$ ABC$ is acutangle.
Prove that

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {4\sqrt {2}abc}{\sqrt {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}}\geq10$

.

[quote="Ligouras"]$ ABC$ is acutangle.
Prove that

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {4\sqrt {2}abc}{\sqrt {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}}\geq10$

.[/quote]

what's "acutangle" mean?

P.S.
when $ a\equal{}b\equal{}c\equal{}1$, ineq dose not hold.

[quote="kuing"][quote="Ligouras"]$ ABC$ is acutangle.
Prove that

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {4\sqrt {2}abc}{\sqrt {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}}\geq10$

.[/quote]

what's "acutangle" mean?

P.S.
when $ a \equal{} b \equal{} c \equal{} 1$, ineq dose not hold.[/quote]

Sorry......

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {4\sqrt {2}abc}{\sqrt {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}}\geq10$

[quote="Ligouras"]If $ a,$ $ b,$ $ c$ are the side-lengths of a triangle, then
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {4\sqrt {2}abc}{\sqrt {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}}\geq10$[/quote]
Actually, the stronger inequality holds
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)}\plus{}\frac{8\sqrt{2}abc}{\sqrt{(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)}} \ge 12.$
[i][b]Proof. [/b][/i]By the AM-GM Inequality, we have
$ \sqrt{2ab(a^2\plus{}b^2)} \le \frac{2ab\plus{}(a^2\plus{}b^2)}{2}\equal{}\frac{(a\plus{}b)^2}{2}.$
Multiplying this and its analogous inequalities, we get
$ \sqrt{8a^2b^2c^2(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)} \le \frac{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}{8},$
from which it follows that
$ \begin{aligned} \frac{8\sqrt{2}abc}{\sqrt{(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)}} &\equal{}\frac{32a^2b^2c^2}{\sqrt{8a^2b^2c^2(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)}} \\ &\ge \frac{256a^2b^2c^2}{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}.\;\;\;\; (1)\end{aligned}$
On the other hand, by the third degree Schur's Inequality, we have
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)} \ge \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc}.\;\;\;\; (2)$
From (1) and (2), we see that it suffices to prove that
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc}\plus{}\frac{256a^2b^2c^2}{(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2} \ge 12.$
Setting $ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc}\equal{}t.$ The above inequality becomes
$ t\plus{}\frac{256}{t^2} \ge 12,$
which is true according to the AM-GM Inequality,
$ t\plus{}\frac{256}{t^2} \equal{}\frac{t}{2}\plus{}\frac{t}{2}\plus{}\frac{256}{t^2} \ge 3\sqrt[3]{\frac{t}{2}\cdot \frac{t}{2}\cdot \frac{256}{t^2}}\equal{}12.$[/i]

[quote="kuing"][quote="can_hang2007"][quote="Ligouras"]If $ a,$ $ b,$ $ c$ are the side-lengths of a triangle, then
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {4\sqrt {2}abc}{\sqrt {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}}\geq10$[/quote]
Actually, the stronger inequality holds
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \plus{} \frac {8\sqrt {2}abc}{\sqrt {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}} \ge 12.$
[i][b]Proof. [/b][/i]By the AM-GM Inequality, we have
$ \sqrt {2ab(a^2 \plus{} b^2)} \le \frac {2ab \plus{} (a^2 \plus{} b^2)}{2} \equal{} \frac {(a \plus{} b)^2}{2}.$
Multiplying this and its analogous inequalities, we get
$ \sqrt {8a^2b^2c^2(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)} \le \frac {(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}{8},$
from which it follows that
$ \begin{aligned} \frac {8\sqrt {2}abc}{\sqrt {(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}} & \equal{} \frac {32a^2b^2c^2}{\sqrt {8a^2b^2c^2(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)}} \\
& \ge \frac {256a^2b^2c^2}{(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}.\;\;\;\; (1)\end{aligned}$
On the other hand, by the third degree Schur's Inequality, we have
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)} \ge \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}.\;\;\;\; (2)$
From (1) and (2), we see that it suffices to prove that
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \plus{} \frac {256a^2b^2c^2}{(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2} \ge 12.$
Setting $ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \equal{} t.$ The above inequality becomes
$ t \plus{} \frac {256}{t^2} \ge 12,$
which is true according to the AM-GM Inequality,
$ t \plus{} \frac {256}{t^2} \equal{} \frac {t}{2} \plus{} \frac {t}{2} \plus{} \frac {256}{t^2} \ge 3\sqrt [3]{\frac {t}{2}\cdot \frac {t}{2}\cdot \frac {256}{t^2}} \equal{} 12.$[/i][/quote]

very nice![/quote]

My Friend can_hang2007 very very nice !!!!! Thanks


trigonometry
AMC
AIME
complex numbers
Cyclotomic Polynomials

Let $P$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have positive imaginary part, and suppose that $P=r(\cos \theta^\circ+i\sin \theta^\circ),$ where $0<r$ and $0\le \theta <360.$  Find $\theta.$

[hide]Divide through by . We get the equation . Let . Then . Our equation is then , with solutions . For , we get . For , we get  (using exponential form of ). For , we get . The ones with positive imaginary parts are ones where , so we have .[/hide]

[quote="joml88"]Let $P$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have [b]positive imaginary part[/b][/quote]

Notice the bold part ;)

Elemenop, what's the exponential form of cos? I've heard of it before but I forgot about it...I had the exact same solution as you up to that point, when I got stuck. :(

Consider
\begin{eqnarray*} e^{i\theta} &=& \cos \theta +i\sin \theta\\ e^{-i\theta} &=& \cos(-\theta)+i\sin(-\theta)\\ &=& \cos \theta-i\sin \theta \end{eqnarray*}

So then \[ 2\cos \theta=e^{i\theta}+e^{-i\theta}\Leftrightarrow \cos \theta =\frac{e^{i\theta}+e^{-i\theta}}2. \]

Oh, thanks. But how does that help? All I got was $\cos \theta =\frac{-1+\sqrt{5}}{4}$... :(

Oh.. :oops: if I knew that I could have solved the problem: http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=4601

I didn't use exponential form, I just did $\text {cis} (\theta) + \text {cis} (-\theta) = 2\cos \theta$. Are there are any other common trig values I should know? All I know is 15, 30, 45, 75, 90. :(

$\sin18=\frac{-1+\sqrt{5}}{4}$ and $\cos36=\frac{1+\sqrt{5}}{4}$ are all the extra ones I know.

How would you proceed from there, fuzzylogic?  Would the roots just be the fifth roots of unity and the sixth roots of unity?

Sorry if this is a very simple question, I'm not good with complex numbers. :oops:

It's all the sixth roots + fifth roots of unity, not including the second or third roots of unity.

Another derivation: Here, $\Phi_n(x)$ is the $n$th Cyclotomic Polynomial (roots are the primitive $n$th roots of unity).

Adding $z\Phi_5(z)$ gives $z^6+z^5+2z^4+2z^3+2z^2+z+1-z\Phi_5(z)=0$.

from here it is clear

$(z^2+1)\Phi_5(z)-z\Phi_5(z)=0$

$\Phi_6(z)\Phi_5(z)=0$

So the desired roots are $\text{cis }\phi$ where $\phi\in\{60^\circ,72^\circ,144^\circ\}$. Thus $\theta=276^\circ$

[hide=Extra]
Roots in radicals:
$z=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i$
$z=\frac{-1+\sqrt{5}}{4}\pm\frac{\sqrt{10+2\sqrt{5}}}{4}i$
$z=\frac{-1-\sqrt{5}}{4}\pm\frac{\sqrt{10-2\sqrt{5}}}{4}i$
[/hide]

We have that
\begin{align*}
z^6+z^4+z^3+z^2+1&=0 \\
z^6+z^4+z^2+1&=-z^3 \\
\frac{z^8-1}{z^2-1}&=-z^3 \\
z^8-1&=-z^5+z^3 \\
z^8+z^5-z^3-1&=0 \\
(z^5-1)(z^3+1)&=0 \\
\end{align*}
Now, finding the roots with positive imaginary part should be an easy exercise. Note that although $z=-1$ and $z=1$ are extraneous solutions, they do not contribute to the answer so they are fine.
The solutions with positive imaginary part are $\operatorname{cis}(72^{\circ}),\operatorname{cis}(144^{\circ}),\operatorname{cis}(60^{\circ})$ whose product is $\operatorname{cis}\left(\boxed{276}^{\circ}\right)$.

Hello, whenever I try to post a solution to this problem, the server times out. Does anyone else have this issue?


Given triangle ABC inscribed circle (O) . $ AX , BY , CZ$ ($ X,Y,Z \in (O)$)are angle bisectors of A , B and C respectively . XY intersects BC , AC at Q , P  .YZ intersects AC , AB at N , M .ZX intersects AB , BC at S , R .
Prove that :$ \frac{2}{3} S(ABC) \leq S(MNPQRS) \leq \frac{2}{3} S(XYZ)$

[hide]any solution without using trigonometry ?[/hide]

Let $ I$ be the incenter.

It is well-known that $ MQ, NR, PS$ are parallel to $ BC, CA, AB$ respectively and are concurrent at $ I$. Also, $ I$ is symmetric to $ A, B, C$ with respect to $ YZ, ZX, XY$.

Hence, the triangles $ MSI, NIP, IRQ$ are similar to triangle $ ABC$.

Therefore, $ S(MSI) \equal{} (\frac{AB}{AB\plus{}BC\plus{}CA})^2 S(ABC)$ similar equality for $ S_{NIP}, S_{IRQ}$.

Then $ S(MSI) \plus{} S(NIP) \plus{} S(IRQ) \equal{} S(ABC) \sum (\frac{AB}{AB\plus{}BC\plus{}CA})^2 \leq \frac{1}{3} S(ABC)$ [1]. (CBS Inequality).

Combining  $ S(MNPQRS) \equal{} S(MSI) \plus{} S(NIP) \plus{} S(IRQ) \plus{} S(IMN)\plus{}S(IPQ)\plus{}S(IRS)$,
$ S(ABC) \equal{} S(MSI) \plus{} S(NIP) \plus{} S(IRQ) \plus{} 2(S(IMN)\plus{}S(IPQ)\plus{}S(IRS))$ and [1] gives the first inequality.


Let a triangle $ABC$ with $2\hat{B}=\hat{C}$ .Also take a point $M$ in the interior of the triangle such that  $AM=AC$ and $BM=MC$ .Prove that $\widehat{MAB}=\frac{\hat{A}}{3}$

Hong Kong TST 1994. It can be solved with Trig Ceva. There also exists a geometrical proof. Search for "Hong Kong TST" in the Geometry section with probably provide a link.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=54461

Best regards,
sunken rock


Guys this problem isn't so hard!
Don't forget about him! :wink:

[quote="Tiks"][quote="Hawk Tiger"][quote="Tiks"]The triangles $ABC$ and $A_{1}B_{1}C_{1}$ are similar and have opposite orientation. On the segment $AA_{1}$ is chosen a point $A'$, such that $\frac{AA'}{A_{1}A'}=\frac{BC}{B_{1}C_{1}}$.The points $B'$ and $C'$ are constructing analogously. Prove that the point $A'$,$B'$ and $C'$ are collinear.[/quote]
Is it$\frac{\overline{AA'}}{\overline{A_{1}A'}}=\frac{BC}{B_{1}C_{1}}$.?[/quote]
Don't you think that $\frac{\overline{AA'}}{\overline{A_{1}A'}}=\frac{AA'}{A_{1}A'}$?[/quote]
I think $\frac{AA'}{A_{1}A'}=\frac{|AA'|}{|A_{1}A'|}$
and I think I have solved the problem but I have not a dictionary now,i can't post my solution. :blush:

We will wait untill you get a dictionary  :lol: !

I have found my math dictionary,My solution:
there is a homothetic transfomation $H(O,k)$ and a axial transfomation$A(l)$
such that$H(A(\triangle A'B'C'))=\triangle ABC$
suppose $A(\triangle A'B'C')=\triangle A_{1}B_{1}C_{1}$
then $A_{1}A'\bot l,B_{1}B'\bot l, C_{1}C'\bot l$
and $A_{1}A'//A''O,B_{1}A'//B''O,C_{1}A'//C''O$


logarithms
calculus
calculus computations

Find a power series for $f(x)=x^{2}\ln{(1+x^{2})}.$

The power series for $\ln (1+x) = x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}\ldots$

Therefore in order to find the power series for $\ln (1+x^{2})$ we take every $x$ and replace it with an $x^{2}$, doubling all the exponents.  Then, to turn it into $x^{2}\ln(1+x^{2})$ we multiply the entire thing by $x^{2}$, adding $2$ to all the exponents.

