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5

I'm the developer of Euclid: The Game. I'm glad you like the game, don't worry about that you can't get this solution in 3 moves. Only very recently this record is set. No-one in the top 10 of the highscores has been able to do this, in fact, from all the scores that are submitted, no-one has ever done this in 3 moves. Some user noted in the comments that ...


3

You seem to be choosing D as the midpoint of BC, which would work. Except that the game doesn't have a midpoint function so you're creating that point - not guaranteeing that it will be the midpoint. Think if you can use circles in some way to find a point equidistant from B and C. If you can do that, you're set!


2

The vectors in vector spaces are abstract entities that satisfy some axioms. For a vector space that is not a set of points, consider the set of all continuous function $[0,1] \to \mathbb R$. The Euclidean spaces $\mathbb R^n$ are examples of vector spaces.


2

While a vector space is something very formal and axiomatic, Euclidean space has not a unified meaning. Usually, it refers to something where you have points, lines, can measure angles and distances and the Euclidean axioms are satisfied. Sometimes it is identified with $\mathbb{R}^2$ resp. $\mathbb{R}^n$ but more as an affine and metric space (you have both ...


2

The geodesics on $H^2$ are, indeed, semi-circles centered on the real line and also vertical half-lines. This is the hyperbolic (curvature=-1) plane, so the only projective axiom which can be violated is the second one - and, indeed, there are lines which do not intersect (e.g., semi-circles with the same center and different radii).


2

Proof. Three circumcircles of $\triangle AFE$,$\triangle BDF$,$\triangle CED$ concur at $M$ (Miquel point) $\angle BB'M+\angle AA'M=\angle BFM+\angle AFM=180^{\circ}$ and then $\square IA'MB'$ is concyclic. Likewise, $\square IB'MC'$ is concyclic and then $I$,$A'$,$B'$,$C'$,$M$ are concyclic. Remark. (1) $I$ need not be incenter. (2) $BF+BD=AC$, ...


2

I am a fan of collisions. Get some simple euclidean shapes (billiard balls, or dice on ice) and show her how to calculate the angles that they'll move in after hitting each other.


2

Good observation skills! Let's chase definitions and then be happy :) Let me write $n = 10, m = 3$ and let your matrix be denoted $A$. Then we calculate! $$ \begin{align} f &= \|A\|_F \;\; \text{ your defn} \\ & = \sqrt{\sum_{i=1}^n\sum_{j=1}^m |A_{ij}|^2}\;\; \text{ defn of Frobenius norm} \\ & = \sqrt{\sum_{i=1}^n\sqrt{\sum_{j=1}^m ...


2

A rotation around an axis is a composition of two symmetries (reflections) wrt planes which intersect at this line at half the angle. If your two rotations are represented as $R_i=S_{i1}\circ S_{i2}$ so that each plane $S_{12}$ and $S_{21}$ contains both axes, then $S_{12}=S_{21}$, and $R_1\circ R_2=S_{11}\circ S_{22}$.


1

$\def\\#1{{\bf#1}}$We can find the axis of $R_2R_1$ by writing each of the rotations as a product of two reflections. We may assume without loss of generality that $\\v_1$ and $\\v_2$ are unit vectors, and are independent (if they are scalar multiples of each other then $\\v_1=\pm\\v_2$ and the problem is easy). Define the following vectors: $$\eqalign{ ...


1

From (1) $\angle PAE$ is the common angle. (2) $\angle ABP = \angle Q = \angle P$ [equal chord, equal angles] + [base angles, isosceles ⊿] ∴⊿APE~⊿ABP Thus, $AP^2 = AB. EA$


1

I remember that we did some geometrical optics in high school. Maybe it is a little to advanced, but maybe you'd like to judge by yourself. So what I concretly remember is: Snells law Lens optics, especially the "thin lense formula" Snells law uses very basic trigonometry (actually only the definition of sin), but you could avoid this by just using ...


1

People already gave some good ideas, but I'll pitch in what could be a more systematic approach Without calculus is somewhat difficult to study physics after Newton. Fortunately a lot was known before him. An account of this can be seen in Rene Dugas' "A History of Mechanics". In this book he explains how people did mechanics in the past, with lots of ...


1

$f$ is an isometry so for all $x,y\in\Bbb R$ we have $$|f(x)-f(y)|=|x-y|$$ so for $y=0$ we have $$|f(x)-f(0)|=|x|$$ hence $$f(x)=\pm x+a,\quad\forall x\in\Bbb R$$ and since $f$ hasn't a fixed point then the only possibility is $$f(x)=x+a,\quad a\ne0$$


1

If $f$ is a non-trivial translation that means $\exists x_0 \in \mathbb{R}$ such that $f(x_0) \neq x_0$. Then $c=f(x_0)-x_0$ is the amount of translation. Now you can show that no $x$ is fixed.


