geometry
geometric transformation
dilation

Define $ A$ and $ B$ to be points inside a circle $ C$. Prove that there is a circle that goes through $ A$ and $ B$ and is contained entirely inside $ C$.

I'd appreciate a hint so I can try to solve this

This is actually cool...You can actually to some extent characterize all such circles.

[hide="hint"]Let $ P$ be a point on $ C$.  Let the line through $ P$ parallel to $ AB$ hit the circle again at $ Q$.  Now let $ PA$ and $ BQ$ meet at $ T$.  What can you say about triangles $ PTQ$ and $ ATB$?  Can you make $ T$ be where you want it to be?[/hide]

[hide="A pretty simple solution"]
Let the $ 2$ points be in circle $ C_1$ with center $ O$, we dilate (reduce the scale of) $ C_1$ with the center fixed at $ O$ until it intersects $ 1$ or $ 2$ points. Let $ C_2$ be the dilation of $ C_1$. If $ C_2$ intersects $ 2$ points, then we are done, for the dilated circle $ C_2$ lies within $ C_1$. But if it only intersects one point, then let $ A$ be the point on $ C_2$ and $ B$ inside circle $ C_2$. We set $ C_3$ to be the circle that is internally tangent to $ C_2$ at $ A$. It's obvious that if we dilate (increase scale of) $ C_3$ with the condition of it still being tangent to $ A$, it will intersect $ B$. So $ C_3$ intersects $ A$ and $ B$, and $ C_2$ is inside $ C_1$. We conclude that $ C_3$ is inside $ C_1$, and we are done.
[/hide]


geometry
circumcircle
inradius
perimeter
trigonometry
geometry unsolved

Let R be circumradius, r inradius and s semi-perimeter of triangle [b]ABC[/b].
Prove: If $2R+r=s$ then [b]ABC[/b] is right-angled triangle.

[color=darkblue][b]The first proof.[/b] $\tan \frac{A}{2}=\frac{r}{s-a}$ a.s.o. ; $\sum \cot\frac{A}{2}=\prod\cot\frac{A}{2}=\frac{s}{r}$ and $\sum \cot\frac{A}{2}\cot\frac{B}{2}=\frac{4R+r}{r}$. Thus, the equation $f(t)\equiv rt^{3}-st^{2}+(4R+r)t-s=0$ have the roots $\cot\frac{A}{2}$, $\cot\frac{B}{2}$, $\cot\frac{C}{2}$. But $f(1)=r-s+4R+r-s=2(2R+r-s)$. Therefore, $ABC$ is right-angle triangle $\Longleftrightarrow$ $\frac{\pi}{2}\in \{A,B,C\}$ $\Longleftrightarrow$ $1\in \left\{\cot\frac{A}{2},\cot\frac{B}{2},\cot\frac{C}{2}\right\}$ $\Longleftrightarrow$ $f(1)=0$ $\Longleftrightarrow$ $2R+r=s\ .$
[b]The second proof.[/b]  I'll use the well-known relations $\sin A+\sin B+\sin C=\frac{s}{R}$ and $\cos A+\cos B+\cos C=1+\frac{r}{R}$. Therefore, $\boxed{\ 2R+r=s\ }$ $\Longleftrightarrow$ $2R-a=(s-a)-r$ $\Longleftrightarrow$ $2R(1-\sin A)=r\left(\cot \frac{A}{2}-1\right)$ $\Longleftrightarrow$ $2R\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)^{2}\sin \frac{A}{2}=r\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)$ $\Longleftrightarrow$ $\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)\left[2\sin\frac{A}{2}\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)-\frac{r}{R}\right]=0$ $\Longleftrightarrow$ $\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)\left[\sin A+\cos A-\left(1+\frac{r}{R}\right)\right]=0$ $\Longleftrightarrow$ $\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)\left[\sin A+\cos A-(\cos A+\cos B+\cos C)\right]=0$ $\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)[\sin A-(\cos B+\cos C)]=0$ $\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)\left(2\sin \frac{A}{2}\cos\frac{A}{2}-2\sin\frac{A}{2}\cos\frac{B-C}{2}\right)=0$ $\Longleftrightarrow$ $\left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)\left(\cos\frac{B+C}{2}-\cos\frac{B-C}{2}\right)=0$ $\Longleftrightarrow$ $\boxed{\ \left(\cos\frac{A}{2}-\sin\frac{A}{2}\right)\left(\cos\frac{B}{2}-\sin\frac{B}{2}\right)\left(\cos \frac{C}{2}-\sin\frac{C}{2}\right)=0\ }$ $\Longleftrightarrow$ $\boxed{\ \prod \left(1-\tan\frac{A}{2}\right)=0\ }$ $\Longleftrightarrow$ $A=\frac{\pi}{2}$ or $B=\frac{\pi}{2}$ or $C=\frac{\pi}{2}$ $\Longleftrightarrow$ $\boxed{\ \frac{\pi}{2}\in \{A,B,C\}\ }\ .$
[b]Remark.[/b] $\boxed{\ 2R+r=s\ }$ $\Longleftrightarrow$ $1+\left(1+\frac{r}{R}\right)=\frac{s}{R}$ $\Longleftrightarrow$ $\boxed{\ 1+\cos A+\cos B+\cos C=\sin A+\sin B+\sin C\ }\ .$[/color]

Thanks, man.  Good use of trig identities and Viet's formulas.


Kent Merryfield and I were just discussing over PM the following difficulty: it seems to be impossible to search for threads with the word "L'Hopital" (or "L'Hospital," or "L'Hôpital") -- I guess that the apostrophe fouls things up.  Is there any workaround for this?  (It doesn't seem to be possible using the asterisk, for example, which doesn't appear to work as it should in the search.)

Also, does the search engine recognize letters like "ô"?

Search is based on MySQL FULLTEXT, so whatever works with that should work here. I was able to find L'Hopital without problem. Can you give the exact search you are using?

I think that this thread describes the problem:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=288713[/url]
EDIT: never mind. I was confused. Ignore this. :blush:

Searching for "L'Hopital" with no author in the Calculus Computations and Tutorials forum returned 2 results.

Searching for "L'Hospital" with no author in that same forum returned 4 posts.

Seraching for "L'Hôpital" with no author in that same forum returned 0 posts.

Searching for any of these globally with Kent Merryfield as author returned 0 posts.

That is not plausible. Each of these searches should return pages and pages of posts.

I was searching for title or contents - the default inclusive setting.


modular arithmetic
number theory solved
number theory

Given a random number square. Prove that its last to digits multiple is an even number
My english is bad...  :blush:
$x^2=\overline{a_na_{n-1}a_{n-2}...a_1a_0}$
Prove that:
$a_1*a_0=2k$ for every natural $x$
Hope now you understand the problem

uh is it true guys?
If $a_0$ is even ==> done  :D
If not then from $x^2\equiv1\pmod{8}$ then $\overline{a_1a_0}\equiv1\pmod{8}$. This implies $a_0=1,9$. Both implies $\overline{a_11}\equiv1\pmod{8}$. So $10a_1$ is divisible by $8$ so $a_1$ must be even!  :!:

[quote="pluricomplex"]This implies $a_0=1,9$. Both implies $\overline{a_11}\equiv1\pmod{8}$. So $10a_1$ is divisible by $8$ so $a_1$ must be even!  :!:[/quote]
Wrong!!!

Anyway, this problem is just a checking of finite cases. To see why the product of the tens digit and the units digit of a perfect square is always even, we let the number be ${(10a+b)}^2 = 100a^2 + 20ab + b^2$, (so, here $b$ is our units digit), taking $\mod 100$ and checking the five cases $b = 1, 3, 5, 7, 9$, we are done.

Is this the only way of solving problems like this one... i mean with $mod$ or there are others ??? BTW dblues thanks for your posts in the other perfect square problems i have posted  :D

man i don't think it's wrong completely. I just forgot the case $a_0=5$. But in that cases $a_1=2$ which is surely even :D

By considering the last $2$ digits, you are only ensured that $\overline{a_1a_0} \equiv 1 \pmod{4}$, since $8 \nmid 100$. For example, for $13^2 = 169$, we have $69 \equiv 5 \pmod{8}$. Thus, we only know that for odd $x$, we have $\overline{a_1a_0} \equiv 1 \pmod{4}$, and not $\overline{a_1a_0} \equiv 1 \pmod{8}$

You also said that "This implies $a_0 = 1$ or $9$", but this is not true as it doesn't follow from the fact that $\overline{a_1a_0} \equiv 1 \pmod{4}$. Rather, you should say instead that since for odd perfect squares, the units digit must be $1, 5$, or $9$, thus we have (by your line of argument) that $\overline{a_11} \equiv 1 \pmod{4}$, which implies that $10a_1 \equiv 0 \pmod{4}$, and so we have $a_1$ is even as desired.


geometry
3D geometry
sphere
trigonometry
Pythagorean Theorem
trig identities
Law of Cosines

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon.  A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres.  The radius of this last sphere is $a+b\sqrt{c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime.  Find $a+b+c.$

cool.  might have made an arithmetic error, but I'm pretty sure its the right idea.

[hide]Cut a horizontal cross section through the centers of the 8 sphere's to find the horizontal distance to the center of the 9th sphere.  Regular octogon $\Rightarrow \frac{200+2(\frac{200}{\sqrt2})}{2} = 100 + 100\sqrt2$.

Then cut a vertical cross section through two sphere centers on opposite sides of the octogon and the center of sphere #9.  Draw the picture and its pretty clear the Pythagorean Theorem will finish it off.
\[ (r-100)^2 + (100+100\sqrt2)^2 = (r+100)^2 \]
I'm getting $r = 75+50\sqrt2 \Longrightarrow 127$?[/hide]

[hide="I think you made an error with the octagons diagonal, or the way you substituted."][/hide]

[hide="Method:"][/hide]

[hide]

We can set half the length of the octagon's diagonal as $n$.

So by Law of Cosines, $200^2 = n^2 + n^2 - 2n^2 \cos{45} \Rightarrow n^2 = \frac{200^2}{2-\sqrt{2}}$.

We then know by the Pythagorean theorem that:

$(R - 100)^2 + n^2 = (R + 100)^2$

And substituting,

$R = 100 + 50\sqrt{2}$

So the answer is $152$.[/hide]

[quote="white_horse_king88"][hide="I think you made an error with the octagons diagonal, or the way you substituted."][/hide]

[hide="Method:"][/hide]

[hide]

We can set half the length of the octagon's diagonal as $ n$.

So by Law of Cosines, $ 200^2 \equal{} n^2 \plus{} n^2 \minus{} 2n^2 \cos{45} \Rightarrow n^2 \equal{} \frac {200^2}{2 \minus{} \sqrt {2}}$.

We then know by the Pythagorean theorem that:

$ (R \minus{} 100)^2 \plus{} n^2 \equal{} (R \plus{} 100)^2$

And substituting,

$ R \equal{} 100 \plus{} 50\sqrt {2}$

So the answer is $ 152$.[/hide][/quote]

How did you make the right triangle?

[quote=IW@IT]can anyone give the picture of this problem??

Thank you[/quote]

Hey, obviously this is much much later but I really struggle to visualize and I also would like at least an attempt at an image, can anyone provide?

It's 1998 AIME problem 10. There are a few diagrams in the solutions.


"An exam has 30 multiple-choice problems. A contestant who answers m questions correctly and n incorrectly (and does not answer 30 - m - n questions) gets a score of 30 + 4m - n. A contestant scores N > 80. A knowledge of N is sufficient to deduce how many questions the contestant scored correctly. That is not true for any score M satisfying 80 < M < N. Find N."

I know that for 30-k questions correct, the possible scores are:

150-5k, 150-5k+1,..., 150-4k.

Where does I go from here?


let $ S \equal{} \{ 1,2,\ldots ,2008 \}$ , find the minimum number of elements $ n$ in a subset of $ S$ so that in every $ n$ element subset of $ S$ exits two numbers $ a$ , $ b$ ($ a$ can be equal to $ b$) such that $ a \plus{} b$ is a power of $ 2$.


Hi Everyone, I have a problem that I thought maybe all of you could help me with!

I have a sentence that is as follows:

"For this specific case study we will assume that the solution is being developed on Windows 2003 with SharePoint which runs on ASP.NET.  Everything that is discussed in this report has an equivalent construct on most platforms. "

When I look at the document it is only putting a single space after the ASP.NET.  I think this is because it thinks that it is another sentence or something.  How do I stop it from doing that?

Thanks a lot!
Jim

I think this is what you want.  I took it from "The not so short introduction to $\LaTeXe$" which is available for free on internet.

[quote]2.6 The Space Between Words
To get a straight right margin in the output, LATEX inserts varying amounts
of space between the words. It inserts slightly more space at the end of a
sentence, as this makes the text more readable. LATEX assumes that sentences
end with periods, question marks or exclamation marks. If a period follows
an uppercase letter, this is not taken as a sentence ending, since periods after
uppercase letters normally occur in abbreviations.
Any exception from these assumptions has to be specified by the author.
A backslash in front of a space generates a space that will not be enlarged. A
tilde ‘~’ character generates a space that cannot be enlarged and additionally
prohibits a line break. The command \@ in front of a period specifies that
this period terminates a sentence even when it follows an uppercase letter.

Mr.~Smith was happy to see her\\
cf.~Fig.~5\\
I like BASIC\@. What about you?

