New answers tagged

-1

Hint: A circle consists of $x$ radians, or $z$ degrees... Out of $12$ hours, $4$ hours represents $p$% of the clock ... I leave the rest to you


0

Let $P$ be a point of the line and $\vec u$ a unit vector along the line direction. If the line is given by its cartesian equation $y=mx+b$ you can take for instance $P=(0,b)$ and $\vec u={(1,m)\over\sqrt{1+m^2}}$. Let $A$ be the point to reflect and define $\vec a=A-P$. The projection of $\vec a$ on the line is $(\vec u\cdot\vec a)\vec u$ so we get for ...


1

Sine rule says :$${BC\over \sin\angle A}={AB\over \sin\angle C}={AC\over \sin\angle B}={AC\over \cos\angle C}\implies \tan\angle C=3/7$$$$\implies \sin \angle C=3/\sqrt{58}$$ Now since you know $\angle C$, you know$\angle B \ \&\ \angle A$ Hence you get $BC$in terms of $AB$. Now apply formula$$Area=4.2=0.5\times BC\times AC\sin\angle C$$ you will get ...


0

Hint: The area of a triangle could be expressed in terms of product of two sides and the sine of the angle in between. http://www.regentsprep.org/regents/math/algtrig/att13/areatriglesson.htm So the angle $A$ can be found. And then that with the given angle equation, reveal all angles. You can then recover the length $|BC|$ by applying the area formula once ...


0

Whether or not the points lie on a graph we have for a set of $n$ points $$ \bar{x} = \Sigma x_i/n, \bar{y} = \Sigma y_i/n, $$ For a triangle $n=3.$


1

Following your direction of proof construction: Setting $\theta = \angle ABC \implies \angle ADC = 2\theta$ Continuing lines $AD$ and $BC$ to find $E$, we find $\angle ECD = \theta $ from parallel line crossing. Then since $\angle EDC = \beta = 180^\circ-2\theta, \angle CED = \theta$ also and $\triangle EDC$ is isosceles so $|ED|=|DC|=p$ Then $\...


1

Solved. Mis-read the question. It asked for the smallest possible perimeter. Therefore solved with optimisation.


0

Use the Pythagorean theorem. Assuming that all of the four triangles are identical, let us denote the long side by $b$ and the short side by $a$. Then $a^2 + b^2=c^2$ where $c$ is length of any of the sides of the inner square. Moreover, we know that $a+b=10$ since that is the length of the bigger square. Thus, $(a+b)^2=a^2+2ab+b^2=100$. Thus $c^2=(a+b)^2-...


2

Once you deduce $DE=p$ Check that $${DE\over AE}={DC\over AB}\implies {p\over p+q}={p\over AB}$$


4

Hint: draw an angle bisector and show that it splits the original trapezoid in a parallelogram and an isosceles triangle. It follows that $AB=\color{green}{p}+\color{red}{q}$.


0

This must be computed numerically. It is not clear from your question if the total length is $240.4851$ or $240.481$. Assuming it is the latter, then it turns out that you need $30$ chords and $$ B=0.325398. $$


2

\begin{align} & V=\int_{0}^{2}{\int_{-\sqrt{2x-{{x}^{2}}}}^{\sqrt{2x-{{x}^{2}}}}{\int_{0}^{{{x}^{2}}+{{y}^{2}}}{dzdydx=}}}\int_{0}^{2\pi }{\int_{0}^{2\cos \theta }{\int_{0}^{{{r}^{2}}}{r\,dzdrd\theta}}} \\ & V=\int_{0}^{2\pi }{\int_{0}^{2\cos \theta }{{{r}^{3}}}}drd\theta =4\int_{0}^{2\pi }{{{\cos }^{4}}}\theta d\theta =16\int_{0}^{\frac{\pi }{2}}...


0

Geometric solution. $f(1)=a+bi$ has the same distance to both (0,0) and (1,0). Then $f(1)=a+bi$ must lie on the line x=0.5: the line across the mid-point between (0,0) and (1,0) and orthogonal to the straight line connecting (0,0) and (1,0). Now consider the triangle of the three points: (0,0), (0.5,0) and (a,b). The angle at (0.5,0) is 90 degree, and the ...


1

Let we start with the $2D$-version first. Given three distinct parallel lines $\ell_1,\ell_2,\ell_2$ in the plane, there is an equilateral triangle $ABC$ with $A\in\ell_1,B\in\ell_2,C\in\ell_3$. It is enough to pick any $A\in\ell_1$, then rotate $\ell_3$ around $A$ by $60^\circ$ clockwise, meeting $\ell_2$ at $B$. The perpendicular bisector of $AB$ will meet ...


0

The one thing that the problem tells us is that this function exists, and has the same special property for any $z$. So let's pick any random $z$. I'm going to pick $2$ because it is easy to work with. Since $f(2)$ must be equidistant from $0$ and $2$, it must be on the line $\Im(z)=1$. Since $|a+bi|=8$, we know that $f(2)$ is also on the circle $|z|=2\...


