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area of any quadrilateral is $\frac{1}{2}d_1(h_1+h_2)$, where $d_1$ is any diagonal and h are its perpendicular distance from other two vertices whose sum, in this case is the other diagonal. you can derive this formula by breaking the quadrilateral into two small triangles.


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Hint: The quadrilateral is a kite. The area of a kite is simply the product of the lengths of the two diagonals, divided by two (why?).


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I've been exposed to a solution to this problem. Since $AB=AD$, $\angle ABD = \angle ADB$. Furthermore, from the proprerties of a cyclic quadrilateral, $\angle ACB = \angle ADB$ and $\angle ACD = \angle ABD$. Therefore, $\angle ACB = \angle ADB = \angle ABD = \angle ACD$. Most importantly, $\angle ACB = \angle ACD$. Note that the area of $ABCD$ is equal to ...


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You can also remark that $$Ax + By + C = 0=\alpha x+ y + \gamma$$ (with $\alpha=\frac AB$,$\gamma=\frac CB$) and then, applying the conditions, $$\alpha x_1+ y_1 + \gamma=0$$ $$\alpha x_2+ y_2 + \gamma=0$$ and you have to solve for $\alpha $ and $\gamma$ two linear equations. Using the classical methods, you get $$\alpha=-\frac{{y_1}-{y_2}}{{x_1}-{x_2}}$$ ...


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Here are a few ways of finding values of $A$, $B$, and $C$. Method 1: A line is the set of points $\vec{x}$ such that $$\vec{a}\cdot \vec{x_0} = \vec{a} \cdot \vec{x}$$ where $x_0$ is an arbitrary point on the line and $\vec{a}$ is a nonzero vector perpendicular to the line. We know that the line will be parallel to the vector $$\vec{d} = \langle x_2-x_1, ...


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The following might work: Prove this for a square (trivial) Then use the result that any convex non-degenerate quadrilateral can be obtained by a perspectival (projective) transform from the square and that the ratios will be preserved.


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$\text{slope} = m = \dfrac{y_2-y_1}{x_2-x_1} \Rightarrow y - y_1 = m(x-x_1) = \dfrac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \Rightarrow (x_2-x_1)y - y_1(x_2-x_1) = (y_2-y_1)x - x_1(y_2-y_1) \Rightarrow -(y_2-y_1)x + (x_2-x_1)y -y_1(x_2-x_1) +x_1(y_2-y_1)=0$. This gives the formula: $A = y_1-y_2$ $B = x_2-x_1$ $C = x_1y_2-x_2y_1$


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This is called solving simultaneous equations and is a matter of algebra. There are two ways to go about this but they are almost the same and have the same spirit. The first way to do it is to solve directly for $A,B,C$ by substituting the points into $Ax+By+C=0$ to get $$ \begin {eqnarray*} Ax_1 + By_1 &=& -C, \\ Ax_2 + By_2 &=& -C. \end ...


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HINT: Set the values of $x,y$ to form two linear simultaneous equation for the unknowns $A,B$ Solve for $A,B$ in terms of $C, x_1,y_1,x_2,y_2$


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Alright, here's a picture with some edges given labels: and here's the equations that the Pythagorean Theorem spits out, assuming $AB=BC=CA=1$: $$a^2+\gamma^2=(1-c)^2+\beta^2$$ $$b^2+\alpha^2=(1-a)^2+\gamma^2$$ $$c^2+\beta^2=(1-b)^2+\alpha^2$$ And guess what! It is impossible to show the desired result that $\alpha+\beta+\gamma=\frac{\sqrt{3}}2$ using just ...


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Wouldn't having all the angles pi/2 just make the figure a regular square because every angle would be 90 degrees?


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If you draw a horizontal from $A$ and verticals from $B$ and $C$ you create a pair of similar right triangles. The length of $AB$ is $d=\sqrt{(Ax-Bx)^2+(Ay-By)^2}$ The length of $AC$ is $d+L$, so $Cx=Ax+\frac {d+L}d(Bx-Ax)$ and the same for $y$


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Hint: You have that $\overrightarrow{AB}/\|\overrightarrow{AB}\|$ is a unit vector. So what is the length of $$L\frac{\overrightarrow{AB}}{\|\overrightarrow{AB}\|} \quad ?$$


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$$x_C=\frac{(x_B)(L+\sqrt{x_b^2+y_B^2})}{\sqrt{x_b^2+y_B^2}}\\y_C=\frac{(y_B)(L+\sqrt{x_b^2+y_B^2})}{\sqrt{x_b^2+y_B^2}}$$ For each coordinate you know, divide by the length of $AB$ and multiply by length $L$. Well, above equations assumed $A$ is at the origin. If $A$ is not the origin, use ...


