New answers tagged

0

They're referring not to the equations in $(1)$, but to the quadratic equations in the line above $(1)$. That's a slight error. Let $A$ be the point of intersection of line $p$ with the $y$-axis. Once they have that $x = y = d/2$, it means that the distances from $A$ to the two points of intersection of $p$ with either circle must be $d/2$ and $3d/2$. It ...


1

Assuming that's C code and the variables are integers, the following will give a $\Delta \in [-180, 180)$: delta = (targetAngle - myAngle + 540) % 360 - 180; If the variables are floating-point, you need to replace % with fmod, but performance-wise it becomes wasteful vs. just checking the ranges as you had it. [EDIT] Short explanation: targetAngle - ...


3

A rounded rectangle of size $2a\times2b$ with rounding radius $r$ is given by $$f(x;a,r) + f(y;b,r) = 1$$ where $$f(x;a,r)=\begin{cases}\left(\frac{|x|-(a-r)}r\right)^2&\text{if $|x|\ge a-r$,}\\0&\text{otherwise.}\end{cases}$$ You want to approximate this with some function of the form $(|x|/a)^p$. Compare derivatives at $x=a$ and you get $p=2a/r$. ...


0

This is all visually based so no proof here. But currently I have, given approximate corner radius $r$ and width $w$, height $h$, let $\alpha = w/2$, $\beta = h/2$, the the equation for a nice-enough looking rounded-corner rectangle, aka a rectircle, is: $$ |\dfrac{x}{\alpha}|^{2\alpha/r} + |\dfrac{y}{\beta}|^{2\beta/r} = 1 \\ \alpha, \beta, r \gt 0 $$ ...


1

This is a very interesting question, and the answer to it comes from a simple observation of the following figure. If I am thinking correctly, the correct equation to your wanted rectangle is not unique, because it can be made more accurate by better constants. However for a rough idea, just try to think of this. The shape of the curve x^2+y^2=1, it is a ...


0

Use the equation $$ A x^2 + B y^2 + D x + E y = 1 $$ With the points given you have three equations $$ \begin{align} A \alpha^2 + D \alpha & = 1 \\ A \beta^2 + D \beta & = 1 \\ B \gamma^2 + E \gamma 7 & = 1 \end{align} $$ I include a skew parameter $\zeta = -B/A$ also to parametrize all the curves and together I get the solution ...


0

Is the locus a circle ? In general, no. It is a multifocal ellipse. More information on this topic can be found here, along with a video showing how to actually draw one.


0

A conic section is characterized by five real degrees of freedom. Knowing three distinct points on it will account for three degrees of freedom (not six since each point might still move along the conic without changing it). Assuming that it's a right hyperbola should account for another degree of freedom, so we should end up with a one-parameter family of ...


1

This is a proof by homogeneous coordinates, projective geometry and computer algebra. I start with the first case. First off, observe that your whole setup is invariant under projective transformations. So without loss of generality you may assume that the original conic is the unit circle, and you may further assume that $P$ is at position $[1:0:1]$. Then ...


0

you checked only the first time, if you check that: $k_1,k_2 \in \mathbb{Z}$ $|\theta_h - \theta_m| = \frac{\pi}{2}+2\pi k_1$ or $|\theta_h - \theta_m| = -\frac{\pi}{2}+2\pi k_2$ $\frac{22}{12} t = \frac{1}{2}+2k_1$ or $\frac{22}{12} t = -\frac{1}{2}+2k_1$ $0 \leq t\leq24$ because we check for 1 day. $0 \leq \frac{22}{12} t \leq 44$ $0 \leq ...


1

The "condition" here is the constraint on $X$ that it's distance from $2x-y=1$ is $10$ units. By imposing the condition, you get the locus of $X$. In this case, the locus is a pair of parallel straight lines parallel to the given line, which you have found out. The reason why there are two lines is that $X$ can lie on either side of $2x-y=1$, and each ...


1

In 12 hours they make 11 overlappings, it is 1 and something 2 and something ... 11 and something. This is because the minute hand moves much faster than the hour hand and for $360^{\circ}+30^{\circ}$. In 12 hours the minute hand will cross $12 \cdot 360^{\circ} = 11 \cdot (360^{\circ} + 30^{\circ})$ so they overlap 11 times. The same goes for $180^{\circ}$ ...


