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7

This is just a stream of thoughts, for now. Assume that such a set of disjoint, closed balls exist. How to build it? If we start placing disks on a regular curve that disconnects $\mathbb{R}^2$, like JMoravitz suggested, its curvature must be bounded away from zero, otherwise by considering the set of tangents to the curve we have that $N_l$ is unbounded. ...


6

The surface of a sphere, $S^2$, is a manifold that locally looks flat (if you get down to level earth, wouldn't you say the Earth looks flat?), yet obviously is not globally flat. What it means, heuristically, is that you can approximate the neighborhood of any point with a flat plane (or, generally with the appropriate $\mathbb{R}^n$), and "do math" on the ...


5

Place the square on a Cartesian coordinate grid. We can choose the units so the square is a unit square. The coordinates of the vertices $B,F,D,E$ are then obvious. (I limited this diagram to only what is necessary for my solution: your diagram has unneeded line segments and not enough labels for the points.) Point $C$, the midpoint of $\overline{BF}$, ...


4

Borrowing @Rory Daulton's notation, construct a second square directly beneath the original, and extend the line $\overleftrightarrow{DC}$ to meet the lower square at vertex $K$. The circle centered at $B$ with radius $BE$ passes through points $E$, $G$ and $K$. Therefore the inscribed angle $\angle EKG$ is half the central angle $\angle EBG$, since both ...


3

Let $S$ denote the area of the triangle, so: $$ S = \frac{ab\sin{(\gamma)}}{2} = \frac{cb\sin{(\alpha)}}{2} = \frac{ac\sin{(\beta)}}{2} $$ And as for $\theta \in[0,\pi]$: $0\leq\sin{\theta}\leq1$, we get as you said: $$ S\leq\frac{ab}{2}\\ S\leq\frac{ac}{2}\\ S\leq\frac{bc}{2}\\ $$ So: $$ 4S \leq 2ab\\ 4S\leq 2ac\\ 4S\leq 2bc\\ $$ And for $a,b,c\in ...


3

Note that if $A$ and $B$ are vectors representing two sides of a triangle, then the third side is represented by vector $C=B-A$. Let the corresponding magnitudes of the sides be $a,b,c$ The cosine rule tells us that $c^2=a^2+b^2-2ab \cos \theta$ Computing the squares of the magnitudes from the components (using Pythagoras) we obtain $$2ab\cos \theta=\sum ...


3

You might consider the following image: Each of the black circles represents the minimum distance points must be away from each other. The blue circle is the unit circle. Notice that moving any single point out of this configuration necessarily either places a point into the unit circle or increases the sum of distances to the origin. Edit: There is ...


3

My solution, for now: We take the line $l$ through $BT\cap AC,AT\cap BC$, $D=l\cap AB$, then $DC$ is tangent to the ellipse at $C$; Let $E=AT\cap DC$ and $\Gamma$ be the circle tangent to $AT,BT$ at $A,B$; Let $F=CT\cap\Gamma$ (one of the two intersections); Let $G$ be the intersection between $AT$ and the tangent to $\Gamma$ at $F$; Let $H$ be one ...


3

This was a fun question to play around with. If $f_X:X\to Z$ and $f_Y:Y\to Z$ are functions, then you could define the "metric" $d:X\times Y\to\mathbb{R}$ by: $$d(x,y) = d_Z(f_X(x),f_Y(y))$$ for any metric $d_Z:Z\times Z\to\mathbb{R}$. Then, as $d_Z$ is a metric, we have for all $x\in X$ and $y,z\in Z$: $d(x,y) = d_Z(f_X(x),f_Y(y))\geq 0.$ Note that ...


3

By the Cartan–Dieudonné theorem, reflections around $(n-1)$-dimensional hyperplanes generate the entire orthogonal group $O(n)$ of $\mathbb R^n$. So, asking that a subset $A \subset \mathbb R^n$ is closed under reflections is the same as asking that it is closed under the natural operation by $O(n)$. Thus, $A$ is a union of $O(n)$-orbits. Now $O(n)$ contains ...


