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2

$$r = \frac {6}{1-\sinθ}$$ $${r-r\sinθ}=6$$ $$\sqrt{x^2 + y^2}-y=6$$ $$\sqrt{x^2 + y^2}=y+6$$ now squaring both side and simplyfying we get $$y = \frac{1}{12} x^2 -3$$


1

Write $$6=r(1-\sin\theta)=r-r\sin\theta=\sqrt{x^2+y^2}-y$$ Then $$6+y=\sqrt{x^2+y^2}\implies12y+36=x^2$$


1

Just multiply by denominator: $$ r = \frac{6}{1-\sin\theta} \Longrightarrow r(1 - \sin\theta) = 6\Longrightarrow r - y = 6 $$ Now $$ \sqrt{x^2+y^2} = 6+y\Longrightarrow x^2 + y^2 = (6+y)^2, $$ or $$ x^2 = 36 + 12y \Longrightarrow y = \frac{x^2}{12} - 3 $$


0

First decide whether you mean the line segments only, or the whole lines through the two points. Then the following method should lead you to the answer: Generate four points at random. There are three ways to pair those points together, and the probability that the lines/line segments intersect will be $0$ or $\frac13$ or $\frac23$ or $1$, depending on ...


9

Consider the following, where we see a full barn (of a curious gravitational nature) with four ladders of length $\color{green}{g}$ leaning against side-$w$ boxes stuck in the corners where floor and ceiling meet walls: Clearly, $$\begin{align} (w+\color{violet}{p})^2 &= W + \color{violet}{P} + 2 \color{green}{G} + 2 \color{blue}{B} + 2 ...


0

Given any convex polyhedron $P$, consider the Minkowski sum of $P$ with $\bar{B}(\epsilon)$, a closed ball centered at $0$ with small radius $\epsilon$. $$P_{\epsilon} \stackrel{def}{=} P + \bar{B}(\epsilon) = \{\; \vec{p} + \vec{q} : \vec{p} \in P, |\vec{q}| \le \epsilon\; \}$$ Let $K$ be the Gaussian curvature on the boundary $\partial P_{\epsilon}$. Let ...


0

$$\exists Z\in\mathbb C:\exists a\in\mathbb R:\forall k\in\{1,2,\dots,n\}: \lvert Z-z_k\rvert=a$$ is indeed a valid definition of cocircularity. It's pretty straight-forward: you ask whether there is a center $Z$ and a radius $a$ such that the points lie on the circle defined by these two.


4

Consider the equation of the line that the ladder makes: $$\mathcal{l} = \{ (x, y) ~:~ y = mx + h \}$$ where $h$ is the height of the ladder on the wall and $m$ is the slope of the ladder. We know that $(1, 1)$ is on the line, so $$1 = m + h$$ And we know that the distance from the x-intersecpt to the y-intersept is $4$. So $$h^2 + (-h/m)^2 = 4^2$$ ...


14

Let $|FB|=x$. By similarity of triangles we then have $|CE|=1/x$. Pythagoras thus gives $$ 4=\sqrt{x^2+1}+\sqrt{1+(1/x)^2}=\sqrt{x^2+1}(1+\frac1x). $$ Squaring this gives us $$ 16=(x^2+1)(1+\frac2x+\frac1{x^2}), $$ but I prefer to move one factor $x$ from the former factor on r.h.s. to the latter, so $$ 16=(x+\frac1x)(x+2+\frac1x). $$ Getting warmer! Write ...


1

Note how the Fibonacci sequence provides a nice way of generating more such triangles, since for any four consecutive elements $S_n, S_{n+1}, S_{n+2},S_{n+3}$ we have: $S_n \cdot S_{n+3} = S_{n+1} \cdot S_{n+2} \pm 1$ (sign alternates for even and odd values of n). Then make one of the two triangles on the "not-quite" hypotenuse have dimensions (width x ...


0

Remember 10 feet = 120 inches. It is directly above the person's head, so just subtract the smaller one from the larger: (120 in) - (75 in). You get 45 inches. To convert between units, you know that 12 inches (12 in) make 1 foot (1 ft). Therefore, multiply your (45 in) by $\frac{1 ft}{12 in}$ (same as multiplying by 1 because you know the distance is the ...