0

Construction:- Let P be a point on BD such that $CP \bot BD$. Join AP $\alpha = \tan ^{–1} (\dfrac {x}{y})$ Both s and t can be found. Applying cosine law to $\triangle ADP$, z can be found. If the paper is then folded according to the description, C’ (vertex C of ABCD) is now directly above P with $\angle \gamma = 90^0$ Then, $AC’ = \sqrt (z^2 + ...


1

Your equation $|\theta_m - \theta_h|= \frac{\pi}{2}$ does not measure the time interval between successive events, but the time between midnight and the first event after midnight. That is why the 'solution' $t=3/11$ seems to vary for other angles $\theta\neq\pi/2,$ and why it is incorrect even for $\theta=\pi/2.$ We do not need to measure the time between ...


0

Line segments $AE$ and $CF$ are altitudes of triangles $BAD$ and $DCB$. When the rectangle is folded as described, line segments $AE$, $EF$, and $CF$ will be edges of a rectangular cuboid with main diagonal $AC$. The length of $AC$ is given by: $$|AC|^2=|AE|^2+|EF|^2+|CF|^2$$ Can you work out these lengths? Hint: The six triangles in the diagram are all ...


0

$\frac 3{11}$ hours is the time it takes the minute hand to gain $90^\circ$ on the hour hand. That will be the time from midnight (or noon) to the first $90^\circ$ point. Having had one case you need the minute hand to gain $180^\circ$ to get to the next one, which takes $\frac 6{11}$ hours.


1

HINT You want to reflect lines $ x= \pm 0.5, y= \pm 0,5 $ about $ y = m x , m= \tan \theta .$ Instead of that consider four among the eight rotated lines $$ \pm x \cos \theta \pm y \sin \theta = 0.5 ; \, \pm x \sin \theta \pm y \cos\theta = 0.5 $$


1

Let $ABCD$ be your square. Let's call $G$ and $H$ the points where $l$ intersects, respectively, the sides $AB$ and $CD$. We have the two trapezia $GBCH$ and $AGHD$ that both have the same area (half of the original square, so $1/2$). We now want to reflect $AGDH$ around $l$, let's call $L$ and $K$ the images of $A$ and $B$ after the reflection. The new ...


1

There are various ways to do this. You could set up an equation for the distance from an arbitrary point $(u,v)$ to the extended line $AB$, and another equation for the distance from an arbitrary point $(u,v)$ to the extended line $BC$. (I can't write the coordinates of the point as $(x,y)$ since you already used those symbols to mean something else.) Solve ...


1

The intersection is a hyperbola, symmetric with respect to the vertical plane passing through the cones vertices. It follows that given $(x_1,y_1)$ you can find another point $(x_2,y_2)$ on the hyperbola by taking the symmetric of $(x_1,y_1)$ with respect to the line passing through $(a_1,a_2)$ and $(b_1,b_2)$: $$ ...


2

Hint: Without loss of generality we can assume that $B$ is the origin and $C$ is a point on the $x$ axis. Changing a bit your notation, let $P=(x_P,\pm b)$ where $|b|$ is the distance from $BC$ (your $y$). If we have $A=(x_A,y_A)$ than the line $BA=r$ has equation $ y_Ax-x_Ay=0$. The distance $a$ of $P$ from this line $r$ is: $$ ...


1

We can, without loss of generality, assume a coordinate system with origin at B and x-axis along line BC. Then line AB can be written as y= mx. Since I am using an xy-coordinate system, I am going to call the distance to the given point, from line AB, "p" rather than "x" and the distance to line BC "q" rather than "y". Let the given point have ...


2

Draw $GH$ parallel to $DE$. GAC equals to the alternate one CBH because cutting by parallels,GCA to its vertically opposite HCB, given $ CA= CB, $ so congruent.


1

Let $H$ be on $DE$ such that $CH\perp DE$. You have that $C$ is midpoint of $AB$, and $CH||AD||BE\Rightarrow DH=EH$. Then $\triangle DHC\simeq\triangle EHC $ because $DH=EH$, $CH$ is common and $\angle DHC=\angle EHC=90^o$.


0

Hint: $$\sin a = 2\sin(a/2)\cos(a/2) = 2\tan(a/2)\cos^2(a/2) = \frac{2\tan(a/2)}{1+\tan^2(a/2)} $$ $$\cos a =\cdots$$


3

Consider (as example) point set of $86$ points: pts = { {-32548, 4941, 1746}, {-32182, -7188, -473}, {-31473, -484, -9790}, {-30188, -7755, 11126}, {-29817, 6257, 12734}, {-29196, 12267, -8910}, {-28393, -13023, -10705}, {-28249, 16917, 3300}, {-26936, 5415, -18260}, {-26336, -19736, 1372}, {-24723, -7559, ...


