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1

Let $\color{red}{C'D'=a}$ in isosceles $\triangle AC'D'$ then the angle of vertex A is given as $$\angle C'AD'=\frac{180^\circ}{5}=36^\circ\implies \angle C'AM=\frac{\angle C'AD'}{2}=18^\circ$$ Now, drop a perpendicular say $AM$ from vertex $A$ to the side $C'D'$ in $\triangle AC'D'$, we get $$\tan\angle C'AM=\frac{\frac{C'D'}{2}}{AM}$$ $$\implies ...


0

This is soulution of my friend. WOLG, we can suppose $B$ is between $A$ and $C$. Let $\overrightarrow{OD}= \overrightarrow{OA}+\overrightarrow{OC}$ and $\overrightarrow{OE}= \overrightarrow{OB}+\overrightarrow{OD}$. We have $\widehat{AOC} \leqslant 180^\circ$, therefore $\widehat{AOD} \leqslant 90^\circ$ and $\widehat{DOC} \leqslant 90^\circ$. Thus $\angle ...


0

When you press the cos button on your calculator it probably takes use of the following definition: cos(x) = $\sum_{0}^{\infty}\frac{(-1)^n x^{2n}}{2n!}$


0

Suppose that the black dot on the rear wheel (where the wheel connects with the coupling rod) was a distance of $1$ from the center of the wheel, and its initial position was directly to the right of the center of the wheel. Given that the wheel had turned and angle of $\theta$ (from its initial position), Vertically, how far above the center of the wheel ...


1

\begin{align} x&=\frac{|AD| \cos\tfrac{\alpha}{2} +|BE|\cos\tfrac{\beta}{2} +|CF|\cos\tfrac{\gamma}{2}}{\sin\alpha+\sin\beta+\sin\gamma} =?\quad (1) \end{align} Consider $\triangle ADC$. We know that it shares the circumradius $R=1$ and the side $|AC|$ with $\triangle ABC$, and we also know all the angles: $\angle CAD=\tfrac\alpha2$, $\angle ...


1

Let $\mathbf{ABC}$ represent the area of triangle $ABC$ $D = \frac 12B + \frac 12C$ $F = \frac 12(A + D) = \frac 12A + \frac 14B + \frac 14C$ $E=(1-\alpha) A + \alpha C$ $E = (1-\beta)B + \beta F = \frac 12 \beta A + (1 - \frac 34\beta)B + \frac 14 \beta C$ $(1-\alpha) A + \alpha C = \frac 12 \beta A + (1 - \frac 34\beta)B + \frac 14 \beta C$ ...


0

If the ratios between the bases are the same, say $k$, the trapezoids are similar. Indeed, both trapezoids $ABCD$ and $A'B'C'D'$ can be consedered as "pieces" of right triangles with acute angles $80$ and $10$ degrees. The triangles have been cut along a segment parallel to the side opposite to the $10$ degrees angle. Let $AB$ and $CD$ the parallel sides, ...


0

Two trapezoids with the same angles are similar if and only the ratio between one particular pair of neighboring sides is the same in two trapezoids. If the trapezoid is not a parallelogram, it is also enough to know that the ratio between the two parallel sides is the same in the two trapezoids.


1

Judging from the slides you provided, I think you have misunderstood the definition. In the Preliminaries section I see the definition: $$\{z \quad | \quad (z-x_0)^TH_{x_0}(z-x_0) \leq r^2\}$$ where $H_{x_0}$ is defined by $$H_{x_0} = \sum \frac{a_i^T * a_i}{(b_i - a_ix_0)^2}$$ where $a_i$ are row vectors of the matrix $A$. However, $H_{x_0}$ is not a ...


0

In this case it is indeed a ball in $\mathbb R^n$ with the usual Euclidean norm with the radius you say and a center at $x_0$.


3

The internal angle bisector of $\widehat{BAC}$ meet the circumcircle of $ABC$ in the midpoint of the $BC$ arc. So, if $O$ is the circumcenter of $ABC$, we have $\widehat{AOD}=\widehat{AOB}+\frac{1}{2}\widehat{BOC}=2\widehat{C}+\widehat{A}$ and: $$ AD\cos\frac{\widehat{A}}{2} = 2R ...


