from the first only by a change in its $z$ coordinate will ...


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The equation of an object is a way of telling whether a point is part of an object -- if you substitute the coordinates of the point into the equation and the equation is true, then the point is on the object; if the equation is not true for that point, then the point is not on the object. There is no $z$ because the z-coordinate is not part of the decision ...


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I'll rephrase the argument by Emanuele Paolini without "$1$-Lipschitz" and projection. Place a bomb inside of the convex polygon. When it goes off, the polygon breaks apart at the vertices, with every side of it flying out in the direction in which it was facing: I drew arrows to show the directions in which two of the sides go. Flying out, each side ...


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The formulas for the lengths of the medians of a triangle given the sidelengths are: $m_a^2 = \dfrac{2b^2+2c^2-a^2}{4}$ $m_b^2 = \dfrac{2c^2+2a^2-b^2}{4}$ $m_c^2 = \dfrac{2a^2+2b^2-c^2}{4}$ Solving for $a,b,c$ in terms of $m_a,m_b,m_c$ gives: $a^2 = \dfrac{8m_b^2+8m_c^2-4m_a^2}{9}$ $b^2 = \dfrac{8m_c^2+8m_a^2-4m_b^2}{9}$ $c^2 = ...


1

put Circle center on $O$, $ABCD $is counter clockwise , $A(\cos{A},\sin{A}),B(\cos{B},\sin{B})$ etc. $AB=2|\sin{\dfrac{B-A}{2}}|,CD=2|\sin{\dfrac{D-C}{2}}|$,$|AB-CD|=2||\sin{\dfrac{B-A}{2}}|-|\sin{\dfrac{D-C}{2}}||$ $B-A=2\alpha,C-B=2\beta,D-C=2\gamma \implies \alpha+\beta+\gamma<\pi,0<\alpha,\beta,\gamma<\pi$ the inequality become: ...


1

Calling $x, y, z$ the coordinates of the source, and $x_i, y_i, z_i$ ($i=1...4$) those of the receivers, the distance $D_i$ between the source and the receiver $i$ is given by $$\sqrt{(x-x_i)^2+(y-y_i)^2+(z-z_i)^2}$$ Considering two receivers $i$ and $j$, the difference $D_i-D_j=D_{ij}$ in the distance from the source is given by $$ ...


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If a convex set is contained in another set, the convex set has smaller perimeter. This can be understood by considering the projection map on the convex set (i.e. the map which sends a point to the closer point on the convex set). This is a map which does not increase distances and sends the perimeter of the enclosing set onto the perimeter of the convex ...


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The claim is false as written. $G = \langle a, b \mid b^2 = 1 \rangle \cong \mathbb{Z} \ast \mathbb{Z}_2$. This group is a free product $A \ast_C B$, where $\mid A/C \mid \geq 3$ and $\mid B/C \mid \geq 2$. ($A=\mathbb{Z}, B=\mathbb{Z}_2, C=\{1\}$). By Stalling's Theorem, this group has infinitely many ends. But $\mathbb{Z}$ has 2 ends. Since ends are a ...


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You can use circular inversion, or just intersect the segment $BI$ with the hyperbola that is the locus of points $P$ for which $PG-PI=BG$. That intersection is the center of the circle you are looking for. If you are familiar with the Descartes circle theorem, you know in advance the radius $r$ of your circle, since, given that $BG=1$, ...


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Presumably, $\pi(z,\ell)=z$? It's not bijective, at least for $n>1$. $$\pi^{-1}(0)=\{(0,\ell)\mid\ell\in\mathbb CP^{n-1}\}$$ It is a bijection when excluding the $0\in \mathbb C^n$ and $\pi^{-1}(0)$ from the sets, because when $z\in\mathbb C^n\setminus\{0\}$, there is exactly one line $\ell$ through zero containing $z$.



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