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0

I have edited your question to make it more specific. Then, the correct format of your formula should be $d= \frac{A x_1 + B y_1 + C}{\pm \sqrt{ A^2 + B^2}}$ The above answers your 3rd question----The substitution $(x = x_1$ and $y = y_1)$ is a must if $(x_1, y_1)$ is the point in question. Note. You can still use the absolute sign form.


0

It is the cylinder length multiplied by the cross sectional area.


2

This is $O(n)$. First, define the deviation of a vertex $p_r$ to be the (signed) angle by which the direction of $p_rp_{r+1}$ differs from the direction of $p_{r-1}p_r$. This can be calculated as $\arcsin z$, where $z$ is the $z$-coordinate of: $\dfrac{(p_{r+1}-p_r)}{|p_{r+1}-p_r|} \times \dfrac{(p_r - p_{r-1})}{|p_r - p_{r-1}|}$ Now, we get a convex set ...


1

In general for a parametized face $\vec F(u,v)$ the area is the integral $$A=\int \left|\frac{\text d\vec F(u,v)}{\text du}\times\frac{\text d\vec F(u,v)}{\text dv}\right|\text du \text dv$$


2

Alternatively to checking oriented area as in Yves Daoust's answer, one could check that the polygon is locally convex at each vertex as follows. Let $p_i$ be a fixed vertex. For each other vertex $w$, show that the angles $\angle p_{i-1}p_i w$ and $\angle wp_ip_{i+1}$ are accute. This can be done arithmetically by using the distances between the points ...


2

It can all be done easily in $O(n)$ operations: for safety, first check that the given polygon can be split in two monotone chains (find the leftmost and rightmost of them and check monotonicity in between, i.e. increasing abscissas); if that fails, the polygon is not convex. Then check that all angles are convex in these chains. This is a simplified ...


0

HINT : Since $\triangle COD:\triangle AOB=10:40=1^2:2^2,$ $$\triangle COD:\triangle AOD=CO:OA=1:2$$ and $$\triangle COB:\triangle AOB=CO:OA=1:2.$$


0

If we choose as origin the bottom-left vertix, and call the radius with the letter $\ r$: $\ X_{2}(c_{x},c_{y})\iff X_{2}(50+c,r-c)$, where $c=r \sin(45°)=\frac{r}{\sqrt2}$


8

Hope the following illustration helps:-


1

Let the perpendicular from $X_2$ to $X_0X_1$ be $X$. Then $\cos a=X_0X/50$ and $\sin a=X_2X/50$. From the way you have defined the coordinates so that moving right increases the horizontal coordinate and moving down increases the vertical coordinate, the coordinates of $X_2$ are $$(50+50\cos a,50+50\sin a).$$


8

If you're talking about a distance function, it has to be non-negative: http://en.wikipedia.org/wiki/Metric_%28mathematics%29. You might want to define a signed distance (which is clearly not a distance anymore, in the axiomatic sense of the term). For instance, in $\mathbb{R}$ you can define the distance from a certain point $x_0$ as $d(x)=\vert ...


2

put your square in a cartesian coordinate system an compute the equation of the incircle in the form $(x-x_M)^2+(y-y_M)^2=R^2$ $M(x_M,y_M)$ is the middlepoint of the incircle with the coordinates $x_M,y_M$


6

Let $\mathcal{H}\colon\mathbf{w}\cdot\mathbf{x}+b=0$ be a hyperplane in $\Bbb{R}^n$, then $$ d = \frac{\mathbf{w}\cdot\mathbf{x}_0+b}{\lVert\mathbf{w}\rVert} $$ gives the signed distance (with respect to the normal vector) between a point $\mathbf{x}_0$ and the hyperplane. $\lvert d \rvert$ gives the "traditional" distance. The signed distance takes into ...


1

What also works is when you assume first assume for every zig zag line that there are two parallel infinite lines and one infinite line intersecting both. Now, for the case of the zig zag lines, every parallel line stops at the intersection with the third line, so you have to subtract 2 regions from $$L_{3n}$$. The third line is bounded in two directions, so ...


