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The $ \cap$ symbol represents an intersection, or the items in common between two lists. The only numbers in common between the three sets given are the positive even numbers $ 20$ and below, which are $ 2$, $ 4$, $ 6$, $ 8$, $ 10$, $ 12$, $ 14$, $ 16$, $ 18$, and $ 20$. Adding them up, $ 2\plus{}4\plus{}6\plus{}8\plus{}10\plus{}12\plus{}14\plus{}16\plus{}18\plus{}20\equal{}2(1\plus{}2\plus{}3\plus{}4\plus{}5\plus{}6\plus{}7\plus{}8\plus{}9\plus{}10)\equal{}2(\frac{10(10\plus{}1)}{2})\equal{}10(11)\equal{}\fbox{110}$.

Given that $ A$ is the set of positive integers, $ B$ is the set of whole numbers less than or equal to $ 20$, and $ C$ is the set of even integers, what is the sum of the numbers in $ A \cap B \cap C$?

[hide="Solution"] A quick and easy way is trying with a sample quadratic equation like: $ x^2\minus{}3x\plus{}2\equal{}0$ with one root double the other. Substituting $ a,b$ and $ c$ into the possible choices options $ \mathbf{(C),(D)}$ and $ \mathbf{(E)}$ don't satisfy the requirements. Then using another sample equation: $ x^2\minus{}6x\plus{}8\equal{}0$ using the same method option $ \mathbf{(A)}$ doesn't satisfy the requirements so the answer is $ \boxed{\mathbf{(B)}}$ [/hide]

For one root of $ ax^2 \plus{} bx \plus{} c \equal{} 0$ to be double the other, the coefficients $ a,\,b,\,c$ must be related as follows: $ \textbf{(A)}\ 4b^2 \equal{} 9c\qquad \textbf{(B)}\ 2b^2 \equal{} 9ac\qquad \textbf{(C)}\ 2b^2 \equal{} 9a\qquad \\ \textbf{(D)}\ b^2 \minus{} 8ac \equal{} 0\qquad \textbf{(E)}\ 9b^2 \equal{} 2ac$

[hide="Alternatively"] Let the roots be $ r$ and $ s$, with $ r=2s$. By Vieta's, we have $ r+s=3s=-\frac{b}{a}$ and $ rs=2s^2=\frac{c}{a}$. So, $ s=-\frac{b}{3a}$. Substituting, $ 2(\frac{b^2}{9a^2})=\frac{c}{a}$, so $ 2b^2=9ac$. $ \boxed{\textbf{B}}$ [/hide]

For one root of $ ax^2 \plus{} bx \plus{} c \equal{} 0$ to be double the other, the coefficients $ a,\,b,\,c$ must be related as follows: $ \textbf{(A)}\ 4b^2 \equal{} 9c\qquad \textbf{(B)}\ 2b^2 \equal{} 9ac\qquad \textbf{(C)}\ 2b^2 \equal{} 9a\qquad \\ \textbf{(D)}\ b^2 \minus{} 8ac \equal{} 0\qquad \textbf{(E)}\ 9b^2 \equal{} 2ac$

Find the max value of $ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}} \plus{} \sqrt [3]{4x \minus{} 3 \minus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}}$ when x € (-1,1). $ \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}\equal{}\sqrt{(1\minus{}x)(4\minus{}x)^{2}}\equal{}[\minus{}1<x<1 ]\equal{}(4\minus{}x)\sqrt{1\minus{}x}$ $ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}}\equal{}\sqrt[3]{4x\minus{}3\plus{}(4\minus{}x)\sqrt{1\minus{}x}}\equal{}\sqrt{1\minus{}x}\plus{}1$ ---> $ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}} \plus{} \sqrt [3]{4x \minus{} 3 \minus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}}\equal{}\sqrt{1\minus{}x}\plus{}1\plus{}1\minus{}\sqrt{1\minus{}x}\equal{}2$

Find the max value of $ \sqrt[3]{4x\minus{}3\plus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}} \plus{} \sqrt[3]{4x\minus{}3\minus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}}$ when x € (-1,1).

If I understood it correctly, you proved it is [b]always[/b] 2, right? $ \sqrt [3]{4x \minus{} 3 \plus{} \sqrt {16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}}} \equal{} \sqrt [3]{4x \minus{} 3 \plus{} (4 \minus{} x)\sqrt {1 \minus{} x}} \equal{} \sqrt {1 \minus{} x} \plus{} 1$ This step was a nice observation :lol:

Find the max value of $ \sqrt[3]{4x\minus{}3\plus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}} \plus{} \sqrt[3]{4x\minus{}3\minus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}}$ when x € (-1,1).

We set $ A \equal{} 4x \minus{} 3$ and $ B \equal{} 16 \minus{} 24x \plus{} 9x^{2} \minus{} x^{3}$ Observing that $ \sqrt [3]{A \pm \sqrt {B}} \equal{} \sqrt [3]{(A \pm \sqrt {B})\cdot1\cdot1}$ then we obtain from AM-GM : $ \sqrt [3]{A \plus{} \sqrt {B}} \plus{} \sqrt [3]{A \minus{} \sqrt {B}} \leq \frac {(A \plus{} \sqrt {B} \plus{} 2) \plus{} (A \minus{} \sqrt {B} \plus{} 2)}{3} \equal{} \frac {2A \plus{} 4}{3} \equal{} \frac {8x \minus{} 2}{3}$ But, $ \frac { \minus{} 10}{3}\leq\frac {8x \minus{} 2}{3}\leq 2$ when x € [-1;1], indeed by transitivity we obtain : $ \sqrt [3]{A \plus{} \sqrt {B}} \plus{} \sqrt [3]{A \minus{} \sqrt {B}} \leq 2$, which is the max, reached when $ x \equal{} 1$.

Find the max value of $ \sqrt[3]{4x\minus{}3\plus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}} \plus{} \sqrt[3]{4x\minus{}3\minus{}\sqrt{16\minus{}24x\plus{}9x^{2}\minus{}x^{3}}}$ when x € (-1,1).

Let $x,y,z$ be nennegative real numbers The following is also nice \[(x+y+z)^{5}\geq 8(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2})?\] I think it is very nice since, $(x+y+z)^{5}-8(x^{2}+y^{2}+z^{2})(x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y)=$ $=\sum_{cyc}x(y+z-x)^{4}+32\sum_{cyc}x^{2}y^{2}z.$ :)

Let $x,y,z$ be nennegative real numbers, find the greatest possible positive value of $k$ such that \[(x+y+z)^{5}\geq k(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x).\] The following is also nice \[(x+y+z)^{5}\geq 8(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2})?\]

Nice solutions, Arqady. For the first one, if we put $y=1$ and $z=0$, it follows by calculus that $\frac{(x+y+z)^{5}}{ (x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x)}$ attains its minimum $k\approx 11.353$ for $x\approx 3.26953$, the only root of $x^{3}-4x^{2}+3x-2=0.$ It only remains to show (but that should not be too hard) that in an extremal case, one of $x,y,z$ must be $0$. :roll:

Let $x,y,z$ be nennegative real numbers, find the greatest possible positive value of $k$ such that \[(x+y+z)^{5}\geq k(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x).\] The following is also nice \[(x+y+z)^{5}\geq 8(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2})?\]

Let $ a,b,c\geq 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$.Prove that: $ \frac {1}{6 \minus{} ab} \plus{} \frac {1}{6 \minus{} bc} \plus{} \frac {1}{6 \minus{} ca}\leq \frac {3}{5}$. $ <\equal{}>16(ab\plus{}bc\plus{}ca) \plus{}a^2b^2c^2 \le 36\plus{}13abc<\equal{}>$ $ a^2(b\plus{}c)\plus{}b^2(c\plus{}a) \plus{}c^2(a\plus{}b)\plus{}\frac{3}{4}a^2b^2c^2 \le a^3\plus{}b^3\plus{}c^3\plus{}\frac{15}{4}abc$( which is true by Schur and $ abc \le 1$); :)

Let $ a,b,c\geq 0$, s.t. $ a\plus{}b\plus{}c\equal{}3$.Prove that: $ \frac{1}{6\minus{}ab}\plus{}\frac{1}{6\minus{}bc}\plus{}\frac{1}{6\minus{}ca}\leq \frac{3}{5}$.

Does there exist a function f such that for all x the following equality holds $ f(f(x)) \equal{} \left\{\begin{array}{l} \sqrt {2009} ,x\in Q \\ 2009,x\in\mathbb{R} \end{array}\right.$ :?: I suppose you mean $ f(x)\equal{}2009$ $ \forall x\in\mathbb R\backslash \mathbb Q$ (and not $ \mathbb R$). If so, no such function exists : Let $ g(x)$ be the function defined thru $ f(\mathbb Q)\equal{}\{\sqrt{2009}\}$ and $ f(\mathbb R\backslash \mathbb Q)\equal{}\{2009\}$ We have $ f(f(x))\equal{}g(x)$ and so $ f(g(x))\equal{}g(f(x))$ and so $ f(2009)\in\{\sqrt{2009},2009\}$ and $ f(\sqrt{2009})\in\{\sqrt{2009},2009\}$ Obviously $ f(x)\neq x$ $ \forall x$ (since $ g(x)\neq x$ $ \forall x$) and so $ f(2009)\in\{\sqrt{2009},2009\}$ and $ f(\sqrt{2009})\in\{\sqrt{2009},2009\}$ implies : $ f(2009)\equal{}\sqrt{2009}$ and $ f(\sqrt{2009})\equal{}2009$ but then $ f(f(2009))\equal{}2009\ne g(2009)\equal{}\sqrt{2009}$ So no such function.

Does there exist a function $f$ such that for all $x$ the following equality holds? \[ f(f(x)) = \begin{cases} \sqrt{2009}, & \mbox{if } x\in \mathbb Q,\\ 2009, &\mbox{if } x\in\mathbb{R} \setminus \mathbb Q. \end{cases} \]

if $ m,n\neq 1$ $ n + m|mn + 1$ $ \iff n + m|n^2 - 1$ $ \iff \exists d\in E = \{k\in\mathbb{N}: \ k|n^2 - 1,k > n\}: \ m = d - n$ $ \iff \exists d\in E = \{k\in\mathbb{N}: \ k|n^2 - 1\}: \ m = max\{d,\frac {n^2 - 1}{d}\} - n$ then : $ {\{(n,m)\in\mathbb{N}^*^2: \ \frac {mn + 1}{m + n}\in\mathbb{Z}^ + \} = \{(n,max\{d,\frac {n^2 - 1}{d}\} - n): d|n^2 - 1,n\neq 1\}\cup\{(n,1),(1,n): n\in\mathbb{N}^*\}}$ for exemple $ d = 1,n\neq 1$ gives $ (n,m) = (n,max\{1,n^2 - 1\} - n) = (n,n^2 - n - 1)$ and $ \frac {(n) + (n^2 - n - 1)}{(n) + (n^2 - n - 1)} = n - 1\in\mathbb{Z}^ +$

It' s an exercise

It' s an exercise Let $ S\equal{}m\plus{}n$. We want $ mn\plus{}1\equal{}n(S\minus{}n)\plus{}1\equal{}0\pmod{S}$ and so $ n^2\equal{}1\pmod{S}$ So, the set of solutions is $ \{(n,d\minus{}n),(d\minus{}n,n)$ where $ d$ is any divisor $ >n$ of $ n^2\minus{}1\}$

It' s an exercise

Let's go... Here is some explanations and a reference: The question amounts wondering which is the distance between the polynomial function $x \mapsto x^4$ and vector space of the polynomial function of degree $3$ for the "norme" infinite on the interval $[-1,3]$ i.e $\|g\|= \sup_{[-1,3]} |g(x)|$ where $g$ is a continuous function on $[-1,3]$ and for which polynomial this distance is reached. It is the Tchebychev's polynomial of degree $4$, brought back to the segment $[-1, 3]$ and standardized so that the coefficient of $x^4$ is equal to $1$. The solution is: $f(x) = x^4 - 4x^3 + 2x^2 + 4x -1$. The extremum correspondent of $f$ is $2$. [i]Ref.: J.P Demailly, Analyse numérique et équation différentielles. (collection pug)[/i] [in french :) ] I hope that it is comprehensible :blush:. If a French wants to help me to correct my English... :(

Let $ f(x)= x^4 + ax^3 + bx^2 + cx +d $ Find $ a,b,c,d $ such that : $ \max_{x\in [-1,3]}/f(x)/ $ get the minimum (/f(x)/ is the absoluted value of f(x))