0

Regard a complex number $z$ as $x+iy$ representing the point $x\choose y$ in the plane. Then $f(z)= (a+ib)z$ corresponds to matrix multiplication by $\pmatrix{a & -b\cr b & a}$ (verify this!). Dividing all entries of this matrix by $8$ calling the corresponding function $g(z)$ we can see that $f(z)= 8g(z)$. Now $g(z)$ corresponds to rotation of ...


3

Path to the solution: A quadrilateral is cyclic iff two opposite angles add to $180^\circ$; Any cyclic quadrilateral can be dissected into $4$ cyclic quadrilaterals; By the previous point, if a cyclic quadrilateral can be dissected into $n$ cyclic quadrilaterals, it can be dissected into $n+3$ cyclic quadrilaterals, too; From the previous point, it is ...


2

Hint: It's easy to see MC=MD, hence points D X and C are located on a circle, with M being the center. Now, we know angle XCD is half of angle DMX (why?)


7

Since $MC=MD=MX$, the points $C,X,D$ lie on a circle centered at $M$ and $$\widehat{XCD}=\frac{1}{2}\widehat{XMD} = \color{red}{13^\circ}.$$


1

Find the $\angle ADM$ from the right-angled triangle. This will help you find $\angle XDC$, as $\angle XDC = 77^{\circ} - \angle CDM = 77^{\circ} - 90^{\circ} + \angle ADM$. Then use Sine Theorem on $\triangle DMX$ to find the length of $\overline{DX}$ and then finally use Sine Theorem on $\triangle DXC$ to find $\angle XCD$. This should be your outline for ...


2

By rotation and stretching. Multiplication by $z=re^{i\theta}$ with $r$ and $\theta$ real corresponds to rotating the plane over $\theta$ radians, and stretching the plane in all directions by a factor $r$. I also find that this video has very nice animations illustrating the geometry of arithmetic on complex numbers.


0

The proof is as simple as pointed out by @ajotatxe. The following is a more constructive proof, and reflects how to put rigor in such an evident statement. It is just some "mathematical technique." Call your convex set $C$ and pick a point $x\in C$. Call $\mathcal{F}$ the set of pairs $(a,b)$ of distinct points $a,b\in C$ such that $a\leq x\leq b$ (notice ...


7

Geometrically, this is the fact that any angle subtended by a diameter is a right angle. Indeed, consider the triangle formed by the points $z$, $1$, and $-1$. Since the segment from $1$ to $-1$ is a diameter of the unit circle, the angle at $z$ is a right angle. The Pythagorean theorem then says $|z-1|^2+|z-(-1)|^2=|1-(-1)|^2$, which is just what you've ...


6

Look at $-1$ and $1$ as the endpoints of a diameter of the unit circle. If $z\not\in\{-1,1\}$ lies on the unit circle, the lines joining $1$ and $z$, $-1$ and $z$ are orthogonal, i.e. the triangle having $1,-1,z$ as vertices is a right triangle. By the Pythagorean theorem, $$ \|1-z\|^2 + \|-1-z\|^2 = 2^2 = 4.$$


0

A convex set is connected (because it is path-connected), and connected sets of $\Bbb R$ are intervals.


1

Thanks for your question. It's sometimes really hard to prove something that seems to be obvious at first glance. I have three ideas: Observe a pythagorean tiling inside yours. Then derive from it that your pattern is indeed a tessellation. You can explain it using the fact that your pattern is invariant to translation by a vector $[2,1]$. Therefore it ...


-1

For the record, the example you gave did not have $5$ congruent parts in the picture, as the relative positions of the six-triangle groups differ from each color, easily seen by the purple and blue mix being completely secluded from both uniform colors, making them neither congruent nor similar shapes to the other groups. As the equilateral triangle has ...


1

Since $C=(0,c)$ The midpoint of $BC$ is $P=(2,1+{c\over 2})$ Now $AP$ is perpendicular to $BC$ so, $${2-c\over 4-0}\cdot {9-(1+{c\over 2})\over 5-2}=-1$$ Solve and you will get $c=4 $ OR $14$


0

In that context of that lecture, they are not talking just about the general topological concept of embedding, but instead a more special metric concept of embedding. A metric embedding is a topological embedding which preserves the metric tensor in an appropriate sense. So yes, you do need to understand topological embeddings, but that's not enough, you ...


0

An embedding between two topological spaces $X,Y$ is a map $f:X \rightarrow Y$ such that the induced map $f:X \rightarrow f(X)$ is an homeomorphism (a bicontinuous function between two topological spaces). An homeomorphism is also an embedding but it is not true the converse. Notice that the reason of why a 2D saddle surface can't be embedded in a 3D ...


1

Why not just apply a circular inversion? If we have $p_0,p_1,\ldots,p_n\in\mathbb{R}^n$ in general position, we may consider $q_1,q_2,\ldots,q_n$ as the images of $p_1,p_2,\ldots,p_n$ under a circular inversion with respect to a unit hypersphere centered at $p_0$. There is a hyperplane $\pi$ through $q_1,q_2,\ldots,q_n$, and by applying the same circular ...