2

The total volume of the two spheres is $$\frac{4 \pi}{3} r_1^3 + \frac{4 \pi}{3} r_2^3.$$ We're looking for a configuration in which the radii of the two spheres are $R$ and $d - R$ for the prescribed distance $d$ and some (large ball) radius $R$; by construction these radii must satisfy $$\frac{4\pi}{3} R^3 + \frac{4\pi}{3} (d - R)^3 = \frac{4 \pi}{3} r_1^3 ...


-1

Consider the figure below: Since the sides of the triangles are perpendicular to the radii of the incircle at the point of tangency, $\angle AGP$, $\angle AEP$, and $\angle BFP$ are right angles. Since two tangents from the same external point to a circle are congruent, $\overline{AG} \cong \overline{AE}$, $\overline{CG} \cong \overline{CF}$, and ...


-1

$AB=4\\ AC=2\\ BC=2\sqrt{3}\\ \angle ACB=90^\circ\\ \angle BAC=\tan^{-1}\left(\dfrac{AC}{BC}\right)=\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)=30^\circ\\ \angle ABC=90^\circ-\angle BAC=90^\circ-30^\circ=60^\circ\\$ The coordinates of $P$ is the sum of the coordinates of the triangle's vertices divided by the sum of the triangle's side lengths. $ A=(0,2)\\ ...


1

The measure of the arc of an inscribed angle is twice the angle. arc AB is the arc defined by the inscribed angle ACB, so ACB must be 70 degrees since arc AB is 140 degrees. We now have two angles of triangle BCE: 70 and 86, so the third is 24 degrees. The length of the arc defined by the inscribed angle CBD is twice its measure, making it 48.


3

Divide the square into $5 \times 5$ squares of side $0.2$m. The diagonal of each square is $\frac {\sqrt 2}5 \lt \frac 27$, so each square can be covered by a disk of radius $\frac 17$ Two mosquitoes per square is $50$, there must be a third one in a square so squash that square.


0

Use the Inscribed Angle Theorem (the measure of an inscribed angle is half the measure of its intercepted arc) to find the measure of arc XY. Then find the measure of arc XZY. Then use the generalization of the Inscribed Angle Theorem that uses tangent lines to find your desired measure of angle YXV.


0

Hint: The hypotenuse of the larger right triangle is $\sqrt{2}$, so the length of the base is $1$. That leaves $\sqrt{2} - 1$ as the base of the smaller triangle. Now notice that the angle opposed to d is $\frac{\pi}{4}$, because the lengths of the opposite and adjacent sides of the larger triangle are the same.


0

HINT : The two right triangles at the bottom are the isosceles right triangles (why?).


3

HINT : Note that $\triangle EFC:\triangle ADE=EF^2:AD^2$ and that $AD=\sqrt 3\times ED=\sqrt 3\times EF$ because $$DE:AE:AD=1:2:\sqrt 3.$$


2

This answer is based on @noneEggs's suggestion. 1: Extend BP to Q such that BP = (0.5) BQ 2: (Can be skipped.) A circle (centered at P and radius = PB) is drawn cutting BP extended at Q. 3: A line through Q is drawn parallel to AB cutting AC (extended if necessary) at R. 4: Join BR. 5: A line through Q is drawn parallel to RB cutting AB at S. 6: Join ...


1

The three "hearts" are similar and the ratio of their lengths is $6:8:10=3:4:5$. So, the ratio of their areas is $3^2:4^2:5^2=9:16:25$. Since $25-16=9$, $A$ and $B$ have the same area.


1

Let C be the area of the dark blue region separating A and B. Then A+B+C is a 10 by 10 square, plus two semicircles of diameter 10: $$ A+B+C = 100 + 25\pi $$ B+C is an 8 by 8 square, plus two semicircles of diameter 8: $$ B+C = 64 + 16 \pi $$ B is a 6 by 6 square, plus two semicircles of diameter 6: $$ B = 36 + 9\pi $$ Then going up the chain, $$ C = 28 + 7 ...


1

Since $\triangle ACD$ is a right triangle with $CD=6$, we have $$AB=AC-BC=\sqrt{6^2-4^2}-2=2\sqrt 5-2.$$


2

Let $ABCD$ be the quadrilateral where $A$ is the upper left point, $D$ is the upper right point. $\triangle ABC$ is an isosceles triangle with $AB=BC$. Here, take a point $F$ on the line $DC$ such that $\angle{FBC}=20^\circ$. Since $\angle{BCF}=\angle{BFC}=80^\circ$, we have $BC=BF$. Then, we know that $\triangle ABF$ is an equilateral triangle, so we have ...