The additional space after periods can be disabled with the command
\frenchspacing
which tells LATEX not to insert more space after a period than after ordinary
character. This is very common in non-English languages, except bibliographies.
If you use \frenchspacing, the command \@ is not necessary.[/quote]

ledurt

$\begin{array}{c}
\begin picture
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x from 0 to 200, y from 0 to 100
\circ ulararc 360 degrees from 50 0 center at 50 50
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axes ratio 3:1 180 degrees from 0 50 center at 50 50
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<3pt>
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axes ratio 3:1 180 degrees from 100 50 center at 50 50
%Error. "setsolid" is a bad command.
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<5pt> [.5,1] from 110 50 to 130 50
%Error. "plot" is a bad command.
140 30 140 70 /
\circ ulararc 360 degrees from 140 0 center at 140 15
%Error. "ellipticalarc" is a bad command.
axes ratio 3:1 180 degrees from 125 15 center at 140 15
%Error. "setdots" is a bad command.
<3pt>
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axes ratio 3:1 180 degrees from 155 15 center at 140 15
%Error. "setsolid" is a bad command.
\circ ulararc 360 degrees from 140 70 center at 140 85
%Error. "ellipticalarc" is a bad command.
axes ratio 3:1 180 degrees from 125 85 center at 140 85
%Error. "setdots" is a bad command.
<3pt>
%Error. "ellipticalarc" is a bad command.
axes ratio 3:1 180 degrees from 155 85 center at 140 85
%Error. "setsolid" is a bad command.
%Error. "arrow" is a bad command.
<5pt> [.5,1] from 155 7.5 to 170 12
%Error. "arrow" is a bad command.
<5pt> [.5,1] from 155 22.5 to 170 18
%Error. "arrow" is a bad command.
<5pt> [.5,1] from 155 77.5 to 170 82
%Error. "arrow" is a bad command.
<5pt> [.5,1] from 155 92.5 to 170 88
%Error. "arrow" is a bad command.
<5pt> [.5,1] from 145 50 to 160 50
\put {} at 180 15
\put {} at 190 85
\put {} at 170 50
%Error. "endpicture" is a bad command.
\\
\mbox{ Figure \#2}\\
\end{array}$


Hello to all members of the AMC forum.  Your new moderator is WarpedKlown1335.  If you wish to ever contact me, my AIM screen name is WarpedKlown51435.  I also post challenge problems on my away messages.  Since this forum is lacking in problems, maybe I will post some soon, but I have soccer tryouts everyday after school so that may be tough.  I'll try my best though.  Good-bye.

Uhh...could you put this in one of the announcements?  We have a lot of clutter at the top of the forum.

Goodbye, announcement.  Too much clutter.


algebra
polynomial
quadratics
algebra proposed

Here is a very beautiful problem:
  If $p$ is a real polynomial that is nonnegative on [-1,1], then we can write $p(x)=A^2(x)+(1-x^2)B^2(x)$ for two real polynomials A,B.

0. Well known trick $(A^2+(1-x^2)B^2)(X^2+(1-x^2)Y^2)=U^2+(1-x^2)V^2$.

1. Consider all pairs of complex conjugated roots and join them into quadratic trinomials of form $q(x)=x^2+2b\alpha x +b^2$, where $|\alpha<1|$. Then we can represent $q(x)=c(x+d)^2+(1-x^2)(c-1)$ for some $c\geq 1$ and $d$.

2. For $x+b$, $b>1$, there is representation $c(x+d)^2+(1-x^2)c$.

3. The same for $-x+b$, $b>1$.

4. A case chasing with real roots, lying outside of $(-1,1)$.

I am interested in your approach.

As usual, I will dissapoint you.  ;) It is exactly the approach I found.

I think this problem is one of the most liked by Moubi  :D

Pierre.

The first one I have seen a similar problem like this one, it was in 1985 at LLG
$P$ is real positive polynomial then $P$ is a sum of two square of real polynomial.

Well, Moubi, you won't believe it, but it is still much used at LLG. I've seen it given to my group collegues in at least two oral examinations. Yet, time passes and here is what I get in 2005 (second problem of the examination):
    Consider polynomials $P,Q$ with simple roots, all real and such that between any two roots of P (strictly) there is at least a root of Q and vice-versa.  Prove that if a,b are not both zero, the polynomial aP+bQ also has simple real roots. Study the reciiprocal assertion. Nice, isn't it?


If $3 = k \cdot 2^r$ and $15 = k \cdot 4^r$, then $r =$

$\text{(A)}\ - \log_2 5 \qquad \text{(B)}\ \log_5 2 \qquad \text{(C)}\ \log_{10} 5 \qquad \text{(D)}\ \log_2 5 \qquad \text{(E)}\ \displaystyle \frac{5}{2}$

[hide]From $15 = k \cdot 4^r$ we obtain $3 \cdot 5 = k \cdot 2^{2r}$. Thus $3 = \frac{k \cdot 2^{2r}}{5} = k \cdot 2^r$. Cancelling $k$ from both sides and multiplying both sides by $5$, $2^{2r}= ({2^r})^2 = 5 \cdot 2^r$. As $2^r \neq 0$ we can cancel $2^r$ from both sides, obtaining $2^r = 5$.

Thus $r = \log_2 {5}$ and the answer is $D$.[/hide]

[hide="Answer"]$\log 3=\log k+r\log 2$
$\log 15=\log 3+\log 5=\log k+2r\log 2$
$\log 5=r\log 2\Rightarrow r=\frac{\log 5}{\log 2}=\log_25\Rightarrow \boxed{D}$[/hide]

[hide]Take the second equation and divide it by the first, giving
$5=2^r$
$\Rightarrow r=\log_2{5}$ $\boxed{D}$[/hide]

Yeah, it is.. didn't see it at first tho... ;)


Let $ ABC$ be an acute-angled triangle. $ C_{1}$ and $ C_{2}$ are two circles of diameters $ AB$ and $ AC$, respectively. $ C_{2}$ and $ AB$ intersect again at $ F$, and $ C_{1}$ and $ AC$ intersect again at $ E$. Also,  $ BE$ meets $ C_{2}$ at $ P$ and $ CF$ meets $ C_{1}$ at $ Q$. Prove that $ AP=AQ$.

[b]Proof.[/b] $ \left\{\begin{array}{c}PA^{2}=AC\cdot AE=bc\cdot\cos A\\\ QA^{2}=AB\cdot AF=cb\cdot \cos A\end{array}\right\|\implies PA=QA=\sqrt{\frac{b^{2}+c^{2}-a^{2}}{2}}$.

[quote="Virgil Nicula"][b]Proof.[/b] $ \left\{\begin{array}{c}PA^{2} \equal{} AC\cdot AE \equal{} bc\cdot\cos A \\
\ QA^{2} \equal{} AB\cdot AF \equal{} cb\cdot \cos A\end{array}\right\|\implies PA \equal{} QA \equal{} \sqrt {\frac {b^{2} \plus{} c^{2} \minus{} a^{2}}{2}}$.[/quote]

Why is it exactly that $ PA^2 \equal{} AC \cdot AE$? I understand the rest but just not that part.

[quote="modularmarc101"]

Why is it exactly that $ PA^2 \equal{} AC \cdot AE$? I understand the rest but just not that part.[/quote]

 [geogebra]2806dcaf8f8f8944f26801071af9694516c9a14f[/geogebra]

Since $ PE$ is the altitude in right $ \triangle APC$, so by similarity $ \frac {PA}{PC} \equal{} \frac {AE}{PA}$ so $ PA^2 \equal{} AC \cdot AE$.

It's easy to see and worth mentioning that, $E,F$ are nothing but the foot of perpendiculars from $B,C$, Now, Let $D$ be the foot of altitude from $A$, and Construct a circle $\omega$ with radius $\sqrt{AH \cdot AD}$
[b]Lemma:[/b] Let $\omega \cap \odot (ABC)=X,Y$, then, $X,Y$ lie on $EF$
[b]Proof:[/b] Perform $\Phi$ (inversion) around $\omega$, and note that $A,X,Y$ remain invariant under $\Phi$, of course, $\Phi : B \longleftrightarrow F$ $ ; C \longleftrightarrow E$ which implies $XYCB$ is cyclic $\implies $ $X,Y \in EF$ $\qquad \blacksquare$

Easy to see, that both, $C_1,C_2$ pass through $D$ in the original diagram, $\Phi :$ $D$ $\mapsto$ $H$, $\longrightarrow \text{Orthocenter}$ in $\Delta ABC$,
Note that $\Phi : C_1 \mapsto CF , C_2 \mapsto BE$, Denote $P^*,Q^*$ as the image of $P,Q$ under $\Phi$, then, notice, $Q$ already lies on $C_1 $ and $CF$, and $Q^*$ will also lie on the intersection of $C_1$ and $CF$ which is $Q$, hence, $P \equiv P^*$ and $Q \equiv Q^*$, hence, $P,Q$ lie on $\omega$ $\implies$ $AP=AQ$


Let $ ABC$ be a triangle and let $ P\in (AB)$ , $ Q\in (AC)$ be two fixed points. For a mobile point $ M\in (PQ)$ denote

 $ \{\begin{array}{c} E\in CM\cap AB \\
\ F\in BM\cap AC\end{array}$ . Prove that $ \frac {QC}{QA}\cdot\frac {FA}{FC} + \frac {PB}{PA}\cdot \frac {EA}{EB} = 1$ ([b]constant[/b]) and the area $ [AEF]$ is maximum iff $ EF\parallel PQ$ .

[b]Remark.[/b] See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=197178]here[/url] a particular case of this problem !

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle and let $ P\in (AB)$ , $ Q\in (AC)$ be two fixed points. For a mobile point $ M\in (PQ)$ denote

 $ \{\begin{array}{c} E\in CM\cap AB \\
\ F\in BM\cap AC\end{array}$ . Prove that $ \frac {QC}{QA}\cdot\frac {FA}{FC} + \frac {PB}{PA}\cdot \frac {EA}{EB} = 1$ ([b]constant[/b]) and the area $ [AEF]$ is maximum iff $ EF\parallel PQ$ .[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the intersection $ N\in AM\cap BC$ .Apply the [b]Menelaus' theorem[/b] to the transversals :

$ \{\begin{array}{cccc} \overline {CME}\ ,\ \triangle ABN\ : & \frac {CN}{CB}\cdot\frac {EB}{EA}\cdot\frac {MA}{MN} = 1 & \Longrightarrow & \frac {NC}{BC} = \frac {EA}{EB}\cdot\frac {MN}{MA} \\
 \\
\overline {BMF}\ ,\ \triangle ACN\ : \ & \frac {BN}{BC}\cdot\frac {FC}{FA}\cdot\frac {MA}{MN} = 1 & \Longrightarrow & \frac {NB}{BC} = \frac {FA}{FC}\cdot\frac {MN}{MA}\end{array}\ \ (*)$

From the well-known property $ \frac {PB}{PA}\cdot NC + \frac {QC}{QA}\cdot NB = \frac {MN}{MA}\cdot BC$ obtain $ \frac {PB}{PA}\cdot\frac {NC}{BC} + \frac {QC}{QA}\cdot\frac {NB}{BC} = \frac {MN}{MA}$ .

Using the relations $ (*)$ obtain $ \frac {PB}{PA}\cdot\frac {EA}{EB}\cdot\frac {MN}{MA} + \frac {QC}{QA}\cdot \frac {FA}{FC}\cdot \frac {MN}{MA} = \frac {MN}{MA}$ , i.e. $ \boxed {\ \frac {PB}{PA}\cdot\frac {EA}{EB} + \frac {QC}{QA}\cdot \frac {FA}{FC} = 1\ }$ .

The second question of this problem is more difficult ! But the upper identity is essentially.[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle and let $ P\in (AB)$ , $ Q\in (AC)$ be two fixed points. For a mobile point $ M\in (PQ)$ denote

 $ \begin{array}{c} E\in CM\cap AB \\
\ \ F\in BM\cap AC\end{array}$ . Prove that $ \frac {QC}{QA}\cdot\frac {FA}{FC} + \frac {PB}{PA}\cdot \frac {EA}{EB} = 1$ ([b]constant[/b]) and $ [AEF]$ is maximum iff $ EF\parallel PQ$ .[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b]

$ \blacktriangleleft 1\blacktriangleright$ Denote the intersection $ N\in AM\cap BC$ . Apply the [b]Menelaus' theorem[/b] to the transversals :

$ \{\begin{array}{cccc} \overline {CME}\ ,\ \triangle ABN\ : & \frac {CN}{CB}\cdot\frac {EB}{EA}\cdot\frac {MA}{MN} = 1 & \Longrightarrow & \frac {NC}{BC} = \frac {EA}{EB}\cdot\frac {MN}{MA} \\
 \\
\overline {BMF}\ ,\ \triangle ACN\ : \ & \frac {BN}{BC}\cdot\frac {FC}{FA}\cdot\frac {MA}{MN} = 1 & \Longrightarrow & \frac {NB}{BC} = \frac {FA}{FC}\cdot\frac {MN}{MA}\end{array}\ \ (*)$

From the well-known property $ \frac {PB}{PA}\cdot NC + \frac {QC}{QA}\cdot NB = \frac {MN}{MA}\cdot BC$ obtain $ \frac {PB}{PA}\cdot\frac {NC}{BC} + \frac {QC}{QA}\cdot\frac {NB}{BC} = \frac {MN}{MA}$ .

Using the relations $ (*)$ obtain $ \frac {PB}{PA}\cdot\frac {EA}{EB}\cdot\frac {MN}{MA} + \frac {QC}{QA}\cdot \frac {FA}{FC}\cdot \frac {MN}{MA} = \frac {MN}{MA}$ , i.e. $ \boxed {\ \frac {PB}{PA}\cdot\frac {EA}{EB} + \frac {QC}{QA}\cdot \frac {FA}{FC} = 1\ }\ (1)$ .

$ \blacktriangleleft 2\blacktriangleright$ Denote $ \{\begin {array}{ccc} a = \frac {PB}{PA} & , & b = \frac {QC}{QA} \\
 \\
x = \frac {EA}{AB} & , & y = \frac {FA}{AC}\end{array}$ . Thus, $ \{\begin{array}{c} \frac {EA}{EB} = \frac {x}{1 - x} \\
 \\
\frac {FA}{FC} = \frac {y}{1 - y}\end{array}$ and the relation $ (1)$ becomes $ a\cdot\frac {x}{1 - x} + b\cdot\frac {y}{1 - y} = 1$ ,

where $ \{x,y\}\subset (0,1)$ . The area $ [AEF]$ is maximum $ \Longleftrightarrow$ $ \frac {[AEF]}{[ABC]}$ is maximum $ \Longleftrightarrow$ $ \frac {EA}{AB}\cdot\frac {FA}{AC}$ is maximum $ \Longleftrightarrow$

$ xy$ is maximum. Using the proposed extremum problem from[/color] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=197263][size=117][color=red][b]here[/b][/color][/size][/url] [color=darkblue]the maximum is touched iff $ \frac {x}{b + 1} = \frac {y}{a + 1}$ $ \Longleftrightarrow$

$ \frac {\frac {EA}{AB}}{\frac {QC}{QA} + 1} = \frac {\frac {FA}{AC}}{\frac {PB}{PA} + 1}$ $ \Longleftrightarrow$ $ \frac {EA}{AB}\cdot\frac {QA}{AC} = \frac {FA}{AC}\cdot\frac {PA}{AB}$ $ \Longleftrightarrow$ $ \frac {AE}{AF} = \frac {AP}{AQ}$ $ \Longleftrightarrow$ $ \boxed {\ EF\parallel FQ\ }$ .