0

I don't want to close this question, so I will post the answer from Mathoverflow here. The credits goes to Joseph O'Rourke. (Thank you very much!) Let me try to sketch a proof of "the evident fact that sum of angles around a vertex of a convex polytope is less than $2\pi$," as phrased by Alexandre Eremenko. Let $v$ be a vertex of the convex polyhedron ...


3

Let's assume that $x$, the side of the equilateral triangle, is a known positive quantity and that side $BC$ is horizontal (or that point $A$ is directly above point $M$). Let's also assume you are using Cartesian coordinates (where increasing the first coordinate means moving right and increasing the second coordinate means moving up). Then $x$ is the side ...


2

Draw the circle of radius $\frac {\sqrt 3}3x$ around $M$. Pick an arbitrary point $A$ on this circle. Then intersect the circle of radius $X$ around $A$ with the first circle to determine $B,C$ as intersection points. Note that $A$ could be picked anywhere on the circle, hence the result is not unique.


1

Note that we are assuming that $BC$ is parallel to the coordinate axis. Now, we can get all the horizontal and vertical distances we need to solve this problem. Let $P$ be the midpoint of $BC$. Note that we have $PC = BP = x/2$ and $MP = (x/2)/\sqrt{3}$. Finally, $MA = x/\sqrt{3}$. This should be enough to get all the coordinates. EDIT: A better method, ...


0

Using $d(p,f1)+d(p,f2)=2a$ produces an insane amount of calculations with roots. It's better to use the general ellipse equation $y^2/a^2+x^2/b^2=1$. Than you substitute $x$ and $y$ by the xy-coordinates of the point. You will get: $$\frac{9}{a^2}+\frac{36\over 4}{b^2}=1$$ But $a^2=b^2+c^2$ and $c=a/2$, then: $a^2=\frac{4b^2}{3}$. Substitute in the equation ...


1

Another version of a): More formally:


3

No, it's not possible, at least not possible in a bi-continuous way. Because by a generalization of a Peano curve, there's a map from the unit interval onto the unit cube in 3-space, the euler-angle parameterization can be reduced to a single-number parameterization...but that's completely useless in practice. Now...why, if you want a "nice" ...


1

For small numbers of cells $3,4,5,..$ there may be a possibility of finding explicitly the optimal partition (the Y 120 degrees partition in the case $n=3$, etc). For large numbers of cells in the partitions, analytical results are not known. In order to find candidates for the optimal partitions, some numerical studies were performed: Cox and Fikkema: ...


1

$|2z-1| = |z-2|$: Rearrange to $|z-\frac12| = \frac12|z-2|$: distance from $\frac12$ is half distance from $2$. So it's a circle (distance from one point is a constant multiple $-$ not $0$ or $1$ $-$ of distance from another point). You know two points on the diameter: $1$ and $-1$. So you know the centre and radius. $|z-8|+|z+8|=20$: Distance from one ...


1

Some fundamental equations: $|z - z_0| = \rho$: circle with center $K(z_0)$ and radius $\rho$. $|z - z_1| = |z - z_2|$: perpendicular bisector of the line segment with end points $K(z_1)$ and $K(z_2)$. $|z-z_1| + |z-z_2| = 2a, \, a>0 \text{ and } |z_1-z_2|<2a $: ellipse with foci $E(z_1),E'(z_2)$ and constant sum $2a$ (recall the definition of the ...


1

A manifold is a space $X$ such that each point $p\in X$ has neighborhoods which are homeomorphic to open euclidean balls. In particular the neighborhoods of all points $p\in X$ "look alike". If the manifold $X$ is a compact topological space then $X$ is called a closed manifold (this wording is somewhat unfortunate, see below). The prime examples of ...


0

Expanding on the above comment, if we substitute $z = c(1-x)$ in the equation of $K$ we get $4x^2 = y^2 + c^2(1-x)^2$, i.e. $$ (4-c^2)x^2 - y^2 + 2c^2 x = c^2 \label{eq:1} \tag{1} $$ Now all we have to do is distinguish a few cases. If $c = 0$ then \eqref{eq:1} becomes $$ 0 = 4x^2 - y^2 = (2x + y)(2x - y) $$ so $E_0 \cap K$ is a degenerate conic, the union ...