0

Since $C=(0,c)$ and $$ AB^2 = (5-4)^2+(9-2)^2 = 50, $$ $$ AC^2 = 5^2+(9-c)^2 = 25+(9-c)^2, $$ we must have $|9-c|=5$, so the solutions are $\color{red}{c=4}$ and $\color{red}{c=14}$.


1

Let the coordinates of $C$ be $(0,c)$, and then express the lengths of $AB$ and $AC$ by using the distance formula. The distance formula: The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is: $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$


3

Hagen von Eitzen's answer gives a neat theoretical approach of this problem. However, I would like to expose a constructive and computational way to find the radius and center of the $(n-1)$-sphere determined by $n+1$ suitable points in $\mathbb{R}^n$. Let $n$ be an integer greater than $1$ and let say $x_i:=(x_{i,j})_{j\in\{1,\cdots,n\}},i\in\{0,\cdots,n\}$...


4

Yes, if the $n+1$ points are in general position, which simply means that the $n+1$ points must not lie in a hyperplane. We can proceed by induction: If $x_0,\ldots, x_{n}$ are our $n+1$ points in general position, then any $n$ of them, for example $x_0,\ldots, x_{n-1}$, certainly lie in a common $(n-1)$-dimensional hyperplane $H$. We can identify $H$ with $...


0

We have found a proof and formalized it using the Coq proof assistant, the details are here: https://hal.inria.fr/hal-01332044


1

Take the general equation of a line in 3D through origin i.e. ${x\over a}={y\over b}={z\over c}$ Now calculate distance of the 3 points from this line and equate them . You will get 2 equations with 3 unknowns. So you will get (say) $a\ \&\ b$ in terms of $c$ And hence you will get your desired axis.


0

You need a parametric form, since the curve is multi-valued both along the horizontal axis and the vertical axis. I started from $$x=\sin(t)\\y=t$$ to get Then I noted that as you go away from the center the points are moved upwards, so I added a term proportional to $x^2$ $$x=\sin(t)\\y=t+3x^2$$ and I get You can now play with some parameters, or the ...


2

Yes, it can be done in some cases. Your example of two vertical and two horizontal lines is one. Following the tic-tac-toe board, there are four corner pieces (like your B), four edge pieces (like your A), and one center piece. If we place the lines symmetrically around the center and make the area of the center piece $\frac 15$ of the disk, the center ...


1

You have to be careful since $\vec{u}+\vec{v}$ lies on the angle bisector of $\vec{u},\vec{v}$ only if $\|u\|=\|v\|$. That said, your proof is fine.


0

In two dimensions, those sectins are proportional to the angle between the enclosing lines. In three dimension the section is proportional to the enclosed solid angle. This can be generalized with the n-dimensional solid angle.


0

Approaching this from a multiple-choice perspective, the area of the semicircle overall is about $$ 36\pi/2 \approx 18\cdot 3{\small\frac 17} = 56{\small\frac 47} = 3\times 18{\small\frac 67}$$ which we need to reduce by the $3$ "ice-cream cones" like $YTCZ$. Then letting setting the radius of the small semicircles $w = |RC| = |RZ| = |YZ|/2$, we see $|YR| =...


1

The side of a triangle $x={12\over \sqrt{3}+1}$ (Why? Check out the comment) Shaded area $=18\pi-3({\sqrt{3}x^2\over 4}+\pi {x^2\over 8})$


1

Ok so $OE$ is 2cm and $AE$ is r. So, $AO=\sqrt{4+r^2}$. Similarly $CE=\sqrt{4+r^2}+2=\sqrt{3r^2}$. Solve it you get answer as $2\sqrt{3}$


1

The Quora "proof" does not actually claim to be a proof of Fermat's Last Theorem - that there is no solution of $$x^n+y^n=z^n$$ for $n>2$. It claims to be a proof that there is no solution of $$x^{p-1}+y^{p-1}=z^{p-1}$$ for $p-1>2$. Thus it would say nothing about (for example) $x^3+y^3=z^3$ or $x^5+y^5=z^5$. So the result only deals with certain ...


-1

$(x - \sqrt{5})^2 + (y-2\sqrt{2})^2 = 10$


0

Usually, we analyze a wave by either showing displacement-position graph of particular time; or displacement-time graph at particular position. Sometimes we use a wave function to describe the displacement for whole space ($x$ in one dimension) at all time. For a sine wave moving to the right, we have $$\phi(x,t)=A\sin \left( \frac{x}{\lambda}-\frac{t}{...


3

As you may know, the equation for a circle with center of $(x_c, y_c)$ and radius of $r$ is $$(x-x_c)^2+(y-y_c)^2=r^2.$$ So, the answer is $$(x-\sqrt{5})^2+(y-2\sqrt{2})^2=10$$



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