[hide="Solution"]Observe that solutions to $ x^2 \plus{} 7x \plus{} 6 \equal{} x \iff x^2 \plus{} 6x \plus{} 6 \equal{} 0$ are also solutions to the given equation*. We now wish to rewrite the equation in the form $ (x^2 \plus{} 6x \plus{} 6)(x^2 \plus{} ax \plus{} b) \equal{} 0$. Equating the constant terms of the original equation and the factored form, we get $ 6^2 \plus{} 7 \cdot 6 \plus{} 6 \equal{} 6b \implies b \equal{} 14$ Similarly, equating $ x^3$ terms of both gives $ 2 \cdot 7 \equal{} 6 \plus{} a \implies a \equal{} 8$. Thus, the original equation can be rewritten as $ (x^2 \plus{} 6x \plus{} 6)(x^2 \plus{} 8x \plus{} 14) \equal{} 0$ We can now solve this using the quadratic formula: $ \boxed{x \equal{} \minus{} 3 \pm \sqrt {3} , \; \; x \equal{} \minus{} 4 \pm \sqrt {2}}$ [i]* The left-hand form of this quadratic is a root equation for the given equation[/i][/hide]

Find all solutions of: $ (x^2 \plus{} 7x \plus{} 6)^2 \plus{} 7(x^2 \plus{} 7x \plus{} 6)\plus{} 6 \equal{} x$

Since $y=2\sin \theta\cos\theta$, it follows that $(x/2)^{2}+(y/x)^{2}=1$. I think the curve looks like this: $\infty$ Very keen guess: [url=http://allyoucanupload.webshots.com/v/2005945213650995389][img]http://aycu34.webshots.com/image/6753/2005945213650995389_rs.jpg[/img][/url]

If $x = 2 \cos \theta$, $y = \sin 2\theta$, how would you eliminate the parameter $\theta$? I am trying to find the points on the curve where the tangent is horizontal and vertical. I figure that the cartesian equation is an ellipse. But I tried this: $x = 2\cos \theta$, $y = 2\sin \theta \cos \theta$, but I cant use the identity $\sin^{2}\theta+\cos^{2}\theta = 1$. I took the derivative and got: $\frac{-\cos 2\theta}{\sin \theta}$ but I dont see how this would help if I am trying to find the specific points. Any help is appreciated. Thanks

Note that since $y = 2\sin \theta \cos \theta$ and $\frac{x}{2}= \cos \theta$, $y = (2\sin \theta )\frac{x}{2}= x \sin \theta$. Therefore, $\sin \theta = \frac{y}{x}$. As a result, $\sin^{2}\theta+\cos^{2}\theta = (\frac{y}{x})^{2}+(\frac{x}{2})^{2}= 1$. That's where mlok's equation came from.

If $x = 2 \cos \theta$, $y = \sin 2\theta$, how would you eliminate the parameter $\theta$? I am trying to find the points on the curve where the tangent is horizontal and vertical. I figure that the cartesian equation is an ellipse. But I tried this: $x = 2\cos \theta$, $y = 2\sin \theta \cos \theta$, but I cant use the identity $\sin^{2}\theta+\cos^{2}\theta = 1$. I took the derivative and got: $\frac{-\cos 2\theta}{\sin \theta}$ but I dont see how this would help if I am trying to find the specific points. Any help is appreciated. Thanks

[hide="Solution"] Let us assume the initial length and breadth to be $100l$ and $100b$ respectively. Initially, the perimeter will be $\\ P_{1} = 2(100l + 100b) \newline \Rightarrow P_{1} = 200(l + b) ...........(1)$ Now since length and breadth is increased by 10%, we get the new perimeter to be $\\ P_{2} = 2(110l + 110b) \newline \Rightarrow P_{2} = 220(l + b) ...........(2)$ Here 110 because, 10% of 100 is 10, adding that to 100 yields 110. Now simply divide 2 by 1 and multiply it by 100 to see how much more it is than the initial stage. $ \\ \frac{P_{1}}{P_{2}} \times 100 = \frac{220 (l+b)}{200(l+b)} \times 100 \newline \newline \Rightarrow \frac{P_{1}}{P_{2}} \times 100 = \boxed {110 \%} $ which is 10% more than the initial perimeter. [/hide]

If the length and width of a rectangle are each increased by $ 10 \%$, then the perimeter of the rectangle is increased by \[ \textbf{(A)}\ 1 \% \qquad \textbf{(B)}\ 10 \% \qquad \textbf{(C)}\ 20 \% \qquad \textbf{(D)}\ 21 \% \qquad \textbf{(E)}\ 40 \% \]

suppose $n$ is the number of students then there are $\frac n 2$ staplers ,$\frac n 3$rulers ,$\frac n 4$glue bottles so $\frac n 2 +\frac n 3 +\frac n 4 =65$ thus we have $\frac{13n}{12}=65$ so $n=60$. I think you might find the middle school section on this site more suited to your needs. :)

In an art class, there were just enough staplers , rulers and glue bottles so that every 2 students had to share a stapler, every 3 students had to share a ruler, and every 4 students had to share a glue bottle. If the sum of the number of staplers, rulers, and glue bottles used by the class was 65, how many students were in the class?

Hello, $ 12=2^{2}3$. Therefore we get as possible solutions: $2^{2}3=12$ $2^{2}*3*5=60$ $2^{2}3^{2}=36$ and $2^{2}3^{2}*5=180$. Thus we have $4$ positive integer factors, which are multiplies of $12$. Best wishes from Leipzig/Germany.

How many positive integer factors of $ 2^2 \times 3^2 \times 5$ are multiples of 12?

[hide="15"]Is that all you got wrong? Let the roots be $ r$ and $ s$, then by Vietas we have $ n \equal{} 4rs$ and $ p \equal{} ( \minus{} r \minus{} s)$, so $ \frac {n}{p} \equal{} \frac {4rs}{ \minus{} r \minus{} s}$, but we also know again by Viets and the fact that m is in both equations that $ 2r \plus{} 2s \equal{} \minus{} rs$, which simplifies the above fraction to $ 8$.[/hide] [hide="21"]First note that we don't have to worry about denominators because when we take $ q/p$ they will cancel. So for our purposes we have $ p \equal{} 10$, because there are 10 ways to choose the number, and $ q \equal{}$ $ 10\choose{2}$$ {4\choose{2}}^2\equal{} 1620$, so the answer is A[/hide]

I need help with the question and please show ur solution the AMC 10B questions 21(http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_21) and question 15 (http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_15)

ok for 15 there are 5 pairs that make 20 dollars : $ (20,(1,5,10,20))$and $ (10,10)$. There are also 4 types of bills and you are choosing 2 of them, which can be done in (4C2) + 4 = 10 ways, so the answer is 1/2. For 22, we simplify to $ \frac{2(n\minus{}1)!}{n\plus{}1}$ which is not reducible iff n + 1 is an odd prime. There are 8 odd primes less than 24 so the answer is $ 24 \minus{}8 \equal{} 16$

I need help with the question and please show ur solution the AMC 10B questions 21(http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_21) and question 15 (http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_15)

For 22, we simplify to $ \frac {2(n \minus{} 1)!}{n \plus{} 1}$ which is not reducible iff n + 1 is an odd prime. There are 8 odd primes less than 24 so the answer is $ 24 \minus{} 8 \equal{} 16$ how did u know that odd primes would not work

I need help with the question and please show ur solution the AMC 10B questions 21(http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_21) and question 15 (http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_15)

For 22, we simplify to $ \frac {2(n \minus{} 1)!}{n \plus{} 1}$ which is not reducible iff n + 1 is an odd prime. There are 8 odd primes less than 24 so the answer is $ 24 \minus{} 8 \equal{} 16$ how did u know that odd primes would not work If $ n \plus{} 1$ is prime, it will not be a divisor of $ (n \minus{} 1)!$. We have to exclude the even prime, $ 2$, because there is a 2 in the numerator, so $ 2$ will always divide evenly. This leaves us with the odd primes.

I need help with the question and please show ur solution the AMC 10B questions 21(http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_21) and question 15 (http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems#Problem_15)

the greater measure is 20 inches, so to be reduced to 5 inches, you divide by 4 the other measure is 16 inches, so we divide that by 4 also, to get 4 now we have our side lengths (4 and 5), so we find the perimeter $ 4\plus{}4\plus{}5\plus{}5\equal{}18$ our answer is 18

A photograph measuring 16 inches by 20 inches is reduced uniformly so that the greater measure becomes 5 inches. What is the number of inches in the perimeter of the reduce photo?

A less random sounding solution: By "uniformly" they probably mean proportionally, so you're dealing with two rectangles which are the same shape but scaled down. Call the big rectangle R and the little rectangle r. The rule of proportionality would tell you: $ \frac{long side of r}{long side of R}\equal{}\frac{short side of r}{short side of R}$. Numerically: $ \frac{5}{20}\equal{}\frac{x}{16}$ $ \frac{1}{4}\equal{}\frac{x}{16}$ $ 4\equal{}x$ (Then the perimeter would be as above, 18)

A photograph measuring 16 inches by 20 inches is reduced uniformly so that the greater measure becomes 5 inches. What is the number of inches in the perimeter of the reduce photo?

if 3/8 must have a cat and we know there are 9 people who have a cat, 9+x+w+z=3/8(32), so 9+x+w+z=12 -> x+w+z=3, but the 3 numbers must be integer positive, so the solution must be x=w=z=1.... very simply!!!!! Not so fast, damianoT! ;) There are 9 people who have [i]only[/i] a cat. Since 12 people have cats, this means that 3 have cats in conjunction with some other pet. The variables $w$,$x$,$y$, and $z$ have to be non-negative integers, but they do not necessarily have to be positive integers. It would be perfectly legimate for one of the variables to have a value of zero. For example, if $w=0$ then that simply means that none of the students own a cat and some other pet that is not a dog (i.e. a cat and a fish, a cat and a bird, etc.) This was a good problem! [hide="Here's how I solved it:"] We have $32 \times \frac{1}{2}=16$ students with dogs, $32 \times \frac{3}{8}=12$ students with cats, $6$ with something else, and $5$ with no pets at all. This means that $32-5=27$ students have at least one pet. We use the Venn diagram to set up the following equations: Total pet equation: $9+10+2+w+x+y+z=27$ Dog equation: $10+x+y+z=16$ Cat equation: $9+w+x+z=12$ Other equation: $2+w+y+z=6$ When we isolate the variables on the left, these equations become: Total: $w+x+y+z=6$ Dog: $x+y+z=6$ Cat: $w+x+z=3$ Other: $w+y+z=4$ Now, we can solve for $w$, $y$, and $x$ by subtracting the Dog, Cat, and Other equations, respectively, from the Total Equation: Total - Dog: $(w+x+y+z) -(x+y+z)=6-6$, which gives us $w=0$. Total - Cat: $(w+x+y+z)-(w+x+z)=6-3$, and so $y=3$. Total - Other: $(w+x+y+z)-(w+y+z)=6-4$, meaning that $x=2$. Substitute these values into any of the four equations above and you'll see that $z=1$. Since $z$ is the variable at the heart of the Venn diagram, it represents the number of students who own a dog, a cat, and some other kind of pet as well. As we now see, there is only $\boxed{1}$ student in that category.[/hide] [hide="Here is the completed Venn diagram for the problem:"] [asy]unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("10",A); label("2",B); label("9",C); label("1",(0,0)); label("0",(B+C)/2); label("3",(A+B)/2); label("2",(A+C)/2); label("Cats",2.4C,C); label("Other Pets",2.4B,B); label("Dogs", 2.4A,A);[/asy][/hide] Tim

Jeremy made a Venn diagram showing the number of students in his class who own types of pets. There are 32 students in his class. In addition to the information in the Venn diagram, Jeremy knows half of the students have a dog, $ \frac{3}{8}$ have a cat, six have some other pet and five have no pet at all. How many students have a cat, a dog and some other pet? [asy]unitsize(50); import graph; pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2); draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2)); label("10",A); label("2",B); label("9",C); label("$z$",(0,0)); label("$w$",(B+C)/2); label("$y$",(A+B)/2); label("$x$",(A+C)/2); label("Cats",2.4C,C); label("Other Pets",2.4B,B); label("Dogs", 2.4A,A);[/asy]

Let $\ f$ be a function such that $\ f(x)=\arctan\left(\frac{1}{x}\right)$.Compute $\int_{0}^{1}f(x)dx$ Let $J: =\int_{0}^{1}\arctan\left(\frac{1}{x}\right) dx$ integrate by parts: let $u=\arctan\left(\frac{1}{x}\right) , du=\frac{-\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}}dx =-\frac{dx}{1+x^{2}}, dv=dx, v=x$ $J=\left[ x\arctan\left(\frac{1}{x}\right)\right]_{x=0}^{1}+\int_{0}^{1}\frac{x}{1+x^{2}}dx = \left[ x\arctan\left(\frac{1}{x}\right)+\frac{1}{2}\log(1+x^{2})\right]_{x=0}^{1}$ $=\frac{1}{2}\left( \frac{\pi}{2}+\log(2)\right)$ -Ben

Let $\ f$ be a function such that $\ f(x)=\arctan\left(\frac{1}{x}\right)$.Compute $\int_{0}^{1}f(x)dx$

This is a very useless problem, IMHO: no use of matrices at all, just take $ B\equal{}af(A)\plus{}f(a)I_n$ for suitable $ a$ (to make $ B$ invertible) of modulus $ 1$. Here, $ f$ means complex conjugate. The converse is obvious. Bad problem.