3

Always, since $p\not\in C_p(M).$


5

Firstly, in figure 1, Using Napier's Analogy in $\Delta ABC$, we have, $$\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cdot \cot \left( \frac{A}{2}\right)$$ Now, in figure 2, Using Sine Law in $\Delta ABM$, we have, $$\dfrac{x}{\sin \left(\frac{A}{2} + \alpha \right)}=\dfrac{c}{ \sin(\angle BMA)}$$ Similarly, in $\Delta ACM$, ...


0

In programming I use the "perspective" matrix. There a lot of formulas but in my case for 3D it is just E matrix of size 4x4 but with m34 != 0 (m34 when m43 means z-translation or vice versa). So it is still a matrix but with some exceptions because if you merge it with usual transform matrices it adds a distortion to objects (in my case it transforms rects ...


1

Let's simplify and particularize this problem by dividing the three complex numbers by $-z_1$, giving us $-1$ (let's call it $w_1$), $w_2=-\dfrac{z_2}{z_1}$, and $w_3=-\dfrac{z_3}{z_1}$. These three points give us a similar triangle to the original, rotated around the origin and shrunk so that the three new points are on the unit circle. When we find the ...


0

In a projective geometry, the points are the one-dimensional subspaces (a point is associated with a line through the origin in the vector space), as you have said. The LINES are the TWO-dimensional vector subspaces (so a line is associated with a plane through the origin). Three points are on a common line if the one-dimensional subspaces they determine ...


1

There are many definitions, but geometrically I believe the unit-circle definition is the best. It allows you to understand easly many trigonometric relations.


1

Construction: Extend AM to cut the circle BXHC at P. Join BP and join CP. $\beta$ is the exterior angle of the cyclic quadrilateral $PCHX$. Therefore, $\alpha = \beta$. $\gamma = \beta$ because they are the angles on the same segment of the cyclic quadrilateral $AHXC’$. $\alpha = \gamma$ implies $AB // CP$. Similarly, $\theta = \delta$ implies $BP // ...


3

I decided to add the comment I made as a solution. I think this is a bit more intuitive for me. The solution posed by Harish is nice as well and I mean nothing by the statement "would it not be easier" since its all about personal preference.


4

Equation of the plane parallel to the given plane: $3x-y-z=0$ is $$3x-y-z+c=0 $$ Where $c$ is an arbitrary constant. Now, using $\color{blue}{\text{distance formula for parallel planes}}$ as follows $$\frac{\left|c-0\right|}{\sqrt{(3)^2+(-1)^2+(-1)^2}}=3$$ $$\left|c\right|=3\sqrt{11}$$ $$c=\pm 3\sqrt{11}$$ Hence, there are two parallel planes at distance ...


1

We have that the circumcircle of $B'HC'$ goes through $A$ and the circumcircle of $BHC$ goes through $A''$, the symmetric of $A$ with respect to the midpoint of $BC$. Moreover, $B,C,B',C'$ are concyclic points, hence if $Y=BC\cap B'C'$, we have that $HX$, i.e. the radical axis, goes through $Y$ (that is not crucial for our proof, but I think it is worth ...


0

You can go directly for the angle $\alpha$ between sides $b$ and $q$ by requiring that the vectors for sides $p$ and $q$ are orthogonal, so their scalar product vanishes: $\vec q = \left( \begin{array}{c} q \sin\alpha\\ q \cos\alpha \end{array} \right), \ \vec p = \left( \begin{array}{c} a - q \sin\alpha\\ q \cos\alpha - b \end{array} \right),$ $\vec p ...


1

given $$ x^2 - 15 xy + y^2 - 7 x - 8 y = 2,$$ multiply through by $4$ and complete the first square, for $$ u = 2x - 15 y - 7. $$ The remaining nonconstant terms are $-221 y^2 - 242y,$ which is kind of bad. $221 = 13 \cdot 17$ is squarefree, so we multiply through by $221$ and use $v = 221 y + 121.$ So far, we have $$ \frac{221u^2 - v^2 + 3812}{884} = 2. ...


