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21

Let $|FB|=x$. By similarity of triangles we then have $|CE|=1/x$. Pythagoras thus gives $$ 4=\sqrt{x^2+1}+\sqrt{1+(1/x)^2}=\sqrt{x^2+1}(1+\frac1x). $$ Squaring this gives us $$ 16=(x^2+1)(1+\frac2x+\frac1{x^2}), $$ but I prefer to move one factor $x$ from the former factor on r.h.s. to the latter, so $$ 16=(x+\frac1x)(x+2+\frac1x). $$ Getting warmer! Write ...


12

Consider the following, where we see a full barn (of a curious gravitational nature) with four ladders of length $\color{green}{g}$ leaning against side-$w$ boxes stuck in the corners where floor and ceiling meet walls: Clearly, $$\begin{align} (w+\color{violet}{p})^2 &= W + \color{violet}{P} + 2 \color{green}{G} + 2 \color{blue}{B} + 2 ...


9

The answer is yes; the given configuration is the unique such configuration up to Euclidean transformations. I advise to keep pencil and paper at hand when checking the proof. EDIT: I have completed the proof, and revised most of the earlier part to make the entire proof more 'streamlined'. Let $P_1,\ldots,P_7$ seven points in the Euclidean plane, such ...


9

Try to figure that out from the attached diagram.


6

We will use the following defintion: Definition. A line $l$ is orthogonal to a plane $\Pi$ if it is orthogonal to every line in the plane, which passes through the intersection of the line $l$ and the plane $\Pi$. So, with this definition, you are asking for a proof of the following statement: Theorem. Every line that is orthogonal to two ...


6

No, because that would imply an infinite sequence of smaller and smaller triangles with the same property: $\hspace{90pt}$ The key to the proof below is this property: For any point $(x,y) \in \mathbb{Z}^2$ we have that $(-y,x)$ and $(y,-x)$ are its ccw and cw rotations around $(0,0)$ by $\frac{\pi}{2}$. This implies that we can rotate points of ...


6

Let $P=(a,b),Q=(c,d)$ be two of the three points on the grid (assumed to be rational). The midpoint between these is $M=(\frac{a+c}{2},\frac{b+d}{2})$. Let $w=\sqrt{(d-b)^2+(c-a)^2}$ the length of a side. A vector in the direction of this side is given by $\vec{u}=\langle c-a,d-b\rangle$ and $\|\vec{u}\|=w$. The vector perpendicular to $\vec{u}$ is given by ...


6

Yes. Normally I like to give some more motivation for an answer, but in this case it's probably hard to say anything suggestive that the illustration below does not, besides perhaps that after experimenting one might guess that the four-point "diamond" configurations are helpful (and probably necessary) in constructing such an arrangement. (I remember ...


5

You can even generalize the law of cosines. $$ \cos\left(\frac{z}{r}\right) = \cos\left(\frac{x}{r}\right) \cos\left(\frac{y}{r}\right) + \cos(\phi) \sin\left(\frac{x}{r}\right) \sin\left(\frac{y}{r}\right). \tag 1 $$ What we have is $$ \begin{array}{rcl} r^2 > 0 &\rightarrow& \textrm{spherical}\\ && ...


4

A restatement of your question: Given a line in $\Bbb R^3$, at how many points may at least one of the coordinates change sign? Certainly a max is three, so the answer to your question is four, once we find such a line. If you join $(-1,-2,-3)$ to $(1,1,1)$, that line will do it. I think a somewhat more interesting question would deal with a ...


4

Consider the equation of the line that the ladder makes: $$\mathcal{l} = \{ (x, y) ~:~ y = mx + h \}$$ where $h$ is the height of the ladder on the wall and $m$ is the slope of the ladder. We know that $(1, 1)$ is on the line, so $$1 = m + h$$ And we know that the distance from the x-intersecpt to the y-intersept is $4$. So $$h^2 + (-h/m)^2 = 4^2$$ ...


4

Let your angle $\angle (h,k)$ be given as angle $\angle BAC$ in this diagram, with $h=\overrightarrow {AB}$ and $k=\overrightarrow {AC}$. Draw the ray $\overrightarrow{AP}$. Place any point $D$ on the bisector of $\angle (h,k)$ and draw the circle centered at $D$ tangent to both rays $h$ and $k$. Let the intersections of circle $D$ with ray ...


3

You can rotate any single vector to its antipodal vector. You cannot rotate all vectors at once to their antipodes by a single rotation, or even by a composition of rotations (for instance, for a single rotation the vectors on the axis won't move at all). The orientation is the simplest explanation why this is not possible. Try for some familiar shape (for ...


3

Okay, let the constant yaw-rate be $\omega$. As the body has a constant linear forward velocity of $v$, it must follow a perfectly circular path. Let us consider the path of the body from $t=0$ to $t=\Delta t$. We see from the figure that: $$\phi_{G} + R = \frac{\pi}{2}$$ $$\theta + 2R = \pi$$ $$\Rightarrow \theta = 2\phi_{G}$$ From the sine rule in ...


3

Let the triangle be $\triangle ABC$. Let $\angle A=60^{\circ}$. Let the incircle touch the triangle on the points $A_1,B_1,C_1$ (opposite to $A,B,C$, respectively). Let $A_1B=a,\,A_1C=b$. Then $C_1B=a,\, B_1C=b$. Let $AC_1=AB_1=x$. Then by the Law of cosines: $$(a+b)^2=(x+a)^2+(x+b)^2-2(x+a)(x+b)\cos 60^{\circ}$$ Solve this quadratic equation. The area is ...