Let $ A$ be a $ n\times n$ matrix with complex elements. Prove that $ A^{\minus{}1} \equal{} \overline{A}$ if and only if there exists an invertible matrix $ B$ with complex elements such that $ A\equal{} B^{\minus{}1} \cdot \overline{B}$.

Still you should see the results (very bad). And actually after thinking about it $ f(A)\equal{}\overline A$ I think and $ f(a)\equal{}\overline a$, and of course just add the condition for $ B$ invertible. Haven't worked out the last condition (B invertible) but I think this might work, because there are finite values of $ a$ for which $ B$ is not invertible, and we just take one that is good.

Let $ A$ be a $ n\times n$ matrix with complex elements. Prove that $ A^{\minus{}1} \equal{} \overline{A}$ if and only if there exists an invertible matrix $ B$ with complex elements such that $ A\equal{} B^{\minus{}1} \cdot \overline{B}$.

[hide="Solution"]Use $a$, $b$, and $c$. Clearly, $c=\sqrt{a^{2}+b^{2}}$. Set up the equation $ab=3a+3b+3\sqrt{a^{2}+b^{2}}$. Solve for the radical and we get $3\sqrt{a^{2}+b^{2}}=ab-3a-3b$. Square both sides to get $a^{2}b^{2}-6a^{2}b-6ab^{2}+9a^{2}+18ab+9b^{2}=9a^{2}+9b^{2}$, or $a^{2}b^{2}-6a^{2}b-6ab^{2}+18ab=0$. Factor to get $ab(ab-6a-6b+18)=0$. Clearly, $a$ and $b$ must be positive, so $ab\neq0$. Divide by it to get $ab-6a-6b+18=0$. Add 18 to both sides and factor to get $(a-6)(b-6)=18$. Since $a,b\in\mathbb{Z}$, $a-6$ and $b-6$ are integers as well. We have the following possibilities for $a-6$ and $b-6$: $\pm1\text{ and }\pm18$ $\pm2\text{ and }\pm9$ $\pm3\text{ and }\pm6$ No negative solution works, because if we have -6, -9, or -18, either $a$ or $b$ will not be a positive integer. Trying out the three that remain, we get $\boxed{7,24,25}$, $\boxed{8,15,17}$, and $\boxed{9,12,15}$. And guess what: THEY ALL WORK![/hide]

Source: Romania 1999 7.1 Determine the side lengths of a right trianlge if they are intgers and the product of the leg lengths is equal to three times the perimeter.

[hide="my way"] in other words we have: $2\triangle = 6s$ $\triangle = 3s$ $rs = 3s$ $r = 3$. Now draw the diagram, since $r = 3$, and tangent lines from the same point are congruent, we deduce that $a+b-6 = c = \sqrt{a^{2}+b^{2}}$. Squaring and using Simon's favorite trick we get $(a-6)(b-6) = 18$ and get the sides lengths: $(9, 12, 15)$ $(7, 24, 25)$ $(8, 15, 17)$ [/hide]

Source: Romania 1999 7.1 Determine the side lengths of a right trianlge if they are intgers and the product of the leg lengths is equal to three times the perimeter.

hello, after squaring both sides of the equation you will get: $ \sqrt {2(5 \minus{} 9\cos(x) \plus{} 4\cos^2(x))} \equal{} 6\cos(x) \minus{} 6 \plus{} 8\cos^2(x)$ with $ \cos(2x) \equal{} 1 \minus{} 2cos^2(x)$. Squaring one more time and setting $ t \equal{} \cos(x)$ gives: $ 32t^4 \plus{} 48t^3 \minus{} 34t^2 \minus{} 27t \plus{} 13 \equal{} 0$. Sonnhard.

Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$

Exactly $ \sqrt {2(5 \minus{} 9\cos(x) \plus{} 4\cos^2(x))} \equal{} 3\cos(x) \plus{}1\minus{} 4\cos^2(x)$ with $ \cos(2x) \equal{} 1 \minus{} 2cos^2(x)$. Squaring one more time and setting $ t \equal{} \cos(x)$ gives: $ (t\minus{}1)(t\plus{}1)(4t\minus{}3)^2 \equal{} 0$.

Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$

$ t \equal{} \cos{x} \equal{} \minus{} 1$ yields $ 2 \plus{} 3 \equal{} 1$ in the original equation, and is therefore a degenerate solution. [hide="Geometric Solution"]Let $ A,B,C,D$ be points in a plane such that $ AB \equal{} AC \equal{} 1$, $ AD \equal{} 2$, and $ AC$ bisects $ \angle BAD$. Let $ x \equal{} \angle BAC \equal{} \angle CAD$. Using the Law of Cosines: $ BC \equal{} \sqrt {2 \minus{} 2\cos{x}}$ $ CD \equal{} \sqrt {5 \minus{} 4\cos{x}}$ $ BD \equal{} \sqrt {5 \minus{} 4\cos{2x}}$ By the triangle inequality, $ BC \plus{} CD \ge BD$ with equality iff $ B,C,D$ are collinear and $ BD$ is the longest side. It is easy to see that $ B,C,D$ are collinear for $ x \equal{} 0^\circ, 180^\circ$, however, $ x \equal{} 180^\circ$ yields $ CD$ as the longest side, and is not a solution. The only other possibility for $ B,C,D$ to be collinear is if $ AC$ is a cevian in $ \Delta ABD$. Using the angle bisector formula: $ BD \equal{} \frac {3}{2}\sqrt {2}$. So $ BC \equal{} \frac {\sqrt {2}}{2}$, and $ CD \equal{} \sqrt {2}$. From the law of cosines, $ \cos{x} \equal{} \frac {3}{4}$. Thus, the solutions to the original equation are $ x \equal{} 0, \pm \arccos{\frac {3}{4}}$ and any equivalent angles. [/hide]

Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$

Squaring both sides and let $ t\equal{}\cos x$ and squaring again gives $ (t\minus{}1)(t\plus{}1)(4t\minus{}3)^2\equal{}0 \left(\minus{}\frac{1}{4}\leq t\leq 1\right)\Longleftrightarrow t\equal{}1,\ \frac{3}{4}$. The answer is $ x\equal{}2m\pi\ ,\ \cos ^ {\minus{}1} \frac{3}{4}\plus{}2n\pi$ for $ m,\ n\in\mathbb{Z}$.

Solve the equation $ \sqrt{2\minus{}2\cos x}\plus{}\sqrt{5\minus{}4\cos x}\equal{}\sqrt{5\minus{}4\cos 2x}.$

[hide="Answer"]Label the intersection of the diagonals $E$. $AE^2+BE^2=11$, $BE^2+CE^2=BC^2$, $AE^2+DE^2=1001$, so we have $11-BE^2=1001-DE^2\Rightarrow BE^2=DE^2-990$. Substituting, $CE^2+DE^2-990=BC^2=CD^2-990$. Drawing the perpendicular $AF$ to $DC$, we have $CF=\sqrt{11}$. $AF^2=BC^2=1001-(CD-\sqrt{11})^2=990-CD^2+2\sqrt{11}=CD^2-990$, so we have $CD^2-CD\sqrt{11}-990=0\Rightarrow CD=10\sqrt{11}$. Therefore, we have $BC^2=CD^2-990=\boxed{110}$.[/hide]

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

[hide]Let the perpendicular from $A$ to $CD$ intersect $BD$ at $F$. Let the two diagonals intersect at $E$, and let $\angle{EAF}=\theta$. Then \[ \angle{EAF} \cong \angle{EBA} \cong \angle{ECB} \cong \angle{EDC} = \theta \] Thus, with all four of these triangles having an angle of theta and a right angle, we have four similar triangles. Let $AE=a$, $BE=b$, $CE=c$ and $DE=d$. Note that \[ \frac{EF}{a}=\frac{a}{b}=\frac{b}{c}=\frac{c}{d} \] So these triangles are in a constant ratio (it can be observed that the ratio of one to the next in the counter clockwise direction is $\cot{\theta}$; let this ratio, also equal to $\frac{b}{a}=x$). So, \[ a^2+d^2=a^2+a^2x^6=a^2(1+x^6)=a^2(1+x^2)(1-x^2+x^4)=1001 \] \[ a^2+b^2=a^2+a^2x^2=a^2(1+x^2)=11 \] Dividing the top by the bottom, \[ 1-x^2+x^4=91 \implies (x^2)^2+(x^2)-90=0 \] Using the quadratic formula, \[ x^2=\frac{1 \pm \sqrt{1+360}}{2}=10, -9 \] Hence, $x^2=10$. And because $BC^2=AB^2x^2$, $BC^2=11(10)=\boxed{110}$[/hide]

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

[hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=699&sid=0410aaa1bd30b9a182e2c716eee02a44[/img] Let $BC=x$. This means that $AC=\sqrt{11+x^2}$. Now notice that $\triangle ABC \sim \triangle BCD$. Therefore, $\frac{BD}{BC}=\frac{AC}{AB} \implies BD=\frac{x\sqrt{x^2+11}}{\sqrt{11}}$. Another valuable piece of info we get from these similar triangles is that $%Error. "cosACB" is a bad command. =\cos ABD=\frac{x}{\sqrt{x^2+11}}$. Now, it seems like applying the law of cosines on $\angle ABD$ will come out nicely. $AD^2=AB^2+BD^2+2(AB)(BD)(\cos ABD)=11-\frac{x^2(x^2+11)}{11}-4=1001$. Simplifying further yields $(x^2+99)(x^2-110)=0$, so $BC=x=\boxed{110}$. [/hide]

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

[hide="Answer"]Label the intersection of the diagonals $ E$. $ AE^2 \plus{} BE^2 \equal{} 11$, $ BE^2 \plus{} CE^2 \equal{} BC^2$, $ AE^2 \plus{} DE^2 \equal{} 1001$, so we have $ 11 \minus{} BE^2 \equal{} 1001 \minus{} DE^2\Rightarrow BE^2 \equal{} DE^2 \minus{} 990$. Substituting, $ CE^2 \plus{} DE^2 \minus{} 990 \equal{} BC^2 \equal{} CD^2 \minus{} 990$. Drawing the perpendicular $ AF$ to $ DC$, we have $ CF \equal{} \sqrt {11}$. $ AF^2 \equal{} BC^2 \equal{} 1001 \minus{} (CD \minus{} \sqrt {11})^2 \equal{} 990 \minus{} CD^2 \plus{} 2\sqrt {11} \equal{} CD^2 \minus{} 990$, so we have $ CD^2 \minus{} CD\sqrt {11} \minus{} 990 \equal{} 0\Rightarrow CD \equal{} 10\sqrt {11}$. Therefore, we have $ BC^2 \equal{} CD^2 \minus{} 990 \equal{} \boxed{110}$.[/hide] Wait, where did you get the substitution of $ \minus{}990$ in the equation $ CE^2 \plus{} DE^2 \minus{} 990 \equal{} BC^2 \equal{} CD^2 \minus{} 990$?

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

[hide]Let $ E$ be the foot of the perpendicular from $ A$ to $ CD$. Note that since $ AC \perp BD$, $ \triangle ABC \sim \triangle BCD \implies \frac {AB}{BC} \equal{} \frac {BC}{CD} \implies BC^2 \equal{} CD \cdot \sqrt {11}$. Also, by the Pythagorean theorem, $ AE^2 \plus{} ED^2 \equal{} AD^2$, but $ AE \equal{} BC$, $ ED \equal{} CD \minus{} \sqrt {11}$, so this simplifies to $ CD^2 \minus{} \sqrt {11} \cdot CD \minus{} 990 \equal{} 0$. Solving ($ CD$ must be positive), $ CD \equal{} \frac {\sqrt {11} \plus{} \sqrt {11 \cdot 19^2}}{2}$, so $ BC^2 \equal{} CD \cdot \sqrt {11} \equal{} \frac {11 \plus{} 11 \cdot 19}{2} \equal{} 110$. [/hide] As for mr.k_74's question, JeseusFreak197 had earlier that $ BE^2 \plus{} CE^2 \equal{} BC^2$, so he substituted for $ BE^2$ in this equation.

