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Since $\{v_1,\dotsc,v_{n-1},v_n\}$ is a basis, so too is $\{v_1-v_n,\dotsc,v_{n-1}-v_n,v_n\}$, and hence $$ \forall 1 \leq k \leq n, \quad \det(v_1-v_n \vert \cdots \vert v_{n-1}-v_n \vert v_k) = \begin{cases} 0, &\text{if $1 \leq k \leq n-1$,}\\ \det(v_1 \vert \cdots \vert v_n) \neq 0, &\text{if $k = n$.} \end{cases} $$ By multilinearity of the ...


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Some books will tell you that they are the same, and some will speak of Euclidean spaces in higher dimensions. It's a matter of convention. However, the space $\mathbb R^3$, when not assigned an inner product, is only a vector space, so that one cannot speak of angles and distances as one would in Euclidean space. And the space $\mathbb R^3$ has an ...


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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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Suppose you lower the bottom face of the cube enough that it no longer intersects the plane. (That is, you replace the cube with a tall square prism.) Then the pentagon becomes a parallelogram $ABFE$, with $C$ and $D$ lying on the edges $BF$ and $FE$ respectively. Clearly $|AB|+|AE|=|BF|+|FE|$, and $|BF|+|FE|\ge|BC|+|CD|+|DE|$.


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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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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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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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From the Trig Ceva theorem it follows that: $$\frac{AF}{FB}\cdot\frac{BD}{DC}=1\tag{1}$$ must hold, or: $$\frac{b \cos A}{a \cos B}\cdot\frac{c}{b}=1,\tag{2}$$ $$\cot A\cdot\frac{\sin C}{\sin(\pi/2-B)}=\cot A\cdot\frac{\sin\widehat{ACB}}{\sin\widehat{FCB}}=1\tag{3}.$$ Since $\widehat{ACB}>\widehat{FCB}$, $\cot A<1$ must hold, so: ...


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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 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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Here the list organised by main subject. Add more if you know some, but add reference to where it comes from. If a proposition falls under more than 2 subjects you may add them under both. Like triangle 5 ( Every triangle can be circumscribed ) and circle 1 ( Given any three points not on a straight line, there exists a circle through them). Lines: ...


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Draw the side given AB draw the angle given A produce the adjacent side which equal to sum of given sides AP connect remaining point from it PB bisect that side PB produce that side until cut AP take the point of intersection C now you have triangle.


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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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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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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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