SMM
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In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.
22 votes

Look at the picture: From $\triangle ABE$ we have $(2+r)^2= 2^2+(2-r)^2$ so $r=1/2$. From $\square ECGF$ we have $CG^2=(1/2+s)^2-(1/2-s)^2= 2s$. From $\square ADGF$ we have $GD^2= (2+s)^2-(2-s)^2= 8s$...

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Rudin Ex 2.2: Proving countability of algebraic numbers with the hint
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9 votes

I believe that this hint helps you to establish that $\mathbb Z[X]$ is countable. For a polynomial $f(X)=a_nX^n+\cdots +a_1X+a_0\in\mathbb Z[X]$, $a_n\neq 0$, define the number $h(f)$ by: $$h(f)= n+|...

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Find angle UFO in the picture attached
8 votes

Consider a regular 36-gon $A_1A_2\ldots A_{36}$ inscribed in a circle of radius $R$. Inscribed angle over any side is $5^\circ$. We can see our configuration as it is shown on the picture. It ...

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Galois group of $x^4 + 2$ over $\mathbb{Q}$
8 votes

Note that $\varepsilon_8^i\sqrt[4]2$, for $i=1,3,5,7$, are roots of this polynomial, where $\varepsilon_8=\frac{\sqrt 2}{2}+i\frac{\sqrt 2}{2}$ is the eighth primitive root of unity. Hence splitting ...

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Through two given points on a circle, construct two parallel chords with a given sum.
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7 votes

Consider the picture: Let $A$ and $B$ be the two points, and $AC$ and $BD$ be the desired chords; let $AC+BD=a$. Assume first that the situation is like on the picture, i.e. $C$ and $D$ are on the ...

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Left action of a group on permutation representation
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7 votes

Note that by the definition $g*\phi$ is in $F(X,\mathbb C)$ defined by $(g*\phi)(x):= \phi(g^{-1}x)$, so the calculation is the following: $$\begin{align} (h*(g*\phi))(x)&= (g*\phi)(h^{-1}x)\\ &...

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Does there exist irreducible polynomials of each degree over arbitrary field?
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7 votes

Let $\bar{\mathbb{Q}}$ be the algebraic closure of $\mathbb Q$. Consider a maximal subfield $F$ of $\bar{\mathbb{Q}}$ not containing $\sqrt{2}$; this is straight-forward by Zorn's lemma. Clearly, $F$ ...

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Groups of order $360$ have a subgroup of order $10$
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6 votes

If $n_5=6$ and $P$ is a Sylow 5-subgroup, you have $|N_G(P)|=60$. Set $H=N_G(P)$, and we find a subgroup of order $10$ in $H$. Your arguments work. In $H$, $n_5$ is either $1$ or $6$. If $n_5=1$, ...

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Show some $\mathbb{X} \subseteq \mathbb{N}^+$ occurs as the finite spectrum of a sentence for this language.
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6 votes

In a language of groups $\mathcal L=\{\cdot, ^{-1},e\}$ take $\phi$ to be the conjunction of: $\cdot$ is associative; $e$ is neutral for $\cdot$; $x^{-1}$ is inverse of $x$; $\forall x(x\cdot x\cdot ...

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Finding determinant of $n \times n$ matrix
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6 votes

If $c_i$ is $i$th column of your second determinant, do $c_n= c_n-c_{n-1}$, $c_{n-1}=c_{n-1}-c_{n-2}$, ..., $c_2=c_2-c_1$ to get: $$\left|\begin{array}{ccccccc} 1-x & x & 0 & 0 & \...

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Equilateral triangles on the sides of a triangle
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5 votes

Denote the points like on this picture: Consider the following composition of rotations: $I= R_{C',60^\circ}\circ R_{A',60^\circ}\circ R_{B',60^\circ}$. The classification of isometries says that $I$ ...

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Parametrizing the square spiral
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5 votes

I am not sure if this answers the question. We can note that the squares of even numbers are on the diagonal of the second quadrant, so if we set: $$\hat n=\max\{2k\mid (2k)^2\leqslant n\},$$ or in ...

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Show that a subgroup of order $N \in \mathbb{N}$ is isomorphic to $\mathbb{Z}/n\mathbb{Z} × \mathbb{Z}/nn'\mathbb{Z}$.
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5 votes

If $K$ is cyclic then $K\cong\mathbb Z/N\mathbb Z$ which you can write as $\mathbb Z/1\mathbb Z\times\mathbb Z/N\mathbb Z$. Assume that $K$ is not cyclic. By the fundamental theorem of finite abelian ...

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Sum of subsets is the whole set?
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4 votes

First note that if $B\subsetneq\mathbb Z_n$ is not contained in a proper subgroup and contains $0$, then $B\subsetneq B\oplus B$: $B\subseteq B\oplus B$ follows since $0\in B$, and $B\neq B\oplus B$ ...

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Definability of subgraphs of random graphs
4 votes

If $N$ is a random graph (countable) then it is a fact that for any finite partition of $N$ at least one of the members of the partition is a random graph (with induced graph structure). Since there ...

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Prove -1 Transformation (A binomial identity):$\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$
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4 votes

Note that for real $\alpha$ and positive integer $q$, $\alpha\choose q$ is defined by: $${\alpha\choose q}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-(q-1))}{q!}.$$ So: $${-p\choose q}= \frac{(-p)(-...