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

[hide=quick] Set the trapezoid on the coordinate plane with $C$ as the origin and $CD$ lying on the $x$ axis. Furthemore denote $CD=e$ and $BC=a$ Then looking at slopes we have $(\frac{a}{e})(\frac{a}{\sqrt{11}})=1$. Thus $a^2=e\sqrt{11}$. Now doing distance formula on $AD$ gives $a^2+(e-\sqrt{11})^2=1001$. Now just substitute and you get a quadratic in $a^2$ which an be easily found to have positive solution $\boxed{110}$[/hide]

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

[hide=quick from this lemma] lemma: if $AB \perp BC$ and $CD\perp BC$, such that $AB=h_{1} , DC=h_{2}$ and $BC=d$ then $d^2=h_{1}h_{2}$ proof of lemma is easy from similarity, then we have from this lemma $x^2=y\sqrt{11}$ and from orthogonal diagonal theorem we have $x^2+1001=y^2+11$, solving for $x^2$ we get $x^2=\boxed{110}$[/hide]

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

[hide=quick from this lemma] lemma: if $AB \perp BC$ and $CD\perp BC$, such that $AB=h_{1} , DC=h_{2}$ and $BC=d$ then $d^2=h_{1}h_{2}$ proof of lemma is easy from similarity, then we have from this lemma $x^2=y\sqrt{11}$ and from orthogonal diagonal theorem we have $x^2+1001=y^2+11$, solving for $x^2$ we get $x^2=\boxed{110}$[/hide] Wow, this is the simplest solution. :o I literally had to use 3 similar triangles and 2 pythagorean theorems to bash out the answer :P

In trapezoid $ABCD,$ leg $\overline{BC}$ is perpendicular to bases $\overline{AB}$ and $\overline{CD},$ and diagonals $\overline{AC}$ and $\overline{BD}$ are perpendicular. Given that $AB=\sqrt{11}$ and $AD=\sqrt{1001},$ find $BC^2.$

This is probably stupid, but whenever I see a problem like this I'm strongly tempted to approximate it by hand. Intuition tells us that $e^\pi$ is probably bigger, so we need to do a low estimate. Call $e = 27/10$ and $\pi = 22/7$. That should still be a small estimate. So we have $3^{66/7} / 10^{22/7}$. That's about $729 * 27 * 1.6$ in the numerator and $1000 * 1.4$ in the denominator. That reduces to $729 * 27 / (5^3 * 7) \approx 104 * 27 / 125 \approx 112 / 5 = 22.4$. After the amount of time that took and the fact that I lost track of the magnitude of my error, I decided not to do the other one. But considering $e^\pi \approx 23.131$ and all of my calculations were mental, 22.4 isn't bad for an intentionally low estimate. :D

Which is bigger, $e^\pi$ or $\pi^e$? [hide="solution"] Let's consider the maximum of $x^{1/x}$. First we take a natural logarithm. This gives $\frac{\ln(x)}{x}$. The maximum of this function is the maximum of the first function since $\ln(x)$ is increasing. The maximum is easily found to be at $x=e$. So now we have: $e^{1/e}>\pi^{1/\pi}$ Raise both sides to the $e\pi$ power. [/hide]

Define : $x\ s.s\ y\Longleftrightarrow sgn(x)=sgn(y)$, i.e. $x=y=0$ or $xy>0\ .$ Here are some examples : $1.\blacktriangleright\frac ab\ s.s\ ab$ for $b\ne 0\ .$ $2.\blacktriangleright 0<a\ne 1\ ,\ \{x,y\}\subset R\Longrightarrow (a^x-a^y)\ s.s\ (a-1)(x-y)\ .$ $3.\blacktriangleright 0<a,b\ne 1\ ,\ x\in R\Longrightarrow (a^x-b^x)\ s.s\ x(a-b)\ .$ $4.\blacktriangleright 0<a\ne 1\ ,\ \{x,y\}\subset (0,\infty )\Longrightarrow \log_a x-\log_a y\ s.s\ (a-1)(x-y)\ .$ $5.\blacktriangleright 0<a\ne 1\ ,\ b>0\Longrightarrow \log_a b\ s.s\ (a-1)(b-1)\ .$ $6.\blacktriangleright (|x|-|y|)\ s.s\ (x^2-y^2)\ ;\ |x|\ s.s\ x^2\ .$ $7.\blacktriangleright (\arcsin x-\arcsin y)\ s.s\ (x-y)\ ;\ \arcsin x\ s.s\ x\ .$ $8.\blacktriangleright (\arccos x-\arccos y)\ s.s\ (y-x)\ ;\ \arccos x\ s.s\ (1-x)\ .$ $9.\blacktriangleright (\arctan x-\arctan y)\ s.s\ (x-y)\ ;\ \arctan x\ s.s\ x\ .$ $10.\blacktriangleright \{x,y\}\subset [0,\infty )\Longrightarrow (\sqrt [n] x-\sqrt [n] y)\ s.s\ (x-y)\ .$ $F.1\blacktriangleright f: D\rightarrow R\ ,\ f\nearrow\Longrightarrow f(x)-f(y)\ s.s\ (x-y)$ for any $\{x,y\}\subset D\ .$ $F.2\blacktriangleright f: D\rightarrow R\ ,\ f\searrow\Longrightarrow f(x)-f(y)\ s.s\ (y-x)$ for any $\{x,y\}\subset D\ .$ [b]Proof of the proposed problem.[/b] $(e^{\pi}-\pi^e)\ s.s\ (\ln e^{\pi}-\ln \pi^e)=(\pi\ln e-e\ln \pi)\ s.s\ (\frac{\ln e}{e}-\frac{\ln\pi}{\pi})=$ $f(e)-f(\pi )$, where $f: (0,\infty )\rightarrow R\ ,\ f(x)=\frac{\ln x}{x}\ .$ But $f'(x)\ s.s\ (1-\ln x)=(\ln e-\ln x)\ s.s\ (e-x)\ .$ Therefore, for any $0<x\ne e\ ,$ $f(x)< f(e)\Longrightarrow f(\pi)<f(e)\Longrightarrow\boxed {e^{\pi}>\pi ^e}\ .$ [b][u]A similar exercise :[/u][/b] $\{a,b\}\subset (0,e^2]\Longrightarrow (a^{\sqrt b}-b^{\sqrt a})\ s.s\ (a-b)\ .$ For example, $3^{\sqrt 5}<5^{\sqrt 3}\ .$ $\{a,b\}\subset [e^2,\infty )\Longrightarrow(a^{\sqrt b}-b^{\sqrt a})\ s.s\ (b-a)\ .$ For example, $3^{\sqrt 5}<5^{\sqrt 3}\ ;\ 512>81^{\sqrt 2}\ .$ [b]Remark.[/b] For any $0<x\ne e^2\ ,\ x^{\sqrt e}<e^{\sqrt x}\ .$

Which is bigger, $e^\pi$ or $\pi^e$? [hide="solution"] Let's consider the maximum of $x^{1/x}$. First we take a natural logarithm. This gives $\frac{\ln(x)}{x}$. The maximum of this function is the maximum of the first function since $\ln(x)$ is increasing. The maximum is easily found to be at $x=e$. So now we have: $e^{1/e}>\pi^{1/\pi}$ Raise both sides to the $e\pi$ power. [/hide]

Your integral can be calculated exactly. Let consider the case when $n=2p$. Then I obtained, assuming that I made no computational mistakes, that: $\lim\int_{0}^{\infty}\frac{(\sin(x))^{2p}}{(x)^{2p}}dx=\lim_{p\rightarrow \infty}\int_{0}^{1}\frac{2p}{(1+4y^{2})(1+42^{2}y^{2})\cdots (1+4p^{2}y^{2})}dy$. Why this limit is 0?. Any ideas?

Find $\int_{0}^{\infty}\frac{(\sin{x})^{n}}{x^{n}}dx,n\in \mathbb{N}$ (this integral is motivated by another problem mentioned here ). Also evaluate; $a_{n}: =\int_{0}^{\infty}\frac{(\sin(x) )^{n}}{x^{n}}dx,$ $\lim_{n\rightarrow \infty}a_{n}=?$

Why this limit is 0?. Any ideas? For $n\ge 2,\ \left|\frac{\sin^{n}x}{x^{n}}\right|\le \frac{\sin^{2}x}{x^{2}},$ and the latter is an absolutely integrable function on $[0,\infty).$ For $x\ne0,\lim_{n\to\infty}\frac{\sin^{n}x}{x^{n}}=0,$ pointwise. Hence by the Lebesgue Dominated Convergence Theorem, we can interchange integral and limit to conclude that $\lim_{n\to\infty}\int_{0}^{\infty}\frac{\sin^{n}x}{x^{n}}\,dx=0.$

Find $\int_{0}^{\infty}\frac{(\sin{x})^{n}}{x^{n}}dx,n\in \mathbb{N}$ (this integral is motivated by another problem mentioned here ). Also evaluate; $a_{n}: =\int_{0}^{\infty}\frac{(\sin(x) )^{n}}{x^{n}}dx,$ $\lim_{n\rightarrow \infty}a_{n}=?$

wlog assume $ x\geq y$ than we have $ x^2 < x^2 \plus{} 3y < x^2 \plus{} 4x \plus{} 4 \equal{} (x \plus{} 2)^2$ so $ x^2 \plus{} 3y \equal{} (x \plus{} 1)^2$ and further $ 3y \equal{} 2x \plus{} 1$ by solving this linear diophan equation we get $ x \equal{} 1 \plus{} 2k$ and $ y \equal{} 1 \plus{} 3k$ for a non-negative integer $ k$ $ y^2 \plus{} 3x \equal{} (1 \plus{} 2k)^2 \plus{} 3(1 \plus{} 3k) \equal{} 4k^2 \plus{} 13k \plus{} 4$ but $ (2k \plus{} 3)^2 < 4k^2 \plus{} 13k \plus{} 4 < (2k \plus{} 4)^2$ for $ k > 5$ so we just need to manually check $ k$ up to five. it turns out to be that $ k \equal{} 5$ and $ k\equal{}0$ are the only solutions. for which we get $ x \equal{} 16$ and $ y \equal{} 11$ giving $ x^2 \plus{} 3y \equal{} 17^2$ and $ y^2 \plus{} 3x \equal{} 13^2$ in the first and $ y\equal{}x\equal{}1$ and $ x^2\plus{}3y\equal{}y^2\plus{}3x\equal{}4$ in the second case

Find all pairs of positive integers $ (x,y)$ such that both $ x^2 \plus{} 3y$ and $ y^2 \plus{} 3x$ are perfect squares.

[hide] To have a right triangle in a circle, you have to have one of the sides a diameter. If $X$ is odd, then this can't happen... If $X$ is even ($X\ge4$), then there are $\frac{X(X-2)}{2}$ ways to choose a right triangle (there are $\frac{X}{2}$ ways to choose a diameter, and $X-2$ ways to choose another point of the circle). Then the probability would be $\frac{X(X-2)}{2\binom{X}{3}}$ (I think). [/hide]

if X points are equally spaced on a circle, find the probability that 3 points chosen at random will form a right triangle.[/code]

If $ R$ is a finite Boolean ring with identity $ 1 \neq 0$ then $ R \cong \mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}$. To see this, you can use the Chinese Remainder Theorem and notice that for $ e \in R$ with $ e^2 \equal{} e$ one has $ R \cong Re \times R(1 \minus{} e)$ where $ e$ and $ 1 \minus{} e$ are the identities for the subrings $ R$ and $ R(1 \minus{} e)$ respectively. On the other hand, if $ R$ is a set of $ n \equal{} 2^k$ elements, you can consider the unique field of $ 2^k$ elements and look at it as an $ \mathbb{F}_2$-vectorspace. Now it clear that we have a Boolean ring, so we're done.