0

I have a robot arm with a gripper. I know the gripper pose (relative to the robot base coordinate system) at any moment. $T_{\text{base}:\text{gripper}}$ At startup, I record the pose of the gripper and set this as the original pose O $T_{\text{base}:\text{gripper}_O}$ Then, the gripper moves to its new pose N, again in the robot base ...


1

There is a natural smooth action of $\text{GL}(n + k)$ on $\text{Gr}(n, k)$; fix a point $v$ to obtain a map$$f: \text{GL}(n + k) \to \text{Gr}(n, k),\text{ }g \mapsto g \cdot v.$$The derivative of this map at zero yields a map of vector spaces$$df_v : \text{Hom}(U, U) \mapsto T_v \text{Gr}(n, k),$$where we denote the total vector space $U = ...


1

Almost certainly, but there is an edge case where it can fail. It is certain that $A,B,C,$ and $O$ are coplanar, but ....


1

In this answer I'm presenting an opinion that's different from the other answers. The other two answers seem to be trying to mathematically explain why one is supposed to integrate $\mathrm ds$ instead of $\mathrm dx$, but in my opinion, mathematics itself is unable to account for this, because they both are mathematical hypotheses to model the real ...


3

One way to see your error is to note that the circumference of a circle is measured in (say) inches, and the surface area of a sphere is measured in square inches. It can’t work out that adding up inches gives you square inches. Your approach is not useless, however. Each circle you’re using contributes an infinitesimally wide strip of the surface area - ...


8

The key thing that's going on here is that you cannot compute the hypotenuse of a right triangle by taking one of the legs of the triangle. Consider the integral that gives arclength of a curve: You're adding up $\Delta s = \sqrt{(\Delta x)^2+(\Delta y)^2}$ when you chop the curve into pieces, not adding up $\Delta x$. When you're computing your integral, ...


2

Well, both of these are true for the ordinary sphere in three dimensions as well, just not as much. More of the volume of a sphere is near the surface than at any other depth. To put it pictorally, if you were to think of a sphere in layers like an onion, the layers near the surface contain more volume simply because they're bigger. On the other hand, if ...


1

Let $O$ be the center of the circle. Take a point $B$ at distance $b$ from O, with $0<b<r=d/2$. Let the line thru $B,$ perpendicular to the segment $OB,$ meet the circle at points A and C. We have $AO=CO=r=d/2.$ The triangles $ABO, CBO$ are right-angled at $B.$ So the length of the chord $AC$ is $$AB+BC=\sqrt {r^2-b^2}+\sqrt {r^2-b^2}=2\sqrt ...


1

You want $\left( \dfrac{|\vec{a} \cdot \vec{b}|}{||\vec{a}||\, ||\vec{b}||} \right) \gt \left(\dfrac{|\vec{a} \cdot \vec{c}|}{||\vec{a}||\, ||\vec{c}||} \right)$, i.e. $\dfrac{|\vec{a} \cdot \vec{b}|}{ ||\vec{b}||} \gt \dfrac{|\vec{a} \cdot \vec{c}|}{ ||\vec{c}||} $ This is equivalent to $\dfrac{(\vec{a} \cdot \vec{b})^2}{\vec{b} \cdot \vec{b}} \gt ...


1

Hint: An alternate approach would be to let the two diagonals intersect at Z, and then to notice that the triangles XAB and XPZ are similar, and that the same holds for triangles YCD and YQZ. Since BD halves PQ, and the latter is parallel and equal to both AB and CD, the result immediately follows.


0

Here's what I think you did: you showed the stuff on the left side was true, and said it implied the right side. But notice the following: $$\lim_{a \to c}(c^2-a^2)\pi = 0 \implies\lim_{a \to c}b^2 \pi = 0$$ For this to happen, you have to have assumed that $c^2-a^2 = b^2$, which is assuming what you want to prove (and thus, you don't have a proof due to ...


0

A suitable affine transformation $T$ of your figure will make two of the parallels vertical and two of the parallels horizontal. This $T$ will multiply all areas with the same factor. Therefore you may assume that two of the parallels are $x=0$, $x=1$, and that two of the parallels are $y=0$ and $y=1$. It follows that two of the vertices are $A=(1,0)$ and ...