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The number of divisors of $0$ in a finite ring
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4 votes

If $e\neq 0$ is a left zero divisor in $A$, then the set of zero divisors contains the left ideal $Ae$. Since this ideal is non-trivial (it contains $e$) and it is not equal to $A$ (it doesn't contain ...

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Do injective homomorphisms preserve generating sets?
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4 votes

No. You can take inclusion map $\{1,-1\}\longrightarrow \{1,-1,i,-i\}$, considered as subgroups of $\mathbb C^\times$. $S=\{i\}$ is a generating set with empty preimage.

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Solving perpendicularity problem not using scalar product of vector
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3 votes

Let $E$ be the midpoint of $AC$. Since $EN$ is a middle line of $\triangle CAD$, we have $EN\parallel AD$ and $EN=AD/2$. Similarly, $EM\parallel CB$ and $EM=CB/2$. In particular, $EN:EM=AD:CB$. Since $...

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Is there a way to put 5 points on the surface of the sphere so that they are indistinguishable?
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3 votes

We will prove that this is not possible. Toward a contradiction, assume that $A,B,C,D,E$ are points on the sphere satisfying the condition, and let $O$ be the center of the sphere. Any isometry which ...

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Subsets of $\mathbb Z/n\mathbb Z$ disjoint with some of its shifts
3 votes

Denote by $X-X=\{x-y\mid x,y\in X\}$. Then: $X$ is disjoint from some of its shift iff $X-X\subsetneq\mathbb Z/n\mathbb Z$. Moreover: $X\cap (a+X)=\emptyset$ iff $a\notin X-X$. For $(\Rightarrow)$...

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Squares on the sides of a regular pentagon
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3 votes

Note that, by the symmetry of the picture, $\angle OQP= \angle OPQ= \angle OPX_4$. Consider $\triangle OQP$ and the circumcircle. We have that the inscribed angle over $OP$, $\angle OQP$, is equal to ...

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Find angle without trigonometry
3 votes

Maybe you can use the following. Consider a regular 9-gon $I_1I_2\ldots I_9$ with center $O$, and find your picture by setting $A=O$, $B=I_3$ and $C=I_5$. Then note that $M$ is the intersection of ...

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For $H$ the orthocenter of $\triangle ABC$, prove $|AH|h_a+|BH|h_b+|CH|h_c=\frac12\left(a^2+b^2+c^2\right)$
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3 votes

If $A',B',C'$ are feet of the altitudes, from $ABA'\sim AHC'$ we have $AH\cdot h_a= AB\cdot AC'$. Similarly, from $BAB'\sim BHC'$, $BH\cdot h_b= AB\cdot BC'$ follows. By adding we have $AH\cdot h_a+...

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Let $G$ be non-abelian simple and $p$ largest prime divisor of $|G|$. If $H<G$, then $|G:H|\geq p$.
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3 votes

As you noted, if $|G:H|=k<p$, by simplicity we can embed $G$ into $\mathbb S_k$. But $G$ has a Sylow $p$-subgroup of order $p^m$ for some $m\geqslant 1$, and $\mathbb S_k$ doesn't have subgroups of ...

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All real functions such that $(x+y)(f(x) - f(y)) = (x - y)f(x + y)$
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3 votes

Let $f$ satisfy the equation. By setting $(x,y)$ to be $(t,1)$, $(t+1,1)$ and $(t,2)$ we get: \begin{align} (t-1)f(t+1)&=(t+1)(f(t)-f(1))\\ tf(t+2)&=(t+2)(f(t+1)-f(1))\\ (t-2)f(t+2)&=(t+2)(...

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A maximal regular ideal of commutative ring is prime?
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3 votes

It seems that $F\times\mathbb Z_4$ is a counterexample for any field $F$. Namely, $I=F\times\{0\}$ is a maximal regular ideal generated by the idempotent $(1,0)$, but $(0,2)^2=(0,0)\in I$ and $(0,2)\...

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The minimal size of generating set of quotient group
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3 votes

Assume that $\{g_1,\ldots,g_n\}$ is a generating set for $G$ of the minimal size. If $Q$ is a quotient group $G/H$ for $H\triangleleft G$, note that $\{g_1H,\ldots,g_nH\}$ is a generating set for $Q$. ...

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Galois extension of $\mathbb{Q}$ with Galois group $\mathbb{Z}/4\mathbb{Z}$ that contains $\mathbb{Q}(\sqrt{3})$?
3 votes

We will prove that such extension doesn't exist. Lemma 1. Equation $X^2=3Y^2+3Z^2$ doesn't have integer solution $(a,b,c)$ such that $a\neq 0$. (In fact, $(0,0,0)$ is the only solution.) Proof. If $(...

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Inverse of a unipotent matrix
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3 votes

Write: $$0= (A-E)^k= \sum_{i=0}^k{k\choose i}(-1)^iA^{k-i}= \sum_{i=0}^{k-1}{k\choose i}(-1)^iA^{k-i}+ (-1)^kE=$$ $$= A\sum_{i=0}^{k-1}{k\choose i}(-1)^iA^{k-i-1}+ (-1)^kE.$$ It follows that $A^{-1}= (...

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