# Tag Info

0

The area of $QRS$ is just four times the area of the pedal triangle of $P$. By Euler's theorem, the area of the pedal triangle just depends on the distance from the circumcenter $O$: $$[QRS]=\frac{R^2-OP^2}{R^2}[ABC]$$ hence the claim is trivial.

1

I'll argue up to the formula, so the "why" will be clear. Let's use this diagram of $\triangle ABC$ with the Cartesian coordinates moved so that points $A$ and $B$ are on the $x$-axis and $C$ has a positive $y$-coordinate. Point $I$ is the incenter, segments $ID$, $IE$, and $IF$ are perpendiculars from $I$ to sides $\overline{BC}$, $\overline{AC}$, and ...

0

it is $l+w$, sum of its length and width.

0

For the ones wondering, I did the following: $$2 \cdot \angle BAP = \angle BM_1P$$ $$2 \cdot \angle BAP = \angle BM_2Q$$ The length of $BP = 2r_1 \cdot \sin(\frac12\angle BM_1P) = 2r_1 \cdot \sin(\angle BAP)$ The length of $BQ = 2r_2 \cdot \sin(\frac12\angle BM_2Q) = 2r_2 \cdot \sin(\angle BAP)$ Conclusion, since $r_1 = r_2$, $BP$ has to be the same ...

3

Hint: Look at the angle $\angle BAP$ as inner angle of the two circles. Then use the fact that the two circles have the same radius.

-1

Andre and MvG already answered this question very well. But still I want to add something from the Master Euclid: If you study the Proposition 37 of Euclid's Book 1, it directly relates to this question. His proposition states that "Triangles which are on the same bases and between the same parallels (lines) are equal to one another." You may be worried ...

1

There is an obvious lower bound: the area of the big circle has to be greater than the total area of the small circles, hence $\pi R^2 \geq \pi x r^2$ gives $R\geq r\sqrt{x}$. On the other hand, one may consider the least $\rho\in\mathbb{R}^+$ such that inside the disk $\|z\|\leq\rho$ there are at least $x$ Eisenstein integers (an hexagonal packing). Then ...

0

Let's look at each in turn, The intersection of a point and a point is a point. Two intersecting points can only result in a point, this is correct. The intersection of a point and a line is a point. A point can only intersect with a line at one point on the line, this is correct. The intersection of a point and a plane is a point. A point can only ...

0

There are three important things that you need to think about: Things can miss each other, i.e. be "disjoint". For example, what is the intersection of the points $(0,0)$ and $(1,1)$? What about the lines $y=x$ and $y=x+1$. If they do intersect, what if they overlap? In two dimensions: the line $y=x$ and the line $x-y=0$ intersect along a line. In three ...

1

The intersection of a point and a point is a point if they are the same. If they are two different points, then the intersection is empty. the intersection of a point and a line, is a point only if the point lays on the line. otherwise the intersection is empty. The intersection of a line and a line, may be a point (if they cross each other but are ...

0

It is fairly hard to visualize the symmetry, because as was written in the other two answers the objects in the Complex case are in general four-dimensional. The symmetry can be displayed however, by dropping one dimension. To augment the two answers given and help you visualize it, here is the case for the two maps: $\exp$, the principal branch of $\log$ ...

0

Let $A_1,B_1,C_1$ be the intersection points of $AM,BM,CM$ with $BC,CA,AB$ respectively,and let $D,E,F$ be the feet of eprepndiculars form $M$ to $BC,CA,AB$ respectively.Then we have $$\cot \angle MAB+\cot \angle MBC+\cot \angle MCA=\dfrac {FA} {FM}+\dfrac {BD} {MD}+\dfrac {CE} {ME}= \\$$ $$=\sqrt{\dfrac {MA^2} {FM^2}-1}+\sqrt{\dfrac {MB^2} ... 1 According to MathWorld, the density for d is$$ \frac{3d^2}{r^3}-\frac{9d^3}{4r^4}+\frac{3d^5}{16r^6}\;. $$Thus the cumulative distribution function that you're looking for is$$ \int_0^d\left(\frac{3d'^2}{r^3}-\frac{9d'^3}{4r^4}+\frac{3d'^5}{16r^6}\right)\mathrm dd'=\left(\frac dr\right)^3-\frac9{16}\left(\frac dr\right)^4+\frac1{32}\left(\frac ...