The final product is $x^{4}-\frac{x^{6}}{2}+\frac{x^{8}}{3}-\frac{x^{10}}{4}\ldots = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{2n+2}}{n}$


calculus
integration
algebra
polynomial
calculus computations

This is a problem from the section on integrating natural logs that's giving me trouble:
$\int\frac{x(x+2)}{x^{3}+3x^{2}-4}dx$
I divided the denominator by $(x+2)$, and that canceled out, giving $\int\frac{x}{x^{2}+x-2}dx$.  After that, though, I can't figure out a substitution that doesn't give a polynomial in the denominator...

Thanks.  I tried factoring the denominator into $(x+2)(x-1)$, but I didn't try splitting the fraction up.  Haven't done stuff like that since Algebra 2...


Why does the integral sign look like the holes in a cello (there are two of them on each side of the bridge- which is the thing that looks like a man with string on top of him)?
Here is a picture of a cello:

http://upload.wikimedia.org/wikipedia/commons/d/da/Cello_bridge.jpg

Even though I play cello, I've never understood why they look alike. Which came first- the hole style or the integral?

This isn't exactly a mathematical question.  The integral sign derives from the long "s" symbol, and Leibniz chose it because it represented the limit of a [b]s[/b]um.  The f-hole is likely completely unrelated.

Like t0r said, the integral sign is really a [url=http://en.wikipedia.org/wiki/Long_s]long "s"[/url] in a italicized form. This symbol looks like an italicized "f", the symbol which appears on the cello and other string instruments, without the horizontal stroke. The relation is purely typographical.


How many numbers are in the following list:

\[ \minus{}4, \minus{}1, 2, 5,\ldots, 32\]

this problem is long time taking in counting, but the solution has really helped me. :lol:

Since this is an arithmetic sequence, we subtract the first term from the last, and divide by the common difference $ d$. $ \frac{32\minus{}(\minus{}4)}{3}\equal{}12$. However, we have under counted by $ 1$ (since the first term is skipped over) so
                                                   $ 12\plus{}1\equal{}\boxed{13}$.

[quote="erethorma"]It would help to explain why/how "we have undercounted" and why/how the first term was "skipped over".[/quote]

Starting from the beginning:
First, call the first term $x$, the number of terms $n$, and the common difference $d$. The last term can thus be expressed as $x+d(n-1)$ (Do you see why we multiply by $n-1$, not $n$?). We know the first term is $-4$, the last term is $32$, and the common difference is $3$. We can then set up an one variable equation: $32=-4+3(n-1)$
[hide="Note (read if you don't see how this is related to gaussintraining's equation):"]
gaussintraining's equation ($ \frac{32-(-4)}{3}=12 $) is just a bit more simplified version of this.

$32=-4+3(n-1) \Rightarrow 32-(-4)=3(n-1) \Rightarrow \frac{32-(-4)}{3}=n-1$

Now do you see why gaussntraining had to add an extra one? (I can't find an easy way to explain why you are skipping one though)
[/hide]
\[36=3(n-1)\]
\[\frac{36}{3}=n-1\] This probably looks very familiar:
\[12=n-1\]
\[n=\boxed{13}\]

I failed and somehow got 11 instead of 13 lol :D

Simple solution:
Count them.
SUPERSTITIOUS

add 7 to all terms: 3,6,9,12,...39 divide by 3 1,2,3,4,...13 so theres 13 hehe also enter 7 as your answer


function
probability
expected value
probability and stats

This looks like the most appropriate place in the forum to ask this.

I recently ran across an interesting problem hypothetical auction in which the highest bidder wins, and pays the average of all the other bids. Private values that are uniformly distributed on [0,1]. The question asks for the symmetric Bayse-Nash equilibrium and best response to truthful bidding.

My intuition is that since the expected value of uniform distribution is 0.5, anyone with a value of v < 0.5 would just bid their value because bidding higher might result in negative surplus. Anyone with a value v > 0.5 would bid 1, because their expected surplus would be v - 0.5, which would be positive. I'm not sure on how to approach the truthful bidding part.

Any comment or suggestion? Moderators please feel free to place this somewhere else more appropriate.

[color=green][Here is fine. The question stays in this forum. - moderator][/color]

I am searching for the Nash-Equilibrium bid function b(v).

Let's guess. Maybe it's linear?
I verify whether it is of the form $ b(v) \equal{} c\cdot v$.
How a bidder think?

If my value is $ v$ and i bid $ b$ then my expected profit is:
$ E \equal{} P($I win$ )(v \minus{} E(\frac {cv_{2} \plus{} cv_{3} \plus{} ... \plus{} cv_{n}}{n \minus{} 1}|$I win$ ))$
Where:
$ P($I win$ )\equal{}P(cv_{2} < b,cv_{3} < b,...,cv_{n} < b)\equal{}\left(\frac {b}{c}\right)^{n \minus{} 1}$
$ E(\frac {cv_{2} \plus{} cv_{3} \plus{} ... \plus{} cv_{n}}{n \minus{} 1}|$I win$ ))\equal{}E(cv_{2}|cv_{2}<b)\equal{}\frac{b}{2}$
It may be simplyfied to:
$ E \equal{} \left(\frac {b}{c}\right)^{n \minus{} 1}(v \minus{} \frac {b}{2})$
I maximize it, choosing $ b$ and get equation:
$ b \equal{} v\frac {2(n \minus{} 1)}{n}$.

It seems plausible to me. Because in the case of $ n \equal{} 2$ it gives $ b \equal{} v$ (truthful bidding) but in that case this auction is just the second price auction.


http://mathworld.wolfram.com/IsodynamicPoints.html

An inversion with a center of one of the isodynamic points maps $ ABC$ to an equilateral triangle so this problem is basically solved...


trigonometry algebra polynomial geometry function

Here is a round 1 for 5th COP problems. You only need to submit the answers. For more information, read the "About COP (Contest of Opportunity)" thread. These are due 9/4/05. Take the problem seriously because you need to get ALL right to qualify to the next round. PM me the answers.

[size=150]Problems[/size]

1. In $\triangle RYE$ where $\angle Y = 90^\circ$, $\overline{RY} = 12$ and $\overline{YE} = 5$. If the square $RACE$ is constructed outside the triangle, compute the distance of segment $\overline{AY}$.

2. When $f(x)$ is divided by $x-1$, the remainder is 5. When $f(x)$ is divided by $x-3$, the remainder is 7. What will be the remainder if $f(x)$ is divided by $x^2-4x+3$?

3. Let $i = \sqrt{-1}$. Compute $m$, in radians at interval $\frac{\pi}{7} < m < \frac{4 \pi}{7}$, so that:

\[\frac{12 \cos (2m) + 12 i \sin (2m)}{12 \cos (m) + 12 i \sin (m)} = i\]

4. Unique polynomial $u(x)$ is defined as below:

$u(n) = (x_1+x_2+x_3 + \cdots x_{n+1})^{n+2}$

Find the number of terms in polynomial $u(7)$ (when it's expanded).

5. We're trying to draw $\triangle ABC$ such that

\[a+b = c+1\\
\cos (C) = 0\]

where $a,b,c$ are all integers and as usual, $a$ is opposite of $\angle A$ and so on. Is this possible? If so, how many such triangles exist? If not, put 0 on your answer.

For #2, you say [b]the[/b] polynomial. However, don't two polynomials have the same roots if one is a multiple of the other?

[b]EDIT:[/b] And #3 is impossible, interval is not big enough.

What do you mean that one is multiple of other?

Can you restate what you mean again?

I mean that $x^2-x+1$ has the same roots as $2x^2-2x+2$. So, I can divide both the original cubic and the answer by any number I want to, so long as all of the terms are divided by the same number, correct?

Yes.

Please pick the most simplest polynomial $h(x)$. That is, choose $x^2-x+1$ instead of $2x^2-2x+2$.

So, are whole numbers considered simpler than fractions?

BTW, see my edit on the first post.

I edited #3. There's nothing wrong with the interval.

I still don't get what's so hard about #1. Make the polynomial to have leading coefficent of 1. Is that answering your question?

Okay, mine does. I'll leave it the way it is. And thanks for the correction on #3.

BTW, #4 has two really cool solutions. :lol:

[b]EDIT:[/b] For clarity: solutions, not answers. ;)

Any restrictions for $f(x)$ on #2? Like, a certain degree, or can it be anything as long as it works?

Nope.

You don't need to find $f(x)$ though. :)

I knew that was something wrong with #3. :). Anyway it's so simple .. but #4  :bomb:

[quote="JesusFreak197"]#4 is fun. :lol:[/quote]

I don't think that. :| Haven't seen before such type of problems.

Did you intend to make this a bit harder than your previous COPs?  It seems like it.

I think it's bit harder.

But like the last one, I'm planning to write COP Round 1 B as well. I haven't actually wrote any problems but I have some in my mind.

I did my best to add all five kinds of topics: combinatorics, trigonometry, geometry, algebra, etc. #2 and #5, in my opinion, aren't that hard. Number 3 can be solved with luck. Number 4, in my opinion, is the HARDEST problem of all.  :)

[quote="Silverfalcon"] Number 4, in my opinion, is the HARDEST problem of all.  :)[/quote]

Yeah! I hate combinatorics :( !

[quote="Andreas"][quote="Silverfalcon"]

4. Unique polynomial $u(x)$ is defined as below:

$u(n) = (x_1+x_2+x_3 + \cdots x_{n+1})^{n+2}$

Find the number of terms in polynomial $u(7)$ (when it's expanded).

[/quote]

I think you want $a^2 + b^2 +2ab$ and not $a^2 + b^2 + ab + ab$ but the first it's not expanded. Right ?

Do we post complete solutions or only answers?
Thanks  :)[/quote]

First one.

Answers.

[quote="chess64"]When's B? I don't think I'll be able to do this one. :([/quote]

Sometime next week I suppose. It depends when I finish writing the problem. I didn't even "start" yet but  :) I'll notify here when I have certain date.

Will I find out soon if my answer to the new #2 was correct?

You can and you will.

But I don't want to give away any results before the deadline. I know you want to see your score but I'll tell all the scores at once.

Also, I only received one answer for COP so far. I know this is only half way but please try to attempt this. I know #4 is irritating one if you don't know how to attack it but as we know, you'll never develop problem solving skills without attacking challenging problems.

Okay, as long as I know that that's the reason. :)

I don't know how to solvve #4 [img]http://www.math-linfo.as.ro/102.gif[/img]

yikes...I may send in my answers. Or may wait for B.

Please submit the answers by tomorrow 11:59:59.

Why's everyone think #4 is so hard?  Number 3 is the only one giving me problems...

Edit: Is it just me or is something wrong with #3.  I can only find one solution and it's out of the range...

Heh, everyone has different ideas about what's hard. If I got it wrong, then #2 was the hard one, since I have no idea how else to solve it... if not, I guess #4 was the hardest, although I didn't have too much trouble solving it.

[quote="eryaman"]Why's everyone think #4 is so hard?  Number 3 is the only one giving me problems...[/quote]

Same... :? What form do we give our answer in? Are we allowed to give a decimal, or should we express at as $f(x)$ for some function $f$ and number $x$? For example, $\sqrt{98664666459}$ or something....


1) What is the name given to words that can be read from left to right or from right to left, but with different meanings?

2) In my own language (Portuguese) I can only remember two such words. What examples are there for other languages?

Those words are called palindromes.

http://www.fun-with-words.com/palin_example.html

'raw' and 'war' is one such example :P

Another pair is 'god' and 'dog'

EDIT: @tokenadult: Great link!

[quote="Carcul"]1) What is the name given to words that can be read from left to right or from right to left, but with [b]different meanings?[/b][/quote]

So you mean that read in either direction they are different (legitimate) words?  I don't know if there's a term for it, but there are a lot of short words like that in English:

- (s)pot and top(s)
- (s)pat and tap(s)
- (s)top and pot(s)
- are and era

Ah thanks. I saw the wikipedia article about palindromes and I got the idea that palindromes where words that could be read in either direction, but with the same meaning. Here are some examples from Portuguese:

- "aroma" and "amora"
- "animal" and "lâmina"
- "amor" and "Roma"
- "assim" and "missa"

I agree with jb05.  Here are English examples of words that fit this definition of "palindrome":

level
noon
mom
dad
racecar (although usually written as two words, it still works)

In the context of this thread, look for the music video "Bob" here:

[url=http://music.yahoo.com/ar-269394-videos--Weird-Al-Yankovic]Weird Al Music Videos[/url]

My favorite: A dog, a panic, in a pagoda!

