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1

The isoperimetric inequality gives the lower bound: $$L_n\geq 2\pi\sqrt{n},$$ and I believe we can achieve: $$L_n\leq C\cdot\sqrt{n}$$ through the following strategy: choose some integers $m_1,\ldots,m_k$ such that: $$0\leq m_1\leq m_2\leq\ldots\leq m_k,\qquad m_1+\ldots+m_k = n$$ and consider concentric regular polygons $\Gamma_1,\ldots,\Gamma_k$, with ...


0

If you have all the sides, you don't need the angles-which do you believe? Taking the sides, you have two equations in two unknowns: one for the distance from $A$ and one from the distance from $B$. So $(x_3-x_1)^2+(y_3-y_1)^2=d_1^2$ and $(x_3-x_2)^2+(y_3-y_2)^2=d_2^2$ You are correct there will be two solutions. With two quadratics you might expect ...


0

(I just want to comment on answer by @varun-iyer but my reputation is too low to do so) Dividing circle in equal pie shape is obviously wrong, OP already gave an example image (smaller circle in middle, then divide the donut shape equally) that always result in smaller perimeter sum for most cases. Using OP's example partitioning, $$L_n = ...


-1

Well, if you have a unit circle with area of $\pi$, and each nth piece having an area $\pi/n$, we can say that the perimeter of each nth piece will be $$\frac{2\pi}{n} + 2$$ So the sum of all the pieces will be: $$(\frac{2\pi}{n} + 2)*n = 2\pi + 2n$$ If we express as a function of n: $$f(n) = 2\pi + 2n$$ Taking the derivative: $$f'(n) = 2$$ $f'(n)$ ...


1

This is a common exercise in basic complex analysis so I will help you by outlining the steps of the proof: 1) Show that the class of functions of the form $\frac{az+b}{cz+d}$ with $ad-bc \neq 0$ send lines to lines and circles to circles. 2) Show that every bijective map $T(z): \mathbb{C} \to \mathbb{C}$ has the form $az+b$ with $a \neq 0$. 3) Extend (2) ...


1

Take one of the corners, say it has coordinates $(x,y)$. Let $\alpha$ be the angle that $(x,y)$ makes with the positive $x$-axis and let $r=\sqrt{x^2+y^2}$. Then $x=r\cos\alpha$ and $y=r\sin\alpha$. Let's say you are rotating by an angle of $\theta$. Call this linear transformation $T_\theta:\mathbb{R}^2\to\mathbb{R}^2$. Then $T_\theta(x,y)$ makes an angle ...


3

As long as you only need a cover, and not a partition, things are easy: Given a pair of foci and any point in the plane, there exists an ellipse as well as a hyperbola through that point with the given foci. There are three special cases. If the point lies on the line segment between the foci, the ellipse will degenerate to that line segment. If the point ...


0

As suggested by the OP, if we can show that the line through D parallel to AB is the perpendicular bisector of AC, then x = … = 15 degrees. The following shows how this can be achieved. Constructions:- (1) Draw the rectangular coordinate system X’DX and Y’DY. (2) Draw a circle using D as center, radius = DB. Then AC = DB = DY = DY’. (3) Join AB and also ...


0

If you project a line segment onto a plane, it'll still be a line segment, hence have no inflections. (Except in the special case where the projection is a single point, i.e., a projection "along the line"; in that case there are no inflections, but the projection is a degenerate segment, i.e., a single point.) If you mean "a piece of a curve", then it's ...


0

A useful technique for dealing with circle packing problems of this type is circle inversion. Since inversion preserves tangency and sends every line and circle to another line or circle (with circles passing through the center of the circle of inversion becoming lines), if we invert with respect to a circle of radius $1$ centered at one of the corners of ...


0

I'm adapting my answer on Stack Overflow. 2D case Just like the dot product is proportional to the cosine of the angle, the determinant is proprortional to its sine. And if you know the cosine and the sine, then you can compute the angle. Many programming languages provide a function atan2 for this purpose, e.g.: dot = x1*x2 + y1*y2 # dot product det ...