$ \blacktriangleleft 3 \blacktriangleright$ Now I"ll ascertain the position of the point $ M$ for which the area $ [AEF]$ is maximum.

Suppose $ EF\parallel PQ$ , i.e. $ \frac {EA}{EP} = \frac {FA}{FQ}$ . Apply the [b]Menelaus' theorem[/b] in $ \triangle APQ$ to the transversals :

$ \{\begin{array}{cc} \overline {BMF}\ : & \frac {BP}{BA}\cdot\frac {FA}{FQ}\cdot\frac {MQ}{MP} = 1 \\
 \\
\overline {CMP}\ : & \frac {CA}{CQ}\cdot\frac {MQ}{MP}\cdot\frac {EP}{EA} = 1\end{array}\ \|\ \bigodot\ \Longrightarrow$ $ (\frac {MP}{MQ})^2 = \frac {BP}{BA}\cdot\frac {CA}{CQ}$ $ \implies$ $ \frac {MP}{MQ} = \sqrt {\frac {a(b + 1)}{b(a + 1)}}$ .

Denote the points $ \{\begin{array}{c} S\in PQ\cap BC \\
 \\
R\in PQ\ ,\ SR = SM\end{array}$ . Apply [b]Menelaus' theorem[/b] in $ \triangle APQ$ to the transversal $ \overline {SBC}$ :

$ \frac {SP}{SQ}\cdot\frac {CQ}{CA}\cdot\frac {BA}{BP} = 1$ $ \implies$  $ \frac {SP}{SQ} = \frac {BP}{BA}\cdot\frac {CA}{CQ}$ , i.e. $ \boxed {\ \frac {SP}{SQ} = (\frac {MP}{MQ})^2\ }$ . The last relation means that

the points $ M$ , $ R$ are conjugate harmonically w.r.t. the points $ P$ , $ Q$ i.e. $ \frac {MP}{MQ} = \frac {RP}{RQ}$ $ \Longleftrightarrow$[/color] [url=http://www.mathlinks.ro/viewtopic.php?search_id=2076928253&t=46146][size=117][color=red][b]( see [u]lemma[/u] from here !)[/b][/color][/size][/url]

[color=darkblue]$ \boxed {\ SM^2 = SP\cdot SQ\ }$ $ \Longleftrightarrow$ $ PS\cdot PQ = PM\cdot PR$ $ \Longleftrightarrow$ $ PM = \frac {PS\cdot PQ}{PR}$ . [b]The construct results from these ![/b]

[b][u]Remark.[/u][/b] Otherwise, denote $ T\in EF\cap BC$ , $ L\in EF\cap AS$ , $ G\in EF\cap AM$ . If $ EF\parallel PQ$ , then $ T\in AR$ , $ LG = LT$ .

From a well-known property that $ T$ , $ G$ are conjugate w.r.t. $ E$ , $ F$ obtain that the division $ (P,Q,M,R)$ is harmonically a.s.o. [/color]


$P$ is an inner point of triangle $ABC$.$K, L, M$ are points of intersection  $CP, AP,BP$ respectively with sides:$AB, BC, CA$. Proove that if it is possible to inscribe circle into $AKPM$ and $KBLP$, then we can also inscribe circle into $PLCM$

can anybody help?

This is a difficlut problem but it becomes very easy if you know this lemma:
[hide="This is the lemma"]
Let be given a quadrilateral $ABCD$.Two line $AB,CD$ intersect at $E$ and two line $BC,AD$ intersect at $F$.Prove that:it is possible to inscribe circle into $ABCD$ if and only if $AE+CF=AF+CE$[/hide]

It reminds me a problem from an old  Turkish Olympiad , if  I'm not mistaken.

 [u]Babis[/u]


Consider the set $A$ containing all positive integers whose decimal expansion contains no $0$, and whose sum $S(N)$ of the digits divides $N$.

(a) Prove that there exist infinitely many elements in $A$ whose decimal expansion contains each digit the same number of times as each other digit.

(b) Explain that for each positive integer $k$ there exist an element in $A$ having exactly $k$ digits.

Both problems use the same fact that divisibility of $n$ by $2^{m}$ or $5^{m}$ depends only on the last $m$ digits of $n$.


[quote="delegat"]If a, b and c are positive real numbers with sum 3 prove the inequality
a/ab+1 + b/bc+1 + c/ca+1>=3/2 :roll:[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%22A&t=21689


In a bag I have the letters aabbcde.  I choose exactly two vowels and two consonants.  If the a's and b's are indistinguishable from eachother, how many differnet combination of 4 letters can I create?

oops...my bad changed thanks..

[quote="madskillz1"]In a bag I have the letters aabbcde.  I choose exactly two [b]vowels[/b] and two consonants.  If the a's and b's are indistinguishable from eachother, how many differnet combination of 4 letters can I create?[/quote]

Edited for correctness  :D
[hide="Solution"]
Only one way to get both the vowels and both the consonants the same, of which there are $\frac{4!}{2!2!}=6$ combinations.
There are three ways to get both vowels the same and both consonants different of which are $\frac{4!}{2!}=12$ combinations of each. There is another way with both consonants same but with different vowels and there are again $\frac{4!}{2!}=12$ combinations.
There are three ways where both the vowels and consonants are different and there are $4!=24$ combinations of each.
$6+(3+1)(12)+(3)(24)=6+48+72=126$.
[hide="Final Answer"]
$126$
[/hide][/hide]

Thanks for the solution but I have another quetison

How do you know how many regular combinations you can create when order does not matter

I can list them out like

aabb
aabc
aabd
aacd
aebb
aebc
aebd
aecd

thats 8, but whats the non brute force way to count the 8 combinations without order


Rhombus $ ABCD$, with a side length $ 6$, is rolled to form a cylinder of volume $ 6$ by taping $ \overline{AB}$ to $ \overline{DC}.$ What is $ \sin(\angle ABC)$?

$ \textbf{(A)}\ \frac {\pi}{9} \qquad \textbf{(B)}\ \frac {1}{2} \qquad \textbf{(C)}\ \frac {\pi}{6} \qquad \textbf{(D)}\ \frac {\pi}{4} \qquad \textbf{(E)}\ \frac {\sqrt3}{2}$

[hide]
Draw an altitude from A to BC. Call this altitude h. It is also the height of the cylinder. We know that $\sin(\angle ABC)=\frac{h}{6}$ which mean that $h=6\sin(\angle ABC)$

The circumfrence of the cylinder is the side length, which is 6. That means the radius is $\frac{6}{2\pi}=\frac{3}{\pi}$. Then the area is $\pi\left(\frac{3}{\pi}\right)^{2}=\pi\left(\frac{9}{\pi^{2}}\right)=\frac{9}{\pi}$

The volume of a cylinder is the area of the base times the height, so:
$h\left(\frac{9}{\pi}\right)=6$
$6\sin(\angle ABC)\left(\frac{9}{\pi}\right)=6$
$\sin(\angle ABC)\left(\frac{9}{\pi}\right)=1$
$\sin(\angle ABC)=\boxed{\frac{\pi}{9}}$
Which is choice A.
[/hide]

tell you people the truth, I drew the diagram out and cutted it in order to visualize it, took me 10 minutes -.-

[quote="Altheman"]i ripped out a piece of paper and folded a rhombus...everybody around me was like wtf[/quote]

 haha exactly the same thing with me.. i ripped a paper out when i was so desperate for an answer

I was lucky I didn't have to do it. I was thinking of it... but here's my solution.

[hide="solution"]The base length of the rhombus is 6, so the circumfrence of the circle is 6. Therefore, the radius is $\frac{3}{\pi}$. We square that and multiply by $\pi$:

$\frac{9}{\pi}$.

Therefore,

$\frac{9}{\pi}\times h =6$

$h=\frac{2\pi}{3}$

So we have a right triangle with two of the sides $\frac{2\pi}{3}$ and 6. So the sine of that angle in mind is $\frac{\frac{2\pi}{3}}{6}=\frac{\pi}{9}$[/hide]

@tjhance: We both have different solutions from the book, but we both have the same correct answer. :)


induction
inequalities
number theory
relatively prime

[quote="N.T.TUAN"]Find all natural numbers n < 1978 with the following property: If m is a natural number, 1 < m < n, and (m, n) = 1 (i.e., m and n are relatively prime), then m is a prime number.[/quote]

Lemma: $ n$ has to be divisible by each prime number $ \le\sqrt {n}.$
Proof: Suppose $ p^2 < n,$ then $ (n, p^2)\neq 1$ which implies that $ p|n.$

So we would need to find all natural numbers $ k$ such that $ \prod_{i \equal{} 1}^k p_i \le p_{k\plus{}1}^2$ ...

[quote="bertram"][quote="N.T.TUAN"]Find all natural numbers n < 1978 with the following property: If m is a natural number, 1 < m < n, and (m, n) = 1 (i.e., m and n are relatively prime), then m is a prime number.[/quote]

Lemma: $ n$ has to be divisible by each prime number $ \le\sqrt {n}.$
Proof: Suppose $ p^2 < n,$ then $ (n, p^2)\neq 1$ which implies that $ p|n.$

So we would need to find all natural numbers $ k$ such that $ \prod_{i \equal{} 1}^k p_i \le k^2$ and I'd say that only $ k \equal{} 2,3$ satisfy this.[/quote]

Don't 6,8,12,18 (to name a few) satisfy the given property? :huh:
(for n=12, we want m<12 and (m,n)=1 so m=5,7,11, which are all prime...)

We want to find all $ n$ wich have no composite relative primes, this means that if a prime $ p<\sqrt{n}$ doesn't divide $ n$ then for all $ pk<n$(k positive integer) $ k$ and $ k$ have a common factor greater than one.

 :)

The problem is well-known.  [hide="The numbers in question"] $ 2, 3, 4, 6, 8, 12, 18, 24, 30$.  It's not hard to prove by induction that once $ n > 49$ we have $ \prod_{i\equal{}1}^{k} p_i > p_{k\plus{}1}^2$ as per bertram's lemma. [/hide]

Yeah, the inductive proof is very easy if you use bertrand's potulate :P

(Since $ \frac{p_{k\plus{}1}}{p_k}<2<p_k,$ so if the inequality given above doesn't hold for some $ k_0$ it won't hold for any $ k\ge k_0$.)

[quote="t0rajir0u"]$ n > 49$ we have $ \prod_{i \equal{} 1}^{k} p_i > p_{k \plus{} 1}^2$ as per bertram's lemma. [/quote]

Isn't that also called Bonse's inequality?

I didn't know that, but after some googling, yes, apparently, it is.


geometry
analytic geometry
calculus
real analysis
real analysis unsolved

You have n points in the plane (you know the coordinates of the points) and you have to find point A so that the sum of distances fron the n pointsto A is minimum. I know that for n = 3 point A is called Fermat's point. This is a computer programming task, and a friend of mine told me it is a classical problem in calculus and you can solve it using derivation, I couldn't solve it myself ... Any help would be appreciated. Thanks

hey do you remember where exactly this problem is from?

Some acm online judge it may be the timus one but I can't tell for sure.


[quote="LoboSolitario"]Given $ \overline{A'C} \equal{} \dfrac{1}{3}\cdot\overline{BC}, \overline{B'A} \equal{} \dfrac{1}{3}\cdot\overline{CA},$ and $ \overline{C'B} \equal{} \dfrac{1}{3}\cdot\overline{AB},$ find $ \dfrac{(MNP)}{(ABC)}$.

Note: $ (XYZ)$ denote the area of the triangle $ XYZ$.

Maybe [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=176080]this[/url] will be useful.  :)

[hide="Answer"]$ \dfrac{1}{7}$[/hide][/quote]

Quoting my 1337 geometry textbook, if we set $ x\equal{}\frac{|BA'|}{|A'C|},y\equal{}\frac{|CB'|}{|B'A|},z\equal{}\frac{|AC'|}{|C'B|}$ the quotient will be given by
\[ \frac{(1\minus{}xyz)^2}{(xy\plus{}x\plus{}1)(yz\plus{}y\plus{}1)(zx\plus{}z\plus{}1)} \equal{} \frac{\left(\frac{7}{8}\right)^2}{\left(\frac{7}{4}\right)^3}\equal{}\frac{1}{7}
\]

If anyone cares about a proof I'll see if I can translate it. ;)

[quote="bertram"]
If anyone cares about a proof I'll see if I can translate it.[/quote]

Would be very nice  :)

I won't provide a formal solution, but in short, using the provided conditions, one could find the ratios of the many small triangles in the diagram by considering triangles with the same height but their bases in the ratio 1:2.  Then the rest of the problem falls into place.

I think I remember seeing this problem from some really early AMC.


A weed killer is to be diluted at a ratio of 1 oz. weed killer to 24 oz. water. At that rate, how many ounces of weed killer should be mixed with 32 ounces of water? Express your answer as a mixed number.

We set up a ratio.  We get $ \frac{x\text{ oz. weed killer}}{1\text{ oz. weed killer}}\equal{}\frac{32\text{ oz. water}}{24\text{ oz.water}}$.  This simplifies to $ x\equal{}\frac{4}{3}\equal{}\boxed{1\frac{1}{3}}$


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

A cute "quickie", as some people call these :):

Is it true that every finite group of even order $>2$ has an automorphism of order $2$?

no. if $f$ were such an automorphism, then $f(e) = e$ and the other elements, of which there are an odd number, could be arranged in pairs ${a, b}$ such that $f(a) = b$ and $f(b) = a$.

An automorphism $f$ has order two if it's different from the identity, and $f\circ f$ equals the identity. It may have [b]some[/b] fixed points other than $1$. In other words, the elements different from $1$ are not necessarily paired up like that.

Another way to phrase the question: is it true that the automorphism group $\mbox{Aut~}G$ of any group $G$ of even order $>2$ has even order?