[quote="jb05"]Palindromes are words that are [i]exactly[/i] the same read backwards. I've never heard of a name for words that are different (even with the same meaning) read backwards.[/quote]
I’ve seen one crossword enthusiast describe such words as “marganas”. A “margana” is an anagram formed by simply reversing the spelling of a word or phrase. :P

do you mean anagrams? they're words that have the same letters in different a different order like "earth" and "heart"

I made a stupid one:

A dog? Apart lull? Ew! A well ultra pagoda!

Famous ones:

Rats live on no evil star.

A man, a plan, a canal, PANAMA!

Madam, I'm Adam.

God's dog.

I know that this is revival, but I have to post this:

[youtube]Nej4xJe4Tdg[/youtube]

How do you make palindromes?
A dog app! Moo doomp! A pagoda!

A rat!
Far away a star, rats a yaw araf!
Tar A.!

Squirrel!
I am a Neopet; I vomit on people.
Elpor P. not I movite.
Poe! Nam ail erriuq!
S!!!!

Art of Problem Solving is awesome!
Neopets just have fur, we have math!
We vomit on rats starnot I move! Whtam eva hewru, feva htsujst epoen, emosew.
Asi gnivlos melbor Pfotra.

Rats!
"No, do not kill my tree. It is a prized possesion to me, a treasure like gold and diamonds. Like conlanging, I cannot live without it."
"I will go find another tree to cut for my castle."
"I value nature as much as I value mathematics."
Ector! They fight in glory!
The first person, who tells the other not to cut his tree, stops fighting, gnith gifs pots.
Eer tsiht ucotton reh toeh tsl let ohwno sreptsi fe.
H. Tyrolg nit H. Gifyeht rot cescitam. E H. Tameul avi sahcum sa erutan eulavi, eltsacy mro.
"F. Tucot eer treh tonad nifogl liwi. Ti tuohti wevil ton nacigni. Gnalnoc e kils D. Nomaid D. Nad Log e kiler us.
A ertaem ot noise S. Sopdez irpas iti. E ertym L. Likton od, on.
Star!

Happy Day! School Time!
Bell rings. Students enter.
But they remain silent.
Today is the No English Contest. A!
"Hola."
"Aloha."
"Tsetnoch silgneon d htsi yado T. T. Nelisnia."
"Merye H.T.T.U. Bretnest."
Ned uts S. Gnirlle. Bemit lo ohcsy dyp.
PAH!

Good, you're trapped. The only way you can escape is to fill a page with a palindrome. Let's see what happens.
By the way, I am a NEOPET!!!!!!!!Muhwahwahwahwa!!!!!!!!!!!!
I vomit on rats! Ha! You vandalized my picture in the magazine!
You will not escape me.
War! Raw war! Run! I've poisoned you! If you don't find the antidote, you will not live!
Do you see eesuoy od?
Evil ton L.L.Iwuoyet odit naeht dnif tnodu oyif, u oyden os i opevi nur. Raw war! Raw! Eme pac seton ll. Iwuoy enizag. Ameht.
Nie! Rutcipym de zilad, nav uo yah. Star! No timovia! Whawhawhawhumte poe namai yaw, eht y bsneppahtahw e esst ele mord nila pahti wegapallifotsi e pacse nac u oyyaw yl no eht depparteruoydoog!!!!!!!!!!!!!!!!!!!!!!!!!!


AMC
AMC 10
geometry
3D geometry
AIME
trigonometry
factorial

132  :(   :(   :(  it is like 10B last year, but only slightly easier. But then, it seems insignificant in improving my score (3 stupid arithmetic mistakes GRRRRRRRRRR). Since I am not going to take the B contest, so I am not going to get another chance to get a perfect score on 10 ever again.

[quote="yif man12"]150! Just barely got a perfect score, since I realized I made 2 mistakes in the last 5 minutes of the test.[/quote]

Same story here.  I realized that I'd forgotten 1 was a perfect cube on #15, and that n = 1 and 16 didn't count on #24.

The test was much, much easier compared to last year.  There weren't any hard problems - just a lot of chances to make mistakes.

112.5. Won't make it to top 1%, I think? Could've been 118.5, except I got the factorial and perfect cube problem wrong...

I'm horrible compared to you I am going to hit my head on the wall now...

By the way, does anyone want to guess what's the top 1% cutoff point is?

I got a 131 in 8th grade! I missed #14 because I didn't recognize the way to solve it easily during the test so I wrote out all of the combinations and forgot to count one. Missed 18 because I forgot ABAA was twice as likely. Omitted 23 and 25. Got the rest right.

I got a 121.5 I think.  I thought I had a 127.5 for a while, since both me and my dad forgot that 1 was a perfect cube in 15.  Hopefully, I didn't make any other stupid mistakes, and I qualified for AIME.

Look at the answers at [url=http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2005-10a/05AMC10AB-answers.html]www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2005-10a/05AMC10AB-answers.html[/url]

The answers are up, so you can compare the answers you wrote in your booklet with the ones on the internet.

I am so mad. I got (I think) all the questions except for #1 because I didn't read the word "each."

... I am furious at myself...

133.5. I not very satisfied though, since I DEFINITELY should have got #22 right. Also, I overestimated the difficulty of #19 and thought there were more complicated geometry involved. Oh well, its time to look on to AIME.  ;)

Darn. I left it in my locker -_-;; .

I'm 100% sure I got a 144 or a 150, though. First or third in the nation...

144.  RAWR.  Haha, well I'm pleasantly happy, but I still have that "Oh, so very very close" mindset right now.

I think i got a 134.5 because I missed the one where ABAA is twice as likely, and I also missed the one where you have the sum of the two values which make it a perfect square! Curse my carelessness!

ack i got a 144 as well, forgot ABAA was weighed twice on number 18.

131, slightly below average relative to my most recent practice test scores. I'm okay with it, but I want a better one on 10B (or 12B, whichever I decide to take).

Can you take AMC 10B again even if you have taken AMC 10A?

Yes, anyone can take both (10th graders and below obviously)

[quote="SirErnest"]Are the 10Bs harder than the 10As?[/quote]

While the exams are not standardized, I don't think the AMC intends one exam to be harder than the other and probably does not know which exams will result in higher scores until they are all graded.

I think A is going to end up being definitely easier than B
ugh.. my friend for 25 just brute forced it using law of cosines in his calculator and then got a decimal approximation and compared the answer choices...

Got 124 - 8th Grade. Qualified for AIME.

Sorry for the very late post - new to AoPS  :)

I'm a little upset with my 121.5 (8th Grade) but I won the high school and qualified for the AIMEs so I'm happy (since all our smart people are in private schools [cough cough Yifan]) anyways when are you allowed to take the Bs? I think all my three wrong were dumb assumptions like multiplying diagonals for the are of a rectangle?? Wow I could have done so much better if I was more careful....

Sorry, but the B's are over.

And I got the exact same score on the exact same test as hockeyadi23, and we're both in 8th grade.  How cool is that?

[quote="paladin8"][quote="qweretyq"]
ugh.. my friend for 25 just brute forced it using law of cosines in his calculator and then got a decimal approximation and compared the answer choices...
[/quote]

that's what i did =)... i used law of cosines and then heron's... it was ugly, but pretty quick. then i noticed i could do it in a few seconds anyway :P[/quote]

Hehe, same here. I didn't really feel like finding a clever way to do it so I just did it that way Turned out pretty well.

You know...I thought I was good at math... Until I came to this website.

YOU GUYS ARE SICK!! AND I LOVE IT!!
 I only got like a 93.. Missed  pretty much every question.. Got some easy ones, guessed some hard ones. I'm looking forward to spending time with this community so that I can qualify for (through AMC10) AIME next year!

gotta doublepost.. dont want you guys thinking the picture is me..

[quote="xdl615x"]"If one studies too zealously, one easily loses his pants." [/quote]

LOL..........but how come??????????


Find all differentiable functions f: R to R such that f'(x)=f(x) for all x in R

$  f'(x)-f(x)=0\Leftrightarrow e^{x}(f'(x)-f(x))=0\Leftrightarrow \frac{e^{x}f'(x)-e^{x}f(x)}{e^{2x}}=0\Leftrightarrow$
$  \left(\frac{f(x)}{e^{x}}\right)'=0\Leftrightarrow\frac{f(x)}{e^{x}}=c\Leftrightarrow f(x)=ce^{x}$


Two players play a game using $923^k$ balls.  Each player removes either $2,3 $or $9$ balls. The player who can't move loses. Which player has a winning strategy, and how?

$ 923\equiv-1\mod 11$

If $ k$ is even, then $ 923^{k}\equiv 1\mod 11$
If $ k$ is odd, then $ 923^{k}\equiv-1\mod 11$

The task is to make the opponent play always from $ 0,1,5$ or $ 6\mod 11$

Suppose that $ k$ is odd. The first player removes $ 9$ balls. The result is $ 1\mod 11$

After this point:
If the opponent removes $ 2$ then remove $ 9$ and vice versa. Then always the opponent will play from $ 1\mod 11$
So the problem is when the opponent removes $ 3$

Let's see all the possible cases

Suppose that there are $ x\mod 11$ balls where $ x = 0,1,5$ or $ 6$

If $ x = 0$ or $ 1$ and the opponent removes $ 3$ then remove $ 3$. Result = $ x-6\equiv 5$ or $ 6\mod 11$
If $ x = 5$ or $ 6$ and the opponent removes $ 3$ then remove $ x-3$. Result = $ x-3-(x-3)\equiv 0\mod 11$

When the balls are $ < 11$ and it's the turn of the opponent, we have:
If there are $ 0$ or $ 1$ balls, the opponent cannot play, so he loses
If there are $ 5$ balls, he can only remove $ x = 2$ or $ 3$. Then remove $ 5-x$
If there are $ 6$ balls, he can only remove $ x = 2$ or $ 3$. Then remove $ 3$. Result = $ 3-x$ which is $ 0$ or $ 1$

So if $ k$ is odd, the first player wins
$ \hline$

If $ k$ is even, then $ 923^{k}\equiv 1\mod 11$. This is a losing position.

As we saw above, everything you do, the opponent will bring you again to a losing position.
So at this case winner is the 2nd player


algebra
polynomial
number theory unsolved
number theory

For $P(x)\in\mathbb Z[X],degP> 1,M,n\in\mathbb N$ prove that the equation :
                $n^x-P(y)=M$
has only finitely many solutions $(x,y)\in \mathbb N^2$.
[Edited by Myth]

Are you sure this has an elementary solution?  :?

Actually,i'm not sure about this easy-looking problem  .I've done some effort to get completely stuck :blush:

Anyway, I think you meant $\deg P>1$, don't you?

BTW, if $P(x)=ax^2+bx+c$, then we can reduce your problem to $4an^x-4aM$ is a perfect square for infinitely many $x$. Take a look at http://www.mathlinks.ro/Forum/viewtopic.php?highlight=a2%5En%2Bb&t=34569

Quite so,I forgot it  :blush:
Do I have any hope ,at least for $M=1$ ? :?

It doesn't matter, since we can always assume $P$ contains all necessary contsants, so we are to solve your problem only with $M=1$. Hmm... What about $P=x^2-1$ and $M=1$. Definitely, we have infinitely many solutions.

[quote="blang"]Let $P=X^2-1$, $M=1$ and $n=2$. Then $(x,y)=(2k,2^k)$, $k>0$ is a solution  :?[/quote]I was first! :)

Thanks a lot,Myth,blang!!
Then ofcourse we have infinitely many solutions for every value of $degP$ with apropriciate $P(x)$.One of the hardest case ,if I don't make stupid mistake again,is $P(x)=x^m,m\in\mathbb N,m>1$.Maybe it's still easy enough to try for you?(Ofcourse $M$ is arbitrary).

Hmm... the case $M=1$ gives the Catalan conjecture, so its possible that it's also very hard for this case

No, here two variables are fixed. Of course, the problem is not easy. I wanted to attach a file with a very interesting article about differences between powers, but I always have the nice message: "the extension array is not allowed". What does this mean?  :(

Postscript, I think. I open it using gsview32.

Convert it to PDF. BTW, it will save some space.

I would, if I knew...  How can one convert from postscript to PDF?

Ekh...
Open your document. Choose Convert in the menu File. Select "pdfwrite" device, resolution 300. Press Ok. Type appropriate name for pdf file (don't forget .pdf extension).

Thank you very much, :blush:  :blush:  :blush:

Though you could say it is in French. I am not so great French man as you.

