This was made to public just today (even though we knew that we made it 3 weeks ago, we were adviced not to make it public knowledge).

[list]
[*]Oleg Ivrii, 11
[*]János Kramár, 11
[*]Dong Uk (David) Rhee, 10
[*]Peng Shi, 10
[*]Jacob Tsimerman, 10
[*]Yufei Zhao, 10[/list]

Leader: Dr. Christopher Small (University of Waterloo)
Deputy Leader: Dr. Edward Wang (Wilfrid Laurier University)
Observer with Leader: Prof. Felix Recio (University of Toronto)

We definitely have a very young team.

[quote="Lagrangia"]I just read these names and I was amazed when I saw them. Are they Canadians? :D :D[/quote]

lol. Canada is known for its cultural diversity. Last year there was only one "native" Canadian. This year, there are none.

Oh there is a KOREAN  CANADIAN Dong Uk (David) Rhee .

I'm KOREAN too


[quote="Euclidean Geometer"]This ones terribly simple, though here u go...

Given any two reals show that there always exists a third real no so that the sum of the squares of the three is equal to the square of their sum. :D[/quote]

Let the third no be

$ \frac{\minus{}ab}{a\plus{}b}$

[quote="alter ego"][quote="Euclidean Geometer"]T
Given any two reals:D[/quote]

however, if the reals chosen are a and -a, , there does not exist a third real satisfying the given condition[/quote]

Ahhh...Yes, I forgot that.. I guess my denominator takes care of that.

Well, yes as alter ego pointed out, that needs be taken care of.
And i think i must mention non zero reals.Its a lil loosely framed cos i posed it myself  :D
There was a Beautiful identity that i was trying to show by this problem,
silly enough, but elegant  :wink:

We have for distinct x,y,z
 1/(x-y) , 1/(y-z) and 1/(z-x) are three numbers that satisfy the conditions of my problem.


[quote="nguoivn"]Given $ a, b, c \ge 0$ and $ k \ge 2$. Prove that:
i) $ \frac {ab(a \plus{} b)}{a^2 \plus{} b^2 \plus{} 2c^2} \plus{} \frac {bc(b \plus{} c)}{b^2 \plus{} c^2 \plus{} 2a^2} \plus{} \frac {ca(c \plus{} a)}{c^2 \plus{} a^2 \plus{} 2b^2} \le \frac {a \plus{} b \plus{} c}{2}$

ii) $ \frac {a^k(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {b^k(c \plus{} a)}{c^2 \plus{} a^2} \plus{} \frac {c^k(a \plus{} b)}{a^2 \plus{} b^2} \ge a^{k \minus{} 1} \plus{} b^{k \minus{} 1} \plus{} c^{k \minus{} 1}$
 :)[/quote]

second one:

$ \frac {a^k(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {b^k(c \plus{} a)}{c^2 \plus{} a^2} \plus{} \frac {c^k(a \plus{} b)}{a^2 \plus{} b^2} \ge a^{k \minus{} 1} \plus{} b^{k \minus{} 1} \plus{} c^{k \minus{} 1}$

\[ \Longleftrightarrow \sum {\frac {{ab\left( {\left( {a \minus{} b} \right)\left( {a^k \minus{} b^k } \right) \plus{} c^2 \left( {a \minus{} b} \right)\left( {a^{k \minus{} 2} \minus{} b^{k \minus{} 2} } \right)} \right)}}{{\left( {b^2 \plus{} c^2 } \right)\left( {c^2 \plus{} a^2 } \right)}}} \ge 0.\]
for $ k\ge2$, whether $ a\le b$ or $ a\ge b$, we also have $ {\left( {a \minus{} b} \right)\left( {a^k \minus{} b^k } \right)}\ge 0$ and $ {\left( {a \minus{} b} \right)\left( {a^{k \minus{} 2} \minus{} b^{k \minus{} 2} } \right)}\ge 0$, so the inequality holds...