A ring is called boolean if $ x^2\equal{}x$ for all $ x$ in $ A$. Prove that one can define a structure of boolean ring of order $ n$ if and only if $ n\equal{}2^k$ for some $ k \in N$.

what you want to prove is $\displaystyle\alpha(G)\le n-\frac{\sum_i d_i}{n}$ where $\alpha(G)$ is the independence number of $G$ and $d_i$'s are the degrees of the vertices of $G$.(the rest follows from the fact that $\chi(G)\ge \frac{n}{\alpha(G)}$) or more simply what you need is that $\frac{\sum_i d_i}{n}\le n-\alpha$. if $I$ is a maximum independent set of size $\alpha$ then note that for each vertex of $I$ the degree is atmost $n-\alpha$;so $\sum_i d_i\le (n-\alpha)(n-1)+\alpha(n-\alpha)$ so that $\frac{\sum_i d_i}{n}\le (n-\alpha)(\frac{n-\alpha+1}{n})$ and since $\frac{n-\alpha+1}{n}\le 1$ we are through.

prove that in graph with $v$ virtex and $\epsilon$ edge : $\frac {v^2}{v^2-2\epsilon} \leq \chi$

$ F(x) \equal{} 0$ has local maximum $ \Longrightarrow F'(x) \equal{} 0$ has 3 distinct real roots $ \alpha ,\ \beta ,\ \gamma\ (\alpha <\beta <\gamma)$. Under this condition, We can rewrite the given inequality as follows. $ \frac{3}{10}\cdot\frac{16}{81}(\alpha^{2}\plus{}\beta^{2}\plus{}\gamma^{2}\plus{}\alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha )^{2}$ $ <\frac{1}{3}(2\gamma\minus{}\alpha\minus{}\beta)(\beta\minus{}\alpha )^{3}< 3\cdot\frac{16}{81}(\alpha^{2}\plus{}\beta^{2}\plus{}\gamma^{2}\plus{}\alpha\beta\plus{}\beta\gamma\plus{}\gamma\alpha )^{2}$

$ F(x)$ is polynomial with real coefficients. $ F(x) \equal{} x^{4}\plus{}a_{1}x^{3}\plus{}a_{2}x^{2}\plus{}a_{1}x^{1}\plus{}a_{0}$. $ M$ is local maximum and $ m$ is minimum. Prove that $ \frac{3}{10}(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}< M\minus{}m < 3(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}$

$ F(x)$ is polynomial with real coefficients. $ F(x) \equal{} x^{4} \plus{} a_{1}x^{3} \plus{} a_{2}x^{2} \plus{} a_{1}x^{1} \plus{} a_{0}$. $ M$ is local maximum and $ m$ is minimum. Prove that $ \frac {3}{10}(\frac {a_{1}^{2}}{4} \minus{} \frac {2a_{2}}{3^{2}})^{2} < M \minus{} m < 3(\frac {a_{1}^{2}}{4} \minus{} \frac {2a_{2}}{3^{2}})^{2}$ Ups... There are too many squares here. :blush: $ \frac {3}{10}(\frac {a_{1}^{2}}{4} \minus{} \frac {2a_{2}}{3})^{2} < M \minus{} m < 3(\frac {a_{1}^{2}}{4} \minus{} \frac {2a_{2}}{3})^{2}$

$ F(x)$ is polynomial with real coefficients. $ F(x) \equal{} x^{4}\plus{}a_{1}x^{3}\plus{}a_{2}x^{2}\plus{}a_{1}x^{1}\plus{}a_{0}$. $ M$ is local maximum and $ m$ is minimum. Prove that $ \frac{3}{10}(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}< M\minus{}m < 3(\frac{a_{1}^{2}}{4}\minus{}\frac{2a_{2}}{3^{2}})^{2}$

Decomposing into columns isn't a very informative way of looking at row operations. Row operations correspond to multiplication on the left by elementary matrices. If you show those latter three statements, you will have proved this in the case that $ A$ is an elementary matrix of one of those three types. The next step is to find an algorithm to decompose $ A$ as a product $ E_1E_2\cdots E_m$ of elementary matrices; then $ \det(A)\equal{}\det(E_1)\det(E_2)\cdots \det(E_m)$ and $ \det(AB)\equal{}\det(E_1)\det(E_2)\cdots \det(E_m)\det(B)$.

det(AB)=det(A)det(B) Maybe there is intuitive proving of this using the following - $ AB \equal{} \left[Ab_1\cdots Ab_n\right]$ following is actually just def of determinant -linear property $ \begin{vmatrix} {r_1}\\ {.}\\ {.}\\ {.} \end{vmatrix} \equal{} \begin{vmatrix} {k \cdot r'_1 \plus{} r''_1}\\ {.}\\ {.}\\ {.} \end{vmatrix} \equal{} k \begin{vmatrix} { r'_1 }\\ {.}\\ {.}\\ {.} \end{vmatrix} \plus{} \begin{vmatrix} { r''_1 }\\ {.}\\ {.}\\ {.} \end{vmatrix}$ - Swapping 2 rows switches the sign of the determinant so Adding a scalar multiple of a row to another doesn't change the determinant - If a single row is multiplied by a scalar r, then the determinant of the resulting matrix is r times the determinant of the original matrix. ... Let me know some proving using $ AB \equal{} \left[Ab_1\cdots Ab_n\right]$

Umm... I think you can't construct a triangle with sides $ 2$, $ 1$, $ 1$. ($ b \plus{} c > a$ isn't true). Oh, no. In my definition a triangle is definited by the condition $ (a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) \ge 0$. ;)

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

I guess it depends on what you call a triangle - it the case of triangle sides being $ 2$, $ 1$, $ 1$, you would only have a line, but I understand this is a degenerative case and as such, may be considered a triangle. Anyway, even if you consider $ (2,1,1)$ a triangle, the statement is still not true - take any triple $ (2t,t,t)$ and all of its permutations and the equality will obviously still hold since the inequality is homogeneous. I think there is something misunderstanding between me and you, sorry for my poor English. Actually, as you said, the equality holds when $ a\equal{}2t,b\equal{}c\equal{}t$ but I just want to write it as $ a\equal{}2,b\equal{}c\equal{}1$. I think it is nothing at all since the main important thing is my inequality. By the way, my inequality is true indeed. I am sure of it. ;)

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

Using the substitutions $ a\equal{}y\plus{}z$, $ b\equal{}z\plus{}x$, $ c\equal{}x\plus{}y$ and $ x\plus{}y\plus{}z\equal{}p$, $ x^2\plus{}y^2\plus{}z^2\equal{}1$, the inequality becomes $ \frac 1{1\plus{}2px\minus{}x^2}\plus{}\frac 1{1\plus{}2py\minus{}y^2}\plus{}\frac 1{1\plus{}2pz\minus{}z^2} \le \frac 9{3p^2\minus{}1}$. We apply EV-Theorem to the function $ f(u)\equal{}\frac 1{u^2\minus{}2pu\minus{}1}$ for $ 0\le u<p$. Since $ g(x)\equal{}f'(x)\equal{}\frac{\minus{}2(x\minus{}p)}{(x^2\minus{}2px\minus{}1)^2}$ is convex, because $ f"(x)\equal{}\frac{24(p\minus{}x)[(x\minus{}1)^2\plus{}p^2\plus{}1]}{(x^2\minus{}2px\minus{}1)^4}>0$, it suffices to consider the cases $ x\equal{}0$ and $ 0<x\le y\equal{}z$. These cases are equivalent to $ a\equal{}b\plus{}c$ and $ b\equal{}c$, respectively.

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

Solution is here: http://www.toanthpt.net/forums/showthread.php?p=66534#post66534 I think it' s not WRONG :D I have studied in detail this solution and I think it is wrong. Now I will explain my reasoning. WLOG we can assume $ a\plus{}b\plus{}c\equal{}3$ We put $ r\equal{}abc$ and $ q\equal{}ab\plus{}bc\plus{}ca$ Our inequality is equivalent to $ 2\sum ab \sum a^4 \plus{}6\sum ab \sum a^2b^2 \leq 18a^2b^2c^2 \plus{}9\sum a^2b^2(a^2\plus{}b^2)$ which becomes after substitution $ f(r)\equal{} 2q((9\minus{}2q)^2\plus{}q^2\minus{}6r) \minus{}9((q^2\minus{}6r)(9\minus{}2q)\minus{}r^2) \leq 0$ Now we use this fact $ 6r \leq 5q\minus{}9$ that is $ \sum (b\plus{}c\minus{}a)(a\minus{}b)(a\minus{}c) \geq 0$ It is true if $ a,b,c$ are sidelenghts of a triangle, by Valenticu-Shur inequality. Putting $ x\equal{}b\plus{}c\minus{}a$ , $ y\equal{}c\plus{}a\minus{}b$ and $ z\equal{}a\plus{}b\minus{}c$ it is sufficient to show that for $ a \geq b \geq c$ $ cz \geq by$ and this is true being equivalent at $ (b\minus{}c)(b\plus{}c) \geq a(b\minus{}c)$ Deriving $ f(r)$ we find that $ f'(r)\equal{} 120(3\minus{}q) \plus{}126 \plus{}8r > 0$ Now we have $ f(r) \leq f((5q\minus{}9)/6) \equal{} (q\minus{}3)^2(112\minus{}315)$ which is not least of zero as $ q leq 3$!!!! So I think this solution is wrong. In the solution at Vietnamese forum I found a mistake in calculation as $ f(r)$ is not equal at $ f(r)\equal{} 2q((9\minus{}2q)^2\plus{}2q^2\minus{}12r) \minus{}9((q^2\minus{}6r)(9\minus{}2q)\minus{}r^2) \leq 0$ in fact this expression is not equal to zero for $ q\equal{}3$ :wink: Have you a simple solution can_hang2007?

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

Solution is here: http://www.toanthpt.net/forums/showthread.php?p=66534#post66534 I think it' s not WRONG :D I have studied in detail this solution and I think it is wrong. Now I will explain my reasoning. WLOG we can assume $ a \plus{} b \plus{} c \equal{} 3$ We put $ r \equal{} abc$ and $ q \equal{} ab \plus{} bc \plus{} ca$ Our inequality is equivalent to $ 2\sum ab \sum a^4 \plus{} 6\sum ab \sum a^2b^2 \leq 18a^2b^2c^2 \plus{} 9\sum a^2b^2(a^2 \plus{} b^2)$ which becomes after substitution $ f(r) \equal{} 2q((9 \minus{} 2q)^2 \plus{} q^2 \minus{} 6r) \minus{} 9((q^2 \minus{} 6r)(9 \minus{} 2q) \minus{} r^2) \leq 0$ Now we use this fact $ 6r \leq 5q \minus{} 9$ that is $ \sum (b \plus{} c \minus{} a)(a \minus{} b)(a \minus{} c) \geq 0$ It is true if $ a,b,c$ are sidelenghts of a triangle, by Valenticu-Shur inequality. Putting $ x \equal{} b \plus{} c \minus{} a$ , $ y \equal{} c \plus{} a \minus{} b$ and $ z \equal{} a \plus{} b \minus{} c$ it is sufficient to show that for $ a \geq b \geq c$ $ cz \geq by$ and this is true being equivalent at $ (b \minus{} c)(b \plus{} c) \geq a(b \minus{} c)$ Deriving $ f(r)$ we find that $ f'(r) \equal{} 120(3 \minus{} q) \plus{} 126 \plus{} 8r > 0$ Now we have $ f(r) \leq f((5q \minus{} 9)/6) \equal{} (q \minus{} 3)^2(112 \minus{} 315)$ which is not least of zero as $ q leq 3$!!!! So I think this solution is wrong. In the solution at Vietnamese forum I found a mistake in calculation as $ f(r)$ is not equal at $ f(r) \equal{} 2q((9 \minus{} 2q)^2 \plus{} 2q^2 \minus{} 12r) \minus{} 9((q^2 \minus{} 6r)(9 \minus{} 2q) \minus{} r^2) \leq 0$ in fact this expression is not equal to zero for $ q \equal{} 3$ :wink: Have you a simple solution can_hang2007? Hello, my friend. :) My solution for it used the old way-SOS. Here is it :)

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

Very nice inequality and very nive proof, can_hang2007! :lol: Even, we can prove that $ S_b \plus{} S_c\geq0.$ :wink: $ S_b \plus{} S_c \equal{} \frac {1}{a^2 \plus{} c^2} \plus{} \frac {2(b^2 \minus{} ac)}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \plus{} \frac {1}{a^2 \plus{} b^2} \plus{} \frac {2(c^2 \minus{} ab)}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \equal{}$ $ \equal{} \frac {1}{a^2 \plus{} c^2} \plus{} \frac {2(c^2 \minus{} ab)}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {1}{a^2 \plus{} b^2} \plus{} \frac {2(b^2 \minus{} ac)}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$ $ \equal{} \frac {3c^2 \minus{} 2ab \plus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2ac \plus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)}\geq$ $ \geq\frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$ $ \equal{} (b \minus{} c)\left(\frac {3b \minus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \minus{} b}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\right) \equal{}$ $ \equal{} \frac {(b \minus{} c)^2(4a^2 \plus{} (b \minus{} c)^2)}{(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\geq0.$

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

$ \frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$ $ \equal{} (b \minus{} c)\left(\frac {3b \minus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \minus{} b}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\right) \equal{}$ I think it is wrong. Can you please explain it to me, arqady. Thank you very much.