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In everyday language (outside of mathematics/geometry) we almost never use "equal" to mean that something is equal to itself. We use it to talk about two or more separate things, never just one. I was taught to say two line segments or two angles are equal if they have the same measure. In the "olden days" when I was coming along, only closed figures could ...


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this is a hunch. i expect the maximum area to occur when the cyclic quadrilateral looks like a kite: the diagonals are orthogonal and the quad is symmetric about $AC$. if you let $AB = AD = 3x, BD = 5x.$ let the diagonal $AC$ and $BD$ meet at $E, AE = y, EC = 6-y.$ we have two equations for $x, y$ they are: $$9x^2 = 25/4 x^2 + y^2, y(6-y) = 25/4 x^2.$$ ...


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Construct a line through the green point perpendicular to one of the two lines. You now have the radius of your circle (distance from the green point to the intersection of that perpendicular line and the other line). Now draw the circle with that radius and you're done. EDIT: I assumed the OP meant construction using a ruler and a compass. This can also be ...


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You should find two things: Center of the circle: it's green points: $G=(x_1,y_1)$. Radius of the circle: it's a distance from a point $G$ to a line $Ax+By+C=0$ (see here): $$r=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$$ Equation of the circle is: $$(x-x_0)^2+(y-y_0)^2=r^2$$


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NOTE: This post has been edited due to @RicardoCruz catching a mistake. I also changed the desired variable from $z$ to $u$, for my own reasons. Here is an approach that leads to a not-so-easy but possible construction. Rotate your circles and line onto a Cartesian frame of reference so line $l$ is horizontal and translate so circle $S_1$ is centered at ...


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Using hyperspherical coordinates $(r,\eta,\theta,\phi)$ the metric is $$g_E = dr^2 + r^2(d\eta^2+ sin(\eta)^2d\theta^2 +sin(\eta)^2sin(\theta)^2 d\phi^2 )$$


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$(A, B)$ is a normal vector for the line, therefore $v = (-B, A)$ is a direction vector and you get all points on the line with $$ (x, y) = P_0 + t \, v = (X_0, Y_0) + t (-B, A), \quad t \in \mathbb R. $$ Now choose $t$ such that the length of $t(−B,A)$ is equal to the given distance $d$, this gives the two points $$ (X_2, Y_2) = (X_0, Y_0) \pm ...


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From the equation $Ax+By+C = 0$, find the slope of the straight line: $$A\cdot{\Delta x} + B\cdot{\Delta y} = 0$$ As long as $B\ne 0$, $\Delta y = -\frac AB \Delta x$. Also it is known that $$d = \sqrt{(\Delta y)^2 + (\Delta x)^2} = \left|\Delta x\right| \sqrt{1+\frac {A^2}{B^2}}$$ So $\Delta y$ and $\Delta x$ can be solved (and there are two pairs of ...


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Here is a link for a beautiful paper by Eisenbud, Harris, and Green: Eisenbud, David; Green, Mark; Harris, Joe: Cayley-Bacharach theorems and conjectures. Bull. Amer. Math. Soc. (N.S.) 33 (1996), no. 3, 295–324. See Theorem CB1, Theorem CB2, and Theorem CB3. The question you asked is actually Theorem CB2.


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The following should fulfill your requirements. It is the subsurface of an elliptical $z$-axis cylinder $(x/R_x)^2 + (y/R_y)^2 = 1$, that is contained within the filled elliptical $y$-axis cylinder $(x/R_x)^2 + (z/R_z)^2 < 1$. $$M:=\{(x,y,z) \in \mathbb{R}^3 : (x/R_x)^2 + (y/R_y)^2 = 1; (x/R_x)^2 + (z/R_z)^2 < 1 \textrm{ and } x>0, z>0\}$$ Rx = ...


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you can find the formula(coxeter) for the radius of the inner sphere of regular solids. the contact point of the plane of the face and the inner sphere is the center of that face.


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Here is the solution.${}{}{}{}{}{}{}{}{}{}{}{}{}$


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Let one of other two sides is x then third side is 25-x. By Pythagoras theorem, $x^2=5^2+(25-x)^2$ $x=13$, and it is the hypotenuse because it is the largest sides, third being 12.