0

Rewrite the equation of the given line in the standard form: $$\frac x4+\frac y3=1$$ Then you know its $x$-intercept is $(4,0)$ and its $y$-intercept is $(0,3)$, so that the $x$- and $y$-intercepts you're seeking are respectively $(8,0)$ ad $(0,9)$, and the equation of the corresponding line is: $$\frac x8+\frac y9=1,\enspace\text{or}\quad9x+8y=72.$$


0

$x$ intercept is $8$, $y$ intercept is $9$


0

'Magnify' the plane 2 times along x direction and 3 times along y direction. The line becomes $3(\frac{x}{2})+4(\frac{y}{3})=12$ i.e. $9x+8y=72$


0

if i understood this right we have $x_0=8$ and $y_0=9$ from the first equation. Let $ax+by=c$ then we get $x=c/a=8$ and $y=c/b=9$ thus our equation is $$x/8+y/9=1$$


2

Step 1: find the x-intercept of the line $3x+4y=12$. Step 2: find the y-intercept of the line $3x+4y=12$. Step 3: calculate $2$ times the x-intercept and $3$ times the y-intercept. Step 4: now you have $2$ points that are on the line. Calculate the equation of the line.


1

Geometric Proof: In this proof, $M$ is not assumed to be inside $ABC$. Let $P$ and $Q$ be the points such that $APBC$ and $MQBC$ are parallelograms. Note that $APQM$ is also a parallelogram. Hence, $AP=BC=MQ$, $BP=CA$, $BQ=MC$, and $PQ=MA$. Consider the quadrilateral $APQB$. We have by Ptolemy's Inequality that $$MC\cdot BC+MA\cdot AB=AP\cdot BQ+PQ\cdot ...


0

I just wrote a program to investigate 5 fold symmetry in 2d, it generates something akin to penrose tiling. here is a pic and a video on YT: i am not convinced that it can't practically be tiled in a square shape, you may have places where only one triangle is wrong in a frame of 1000 of them or something, else you could find a place with good 4 fold ...


1

Distance $PQ$ is needed to give complete answers, but here is an idea to solve the problem. Call $d=PQ$. We want to know the number of solutions of this system: $$\begin{cases}a^2+b^2=d^2\\ab=2x\end{cases}$$ If we add and substract the first equation and the double of the second one, we get $$\begin{cases}(a+b)^2=d^2+4x\\(a-b)^2=d^2-4x\end{cases}$$ This ...


0

For generalization of the Pell equation question arises constantly. So writes the Pell equation in General form. $$Ap^2-Bs^2=1$$ If we know any solution of this equation. $( p ; s)$ If we use any solutions of the following equation Pell. $$x^2-ABy^2=1$$ Then the following solution of the desired equation can be found by the formula. $$p_2=xp+Bys$$ ...


2

in right triangle, we have $$\tan \theta=\frac{3}{2}$$ $$\implies \color{red}{\theta=\tan^{-1}\left(\frac{3}{2}\right)}$$ Let, $R$ be the radius of circle then the length of given arc $$=\text{(aperture angle)}\times \text{(radius of circle)}=12$$ $$\implies \tan^{-1}\left(\frac{3}{2}\right)\times R=12$$ $$R=\frac{12}{\tan^{-1}\left(\frac{3}{2}\right)}$$ ...


1

Hints: Note that, since $z_1$,$z_2$, and $z_3$ all have equal magnitude, then they all lie on a common circle centred at the origin. Observe that this circle is the circumcircle of your triangle. Now just create a line which goes through the point A and the mid-point of the opposite side. Can you continue from here?


0

Here's a quick geometric sketch. I found the answer by using the similar triangles shown in the sketch. You can use the relationships between the lines as ratios of the height of the base of the triangle to the total height and length of the side to the total length, and then apply the same ratios to the new triangle.


3

We have: $\tan \theta = \dfrac{3}{2}, \tag{1}$ whence $\theta = \arctan(\dfrac{3}{2}): \tag{2}$ now $12m = r\theta$ $= r \arctan(\dfrac{3}{2}), \tag{3}$ where $r$ is the radius; thus $r = \dfrac{12m}{\arctan(\frac{3}{2})}; \tag{4}$ the side of the square is $2r$; the area is thus $4r^2 = (2r)^2$ $= \dfrac{576m^2}{\arctan^2(\frac{3}{2})}; \tag{5}$ ...


4

The radius of the circle is half the length of a side of the square. The arc length formula might be useful $$s=r\theta$$ Since we are given an arc length. We just need the corresponding angle in radians. That is where the other two side lengths. $$tan(\theta)=3/2$$ $$r=12/tan^{-1}(3/2)$$ $$A=l^2=(2r)^2$$ This answer will be in square meters so you ...