[Edit: I forgot to type "not"]

Well, I simply forgot. But I bet you will understand! And yes, you are right, I'm so great french man exactly as you are.  :(

Of course, it was my foolish typo. Sometimes I forgot to write "not". Read again.


inequalities
trigonometry
geometry proposed
geometry

In a triangle ABC, Prove that
(1/(sinA*sinB))+(1/(sinB*sinC))+(1/(sinC*sinA)) ≥ 4

simpler:

$ \frac {1}{\sin A \cdot \sin B} \plus{} \frac {1}{\sin B \cdot  \sin C} \plus{} \frac {1}{\sin C \cdot \sin A} \equal{}$ $ \ \frac {\sin A \plus{} \sin B \plus{} \sin C}{\sin A \cdot \sin B \cdot \sin C} \equal{} \frac{2R}{r} \ge 4$


If $\ a,b,c>0$, then the inequality is true:
$\sum\frac{ab}{c(c+a)}\geq\sum\frac{a}{c+a}$

It's not clear to me what you mean by sums on left and right. Please write all the terms on both sides.

our ineq is equivalent to the below:
(if i did't emphasize,$\Sigma$means cyclic symmetric sum)
$\sum{a^{2}b^{2}(a+b)(b+c)}\ge\sum{a^{2}bc(a+b)(b+c)}$
expand we obtain:
$\sum{a^{4}b^{2}}+\sum{a^{3}b^{3}}\ge\sum{a^{3}bc^{2}}+3a^{2}b^{2}c^{2}$
since $a^{4}b^{2}+b^{4}c^{2}\ge2a^{2}b^{3}c$
sum them all we could prove it.

the inequality is true also if you just take cyclic sums  :wink: (see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47342 for the cyclic case)

[quote="stancioiu sorin"]If $ x,y,z > 0$, then the inequality is true:

$ \frac {xy}{z(z \plus{} x)} \plus{} \frac {yz}{x(x \plus{} y)} \plus{} \frac {zx}{y(y \plus{} z)}\geq \frac {x}{z \plus{} x} \plus{} \frac {y}{x \plus{} y} \plus{} \frac {z}{y \plus{} z}.$[/quote]Using the arithmetic-mean - geometric-mean inequality,

$ xyz(y \plus{} z)(z \plus{} x)(x \plus{} y)\left[\sum_{cyc}\frac {xy}{z(z \plus{} x)} \minus{} \sum_{cyc}\frac {x}{z \plus{} x}\right]$

$ \equal{} \sum_{cyc}y^2z^4 \plus{} \sum_{sym}y^3z^3 \minus{} \sum_{cyc}xy^2z^3 \minus{} 3x^2y^2z^2$

$ \equal{} \sum_{cyc}\left(\frac {2y^3z^3 \plus{} z^3x^3}{3} \minus{} xy^2z^3\right) \plus{} \sum_{cyc}y^2z^4 \minus{} 3x^2y^2z^2\geq 0.$


I entered a list of numbers into L1

Stat -> edit

and then, I entered the data into the list.

However, I need to rename L1 to something different: EXT

I can't figure out how to rename the list.
Does anyone know?
Is there a workaround?

For what I understand, it seems like I'm screwed, because each tutorial tells me I need to create a list with a name before I enter the data. Is there any possible way to create a name for this data?


SAY HYPOTHETICALLY UR GOING TO A MATH CONTEST LET'S SAY THIS IS HYPOTHETICALLY PRINCETON WHICH COLOR WOULD U RATHER WEAR FOR TEAM SHIRTS.

SEE ME AND ARNAV COULDN'T DECIDE SO WE WERE LIKE W/E LET'S LET AOPS DECIDE RIGHT...

finally a worthwhile topic on this forum

for those of you who don't know your colors

unc blue is this:

http://mtimages.cstv.com/unc/slog%20blue%20guys.jpg

hot pink is this:

http://www.telefon.de/images/big/motorola_razr_v3_hot_pink_back.jpg

yellow.    yellow. That's the color that two teammates got to the countdown on the same year.

OMG THE NEW JERSEY MATHCOUNTS TEAM SHIRTS WERE GOING TO BE HOT PINK AND SAY WE > YOU (I DESIGNED THEM) BUT THE PRINTER COULDN'T PRINT THEM BECAUSE OF ISSUES AWIRJAOWIJODAJPSDOIGJAPOFIJASOPIFJDSPOVJZPOVJDZOPJSZG

I WOULD CHOOSE PINK

Treething..Jersey(forgive me i prob. should know that). Sorry I was born there and no longer am there either... GO HOT PINK!!!!!

[quote="Iversonfan2007"][quote="dts"]You know... I can't speak for last year, because I've since moved, but two years ago, hot pink T-shirts almost happened for the North Carolina ARML team.[/quote]

they DID happen this year though[/quote]

Ah.

The year before, though, I was one of those people who, through their opposition, prevented that...

HOT PINK!

i find it weird how the second blue choice and the second pink choice aren't chosen

[quote="MysticTerminator"][quote="K81o7"]don't wear team shirts.  Paint them on if you want to.  Go like the guys in the first picture.[/quote]

don't spam our threads. Thank you.[/quote]

OK.

[quote="pianoforte"]

i find it weird how the second blue choice and the second pink choice aren't chosen[/quote]

because they are actually death traps! choose on of them and see what happens!!  :o

Haha it looks like we'll be seeing you guys in hot pink shirts... I guess we won't have any trouble recognizing your team. :D Actually I've seen Arnav before so I could probably recognize you by him anyway...

nefarious

[quote="Xevarion"]Haha it looks like we'll be seeing you guys in hot pink shirts... [/quote]

eh i don't know it's pretty close and the voting won't end till December anyways...

the two answer thing inspired me..further..

[quote="MysticTerminator"][quote="K81o7"]don't wear team shirts.  Paint them on if you want to.  Go like the guys in the first picture.[/quote]

don't spam our threads. Thank you.[/quote]

I wasn't joking.


[quote="hien"]Solve the inequation with variable $x$:
$\sqrt{x-1}+x-3\ge \sqrt{2(x-3)^{2}+2x-2}$[/quote]
It give $((x-3)-\sqrt{x-1})^{2}\le 0$. Therefore $x^{2}-7x+10=0$, or $x=5$.

Applying the Cauchy-Schwart's inequality we get:
$\sqrt{x-1}+(x-3)\le\sqrt{2(x-3)^{2}+2(x-1)}$
Hence we also get roots.


[quote="bkunchanapalli"]1.Factorize  without    removing     brackets    at    any    stage

   $2y(y+z)-(x+z)(x+y)$

2.factorize    the     polynomial    in   x

   $a^5$(b-c)+$b^5$(c-a)+$c^5$(a-b)

doubt:what is meant by linear factor in a polynomial

3.let  p(x)  be   a polynomial   whose degree    is 2006    and    has    real  coefficients

   and p(x)     is    greater      then    or     equal    to    1 then

   (i)it   should   have     at     least   one  complex    root
   (ii)it   should   have   2006   distinct    roots
   (iii)it should    have   1003   quadratics  as    a product    none    of    which   has   linear    factors
    (iv)it   should    not    have    a factor  of $(x-k)^{2005}$[/quote]

can  anyone please   solve   the   first  problem  of   factorization

[quote="luimichael"]$\!\(\((a - b)\)\ \((a - c)\)\ \((b - c)\)\ \((a^3 + a^2\ b + a b^2 + b^3 + a^2 c + a b c + b^2 c + a c^2 + bc^2 + c^3)\)\)$[/quote]

thanks .doubt:

linear expansion of 1st degree is a+b+c
is there any formula of method to find linear expansion of nth degree

can you solve the third sum of polynomial

For question 3.

i) If you plot the graph of y = P(x), then the whole curve lies above y = 1, which implies that P(x) has no real roots.
By the Fundamental theorem of algebra, every non-constant polynomial should have at least one root. Now, this root is not real, so it must be complex.

ii) I think this statement is not true in general. For example, if we let $P(x)=(x^2+1)^{1003}$. Then P(x) satisfies the given conditions but it has repeated roots.

iii) Because P(x) is a polynomial with real coefficients, all it's complex roots will occur in conjugate pairs.
For each pair of conjugate roots, say, $z$ and $\bar z$, we may consider
$(x-z)(x-\bar z)=x^2-(z+\bar z)x+z \bar z =x^2-2Re(z)+|z|^2$, which is a quadratic factor with real coefficients. There are totally 1003 pairs of conjugate roots and hence 1003 quadratic factors with real coefficients.
"There are no linear factors" is not clear. I would rather say, "there are no [b]real [/b]linear factors."

iv) In view of iii), $P(x) = q_1(x)q_2(x)...q_{1003}(x)$, where each $q_r(x)$ is not possible to be splitted into linear factors. The statement is clearly true.


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

For each complex matrix $A$, denote $P_A$ is the characteristic polynomial of $A$ and $P_A$ can be expressed:
\[ P_A(x)=(-1)^n(x^n+f_1(A)x^{n-1}+...+f_{n-1}(A)x+f_{n}(A))  \]
where $f_i(A)$ is a polynomial of entries of matrix $A$. (ofcourse $f_i$ fixed for all $A$)
a) prove that for every $A,B$ complex matrices, $f_i(AB)=f_i(BA)$ for all $i$
b) Suppose $Q: A |--\to Q(A)$ is a polynomial of entries of $A$, such that $Q(AB)=Q(BA)$ for all $A,B$, then there is $F\in C[x_1,x_2,...,x_n]$ polynomial such that $Q(A)=F(f_1(A),f_2(A),...,f_n(A))$ for all $A$.

I believe [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial&t=44770]this[/url] is a proof for (b) for infinite fields in general.


trigonometry
algorithm
linear algebra
matrix
integration
vector
parameterization

$ \left\{\begin {array}{1} \frac {dx}{dt} = x + y + z \\
 \\
& \frac {dy}{dt} = - 2y + t \\
 \\
& \frac {dz}{dt} = 2z + \sin t \end{array} \right.$

$ y(t) \equal{} \mathrm{C}_1e^{ \minus{} 2t} \plus{} \sum_{i \equal{} 2}^{\infty} \frac {( \minus{} 2)^{i \minus{} 2}}{i!}t^i \equal{} \mathrm{C}_1e^{ \minus{} 2t} \plus{} \frac {1}{4}\sum_{i \equal{} 2}^{\infty} \frac {( \minus{} 2t)^{i }}{i!} \equal{} (\mathrm{C}_1 \plus{} \frac {1}{4})e^{ \minus{} 2t} \plus{} \frac {1}{2}t \minus{} \frac {1}{4}$

$ z(t) \equal{} \mathrm{C}_2e^{2t} \minus{} \frac {1}{5}\cos t \minus{} \frac {2}{5} \sin t$

$ \Rightarrow x(t) \equal{} \frac {3}{10}\cos t \plus{} \frac {1}{10}\sin t \minus{} \frac {1}{3}(\mathrm{C}_1 \plus{} \frac {1}{4})e^{ \minus{} 2t} \plus{} \mathrm{C}_2e^{2t} \plus{} \mathrm{C}_3e^t\minus{}\frac{1}{2}x\minus{}\frac{1}{4}$

by inspection $ z(t)\equal{}Ae^{2t}\minus{}\frac{2\sin t \plus{}\cos t}{5}$
$ y(t)\equal{}Be^{\minus{}2t}\plus{}t\minus{}\frac{1}{2}$

$ L[x]\equal{} x'\minus{}x\equal{}y\plus{}z$ so

$ L[Ce^t]\equal{}0, L[a_re^{rt}]\equal{}(r\minus{}1)a_re^{rt}, L[c\sin t \plus{}d\cos t]\equal{}\minus{}(d\plus{}c)\sin t \plus{}(c\minus{}d)\cos t$
$ L[et\plus{}f]\equal{}(e\minus{}f)\minus{}et$

so we must have
$ x(t)\equal{}Ce^t\plus{}Ae^{2t}\minus{}\frac{B}{3}e^{\minus{}2t}\minus{}t\minus{}\frac{1}{2}\plus{}\frac{\sin t\plus{}3\cos t}{10}$

[u]Remark.[/u] There are many ways to solve this problem: (1) use Putzer's algorithm find a fundamental matrix for $ \Phi(t)$ for the homogenuous system, and then the general solution will be $ (x\ y\ z)^{t} \equal{} \Phi(t) c \plus{} \Phi(t)\int\Phi^{ \minus{} 1}(t)(0\ t\ \sin t)^t\ dt$; (2) diagonalize the matrix of the homogenuous system (if diagonalizable), and then solve the new first order systems and multiply solution vector of the new system with the matrix formed by the matrix with the eigenvectors in the proper order; and (3) find the fundamental matrix for the homogenenous system, and then use the method of variation of parameters for matrices.


