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5

Let $O$ be the incenter of $\triangle ABC$, and let $X$, $Y$, $Z$ be its excenters. Let $O^\prime$ be the incenter of the target $\triangle A^\prime B^\prime C^\prime$. The keys to this solution these observations: $\overleftrightarrow{XA^\prime}$, $\overleftrightarrow{YB^\prime}$, $\overleftrightarrow{ZC^\prime}$ concur at $O^\prime$ $O^\prime$, as well ...


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 ...


4

The key is that two reflections with respect to two non parallel lines is a rotation, where the rotation angle is twice the angle between the two non parallel lines. Consider image 1. A radial line is associated with an angle $\xi$ and we can denote a radial line as $r(\xi)$. A reflection with respect to a radial line with associated angle $\xi$ can be ...


3

Polynomial proof Here is a rather ugly computational proof. I'm using homogeneous coordinates in the projective plane $\mathbb{RP}^2$. W.l.o.g. the incircle is the unit circle, represented by the matrix $$M=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}$$ For some parameters $a,b,c\in\mathbb R$, the vectors $$ ...


3

You know the angle between the red and green sides, call that $\beta.$ From the law of cosines, you know the length of the blue side, call that $b.$ Call the read and green sides $r,g.$ Call the opposite angles $\rho, \gamma. $ You know these from the law of sines. The purple line splits the angle $\beta$ into two known pieces, call those $\beta_r$ on the ...


2

the laws of sine and cosine for triangles would allow you to compute this. The law of sines says that in a triangle with side length $a, b, c$ and angles $\alpha, \beta, \gamma$ (where $\alpha$ is the angle opposite to $a$ and so on) you have the equality $$\frac{\sin(\alpha)}{a} = \frac{\sin(\beta)}{b} = \frac{\sin(\gamma)}{c}$$ while the law of cosines ...


2

No. Let $$U=\bigcup_{n=1}^\infty B\left((2n+2^{-n},0),1\right)$$ Then $U$ is connected, but for $0<\epsilon<1$, we find that for all $n$, the ball $B((2n+2^{-n}),1-\epsilon)$ is contained in $U_\epsilon$, but for sufficiently large $n$ there is a gap between the $n$th and the $(n+1)$st ball. Indeed, by Pythagoras the width "neck" between consecutive ...


2

Let $M$ be the midpoint of $CH$. Furthermore, let $P$ be a point on (the extension of) $CH$ such that $SP$ is perpendicular to $CH$. Because $\angle CHS - \angle CSB = 90^{\circ}$, it is straightforward to show that the triangles $\triangle SBC$ and $\triangle SPH$ are similar. Consider spiral homotheties $$B(90^\circ \text{counter ...


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

If you look at the vector equation of a plane $$P:\; \mathbf{p}+t\mathbf{u}+s\mathbf{v}$$ it's simple: you need two vectors, $\mathbf{u},\mathbf{v}$, (linearly independent) and a point, $\mathbf{p}$. Given three points, we can take any two of them (assuming they're non collinear, which of course is necessary) and write out their coordinate differences ...


2

Consider the two pairs of similar right triangles (red and blue) in the diagram. Writing $C$ for either $C_{e}$ or $C_{i}$, we have $$\frac{|\overline{AC}|}{a} = \frac{|\overline{BC}|}{b} \qquad\to\qquad a\;|\overline{BC}| = b\;|\overline{AC}|$$ Since $A$, $B$, and $C$ are collinear, we can write coordinate-vector equations $$a\;(C-B) = \pm\;b\;(C-A)$$ ...


1

If you have two points, $A$ and $B$, then you can consider the linear combination $$P=A+\lambda(B-A)=(1-\lambda)A+\lambda B$$ It will result in a point on the line $AB$ which satisfies $$\frac{\lvert A,P\rvert}{\lvert A,B\rvert}=\lambda\quad\implies\quad \frac{\lvert A,P\rvert}{\lvert B,P\rvert}=\frac{\lambda}{1-\lambda}$$ You can also reverse this, and ...


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 – ...


1

(The following solves a simpler problem; see Blue's comment.) Since both triangles have the same circle as incircle it is enough to verify that their angles agree. To this end connect the center $O$ of $ABC\,$'s incircle with the chosen tangency point $P$ and look at the angles formed between $OP$ and the angle bisectors of $ABC$. These angles reappear ...


1

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

Euclidean distance is exactly what we get when we consider the modulus of a complex number: $$|a+bi| = \sqrt{a^2 + b^2}$$ so it makes complete sense to use Euclidian distance, and this is in some sense the most natural distance to choose, since $\mathbb C$ is isomorphic to $\mathbb R^2$.


1

You can certainly use the Euclidean idea of distance between two complex numbers $$ |z_2 - z_1| = \sqrt{(x_2-x_1)^2 + (y_2 - y_1)^2} $$ However, the complex plane contains more algebraic structure than plain euclidean space $R^2$ so calling complex numbers "part of the euclidean space" is not quite accurate. Here is a link to another question with a ...


1

Wlog $r=1$. The average is independent of the direction of the special point, so take the average over all directions: the answer is the mean distance between a random point on a sphere and a random point inside it. That distance depends only on how far away the point is from the centre. To calculate the mean distance between a specific point inside and the ...


1

The direct integration route isn't actually that bad, and obtains in fewer words the same triple integral computed by Kirill. We want to compute $\int_S dV \, |\hat{z}-\mathbf{r}|$ where $S$ is the sphere of radius $r$. By the law of cosines, we can write the separation as $$ |\hat{z}-\mathbf{r}| = \sqrt{r^2-2r \rho \cos{\theta}+\rho^2} $$ where $\theta$ is ...


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

For a vector-based approach, let the two endpoints be given by position vectors $\mathbf{x}_1,\mathbf{x}_2$ and the intersection point by $\mathbf{x_3}=s\hat{n}$ where $\hat{n}$ is the direction of the purple line. The desired parameter is the length $s$. To find this, we parameterize the blue line as $\mathbf{y}(t) = (1-t) \mathbf{x}_1+t \mathbf{x}_2$. ...


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

This has to do with the nontrivial fact that the center of gravity of the vertices of a triangle coincides with the center of gravity of its area. (The center of gravity of the vertices is their arithmetic mean, and the center of gravity of the area has to lie on all three medians, by physical considerations. Both approaches lead to the same point.) Now to ...


1

You're going about it the wrong way. Try this: Divide through by $a$, to get an equation of the form $z^2+Bz+C = 0$, which has the same roots as the original equation. (This step is not strictly necessary, but it simplifies the algebra.) Take the two roots $\alpha, \beta$ that you know about (if the equation has only one root $\alpha$, set $\beta = ...


1

You can use circular inversion, or just intersect the segment $BI$ with the hyperbola that is the locus of points $P$ for which $PG-PI=BG$. That intersection is the center of the circle you are looking for. If you are familiar with the Descartes circle theorem, you know in advance the radius $r$ of your circle, since, given that $BG=1$, ...



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