1

Note that $x^2 - 3 x y + y^2$ has the improper automorphism $(x,y) \mapsto (y,x).$ Given a discriminant $\Delta=5$ positive and not a square, solve $\tau^2 - \Delta \sigma^2 = 4.$ The one with smallest variables is the fundamental solution, in this case $3^2 - 5 \cdot 1^2 = 4.$ The generator of the oriented automorphism group of $A x^2 + B xy + C y^2$ is ...


-1

Here is a counterexample of the argument, showing that the argument is not generally valid: $BC=BC$ (common side) $\angle ACB = \angle BCD$ (common angle) $\angle ABC=\angle BDC$ (= $90^\circ$) By the same logic, $\triangle ABC \cong \triangle BDC$. But they aren't. They are only similar triangles. The key to the argument's apparent validity is that ...


1

I have a little more:) $$ S = 4R^2(\sin A + \sin B + \sin C)\sin\frac A2\sin\frac B2\sin\frac C2\\ S = \frac{r^2}{4}\frac{\sin A + \sin B + \sin C}{\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}\\ S = r^2\left(\cot\frac A2+\cot\frac B2 + \cot\frac C2\right)\\ S = r^2\cot\frac A2 \cot\frac B2 \cot\frac C2\\ S = 2p^2 \frac{\sin A\sin B\sin C}{(\sin A+\sin B+\sin ...


1

The proof is correct. $\angle B = \angle C$ is very vague. He should have written $\angle ABC = \angle ACB$. But we can see that $\angle ACD = \angle ACB = \angle ABC = \angle ABD$.


0

Using the law of sines, one can get $$s=\frac{1}{2}bc \sin(\alpha)=\frac{1}{2}bc \frac{a}{2R}=\frac{abc}{4R}$$ where $R$ is the radius of the circumscribed circle.


1

Extending $\overline{FG}$ crosses $\overline{CE}$ at it midpoint, $F^\prime$ and hits the circle again at $G^\prime$. Clearly, $\overline{F^\prime G^\prime}$ is the side of a triangle congruent to $\triangle FGH$. Define $p := |\overline{DF}| = |\overline{EF}| = |\overline{FF^\prime}|$ (where the last equality follows from the Triangle Midsegment Theorem) ...


0

I would use computer science and the concept of classes and objects. Let Shape be a class, say: class Shape(shape) [Array pts]; Then define Circle as a subclass of Shape: class Circle of Shape [pts={center,radius}]; Then your circle is an object: myCircle = new Circle({0,0},15); Links Object Class


1

You need more than identical gradients for a tangent line, you need identical lines, which also means identical $y$-intercepts (or $x$-intercepts if the line is vertical). That will give you two non-linear equations in the two variables $x$ and $y$, which you can solve to find the anchor point of each tangent line. I think you would be better off using a ...


2

It is impossible to calculate your figure, because the information you have does not completely determine the figure. Your fear about "multiple solutions" is correct--in fact, infinitely many of them. You need more information to get a unique solution--the "unknown's" in your drawing would do. Here are two diagrams, both satisfying your conditions and ...


0

This question is somewhat related to this question, with the main difference that you also have an translation. Lets first make sure that the notation is clear: A vector $\vec{v}$ can be represented in the initial Cartesian coordinate system as, $$ \vec{v} = \begin{bmatrix} \vec{e}_x \\ \vec{e}_y \\ \vec{e}_z \end{bmatrix} \cdot \begin{bmatrix} v_x \\ ...


1

HINT: let $AB$ make angle $\theta$ with the bottom edge, and let the bottom edge of the whole paper be $a$. Let $y=AB$. Then it should be possible to obtain the following expression for $y$ as a function of $\theta$: $$y=\frac {a}{2\sin^2\theta\cos\theta}$$ which can be differentiated to find the minimum.


-1

If your lines and rectangles are fairly predictable, a fast techinque is to use the rectangle as aligned to the axis, and rotate the line segment accordingly. Parameterize the line, and see if any point lies inside the rectangle.


0

translate/rotate the rectangle and the line segment so that the rectangle becomes axis parallel, with a corner at the origin; perform the region discussion as in the Cohen-Sutherland Line Clipping algorithm. In case of a trivial reject or trivial accept, you are done. Otherwise, you will need to plug the coordinates of a corner in the implicit equation of ...



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