I think dreams are a very interesting topic to elaborate on, because analyzing your own dreams can really tell a lot about yourself.  What are your dreams like?

Do your dreams ever repeat?

Are they in black and white, or color?

Are you constantly alone or surrounded by a lot of people?

Do you ever fly? And if you have flown in your dreams, how high have you gotten?

Have you ever died in your nightmares?

Are you often around bodies of water, or are you chiefly on land?

My dreams are always in color, and I fly often too, maybe 50 stories up, and i'm constantly alone, flying over the ocean with a sense of inner peace.  In my nightmares, I'm always running, alone by myself....... I'm not sure if its away from something, but I'm running and searching for something, or someone, and i don't know what it is.  I'm lost, yet i know exactly where i'm going... It makes no sense, but this setting always describes how my nightmares begin.

ok, no i have never had repeats

theyre in color(we dont live in the 1930s! :) )

both

i flew and i fell

if u die in yr dreams, u die in real life

land

well, no, not necessarily.  my friend has died many times in his dreams, and every time he died, he woke up. except in one where he said he had an afterlife  :o  :D

no, youre supposed to die if yr in deep sleep, b/c yr heart beats so fast it explodes, i think... :huh:

key word..."said"

lol dreaming that you die doesn't kill you...

My dreams are often abstract and the events jump from setting to setting.  I can rarely recall an entire dream in order.  Usually immediately after waking up I notice how illogical my recent thoughts were and my recollection falls apart (I have read that the brain is designed to forget dreams to protect sanity).  I've had dreams centered around drug use (generally after quiting something).  It is interesting that my brain can simulate a "high" in a dream perfectly even if I haven't used any drugs in months.  This happened once and my thought as soon as I woke up was "NOOOOOOOO why did I do that?" and then I realized it didn't actually happen.  I've had dreams that things I fear may happen actually do.  Sometimes I just dream abstract things involving colors, patterns, and impossible events that I can't use words to describe.

I have had dreams that repeat (atm I only recall one nightmare that repeated when I was young).

I don't think I've ever had a dream without color.

There are usually many people in my dreams.

Surprisingly I don't recall ever dreaming about flying.

I have never died in my dreams.

I'm usually on land.

um... im pretty sure it does. im not sure, my ss teacher said *slaps head* and, as usual, i believed him...

see, thats easy to say when u havent actually died in yr dreams, and i dont think u wanna...

i had a nightmare that repeated once. i entered a house and then descended the stairs and each time i went down a floor, there was another flight of stairs going down. and each floor looked exactly the same, except it was darker. then i went into the only room that I saw, and i saw a messy bed, a small TV and a doll carelessly thrown on top of it.
Except the doll was bloody, its hair was matted and its clothes were tattered. and its limbs were twisted and shriveled up... and wherever i was in the room, the doll's eyes followed me.  in real life, this doll is my american girl doll, but after i woke up, i haven't played with it since...

fortunately the second time i had this dream i didn't go in the room because i remembered that there was a bloody doll inside

flying is a different matter. it's simple and it feels best when you're flying over the ocean. all you do is jump, believe that you can fly, and you fly. in most of my flying dreams i don't have any wings, but in one i had giant white feathery wings. in another i had a hoverboard  :rotfl:

my night mares are usually the same, and they are colorless, and its really abstract, i can't describe the dream to you, but its probably grayish, and some of my dreams, i can't open my eyes, or i can't see.
and a lot of my dreams comes from the real life, like the image i saw on one of the quarters, the crater lake that's in some states, i dreamed i was sailing through it, and then enters a tunnel, see a mini tiger, and see the tiger's cub, i spoke to me, and then attacked me, and then i keep sailing, i think part of the dream about the tunnel came from harry potter movie 2, and then i ended in a very green place where soldiers dressed in green surrounds, me, and then i don't know what happened next, i probably just ended.

i love dreams, they are soooo much fun, except the nightmares,

sometimes, i feel really sad when the dreams are over.

i never flew though   i want to

maybe tonight i will.

a lot of times when i'm dreaming, everything makes complete sense. then after i wake up, and if i remember it, it doesn't make sense at all. and you know that time in the mornings right after you wake up, and you're in that limbo between conciousness and unconciousness?  at that moment, everything from your dream is preserved in a crystalline clear moment... then you wake up and get back to the real world -_-

i like dreams too (except nightmares, especially ones that repeat). when i'm in happy dreams, i'm constantly alone, and i'm flying, out over some deserted stretch of land or a vast expanse of water. when i have bad dreams (not necessarily nightmares, but pretty crappy ones) its always when i'm around a lot of people. but sometimes in happy dreams i'm flying with someone. it's wonderful  :)

In a really funny dream, i said "well, this is a dream anyway, so i can do whatever i want!" and then started floating in midair and stuff. But most of the time, they don't make a lot of sense.

I don't remember dreams repeating, but it might have happened before.

Color.

Sometimes there are people around me, other times I'm alone.

No, unless you count falling off a roller coaster flying. It wasn't really falling, it was like..... floaty  :|

No, i don't think I've died before.

I'm mainly on land.

New question: Can you feel pain in your dreams?

Yes, I think I can. The ones i remember are when I got sand in my eyes and when I got shot by some kind of gun that doesn't exist.

[quote="Nerd_of_the_Ages"]In a really funny dream, i said "well, this is a dream anyway, so i can do whatever i want!" and then started floating in midair and stuff. But most of the time, they don't make a lot of sense.[/quote]Sounds like a [url=http://en.wikipedia.org/wiki/Lucid_dream]lucid dream[/url] --> [quote]Reality testing

Reality testing (also referred to as reality checking) is a common method used by people to determine whether or not they are dreaming. It involves performing an action with results that will be different if the tester is dreaming. By practicing these tests during waking life, one may eventually decide to perform one while dreaming, which will usually fail, hopefully letting the dreamer realize that he/she is dreaming. Common reality tests include:

    * Reading some text, looking away from the text, and reading it again - the text will probably have changed.
    * Looking at one's watch (remembering the time), looking away, and looking back. As with the text, the time will probably have changed at the second glance.[19]
    * Flipping a light switch. Light levels rarely change in dreams.
    * Looking into a mirror; in dreams, reflections from a mirror often appear to be blurred, distorted or incorrect.[20]
    * Plugging one's nose shut, and attempting to breathe through it. It is usually possible to breathe while doing this because the tester is not actually plugging one's nose in real life.
    * Looking at one's hands one or more times. Hands may look distorted, or grow additional fingers in a dream.
    * Gripping and stretching a finger. In a dream, body image can become distorted, and pulling a finger can elongate it.
    * Jumping into the air. Gravity is often distorted in a dream state and floating, flying or "moon jumping" will occur.
    * Looking around and seeing everything blurred, such as if you were underwater.
    * Being able to move through solid objects like walls with minimal resistance
[/quote]  After getting home from school one day I had work in 2 hours and decided to nap.  I set the alarm for only 20 minutes later.  I was so tired that I immediately fell asleep and started dreaming, but for some reason I was aware of the dream and I was dreaming about being at work.  Everything smelled like work, I could look at ANYTHING at work in great detail.  It seemed like my mind shouldn't be able to produce such a perfect simulation of reality.  Even background noises were present as they usually are, sounds I don't ever normally think about.  Even my mood transformed to the way I feel when I am working.  I couldn't make myself wake up even though I was well aware that it was a dream.  My alarm rang only 20 minutes after I began sleeping and I remembered all of it as if it actually happened.

to nerd: no, i've never felt physical pain in my dreams, but i've felt mental pain countless times.

another new question: in your dreams, do you sometimes recover items that you've lost?

i have, once i lost this beautiful headband, and in several of my dreams after i lost it I had it in my hands, and the texture of it was so vivid, and i put the headband on. those were happy dreams, until i woke up and realized that i didn't have that headband anymore  :(

to xombie: unfortunately for me, i wouldn't have the patience to do these reality checks in my dreams, in my dreams i rarely ever pay attention to myself, and often, the things i do in dreams are way beyond things i would do in real life (like looking at my watch versus flying)

in my dreams, i'm not always an actual pereson, it's more like i'm a spirit, kind of. i can always see my body in my dreams when i hold something in my hands but i never pay attention to myself. everything revolves around the things around me. but my body has no meaning in my dreams. it's more like the soul that matters, and the body is just the container of the soul.

one thing special about my dreams is that weeks later half of them come to pass when I'm least expecting it and I'm like Whoah I saw this before. The key is when I'm least expecting it.

your dreams repeat too! wow. for a long time i thought i was the only one.

the good thing about repeating nightmares is that the second time, you know not to make stupid mistakes and get yourself in danger, or killed. and the good thing about repeating good dreams is that you can live the whole thing over again  :wink: .  except each time one of my dreams repeat, something about it is different. usually it starts out the same way, except the plot twists into a senseless haze of chaos

[quote="Nerd_of_the_Ages"]

New question: Can you feel pain in your dreams?

.[/quote]

i feel pain sometimes in my dream, but when i wake up, i usually find my self in pain too, like if i slept on something, or my neck got into an uncomfortable position, but the pain isn't as extreme as the one in the dream.

I sometimes think that I've had a dream before and try to change the outcome, but I don't know if the dream really is a repeat.

Color

Both

Once, maybe 50 feet, but it felt more like the world was moving relative to me.

No

Land, but never flat open spaces. i see alot of cities, never resembling real cities, or suburbs, or inside (cathedrals, odd houses with trapped door, once a frekin mansion with like 50 ft windows, schools,underworld...)

pain: i remember dreaming that i had a second row of lower teeth jutting outwards, and purple  blood spewing out into a sink because a tooth fell out: i think it was because of my orthodontics

last night i dreamt like i end up in the future, where there's some school in south america, like the battle school in Ender's game, and the kids were produced by budding, and there was this touch screen computer map thing, and i was trying to get to a city. saharan africa was green, and the mediterranean had shrunk. at some point, i thought i was reading a book.

[quote="pianoforte"]
pain: i remember dreaming that i had a second row of lower teeth jutting outwards, and purple  blood spewing out into a sink because a tooth fell out: i think it was because of my orthodontics

[/quote]
i've had my second row of lower teeth pulled out, except there was no blood.

One time I had this dream where I was in some random building somewhere (I think it was in a store). Then, in a few days, I found myself in the same exact building, in which I had never been in before. Once I realized it, I thought I was dreaming again, but obviously I wasn't. So then for the rest of the day I tried explaining to myself how I could possibly know exactly how it looked like before I had ever been there. I even got some really random details right... :o

[quote="Nerd_of_the_Ages"]hmm......

im pretty sure you're not some kind of phycic prophet.

You've probably seen this place before a long time ago, but just don't remember it.[/quote]
That was one of my arguments, but I'm 100% sure I've never even been in that city. Being psychic would be sweet but I highly doubt I am... Or am I? :o :roll:

Ok here's the most boring dream I ever had: I'm sitting in a bus, looking out a window. There are no other cars on the road and nothing special, just some random houses and stuff. Every once in a while the bus rides past a bus stop, and waiting there is the same woman. Every single time. She's just sitting on the bench, and she watches the bus go by... She never gets on the bus, just sits there at every single stop.

Fortunately the last time I had that dream was like 6 years ago. :)

once i had my head chopped off, no twice XD. Once I woke up immediately after the blade got 3/4 of the way, the second time, this monkey hit the guy that was chopping my head off 1/2 way... and I got stitches... I was flying in the monkey one, in this BIG cave

I was just about to eat dinner.......

I just had a dream about Pokemon. PM me for the details :lol:


Prove that $ n^{11}\minus{}n$ is divisible by $ 11$

hello, here you will find the proof of this theorem
http://en.wikipedia.org/wiki/Proofs_of_Fermat's_little_theorem
Sonnhard.

[quote="FelixD"]That's nothing else than Fermat's little theorem. :wink:[/quote]

Yes,that was my solution too :D , but I was wondering If we need to show also the proof of "[b]Fermat's little theorem"[/b] while we use it to solve problems during any math competition exam ?


Sorry for that I could find the problen is wrong.

I will post another problem later.

kunny

Let $ a,\ b,\ c$ be positive numbers. Prove the following inequality.

$ (ab\plus{}bc\plus{}ca)^2\left\{\frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\right\}\leq \frac{5}{2}(a^2\plus{}b^2\plus{}c^2)$.

[quote="kunny"]Let $ a,\ b,\ c$ be positive numbers. Prove the following inequality.