4

The optimal untouched area is when $b=h$ (ie, when $B H$ is a square), where $$\text{ area}=b^2 \left(-2 \sqrt{3}-\sqrt{7}+\frac{2 \pi }{3}+7-4 \csc ^{-1}\left(\frac{4}{\sqrt{7}}\right)\right):$$ Here, $B:H=1:1$ The area $=0$ for $h\geq\left(\sqrt{7}-2\right) b$ - or, of course, for $b\geq\left(\sqrt{7}-2\right) h:$ Here, $B:H=1:\left(14+3 ...


0

Note well that the answer is necessarily positive only in finite dimensions, I.e. Balls in R^n. In infinite dimensions, it depends on the topology used. For example, in infinite dimensional Hilbert space, the unit ball is non-compact in the topology induced by the norm. This can be seen as a basis can be viewed as a sequence without accumulation point. If ...


0

Assuming that the sides of the polygon are axis aligned: If you include the extra condition that the vertices are in general position (no three have the same $x$ coordinate and no three have the same $y$ coordinate), then given a set of points, there is a unique rectilinear polygon that can be constructed from them. In that case, you can order the points ...


1

Just as in your question How to transform (rotate) this hyperbola?, it appears you've omitted the multiplication symbol between the parenthesized factors in the second plot.


1

The problem can be solved using parametric equations. Since you know that both equations pass through the point, $(1,4)^T$, convenient parameterizations for the lines are: $\left(\begin{array}{c}x_1\\y_1\end{array}\right) = \left(\begin{array}{c}1\\7\end{array}\right)s+\left(\begin{array}{c}1\\4\end{array}\right)$ and ...


0

The basic idea is to chop up the convex polytope into a bunch of simplices, then to pick one simplex at random (fairly) and then to sample within the simplex. The hardest part is determining the vertices of the polytope (which you need to triangulate it), since you have a facet-based description (inequalities) rather than a vertex-based description, and the ...


2

You have the vertex $A$ and the equations for the congruent sides. So you can compute an apex angle $\theta$ with the dot product of two vectors from $A$. Then, using that angle (or its complement) you can calculate the length of the congruent side $d$ from $$\frac{d^2 \sin \theta}{2} = 5.$$ This gives you the length to go away from $A$ along the lines, ...


0

The vectors given are positions of the triangle's vertices. Vector $$\vec a = 3i−2j+5k$$ represents a vertex $A$ whose coordinates are $(3, -2, 5)$. Use $$distance = \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$$ for each pair of vertices to calculate the triangle sides' lengths, then add them to obtain a perimeter. Once you know the sides, use the ...


1

The side vectors of the triangle are given by the differences of the position vectors of the vertices. For example $$\vec a -\vec b = 2i+4j-k$$ is one of the sides whose length is $\sqrt{4+16+1}=\sqrt{21}.$ Can you do the same with the other two edges? The following figure may help to better understand the solution: Regarding the angles: Let's ...


0

The angle bisector from $A$ meet the circumcircle of $ABC$ in the midpoint of the $BC$-arc, hence, assuming the origin is in the circumcenter of $ABC$: $$ z_4 = \sqrt{z_2 z_3}.$$


1

$$ax + by + c = 0 \implies y = -\frac{ax+c}{b}$$ So the line intersects the $x$-axis at $\displaystyle \left(-\frac{c}{a}, 0\right)$ and intersects the $y$-axis at $\displaystyle \left(0, -\frac{c}{b}\right)$. It is easy to see that our area is given by $$A = \frac{c^2}{2|ab|}$$ To see that, see below. Excuse my bad drawing skills, but the diagram ...


0

we get the $x$ intersection as $x=-\frac{c}{a}$ and $y=-\frac{c}{b}$ thus our area is given by $$A=\frac{c^2}{2|ab|}$$


5

First of all let us consider the area swept after a one flip (the white area in your diagram and the cyan area in mine) by the rectangle $ABCD$ with base $b=AB$ and height $h=AD$ on a line (in other words, the corners of the outside rectangle aren't taken into account, yet): It can be seen that this area is composed of the circular segment $DD'$ and ...