:( i've been making errors in even the most basic of thinking lately. anyway...

if the group has even order, it has an element, $a$, of order 2. then $f(x) = ax$ is an automorphism of order 2. (so yes, the answer's yes, not no, heh)

$f(f(x)) = f(ax) = aax = x$ and $f$ leaves in fact no points fixed because $a\neq 1$ so it's certainly not the identity

Suppose $|G| = 2^r \cdot s$ with $s > 2$ and $r \geq 1.$
Then there is a prime $p > 2$ such that $p | \# G$.
Hence there is a cyclic subgroup $X$ of $G$ of order $p$.
Then $\mbox{Aut} ( X ) \approx \left( \mathbb{Z}/ p\mathbb{Z} \right)^{\times}$, hence $| \mbox{Aut} ( X ) | = \varphi ( p ) = p - 1$.
Hence $| \mbox{Aut} ( X ) |$ is even, therefore there is an $\sigma \in \mbox{Aut} ( X )$ such that $\sigma^2 = \mbox{Id}$.
We can extend $\sigma : X \longrightarrow X$ to an automorphism $\sigma' : G \longrightarrow G$ by definining $\sigma' ( x ) = \left\{\begin{array}{l} \sigma ( x ), \mbox{ if } x \in X\\ x, \mbox{ otherwise} \end{array}\right.$
The clearly $\sigma \in \mbox{Aut} ( G )$ and $\sigma^2 = \mbox{Id} .$

Suppose $|G| = 2^r$ with $r > 1$.
Proceed by induction on $r.$
For $r = 2$ the claim is clear. (We have $\left. G = (\mathbb{Z}_4, + \right)$ or $\left. G = (\mathbb{Z}_4, \cdot \right) \approx \mathbb{Z}_2 \times \mathbb{Z}_2$).
Suppose the claim true for $r < n$ and $|G| = 2^n$.
If $G$ cyclic, then $\left. | \mbox{Aut} ( G ) | = \varphi ( 2^n \right) = 2^{n - 1}$.
Hence there is an $\sigma \in \mbox{Aut} ( G )$ such that $\sigma^2 = \mbox{Id}$.
If $G$ not cyclic, then there is a subgroup $X$ of $G$ with $|X| > 1$.
But $\# ( X ) | \# ( G )$, therefore $|X| = 2^k$ for some $1 \leq k < n$.
By induction hypothesis, there is an $\sigma \in \mbox{Aut} ( X )$ such that $\sigma^2 = \mbox{Id}$.
As before we extend $\sigma : X \longrightarrow X$ to an automorphism $\sigma' : G \longrightarrow G$, with $\sigma'^2 = \mbox{Id}$.

Warning ! I must say I didn't carefully check the induction step $r=2$.
So this is possibly where the counterexample lies...

There are plenty of examples to show that the construction in the first paragraph doesn't work. Let's say $G=\mathbb Z/6\mathbb Z=\{\bar 0,\bar 1,\bar 2,\bar 3,\bar 4,\bar 5\}$ (where $\bar a$ is the residue of $a$ modulo $6$), for instance, and $X=\{\bar 0,\bar 2,\bar 4\}$ is the unique subgroup of $G$ of order $3$. Let $\sigma$ be the automorphism of $X$ which sends each $x\in X$ to $-x$. Then clearly, the extension $\sigma'$ that you defined is not even a homomorphism: $\sigma'(\bar 1)+\sigma'(\bar 1)=\bar 2$, while $\sigma'(\bar 1+\bar 1)=\bar 4$.

Automorphisms can't just be extended like that from a subgroup of a group to the entire group, you know.

Yes, I just realized I was wrong with that.
Homever, is it true that if $X$ is a subgroup of $G$, then $\#\mbox{Aut}( X) | \#\mbox{Aut}( G)$ ?
That would "save" my proof... EDIT: No, actually that would solve the problem  :P

I found the source of the problem and modified the source field of my first post. Another question asked in the same problem:

Is it true that every $p$-group ($p$ is a prime) of order $>2$ has an automorphism of order $2$?

Grobber, do you have a solution for this one?

I think that the asnwer to the initial problem is no. And i have to construct a group of odd order, which group of automorphisms is also of odd order.
Is you answer the same, and could you please write what ideas did you use.Thanks.

[quote="grobber"]Yup (unless I made a mistake in there somewhere; anyway, I [b]think[/b] I do) :)[/quote]
Grobber, could you please post ypu solution, if you not to spoil the problem for others, you can post it as hidden or pm me, if it is not difficult for you.
Thanks


geometry
angle bisector
geometry unsolved

Sides $ [AB] , [BC],$ and  $ [AC]$  of triangle $ ABC$, are tangents to incircle with center $ I$ in points $ C' , A'$ and, respectively, $ B'$ .If $ B''$ is the middle point of side $ [AC]$  ,prove that sides $ B'I, A'C'$ and $ BB''$ are concurrent.

[quote="shoki"][quote="Syler"][quote="shoki"]it is posted before ,anyway there exists a generalization for this problem:
Let $ P$ be a point on the angle bisector of $ \angle A$ and let $ A',B',C'$ be the projection of $ P$ on $ BC,AC,AB$,respectively.
let the intersection of $ PA'$ and $ B'C'$ be $ N$.then $ AN$ passes through the midpoint of $ BC$ called $ M$.
to prove this we can use sinuses or drawing the perpendicular from $ N$ to $ NP$ which intersects  $ AB,AC$  at points called $ C'',B''$ and then proving that $ N$ is the midpoint of $ B''C''$.[/quote]

Thank you,but what does mean "drawing the perpendicular from $ N$ to $ NP$"?[/quote]
my english is very bad,sorry ...
it means that u draw a line perpendicular to $ NP$ and passing through $ N$.[/quote]

Could you make the drawing?


We know that

$ cos(\pi)\equal{}\minus{}1$

$ cos(\frac{\pi}{2})\equal{}0$

$ cos(\frac{\pi}{3})\equal{}\frac{1}{2}$

$ cos(\frac{\pi}{4})\equal{}\frac{\sqrt{2}}{2}$

$ cos(\frac{\pi}{5})\equal{}\frac{1\plus{}\sqrt{5}}{4}$

$ cos(\frac{\pi}{6})\equal{}\frac{\sqrt{3}}{2}$
...

Is it possible to find a formula for $ cos(\frac{\pi}{n})$ where n - positive integer. ?

Perhaps the [url=http://www.artofproblemsolving.com/Wiki/index.php/Trigonometric_identities#Double_Angle_Identities]double/half angle identities[/url]?

No. We can write down a polynomial equation that it satisfies, but we can't express the solution with real radicals for most $ n$.
Here's $ n \equal{} 7$:
$ z \equal{} e^{i\pi/7},e^{3i\pi/7},e^{5i\pi/7},e^{ \minus{} i\pi/7},e^{ \minus{} 3i\pi/7},e^{ \minus{} 5i\pi/7}$ are the six non-real roots of $ z^7 \equal{} \minus{} 1$, which then satisfy $ z^6 \minus{} z^5 \plus{} z^4 \minus{} z^3 \plus{} z^2 \minus{} z \plus{} 1 \equal{} 0$. Divide by $ z^3$ and substitute $ w \equal{} z \plus{} \frac1z$: $ w^3 \equal{} z^3 \plus{} 3z \plus{} 3z^{ \minus{} 1} \plus{} z^{ \minus{} 3}$ and $ w^2 \equal{} z^2 \plus{} z \plus{} z^{ \minus{} 2}$, so
$ z^3 \plus{} z^{ \minus{} 3} \minus{} z^2 \minus{} z^{ \minus{} 2} \plus{} z \plus{} z^{ \minus{} 1} \minus{} 1 \equal{} w^3 \minus{} z^2 \minus{} z^{ \minus{} 2} \minus{} 2z \minus{} 2z^{ \minus{} 1} \minus{} 1$
$ \equal{} w^3 \minus{} w^2 \minus{} 2z \minus{} 2z^{ \minus{} 1} \plus{} 1 \equal{} w^3 \minus{} w^2 \minus{} 2w \plus{} 1$
and $ w \equal{} 2\cos\frac {k\pi}{7}$ satisfies $ w^3 \minus{} w^2 \plus{} 2w \plus{} 1 \equal{} 0$ for $ k \equal{} 1,3,5$. $ \cos\frac {\pi}{7}$ is the largest of the three roots.


Pro1: let $ a,b,c,r \in R$; $ a,b,c,r\geq0$ and $ ab \plus{} bc \plus{} ca \equal{} 3$ prove that:
$ \frac {a^r}{(b \plus{} c)^2} \plus{} \frac {b^r}{(a \plus{} c)^2} \plus{} \frac {c^r}{(a \plus{} b)^2} \geq \frac {3}{4}$

Pro2: let $ a,b,c, \in R$; $ a,b,c,\geq0$  prove that:
$ \frac {a^{b \plus{} c}}{(b \plus{} c)^2} \plus{} \frac {b^{c \plus{} a}}{(a \plus{} c)^2} \plus{} \frac {c^{a \plus{} b}}{(a \plus{} b)^2} \geq\frac {3}{4}$

Pro3: let $ a,b,c, \in R$; $ a,b,c,\geq0$  prove that:
$ \frac {2a^{b \plus{} c} \minus{} 1}{(b \plus{} c)^2} \plus{} \frac {2b^{c \plus{} a} \minus{} 1}{(a \plus{} c)^2} \plus{} \frac {2c^{a \plus{} b} \minus{} 1}{(a \plus{} b)^2} \geq\frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{ab \plus{} bc \plus{} ca}$

Although the problem looks VERY wierd..


Substitute $ \frac {c^2 \plus{} b^2 \minus{} a^2}{2c} \equal{} a \cos A$.

That hold if $ a$ and $ c$ are positive, right?


Let $ a,b,c,x,y,z>0$ s.t. $ a\plus{}x\ge b\plus{}y\ge c\plus{}z$ and $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z$. Prove that:

$ ay\plus{}bx\ge ac\plus{}xz$

[quote="Inequalities Master"]Let $ a,b,c,x,y,z > 0$ s.t. $ a \plus{} x\ge b \plus{} y\ge c \plus{} z$ and $ a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z$. Prove that:

$ ay \plus{} bx\ge ac \plus{} xz$[/quote]
Introducing $ s: \equal{}a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z$ and eliminating $ b$ and $ y$, the second condition $ b \plus{} y\ge c \plus{} z$ can be written as

$ 2(s\minus{}c\minus{}z)\ge a\plus{}x$.

The ineq to prove reduces to $ (a\plus{}x)(s\minus{}c\minus{}z)\ge2ax$, which is obvious by AM-GM.
As far as I can see, the first conditon isn't needed.  :lol:


[quote="filletwho"]This is a question on the proof, but i think it is general enough. Is repetition in proofs good? Or maybe summarizing what has been said?[/quote]

Repetition has a certain poetry, but we mathematicians prefer shortcuts. If you are writing up several cases that use the exact same technique, you write up the first case (call it n=1) in particular, giving full details, and then you get to say, "Likewise, in the second case, where n=2, we obtain the answer 42," skipping the details because it is obvious that we can find the answer 42 by just replacing n=1 with n=2 in all the calculations. We mathematicians recognize the phrases "Likewise" and "In the same manner" as indicating that the proof simply uses the same technique again.

If the second case is not exactly a repetition, for example, there is an extra step because n is even instead of odd, you may say, "Likewise, in the second case, where n=2, we obtain the answer 42. However, because n is now an even number, we have to have divide the intermediate result d by 2 because we need d to be an odd number. d is not a multiple of 4 because n is not a multiple of 4." (No, I do not have a specific proof involving n and d in mind. I am just making up sentences that sound like many sentences that I have written.) If you fail to mention the differences between the first and second cases when extra steps are involved, that failure will count as a hole in your proof and you will lose points.

Summarizing a technique should be attempted only if you have very good writing skills. A summary of a rigorous proof loses rigor. A grader might mistake such a summary for a failed proof.

Erin Schram
USAMTS grader

So then, let us say that we used brute force on a problem, going through a set of numbers and testing them until we got favorable answers. Do we need to show our work for that, or can we just way we have done it and it works, here is why?

[quote="ZennyK"]If you can explain why brute force yields the correct answer with out showing the work, then you probably don't need to use brute force anyway.[/quote]
Well, I think he meant, if you were to say something like "I tried all values by a brute-force method and found the one that worked", do you have to show all of the brute-force solutions, or just state that you found the one that worked (if I'm interpreting his question correctly).

I did a brute force method for question 2 of round 2, explained what I did, and omitted all the results that weren't pertinant.  It certainly wasn't elegant, so don't do it if you want commendation, but I did receive 5 points.


geometry
circumcircle
conics
geometric transformation
dilation
power of a point
radical axis

The line that pass through the circumcenter of $ ABC$ intersect $ AB,AC$ at $ C_1,B_1$ respectively.prove that the  two circles that $ BB_1$ and $ CC_1$ are their diameter intersect each other at two point that one of them is on the circumcircle of $ ABC$ and the other one is on nine-point circle of $ ABC$.

Denote P=CO∩LC1  Q=BO∩LB1  By pascal threom we know ABCLPQ are on the same conic
The conic is the circumcircle of ⊿ABC  =>  L is on the circumcircle of ⊿ABC
∠FKE=∠LC1B+∠LB1C=1/2(∠POQ+∠BOC)= 2∠A=180-∠FDE
D is on the nine-point circle of ⊿ABC

Let $ CH$ cut $ (ABC)\equal{}\{F'\}$
By [b]Pascal[/b]'s theorem we have L lies on $ (ABC).$
Since $ HC.HF\equal{}HB.HE$ then H lies on radical axis of $ (BB_1)$ and $ (CC_1)$
Let $ HK$ cut $ (ABC)\equal{}\{K'\}$
$ \Rightarrow HK.HL\equal{}HC.HF\equal{}1/2HC.HF'\equal{}1/2HL.HK' \Rightarrow HK\equal{}1/2HK'$
The dilation $ D^{\frac{1}{2}}_{H}: K'\rightarrow K$ then K lies on Nine-point circle of triangle ABC.


geometry
trigonometry
AMC
AIME
USA(J)MO
USAMO
similar triangles

Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD}$, respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}$. The length of a side of this smaller square is $\displaystyle \frac{a-\sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c$.

Ok, I think [b]Towersfreak2006[/b] (who originally pointed out this ambiguity) and I have found a flaw (since they didn't mention that a,b,c are positive integers greater than 1). It's easy to see that $\angle BAE=15$ so $BE=\tan{15}=2-\sqrt{3}$ Let $BE$ be the side of the square (where point E is our vertex on AE). Then we have \[ s=\frac{2-\sqrt{3}}{1}  \] where $(a,b,c)=(2,3,1)$ so $a+b+c=\fbox{006}$

I think the intended answer was \[ \frac{x}{1-x}=\tan{15} \Leftrightarrow x=\frac{3-\sqrt{3}}{6}  \] where $x$ is the side of the square so $(a,b,c)=(3,3,6)$ and $a+b+c=\fbox{012}$

Er thanks. A bit hesitant to question something like the AIME...