1

$\def\vec#1{\overrightarrow{#1}}$ From $\triangle OBC$, by squaring $\vec{BC}=\vec{OC}-\vec{OB}$ we get $a^2=2R^2-2\vec{OB}\cdot\vec{OC}$, or $$ 2\vec{OB}\cdot\vec{OC}=2R^2-a^2 $$ and similarly $$ 2\vec{OC}\cdot\vec{OA}=2R^2-b^2,\quad 2\vec{OA}\cdot\vec{OB}=2R^2-c^2. $$ It follows that $$ \left\Vert\vec{OA}+\vec{OB}+\vec{OC}\right\Vert^2=9R^2-(a^2+b^2+c^2) ...


1

Sincne $\sin^2A+\sin^2B+\sin^2C=2+2\cos A\cos B\cos C$ and $a=2R\sin A ,b=2R\sin B, c=2R\sin C$ in a triangle: $$a^2+b^2+c^2=4R^2(\sin^2A+\sin^2B+\sin^2C)=4R^2(2+2\cos A\cos B\cos C)$$ Since, $\cos A\cos B\cos C\leq \frac18$ or, $\sin^2A+\sin^2B+\sin^2C\leq\frac94$ both of which can be proved independently. $$a^2+b^2+c^2\leq9R^2$$


1

The "$\alpha$" triangle on the left of Figure 3 is an "$\alpha$" triangle from Figure 1b, but scaled by $yz$; so, instead of side-lengths $r$, $x$, $w \;(=\sqrt{r^2+x^2})$, it has side-lengths $r\cdot yz$, $x \cdot yz$, $w \cdot yz$. Likewise, the "$\alpha$" triangle in the upper-right of Figure 3 is scaled by $rz$ to have side-lengths $x\cdot rz$, $r \cdot ...


0

Just a hint : You can always make a parallelogram out of A and P, where A is a vertex and P is the center of the parallelogram itself


0

Yes, your idea for how to start this problem is a good one. Since you have explicit formulas for $x\wedge y$ and $x\cdot y$, you can apply your specific vectors given here to write down a polynomial equation in $m$; solve it and you're done.


2

It's called a "spherical shell".


1

Hint: Do triangles and quads separately. I'm assuming that degenerate triangles (all three verts on one line) aren't counted. So every triangle must have 2 verts on one line, and one on the other. How many ways are there of picking 2 verts on the first line? $N(N-1)/2$. And for each of these, you can pick any vert on the other line. So for triangles like ...


1

Let $P$ be the intersection of $DS$ and $AD'.$ Then $\angle PD'S = \angle AD'E \cong \angle ADE' = \angle ADP,$ using the fact that $\Delta AD'E \cong \Delta ADE'.$ Also $\angle APD \cong \angle SPD'$ since they are a pair of vertical angles. Two angles of $\Delta APD$ are congruent to corresponding angles in $\Delta SPD',$ so the two triangles are similar, ...


1

If you have already shown that $AED'$ and $AE'D$ are congruent, then observe that you can pivot the first triangle about $A$ to get the second. The angle of rotation is $\angle EAE'$, or $90^\circ$, because the edge $AE$ is sent to $AE'$. Therefore every corresponding pair of edges between the two triangles is perpendicular, including in particular $DE'$ ...


2

Here's a terse proof with complex numbers. Take $A$ to be the origin and let the coordinates of $D,E,D',E'$ be $iz,w,z,i w$ respectively. Then be construction we have isosceles triangles formed by the points $\{0,z,iz\}$ and $\{0,w,iw\}$. The angle formed by $DE'$ and $D'E$ is then found as the argument of $\dfrac{z-w}{i(z-w)}=-i$, and is thus $90^\circ$.


5

Here is another way to prove without using the fact that $S$ lies on intersection of circles : $$ \begin{align} &\angle ADD' = \angle AD'D = 45^{\circ} \\~\\ \end{align} $$ $$\begin{align}& \overline{AE} \cong \overline{AE'} \\ &\angle EAD' \cong \angle E'AD \\ &\overline{AD'} \cong \overline{AD} \\ &\implies \triangle AED' \cong ...