5

a) All pairwise distances are different. --edit: this argument is insufficient -- b) To be shot $k$ times, a gunman must be the closest member to $k$ others. Thus, what is the maximum packing density at which point no more gunmen can be added without drawing fire? c) Intersecting trajectories would require that distances $AC < BD$, and form a ...


1

Hints (a) Let each shooting be a directed edge. Then prove that any cycle must be a 2-cycle. Then prove that removing all cycles results in a nonzero number of paths. (b) Maciek gave a more or less full answer already so there is no hint left to give. (c) It is indeed a geometric fact.


13

Suppose that two gunmen $(A, B)$ shoot a special individual $C$. Then if you consider the triangle $ABC$, the longest side is $|AB|$, so the $\angle ACB > 60^\circ$. And you cannot put 6 people around somebody where all the angles are greater than $60^\circ$.


1

In a triangle $ABC$, the angles of the incentral triangle are just $\frac{A+B}{2},\frac{A+C}{2},\frac{A+B}{2}$. If we define the variance of the angles as: $$ \text{Var}(A,B,C) = (A-B)^2+(B-C)^2-(C-A)^2 $$ then we have that the variance of the incentral triangle is exactly one fourth of the variance of the original triangle. The claim readily follows.


5

Your equation can be transformed to $(x+y+z-3)(x+y+z+2)=0$, so $x+y+z-3=0$ or $x+y+z+2=0$, so it's a union of two parallel planes.


1

maybe this could be helpfull for you: http://mathworld.wolfram.com/CassiniOvals.html You want to have $$g(y,r_b,r_c)=x \Leftrightarrow \nabla f(x,y,r_b,r_c)=\delta$$ for given $$y,r_b,r_c,\delta$$ I think there is no analytic solution for g, so you have to solve it numericly. For example you can use something like ...


-1

$\frac{\theta}{360}\times 2\pi (r) + 2r$ That would be the length of the arc in the first part and twice the radius in the second part.


0

Verbally stated, x as well as y each must lie between their respective extreme points to have an intersection. And symbolically $$ (x-x_1)( x_2 -x) > 0,\; AND \, (y-y_1)(y_2-y) > 0 $$ where $ \ge $ sign valid if intersection occurs at one extremity of a line.


0

Hint:Prove that angle EDC is 90 by using similar triangles.


0

We see that two points $P_1$, $P_2$ are an the same (opposite) sides of the line $QR$ if an only if the $\it{oriented}$ areas $\text{Area}(P_1 QR) $ and $\text{Area}(P_2 QR)$ have the same (opposite) sign. The expression that appear are just determinants. This is easily generalized in $n$-space, when inquiring whether two points $P_1$, $P_2$ are on the ...


1

It all depends on what you define as "steep". I would say that the steepness of the slope is mathematically more closely approximated by the absolute value of the gradient, not the value itself. That way, the direction you are traveling in (up or down) does not matter and the line $y=10x$ is equally steep as $y=-10x$ (because if you are going up one it's ...


0

Here is a single equation that gives you an idea of when a point lies inside , outside or on the boundary of a regular octagon of radius 15. You can easily write in your test a meaningful and useful expression .( I have written out the equation for the boundary of an octagon, inside use the equation in the form of an inequality , and the same for points ...


1

First, let me rewrite the equations without changing the result: $$\begin{eqnarray} r_t &=& r_b - r_\Delta \\ r(z) &=& r_b - r_\Delta z \\ x(z) &=& r(z) \cos(z) \\ y(z) &=& r(z) \sin(z) \\ d(z) &=& \frac{r_\Delta (1-z) + r_b - r_\Delta}{r_b} = \frac{r(z)}{r_b} \\ \end{eqnarray}$$ The equations for $x$ and $y$ show ...


2

We need to prove the following: when two vertices of a fixed triangle slide along two arms of a fixed angle, the locus described by the third vertex is an ellipse. In the figure all the letters excepting $t$ and $P(x,y)$ are data of the problem; the angle $t$ determines the position of the vertex $P(x,y)$ so we choose $t$ as a parameter to be eliminated in ...



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