0

Situation $2$: For each selection of any $4$ points on the circumference, you can draw the diagram you have. The line segment that joins the adjacent circumference points, could instead join any of the $4$ pairs of adjacent circumference points, so we have $4$ different triangles for each choice of $4$ circumference points. There are $\binom{n}{4}$ ways to ...


0

This is not true. You need $C^{1,1}$ for the interior ball condition, $C^1$ does not suffice. Counterexample: $\Omega = \{(x,y)\in\mathbb R^2 : y>|x|^p\}$ with $1<p<2$. Make $\Omega$ bounded by capping it off somehow, the issue is local, at $(0,0)$. There is no interior disk that touches $(0,0)$. Indeed, it would have a horizontal tangent there, ...


1

Let's make sure we understand splitting, before we talk about joining. To subdivide a Bézier curve into two, you use the deCasteljau algorithm, as illustrated in the figure below. Suppose we are given a curve defined by four control points $A$, $P$, $Q$, $G$, and a splitting a parameter value $u \in [0,1]$ (which you called $t_{\text{cut}}$). To split ...


12

Assume that the side of the square $\overline{AB}=1$. Consider the diagram $\hspace{3cm}$ By symmetry, $\overline{EC}=\overline{CD}$; therefore, $\overline{CD}=1/2$. Since $\overline{AC}=1$ and $\overline{AD}\perp\overline{CD}$, we have that $\angle CAD=\pi/6$ ($30$-$60$-$90$ triangle). Similarly, $\angle GAF=\pi/6$, leaving $\angle CAG=\pi/6$. Since base ...


0

Converting comments to answer, as requested. The perpendicular from $O$ to $\overline{KL}$ must also be perpendicular to $\overline{MN}$. (Why?) Let $P$ and $Q$ be the points where the perpendicular crosses these segments. Show that $\overline{OP}\cong\overline{OQ}$. (Hint: The perpendicular from $O$ to $\overline{AB}$ meets the segment at its midpoint, ...


1

You can use physical ideas to help construct a simple counterexample to show that unless additional conditions are imposed, like positive definiteness etc, the symmetric part can represent quite different things. Consider (for simplicity) a 2D body (a unit square with one vertex at the origin) in a fixed cartesian basis, and subjected to the following ...


0

I have also made some graphics like this where I needed some kind of interpolation. I often ended up trying to find a general formula that includes the initial case as well as the target as special cases where each just some parameters have to be changed. For example for the function $z \mapsto z^2$ The more general function would be $f_c(z) = z^c$ and for ...


1

I prefered 3D coordinates system to prove the lemma. In my opinion , this tool is the most elementary tool for that problem. The projection area $Q$ on $xy$ plane and bordered by $(OED)$ can be written as $$\mathbf{M_{2}}=\begin{bmatrix}0 & 0 & 1\\x_1 & y_1 & 1\\x_2 & y_2 & 1\end{bmatrix}$$ $Q=\frac{1}{2} ...


1

Let R be Radius & D its Diameter R = 5 ; D = 10 $$ Total Area of the Big Square = D^2 = 100 $$ $$ Area of the Circle = \pi \div 4 \times D^2 = 78.54 $$ $$ Area of Other than Circle = 100 - 78.54 = 21.46 $$ $$ Area of 4 Petals = 78.54 - 21.46 = 57.08 $$ $$ Area of 1 Petal = 57.08 \div 4 = 14.27 $$ $$ Area of \frac 1 4 \times Square = R \times R = 25 ...


2

Consider the quarter-circle of radius $r$ and a $\pi\over 2$ rotation of it where the two arcs share two common corners and each shares three corners with the square of side length $r$. Call the area not covered by quarter-circle pieces a "counter-arc" (a triangle with missing arc pieces). The area covered by quarter-circles can be broken up into ...