0

Let us orient the triangle with the vertex $C$ at the top and the base $AB$ at the bottom, and set $d=\frac{1}{1+\sqrt{3}}$. Now let us build a quadrilateral inside the triangle by identifying: point $M$ on the side $AC$, so that $MC=d$; point $N$ on the side $BC$, so that $NC=d$; points $P$ and $Q$ by drawing two vertical segments of length $d$ from $N$ ...


1

For $w$, $x$, $y$, $z$ between $0$ and $\pi$, the relation $$\sin w \;\cos x \;+\; \sin y \; \cos z \;=\; 2$$ holds if and only if each term ---and each factor of each term--- is itself $1$. Specifically, $$w = y = \frac{\pi}{2} \quad\text{ and }\quad x = z = 0$$ Now, note that any two angles of a quadrilateral have a sum and (absolute) difference between ...


1

From the following figure it can immediately been read off that when the point of contact $Q$ moves counterclockwise by the angle $\phi$ then the inner circle (as a rigid structure) rotates by the amount $2\phi$ clockwise. Therefore after a full turn of $Q$ the point $P$ has made $2$ turns.


2

You are correct that the Angle Bisector Theorem gives you that $|AC| = 2|AB|$. Let's take that as our jumping-off point, assigning $x$ and $y$ to lengths as shown: (I'm using "$x$" instead of "$1$", so that we can follow the value through the formulas better. "$1$"s are so easy to lose!) By the Law of Cosines in $\triangle ABD$: $$\begin{align} y^2 ...


2

Let $\alpha=\angle ABC$. Since $AD$ is the bisector of $\angle CAB$, we have $$AB:AC=BD:CD=1:2\Rightarrow AC=2AB.$$ So, since $$\angle ACB=180^\circ-3\alpha\Rightarrow \sin(\angle ACB)=\sin(3\alpha)=3\sin \alpha-4\sin^3\alpha,$$we have with $\sin\alpha\not=0$, by the law of sines, $$\begin{align}\frac{AB}{\sin (\angle ACB)}=\frac{AC}{\sin(\angle ...


2

HINT: Using Werner Formulas, $$\sin A+\sin B+\sin C+\sin D=4$$ Now for real $x,\sin x\le1$ Can you take it home from here?


1

Presumably, we leave $L_1$ fixed and seek the locus as $L_2$ varies. Let $T$ be the point where $PS$ meets $QR$. We're given that $RS \perp PQ$. Also, $RP\perp QS$ (since the $\overleftrightarrow{RP}$ is the $x$-axis, and $\overleftrightarrow{QS}$ is the $y$-axis). This says that $R$ lies on two altitudes of $\triangle PQS$; since $R$ must then also lie ...


0

Let us write the equation of line $L_1$ as $y= x \tan {\theta} +j$. Using this notation, $\theta$ represents the angle identified at the crossing of the line with the $x$-axis. Similarly, considering that $L_2$ is perpendicular to $L1$ and then its slope is the negative inverse of that of $L_1$, we can write $L_2$ as $y= -x \cot {\theta} +k$. The ...


1

Let $H$ complete parallelogram $\square ABHG$. Also, let $J$ and $K$ complete "double-size" parallelogram $\square ABKJ$ as shown; note that diagonal $BJ$ contains segment $BE$. Since $GD\parallel AB$, we have $\triangle GDF \sim BAF$. (Why?) Since $|GD|= \frac12|AB|$, we have $|DF| = \frac12|FA| = \frac13|DA|$; thus, the area of $\triangle GDF$ (with ...


0

Asking the question "why this optimization problem cannot be solved analytically" seems to me problematic. For one thing, I'm unaware if anyone has even attempted to prove that a problem of this form cannot be solved analytically. [EDIT: I am now! See Semiclassical's link in the comment below; it does happen to be proven in this particular case that no ...


0

Some hints: Let $M$ be he midpoint of $AD$. Since the triangles $\triangle(AEM)$ and $\triangle(DEM)$ have equal sides they are congruent. It follows that $\angle(AME)=\angle(DME)={\pi\over2}$, and that $E\vee M$ is an angle bisector at $E$.



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