3

How are you at three-dimensional integrals? Let the first point be at $(0,0,z)$. There are two spheres: One of radius $r$, centered at the origin, and one of radius $d$, centered at $(0,0,z)$. I would let $r=1$ to remove one letter from your calculations. The spheres' intersection is symmetric about the $z$ axis, which should help you calculate the volume ...

1

You could use $x = \cos t |\cos t|$, $y = \sin t |\sin t|$, $0 \le t \le 2\pi$.

0

From the implicit equation, $$y=\pm(1-|x|)$$with $x\in[-1,1]$. Let $x$ run from $-1$ to $1$, following the lower branch and using a parameter $t$: $$t\in[-2,0]\to x=1+t,y=|x|-1=|1+t|-1.$$ Then let $x$ run from $1$ to $-1$, following the upper branch: $$t\in[0,2]\to x=1-t,y=1-|x|=1-|1-t|.$$ We can condense in a single expression $$t\in[-2,2]\to ... 0 The most natural parametrization is probably \gamma:[0,2\pi)\to\mathbb{R}^2 given by:$$ \gamma(t)=\left(\frac{\cos\theta}{|\sin\theta|+|\cos\theta|},\frac{\sin\theta}{|\sin\theta|+|\cos\theta|}\right) $$i.e. the re-scaled circle. 2 Using the inscribed angle theorem,$$\begin{align}\angle{APB}&=180^\circ-(\angle{PAB}+\angle{PBA})\quad(\leftarrow \text{from $\triangle{APB}$})\\&=180^\circ-(180^\circ-\angle{BDA}-\angle{PAD})\quad(\leftarrow \text{from $\triangle{ADB}$})\\&=\angle{BDA}+\angle{PAD}\\&=\angle{BDA}+\angle{CAD}\\&=\frac{1}{2}\angle{AMB}+\frac ...

1

In the first place: Don't flip the plot, because it mixes up the names of the variables. When studying $f$ and $f^{-1}$ at the same time you see the graph of $f^{-1}$ in the original $f$-plot when you tilt your head $90^\circ$ sideways. For functions $f:\>{\mathbb C}\to{\mathbb C}$ and their inverses $f^{-1}$ there is an analogous mischievous symmetry; ...

0

$\newcommand{\Cpx}{\mathbf{C}}$If $g:\Cpx \to \Cpx$ is a function, you can reflect the graph $w = g(z)$ across the (complex) line $w = z$ to get $z = g(w)$, or $w = g^{-1}(z)$. (Of course, $g^{-1}$ is generally multiple-valued, particularly if $g$ is entire.) Oh right, there's that small obstacle of living in a three-dimensional universe, where the graph of ...