1

Let $f$ be an isometry of $\mathbb R$ that is not the identity. Then there exists $a\in\mathbb R$ with $b:=f(a)\ne a$. Let $c=f(b)$. Then for any $x\in \mathbb R$ we have both $$\tag1 |f(x)-b|=|f(x)-f(a)|=|x-a|$$ and $$\tag2|f(x)-c|=|f(x)-f(b)|=|x-b|.$$ For all $x\in \mathbb R$, we have ...


1

Let $\vec{v}_p=(x_p,y_p,z_p)$ and $\vec{v}_r=(x_r,y_r,z_r)$. Denote $\vec{v}_{p'}=(x_{p'},y_{p'},z_{p'})$ the coordinates of P after rotation. Then we have: $$(\vec{v}_{p'}-\vec{v}_r)=R_x(\phi)R_y(\theta)R_z(\psi)(\vec{v}_{p}-\vec{v}_r)$$ So $$\vec{v}_{p'}=\vec{v}_r+R_x(\phi)R_y(\theta)R_z(\psi)(\vec{v}_{p}-\vec{v}_r)$$


1

Which information are you trying to extract? Assuming the question as given actually gave that x is the distance from the bottom of the ladder to the bottom of the wall, there are 4 pieces of information: 1) There is a 10 foot wall 2) The wall if 5 feet from the building 3) The ladder touches both the wall and the building 4) The distance from the bottom ...


1

By the Pythagorean Theorem, we get: $L=\sqrt{y^2+(x+5)^2}$. Now, we have to find a way to express $y$ in terms of $x$. Notice that the two triangles in the diagram are similar triangles. This means that the ratio of two sides in the first triangle has to be equal to the ratio of the corresponding sides in the second triangle: ...


1

A few thoughts: Resolving vectors into components is all about trigonometric ratios, which in turn are all about similar triangles. Inverse-square laws (gravity, electrical force) can be interpreted geometrically in terms of the surface area of a sphere. Planetary orbits and ellipses. I am sure I will think of more and will add to this list as they occur ...


1

Simple - probably not. Not-so-simple... By solving a quadratic equation, you can determine the two points of intersection of the circles $X$ and $Y$. Then the Ptolemy's theorem gives the condition that one of those 2 points lies on the circle $Z$. PS. Suppose the 3 circles intersect at $O$. An inversion centered in $O$ turns the three circles into three ...


1

Here is one approach to obtain a polynomial condition. For a point $(x,y)\in\mathbb R^2$ you can compute a “circle vector” as follows: $$(x,y)\mapsto(1+x^2+y^2, 1-x^2-y^2, 2x, 2y)$$ Do this for all nine points. Write down the resulting vectors for set $X$ as the rows of a $3\times 4$ matrix. Then compute all $3\times3$ determinants you can obtain by ...


1

Hint If it is a right triangle, you would have $a^2+b^2=c^2$. Here you are given $a=m^2-n^2$,$b=2mn$,$c=m^2+n^2$. Replace and see if it holds.


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Here is the sketch of the proof: Consider what happens when the tangent point moves along the incircle. Prove that the target triangle rotates in the opposite direction (around a different center). Prove the statement (that the target triangle is congruent to $ABC$) when the tangent point lies on one of the sides, e.g., on $AB$.


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Lemma. Given $\overline{XP} \cong \overline{XQ}$ and $\overline{XY}\cong\overline{ZQ}$ (and $\overline{XZ} \cong\overline{YP}$) as in the diagram, let $K$ be isosceles $\triangle XPQ$'s circumcenter (which necessarily lies on the bisector of $\angle X$). Then $X$, $Y$, $Z$, $K$ are concyclic. Proof of Lemma. Evidently, $\triangle XYK \cong \triangle ...


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Your hypothesis is correct (except at the origin). I doubt, in general, that this will drastically improve your algorithm's efficiency, though. I could be wrong... Here's an improved algorithm: Let $i$ be the index for which $P_i$ is maximal. Then $e_i$ (the $i$th vector in your set, ordered the way you wrote it) is the closest point. Proof: Compute the ...


1

Refer to the figure. 2x = 14y = 7z (given) After dividing throughout by 14, we have x/7 = y/1 = z/2 = k for some non-zero k. Then, x = 7k; y = k; and z = 2k [note:- all measures are in degrees] Angle DBC = [1] = z = 2k [2] = 180 – 2z = 180 – 4k [3] = 180 – z – [1] – y – x = … = 180 – 12k [4] = (180 – [3]) / 2 = … = 6k [5] = [4] = 6k [6] = x – ...


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Since $AB$ is the diameter of the 2nd circle, $\angle AGB=\frac{\pi}{2}$. This means that $AG \perp BG$ which is the radius of the 1st circle.


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It suffices to show that the line $(BG)\perp (AG)$. Notice that the triangles $AEG$ and $EBG$ are isosceles so $$\angle{EAG}=\angle EGA$$ and $$\angle EGB=\angle EBG$$ so we have $$\angle AGB=\angle AGE+\angle EGB=\frac12\angle AEB=\frac{180^°}{2}=90^°$$ hence the line $(AG)$ is tangent to the circle on $G$.



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