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

$ \frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$ $ \equal{} (b \minus{} c)\left(\frac {3b \minus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \minus{} b}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\right) \equal{}$ I think it is wrong. Can you please explain it to me, arqady. Thank you very much. Yes, you are right, manlio! It can be $ \frac {3c^2 \minus{} 2bc \minus{} b^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)} \plus{} \frac {3b^2 \minus{} 2bc \minus{} c^2}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \equal{}$ $ \equal{} (b \minus{} c)\left(\frac {3b \plus{} c}{(a^2 \plus{} b^2)(b^2 \plus{} c^2)} \minus{} \frac {3c \plus{} b}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\right)$ and it's nothing. :( Thank you!

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :) only three lines with our trick!

Let $ a,b,c$ be the side lengths of a triangle. Prove that \[ (ab\plus{}bc\plus{}ca)\left( \frac{1}{b^2\plus{}c^2}\plus{}\frac{1}{c^2\plus{}a^2}\plus{}\frac{1}{a^2\plus{}b^2}\right) \le \frac{9}{2}.\] Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}2,b\equal{}c\equal{}1$ and its permutations. :)

In a different topic, someone asked why the following is true: A_1 is disjoint from at least 50 subsets, so it has at most $\frac{n}{51}$ elements. In fact, the reason is simple: Since the vertex $A_{1}$ of the graph $G$ has degree at most 50, there are at most 50 subsets $A_{i}$ (with $i\neq 1$) which are not disjoint from $A_{1}$. Thus, the remaining at least 50 subsets $A_{i}$ (with $i\neq 1$) are disjoint from $A_{1}$. So the set $A_{1}$ is disjoint from at least 50 of the subsets $A_{i}$. Thus, $A_{1}$ is disjoint from the union of these 50 subsets. But, according to the problem, the union of 50 subsets $A_{i}$ has more than $\frac{50}{51}n$ elements. Hence, since the set $A_{1}$ is disjoint from this union, it has less than $n-\frac{50}{51}n=\frac{1}{51}n$ elements. By the way: nice problem, Harazi! Darij

Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements. Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element.

My idea is slightly different. Define the graph $ G$ like Harazi's solution. Suppose that $ G$ is triangle-free. I will prove that all vertices in $ G$ are of degree at most $ 50$. If this is false, then there is a vertex $ v$ adjacent to at least $ 51$ other vertices: $ v_{1},v_{2},\dots,v_{51}$. Consider the sets represented by $ v_{2},v_{3},\dots,v_{51}$. The union of these sets has more than $ 50n/51$ elements. Hence, by the Pigeonhole Principle, one of them, say $ v_{51}$ has more than $ n/51$ elements. Consider $ v_{1},v_{2},\dots,v_{50}$. The union of them has more than $ 50n/51$ elements. Therefore, some vertex in $ v_{1},v_{2},\dots,v_{50}$ must be adjacent to $ v_{51}$; WLOG let it be $ v_{1}$. Thus, $ vv_{1}v_{51}$ is then a triangle, contradicting our assumption. The rest is actually the same. In fact, there are at least (not necessarily disjoint) $ 18$ triangles, if I'm not mistaken. Can you please check my argument? Suppose we have only at most $ 17$ triangles: $ a_{1}b_{1}c_{1}$, $ a_{2}b_{2}c_{2}$, ..., $ a_{k}b_{k}c_{k}$, where $ k\leq 17$. Hence, there are at least $ 101\minus{}3\cdot17 \equal{} 50$ vertices not in these triangles. Let's call these vertices "free vertices." Similar to the previous part, a free vertex $ v$ can have at most degree $ 50$ (otherwise, there must be another triangle). Therefore, at least $ 50$ vertices are not connected to $ v$ and thus, $ |v| < n/51$. This is true for all free vertices. Hence, the union of their corresponding sets has less than $ 50n/51$ elements, a contradiction.

Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements. Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element.

The bounds are incredibly weak. This question was used at a UK selection camp, and I produced the following solution, which disregards $ A_{101}$ entirely. Let $ S_1 \equal{} A_1\cup A_2 \cup ... \cup A_{25}$ and $ S_2,S_3,S_4$ three more unions of 25 $ A_i$ such that all 100 are covered. Let $ T_{i,j} \equal{} X \backslash (S_i \cup S_j)$. The question tells us that $ |T_{i,j}| < \frac {|X|}{51}$, so the union of all $ T_{i,j}$ has size less than $ \frac {6|X|}{51}$, so there are lots of elements of $ X$ belonging to no $ T_{i,j}$. But each of these must be in at least three $ S_i$, and therefore be common to at least three $ A_i$, so there exist three $ A_i$ containing such a point, and thus with nonempty pairwise intersections. The weakness of the bounds actually leads to a similar argument involving only 64 $ A_i$. Joseph Myers then managed to show (using Freddie Manners', another UK student's, method) that with the strong condition of needing a common point to 3 different $ A_i$, that 59 $ A_i$ is the maximum number of $ A_i$ required for such a point to exist, and that this bound is sharp (there's a counterexample for 58).

Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements. Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element.

Hi all. Sorry for digging this very old post back up, but I think there's one method that hasn't been covered by the above discussion, which is pretty simple (I think) but powerful enough to give the sharp bound mentioned by Ilthigore 2 posts up (3 years ago?!). I'm guessing that this is probably the method that Freddie Manners used as well. Consider an incidence matrix with columns labelled $A_1, A_2, \ldots, A_{101}$ and rows labelled $x_1, x_2, \ldots, x_n$, where the $x_j$ are the elements of set $X$. Mark a '1' in column $A_i$ and row $x_j$ if $x_j \in A_i$, and '0' otherwise. Assume on the contrary that there does not exist three distinct $A_i$ with all three pairwise intersections nonempty. Now we count the number of 50-tuples in the rows of the table which contain all '0's. So for example a row with 50 '0's count as 1 50-tuple, and a row with 51 '0's count as 51 50-tuples. On one hand, since each row contains at most two '1's, each row contains at least 99 '0's. Hence the number of 50-tuples is at least $\binom{99}{50}n$. On the other hand, for each choice of 50 $A_i$, their union contains at least $\frac{50}{51}n$ elements of $X$. Hence if we consider the columns corresponding to these 50 $A_i$, there are at most $\frac{1}{51}n$ rows whose intersection with the 50 columns consists of all '0's. Thus by summing over all choices of 50 $A_i$, the number of row '0' 50-tuples is less than $\frac{1}{51}\binom{101}{50}n$. Comparing these two bounds, we get \[\binom{99}{50} > \frac{1}{51}\binom{101}{50} \iff \frac{1}{51}\frac{101 \cdot 100}{(101 - 50) \cdot (100 - 50)} > 1\]. The last inequality is obviously false. In fact, it is still false if we replaced all occurences of 101 by 59. Hence our initial assumption was false and we are done.

Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements. Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element.

@hello123: Please be more precise. I think you are confusing sets and elements, but I am not sure because your posting is too brief. @angyansheng: A little correction: On the other hand, for each choice of 50 $A_i$, their union contains at least more than $\frac{50}{51}n$ elements of $X$. Hence if we consider the columns corresponding to these 50 $A_i$, there are at most less than $\frac{1}{51}n$ rows whose intersection with the 50 columns consists of all '0's. Thus by summing over all choices of 50 $A_i$, the number of row '0' 50-tuples is less than $\frac{1}{51}\binom{101}{50}n$. Comparing these two bounds, we get \[\binom{99}{50} > \frac{1}{51}\binom{101}{50} \iff \frac{1}{51}\frac{101 \cdot 100}{(101 - 50) \cdot (100 - 50)} > 1\]. The first $>$ should be an $<$.

Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements. Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element.

Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements. Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element. [b][color=#f00]Claim:[/color][/b] Atleast $50$ sets have greater than $\frac{n}{51}$ elements. [b][color=#f00]Proof:[/color][/b] Trival by PHP. Obviously two of them have an intersection or else $51=101$, contradiction so some two them $A_X=B$ and $A_Y=C$ have an non empty intersection. So FTSOC assume that for all $A_i$ we must have the intersection of $B$ and $C$ empty then there are $101-2=99$ possibilities for $A_i$ and obviously atleast 50 of them (by PHP) have an empty intersection with $B$ and by by the claim each have $\frac{n}{51}$ elements a total of $n$ elements when we sum up or implying all the other sets are empty, contradiction.

Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements. Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element.

It is known that $X$ is a set with $102$ elements. Suppose $A_1, A_2, ..., A_{101}$ is a set of subset of $X$ such that the sum of every $50$ of them has more than $100$ elements. Prove that there are $1 \le i <j <k \le 101$ such that $A_i \cap A_j$, $A_i \cap A_k $and $A_k \cap A_j$ are not empty. [url=https://artofproblemsolving.com/community/c4h2685439p23297220]Indonesia 2014 [/url]

Let $n\geq 2$ be a positive integer, and $X$ a set with $n$ elements. Let $A_{1},A_{2},\ldots,A_{101}$ be subsets of $X$ such that the union of any $50$ of them has more than $\frac{50}{51}n$ elements. Prove that among these $101$ subsets there exist $3$ subsets such that any two of them have a common element.

Let $ Z$ be the total number of points and set $ Z \equal{} X_1 \plus{} ... \plus{} X_5$ where $ X_i$ is the number of points shown on the ith die. The probability generating function of each $ X_i$ is $ G_{X_i}(s) \equal{} \frac {s \plus{} ... \plus{} s^6}{6}$. Since the $ X's$ are independent: $ G_Z(t) \equal{} \frac {(s \plus{} ... \plus{} s^6)^5}{6^5} \equal{} \frac {s^5(1 \minus{} s^6)^5}{6^5(1 \minus{} s)^5}$. (*) Expand in (*) $ (1 \minus{} s^6)^5$ with the binomial theorem and $ (1 \minus{} s)^{ \minus{} 5}$ with the binomial series ( http://mathworld.wolfram.com/BinomialSeries.html ). In (*) the coefficient of $ s^{15}$ is the required probability.

I'm not really sure how to go about this, so any help would be appreciated! Suppose you throw a six-sided die five times. Find the probability that the sum of the outcomes of the throws is 15 using generating functions.

$ \frac {s^5(1 \minus{} s^6)^5}{6^5(1 \minus{} s)^5}\equal{}\frac{1}{6^5}s^5\left({5 \choose 0}\minus{}{5 \choose 1}s^6\plus{}...\right)\left({\minus{}5 \choose 0}\minus{}{\minus{}5 \choose 1}s\plus{}...\right)$. The coefficient of $ s^{15}$ is $ \frac{1}{6^5}\left({5 \choose 0}{\minus{}5 \choose 10}\minus{}{5 \choose 1}{\minus{}5 \choose 4}\right)$.

I'm not really sure how to go about this, so any help would be appreciated! Suppose you throw a six-sided die five times. Find the probability that the sum of the outcomes of the throws is 15 using generating functions.

[hide="See if you know the answer."]$ x^3 \minus{} x \equal{} x(x^2 \minus{} 1)$ $ \Rightarrow x(x \minus{} 1)(x \plus{} 1)$ As it is three consecutive number so it is divisible by 3! and hence divisible by 3.[/hide]

If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before. If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3.

brute force it by observing it in mod 3. This is essentially what you do when you claim that three consecutive numbers must contain a multiple of $ 3$; I would hardly consider it brute forcing. :wink: Another problem very similar to this that is solved along the same lines is: (Zeitz) Show that $ n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)$ is never a perfect square for all natural $ n$.

If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before. If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3.

If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before. If $ x$ is an integer, prove that $ x^3 \minus{} x$ is divisible by 3. By fermat, we have $ x^3\equiv x\pmod{3}\Longleftrightarrow 3|(x^3\minus{}x)$

If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before. If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3.

Another problem very similar to this that is solved along the same lines is: (Zeitz) Show that $ n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)$ is never a perfect square for all natural $ n$. Could someone help me out on this one? :oops: The only way I know is to expand $ n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)\plus{}1$ and prove that it is always a square by factoring... but that isn't quite "solved along the same lines".

If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before. If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3.