3

Let $$[AMX]=a,[XMBY]=b,[BYN]=c$$ $$[AXTQ]=d,[XYZT]=e,[YNCZ]=f$$ $$[QTD]=g,[TZPD]=h,[ZCP]=i.$$ From $$[BCP]=[BPD]\quad \text{and}\quad[AMD]=[BMD],$$ $$[ABCD]=2(c+f+i)+2(a+d+g)\tag1$$ From $$[ABN]=[ANC]\quad\text{and}\quad[ACQ]=[QCD],$$ $$[ABCD]=2(a+b+c)+2(g+h+i)\tag2$$ From $(1)(2)$, $$2(c+f+i)+2(a+d+g)=2(a+b+c)+2(g+h+i)\Rightarrow f+d-b-h=0\tag3$$ From ...


3

If the rows or columns of a matrix $M$ are unit vectors (in the usual Euclidean norm), then $\det M\le 1$. One geometric interpretation of the determinant is the (signed) volume of the parallelotope ($n$-dimensional generalization of the parallelepided) spanned by the column vectors or row vectors of the matrix. The sign tells you whether the row/column ...


3

Construction: Let the angle bisector of $\angle BAC$ cut $BD$ at $X$. Then, Join $CX$. $X$ is actually is the in-center of $\triangle ABC$. Thus, $p = q$ $r = t + q = \dfrac {2t + 2q }{2} = \dfrac {180 - 48}{2} = 66^ 0 = \angle CAD$ This means $XADC$ is a cyclic quadrilateral Therefore, by angle in the same segment, $x = \angle CAX = \dfrac {48}{2} = ...


3

No such construction is possible in general, because it wold amount to a compass-and-straightedge solution of a sextic that is generally irreducible. For example, there is no compass-and-straightedge construction of a triangle with $a=t_a=1$ and $B = 60^\circ$. Let $a,b,c$ be the sides of the triangle. We're given $a$, $$ \cos B = \frac{a^2+c^2-b^2}{2ac} ...


2

I don't see any way other than ray-tracing. The rays reflect off the black rectangles and are refracted (presumably using Snell's law) by the white rectangles. I noticed that either ray 2 or 3 near the bottom right seem to have been absorbed by a white rectangle. It would be an interesting exercise to program this given a set of rectangles and the index ...


2

Somehow I have a copy of the official solutions, see here. I have not actually read the official solutions yet myself past a quick glance, so I would like to talk a bit only about the solutions on the AoPS Wiki you linked (mostly their "Solution 2"). While Solutions 1 and 3 there are genuinely semi-bashes, Solution 2 is really synthetic in nature. I am ...


2

You have $\log A=\log a + \log b - \log 4, \frac {dA}A=\frac {da}a+\frac {db}b, \frac {dA}A=4\%-3\%=1\%$


2

Let $s:=\frac{a+b+c+d}{2}$. By Bretschneider's Formula, $$S=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\frac{\alpha+\gamma}{2}\right)}\,,$$ where $\alpha$ and $\gamma$ are two opposite angles of the given convex quadrilateral. Hence, $$S\leq \sqrt{(s-a)(s-b)(s-c)(s-d)}\;\leq \left(\frac{(s-a)+(s-b)}{2}\right)\left(\frac{(s-c)+(s-d)}{2}\right)\,,$$ where ...


2

I asked my lecturer today, and he said that the generalizations are as follows An isometry on $\mathbb{R}^3$ that fixes four non-coplanar points, where no three are collinear, is the identity. An isometry on $\mathbb{R}^3$ that fixes three non-collinear points is a reflection or the identity. An isometry on $\mathbb{R}^3$ that fixes two points is ...