I don't know...I just had issues doing it the first time. X_X

[quote="DPopov"]Well, I'm convinced. What does everyone else think?[/quote]
Agree with you! At first, I thought the other vertex is just E(no error)! However, not that easy, so...
But indeed the question has flaw!!!

Indeed, it does not say it is inscribed.

[quote="calc rulz"]They should run through and see how many people actually found the answer to be 6.

They should also see whether it seems anyone who said 6 was simply guessing, or whether there were people who say, got most of the other problems right but got problem 6 wrong due to giving an answer of 6.[/quote]

NO, don't delay the test grading by another month! :)  :lol:

But seriously, probably no one answered 6 since iif they can get 6, they can probably get the "correct" answer..... :?

I agree that the wording of the problem was unclear.  However, I think it is unlikely that many people answered $6$.

When I solved this problem, I first found $BE$, and then used similar triangles to solve for the side length of the square.

If I were taking the test, and I noticed that the problem was unclear about which vertex of the square was on $\overline{AE}$, I would still be reluctant to answer $6$.  I do not believe there has ever been an AIME question that said, "The answer can be expressed in the form $\frac{a-\sqrt{b}}{c}$" when the answer could actually be expressed as simply $a-\sqrt{b}$.

Also, if the other vertex of the square was actually on $E$, the problem would be about twice as easy as it would be if the vertex was not on $E$.

In my opinion, not only does an answer of $6$ "seem" wrong, but it also simplifies the problem by a factor of $2$.  If I took the AIME II, I would definitely not answer $6$ unless I didn't see the other possibility.

But for the the sake of those people that answered $006$ I hope that they decide to award credit for both answers, even if it delays the results  ;)

I too, got my answer as $\frac{2-\sqrt{3}}{1}$ becuase I thought that the vertex was at E, with the square extending outwards, but when I added the three numbers, I did $2+3=\boxed{005}$ and totally forgot about the 1.  So even if credit will be awarded to 006, I still wouldn't get it thanks to my retarded mistake.

Well at least other people in your situation (who didn't mess up adding) will be able to get credit for their solution. Aha, I wonder if this is why they didn't release the solutions today.

[quote="DPopov"]Well, I'm convinced. What does everyone else think?[/quote]
Yeah, it is definitely a flaw, i agree

After consultation with David Hankin, Chair of the AIME Committee that prepared the 2006 AIME II, we allow both 006 and 012 as correct answers.  All AIME II papers have been scored accordingly.

-- Steve Dunbar
AMC Director

We'll settle for 20 points on USAMO, one point apiece from all of those people who wouldn't have made it :P

Yikes....

Now that I think about it more... the problem is [i]really really[/i] flawed.

One could take the other vertex to be not $E$, but $A$.

That would make the new square the original square, side length 1.

Anything of the form $\frac{n - \sqrt{1}}{n-1}$ satisfies their conditions. The sum is $2n$ with $n$ at least 2. So.... any even number greater than or equal to 4 is technically a valid answer.

But... allowing any even answer is just silly. I doubt anyone thought the vertex was $A$, and even if someone did...
Who writes 1 as $\frac{314-\sqrt{1}}{313}$? :)
I don't think it should be counted.... that'd be ridiculous. Especially now that the qualifiers are decided on.... something like this would greatly mess up the floor and index values.

[quote="Dnas"]Yikes....

Now that I think about it more... the problem is [i]really really[/i] flawed.

One could take the other vertex to be not $E$, but $A$.

That would make the new square the original square, side length 1.

Anything of the form $\frac{n - \sqrt{1}}{n-1}$ satisfies their conditions. The sum is $2n$ with $n$ at least 2. So.... any even number greater than or equal to 4 is technically a valid answer.
[/quote]
I agree, it is [i]really really[/i] flawed. [b]It definitely should be thrown out or any even number should be counted as the right answer[/b]. This is a very badly worded question. Ughh

They do specify that it is a "smaller" square.

[quote="asmi"]how do we know angle BAE equals 15?[/quote]

Angle EAF is 60, since the triangle is equilateral.  So BAE = 30/2 = 15, since FAD is also 15.

[hide=Not that hard]
Observe that the diagonal going from $A$ to $C$ is a perpendicular bisector of the side $EF.$
Proof: A square is symmetric with respect to its diagonal. Thus points $E$ and $F$ will also be a reflection of each other with respect to diagonal $AC.$

Note that since angle $FAC=30^circ$, $DAF=15^circ.$
Using the tangent subtraction identity, tan(15)=$2-\sqrt{3}.$
Thus DF=2-√3

We can now write an equation using similarity properties.
Let s be the side of the small square.

Then:
$\dfrac{s}{1-s}=\dfrac{2-\sqrt{3}}{1}.$
Simplifying, we see that $s=\dfrac{3-\sqrt{3}}{6}.$
Adding, we get $3+3+6=12.$

[\hide]

[hide=Solution]
Let the vertex of the smaller square of $AE$ be $G$. Let the foot of the perpendicular from $G$ to $AB$ be $H$. Let $GH=x$. Notice that $HB=GH=x$ and $AH=\frac{x}{\tan \frac{90^\circ-60^\circ}2}=\frac{x}{2-\sqrt3}=x(2+\sqrt3)$. Therefore, $1=AB=AH+HB=x(2+\sqrt3)+x=x(3+\sqrt3)$. Therefore, $x=\frac{3-\sqrt3}{6}$. Therefore, the answer would be $3+3+6=\boxed{012}$.
[/hide]


Let $ AA_1,BB_1,CC_1$ be the altitudes of acute triangle $ ABC$, and let $ X$ be an arbitrary interior point. Let $ M,N,P,Q,R,S$ be the feet of the perpendiculars from $ X$ to $ AA_1,BC,BB_1,CA,CC_1,AB$ respectively. Prove that $ MN,PQ,RS$ are concurrent.

See at http://www.mathlinks.ro/Forum/viewtopic.php?t=126138

This problem has a beautiful story for me dear [b][size=100]Carlez Devos[/size][/b]. Could you please tell me, where did you get it?

Thank you, Kostas Vittas.


Without taking derivatives of both the denominator and the numerator, find:

$lim_{x\rightarrow2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$

Please do not post any calculus in the Getting Started forum anymore.

Well try multiplying the conjugate of the bottom to get rid of the sqare roots.  But, if the methods don't work, use L'Hopital's rule.


Let $ x_1,x_2,x_3,...,x_{1998}$ be natural numbers  and $ {x_1}^2 \plus{} {x_2}^2 \plus{} ... \plus{} {x_{1997}}^{2} \equal{} {x_{1998}}^{2}$.Prove that at least one of these numbers is even.

[hide]If all the numbers on the LHS are odd, then it is $ \equiv 5 \pmod{8}$, not a quadratic residue.[/hide]


[img]http://www.pokeplushies.com/images/adoptables/1414013.gif[/img] [url=http://www.pokeplushies.com/feed/1414013]Click here to feed me a Rare Candy![/url] [url=http://www.pokeplushies.com]Get your own at Pokeplushies![/url]

[img]http://www.pokeplushies.com/images/adoptables/1459970.gif[/img]
[url=http://www.pokeplushies.com/feed/1459970]Click here to feed me a Star Fruit![/url]
[url=http://www.wwwdotcom.com]Don't Get your own at Flyffables![/url]

While you're at it...
[img]http://www.pokeplushies.com/images/adoptables/1457214.gif[/img]
[url=http://www.pokeplushies.com/feed/1457214]Click here to feed me SOULS! I...MUST...HAVE...SOULS![/url]
[url=http://www.flyffables.com](Don't) Get your own at Flyffables![/url]

PLEASE LEVEL MINE UP TOO!

[quote="algebra2000"]u cheated...u even said so

EDIT: also, it's 11:36 PM where I am...how r u online at a time like this?[/quote]

where are you?


In triangle $ ABC$, angle $ BAC \equal{} 2 \times$ angle $ ABC$. Prove that $ a^{2} \equal{} b(b \plus{} c)$.

[hide=" Hint"]

sine rule

[/hide]

because a/sinA = b/sinB, i.e. a/Sin2B= a/2sinBcosB=b/SinB   from law of SIN
we have: a=2bcosB    (1)

b**=a**+c**-2acCosB, i.e. cosB=(a**+c**-b**)/2ac  (2)  from law of COS

thus: a=2b(a**+c**-b**)/2ac, then you can prove it.

Thanks lilyidi.

I had the following solution:

$ \frac{a}{\sin 2\theta} \equal{} \frac{b}{\sin \theta} \equal{} \frac{c}{\sin (180 \minus{} 3\theta)}$

$ \frac{a}{2\sin \theta \cos \theta} \equal{} \frac{b}{\sin \theta} \equal{} \frac{c}{3\sin \theta \minus{}4\sin^{3} \theta}$

$ a\equal{}2b\cos \theta$

$ a^{2} \equal{} 4b^{2} \cos^{2}\theta$

$ c\equal{}(3\minus{}4\sin^{2}\theta)b$

$ b(b\plus{}c) \equal{} b(b\plus{} b(3\minus{}4\sin^{2} \theta)) \equal{} b^{2}(4 \plus{}4\sin^{2} \theta) \equal{} 4b^{2}\cos^{2}\theta$

Therefore $ a^{2} \equal{} b(b\plus{}c)$.


limit
real analysis
real analysis unsolved

$ 0 < a_1 < a$,  $ a_{n \plus{} 1} \equal{} a_n(a \minus{} a_n)$, compute   $ \lim_{n\rightarrow\infty}a_n$

Define $ f(x) \equal{} x(a\minus{}x)$; then $ a_{n\plus{}1} \equal{} f(a_n)$, so $ a_{n\plus{}1} \equal{} f^n(a_1)$.

The equation $ f(x) \equal{} x$ has roots $ 0$ and $ a\minus{}1$. It easily follows that for $ a \leq 1$ we have $ \lim_{n \to \infty} a_n \equal{} 0$, while for $ a > 1$ we have $ \lim_{n \to \infty} a_n \equal{} a\minus{}1$.


Also the Physics forums.
Hope Valentin sees our discussion down the topic [url=http://www.mathlinks.ro/viewtopic.php?t=155483]here[/url]

[quote="Immanuel Bonfils"]Look to this post, "Physics and the new Portal edition", just here below...[/quote]

I'm not sure what that has to do with this. That concerns inclusion on the portal, this is about new moderators.

Come on, someone stick up for the little guy.

[quote="BanishedTraitor"]Despite the fact we have very few active posters, I think we need a moderator (as you can see, our 'current' moderator is not very active.....).[/quote]

yes you are correct. Carcul,paradesi and you should be the mods for that section

Still no answers from anyone. Can someone at least give us a yes or no (possibly with explanation) or even just suggestions because none of the members know what a moderator is capable of.

i think there are sufficiently many people now a days visiting the Physics and chemistry forums  :)

HUGE BUMP......

Yeah, hi again. Want to revive this topic seeing we have no replies from the Administrators. So can we have a new moderator for the Chemistry forum? I mean the stuff in their are a bit messy but they are all good! Like we want to make a collection of concepts in Chemistry. Also Carcul posts some very good problems but over time they get a bit messy and neglected.


inequalities
geometry
circumcircle
trigonometry
inequalities unsolved

I'm having some trouble with the following problem. Any help would be greatly appreciated :) .
The circumcircle [i]k[/i] of acute triangle ABC has radius [i]r[/i]. The bisectors of the angles of the triangle intersect the circle again in the points A', B', & C'. If P & Q are the areas of triangle ABC and triangle A'B'C', respectively, prove the inequality
16Q^3≥27r^4P.

The first step seems to be expressing $P$ in terms of $r$.  If the angles $AOB$, $BOC$, and $COA$ are $a$, $b$, $c$, then we have $P=\frac{r^2}{2}(\sin{a}+\sin{b}+\sin{c})$.  We want to find the maximum value of $P$ in terms of $r$, to strengthen the inequality.  Notice that here, \[ \sin{a}+\sin{b}+\sin{c}=\sin{a}+\sin{b}-\sin{(a+b)} \] \[ =\sin{a}+\sin{a}\cos{b}+\sin{b}+\sin{b}\cos{a} \] I don't know how to prove the maximum of that is $\frac{3\sqrt{3}}{2}$...
Edit: Okay, now I see how to prove $\sin{a}+\sin{b}+\sin{c} \le \frac{3\sqrt{3}}{2}$.  First, we see that all of the angles are less than $\pi$.  Thus, they are on a concave down part of the sine curve.  Using Karamata's inqeuality, the sum is maximized when the three are equal.
Once you've done that, strengthen the inequality to one creating a relationship between $r$ and $Q$, which shouldn't be too hard to prove...(ie you get $\frac{r^2}{Q} \le \frac{4}{3\sqrt{3}}$)

[quote="Inspired By Nature"]Sorry, but what's Karamata's inqeuality? :oops:[/quote]
    Maybe better is to say it $\ Inequality\ of\ Hardy-Littlewood-Polya$.


MATHCOUNTS
FTW
Vieta
algebra
polynomial
sum of roots

Hi there,

In several MATHCOUNTS contests, not to mention some problems in FTW, I have seen problems like:

   Find the sum of all the possible solutions $ x$ to $ |7x \plus{} 2| \equal{} 8$.

How do I solve this type of problem?

$ |7x\plus{}2|$ implies either $ 7x\plus{}2$ or $ \minus{}(7x\plus{}2)$.  We just solve the two equations:

$ 7x\plus{}2\equal{}8 \implies 7x\equal{}6 \implies x\equal{}\frac{6}{7}$

or

$ \minus{}7x\minus{}2\equal{}8 \implies \minus{}7x\equal{}10 \implies x\equal{}\minus{}\frac{10}{7}$

Adding these yields $ \boxed{\minus{}\frac{4}{7}}$.

In general, when given an expression $ |z|$, substitute both $ z$ and $ \minus{}z$ as it, and solve both.

Thank you! Now these crazy problems will haunt me no longer...

In general, if c>0, then the sum of the real solutions x to the equation |ax+b|=c is -2b/a. Try to derive that.

You mean $ \minus{}\frac{2b}a$, right? :P

Another way to deal with these is to square both sides to get, in this example, $ 49x^2\plus{}28x\plus{}4\equal{}64\implies49x^2\plus{}28x\minus{}60\equal{}0$, which, by Vieta's, has sum of roots $ \minus{}\frac{28}{49}\equal{}\boxed{\minus{}\frac47}$. The aforementioned formula can also be derived easily from this (try squaring both sides of $ |ax\plus{}b|\equal{}c$).