I don't think that's the right solution This is what the book says but I don't understand. [hide="Solution"] $ x^{100} \minus{} 2x^{51} \plus{}1\equal{} (x^2 \minus{} 1)q(x) \plus{} ax \plus{} b$. Putting $ x \equal{} 1$ into this relation gives $ b \equal{} 0$. Putting $ x \equal{} \minus{} 1$ gives $ a \equal{} \minus{} 4$. Thus the remainder is $ \minus{} 4x$ [/hide]

Find the remainder on dividing $ x^{100} \minus{} 2x^{51} \plus{} 1$ by $ x^2 \minus{} 1$

I don't think that's the right solution This is what the book says but I don't understand. [hide="Solution"] $ x^{100} \minus{} 2x^{51} \plus{} 1 \equal{} (x^2 \minus{} 1)q(x) \plus{} ax \plus{} b$. Putting $ x \equal{} 1$ into this relation gives $ b \equal{} 0$. Putting $ x \equal{} \minus{} 1$ gives $ a \equal{} \minus{} 4$. Thus the remainder is $ \minus{} 4x$ [/hide] Check the substitution. Putting $ x\equal{}1$ gives $ a\plus{}b\equal{}0$ and putting $ x\equal{}\minus{}1$ gives $ \minus{}a\plus{}b\equal{}\minus{}4$. Solve that simple system, and you get what I said.

Find the remainder on dividing $ x^{100} \minus{} 2x^{51} \plus{} 1$ by $ x^2 \minus{} 1$

$ f(x)\equal{}x^{100}\minus{}2x^{51}\plus{}1\equal{}(x^2\minus{}1)Q(x)\plus{}l(x)$, $ l(x)\equal{}ax\plus{}b$ $ \Longleftrightarrow f(x)\minus{}l(x)\equal{}(x\plus{}1)(x\minus{}1)Q(x)$, which means that $ y\equal{}f(x)$ has intersection points with $ y\equal{}l(x)$ at $ x\equal{}\pm 1$, or $ y\equal{}l(x)$ passes through two points $ (\minus{}1,\ 4),\ (1,\ 0)$, yielding $ l(x)\equal{}\minus{}2x\plus{}2$.

Find the remainder on dividing $ x^{100} \minus{} 2x^{51} \plus{} 1$ by $ x^2 \minus{} 1$

We can now find the relation between u and v $\sin\theta \frac{u'(r\cos\theta)}{u(r\cos\theta)} = \cos\theta \frac{v'(r\sin\theta)}{v(r\sin\theta)}$ where $(r, \theta)$ is the radial coord. it seems to me that we r in need of another differential equation to completely specify the forms of u and v.

Hi all, How can we prove that the unique 2D function which is separable and circularly symmetric is a 2D Gaussian, i.e. $G(x, y) = a e^{\frac{x^2+y^2}{b}}$. Thanks in advance.

Suppose $|f'| \leq M$. Given $\epsilon > 0$. Since $\{a_n\}$ is Cauchy, then there exists $N$ such that for all $n,m>N$, $|a_n-a_m| < \frac\epsilon M$. Then $|f(a_n)-f(a_m)| = |f'(\xi)|\cdot|a_n-a_m| < M \cdot \frac\epsilon M = \epsilon$. So $f\{(a_n)\}$ is Cauchy too. If $f'$ is unbounded, take $f(x)=\frac 1x$ for example. Let $a_n=\frac 1n$, then obviously it is Cauchy, however, $f(a_n)=n$ is not.

Let $f(x)$ be the real- valued function which is differentiable on the opened-interval $(0,1).$ If $f'(x)$ is bounded, then prove that for the Cauchy sequence $\{a_n\}_{n=1}^{\infty}$ in the opened-interval $(0,1),\ f(a_n)$ will become Cauchy sequence as well.How about in case that $f$ is simply differentiable?

The integral: $\int_{1/2}^{1}\int_{0}^{\sqrt{1-x^2}}dydx$ (LaTeX isn't hard, trust me. Mostly you just put computations between 2 $\$$) The line $x=\frac{1}{2}$ is $r cos\theta=\frac{1}{2}$ Hope that helps! :D

Hi can someone help me with the double integration: Sorry I dont know how to use the LATEX code. it is a double integration The first integration goes [b]from 1/2 to1[/b]. the second integration goes from 0[b] to sqrt(1-x^2)[/b]. there is no function after ,just[b] dy dx[/b]. I am asked to calculate the integral using polar coordinates. I dont know how to find the bounds of integration corresponding for r. Thank you B

The integral: $\int_{1/2}^{1}\int_{0}^{\sqrt{1-x^2}}dydx$ (LaTeX isn't hard, trust me. Mostly you just put computations between 2 $\$$) The line $x=\frac{1}{2}$ is $r cos\theta=\frac{1}{2}$ Hope that helps! :D All righty! It makes sense now B

Hi can someone help me with the double integration: Sorry I dont know how to use the LATEX code. it is a double integration The first integration goes [b]from 1/2 to1[/b]. the second integration goes from 0[b] to sqrt(1-x^2)[/b]. there is no function after ,just[b] dy dx[/b]. I am asked to calculate the integral using polar coordinates. I dont know how to find the bounds of integration corresponding for r. Thank you B

$a_{n+1}=\frac{1}{3}a_n+\frac{1}{3}\Longleftrightarrow a_{n+1}-\frac{1}{2}=\frac{1}{3}\left(a_n-\frac{1}{2}\right)$, thus $a_n=\left(\frac{1}{3}\right)^{n-1}\left(x-\frac{1}{2}\right)+\frac{1}{2}$. From $a_n-\frac{1}{6}=\left(\frac{1}{3}\right)^{n-1}\left(x-\frac{1}{2}\right)+\frac{1}{3},\ a_n+\frac{1}{6}=\left(\frac{1}{3}\right)^{n-1}\left(x-\frac{1}{2}\right)+\frac{2}{3}$. If $[x]=m$,then $x-1<[x]\leq x$,so since $\left(a_n-\frac{1}{6}\right)-1<\left[a_n-\frac{1}{6}\right]\leq a_n-\frac{1}{6}$ and $\left|\frac{1}{3}\right|<1$, we have $\sum_{n=1}^{\infty}\left( a_n-\frac{1}{6}\right)=\frac{x-\frac{1}{2}}{1-\frac{1}{3}}-\infty=-\infty$, similarly $ \sum_{n=1}^{\infty}\left( a_n+\frac{1}{6}\right)=\frac{x-\frac{1}{2}}{1-\frac{1}{3}}-\infty=-\infty$,... :?

The sequence $(a_n)$ is given by $a_1=x\in R$ and $3a_{n+1}=a_n+1$ for $n\geq 1$. Set $A=\sum_{n=1}^\infty [a_n-\frac 16]$ and $B=\sum_{n=1}^\infty [a_n+\frac 16]$. (where $[x]$ denote the greatest integer less that or equal to $x$).Compute the sum $A+B$ in terms of $x$. :lol:

I think you can actually determine when the first player wins and when the second does, but for a quick proof of what you want here, assume $B$ (the second player) only wins when $n=n_1,\ldots,n_k$. Now, by the Chinese Remainder Theorem, you can choose some large $n$ (larger than all $n_i$) such that $n-n_i\equiv p_i\pmod{p_i^2},\ \forall i\in\overline{1,k}$, where $p_i$ are distinct primes. There's no perfect square $u^2$ such that $n-u^2\in\{n_i\ |\ i=\in{1,k}\}$, so no matter what $A$ (the first player) does, after his first move they end up in a position favorable to the first player that moves, and that player is now $B$. We have a contradiction.

two player play a game on the board .first there exists number $n$ in positive integer on board .after that first player subtract $x^2$ such that x^2<n .(x is positive integer)then second player keep on this rule .each person who reach 0 win. prove there exist infinitive many $n$ such that second player has strategy to win. :?:

Is the following considered a valid proof? $\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by: $x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ Prove that $\lim x_{n}= \sqrt{\alpha}$. . 1)First show that this sequence convergences. Once you show that then $\lim x_{n}= \lim x_{n+1}$ 2)Second show that $\lim x_{n}\not = 0$. Now you can find its limit. So if you let $L=\lim x_{n}= \lim_{x+1}$ you have, $L = \frac{1}{2}\left( L+\frac{\alpha}{L}\right)$ So $L^{2}= \alpha$. That means $L = \pm \sqrt{\alpha}$. Argue that it cant be negative. So $\lim x_{n}= \sqrt{\alpha}$

Is the following considered a valid proof? $\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by: $x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ Prove that $\lim x_{n}= \sqrt{\alpha}$. By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\alpha}$, it must be that $x_{n}\sqrt{\alpha}> \alpha$, so $\sqrt{\alpha}> \frac{\alpha}{x_{n}}$. And $\frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ is the mean between $x_{n}$ and $\frac{\alpha}{x_{n}}$, and thus, it must be that $\lim x_{n}= \sqrt{\alpha}$.

Have you shown monotonicity? I can see how you might argue that, but it's not obvious - and if it's not obvious, it's incumbent on you to explain how it follows. A proof has an audience - you must reach out to your reader and explain yourself. Yes, I understand this. But I'm certain that my argument for monotonicity is solid, it's just the limit which I wasn't entirely sure of. But without further ado: To show monotonicity, it is sufficient to show that: $x_{n}-x_{n+1}> \frac{\alpha}{x_{n+1}}-\frac{\alpha}{x_{n}}$ To see this, note: $x_{n}-x_{n+1}= \frac{\alpha(x_{n}-x_{n+1})}{\alpha}> \frac{\alpha x_{n}-\alpha x_{n+1}}{x_{n+1}x_{n}}$ The last step uses the fact that $x_{n}> \sqrt{\alpha}$ for all $n$. So $\frac{\alpha x_{n}-\alpha x_{n+1}}{x_{n+1}x_{n}}= \frac{\alpha}{x_{n+1}}-\frac{\alpha}{x_{n}}$, and this concludes the proof.

Is the following considered a valid proof? $\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by: $x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ Prove that $\lim x_{n}= \sqrt{\alpha}$. By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\alpha}$, it must be that $x_{n}\sqrt{\alpha}> \alpha$, so $\sqrt{\alpha}> \frac{\alpha}{x_{n}}$. And $\frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ is the mean between $x_{n}$ and $\frac{\alpha}{x_{n}}$, and thus, it must be that $\lim x_{n}= \sqrt{\alpha}$.

Essentially, you've said that because $y_{n}>b,$ and $z_{n}<b,$ and $y_{n+1}$ is the mean of $y_{n}$ and $z_{n},$ then $y_{n}$ tends to $b$. That's not convincing. In particular, you haven't given a good reason for the sequence to converge at all. Actually, I would give a few points for that (say, 3 out of 10). The reason is that the claim, though lacking a proof, is correct and pertinent to the problem. Indeed, once we know that all $y_{n}>b$ and all $z_{n}<b$, we can say that $y_{n+1}-b\le \frac{1}{2}(y_{n}-b)$, so $y_{n}-b\le C2^{-n}$, which is enough to justify convergence to $b$. :)

Is the following considered a valid proof? $\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by: $x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ Prove that $\lim x_{n}= \sqrt{\alpha}$. By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\alpha}$, it must be that $x_{n}\sqrt{\alpha}> \alpha$, so $\sqrt{\alpha}> \frac{\alpha}{x_{n}}$. And $\frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ is the mean between $x_{n}$ and $\frac{\alpha}{x_{n}}$, and thus, it must be that $\lim x_{n}= \sqrt{\alpha}$.

Same question, different proof. I'm suspicious of my proof because I don't use the fact that the sets are bounded. But it says that the diameter of the metric spaces approach 0, which I do use, and is this in essence the same as bounded? If $\{E_{n}\}$ is a sequence of closed and bounded sets in a complete metric space X, if $E_{n}\supset E_{n+1}$ and if $\lim_{n \rightarrow \infty}diam(E_{n}) = 0$, then $\cap_{i=1}^\infty E_{n}$ consists of exactly one point. Proof: Let $\{x_{n}\}$ be a sequence where $x_{i}\in E_{i}$ for all $i \geq 1$. Let $\epsilon > 0$ then since $\lim_{n \rightarrow \infty}diam (E_{n}) = 0$, there exists some $N$ such that $diam( E_{n}) < \epsilon$ for all $n \geq N$. So $d(x_{n}, x_{m}) < \epsilon$ for all $n, m \geq N$. Then $\{x_{n}\}$ is a cauchy sequence and converges to some point $p$. For any $\delta > 0$, $d(x_{n}, p) < \delta$ so $N_{\delta}(p) \cap E_{n}$ contains $x_{n}$. If $x_{n}= p$ for all $n \geq N$ for some $N$, then $p \in \cap_{n=1}^{\infty}E_{n}$. Otherwise, for all $n$, there exists an $m \geq n$ such that $p \neq x_{m}$. So $p$ is a limit point of $\cap_{n=1}^{\infty}E_{n}$, and so $p \in \cap_{n=1}^{\infty}E_{n}$. Now assume that $q \neq p$ is in $\cap_{n=1}^{\infty}E_{n}$. But $d(q, p) \neq 0$, so let $d(q, p) = \epsilon$. Then there exists some $N$ such that $diam(E_{N}) < \epsilon$, so $q \not\in \cap_{n=1}^{\infty}E_{n}$.