3

Theorem: A triangle is embeddable in $\mathbb{Z}^2$ if and only if all its angles have rational tangents. But $\tan(\pi/3)=\sqrt{3},$ so an equilateral triangle is not embeddable. Reference: Triangles with Vertices on Lattice Points Michael J. Beeson, The American Mathematical Monthly, Vol. 99, No. 3 (Mar., 1992), pp. 243-252


3

Hints: Your ellipse is not centered at the origin. Substitute $\phi=0$ and $\phi=\pi$ to get the ends of the major axis. You get $\left(-\frac{es}{1+e},0\right)$ and $\left(\frac{es}{1-e},0\right)$ as the left and right axis ends. The center of the ellipse is halfway between those two points. The standard equation of an ellipse centered at point $(h,k)$ is ...


3

From your link: V.23. Perturbed ratios ex aequali. If u : v = y : z and v : w = x : y, then u : w = x : z.


3

Let's assume that $x$, the side of the equilateral triangle, is a known positive quantity and that side $BC$ is horizontal (or that point $A$ is directly above point $M$). Let's also assume you are using Cartesian coordinates (where increasing the first coordinate means moving right and increasing the second coordinate means moving up). Then $x$ is the side ...


2

Draw the circle of radius $\frac {\sqrt 3}3x$ around $M$. Pick an arbitrary point $A$ on this circle. Then intersect the circle of radius $X$ around $A$ with the first circle to determine $B,C$ as intersection points. Note that $A$ could be picked anywhere on the circle, hence the result is not unique.


2

Let $\alpha, \beta, \gamma$ be the barycentric coordinate of $P$ with respect to $\triangle ABC$. i.e. $$\vec{P} = \alpha\vec{A} + \beta\vec{B} + \gamma\vec{C}\quad\text{ with }\quad \alpha + \beta + \gamma = 1$$ Since $P$ is inside $\triangle ABC$, we have $\alpha, \beta, \gamma > 0$. It is not hard to see $$\frac{PD}{AD} = \alpha,\quad\frac{PE}{BE} = ...


2

Analytic geometry is not generally axiomatized in the way you seem to be asking. Instead, one proceeds as follows. First, the real numbers $\mathbb{R}$ are axiomatized, as you learn in any advanced calculus or real analysis course. Second, the Euclidean plane $\mathbb{E}^2$ is defined as the Cartesian product $\mathbb{R} \times \mathbb{R}$, and lines are ...


2

$ \renewcommand{\v}{\mathbf{v}} \newcommand{\vo}{\mathbf{v_0}} \renewcommand{\p}{\mathbf{p}} \renewcommand{\d}{\mathbf{d}} \renewcommand{\a}{\mathbf{a}} \renewcommand{\b}{\mathbf{b}} $ First, WLOG assume $\p = \a$, so that $\v = \p + \d t = \a + \d t$. Then the distance between $\v $ and $\a$ is $$ \|\v - \a\| = \|\p + \d t - \a\| = \|\a + \d t - \a\| = ...


2

Since $BA'=\frac{1}{3}BC$ and so on, we have: $$ [AB'C']=[A'BC']=[A'B'C] = \frac{1}{3}\cdot\frac{2}{3}[ABC] \tag{1}$$ so the area of $F$ is just one third of the area of $ABC$. The latter can be computed through Heron's formula: $$ [ABC]=\frac{1}{4}\sqrt{108\cdot 6\cdot 48\cdot 54}=324, \tag{2}$$ so: $$ F = \color{red}{108}.\tag{3}$$


2

Indeed. Let $\textbf{p}_1$ and $\textbf{p}_2$ be points on line $A$ and $\textbf{q}$ be a point on line $B$. Then $d(A,B)$, the distance between lines $A$ and $B$ can be quickly found using the expression ...


2

As I said in my comment above, one can define: $\alpha=1/BC$, $\beta=1/AC$, $\gamma=1/AB$ so that we are required to find the point $P$ such that $\alpha PA + \beta PB + \gamma PC$ attains its minimum value. Following the method outlined here let's denote by $\vec{i}$, $\vec{j}$ and $\vec{k}$ the unit vectors along $\vec{PA}$, $\vec{PB}$ and $\vec{PC}$ ...


2

Assuming closed balls are allowed, $(x_i, x_j)$ have the property if there is $x$ such that $x_i$ and $x_j$ are the two closest points to $x$ in your set, i.e. $\max(|x-x_i|, |x-x_j|) < |x-x_k|$ for all $k \notin \{i,j\}$. Note that if $x$ works here, so does any point in some neighbourhood of $x$. Thus for simplicity we can restrict our attention to the ...


2

According to the given, $\alpha + \beta = 180^0$. Construction: 1) Through P, draw PM // BC. 2) Join PB. 3) Through C, draw CM // PB. 4) Join DM. Then, PBCM is // gm. Hence, PM = BC. In the quadrilateral ADMP and //gm ABCD, from AD = BC = PM and AD // BC // PM, we can say that ADMP is a // gm. This means ∠APB is successfully translated to ∠DMC. ...


2

Here's a result for $n=2$. As discussed at Bounds for the size of a circle with a fixed number of integer points, a circle of radius $R$ centred at the origin passes through at most $4(2\log_5R+1)$ integer lattice points. Our circles need not be centred at the origin, but we can get an upper bound for the denominator of the rational coordinates of their ...


2

This should be the Napoleon-Barlotti theorem.



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