16

You "curvilinear square" just cuts the quarter-circles in thirds, so the distance between two adjacent vertices is $2l\sin\frac{\pi}{12}=\frac{\sqrt{3}-1}{\sqrt{2}}l$, given that $l$ is the length of the side of the original square. So the area of the "circular square" is given by $(2-\sqrt{3})l^2$ plus four times the area of a circular segment. The area of ...


0

this is geo method to prove that only one solution for this question. Assume we already know the solution of $ x=15°$ then if $B< 45°$ ,see picture below: note $\angle BDC= \angle DCA$, so $AC$ tangent $BDC$ circumcircle $K$ ,and $BD=AC$ which means $D$ is intersect point of two circles.(ie, orange and blue) $M$ is midpoint of $BC, A'M \perp BC ...


1

This is NOT a solution. It is just some of the interesting findings in the course of seeking the required solution. I have to give that up because I cannot think of any method of incorporating the given 5x into my work. I just hope that someone can have an elegant geometric solution and some of my findings maybe useful.


7

As stated in the comments, due to the Descartes circle theorem four mutually tangent circles satisfy the identity: $$ 2(\kappa_1^2+\kappa_2^2+\kappa_3^2+\kappa_4^2)=(\kappa_1+\kappa_2+\kappa_3+\kappa_4)^2$$ where $\kappa_i$ is the curvature of the circle $\Gamma_i$, i.e. the reciprocal of the radius. By setting $\kappa_n=\frac{1}{R(n)}$, Vieta jumping ...


6

I believe your recurrence should have the form $$R(n) = \frac{\left(1 - 2 \sum_{k=1}^{n-1} R(k)\right)^2}{4\left(1 - \sum_{k=1}^{\color{red}{n-1}} R(k)\right)}.$$ To solve this, let $S(n) = \sum_{k=1}^n R(k)$, and solve the above expression for $S(n-1)$: $$S(n-1) = \frac{1}{2}\left(1 - R(n) - \sqrt{R(n)(2+R(n))}\right),$$ where we take the smaller root ...


5

Domains are open, so not compact. If you throw in the boundary to make it compact, it will no longer be a manifold (though it will be a manifold with boundary, since $\Gamma$ is smooth).


1

Write the equations of the given lines in the form $l=0$. Then consider the line $$(x+2y-5)+\lambda(x-3y-7)=0.$$ For every $\lambda$ this represents a line passing through the intersection of the given lines (with the exception of the second line itself). This can be rewritten as $$x(1+\lambda)+y(2-3\lambda)-(5+7\lambda)=0.$$ From this you can compute ...


1

Using the feedback given in the comments, I've got a proposal for a solution. I must say that I am not very certain of whether it is correct or not. Let $0<x<y<1$ be the points at which the "stick is broken", and so $x, y-x, 1-y$ are the lengths of the three segments. For a triangle to be formed, the sum of any two sides must be greater than the ...


0

does.fixed length mean independemt of side of triangle. if i assume an equilateral triangle as T and point A as centroid then the sum you are talkimg about would be root(3)/2 times side square (dependent on side)


2

Per light of @Blue's last comment, here's a (very brief) solution: Since $E$ is the midpoint of $E'E''$ and $\angle A=90^\circ$, $\triangle AEE'$ is isosceles at $E$. It follows that $\angle AEE''=2 \angle EAE'$, equivalently $E$ trisects the small arc $AC$. Similarly, $F$ trisects the small arc $AB$ and $D$ trisects the large arc $AC$ (and $AB$). That ...


1

I suspect that this explanation be a bit too imprecise, but I think it can be made to work. (In what follows, it may help to consider the image of a triangular helix joint shown here.) Take two consecutive segments of your helix, and consider the isosceles triangle with them as its legs. Draw the line in 3D which bisects the joint angle, and repeat this ...


9

This the first example that came to my mind, using four identical $45-45-90$ triangles. They obviously have the same area, and the perimeters of both are "two legs and two hypotenuses," it's just that the one hypotenuse is bent into the v shaped notch in the second picture.