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$\vec{c}^2$ is an alternative notation for $\vec{c}\cdot\vec{c}=||\vec{c}||^2$ A motivation for this is that the formulas $$||\vec{a}+\vec{b}||^2=||\vec{a}||^2+2\vec{a}\cdot\vec{b}+||\vec{b}||^2$$ $$||\vec{a}-\vec{b}||^2=||\vec{a}||^2-2\vec{a}\cdot\vec{b}+||\vec{b}||^2$$ and $$(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=||\vec{a}||^2-||\vec{b}||^2$$ now ...


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I think you are wise to focus on $\vec{c} \cdot \vec{c}$, rather than $(\vec{a} + \vec{b})^2$. Let $\vec{c}=(2,1)$. Then $$\vec{c} \cdot \vec{c} = (2,1) \cdot (2,1) = 2 \cdot 2 + 1 \cdot 1 = 5 \;,$$ which is the square of the length of the vector $(2,1)$, i.e., that vector has length $\sqrt{5}$. The dot-product multiplication is "component-wise," i.e., ...


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Typically, an aspect ratio is defined as width/height. To convert between the two formats, you need to take the reciprocal, which is done by dividing $1$ by the number. So if one format gives a ratio of $.75$, the ratio for the other format would be $\frac{1}{.75}\approx1.33$.


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This is the correct solution aspect ratio = original width ÷ original height because Explanation: The aspect ratio of an image describes the proportional relationship between its width and its height. It is commonly expressed as two numbers separated by a colon, as in 16:9. For an x:y aspect ratio, no matter how big or small the image is, if the width ...


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Usually, geometry consists of an underlying topological space (a manifold, for example) and some structure on this space. The structure is an analogy of some tool -- such as a ruler or compass -- that enables you to see more than what the topology sees. It might be something that enables you, for example, to "measure angles and distances" (Riemannian ...


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The answer to your question, "Is there a thorough and generally agreed upon definition of a geometry", is negative: There is no such definition. For instance, Klein's viewpoint (from 1872), was outdated by the time it was proposed, as it did not cover the emerging Riemannian geometry which was (at the time) in its infancy, as well as algebraic geometry ...


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There are several sources which provide a rigorous introductory axiomatic treatment of 3-dimensional geometry, including: The book "Foundations of Three-Dimensional Euclidean Geometry" by I.Vaisman. The book "Non-Euclidean Geometry", by H. S. M. Coxeter. (If you look hard enough, you might find a pdf or djvu file freely, alas illegally, available online.) ...


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In classical Euclidean geometry, there is no demonstration--it is an axiom. Euclid's First Postulate is that a straight line segment can be created between any two points. However, if you want to say that there is a line (not a line segment) that intersects two points, that can be proven simply. Euclid's Second Postulate is that any line segment can be ...


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I've labeled points $B$ and $C$ and angles $a$ and $b$ (in radians) in this diagram: The arc subtended by the angle $b$ is just $br$, and this lets us find the length of $\overline{AB}$ and $\overline{AC}$ as $$br + \frac{l}{2}$$ Now we can use the Pythagorean theorem to find the length of $\overline{AO}$ in terms of $b$, $l$, and $r$: $$\sqrt{\left(br ...


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Let $T$ one of the tangency points, $\alpha=\angle OAT$, $x=OA$. Note that $AT$ and $OT$ are perpendicular. Then $$2\pi r+l=2[(\pi-\alpha)r+\sqrt{x^2-r^2}]$$ but $\alpha=\arccos(x/r)$. We see that $x$ are inside a square root and inside an $\arccos$ in the same equation. I don't think that this equation can be solvable by algebraic means.


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If the distance between the nail and the center of the disk is $d$, then the length of the string is given by: $$ 2\sqrt{d^2-r^2}+2\left(\pi-\arccos\frac{r}{d}\right),$$ so, in order to find $d$ given $l$ and $r$ you have to solve a trascendental equation. Newton's method is well-suited for such a task.


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Permit me the liberty of renaming your variables. We will consider three unit vectors $u$, $v$, and $w$, whose inner products are $u \cdot v = c$, $u \cdot w = b$, and $v \cdot w = a$. Consider the Gram matrix of the vectors $u$, $v$, and $w$, namely the matrix of dot products $$G = \begin{bmatrix} u \cdot u & u \cdot v & u \cdot w \\ v \cdot u ...