0

If three points $\mathrm{A}_1(a_1,\ b_1),\ \mathrm{A}_2(a_2,\ b_2),\ \mathrm{A}_3(a_3,\ b_3)$ are known, the area can be calculated. $$\triangle\mathrm{A_1A_2A_3}= \frac{1}{2}|x_1y_2+x_2y_3+x_3y_1-x_2y_1-x_3y_2-x_1y_3|$$ The formula can be proved by the fact $\operatorname{distance}(ax+by+c=0,\ (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$. Let's ...


1

There are four constants in a "parallel" ellipse so you need four constants to write such an ellipse. For the "parallel" or better "axes parallel" ellipse ( $x,y$ axes are parallel to directions of lines $a,b$ ) the coefficient for $xy $ term vanishes. Comparing term by term in the equation establishes equivalence of two ways of representation: $$ ...


2

Consider a point $P$ on the circle and the diameter going through this point. Let $N$ the other end of the diameter. Let $M=N$. The angle between $\vec{PN}$ and $\vec{PM}$ is zero and the length of the chord is $d$. now if you increase this angle continuously from $0$ to $\frac{\pi}{2}$ by moving $M$ along the circle, the chord will go from $d$ to $0$ and ...


13

This comes down to the intermediate value theorem. On the circle $x^2+y^2=1$, the chord from $(1,0)$ to $(\cos\theta,\sin\theta)$ has length $2\sin\dfrac\theta2$. You can see that by means of the usual "distance formula". As $\theta$ goes from $0$ to $\pi$ (or from $0^\circ$ to $180^\circ$ if you like), the chord goes from $0$ to $2$ and the chord is a ...


5

Yes. There are several ways to prove it. From the axioms for Euclidean geometry: draw a circle with radius $x$ and center on your circle; see where the circles intersect. (Euclid's axioms are a little incomplete - they may not actually guarantee the intersection.) An argument from continuity: start at one end of the diameter of your circle and move around ...


11

Open a compass with the desired length $x$, place the needle on the circumference and draw an arc that intersects the circle. Can you imagine that you'll never meet it ?


3

The answer depends on which axioms you have. If you have any axioms that give you a mesure such that for every real positive number exist a line of that lenght the answer is yes. You can see Hilbert geometry (moder versione of euclidean geometry) and his axioms about continuity and mesure


3

Hint:$$\text{ chord length } = 2r\sin\bigg({\frac{\theta}{2}}\bigg)$$ where $r$ is the radius and $\theta$ is the angle subtended at the center by the chord. Note the continuity of the RHS, now use the Intermediate Value Theorem.


1

Here's a strategy: The area are related to diagonals in the following way: $A=\frac{d_1d_2}{2}$. Now note that the diagonals ($d_1$ and $d_2$) of the rhombus are orthogonal, as seen in the figure below. You can use this to relate the length of a side of the rhombus to the diagonals using Pythagoras theorem. Then you have two equations with two variables, ...


1

After proving that $BP\parallel QD$, you may first consider $\triangle ADY$ and then $\triangle CBX$. In $\triangle ADY$, $AX=XY$ by midpoint theorem. Again, in $\triangle CBX$, $CY=XY$. Hence $BP$ and $QD$ trisect $AC$ at $X$ and $Y$.


1

The quaternions for rotations $R$ and $S$ are: $$t_R=\cos \frac{\pi}{4}+\frac{j+k}{\sqrt 2}\sin \frac{\pi}{4}$$ $$t_S=\cos \frac{\pi}{4}+\frac{i-j}{\sqrt 2}\sin \frac{\pi}{4}$$ The product rotation $S \circ R$ would then be given by $t_Rt_S$, which is easily calculated by the rules of quaternion multiplication to be $$t_Rt_S=\frac{3}{4} +\frac{\sqrt ...


2

There is a caltrop on 76 points. Points 1, 13, 25, 29, 41, and 53 are as follows: {0.0833`, 0.0833`, 0.4930122817942774`} {0.32530527130128584`, -0.20709494964790603`, 0.32530527130128584`} {0.28875291001058745`, 0.28875291001058745`, 0.28875291001058745`} {-0.2142`, 0.40369721678726284`, -0.2142`} {-0.07272969962634213`, 0.35355339059327373`, ...



Top 50 recent answers are included