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21

$$ab+cd=ab(c^2+d^2)+cd(a^2+b^2)=(ad+bc)(ac+bd)=0$$


9

I assume $a,b \in \mathbb{R}$. Since $a^2+b^2 = 1$, we have $-1 \leq a \leq 1$ and likewise $-1 \leq b \leq 1$. Let us take $a = \cos(\alpha)$ and $b = \sin(\alpha)$ without loss of generality. Similarly, $c = \cos(\beta)$ and $d = \sin(\beta)$. We have $ac + bd = \sin(\alpha) \sin(\beta) + \cos(\alpha) \cos(\beta) = \cos(\alpha - \beta) = 0$. You have ...


6

$$\sum_{n=2}^{+\infty}\frac{1}{j^n}=\frac{1}{j(j-1)},\tag{1}$$ hence the problem boils down to computing: $$\begin{eqnarray*} \sum_{j\geq 2}\frac{1}{j(j-1)j!}&=&\sum_{j\geq 2}\frac{1}{(j-1)j!}-\sum_{j\geq 2}\frac{1}{j\cdot j!}\\&=&\frac{1}{2}+\sum_{j\geq 2}\left(\frac{1}{j\cdot(j+1)!}-\frac{1}{j\cdot ...


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 $\overline{BF}$, ...


5

You can interptet $(a, b)$ and $(c, d)$ as two orthogonal vectors that lie on the unit circle. Converting this into polar coordinates, this means there are angles $\phi$, $\theta$ such that $(a, b) = (\cos(\phi), \sin(\phi))$, $(c, d) = (\cos(\theta), \sin(\theta))$ and $|\phi - \theta| = \frac{\pi}{2}$. Now observe that $$ab + cd = \cos(\phi)\sin(\phi) + ...


4

Hint : Every natural number can be written in the form $2^{m-1}(2n-1)$ with unique positive integers $m,n$.


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 ...


4

Hint: If $a$ lines are parallel in one direction, the other $100-a$ lines which are in different directions create $a(100-a)$ intersections.


3

I think there is a typo as I read that book and own one and some typos were spotted by me as well. It should be: $a^2+\sqrt{a}+\sqrt{a} \geq 3\sqrt[3]{a^2\sqrt{a}\sqrt{a}} = 3a$


3

We have $\lim\limits_{z\to\infty} |f(z)|=\infty$ and the function $|f|$ is continous, so $|f|$ has a minimum at some point $z_0$ in the set $|z|\ge1$. Our goal is to prove that $z_0$ is located on the unit circle; in other words, $z_0$ cannot be outside the circle. Let the two roots be $u$ and $v$. They are in the closed unit disk: ...


3

Let $D$ denote the random variable telling how many rolls Dave took to roll a $6$, and let $L$ denote the variable telling how many rolls Linda took. Then what you calculated were $P(D=k)$ for certain values of $k$. For example, the probability that Dave took $5$ rolls to roll a $6$ is $$\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac16$$ What you ...


3

Let $D, L$ be the obvious random variables. Then the desired probability is \begin{align} P & = P(D = 1, L = 1) + P(D = 1, L = 2) + P(D = 2, L = 1) \\ & + P(D = 2, L = 2) + P(D = 2, L = 3) + P(D = 3, L = 2) \\ & + P(D = 3, L = 3) + P(D = 3, L = 4) + P(D = 4, L = 3) \\ & + \cdots \end{align} Each row constitutes an element in a ...


2

If $D$ and $L$ denote the number of rolls needed by Dave and Linda and $E$ denotes the event then:$$P(E)=P(D=L)+P(D=L-1)+P(D=L+1)$$ This with $P(D=L-1)=P(D=L+1)$ on base of symmetry, and with: $$P(D=L)=\sum_{k=1}^{\infty}P(D=k\wedge L=k)=\sum_{k=1}^{\infty}P(D=k)P(L=k)$$For $P(D=L-1)$ you can also find such an expression. edit: ...


2

A classical problem in simultaneous approximation. Let $a_n=(\lambda n\pmod{\pi},\mu n\pmod{\pi})\in\mathbb{T}$ and let we fix some huge $N$. Let $Q_n\subset\mathbb{T}$ be a square centered in $a_n$ having side length $\frac{\pi}{\sqrt{N}}$. Among $Q_1,\ldots,Q_N$, at least two squares have to overlap, since the sum of their Lebesgue measures equals ...


2

Here is perhaps a quick way. By the AM-GM-HM inequality, it must always be the case that $H_n < H_{n+1} < G_{n+1} < A_{n+1} < A_n$, which resolves the sequences of $H_n$ and $A_n$. It is also easy to see that $A_{n+1} H_{n+1} = A_n H_n$, so we also readily get $G_{n+1} = G_n$.