$ (ab \plus{} bc \plus{} ca)^2\left\{\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right\}\leq \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2)$.[/quote]

By AM-GM and Cauchy - Schwarz inequalities, we have:

$ LHS \equal{} \sum_{cyc}{ \left(c \plus{} \frac{ab}{a\plus{}b}\right)^2} \le \sum_{cyc}{ \left(c \plus{} \frac{a\plus{}b}{2}\right)^2} \le 2\left(\sum_{cyc}{ c^2 \plus{} \frac{(a\plus{}b)^2}{4}}\right) \le \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2)$

Exactly same as my solution.  :lol:  But the equality doesn't hold.  :maybe:

$ 2\left(\sum_{cyc}{ c^2 \plus{} \frac {(a \plus{} b)^2}{4}}\right)$

Oh, yes, that should be $ \sum_{cyc} 2\left(c^2\plus{}\frac{(a\plus{}b)^2}{4}\right)$.

[quote="Hong Quy"][quote="NguyenDungTN"][quote="kunny"]Let $ a,\ b,\ c$ be positive numbers. Prove the following inequality.

$ (ab \plus{} bc \plus{} ca)^2\left\{\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right\}\leq \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2)$.[/quote]

By AM-GM and Cauchy - Schwarz inequalities, we have:

$ LHS \equal{} \sum_{cyc}{\left(c \plus{} \frac {ab}{a \plus{} b}\right)^2} \le \sum_{cyc}{\left(c \plus{} \frac {a \plus{} b}{2}\right)^2} \le 2\left(\sum_{cyc}{ c^2 \plus{} \frac {(a \plus{} b)^2}{4}}\right) \le \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
$ 2\left(\sum_{cyc}{ c^2 \plus{} \frac {(a \plus{} b)^2}{4}}\right) \le \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2)$
i think  it's  wrong,  if  i think wrong  explain for me  why?[/quote]
$ 2\left(\sum_{cyc}{ c^2 \plus{} \frac {(a \plus{} b)^2}{4}}\right) \le \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2) \leftrightarrow (a\plus{}b\plus{}c)^2 \leq 0$

[quote="Hong Quy"][/quote][quote="Hong Quy"][quote="NguyenDungTN"][quote="kunny"]Let $ a,\ b,\ c$ be positive numbers. Prove the following inequality.

$ (ab + bc + ca)^2\left\{\frac {1}{(a + b)^2} + \frac {1}{(b + c)^2} + \frac {1}{(c + a)^2}\right\}\leq \frac {5}{2}(a^2 + b^2 + c^2)$.[/quote]

By AM-GM and Cauchy - Schwarz inequalities, we have:

$ LHS = \sum_{cyc}{\left(c + \frac {ab}{a + b}\right)^2} \le \sum_{cyc}{\left(c + \frac {a + b}{2}\right)^2} \le 2\left(\sum_{cyc}{ c^2 + \frac {(a + b)^2}{4}}\right) \le \frac {5}{2}(a^2 + b^2 + c^2)$[/quote]
$ 2\left(\sum_{cyc}{ c^2 + \frac {(a + b)^2}{4}}\right) \le \frac {5}{2}(a^2 + b^2 + c^2)$
i think  it's  wrong,  if  i think wrong  explain for me  why?[/quote][quote="Hong Quy"]
$ 2\left(\sum_{cyc}{ c^2 + \frac {(a + b)^2}{4}}\right) \le \frac {5}{2}(a^2 + b^2 + c^2) \leftrightarrow (a + b + c)^2 \leq 0$[/quote]

Sorry, I have the same idea with minsoens, but when I typed it, I didn't check my calculation.

[quote="minsoens"][quote="NguyenDungTN"]By AM-GM and Cauchy - Schwarz inequalities, we have:

$ LHS \equal{} \sum_{cyc}{\left(c \plus{} \frac {ab}{a \plus{} b}\right)^2} \le \sum_{cyc}{\left(c \plus{} \frac {a \plus{} b}{2}\right)^2} \le 2\left(\sum_{cyc}{ c^2 \plus{} \frac {(a \plus{} b)^2}{4}}\right) \le \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2)$[/quote]I believe the following is correct:
$ LHS \equal{} \sum_{cyc}{\left(c \plus{} \frac {ab}{a \plus{} b}\right)^2} \le \sum_{cyc}{\left(c \plus{} \frac {a \plus{} b}{4}\right)^2} \le \sum_{cyc}2\left({ c^2 \plus{} \left(\frac {a \plus{} b}{4}\right)^2}\right) \le \sum_{cyc}2\left({ c^2 \plus{} \frac {a^2 \plus{} b^2}{8}}\right) \le \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2)$[/quote]

That's right. :lol:

[quote="stergiu"][quote="kunny"]Let $ a,\ b,\ c$ be positive numbers. Prove the following inequality.

$ (ab \plus{} bc \plus{} ca)^2\left\{\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right\}\leq \frac {5}{2}(a^2 \plus{} b^2 \plus{} c^2)$.[/quote]

[color=red][b]NOTE[/b][/color]

 [i]I think, if I have not made a mistake in the operations, that the problem can be improved to  :[/i]

'' Prove the following inequality.

$ (ab \plus{} bc \plus{} ca)^2\left\{\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right\}\leq \frac {9}{4}(a^2 \plus{} b^2 \plus{} c^2)$. ''

But , proving this , I used the inequalities

    a) $ ab \plus{} bc \plus{} ca \leq (a^2 \plus{} b^2 \plus{} c^2)$  and   b)     $ \frac {ab}{a \plus{} b} \leq \frac {a \plus{} b}{4}$

 and not the inequality  $ (a \plus{} b)^2 \leq 2(a^2 \plus{} b^2)$ , used in the proof above. We have then equality for $ a \equal{} b \equal{} c$.

 [u]Babis[/u][/quote]

I tried with your hint, but I obtained again $ {5/2}$

Can you please post your proof?

Thank you very much.

[quote="stergiu"]NOTE

[i]I think, if I have not made a mistake in the operations, that the problem can be improved to :
[/i]
'' Prove the following inequality.$ (ab \plus{} bc \plus{} ca)^2\left\{\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right\}\leq \frac {9}{4}(a^2 \plus{} b^2 \plus{} c^2)$"

I tried with a few nos. and the inequality seems to be true. I'm eager to see the solution too! :)

[b][P.S. Will the inequality remain true if we use $ \frac{13}{8}$ instead of $ \frac{9}{4}$][/b]

[quote="stergiu"]'' Prove the following inequality.

$ (ab \plus{} bc \plus{} ca)^2\left\{\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right\}\leq \frac {9}{4}(a^2 \plus{} b^2 \plus{} c^2)$. ''
[u]Babis[/u][/quote]We have $ (ab\plus{}bc\plus{}ca)^2\left(\frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\right)$
$ \leq \frac{1}{2}(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)(a^2\plus{}b^2\plus{}c^2)\left(\frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\right)$
$ \equal{}\frac{1}{4}((a\plus{}b)^2\plus{}(b\plus{}c)^2\plus{}(c\plus{}a)^2)(a^2\plus{}b^2\plus{}c^2)\left(\frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}\right)$
$ \leq \frac{9}{4}(a^2\plus{}b^2\plus{}c^2)$

[quote="minsoens"]

$ ((a \plus{} b)^2 \plus{} (b \plus{} c)^2 \plus{} (c \plus{} a)^2)\left(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right)\leq 9$[/quote]

It is wrong!! :wink:

[quote="stergiu"]NOTE

[i]I think, if I have not made a mistake in the operations, that the problem can be improved to :
[/i]
'' Prove the following inequality.$ (ab \plus{} bc \plus{} ca)^2\left\{\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}\right\}\leq \frac {9}{4}(a^2 \plus{} b^2 \plus{} c^2)$"[/quote]

[b]HINT[/b]

Well , till here all are the same :

$ LHS \equal{} \sum_{cyc}{\left(c \plus{} \frac {ab}{a \plus{} b}\right)^2} \le \sum_{cyc}{\left(c \plus{} \frac {a \plus{} b}{4}\right)^2}$

But now I expand the identity and I use the fact that $ ab \plus{} bc \plus{} ca \le a^2 \plus{} b^2 \plus{} c^2$.

 After simplyfying  - easy work - we take the result.

 In this form of the inequality we have equality if $ a \equal{} b \equal{} c$

[u] Babis[/u]


geometry
geometric transformation
reflection
trigonometry
geometry unsolved

Let $\left(O\right)$ be a circle, and let $d$ be a line which doesn't intersect $\left(O\right)$. Consider two arbitrary points $M$ and $N$ on the line $d$ such that the circle with diameter $MN$ is externally tangent to the circle $\left(O\right)$ . Prove that there exists a fixed point $P$ such that the angle $\angle MPN$ is constant.

Any ideas ? Can you guess where $P$ is ?

Very nice problem, Romano ! Thanks for it ! What is your source for this problem because I wish to congratulate the its author.[hide="Here is the my solution."][b]Denote:[/b] $A\in d,\ OA\perp d$; the circle $w=C(O,R)$; the circle $\alpha=C(A,r)$; $\{B,C\}\subset d\cap \alpha$ (the point $B$ is in the leftside of the point $A$); $D\in d,\ AD=R,\ C\in (BD)$; $2\phi=m(\widehat {BPC})$; $u=m(\widehat {APM})$, $v=m(\widehat {APN})$; the middlepoint $I$ of the diameter $[MN]$, $IM=IN=x$,  $AM=y$. If the point $P$ exists, then $P\in OA$ (or the its reflection w.r.t. the line $d$) so that $\boxed {\ BP^2=BC\cdot BD\ }\ (*)\Longrightarrow$
$AP^2=BP^2-AB^2=BC\cdot BD-AB^2=2r(R+r)-r^2=$$2Rr+r^2\Longrightarrow\boxed {AP=\sqrt{2Rr+r^2}}$
$\tan \phi =\frac{AB}{AP}=\frac{r}{\sqrt{2Rr+r^2}}\Longrightarrow\tan \widehat {BPC}=\boxed {\ \tan 2\phi =\frac{\sqrt{2Rr+r^2}}{R}\ }$.
The circle $C(I,x)$ is exterior tangent to the circle $w$ $\Longleftrightarrow \boxed {\ y^2+2xy+2Rr+r^2=2Rx\ }\Longrightarrow$
${\tan u=\frac{AM}{AP}=\frac{y}{\sqrt 2Rr+r^2}}\tan v=\frac{y+2x}{\sqrt{2Rr+r^2}}$ $\Longrightarrow\tan \widehat {MPN}=\tan |u-v|=$
$\frac{2x\sqrt{2Rr+r^2}}{y^2+2xy+2Rr+r^2}=\frac{\sqrt{2Rr+r^2}}{2Rx}=\frac{\sqrt{2Rr+r^2}}{R}=\tan 2\phi ,$ i.e. the angle $\widehat {MPN}$ is constant.

[b](*) Remark.[/b]
[u]Question.[/u] Why the fixed point is the point $P$ defined by the relation $BP^2=BC\cdot BD\ ?$
[u]Answer.[/u] $I\rightarrow\infty\Longleftrightarrow m(\widehat {MPN})\rightarrow m(\widehat {ADP})\Longleftrightarrow m(\widehat {ADP})=m(\widehat {BPC})$ $\Longleftrightarrow$ $BP^2=BC\cdot BD\ .$ [/hide]

Nice solution , Virgil ! Indeed i don't know the source of this problem . I got it from my friend's notebook .


Let $n$ $>$ $1$ , such that $2^{n}$ + $2$ divisible by $n$.
Prove that $n$ is a even number

If $n|(2^{n}+2)=2(2^{n-1}+1)$ and n is odd then $n|2^{n-1}+1$. Last problem posted before and I prove, that it had only one solutione n=1.
Therefore $n|(2^{n}+2)\Longrightarrow n=1 \ or \ n \ is \ even$.


Prove that if $ a,\ b, \ c$ are three positive real numbers then \[ \sum_{cyc}\frac{b\plus{}c}{a}\ge3\plus{}\frac{\left(a^{2}\plus{}b^{2}\plus{}c^{2}\right)\left(ab\plus{}bc\plus{}ca\right)}{abc\left(a\plus{}b\plus{}c\right)}\] .

[quote="arqady"][quote="Mîndrilă Claudiu"]Prove that if $ a,\ b, \ c$ are three positive real numbers then
\[ \sum_{cyc}\frac {b \plus{} c}{a}\ge3 \plus{} \frac {\left(a^{2} \plus{} b^{2} \plus{} c^{2}\right)\left(ab \plus{} bc \plus{} ca\right)}{abc\left(a \plus{} b \plus{} c\right)}
\]
.[/quote]
It's obvious after expanding.[/quote]
Yes,arqady.
\[ \leftrightarrow (a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)\ge 6abc(c\plus{}b\plus{}c)\plus{}(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)
\]
\[ \leftrightarrow (ab\minus{}bc)^2\plus{}(bc\minus{}ca)^2\plus{}(ca\minus{}ab)^2\ge 0
\]


geometry
parallelogram
ARML
rectangle
analytic geometry

Source: TJ ARML practice

There are $8$ parallel lines in a plane, and another $n$ lines, parallel to each other but not the other $8$ lines. If $420$ parallelograms are formed by their intersection, calculate $n$.