However, it is very important to note that the above techniques and formulas only apply to $ c>0$.

If $ c\equal{}0$, then there is only one solution (which is thus the sum), $ \minus{}\frac ba$.

If $ c<0$, then there are no solutions.

OK, thanks guys. I think I'll stick with the $ \minus{}\frac{2b}{a}$ method.  :wink:

However, if you come across a problem such as finding solutions to an equation such as $ x^2\plus{}6|x|\plus{}9\equal{}0$, you won't know what to do.  Also, if it asks you to FIND both solutions, you'll be stumped.

Right, makes sense. But in MATHCOUNTS, mostly the only type of problem I see with absolute values is of the form $ |ax \plus{} b| \equal{} c$ where $ a \not\equal{} 0$. But I'll keep your method in mind, in case I ever see something that doesn't have a shortcut like that one.

[quote="AIME15"]However, if you come across a problem such as finding solutions to an equation such as $ x^2 \plus{} 6|x| \plus{} 9 \equal{} 0$, you won't know what to do.  Also, if it asks you to FIND both solutions, you'll be stumped.[/quote]

The solution set is null, since $ x^2 \plus{} 6|x| >\equal{} 0$


Why cannot a lens disperse a ray of white light even though it is made up of many prisms?

In fact, lenses do disperse  :lol: .

[quote]Camera lenses

Chromatic dispersion can be a problem in optical equipment like cameras, microscopes and telescopes. Since a lens is similar to a prism, it will disperse the light into a spectrum. This can be exaggerated when there are several lenses in the system. But you don't want a blurry image where the system focuses different colors of light at different spots. Such blurring is called chromatic aberration.

The solution to that problem is to combine a positive lens that has one index of refraction with a negative lens of a different type of glass and a different index of refraction, such that the dispersion of the two negate each other.[/quote]

See also http://en.wikipedia.org/wiki/Chromatic_aberration


Four dice, coloured red, blue, yellow and green, are rolled. In how many different ways can the product of the numbers rolled equal $ 36$ (where, for example, a red $ 4$ is considered to be different from a blue $ 4$)?

Four dice, coloured red, blue, yellow and green, are rolled. In how many different ways can the product of the numbers rolled equal 36 (where, for example, a red  4is considered to be different from a blue )?
36 is come from
1x2x3x6
1x3x3x4
1x1x6x6
2x3x2x3
each of them has 4! ways
so the answers is 4 x 4!
Am I correct?


Determine the primes $ a,b,c$ so that: $ N\equal{}(a^2\minus{}bc)(b^2\minus{}ca)(c^2\minus{}ab)$ is a prime.

Is the problem right?
If Determine the [b]positive integers[/b] a,b,c so that: N=(a^2-bc)(b^2-ca)(c^2-ab) is a prime.
$ (a, b, c)\equal{}(1, 1, 2), (1, 2, 1), (2, 1, 1)$.

because, $ p \in Prime, 6 \not| p^2\minus{}1 \Rightarrow p\equal{}2, 3$.
But,
$ (a, b, c)\equal{}(2, 2, 2) : 0 \not \in Prime$,
$ (a, b, c)\equal{}(2, 2, 3), (2, 3, 2), (3, 2, 2) : 20 \not \in Prime$,
$ (a, b, c)\equal{}(2, 3, 3), (3, 2, 3), (3, 3, 2) : \minus{}45 \not \in Prime$,
$ (a, b, c)\equal{}(3, 3, 3) : 0 \not \in Prime$.


do you mean

$5^{\sqrt{x^2}}$?

then it is just $5^{|x|}$


NADAL DEFEATS FEDERER 7-5, 3-6, 7-6, 3-6, 6-2

GO NADAL!!!!

WHO SAW THE MATCH?

Pretty beast...didn't Federer end up crying or something?


geometry
trigonometry
AMC
AIME
analytic geometry
number theory

I am strong in algebra, number theory, and trig and believe I can solve most of those problems on the AIME. I feel weak in counting and geometry. I can solve about the first 3 counting problems and the first 5 geometry problems. Geometry seems to be a more tested subject. Would I be better off preparing for Geometry or Counting?

Geometry would be more useful to study and my reasoning for this is you can bruteforce a counting question easier than you can a geometry question.

If you concentrate on geometry then it'll greatly improve your AMC score, and your AIME score.

[quote="Calculus"]Would I be better off preparing for Geometry or Counting?[/quote]

Both

[quote="Pakman2012"]Geometry would be more useful to study and my reasoning for this is you can bruteforce a counting question easier than you can a geometry question.  [/quote]

Hmmm really? First of all, there are only about 5 geometry and a bit over 3 counting on the AIME, so how are you weak in those subjects? Second of all, you can easily (relatively speaking) brute force most geometry problems using trig or coordinates (indeed, some even require the said methods). For counting questions, this is not like the AMC--most of the time you would have to brute force many, many cases to arrive at an answer. Even then, the answer will be prone to silly mistakes, and you won't be able to easily check them like you can with geometry.

That said, geometry is still the more "studyable" subject, so it really depends on the specifics of your own situation. In typical Silverfalcon style, I am going to tell you that it's up to you to figure out how to improve. After all, you know yourself better than anyone else.

[quote="probability1.01"][quote="Pakman2012"]Geometry would be more useful to study and my reasoning for this is you can bruteforce a counting question easier than you can a geometry question.  [/quote]
 In typical [b]Silverfalcon style[/b], I am going to tell you that it's up to you to figure out how to improve. After all, you know yourself better than anyone else.[/quote]

Haha.. I have my own style now.  :rotfl:  :rotfl:

As for counting versus geometry, I agree fully with probability1.01. It is true that counting is just basically a process of thinking, it's not easy to brute force. The problem [i]sounds[/i] easy but when you actually have to attack them, the evils come out of the problem and slaughter you. :D For me, geometry was the topic that I greatly improved this year over couple weeks. I began to work hard on later geometry and let me tell you, I just practice, practice, practice, practice, practice, and practice until you get tired of me saying word "practice."  :D


Leaderboard:

Scrambled....3
Tare.............1
mathfiend....1

1. A donut shop sells 5 cream-filled and 7 jelly-filled donuts for
$ 3.64, and 3 cream-filled and 9 jelly-filled donuts for $ 3.72. What is the cost of 1 cream-filled donut?

2. If the numbers x=2^106, y=4^54, z=8^35 were arranged in decreasing numerical order, the ordering would be what?

3. If the length of a rectangle is doubled and the width is tripled, the area of the rectangle has been multiplied by what?

4. If x^0.3=10, what is x^0.4?

5. 2 squares are placed, one on top of the other, with their centers coinciding but with the top one rotated through 45 degrees relative to the bottom one. If the squares are 1 meter X 1 meter each, what is the volume of the protion of the lower square still visible? (the squares are opaque)

Having problems with the site.  I'll be back when it starts working for me.

Tare gets #'s 1, 2, & 4
mathfiend gets #3
#5 is still out there.

I create a new thread to separate rounds.  Otherwise, the threads get super long and are hard to correct.

Look at past UNL Math Day threads & you'll get a good idea of how they are run.

3-2*:rt2:? I'm not sure what you meant...

OK, Tare gets #5.

Leaderboard for this round:
Tare-4
mathfiend-1
for the whole thing so far:
Tare-5
scrambled-3
mathfiend-2

Next round TBA...

I told you guys there's no competition :) </just kidding!>

i was present for 2 minutes of this round


function
logarithms
algebra unsolved
algebra

[quote="unknown_pattern18"]Solve for b in terms of n and t: or in other words make b the subject of formula
bn^b=t[/quote]
I think the function $t=f(b)=bn^b$ is not defined for non-integer negative $n$.
Even with positive $n$, I think the function is not bijective and $b=f^{-1}(t)$ is not a function (for negative $t$, there can be $0$, $1$ or $2$ different values $f(b)=t$.
And, even if you want integer, positive $n$ and $t$, I don't think there's a closed expression for $b=f^{-1}(t)$ (not that I can prove it).

is it like dividing a number by zero which is not defined?
i dont think there is somethings in maths that doesnt have a solution.
i am sure there must be a way to make b the subject .

Sometimes, the solution is that there is no solution, at least in a certain set . Which is a very important and useful information. But we should have decided the set beforehand, for instance I supposed you meant in $\mathbb{R}$, but with $\mathbb{C}$ there would be more solutions.
[hide="Click here for more"]
I will compare your question with this:
Solve for $b$ in terms of $t$: $t=b^2$.
If we work in $\mathbb{R}$, the solution would be:
If $t>0$, then there are two solutions, $b=+\sqrt{t}$, and $b=-\sqrt{t}$,
so $b=f(t)$ is not a function, because for every $t\ge 0$ there is more than one $b$.
If $t=0$, there's only the solution $b=0$.
If $t<0$, then there are no solutions because $\sqrt{t}$ is not defined in $\mathbb{R}$ for $t<0$.
But if we work in $\mathbb{C}$, there would always be two complex solutions (except for $t=0$, $b=0$).
[/hide]
And then, there are solutions that exist, but can't be expressed with an easy function. And I think (although I'm not good enough to prove it) that that's what happens with your problem.

how is the title relevant?
btw, 1+1=2 is proveable if you define integers based on conways definition of surreal numbers.

bn^b=t comes from solving 1+1,if bn^b=t if b cant become the subject of the formula than i dont believe that 1+1 equals to 2

who said that 1+1=2? everyone knows that 1+1 is actually [spoiler removed]! who doesn't agree is a n00b  :rotfl:

scnr.

Seriously: $bn^b=t$ is equivalent to $be^{\ln n\cdot b}=t$ (since $n^b=\left(e^{\ln n}\right)^b=e^{\ln n\cdot b}$), or to $\ln n\cdot b\cdot e^{\ln n\cdot b}=\ln n\cdot t$. Now, for every x, the solution y of the equation $ye^y=x$ is called [url=http://mathworld.wolfram.com/LambertW-Function.html]the LambertW function of x[/url], and denoted by $y=\text{LambertW }x$. So our equation $bn^b=t$, being equivalent to $\ln n\cdot b\cdot e^{\ln n\cdot b}=\ln n\cdot t$, becomes $\ln n\cdot b=\text{LambertW }\left(\ln n\cdot t\right)$, so the solution is $b=\frac{\text{LambertW }\left(\ln n\cdot t\right)}{\ln n}$.

If this could be written in a more elementary way, why would you think the LambertW function would be defined then?

  darij

[quote="gopherhole112"]how is the title relevant?[quote]
It gets your attention so that you with look at this topic.

In all honesty:

[hide="I cannot solve 1+1..."]... since it's not an equation. ;)[/hide]


How many different lines pass through at least two points in this 3-by-3 grid of lattice points shown?

[asy]size(30);
dot((0,0));
dot((1,0));
dot((2,0));
dot((0,1));
dot((1,1));
dot((2,1));
dot((0,2));
dot((1,2));
dot((2,2));[/asy]


geometry
perpendicular bisector
geometry unsolved

[quote="sunchips"]In triangle $ABC$, $N$ is the nine point center. $N'$ is the isogonal conjugate of $N$. Let $A'$ be the intersection of $BC$ with the perpendicular bisector of $AO$, where $O$ is the circumcenter.

Similarly defining $B'$, and $C'$, prove that $A'$, $B'$, and $C'$ lie on a straight line perpendicular to $N'O$[/quote]

Discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=1395 .

  darij


trigonometry
probability
AMC
AIME
geometry
power of a point

[quote="zanttrang"]Darn, I couldn't remember my Newton sums for #11.  Was the correct answer $S_{14} = -204 = - 2^2 \cdot 3 \cdot 17$, giving an answer of $ 2 + 1 + 1 = 4$?

Also, for #17, how can you rigorously prove that $a = 39$ works?  I was able to find (after the end of the test, grrr) by plugging in $x=1$ that we must have $a \equiv 39 \ (\text{mod} \ 55)$, and a quick table check verifies that $a = 39$ works for the first several $x$, but I'm not sure how to prove that it works for all integer x (without an extremely lengthy and messy proof by induction).[/quote]

55 divides the thingy iff 5 and 11 do. try it now !!!!!:lol::lol::lol:!!!!

darn I knew I shouldn't've gone with E for 23 BOOOOOOO because it doesn't mean anything BOOOOOOOO

curiousness! for the angle bisector question I kept getting 2/7 so I finally  just picked the answer closest to that

[quote="no_one_2000"]Man, am I dumb!  I actually got a=39 for #17 and just NOW realized that it was asking for the sum of the digits.  No wonder none of them worked!

:wallbash_red:

Eh, and how did you do #6 and #13?[/quote]

#13 is tricky:  note that the problem simplifies to

$\sin 2A = 2 + \sin 2B$.  In order for this to mean anything, we must have $\sin 2B = -1$, meaning that $\sin 2A = 1$.  And when $\sin 2A = 1$, we have $\tan 2A$ is undefined (I wish I'd seen that trick during the stupid exam!)

#6 is pretty straightforward:

There are 4 choices for the first number, 3 choices for the second, and 2 choices for all the numbers after that (because you don't want the number to be either of the two numbers before it).  So, the number of arrangements is $(4)(3)(2)(2)(2)(2) = 192$, out of $4^6 = 4096$ possible arrangements, leading to a probability of $\frac{3}{64}$, or choice c.

I'll go ahead and post the solution packet but I don't yet have the answer for #25.  If someone could post a good solution for that one I will be most thankful ;)

25. blah blah blah

Let $[BDF] = x, [CEF] = y$. Then $[ADFE] = 5 - x - y$. We know $2 + y = [BCE] =$ $\frac{1}{2} BC \cdot CE \cdot \sin(\frac{\pi}{3}) =$ $\frac{1}{2} AC \cdot AD \cdot \sin(\frac{\pi}{3}) = [DAC] =$ $5 - x \Rightarrow y = 3 - x$. Now, $\frac{3-x}{5-x} = \frac{[CEF]}{[CAD]} =$ $\frac{CF}{FD} \cdot \frac{CE}{EA} = \frac{[BCF]}{[BCD]} \cdot \frac{[BCE]}{[BCA]} =$ $\frac{2}{2+x} \cdot \frac{5-x}{7} \Rightarrow x = \frac{1}{3}, \frac{8}{3}$, so $\frac{AD}{DB} = \frac{[ADC]}{[DBC]} = \frac{5-x}{2+x} = 2, \frac{1}{2}$. Now note that $DB = AB - AD = AC - CE = AE$, so as $AD > AE$, we have $AD > DB$, so $\frac{AD}{DB} > 1$, so $\frac{AD}{DB} = 2$. Answer choice ...um, I forget. A?