2

Remember that $\Bbb P^n(\Bbb C)$ is the quotient of $\Bbb C^{n+1}\setminus\{0\}$ under the parallelism equivalence of vectors in $\Bbb C^{n+1}$ and there's a quotient map $$ \pi:\Bbb C^{n+1}\setminus\{0\}\longrightarrow\Bbb P^n(\Bbb C) $$ under which the linear subspaces in the target correspond under inverse image to vector subspaces in the source space ...


1

My solution: Let $U,V \sim \mathcal{U}[0,1]$ be 2 sides of your triangle, and $1-u-v$ being the other one (as by definition they add up to 1). Not every pair $(u,v)$ is acceptable, as they need to form a triangle. The condition for that is that the biggest vertex is smaller than the sum of the two other sides, that is: $$ \max(u,v,1-u-v) > 1 - ...


0

Angles do not usually have to be an integral number of degrees, so if you are requiring that you should say so. In that case, for the triangle you are fine. For the quadrilateral, once the angle gets larger than $180$ you have a concave quadrilateral, which is not always allowed, but if you accept them, you are fine. For more sides, angles greater than ...


0

I think, your guess is wrong. You guess smallest possible angle is 1 and largest possible angle is 178 for triangle. Triangle with this angles have angles 1, 1, 178 as you know the sum of all angles of a triangle is 180. We can take an angle which is less than 1. Let this angle be 0.1. Then the sum of other two angles must be 179.9, these angles can be 179 ...


1

The set of values which are valid angle measurements does not have a minimum or a maximum if we are talking about non-degraded convex polygons. It does have an infimum and supremum - in case of a triangle the infimum is 0 and the supremum is 180.


0

Well that is not the case . The largest possible value of an interior angle could be is 180 . After that I think , it is denoted as an exterior angle . Just think about it a little bit . Also if n is supposed to be 5 then largest angle according to the above formula is 536. Not possible .


1

Since the Heron's formula gives: $$ A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)},$$ assuming that the random choice is made accordingly to the broken stick model (notice that not every choice for the side lengths satisfies the triangle inequality, hence for such cases we set the area as zero): ...


1

If you know that the two curves you have resulted from splitting a single initial curve, then you don't need both these curves; knowing either and the cut position is enough to restore the full initial curve. Have a look at this post on Stack Overflow. It discusses how you cut a curve. What you want to do is do the reverse. To find control points $Q_i$ for ...


1

Let $d$ be the length of one side of the polygon, which is what we want to find (e.g. $EH$ in the second picture). Also let $\theta = \angle{EAH}$. Then, with $R$ the radius and $L$ the inner arc length ($EH$ for example), we find $\theta$ by dividing $L$ by the circumference of the circle: $\theta = \dfrac{L}{2\pi R} 2\pi = \dfrac{L}{R}.$ $d = 2R ...


1

Let be $X_1$ the center point of first circle $X_2$ of the second. $V_1$ the intersection point on the first circle $V_3$ of the second. $d$ the distance between $X_1$ and $X_2$. $r_i$ the radii. The angle $\alpha$ between $V_1X_1X_2$ is given by $\cos\alpha=\frac{R_1-R_2}{d} $ Thus $V_1$ is given by $X_1 + R_1*\hat e_\alpha$ where $e_\alpha$ points in the ...


0

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 ...


3

We try to model the family of lines and then try to infer the envelope. The guiding lines (left and right arms of the V shape) are $$ g(t) = u_g \, (1-t) + v_g \, t \quad h(t) = u_h \, (1-t) + v_h \, t $$ for $t \in [0, 1]$, where $u$ is the start point and $v$ the end point of that line. A line $f_r$ of the family starts on $g$ and ends on $h$: $$ f_r(t) ...


3

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.


0

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


0

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 ...


35

┌─┐ ┌┐ │ │ │└─┐ │ └─┐ │ └┐ └───┘ └───┘ Edit: I like the above figures because they're easy to generalize to many sides. But if it's unclear that they have the same area, here's another pair: the L and T tetrominoes. You can imagine sliding the square on the right side up and down relative to the 1x3 bar on the left side; this operation ...


0

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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