2

Consider the student with the most friends. If he's friend with at least 12 people, your problem is solved. If he is not, there is another student who's not his friend. Let's name those two students A and B. A having strictly less than 12 friends, there is at least 12 people different from B he's not friend with. Any of these people are not friend with A, ...


2

As others have pointed out, you need to start out by partitioning the lines into groups of parallel ones. Assume that the sizes of the groups are $n_1,n_2,\ldots, n_k$. In other words, there is a group of $n_1$ lines parallel to each other but non-parallel to the rest, another group of $n_2$ lines parallel to each other but non-parallel to the rest et ...


2

Now that the question has finally been quoted properly, it can be answered. Let $x$ and $y$ denote the total amount of milk and coffee, respectively, in ounces. Then we have the linear system of equations $$ \frac17x+\frac2{17}y=8\;,\\ x+y=8n\;, $$ where $n$ is the number of family members. Solving for $x$ and $y$ yields $$ ...


2

Say you have a column $C_i$ with negative sum, and you change the sign of all elements in that column. This must increase the total sum of all the elements in the grid (hereby just called the "total sum"), since all the other columns are unaffected, and the sum of all the column sums is the same as the total sum. The same thing goes for rows, of course. ...


2

f(999)=f(37)*f(3)*f(3)*f(3) Now, f(3) or f(37) cannot be 1. Because, if, for example, f(3)=1, then f(999)=f(37) which implies 999=37 (because the function is one-to-one). Therefore, f(3) can have lowest value 2. And hence f(37) can have lowest value 3. Thus, f(999)=3*2*2*2=24.


2

You have to look for a number $k$ such that between $k^3$ and $(k+1)^3$ there are exactly $70$ multiples of $k$. Since $k^3$ and $(k+1)^3-1$ are multiples of $k$ this gives $$[(k+1)^3-1]-k^3=69k$$


1

It is a consequence of Chebychev's inequality: $$ \sum_{cyc}\sqrt{\frac{a+b}{c}}≥2\sum_{cyc}\sqrt{\frac{c}{a+b}}\iff\sum_{cyc}\frac{a+b-2c}{\sqrt{c(a+b)}}≥0 $$ Since the $a+b-2c$ and $\frac{1}{\sqrt{c(a+b)}}$ are ordered in the same way, we can apply Chebychev's inequality to obtain: $$ ...


1

Small variation of the answer of loup blanc that does not mention minimal polynomials, and produces an actual annihilating polynmial of degree $r+1$. Note that is is irrelevant that the vector space is over$~\Bbb C$. The image space $W$ of $M$ (or column space if you prefer) has dimension $r$ by definition of the rank. The characteristic polynomial $Q$ of ...


1

Let $2$ distinct positive no. be $a$ and $b\;,$ Here $A_{1}\;,G_{1}$ and $H_{1}$ are $\bf{A.M}\;,\bf{G.M}$ and $\bf{H.M}$ of these two positive no., So $$\displaystyle A_{1} = \frac{a+b}{2}\;\;,G_{1} = \sqrt{ab}\;\;,H_{1} = \frac{2ab}{a+b}$$. And given $A_{n}\;,G_{n}$ and $H_{n}$ are $\bf{A.M}\;,\bf{G.M}$ and $\bf{H.M}$ of no.,s $A_{n-1}$ and $H_{n-1}.$ ...


1

Let $[ABC]$ be the area of $\triangle{ABC}$. Let $[ADF]=x,[AEF]=y$. Then, from $$AD:DB=[ADF]:[DFB]=[ADC]:[CDB],$$ one has $$x:9=x+y+15:9+12\tag1$$ Also, from $$BF:FE=[BFA]:[FEA]=[BFC]:[CFE],$$ one has $$x+9:y=12:15\tag2$$ Solving $(1)(2)$ gives $$x=315,y=405.$$ Hence, $$[ADFE]=x+y=315+405=\color{red}{720}.$$


1

HINT : We have $$\frac 1n\lt\frac mk\lt n\iff \frac mn\lt k\lt mn\tag 1$$ Setting $\lfloor\frac mn\rfloor=s$ gives $$(1)\iff s+1\le k\le mn-1$$with $s\le\frac mn\lt s+1$. Hence, one has $50=(mn-1)-(s+1)+1$.


1

Let the vectors be $u_k = (x_k,y_k)$ for $1\leqslant k \leqslant 2n$. Consider first the case where $x_k \geqslant 0$ for all $k$. Using $\lvert y_k\rvert \geqslant 1 - \lvert x_k\rvert$ for unit vectors, we have $$\sum_{k = 1}^{2n} y_k \geqslant \sum_{k = 1}^{2n} (1 - x_k) = 2n - \sum_{k = 1}^{2n} x_k,$$ so $\sum\limits_{k = 1}^{2n} y_k$ is a ...



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