[quote="no_one_2000"]Man, am I dumb!  I actually got a=39 for #17 and just NOW realized that it was asking for the sum of the digits.  No wonder none of them worked!

:wallbash_red:

Eh, and how did you do #6 and #13?[/quote]

Haha, I haven't noticed until you said it. I got 39 as well and spent 10 minutes redoing the problem to see where I went wrong. I ended up just leaving it blank.
Same thing with number 7, except on that one I kept thinking 17^2=189. Funny how ya think.

[quote="no_one_2000"]Hey, on #8, it should be \frac45 :)  Just a little typo.

Actually never mind.  You fixed it before I made this post. lol[/quote]

I posted the most updated  version of the solutions (I forgot to recompile it  :blush: ).  I caught that error on #8 on the new version, I believe.  But thanks for pointing it out.  Please point out any other typos and such if you find them.

Well if it makes you feel any better, I didn't even get an answer for #7.  When I first read it, I thought, "I wonder if I can use the Power of a Point theorem on this!" (Since that's one of the more recent theorems I learned) But I still didn't see the answer... lol

[quote="MysticTerminator"][quote="no_one_2000"]Thanks, zanttrang.  I didn't notice that trick on #13 either, that's really clever.  Good problem.

For #6, I apparently didn't read the question very well.  I thought any of the digits couldn't equal any of the [i]adjacent[/i] digits, not the two before it.  (So I had 4*3*3*3*3*3)  I left it blank because of that.

I guess I need to learn to read.  But hey, 10th place isn't that bad, right?[/quote]

actually, the way the TeX read, your version is correct.[/quote]

Well, for $x_i\neq x_{i+1}\neq x_{i+2}$, my version could be interpreted as correct, I guess, but since it was verbally specified right above, I'd still say I'm wrong.  But yeah, that is why I got the answer I did.

Thank you Arnav!  Now the solution packet is compete.

wow, that was pretty bad >_< i didn't notice half the things i should have, so i only answered 16...  misread one, and took the bait for #22 :blush:. anyway, thanks to joml88 and rep123max for making the test.  some of the problems were really nice.

[quote="yif man12"]
I kept thinking 17^2=189. Funny how ya think.[/quote]
Hahah!!!! I did that too!  :lol: Eh.  11th place, not too bad I guess.

[quote="Rep123max"]There were 18 participants in the 2005 Mock AMC A:

1234567890
Alan
Angelaj92391
D343seven
Eryaman
Furious
H_S_Potter2002
Hmoonster
Lucky707
No_One_2000
Optimosis
Pkerichang
Sky9073
Tarquin
Towersfreak2006
Wizard
Yif man 12
Zanttrang

The top 5 scorers were:
1. Optimosis
2. Towersfreak2006
3. Yif man 12
4. Alan
5. Zanttrang

And finally, here are the scores:
127.5
120.5
116
106.5
105.5
103
102
102
101.5
101
98.5
92.5
90
83
82
80.5
80.5
78

There is a solutions packet that was compiled by Joseph, he will probably post it soon.[/quote]

NO!!!!!!!!!!!!!!
I got a 103

are you telling me that if i didnt bubble number 7 wrong, I'll get into top 5????

I'm number 6!!!!!

  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:

er...run by me one more time where you were bubbling in answers?

Whew ... good stuff ... (I got 102 ... 12 right, 1 wrong, 12 skipped ... usually I'm 10 right, 3 wrong, 12 skipped so yeah)

Did anyone else skip #4 due to not understanding the question?

lol i just did it now from 3ish to 4ish am lazily and got 129

edit:  my favorite problem was #24, anyone see recursion solution?

Suppose s(n) is the number of ways, it is easy to see s(n+1) = 3s(n) - 1, because for each tuple of two sets we can add an element to neither, one, or the other, except in the case of ({}, {}) where we are double counting once.

Since we have no time to solve the recurrence, we quickly check all the formulas for their adherence to s(n+1) = 3s(n) - 1.  bam!  the first answer works.

[quote="pkerichang"]
NO!!!!!!!!!!!!!!
I got a 103

are you telling me that if i didnt bubble number 7 wrong, I'll get into top 5????

I'm number 6!!!!!

  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:[/quote]
there's no reason to get all upset. it's just a practice. be happy you didn't get 13th place.  :D
i should have done better but on a few problems i knew what to do but i just didn't do it right (i.e. the partition problem). now, there's the mock b to prepare for. i'm going for 12th. hehe.

[quote="zanttrang"]Darn, I couldn't remember my Newton sums for #11.  Was the correct answer $S_{14} = -204 = - 2^2 \cdot 3 \cdot 17$, giving an answer of $ 2 + 1 + 1 = 4$?[/quote]For this special case, there is also a natural way for those who don't know/forgot Newton's sums although the reasoning is essentially the same. The problem was:

[quote]
Let $S_n$ equal the sum of the $n$th powers of the
roots of $x^{14}-3x^2+x-17=0$. Compute the sum of the
exponents in the prime factorization of $S_{14}$. [/quote]
Consider $\{r_i^{14}\}_{i=1}^{14}$.
Each $r_i$ satisfies $(r_i)^{14}-3(r_i)^2+(r_i)-17=0$ or $(r_i)^{14}=17-(r_i)+3(r_i)^2$.
Hence, $\sum_{i=1}^{14}(r_i)^{14}=\sum_{i=1}^{14}17-(r_i)+3(r_i)^2$.
$\sum_{i=1}^{14}(r_i)^{14}=\sum_{i=1}^{14}17-\sum_{i=1}^{14}(r_i)+3\sum_{i=1}^{14}(r_i)^2$.
$\sum_{i=1}^{14}17=14*17$
$\sum_{i=1}^{14}(r_i)$ is just the sum of the roots, or the opposite coefficient of $x^{16}$ divided by the coefficient of $x^{17}$, which is just $0$.
$\sum_{i=1}^{14}(r_i)^2=\left(\sum_{i=1}^{14}(r_i)\right)^2-2\left(\sum_{1\le i\le j\le14}r_ir_j\right)$
Again, the sum of the roots taken one at a time is zero, so the square of the sum of the roots is zero. That is, $\left(\sum_{i=1}^{14}(r_i)\right)^2=0$.
$\left(\sum_{1\le i\le j\le14}r_ir_j\right)$ is the sum of the roots taken two at a time, or the coefficient of $x^{15}$ divided by the coefficient of $x^{17}$, or just $0$.
Hence, $2\left(\sum_{1\le i\le j\le14}r_ir_j\right)=0$.

Finally, we have $\sum_{i=1}^{14}(r_i)^{14}=14*17-0+3\left((0)^2-2(0)\right)=14*17$.

Well, to make people feel better, remeber that even the 15th place person got into the AIME last year.

hey, poo! #23 was on the canadian national olympiad 1997. what the heck?

Wow - when you guys made it difficult, I didn't take it that an AMC 11 'ish' would contain olympiad problems. But, eh, wut the heck you know?

It's a little different when it's multiple choice.  But still, it [i]was[/i] difficult. :)

Lucky707' and Singthesorrow's Mock AMC last year had an IMO problem.

Alternate solution to #25(which generalize the result for any isosceles triangle ABC with AB=AC).

Let line AF intersect BC at G, let $\frac{AD}{BD}=x$, since AB=AC, AD=CE implies $\frac{CE}{AE}=x$, thus by Ceva's theorem $\frac{AD*BG*CE}{BD*CG*AE}=1$, or $\frac{BG}{CG}=\frac{1}{x^2}$. Now by Menelaus Theorem we have
$\frac{AD}{BD}*\frac{BC}{BG}*\frac{GF}{AF}=1=x(1+\frac{1}{x^2}) \frac{GF}{AF}$. Since $\frac{[BFC]}{[ABC]}=\frac{2}{7}=\frac{FG}{AG}$, so $\frac{FG}{AF}=\frac{2}{5}$.

so $x(1+\frac{1}{x^2}) \frac{2}{5}=1$, or $2x^2-5x+2=0$ $x=2, \frac{1}{2}$ and we take 2.

Where can I get the AMC 10 2005A contest questions?

You can see the last 5 questions (with answers and solutions) at the Mathjam archive. [url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php[/url]


hello, to the inequality (i). I don't need Jensen it is simple algebra. Your inequality is equivalent with
$ 4(a^3\plus{}b^3) \geq a^3\plus{}3a^2b\plus{}3ab^2\plus{}b^3$
$ 3(a^3\plus{}b^3) \geq 3a^2b\plus{}3ab^2$
$ 3(a\plus{}b)(a^2\minus{}ab\plus{}b^2) \geq 3ab(a\plus{}b)$
since $ a,b>0$
$ (a\minus{}b)^2\geq0$
Sonnhard.

[hide="Jensen"]
Since $ f \left( x \right) \equal{} x^3 \rightarrow f ' \left( x \right) > 0 \ \forall x>0$, Jensen's can be used.

By Jensen's
$ f \left( a \right) \plus{}  f \left( b \right) \plus{}  f \left( c \right) \geq f \left( a \plus{}b \plus{} c \right)$.

Hence, the inequalities given in the problem are obviously true.
[/hide]

[quote="AstroPhys"][hide="Jensen"]
Since $ f \left( x \right) \equal{} x^3 \rightarrow f ' \left( x \right) > 0 \ \forall x > 0$, Jensen's can be used.

By Jensen's
$ f \left( a \right) \plus{} f \left( b \right) \plus{} f \left( c \right) \geq f \left( a \plus{} b \plus{} c \right)$.

Hence, the inequalities given in the problem are obviously true.
[/hide][/quote]

You need $ f''(x)\ge0$ for $ f(x)$ to be convex. Also, you're applying Jensen incorrectly. The correct statement is $ f(a)\plus{}f(b)\plus{}f(c)\ge3f(\dfrac{a\plus{}b\plus{}c}{3})$.

[quote="Machodog"]a, b, and c represent positive real numbers.

(i) Prove that 4(a^3+b^3) >= (a+b)^3
(ii)Prove that 9(a^3+b^3+c^3) >= (a+b+c)^3[/quote]

The following identities are true for all $ a,b$ and $ c$:

\[ 4\,(a^3\plus{}b^3)\minus{}(a\plus{}b)^3 \equal{} 3\,(a\plus{}b)\,(a\minus{}b)^2\]

\[ 9\,(a^3\plus{}b^3\plus{}c^3)\minus{}(a\plus{}b\plus{}c)^3 \equal{} \frac{27}{4}(a\plus{}b)\,(a\minus{}b)^2\plus{}\frac{1}{4}(5\,a\plus{}5\,b\plus{}8\,c)\,(a\plus{}b\minus{}2\,c)^2\]

This clearly proves that for non-negative $ a,b,c$ left part is no less then zero.

This method can be applied to much harder inequalities. If you want to be able to find such identities yourself, see http://www.mathlinks.ro/viewtopic.php?p=1156480#1156480

Have a nice day!
Sergey Markelov

Is this how you Jensen b) ?

$ f(x)\equal{}x^3$ is convex for positive $ x$, so
$ f(3a) \plus{} f(3b) \plus{} f(3c) \geq 3f(a\plus{}b\plus{}c)$, giving us our inequality.

Bah, okay, I have to admit I messed up using Jensen's, but the $ f ' \left( x \right)$ was a mistake.  :P

Hey guys thanks for all the replies.

I solved (i) like Sonnhard, but i don't know jensen's inequality yet, so is there any other easier way of proving it?

This question was on british maths olympiad round 1, so i'm sure there is an easier solution which doesn't require jensen's inequality.

How about this one?

[quote="Machodog"]a, b, and c represent positive real numbers.

(i) Prove that $ 4(a^3 \plus{} b^3)\geq (a \plus{} b)^3$.
(ii)Prove that $ 9(a^3 \plus{} b^3 \plus{} c^3)\geq (a \plus{} b \plus{} c)^3$.[/quote]

(i) $ 4(a^3 \plus{} b^3)\geq (a \plus{} b)^3\Longleftrightarrow 4(a^2 \minus{} ab \plus{} b^2)\geq (a \plus{} b)^2\ \because a > 0,\ b > 0,\ c > 0$
$ \Longleftrightarrow 3(a \minus{} b)^2\geq 0$. $ Q.E.D.$

(ii) From the result of (i), $ 4(a^3 \plus{} b^3)\geq (a \plus{} b)^3,\ 4(b^3 \plus{} c^3)\geq (b \plus{} c)^3,\ 4(c^3 \plus{} a^3)\geq (c \plus{} a)^3$,

Adding them up gives $ 8(a^3 \plus{} b^3 \plus{} c^3)\geq (a \plus{} b)^3 \plus{} (b \plus{} c)^3 \plus{} (c \plus{} a)^3$.

$ \therefore 9(a^3 \plus{} b^3 \plus{} c^3)\geq a^3\plus{}b^3\plus{}c^3 \plus{} 3(a \plus{} b)(b \plus{} c)(c \plus{} a)\ A.M. \minus{} G.M.$

$ \Longleftrightarrow 9(a^3 \plus{} b^3 \plus{} c^3)\geq (a \plus{} b \plus{} c)^3$. $ Q.E.D.$

For the 2 variable one:

$ a^3 \plus{} 3a^3 \plus{} 3b^3 \plus{} b^3 \geq a^3 \plus{} 3a ^2 b \plus{} 3a b^2 \plus{} b^3$
$ \iff a^3 \plus{} b^3 \geq a^2 b \plus{} a b^2$

Which is easily shown via the [url=http://en.wikipedia.org/wiki/Rearrangement_inequality]Rearrangement Inequality[/url] with the two lists $ (a,b)$ and $ (a^2, b^2)$.

Same approach works for higher numbers of variables.


Find all three digit perfect squares whose digits (in base 10) are also perfect squares.

nice solution. :D mine was longer. i think this cannot be solved in other ways except for trial. :maybe:


[quote="mdk"]Let O be the intersection of the diagonals of the convex quadrilateral ABCD. Prove that if the perimeter of the triangles ABO, BCO, CDO, DAO are equal, then ABCD is a rhombus.[/quote]

see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=53470&hilit=rhombus


Diophantine equation
number theory
algebra
equation
polynomial
IMO Shortlist

Find all pairs of solutions $(x,y)$:

\[ x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1). \]

THis must be a joke! Problem 6? If my computations aren't wrong, then denoting $ s=x+y$ and $ p=xy$ we find that $ 2p(s-4)=s^3-8s^2+8$. Thus, it follows that $ s-4| s^3-4s^2-4(s^2-16)+56$ and so $ s-4| 56$. Also, s must be even. Then one can only check for the cases, find p and find x and y.Bleah!