Is the following considered a valid proof? $\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by: $x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ Prove that $\lim x_{n}= \sqrt{\alpha}$. By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\alpha}$, it must be that $x_{n}\sqrt{\alpha}> \alpha$, so $\sqrt{\alpha}> \frac{\alpha}{x_{n}}$. And $\frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ is the mean between $x_{n}$ and $\frac{\alpha}{x_{n}}$, and thus, it must be that $\lim x_{n}= \sqrt{\alpha}$.

Is the following considered a valid proof? $\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by: $x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ Prove that $\lim x_{n}= \sqrt{\alpha}$. By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\alpha}$, it must be that $x_{n}\sqrt{\alpha}> \alpha$, so $\sqrt{\alpha}> \frac{\alpha}{x_{n}}$. And $\frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ is the mean between $x_{n}$ and $\frac{\alpha}{x_{n}}$, and thus, it must be that $\lim x_{n}= \sqrt{\alpha}$. The logic of the above is that $x_{n}$ is a decreasing sequence, each term exceeding the alleged limit. That does not show that the limit exists or equals the value claimed. It does show that if a limit exists (and it might not exist), it is greater than or equal to $\sqrt{\alpha}$. Any explicit proof that the limit exists, that is an argument using some inequalities and not the abstract principle about bounded monotonic sequences, should also show that the limit is the square root, as claimed. A purely abstract proof of existence would require a further calculation, substituting $L = \lim x_{n}$ in the recursion, to show that $L^{2}= \alpha$, so that $L$ is one of the square roots. It cannot be the negative square root because we have seen already that the positive root is a lower bound to whatever the limit might be.

Is the following considered a valid proof? $\alpha > 0,$ choose $x_{1}> \sqrt{\alpha}$ and let the sequence be defined by: $x_{n+1}= \frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ Prove that $\lim x_{n}= \sqrt{\alpha}$. By induction, it can be shown that $x_{n}> \sqrt{\alpha}$. Furthermore, since $x_{n}> \sqrt{\alpha}$, it must be that $x_{n}\sqrt{\alpha}> \alpha$, so $\sqrt{\alpha}> \frac{\alpha}{x_{n}}$. And $\frac{1}{2}(x_{n}+\frac{\alpha}{x_{n}})$ is the mean between $x_{n}$ and $\frac{\alpha}{x_{n}}$, and thus, it must be that $\lim x_{n}= \sqrt{\alpha}$.

we suppose that $ h\equal{}a\minus{}a'>0,g\equal{}b\minus{}b'$; $ f(x\plus{}h)\minus{}f(x)\equal{}f(x\plus{}h)\plus{}f(a\minus{}x\minus{}h)\minus{}(f(x)\plus{}f(a'\minus{}x))\equal{}b\minus{}b'\equal{}g$ now for $ r\in[0,h[: \ \forall x\in S_r\equal{}\{nh\plus{}r: \ n\in \mathbb{N}\}: \ f(x)\equal{}kx\plus{}m(r)$ where $ k\equal{}g/h$ and $ m: [0,h[\to \mathbb{R}$ $ f(x)\plus{}f(a\minus{}x)\equal{}b\equal{}kx\plus{}m(r)\plus{}k(a\minus{}x)\plus{}m(r')$ where $ x\equal{}nh\plus{}r,a\minus{}x\equal{}n'h\plus{}r'$ and $ r,r'\in[0,h[$ NOW we must pove that $ |m|<C$

Let $ a,b,a',b'$ be real numbers such that $ a\neq a'$ and $ f: \mathbb R\to\mathbb R$ be a function satisfying \[ f(x)\plus{}f(a\minus{}x)\equal{}b\qquad\text{and}\qquad f(x)\plus{}f(a'\minus{}x)\equal{}b'\] for all $ x\in\mathbb R$. Show that there exist real numbers $ k,c,C$ such that \[ |kx\plus{}c\minus{}f(x)|\le C\] holds for all $ x\in\mathbb R$.

Given a set $ \{1,2,3,....n\}$ of natural numbers, find the number of triangles possible with distinct sides. Use the fact that in a triangle with sides $ a,b,c$ that $ a \plus{} b > c$ $ b \plus{} c > a$ $ c \plus{} a > b$

Given a set $ \{1,2,3,....n\}$ of natural numbers, find the number of triangles possible with distinct sides.

The probability that you get $ 60 \%$ or higher is $ \frac{1}{2} \left (1\minus{}\binom{10}{5} \times \frac{1}{1024} \right )\equal{} \frac{1}{2} \left ( \frac{193}{256} \right )\equal{}\boxed{\frac{193}{512}}$.

If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question?

Let's first assign some variables: $ R$=getting question correct $ N$=getting question incorrect First, let's find the probability fraction, then convert it to a percent. I'll present two ways to solve this problem; the first is for those who know combinatorics, and the second is to clarify for those that don't know as much about probability. Okay, since it is as likely for one to get a question right as wrong, we have probabilities of $ R \equal{} N \equal{} 0.5$, and there are only $ 2$ variables, we have a normal binomial distribution, and since a probability of $ 0.5$ makes the right side of the binomial distribution pointless (in the formula $ (R \plus{} N)C(R)*R^(0.5)*(1 \minus{} R)^(0.5)$, the part after the main combination is redundant because $ R \plus{} (1 \minus{} R)$ always equals $ 1$, and thus the right-hand side is the same in all calculations of choices of $ R$ and $ N$), we could just evaluate the left side out of total possible to obtain the answer, which is $ \frac {10C6 \plus{} 10C7 \plus{} 10C8 \plus{} 10C9 \plus{} 1}{2^{10}} \equal{} \frac {193}{512}$. For those who might not know what a binomial distribution is, we could think: Probability=Successes/Possible How many ways are possible? Well, we have $ 2$ possible outcomes for each question, $ R$ or $ N$, and we have $ 10$ questions, so therefore, there are $ 2^10$ possible different outcomes. Now for successes. Since the extremes are usually the easiest, let's consider the extremes first: In order to have $ 10$ of $ R$ (correct), we must have the right answer every time, as we can have no incorrect answers. Thus, there is only [u]$ 1$[/u] way to get all $ 10$ questions right. How about just $ 9$ right? Well gee, thinking in the opposite world, $ 9$ right means $ 1$ wrong, so we need exactly $ 1$ wrong to have $ 9$ right. There are ten questions as choices to get wrong. Thus, there are [u]$ 10$[/u] ways to get $ 9$ right. How about $ 8$? Well, thinking about the opposite again, we need exactly $ 2$ wrong to get $ 8$ right. Well, we have $ 10$ choices for the first problem wrong and $ 9$ choices for the second problem wrong. But then, notice that we have counted each choice twice, for example, on the test, there is no distinguishable difference between getting #4 and #7 wrong and getting #7 and #4 wrong. Thus, we divide our computed answer of $ 9\times10 \equal{} 90$ by $ 2$ to get [u]$ 45$[/u] ways to get $ 8$ correct. Repeating the process we did for getting $ 7$ right and getting $ 6$ right, we find there are [u]$ 120$[/u] and [u]$ 210$[/u] ways, respectively. (This is known as combinatorics, by the way). Since we have figured out all five cases of passing, we finally know our total number of successes: $ 1 \plus{} 10 \plus{} 45 \plus{} 120 \plus{} 210 \equal{} 386$. Thus, the probability, or ratio of successes over possible, is $ \frac {386}{1024}$, which reduces to $ \frac {193}{512}$. We can finally convert our answer to a percent! Making the fraction into a percent, $ \frac {193}{512}$ is approximately $ \fbox{38}\%$. If you actually successfully carried out the second method of solving this problem in an actual FTW game, I declare that you are certainly amazing. Thus, my second solution is given only as an explanation for solving the problem; in the real game, just know probability and use the first method. EDIT 1: LOL I just realized that AIME15 posted a solution, an incorrect solution, while I was writing mines. EDIT 2: Whoopsies! Thx for catching the error, math154! I fixed it now.

If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question?

AIME15's solution is perfectly rigorous and much slicker. By symmetry, it is clearly equally likely for a person to get $ k$ questions right and $ 10 \minus{} k$ questions wrong as it is for a person to get $ k$ questions wrong and $ 10 \minus{} k$ questions right. However, we must be careful to account for the fact that we can't just directly divide by $ 2$, since $ 50\%$ could occur. After subtracting this probability from $ 1$, we can divide by $ 2$ to distinguish between the cases where $ 6\le k\le10$ and $ 6\le10 \minus{} k\le10$.

If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question?

Excuse me, math154: 1) please notice that not everybody who goes to this website would know the combinatorics of probability, and those who do know it could always look at my first solution 2) AIME15 does indeed have an incorrect solution, as the question asked for a percent, and AIME15 gave a fraction Sorry, I did not mean for my comment to be offensive. Also, I believe that your calling AIME15's solution incorrect without further distinction between [i]answer[/i] and [i]solution[/i] was ambiguous (plus, the solution is always more important than the answer :wink: ), which prompted my post with more explanation of his post (I thought you thought his content was incorrect). Similarly, I could say something like, "r15s11z55y89w21's solution was incorrect because it should be $ \frac {10C6 \plus{} 10C7 \plus{} 10C8 \plus{} 10C9 \plus{} 1}{2^10} \equal{} \frac {193}{512}$, not $ \frac {10C6 \plus{} 10C6 \plus{} 10C8 \plus{} 10C9 \plus{} 1}{2^10} \equal{} \frac {193}{512}$." And what the heck (oops, sorry for cussing, mods) are the combinatorics of probability? :wink: :P Please do not take this facetious post too seriously.

If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent change, to the nearest whole percent, of passing if you guess on each question?

[hide]Factor $x^3-2x^2-x+2=(x^2-1)(x-2)=(x+1)(x-1)(x-2)$. Let $R(x)$ be the remainder, so it has degree 2 or less, and let $P(x)=x^{100}-4x^{98}+5x+6$. $R(1)=P(1)=1-4+5+6=8$, $R(-1)=P(-1)=1-4-5+6=-2$, and $R(2)=P(2)=5*2+6=16$. Let $R(x)=ax^2+bx+c$, so that $a+b+c=8$, $a-b+c=-2$, $4a+2b+c=16$. Subtracting the second equation from the first gives $2b=10\Rightarrow b=5$. Then $a+c=3$, $4a+c=6$. Subtracting, $3a=3\Rightarrow a=1\Rightarrow c=2$. The remainder is then $x^2+5x+2$.[/hide]

Find the remainder when $\displaystyle x^{100}-4x^{98}+5x+6$ is divided by $\displaystyle x^3-2x^2-x+2$.

[hide]Factor $x^3-2x^2-x+2=(x^2-1)(x-2)=(x+1)(x-1)(x-2)$. Let $R(x)$ be the remainder, so it has degree 2 or less, and let $P(x)=x^{100}-4x^{98}+5x+6$. $R(1)=P(1)=1-4+5+6=8$, $R(-1)=P(-1)=1-4-5+6=-2$, and $R(2)=P(2)=5*2+6=16$. Let $R(x)=ax^2+bx+c$, so that $a+b+c=8$, $a-b+c=-2$, $4a+2b+c=16$. Subtracting the second equation from the first gives $2b=10\Rightarrow b=5$. Then $a+c=3$, $4a+c=6$. Subtracting, $3a=3\Rightarrow a=1\Rightarrow c=2$. The remainder is then $x^2+5x+2$.[/hide] Very nice, the key thing to remember is that $\text {deg } r < \text {deg } g$.

Find the remainder when $\displaystyle x^{100}-4x^{98}+5x+6$ is divided by $\displaystyle x^3-2x^2-x+2$.

Answer $ 3(xy+y-x-y)-2xy=2$ $ xy-3x=2$ $ x(y-3)=2$ When $ x$ equals $ 2$ or $ 1$ (only positive divisors of 2) then $ y-3$ respectively equals $ 1$ and $ 2$ Hence pairs $ (x,\,y)$ are: $ (2,\,4)$ and $ (1,\,5)$

Find all pairs of nonnegative integers x and y satisfying the equation: 3[y(x+1)-x-y]-2xy=2

Answer $ 3(xy+y-x-y)-2xy=2$ $ xy-3x=2$ $ x(y-3)=2$ When $ x$ equals $ 2$ or $ 1$ (only positive divisors of 2) then $ y-3$ respectively equals $ 1$ and $ 2$ Hence pairs $ (x,\,y)$ are: $ (2,\,4)$ and $ (1,\,5)$ Your solution is correct and is the same as mine :lol:

Find all pairs of nonnegative integers x and y satisfying the equation: 3[y(x+1)-x-y]-2xy=2