[hide]i made a diagram and solved it the brute force way. i am pretty sure it's 5, but it might be 6. i am about 85% sure it's 5, though.[/hide]

[hide] well in order to get the top and bottom of a parrellellogram its 8C2 or 28 and the same would apply to the n lines to get the left and right sides but since 420 / 28 isnt an integer that doesnt make sense.  So I fail[/hide]

[quote="srulikbd"][hide="not usre"]$(8-1)(n-1)=420$ thus, $n=61$?  :? [/hide][/quote]

don't forget to count the larger parallelograms, like the 2x1 and the 3x1 and stuff.

[quote="dwx314"][hide]When counting rectangles in a grid, you take the lC2 and wC2 and multiply them together.

So wouldn't the parallelograms be the same?

If they are:

7C2 (8 parallel lines) = 28
420/28=15
15= 5C2

>>6<< lines

P.S.: who said that 420/28 wasn't an integer?[/hide][/quote]

nateharman1234 did.

[hide]I also got 6 as well[/hide]

[hide]
The intersections between the lines could kind of be thought of as a coordinate grid. Also each unique paralellogram is defined by a bottom-left and top-right corner. Just say the x-coordinate is on the 8 parallel lines and the y-coordinate is on the n parallel lines.

You need to choose 2 x-coordinates and 2 y-coordinates to define a paralellogram. So, there are $\binom{8}{2} \binom{n}{2} = 28 \cdot \frac{n(n-1)}{2} = 14n(n-1)$ possible sets of points, in other words that many possible sets of parallelograms.

Set it equal to the given and solve:
\begin{eqnarray*} 14n(n-1) &=& 420 \\ n(n-1) &=& 30 \\ n^2-n+\frac{1}{4} &=& 30+\frac{1}{4}=\frac{121}{4} \\ \left(n-\frac{1}{2}\right)^2 &=& \left(\frac{11}{2}\right)^2 \\ n &=& \frac{1}{2} \pm \frac{11}{2} = \{6, -5\} \end{eqnarray*}

The only answer that makes sense is $\boxed{6}$.
[/hide]

woah my bad, the calculator on my computer gave me 420/28 as 52.5 and it kept doing it too just now. It must be messed up, ill have to check what the settings are.


calculus
integration
function
real analysis
real analysis unsolved

This seems easy but I want to make sure my answer's right:

Find all functions on $[0,1]$ such that $f$ is continuous on $[0,1]$, and
$\int_{0}^{x}f(t) dt = \int_{x}^{1}f(t) dt$
for every $x \in (0,1)$.

Differentiate (Fundamental Theorem of Calculus): $f(x)=-f(x),$ so only the zero function.


calculus
integration
trigonometry
limit
function
calculus computations

[quote="Abhishek"]Can someone help me with these integrals ?

1. Integral of sinx/(x^1/2) dx {from 0 to infy.} i.e.  [sinx over Sqrt.x]

2. Integral of cos(x^2) dx  {from 0 to infy.}

3. Integral of sin^5 (x)/x^6 {from 0 to infy.}

I think that the powers are large for the last problem, but I just wanted to know how to approach these problems.[/quote]

[hide]

1) $\int_0^{\infty} \frac{\sin x}{\sqrt{x}} \, dx$

We have that $\int_0^{\infty} \frac{e^{-xt}}{\sqrt{t}} \, dt =\frac{\sqrt{\pi}}{\sqrt{x}}$

Multiplying by $\sin x$ and integrating from $0$ to $\infty$. we have
\[ \int_{0}^{\infty} \frac{\sin x}{\sqrt{x}} \, dx = \frac{1}{\sqrt{\pi}} \int_0^{\infty} \int_0^{\infty} \frac{e^{-xt} \, \sin x}{\sqrt{t}} \, dt \, dx \]

Since $\lim_{x \rightarrow \infty} \left| \frac{e^{-xt} \, \sin x}{\frac{1}{x^2+1}} \right| = 0$ and $\int_0^{\infty}\frac{1}{x^2+1} \, dx$ converges so we have
the integran absolutely converges, and FUbini's theorem can be applied.

so $\int_0^{\infty} \frac{\sin x}{\sqrt{x}} \, dx = \frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{t}} \left(\int_0^{\infty} e^{-xt} \sin x \, dx \right) \, dt$

It can be shown that the integral $\int_0^{\infty} e^{-xt} \, \sin x \, dx=\frac{1}{t^2+1}$

So we have
\begin{eqnarray*} \int_0^{\infty} \frac{\sin x}{\sqrt{x}} \, dx &=& \frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{t}} \frac{1}{t^2+1} \, dt \\ &=& \frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{u^{\frac{-3}{4}}}{2(u+1)} \, du \\ &=& \frac{1}{\sqrt{\pi}} \frac{\pi}{2 \, \sin \frac{\pi}{4}} \\ &=& \frac{\sqrt{\pi}}{2} \sqrt{2} \end{eqnarray*}

2) $\int_0^{\infty} \cos x^2 \, dx = \int_0^{\infty} \frac{\cos u}{\sqrt{u}} \, du$

Using similar argument in part (1) and $\int_0^{\infty}e^{-xt} \cos x \, dx = \frac{t}{t^2+1}$
then
\begin{eqnarray*} \int_0^{\infty} \frac{\cos x}{\sqrt{x}} \, dx &=& \frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{1}{\sqrt{t}} \frac{t}{t^2+1} \, dt \\ &=& \frac{1}{\sqrt{\pi}} \int_0^{\infty} \frac{u^{\frac{-1}{4}}}{2(u+1)} \, du \\ &=& \frac{1}{\sqrt{\pi}} \frac{\pi}{2 \, \sin \frac{3\pi}{4}} \\ &=& \frac{\sqrt{\pi}}{2} \sqrt{2} \end{eqnarray*}

3) The safe method to do part (3) is to reduce the power of sine function and using similar method like part (1). I still don't do it but i thinks its work  :)
[/hide]


probability
algebra
polynomial
quadratics
conics
parabola
system of equations

1. If $ x^3 + x^2 + ax + b$ is divisible by $ x^2 + x + 1$, then what is the value of $ b - a$?

2. Let $ f(x) = ax^2 + bx + c$ and supposed that $ (5,16)$ is the vertex and $ f (1) = 32$. Find the value of c.

3. A regular deck of playing cards consists of four categories called suits. Each suit consists of 13 cards ranked using 3 face symbols (called King, Queen and Jack), a letter A (called Ace) and numbers 2, 3, . . . , 9. Suppose the value of each face card is 10. The value of Ace is 11 and the values of all remaining cards are the numbers they bear. What is the probability of drawing 3 cards without replacement whose sum is less than 12 and where each successive card drawn has a value less than the previous one?

4. How many integer solutions does $ 3x^{2} - y^{2} = 20 - 2xy$ have?

5. Given that the quartic polynomial $ p(x) = x^4 + Ax^3 + Bx^2 + Cx + D$ has repeated non-zero
real roots at a and b. What is the value of $ A^2 - 2B + \frac {C}{b} - \frac {2D}{a^2}$?

6. $ \ln(x^2 + 5x) - \ln(x - a - 3) = 0$ has only one solution for x. Find all value(s) of a.

[hide]2. $ x^{2} \minus{} 10x \plus{} 41$  $ c \equal{} 41$[/hide]

Hint for 2: remember that $ \frac { \minus{} b}{2a} \equal{} 5$ and try solving as a system of equations also remember that $ a \plus{} b \plus{} c \equal{} 32$

[hide="#1?"]i pretty much tried to find the complementary factor. since $ x^2\plus{}x\plus{}1$ divides the expression, then the complementary factor has to be of the form $ (x\plus{}c)$ for some constant c. the product of the two expressions has to be $ x^3\plus{}x^2\plus{}ax\plus{}b$. multiplying it out you get the expression is of the form $ x^3\plus{}x^2\plus{}x\plus{}cx^2\plus{}cx\plus{}c$ and the only time this is true is when $ c\equal{}0$? i think i messed up somewhere. [/hide]

[quote="boo"]1. If $ x^3 + x^2 + ax + b$ is divisible by $ x^2 + x + 1$, then what is the value of $ b - a$?

2. Let $ f(x) = ax^2 + bx + c$ and supposed that $ (5,16)$ is the vertex and $ f (1) = 32$. Find the value of c.

3. A regular deck of playing cards consists of four categories called suits. Each suit consists of 13 cards ranked using 3 face symbols (called King, Queen and Jack), a letter A (called Ace) and numbers 2, 3, . . . , 9. Suppose the value of each face card is 10. The value of Ace is 11 and the values of all remaining cards are the numbers they bear. What is the probability of drawing 3 cards without replacement whose sum is less than 12 and where each successive card drawn has a value less than the previous one?

4. How many integer solutions does $ 3x^{2} - y^{2} = 20 - 2xy$ have?

5. Given that the quartic polynomial $ p(x) = x^4 + Ax^3 + Bx^2 + Cx + D$ has repeated non-zero
real roots at a and b. What is the value of $ A^2 - 2B + \frac {C}{b} - \frac {2D}{a^2}$?

6. $ \ln(x2 + 5x) - \ln(x - a - 3) = 0$ has only one solution for x. Find all value(s) of a.[/quote]
[hide="4"] Flip $ y^2$ and $ 20$ to the other sides and add $ x^2$ to get that $ 4(x^2 - 5) = (x - y)^2$, $ (2x)^2 - (x - y)^2 = 20$, $ (3x - y)(x + y) = 20$.  So, we must factorize 20 into two factors so that their sum is 4x, and thus their sum is divisible by 4.

The only possibilities that work are $ 3x - y = 10$ and $ x + y = 2$, $ 3x - y = 2$ and $ x + y = 10$, $ 3x - y = - 2$ and $ x + y = - 10$, $ 3x - y = - 10$ and $ x + y = - 2$ all work.  We find that the solutions for $ (x, y)$ are $ \boxed{(3, - 1), (3, 7), ( - 3, - 7), ( - 3, 1)}$.  [/hide]

[hide="Number 1"]
By basic divisibility, $ P(x) | Q(x) \implies P(x) | Q(x) \minus{} R(x)P(x)$

Hence, $ x^2\plus{}x\plus{}1$ must divide $ (x^3\plus{}x^2\plus{}ax\plus{}b) \minus{} x(x^2\plus{}x\plus{}1) \equal{} (a\minus{}1)x\plus{}b$

Since the above is a linear, a quadratic cannot always divide it unless the linear is equal to 0.

Hence, $ (a,b) \equal{} (1,0)$, and $ b\minus{}a \equal{} \minus{}1$

[/hide]

[hide="#5"]$ p(x)\equal{}[(x\minus{}a)(x\minus{}b)]^2\equal{}[x^2\minus{}(a\plus{}b)x\plus{}ab]^2$, so after expanding we get $ p(x)\equal{}x^4\minus{}2(a\plus{}b)x^3\plus{}[(a\plus{}b)^2\plus{}2ab]x^2\minus{}2ab(a\plus{}b)x\plus{}a^2b^2$

(I hope my math is correct.)

Now just substitute $ A\equal{}\minus{}2(a\plus{}b), B\equal{}(a\plus{}b)^2\plus{}2ab, C\equal{}\minus{}2ab(a\plus{}b), D\equal{}a^2b^2$ to get $ \minus{}2ab$.[/hide]

[hide="#2"]
For f(1), a+b+c=32.
Since the vertex of a parabola $ ax^2 \plus{} bx \plus{} c$ is at $ (\frac { \minus{} b}{2a},f(\frac { \minus{} b}{2a}))$, we find $ b \equal{} \minus{} 10a$, and $ 25a \plus{} 5b \plus{} c \equal{} 16$. We can solve, and find that $ c \equal{} 41$.[/hide]
For #5, you could also use kth symmetric sums.

[quote="i_like_pie"]6. Should it be $ \ln(x^2 \plus{} 5x) \minus{} \ln(x \minus{} a \minus{} 3) \equal{} 0$?[/quote]
Yes, sorry about that.