0

Let $A=(0,1)$, $B=(1,0)$, $C=(0,-1)$, $D=(-1,0)$. The locus $|x|+|y|=1$ is the square with vertices $ABCD$. Define $f:[0,1]\rightarrow\Bbb R^2$ as follows: $$f(t)=\begin{cases} (1-4t)A+4tB & \text{if 0\leq t\leq\frac14} \\ (2-4t)B+(4t-1)C & \text{if \frac14\leq t\leq\frac12} \\ (3-4t)C+(4t-2)D & \text{if \frac12\leq t\leq\frac34} \\ ... 2 You observed that all points (x, y) satisfying |x| + |y| = 1 is a path forming a diamond. Since paths are 1-dimensional lines, you can describe the path as a function of one variable, as (x(t), y(t)). One option for defining (x(t), y(t)) is to start at (0, 1) at t=0, and move counter-clockwise around the origin, with t=1 being at (1, 0), ... 5 Assume we want to find an equation of the form r(\theta)>0 that will describe the square, where:$$x= r(\theta)\cos(\theta),\quad y=r(\theta)\sin(\theta)$$We have that:$$|r(\theta)\sin \theta|=1-|r(\theta)\cos \theta|$$Or factoring out r(\theta) (since we know r>0):$$r(\theta)= \frac{1}{|\cos\theta|+|\sin\theta|}$$To visualize x(\theta) ... 0 x = t, \; y = \pm \big|1 - |t|\big| such that -1 \leq t \leq 1. 2 Starting at (1, 0) and going counter-clockwise around the origin, something like$$ \cases{x(t) = |t-2| - 1\\y(t) = |t-3| - |t-1| + t -2},\quad t\in[0,4] $$should work. 0 Your problem is with “the line perpendicular to this line”. There are multiple lines perpendicular to a given line. You chose one through one of the two defining points, but that's an arbitrary choice. Instead, I'd explicitely take the line through the center. If your original line has direction i, one perpendicular direction is i\cdot i=-1, so the line ... 2 I don't see any way other than ray-tracing. The rays reflect off the black rectangles and are refracted (presumably using Snell's law) by the white rectangles. I noticed that either ray 2 or 3 near the bottom right seem to have been absorbed by a white rectangle. It would be an interesting exercise to program this given a set of rectangles and the index ... 1 If you think of a quantity as something like a distance between two points, there are some similarities with angles, for they also have a numerical value, but there are also some differences. First of all, an angle is usually defined as the length of a fraction of a circumference with the angle's vertex in it's center, but you can't take measurements of the ... 0 Try giving the base unit length (the problem is perfectly well defined up to similarity so this is just for convenience). Then you can use repeated applications of the sine rule to express the various line segments in terms of sines and cosines. If you work your way around clockwise you get the right angled triangles to work with which makes things nicer. ... 0$$TX^2 = OX^2-OT^2 =OM^2-OT^2 = (OQ+QM)(OP-QM)-OT^2$$but since OP\cdot OQ = OT^2,$$ TX^2 = QM(OP-OQ)-QM^2 = 2 QM^2-QM^2 = QM^2 $$as wanted. 1 Everything follows from:$$ BC=2R\sin A,\qquad AH = 2R\cos A, 2\Delta = R^2\left(\sin(2A)+\sin(2B)+\sin(2C)\right), AD = \frac{2\Delta}{a} = \frac{bc}{2R},  AD\cdot AH = bc\cos A = \frac{b^2+c^2-a^2}{2}.1 At the point where the distance is minimal, changes of t in either direction increase the distance. This means that the tangent direction has to be orthogonal to the radius. If you have \begin{align*} x(t) &= at^2 + bt + c & \dot x(t) &= 2at + b \\ y(t) &= dt^2 + et + f & \dot y(t) &= 2dt + e \end{align*} then the vectors are ... 1 As a composition of primitive transformations Let's build this up step by step. Move P into the origin. So you want to reflect z-P in the line spanned between the origin and P'-P. Normalize the direction “vector” of that line to\frac{P'-P}{\lvert P'-P\rvert}$$Rotate things so that that direction vector becomes the real axis. Since the number ... 1 Assuming no special properties, in particular no coonvexity guarantees, you could always compute the signed area of the projected polygon using the shoelace formula. If that is negative, swap your basis vectors. If your polygon is convex, then use two edges for v_0 and v_1, i.e. make the numerator e.g. p_0-p_1 and p_2-p_1. Since the angle between ... 1 The Beltrami-Klein model is an accurate depiction of what it would look like in hyperbolic space. To be a bit more precise, if you live in 3-dimensional hyperbolic space \mathbb{H}^3, and if P \subset \mathbb{H}^3 is a 2-dimensional hyperbolic plane tiled in red and white triangles with angles \pi/2,\pi/3,\pi/7 as in the picture shown in the question, ... 