2

I think the most obvious definition is that $d(x,y)$ is a restriction of a metric on $X\cup Y$ to $X\times Y$. Let's look what properties of $d(x,y)$ we can derive from that. In the following I assume for simplicity that $X\cap Y=\emptyset$. Let's look at the metric axioms, and what follows: $d(x,y)\ge 0$ with equality if and only if $x=y$. Since $X$ ...


2

From a mathematical point of view: Note that if $X\neq Y$ you will have trouble to check the conditions $\tilde d(x,x)=0$ and $\tilde d(x,y)=\tilde d(y,x)$... Of course if $X\subset Y$ or $Y\subset X$ you might consider $\tilde d$ as the restriction of a metric but I'm not sure that it is was you are asking. From an intuitive point of view: We should keep ...


2

For part (A) your reasoning is correct. A line with direction vector $\mathbf{v}$ is parallel to a plane with normal vector $\mathbf{n}$ if and only if $\mathbf{v}$ is orthogonal to $\mathbf{n}$, that is, if and only if $\mathbf{v}\cdot\mathbf{n} = 0$. In this case the direction vector of the line is the same as the normal vector of the plane and so they ...


2

Let $F$ be the point such that $DF||BC$. Clearly $BCDF$ is cyclic and since $BD$ is the angular bisector of $\angle CBF$, this implies $CD=DF$. Let $E$ be the point on $BC$ such that $CE=DE$. Note that triangles $ECD$ and $AFD$ are similar and since $CD=DF$, they are congruent. Hence $CE=AD$ and the condition implies $BC+BD=AD=CE=BC+BE$ i.e. $BD=BE$. So ...


2

The main observation is that R is the center of an excircle of $\triangle APQ$. Also note that because $\angle BAR = \angle CAR$ we have BR=CR s.t. R can also be defined as the intersection of the bisector of $\angle A$ and the bisector of BC. Let R' be the intersection of the angle bisectors of $\angle QPB$ and $\angle PQC$. Since R' is the center of an ...


2

Hint: Let $H_A$, $H_B$, and $H_C$ be the reflections of $H$ about $BC$, $CA$, and $AB$, respectively. Prove that $H_A$, $H_B$, and $H_C$ are on the circumscribed circle of $ABC$.


1

Since $\widehat{AHB}+\widehat{ACB}=\pi$, $AHB$ and $ABC$ have the same circumradius by the sine theorem, since they have $AB$ in common and $\sin\widehat{AHB}=\sin\widehat{ACB}$.


1

AB+AD = BC. ADB = 3x+x = 4x. ABD = 180-(2x+4x) =180-6x. $\frac{\sin(2x)}{BC} =\frac{\sin(180-3x)}{AC} =\frac{\sin(3x)}{AC} =\frac{\sin(x)}{AB} $ $\frac{\sin(4x)}{AB} =\frac{\sin(2x)}{BD} =\frac{\sin(6x)}{AD} $ Therefore $AD =\frac{AB \sin(6x)}{\sin(4x)} $ and $AB =\frac{BC \sin(x)}{\sin(2x)} $, so $AD =\frac{\frac{BC \sin(x)}{\sin(2x)} ...


1

$AB∦AC $ would be sufficient, you don't need all three statements. Alternatively, $\overrightarrow{AB}\times\overrightarrow{AC}\neq0$, or any such permutation


1

The target function is continuous and we can restrict the domain to points in the compact annulus $1\le r\le 3$. Hence the minimum of the function is attained. Therefore it suffices to show for any configuration that is not a regular decagon centered at $O$ and with side length $1/\sqrt 2$, there exists a better configuration. This way you can readily show ...


1

I think you nicely understoof the minimal stable sets for a), i.e. the sets $A_i$ from which you can not remove anything if you want to keep it stable. It is possible to describe some unions of such sets nicely, but if you want to describe all stable sets, you have to go to that level you used. What's the difference between stable and invariant? With a ...



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