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This is for a tent which has a rectangular base $a \times b$ and with the center pole holding up the tent having heights $h_1$ and $h_2$ at the two ends which are at the centers of the two rectangle sides labeled $a$. The height at a point $x$ units from the $h_1$ end is (using the coordinate $0$ at that pole and then $x$ going along the midline of the ...


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Let's move everything so that the center of the sphere is in the origin. Then the sphere is simply $$x^2+y^2+z^2=r^2$$ and the cone becomes $$\left(x+P_x\right)^2+y^2=c^2\left(z+P_z\right)^2$$ and since you don't want a double cone, you also want $$z+P_z>0\quad.$$ The $y=0$ plane intersects the cone in two lines, namely $$x=\pm c(z+P_z)-P_x$$ and ...


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It is well know $$\angle QPM=\angle F_{2}PM,\Longrightarrow PQ=PF_{2},QM=F_{2}M$$ so $$OM//F_{1}Q\Longrightarrow OM=\dfrac{F_{1}Q}{2}=\dfrac{F_{1}P+F_{2}P}{2}=a$$ so $M,N\in\bigodot O$ so $F_{1}M'=F_{2}M$ so $$F_{1}N\cdot F_{2}M=F_{1}N\cdot F_{1}M'=F_{1}A\cdot F_{1}B=(a+c)(a-c)=b^2$$


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WLONG, we can represent any point on the given Ellipse as $x=a\cos A,y=b\sin A$ So, the equation of the tangent at $(a\cos A,b\sin A)$ will be $\dfrac{x\cdot a\cos A}{a^2}+\dfrac{y\cdot b\sin A}{b^2}=1$ $$\iff bx\cos A+ay\sin A-ab=0$$ The coordinates of the foci are $(\pm ae,0)$ where $e$ is the eccentricity So, the distances will be $$\frac{|\pm ...


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Equations in which are present polynomial terms and trigonometric functions do not shaow analytical solutions and only numerical methods could be used. For your specific case, since $-1 \leq \sin(\theta) \leq 1$, you can easily detect that the solution is between the solutions of $$156 \theta-36=554.8$$ $$156\theta+36=554.8$$ that is to say $$3.32564 \lt ...


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Hints: The hypotenuse of a right triangle is a diameter of the circumcircle. If you draw two tangent segments to a circle from a point outside the circle, then those two segments have the same length. A Google Image search for "right triangle incenter" yields this diagram which is quite helpful:


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If $X(N)$ has genus $g$, then $\pi_1(X(N))^{ab} = \mathbb Z^{2g}$, and $\pi_1(Y(N))^{ab} = \mathbb Z^{2g + s -1}$, where $s$ is the number of cusps. Since $\pi_1(Y(N)) = \Gamma(N)$ for $N\geq 3$, it must be that $\Gamma(N)^{ab} = \mathbb Z^{2g+s-1}$. In particular, $\Gamma(N)$ cannot be generated by less than $2g+s-1$ elements.


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Turned out not to be too hard to answer this after all. I found the following inequality: $$u_1 u_2 - \sqrt{(1-u_1^2)(1-u_2^2)} \le U \le u_1 u_2 + \sqrt{(1-u_1^2)(1-u_2^2)}$$ This can be found by setting $\mathbf{V}=\{0,0,1\}$ and writing the other two vectors as $$\mathbf{v}_i = \{\sin{\theta_i}\cos{\theta_i}\,,\, \sin{\theta_i}\sin{\theta_i}\,,\, ...


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Let' say that $A(x_0,y_0)$ is the point of intersection of the tangent $$\epsilon:\begin{array}[t]{l}\frac {x} {a^2} x_0+\frac y {b^2}y_0=1\\(b^2 x_0) x+(a^2 y_0)y -a^2b^2=0\end{array}$$ and the ellipse. The distance between $F_1(-c,0)$ and $\epsilon$ is: $$d_1=\dfrac{\left|-c b^2 x_0-a^2b^2\right|}{\sqrt{b^4x_0^2+a^4y_0^2}}$$ Respectively, the distance ...


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Here's where you can find it in Euclid: http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI13.html The stack exchange content police tell me that links can go stale and insist I say more. So just in case the human race forgets proposition 13 of book VI of Euclid's Elements, I must explain that it tells you how to construct a geometric mean. I.e., ...


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The ancients didn't actually think of it as "finding a square root of a number", so they didn't prove that. What they did was "find a square that has the same area as this rectangle", and/or "given line segments AB and CD, find a line segment EF such that the ratio of AB to EF is the same as the ratio of EF to CD". The latter task can be proved using ...



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