Everybody says that there are 2 books explaning how to construct the 257-gon and the 65537-gon (both regular).
How would you do it?  I searched on Google.  Nothing found

http://mathworld.wolfram.com/257-gon.html

could one of the references be it?

funcia who does not own those books :lol:

You can explicitely express the coordinates (when inscribed in the unit circle) in terms of square roots and rational numbers (those expression can be found by stupid calculations of gaussian sums).
And it's clear how to construct rational numbers and square roots.


ratio
calculus
integration
analytic geometry
graphing lines
slope
floor function

OKay, here are my questions.

(1) What is the ratio of $ x$ to $ y$ given that $ 4(5x \plus{} 3y) \equal{} 3( x \plus{} 2y )$?

[hide]$ \frac { \minus{} 6}{17}$[/hide]

(2)What is the sum of all the positive whole number divisors of $ 210$?

[hide]$ 576$[/hide]

(3)What is the largest possible integral value of $ n$ if $ 11^n$ is a factor of $ 100!$?

[hide]$ 9$[/hide]

(4)The sum of a list of positive integers is 20.  What is the maximum possible product of the integers in this list?

[hide]$ 1458$[/hide]

(5) In slope-intercept form, what is the equation of the line $ y\equal{}3x\minus{}3$ after it has been reflected through the origin?

[hide]$ y \equal{} 3x \plus{} 2$[/hide]

[hide="Number 1"]
Distribute to get $ 20x\plus{}12y\equal{}3x\plus{}6y$.  Manipulate the variables to get $ 17x\equal{}\minus{}6y$.  Divide by $ 17$, so $ x\equal{}\boxed{\minus{}\frac{6}{17}y}$, which is the desired ratio.
[/hide]

[hide="Number 3"]
There are $ \left \lfloor \frac{100}{11} \right \rfloor\equal{}9$ numbers that are divisible by $ 11$ less than $ 100$, so $ \boxed{9}$ is the answer.
[/hide]

[hide="Number 4"]
We want as many terms that are close together as possible, barring $ 1$.  If we use $ 2$'s, we have $ 2^10\equal{}1024$.  If we use $ 3$'s and $ 2$'s, we have $ 3^6 \cdot 2\equal{}1458$.  If we use $ 4$'s, we have $ 4^5\equal{}1024$.  If we use $ 5$'s, we have $ 5^4\equal{}625$.  The higher the number gets, the smaller the product gets.  Thus, the largest possible value is $ \boxed{1458}$.
[/hide]

[quote="AIME15"][hide="Number 1"]
Distribute to get $ 20x \plus{} 12y \equal{} 3x \plus{} 6y$.  Manipulate the variables to get $ 17x \equal{} \minus{} 6y$.  Divide by $ 17$, so $ x \equal{} \boxed{ \minus{} \frac {6}{17}y}$, which is the desired ratio.
[/hide][/quote]

Ah....I was doing so it would look like this: $ z\equal{}\frac{x}{y}$.

[hide="Number 2"]
We could find the factors and add 'em up, but that's boring, isn't it? So we'll try to find a better method.

First, prime factorize 210. We get $ 2 \cdot 3 \cdot 5 \cdot 7$. Thus the sum of the factors is simply $ (2^0 \plus{} 2^1)(3^0 \plus{} 3^1)(5^0 \plus{}5^1)(7^0 \plus{} 7^1) \equal{} 576$.
[/hide]

EDIT: I put the 1+5 there, but for some reason it's not there anymore...

[quote="isabella2296"][hide="Number 2"]
We could find the factors and add 'em up, but that's boring, isn't it? So we'll try to find a better method.

First, prime factorize 210. We get $ 2 \cdot 3 \cdot 5 \cdot 7$. Thus the sum of the factors is simply $ (2^0 \plus{} 2^1)(3^0 \plus{} 3^1)(7^0 \plus{} 7^1) \equal{} 576$.
[/hide][/quote]

Actually, we get $ (2^0\plus{}2^1)(3^0\plus{}3^1)(5^0\plus{}5^1)(7^0\plus{}7^1) \equal{} 3 \times 4 \times 6 \times 8 \equal{} 576$.

[quote="ernie"][quote="isabella2296"][hide="Number 2"]
We could find the factors and add 'em up, but that's boring, isn't it? So we'll try to find a better method.

First, prime factorize 210. We get $ 2 \cdot 3 \cdot 5 \cdot 7$. Thus the sum of the factors is simply $ (2^0 \plus{} 2^1)(3^0 \plus{} 3^1)(7^0 \plus{} 7^1) \equal{} 576$.
[/hide][/quote]

Actually, we get $ (2^0 \plus{} 2^1)(3^0 \plus{} 3^1)(5^0 \plus{} 5^1)(7^0 \plus{} 7^1) \equal{} 3 \times 4 \times 6 \times 8 \equal{} 576$.[/quote]

Isn't that the same as izzy's post?  I forgot the way to find the sum using the primes and so I had to use the boring way of finding all the factors and adding them up.  I probably mis calculated or missed a few factors.

Yeah, I don't understand "reflected through the origin".

[quote="AIME15"]But the origin is a point...

Ahh...we're reflecting a line.

Wouldn't it just be y=3x+3, then?[/quote]

Hm....I've tried it, I know the slope is $ 3$.  BTW: Your answer is wrong, its $ \plus{}2$ not $ \plus{}3$.  I think the main problem is that reflecting to the origin sounds weird, AIME's answer is reflecting through the line $ y\equal{}x$.

[quote="aleph0"]No, AIME15's answer was rotating through the origin.[/quote]

So why does the answer say $ 2$?  I see why it would be $ y\equal{}3x\plus{}3$, but that just doesn't correspond to the answer.  (Unless we AoPSers are smarter than the answers)

Reflection through a point is the same as a 180 degree rotation about that point (according to Vol. 1), and under a 180 degree rotation about the origin, $ (x,y)$ is mapped to $ (\minus{}x,\minus{}y)$. Substituting $ \minus{}y$ for $ y$ and $ \minus{}x$ for $ x$, we get $ \minus{}y\equal{}\minus{}3x\minus{}3$, or $ \boxed{y\equal{}3x\plus{}3}$, which still isn't the given answer. :(

[quote="ThinkFlow"]Reflection through a point is the same as a 180 degree rotation about that point (according to Vol. 1), and under a 180 degree rotation about the origin, $ (x,y)$ is mapped to $ ( \minus{} x, \minus{} y)$. Substituting $ \minus{} y$ for $ y$ and $ \minus{} x$ for $ x$, we get $ \minus{} y \equal{} \minus{} 3x \minus{} 3$, or $ \boxed{y \equal{} 3x \plus{} 3}$, which still isn't the given answer. :([/quote]

I'm guessing now that the answer key was wrong.

[quote="AIME15"]What's the source?

Also, recheck it, you might've made a typo.[/quote]

I've checked it a million times.

Source:  When cleaning up my room, I found a stash of papers underneath my bed.  I don't know where they are from....or what they are doing there...

I think number 5 is

$ y \equal{} \minus{}\frac{1}{3}x \plus{} 1$

Because everything is flipped...


$x,y,z\geq 0,x+y+z=1$ Prove that：
$\sum{\sqrt{\frac{1-x}{1+x}}}\leq 1+\frac{2\sqrt{3}}{3}$

The fraction $\sqrt\frac{1-c}{1+c}$  becomes maximum=1  when c=0 .So it suffices to prove that with $a+b=1$
\[ \sqrt{\frac{1-a}{1+a}} + \sqrt{\frac{1-b}{1+b}} \leq \frac{2 \sqrt{3}}{3} .  \]

Using the fact that $a+b=1$, we rewrite the inequality:

\[ \Leftrightarrow \sqrt{\frac{1-a}{1+a}} + \sqrt{\frac{a}{2-a}} \leq \frac{2 \sqrt{3}}{3}  \]

\[ \Leftrightarrow \frac{\sqrt{(1-a)(2-a)}+\sqrt{a(1+a)}}{\sqrt{(1+a)(2-a)}} \leq \frac{2 \sqrt{3}}{3}  \]

\[ \Leftrightarrow \sqrt{(1-a)(2-a)}+\sqrt{a(1+a)} \leq \sqrt{(1+a)(2-a)} \cdot \frac{2 \sqrt{3}}{3}  \]

(all is positive, so square it)

\[ \Leftrightarrow (1-a)(2-a)+a(1+a)+2\sqrt{(1-a)(2-a)a(1+a)} \leq (1+a)(2-a) \cdot \frac{4}{3}  \]

\[ \Leftrightarrow 6 \sqrt{(1-a)(2-a)a(1+a)} \leq 10a(1-a)+2  \]

\[ \Leftrightarrow 3 \sqrt{(1-a)(2-a)a(1+a)} \leq 5a(1-a)+1  \]

(again all is positive, so square again)

\[ \Leftrightarrow 0 \leq 1-8a+24a^2-32a^3+16a^4=(2a-1)^4 ,  \] which is trivially true

Thanks
But I don't know how to prove the sum will become maximum when c=0
yet.
Could you explain it?

We have $\sqrt{\frac{1-c}{1+c}}\leq 1$  and the equality occur when c=0 .

BTW , my solution is the same as here
http://www.mathlinks.ro/Forum/viewtopic.php?p=298435#298435

And there see the explanation of Soarer ;)


There are two urns each containing an arbitrary number of balls. We are allowed two types of operations:
(a) remove an equal number of balls simultaneously from both the urns,
(b) double the number of balls in any of them
Show that after performing these operations finitely many times, both the urns can be made empty.

[hide]
Let the number of balls in urn $A$ be $a$ and in $B$ be $b$ such that $a < b < 2a$. If this isn't true, double the number of balls in $A$ repeatedly until it is.

Let $b = a+k$ where $k < a$. Take $a-k$ balls away from each urn. So $A = a \rightarrow k$ and $B = b \rightarrow 2k$. Then double $A$ to get $A = k \rightarrow 2k$ so then $A$ and $B$ have the same number of balls. Remove $2k$ from each and they both have none.[/hide]

This is somewhat like paladin8's
[hide="An even easier solution"]
let the number of balls in one urn be $a$ and in the other urn, be $b$.

[b]Case 1[/b]: $a=b$
Just subtract a=b from both urns.

[b]Case 2[/b]: $a>b$
Subtract  $2b+a$ from both urns, and then double the number of balls in urn b. Finally, subtract what ever number of balls you have in anyone of the urns (they both have the same number of balls which is 2a-2b)

[b]Case 3[/b]: $a<b$
Very similar to case 2. Subtract  $2a+b$ from both urns, and then double the number of balls in urn a. Finally, subtract what ever number of balls you have in anyone of the urns (they both have the same number of balls which is 2b-2a)

You may be wondering where I got that $2a+b$ or $2b+a$. Well, here is how:
WLOG, assume $a<b$. I want the number in $b$ to be double the number in $a$. So
$2(a-x)=b-x$
or
$x=2a-b$
I just set $a=b$ and $b=a$, to get the answer for the case of $b<a$ which is $x=2b-a$
[/hide]

[quote=kk108][quote=][hide="An easier solution"]
If one urn A has m balls and urn B has n,
WLOG, assume $m > n$
Take out $n - 1$ balls from each urn.
now urn A has $m-n+1$ balls and urn B has 1 ball
[b][u][i]Now, double urn B and remove 1 ball from both
Keep doing this until both have 2[/i][/u][/b]
then remove 2 from each
and we are done![/hide][/quote]

How do you know it will end up in both having 2 ? It's not clear to me .[/quote]

It is very easy to see... At every move, balls in urn B is doubled, then, 1 ball is removed from each urn and this goes on till both urns have 2 balls ( in fact, the first step in the move has to be performed) and here, you go.

[quote=faultlesslyfaulty][hide="An easier solution"]
If one urn A has m balls and urn B has n,
WLOG, assume $m > n$
Take out $n - 1$ balls from each urn.
now urn A has $m-n+1$ balls and urn B has 1 ball
Now, double urn B and remove 1 ball from both
Keep doing this until both have 2
then remove 2 from each
and we are done![/hide][/quote]

In fact, this process is much easier and elegant... But I can't say it as the easiest "cause there may be a solution even more elegant.

[quote=kk108][quote=][hide="An easier solution"]
If one urn A has m balls and urn B has n,
WLOG, assume $m > n$
Take out $n - 1$ balls from each urn.
now urn A has $m-n+1$ balls and urn B has 1 ball
[b][u][i]Now, double urn B and remove 1 ball from both
Keep doing this until both have 2[/i][/u][/b]
then remove 2 from each
and we are done![/hide][/quote]

How do you know it will end up in both having 2 ? It's not clear to me .[/quote]

Still, if it's not clear, then, just carry forward the two steps until both urns have 1 ball each. In that case, generalization clearly holds.