0 Assume that \frac{DB}{BC}=\alpha. Then \frac{DC}{BC}=(1-\alpha) and by Thales' theorem:$$ \frac{HB}{AB}=\frac{KA}{AC}=\alpha,\qquad \frac{HA}{AB}=\frac{KC}{AC}=(1-\alpha) $$hence:$$ HA\cdot HB+KA\cdot KC = \alpha(1-\alpha)\left(AC^2+AB^2\right) = \alpha(1-\alpha)BC^2 = BD\cdot DC by the Pythagorean theorem. 2 In order to make the Moebius band M smooth at the creases in your figure we have to take care that the tangent planes are vertical there. We begin by producing an S-shaped strip S making up a third of the total surface. This strip will project vertically onto a trapezoid with two sides parallel to the y-axis and having width 2h in the x-direction. ... 0 It is easy to give an upper bound for this problem: The total area of the triangles must not exceed the area of the circle. \begin{align} n \, A_t & \le A_c \iff \\ n \left( \frac{1}{2} b \, h \right) & \le \pi x^2 \iff \\ n \left( \frac{1}{2} \left( 1 \sqrt{1-\left(\frac{1}{2}\right)^2} \right) \right) & \le \pi x^2 \iff \\ n ... 1 As I understand you want certain vertices to leave trails as they change their position. You can do it by right-clicking on the vertex and checking the "Trace On" option. Now the vertex will leave traces of it's colour on the canvas. 0 In eq. 1) the second term is a vector distance (P1-P0) which is then factorized by a/d so that P2 lies in the line connecting P0 and P1 at a distance a/d starting at P0. This equation is not used for calculations anyway. To better understand equation 2 you could draw the axis x and y in anyway you want and you'll find out that (y1-y0)/d corresponds to ... 1 Here is a sketch of these proofs. The details of this are described in great detail in Michael Artin's Algebra - a book well-worth reading. 1a) For the 2\times 2 case: If A is orthogonal, then multiply by a suitable reflection to ensure that \det(A) = 1. Then note that A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$where the vector ... 0 Plug in four given quantities and verify dependence of volume on radius as "cubed".$$ \frac {V_1}{V_2}= {\frac {r_1^3}{r_2^3}} 0 Since \begin{align} X_i&=A\cdot(1-t_i)+D\cdot t_i, \\ Y_i&=B\cdot(1-t_i)+C\cdot t_i, \quad t_i=\tfrac{i}4,\quad i=1,2,3, \end{align} \begin{align} \vec{X_iY_i} &= Y_i-X_i= (B-A)\cdot(1-t_i)+(C-D)\cdot t_i \\ &= \vec{BA}\cdot(1-t_i)+\vec{CD}\cdot t_i \\ &= -\vec{AB}\cdot(1-t_i)-\vec{DC}\cdot t_i \\ \vec{X_iY_i} &= ... 0 \vec{AB}+\vec{BC}+\vec{CD}+\vec{DA}=0 \vec{a}+\vec{BC}-\vec{b}+\vec{DA}=0 \vec{BC}+\vec{DA}=\vec{b}-\vec{a} X_{1}Y_{1}=\frac{DA}{4}+a+\frac{BC}{4}=a+\frac{b-a}{4}=\frac{3a}{4}+\frac{b}{4} X_{2}Y_{2}=\frac{DA}{2}+a+\frac{BC}{4}=a+\frac{(b-a)}{2}=\frac{a}{2}+\frac{b}{2} ... 0 Arithmetic progression of vectors in the plane with a common vector difference  \dfrac{b-a}{4} : 4 a /4 , (3 a +b)/4, (2 a + 2 b)/4 , ( a +3 b)/4 , 4 b /4. $$If instead you had asked for  \vec{AD}, \vec{BC},  then you get same quarter division intermediate grid points writing  a \rightarrow c ,b \rightarrow d.  0 Frankly, I'm not sure that I understand the question either. Are you sure that you've stated it exactly as it is asked? My best guess is as follows: We know$$ V_1 = \frac{4}{3} \pi r_1^3 $$and$$ V_2 = \frac{4}{3} \pi r_2^3 $$so, dividing the first by the second, we obtain$$ \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 $$Perhaps that is the ... 0 Here, we are to verify the formula for volume of sphere (radius is known)$$V=\frac{4}{3}\pi r^3$$& for the radius of sphere (volume is known)$$r=\sqrt[3]{\frac{3V}{4\pi}}$$Sphere 1:for given radius r=2.4\ cm, Volume$$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (2.4)^3\approx 58\ cm^3$$Which is equal to the given volume of the sphere. similarly, for ... 1 ETA: The below assumes that it's \mathbf{a} and \mathbf{b} that are fixed, not A, B, and C. I may have tried to get too cute with this; perhaps the least confusing way to do this is to represent each of the points on the quadrilateral with x-y coordinates (A being at the origin, without loss of generality), and then express each of the desired ... 1 Hint: If ABC are the vertex of the triangle and M,N are the midpoint of AB and AC note that :$$ \dfrac{AM}{AB}=\dfrac{AN}{AC}=\dfrac{1}{2} \quad and \quad \angle MAN =\angle BAC  and use the criterion of similarity (3) here to show that $ABC$ is similar to $AMN$. You can do the same for the other triangles.

3

First, we should note that a very similar question has already been asked here, and several interesting answers were given. But because the question keeps coming up, I'm going to go out on a limb and suggest that there still might be room for a more complete list of reasons why hyperbolic geometry is important in its own right. It's hard to know where to ...

Top 50 recent answers are included