Αυτό είναι το δώρο μου στις ανισότητες για τα παιδιά που θα πάνε στην ΙΜΟ.
[hide]φυσικά δεν είναι δικές μου :roll: [/hide]

$ 1)$
Για $ a,b,c$ θετικούς πραγματικούς έτσι ώστε $ abc\equal{}1$ ν.δ.ο.:
$ (a\plus{}b)(b\plus{}c)(c\plus{}a)\plus{}7\ge 5(a\plus{}b\plus{}c)$.

$ 2)$
Για $ a,b,c$ θετικούς πραγματικούς έτσι ώστε $ abc\equal{}1$ τότε νδο:
$ a^3\plus{}b^3\plus{}c^3\plus{}15 \ge 6\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \right)$

 :)

[quote="silouan"]Για την πρώτη δες εδώ http://www.mathlinks.ro/viewtopic.php?search_id=1235261040&t=191102

Η δεύτερη λύνεται επίσης με την ίδια μέθοδο  :wink:[/quote]
Για κάτι τέτοιες λύσεις νιώθω πραγματικά τυχερός που βρίσκομαι στο Forum :lol: .Eυχαριστώ silouan,με βοήθησε να καταλάβω κάποια πράγματα στους mixing variables η λύση σου....

Την δεύτερη την προσπάθησα 2 ώρες με mixing variables και τελικά νομίζω ότι την έλυσα :D .....

Έστω όπου $ a \equal{} x^2,b \equal{} y^2,c \equal{} z^2$.
Θεωρούμε την:
$ f(x,y,z) \equal{} x^6 \plus{} y^6 \plus{} z^6 \minus{} 6\left(\frac {1}{x^2} \plus{} \frac {1}{y^2} \plus{} \frac {1}{z^2}\right) \plus{} 15$.
Έστω $ x \equal{} min\{x,y,z\}$.Άρα $ x\le 1$ kai $ yz \ge 1$
Θεωρούμε την $ f(x,\sqrt {yz},\sqrt {yz}) \equal{} x^6 \plus{} 2y^3z^3 \minus{} 6\left(\frac {1}{x^2} \plus{} \frac {2}{yz}\right) \plus{} 15$.
Έχουμε:
$ f(x,y,z) \minus{} f(x,\sqrt {yz},\sqrt {zx}) \equal{} y^6 \minus{} 2y^3z^3 \plus{} z^6 \minus{} 6\left(\frac {1}{y^2} \minus{} \frac {2}{yz} \plus{} \frac {1}{z^2}\right) \equal{} (y^3 \minus{} z^3)^2 \minus{} 6\left(\frac {(y \minus{} z)^2}{y^2z^2}\right) \equal{} (y \minus{} z)^2\left((y^2 \plus{} yz \plus{} z^2)^2 \minus{} \frac {6}{y^2z^2}\right) \ge (y \minus{} z)^2\left(9y^2z^2 \minus{} \frac {6}{y^2z^2}\right) \ge 0$ γιατί $ yz \ge 1$.

Επομένως αρκεί νδο:
$ f(x,\sqrt {yz},\sqrt {yz}) \equal{} f(x,\sqrt {\frac {1}{x}},\sqrt {\frac {1}{x}})\ge 0$.
Alla:
$ f(x,\sqrt {\frac {1}{x}},\sqrt {\frac {1}{x}}) \equal{} x^6 \plus{} \frac {2}{x^3} \minus{} 6\left(\frac {1}{x^2} \plus{} 2x\right) \plus{} 15 \ge 0 \Leftrightarrow x^9 \minus{} 12x^4 \plus{} 15x^3 \minus{} 6x^2 \plus{} 2 \ge 0$.
Αλλά από horner αυτό γράφεται:
$ (x \minus{} 1)^2(x^7 \plus{} 2x^6 \plus{} 3x^5 \plus{} 4x^4 \plus{} 5x^3 \minus{} 6x^2 \minus{} 2x \plus{} 2)\ge 0$ με $ x \le 1$.
Aλλά: $ 2(2 \minus{} 2x \minus{} 6x^2 \plus{} 5x^3 \plus{} 3x^4) \ge 0\Leftrightarrow  (2 \minus{} 3x)^2(1 \plus{} 2x \plus{} \frac {3}{4}x^2) \plus{} x^3(1 \minus{} \frac {3}{4}x) > 0$.Done!!!
(αυτή η τελευταία παραγοντοποίηση έγινε με τη βοήθεια του Vasc :lol: ).

Ελπίχω να μην έχω κάνει κάποιο λάθος....Παρακαλώ silouan να την ελέγξεις και αν είχες διαφορετική λύση στο μυαλλό θα ήθελα να την δω.....

Η λογική αυτή είναι ! Μια γρήγορη ματιά λέει ότι και οι πράξεις είναι σωστές. Απλά πολλές φορές εμφανίζεται πρόβλημα και στο πρώτο βήμα της διαδικασίας όπου θα πρέπει να κάνουμε κάτι έξυπν όπως ο Vasc.


AMC
AMC 12
AMC 12 A
AMC 10
AMC 10 B
USA(J)MO
USAMO

I made some mistakes in AMC12A and only get 115.5, and after I saw the discussions on the forum I decided to take AMC10B and this time I got 150!
   Does that mean my USAMO index will consist of my AMC10B score instead of my AMC 12 A score? How many points do I need for AIME to get into USAMO?
   Suggestion for 9 or 10th graders who want to make USAMO, you should definitely take AMC 10, it's much easier.

 :lol:  :lol:

Yes, your index will use your AMC 10B score. You'll probably need a 6 on AIME.

How do you go about calculating what someone would need to make USAMO? For example, I got a 114 on the 10A and a 138 on the 10B.

The cutoff is not exactly the same every year but it is usually around 200.


suppose (C) is a circle inwich its senter is inaccessible.only by using compass find the center of (C).

maybe i should change it to compas :blush: .

the solution of this problem is in one of my books and i think it is really nice.it is so long so i'll send it later. :D (napelon was so cool in math you knew that?)

Yep, that's good old Napoleon Bonaparte. He wasn't a bad mathematician - for example, look at the Napoleon theorem :)

According to Mohr-Mascheroni theorem, all constructions possible with a compass and straightedge are possible with a compass alone (see http://homeweb2.unifr.ch/hungerbu/pub/Mohr/mohr.ps)

This said, constructions are always awfully tricky.

Indeed this problem was solved by Napoléon ! Pierre Simon de Laplace (who had been a teacher of Napoléon, as well as a famous mathematician) is said to have had these words : "We expected everything from you, general, except a lesson of geometry"  ;)

thanks alex and  tutu,,,very use full links :)

[quote="Peter VDD"]napelon = napoleon? Anyway I'm curious to see your solution.[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=57006

[quote="Ezbakhe Yassin"][quote="K81o7"]Am I thinking of the right problem?  As in you have a circle and must determine its center?  If so, I have a really easy solution...[/quote]

with the use only of a compas ;)[/quote]

You can't draw straight lines with a compass?

we can live wthout using ruller.

because compass will do all the doings of ruller.

what is the important works of ruller?

not drawing a straight line.

finding the intersections of the two line.

[quote="Arne"]That's the point ;) It can be done using a compass alone, if I'm not mistaken.[/quote]

exactly...

ure right... ;)

[quote="K81o7"]Am I thinking of the right problem?  As in you have a circle and must determine its center?  If so, I have a really easy solution...[/quote]

it seems that u have a nice solution for it...

then post it plz... ;)

yes yes yes...

exactly yetti and ashegh solution...

[quote="K81o7"][quote="Altheman"]choose 3 points on the circle, and use the construction that finds the circumcenter of that triangle...i think it uses inversion...[/quote]

That's what I was thinking...except i was under the impression that one can draw straight lines with a compass, which makes its pretty trivial.  We would just take the three points, draw the three perpendicular bisectors, and we would be done.  How does one do it without a straightedge?[/quote]

any body?... ;)


calculus
integration
trigonometry
function
trig identities
calculus computations

Evaluate $ \int_{\frac {\pi}{4}}^{\frac {3}{4}\pi } \cos \frac {1}{\sin \left(\frac {1}{\sin x}\right)}\cdot \cos \left(\frac {1}{\sin x}\right)\cdot \frac {\cos x}{\sin ^ 2 x\cdot \sin ^ 2 \left(\frac {1}{\sin x }\right)}\ dx$

Last Edited, Sorry

kunny

A bit of clever trig manipulation gets me to
$ \int_{\frac {\pi}{4}}^{\frac {3}{4}\pi } \csc^2 x \cdot \cot(\csc(x)) \cdot \cos(\csc(\csc(x))) dx$
I think you can make the substitution
$ u \equal{} \csc(\csc(x))$
$ du \equal{} \cot(x)*\csc(x)*\cot(\csc(x))*\csc(\csc(x)) dx$
Well, it seemed like it would work :/
Perhaps $ u \equal{} \csc(x), du \equal{} \minus{}\csc x \cot x dx$ is a better choice.
I'm pretty sure that the trig identities are right, but I'm not sure what substitution I should make to have it all work out.

kunny, i can't calculate your original integral  :(  , but i can do the following (see the extra term in [u][b]box[/b][/u] below)

$ \textbf I \;\equal{}\;\int_{\frac {\pi}{4}}^{\frac {3}{4}\pi } \cos \left(\frac {1}{\sin \left(\frac {1}{\sin x}\right)}\right)\cdot \cos \left(\frac {1}{\sin x}\right)\cdot \boxed{\cos x}\cdot \frac {1}{\sin ^ 2 x\cdot \sin ^ 2 \left(\frac {1}{\sin x }\right)}\quad \textbf dx$

since using the substitution

$ y \;\equal{}\; \sin \left(\frac {1}{\sin \left(\frac {1}{\sin x}\right)}\right)$

gives

$ \dfrac{\textbf dy}{\textbf dx} \;\equal{}\; \cos \left( \frac {1}{\sin \left(\frac {1}{\sin x}\right)}\right) \cdot \cos \left(\frac {1}{\sin x}\right)\cdot \cos x\cdot \frac {1}{\sin ^ 2 x\cdot \sin ^ 2 \left(\frac {1}{\sin x }\right)}$

and in that case, i get

$ \textbf I \;\equal{}\;2\; \int_{\frac {\pi}{4}}^{\frac {\pi}{2}} \cos \left(\frac {1}{\sin \left(\frac {1}{\sin x}\right)}\right)\cdot \cos \left(\frac {1}{\sin x}\right)\cdot \cos x\cdot \frac {1}{\sin ^ 2 x\cdot \sin ^ 2 \left(\frac {1}{\sin x }\right)}\quad \textbf dx$

$ \equal{}\; 2\; \int_{ \cos \left( \text{cosec}\; \sqrt 2\right)}^{ \cos \left( \text{cosec}\;1\right)} \;\; \textbf dy \;\equal{}\; 2\; \left[ \cos \left( \text{cosec} \;1\right) \;\minus{}\;  \cos \left( \text{cosec}\; \sqrt 2\right)\right]$

also sorry i can't be on the forums that much anymore   :(  , since i am teaching a class on [b][u]special functions[/u][/b]

using ( [url=http://www.amazon.com/Course-Analysis-Cambridge-Mathematical-Library/dp/0521588073/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1196403297&sr=8-1]book 1[/url]  and [url=http://www.amazon.com/Treatise-Functions-Cambridge-Mathematical-Library/dp/0521483913/ref=pd_sim_b_img_4]book 2[/url] ), and it is taking up way toooo much of my time.........   :help:

Hello, misan. :)  We have not seen you here recently, so I wonder what does it happen to you.

You are very busy, aren't you.

P.S. I have books 2, too. I have finished reading the book before, but I have almost forgotten of all it.

Especially Bessel function is very important for hydro dynamics.

[color=blue]I am verry sorry to take troubles to you, DiscreetFourierTransform and misan.[/color]

I made a big typo.

kunny


Let $ a,b,c$ be reals s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}2$.Prove that:
$ |a^3\plus{}b^3\plus{}c^3\minus{}abc|\leq2\sqrt{2}$

We prove that $ (a^3\plus{}b^3\plus{}c^3\minus{}abc)^2\leq (a^2\plus{}b^2\plus{}c^2)^3$
$ \Leftrightarrow \sum_{cyc} a^6\plus{}a^2b^2c^2\plus{}2\sum_{cyc} a^3b^3\minus{}2\sum_{cyc} a^4bc\leq \sum_{cyc} a^6\plus{}3\sum_{sym} a^4b^2\plus{}6a^2b^2c^2$
$ \Leftrightarrow 3\sum_{sym} a^4b^2\plus{}2\sum_{cyc} a^4bc\plus{}5a^2b^2c^2\minus{}2\sum_{cyc} a^3b^3\geq 0$
$ \Leftrightarrow \sum_{cyc}(a^2b\minus{}ab^2)^2\plus{}\sum_{cyc}(a^2b\plus{}a^2c)^2\plus{}\sum_{sym} a^4b^2\plus{}5a^2b^2c^2\geq 0$
Equal when $ a\equal{}b\equal{}0,c\equal{}\pm \sqrt{2}$

another solution:

\begin{eqnarray*}|a^3 + b^3 + c^3 - abc| = |(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)| = \left|(a + b + c)\left(\frac {6 - (a + b + c)^2}2\right)\right|\end{eqnarray*}

now let $ x = |a + b + c|$ so $ 0\leq x\leq \sqrt {6}$ so we have:

$ |a^3 + b^3 + c^3 - abc|\leq 2\sqrt 2$

$ \iff\frac {x(6 - x^2)}2\leq 2\sqrt 2$

$ \iff x^3 + 4\sqrt 2\geq 6x$

which is true by AM-GM:

$ x^3 + 2\sqrt 2 + 2\sqrt 2\geq 3\sqrt [3]{8x^3} = 6x$

QED

[quote="Erken"][quote="BaBaK Ghalebi"]another solution:

$ \begin{eqnarray*}|a^3 + b^3 + c^3 - abc| = |(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)|$

[/quote]
But this is not true :maybe:[/quote]

sorry,I misread the problem,I must have confused it with $ |a^3 + b^3 + c^3 - 3abc|$.