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There is no simple equation By that I mean that there is no "nice" function $f(x,y)$ for which $f(x,y)=0$ would determine a rectangle. Of course such a function exists, it's just not useful or nice to look at. For example, for the square with edges $(-1,-1),(-1,1),(1,-1),(1,1)$, you can take the function $$f(x,y) = \max\{|x|,|y|\} - 1$$ which yields the ...


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Use Pythagoras: $$ a^2 + b^2 = c^2 $$ Pythagorean theorem $$a^2+12^2=20^2$$ $$a^2 + 144 = 400$$ Subtract 144 from both sides. $$a^2 = 256$$ you only take the positive answer. $$a = \sqrt{256}=16$$


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Consider circumscribed circle and it's radius $R$. By inscribed angle theorem you get, that $|c|=|R|$, where $c$ is third side of your triangle $a=b=10$. Now you have formula $\displaystyle S=\frac{abc}{4R}$, where $S$ is area of triange. So: $$S=\frac{10 \cdot 10 \cdot c}{4R}=\frac{100}{4}=25$$


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There's an interesting property that you can use for this: a triangle inscribed in a circle, with two of its points at opposite ends of a diameter of the circle, is a right triangle. Also, a line tangent to a circle at a particular point C, makes a right angle to the radius line through C. Try drawing a circle with diameter AB.


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First set $\theta = 2x-10$ and find the domain of $\theta$. The law of cosines states $$ \begin{align} 9^2 & = D^2 + L^2 -2 D L \cos \theta \\ 10^2 & = D^2 + L^2 -2 D L \cos 38° \end{align} $$ where $D$ is the common edge, and $L$ is the equal outside edge (marked with $\parallel$) Now consider the ratio $\lambda = \frac{L}{D}$ with the second ...


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You can form a circle of any radius about any point. If you used the vertex between the equal legs as the center, and a radius equal to the length of these legs, the ends of these legs would necessarily fall on the edge of said circle. But if you pick a vertex between unequal legs, it will of course be impossible to get the end points to fall on a circle. ...


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If a line from $A$ intersects a circumference in two points $P$ and $Q$ then it holds that $AP·AQ = {AT}^2$ where $T$ is a point of the circumference so that $\overline{AT}$ is tangent to the circumference. This is called power of a point Create such a line (for instance $\overline{AB}$). Name the points of intersection with the circumference $P$ and $Q$ ...


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I prefered 3D coordinates system to prove the lemma. In my opinion , this tool is the most elementary tool for that problem. The projection area $Q$ on $xy$ plane and bordered by $(OED)$ can be written as $$\mathbf{M_{2}}=\begin{bmatrix}0 & 0 & 1\\x_1 & y_1 & 1\\x_2 & y_2 & 1\end{bmatrix}$$ $Q=\frac{1}{2} ...


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I don't follow your argument about the tighter lower bound on the angle. The way I read it, the angle $2x-10$ can indeed become zero, namely when the absolute value of the difference in edge lengths becomes $9$: $$\lvert AB-AC\rvert=9\implies\angle ACB,\angle CBA\approx71°\pm58.316971°$$ I'm more concerned about the upper bound. For reasons of symmery (and ...


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WLOG, draw in one of the diagonals, $d$. This will form two triangles using parts of $d$, $L$, and the sides of the parallelogram. Using properties of parallel lines, these triangles must be similar. Moreover, your discovery that the trapezoid bases must be equal implies that these triangles are actually congruent, and therefore $L$ intersects the diagonal ...


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Try drawing an altitude from one of the $75$ degree angles to the opposite 10-unit side. You should be able to determine the length of this altitude using a 30-60-90 triangle.


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Let the barycentric coordinates be $(u,v,w)$ with as usual $u+v+w=1.$ If your point is interior to (or on the edge of ) the triangle we also have $u,v,w \ge 0.$ Now if we choose to map to the equilateral triangle in the $x,y$ plane with vertices $A'=(-1,0),\ B'=(1,0),\ C'=(0,\sqrt{3})$ then the barycentric point $X:(u,v,w)$ gets mapped into $u\cdot ...


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Yes, the theorem you quoted is missing an injectivity assumption. It is possible to change variables by a non-injective map, but then one must account for multiple preimages in the integral on the right (multiply $f(x)$ by the number of preimages of $x$ under $\phi$). As is, the following is a counterexample: let $n=2$, let $A$ be the annulus described ...


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Call the intersection points of one (of the two) adjacent lines with the circle $(a,b)$ and $(c,d)$ (4 parameters). You know the distance between these points and the midpoints of the circle (two equations). The diameter connecting the midpoint of one of the circles and the intersection point of the adjacent point with that circle is perpendicular to the ...


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Let be $X_1$ the center point of first circle $X_2$ of the second. $V_1$ the intersection point on the first circle $V_3$ of the second. $d$ the distance between $X_1$ and $X_2$. $r_i$ the radii. The angle $\alpha$ between $V_1X_1X_2$ is given by $\cos\alpha=\frac{R_1-R_2}{d} $ Thus $V_1$ is given by $X_1 + R_1*\hat e_\alpha$ where $e_\alpha$ points in the ...


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A curious equation : $$\lim_{n \to \infty}\left({\left(\frac{x}{a}\right)^{2n} + \left(\frac{y}{b}\right)^{2n}}\right) = 1$$ For $-a<x<a$ then $\frac{x^2}{a^2}<1$ then $\left(\frac{x^2}{a^2}\right)^\infty=0$. Hense $\frac{y^2}{b^2}=1$ . So $y=b$ or $y=-b$ which gives the upper and the lower sides of the rectangle. For $-b<y<b$ then ...


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For a rectangle centered at $(a,b)$ with $l=$ side length parallel to the $x$-axis and $m=$ side length parallel to the $y$-axis, the equation is $$\text{Max} \biggl( \frac{2 \, |x-a|}{l}, \frac{2 \, |x-b|}{m} \biggr)= 1 $$


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Let $D$ be the incenter of the triangle $ABI$ Let's draw the picture: Thus $f(D)=f(I)f(A)f(B)=f(A)^2f(B)^2f(C)$ Now, consider pictures symmetric pictures with respect to the line $IC$. We get points $A',B',D'$ for which we have 1) $f(A')f(B')=f(A)f(B)=f(I)/f(C)$ 2) $f(D')=f(A')^2f(B')^2f(C)=f(A)^2f(B)^2f(C)=f(D)$ Thus, we have so proved that ...


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You aren't missing anything, (0.1) is the expression of the dot product with respect to an orthonormal basis. If the basis is not orthonormal you get (0.3). Notice that you do not just need the vectors to be pairwise orthogonal, but also of unit norm in order for (0.1) to hold.


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First, the idea of "transformations" is not something that was added to Euclidean geometry; rather, the concept of Euclidean geometry as the study of invariants of Euclidean space under affine transformations is simply a more modern take on the five axioms. There are certainly studies of the underlying axiomatic system of Euclidean geometry, in fact it is ...


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The triangle is isosceles so the other two angles are the same. Since all the angles in a triangle add up to $180^\circ$ then those angles are $70^\circ$ each. Now, $x$ is the suplementary angle to one of them i.e. $70^\circ + x = 180^\circ$ therefore $x = 110^\circ$



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