[quote="nguoivn"]Given $ a, b, c \ge 0$ and $ k \ge 2$. Prove that:
i) $ \frac {ab(a \plus{} b)}{a^2 \plus{} b^2 \plus{} 2c^2} \plus{} \frac {bc(b \plus{} c)}{b^2 \plus{} c^2 \plus{} 2a^2} \plus{} \frac {ca(c \plus{} a)}{c^2 \plus{} a^2 \plus{} 2b^2} \le \frac {a \plus{} b \plus{} c}{2}$
 :)[/quote]
This is not true, nguoivn. A counterexample is $ (a,b,c)\equal{}(0.5,1,1).$

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c \ge 0$ and $ k \ge 2$. Prove that:
i) $ \frac {ab(a \plus{} b)}{a^2 \plus{} b^2 \plus{} 2c^2} \plus{} \frac {bc(b \plus{} c)}{b^2 \plus{} c^2 \plus{} 2a^2} \plus{} \frac {ca(c \plus{} a)}{c^2 \plus{} a^2 \plus{} 2b^2} \le \frac {a \plus{} b \plus{} c}{2}$
 :)[/quote]
This is not true, nguoivn. A counterexample is $ (a,b,c) \equal{} (0.5,1,1).$[/quote]
Sorry, I think the easier is trues  :maybe: :
$ \frac {ab(a \plus{} b)}{2a^2 \plus{} 2b^2 \plus{} 5c^2} \plus{} \frac {bc(b \plus{} c)}{2b^2 \plus{} 2c^2 \plus{} 5a^2} \plus{} \frac {ca(c \plus{} a)}{2c^2 \plus{} 2a^2 \plus{} 5b^2} \le \frac {a \plus{} b \plus{} c}{4}$


inequalities
calculus
derivative
real analysis
real analysis theorems

How to prove (1+x/n)^n is increasing wrt n for x>0??
ie given fn(x) = (1+x/n)^n => fn(x) < fn+1(x) for every x>0
also how prove fn(x) decreasing for x<0?(i believe it's exactly) the same technique..

Suppose $a_n = (1+\frac xn)^n$

$\frac{a_{n+1}}{a_n} = \frac{(1+\frac x{n+1})^{n+1}}{(1+\frac xn)^n} = \left(\frac{1+\frac x{n+1}}{1+\frac xn}\right)^n\left(1+\frac{x}{n+1}\right)$

$\frac{1+\frac x{n+1}}{1+\frac xn} = \frac{(n+x+1)n}{(n+1)(n+x)} = 1-\frac{1}{(n+1)(n+x)}$
hence
$\left(\frac{1+\frac x{n+1}}{1+\frac xn}\right)^n = \left(1-\frac{1}{(n+1)(n+x)}\right)^n \geq 1-\frac{n}{(n+1)(n+x)} = \frac{(n+1+x)n}{(n+1)(n+x)}$ (here we use Bernoulli's Inequality : $(1+x)^\alpha \geq 1+\alpha x$ for $x\geq -1$ and $\alpha > -1$. Here we only meet the case that $\alpha$ is a positive integer and you can prove the inequality by induction).

So
$\frac{a_{n+1}}{a_n}\geq \frac{(n+1+x)n}{(n+1)(n+x)}\left(1+\frac{x}{n+1}\right) = \frac{n+2n^2+n^3+2nx+2n^2x+nx^2}{n+2n^2+n^3+2nx+n^2x+x} > 1$

Another way, which is much quicker, is to use AM-GM inequality as follows:

$\sqrt[n+1]{(1+\frac{x}{n})^n} = \sqrt[n+1]{(1+\frac{x}{n})(1+\frac{x}{n})\cdots (1+\frac{x}{n})\cdot 1} \leq \frac{1}{n+1}\left((1+\frac{x}{n})+\cdots+(1+\frac{x}{n})+1\right) = \frac{1}{n+1}\left(n\cdot(1+\frac xn)+1\right) = \frac{n+x+1}{n+1} = 1+\frac{x}{n+1}$

Another way (since this is the calculus forum, after all):  Assuming $f_n$ is positive, it increases or decreases with $n$ precisely when its log does.

$\frac{d}{dn} \log(f_n(x)) = \frac{d}{dn} (n \log(1+x/n))$
$=\log(1+x/n)+\frac{n}{1+x/n} \frac{-x}{n^2}$
$= \log(1+x/n)-1+\frac{n}{n+x}$

This last expression is 0 for $x=0$, and its derivative with respect to $x$ is

$\frac{1/n}{1+x/n} - \frac{n}{(n+x)^2} = \frac{x}{(n+x)^2}$, which is positive for positive $x$ and negative for negative $x$.

Thus the derivative of $\log(f_n)$ with respect to $n$ is positive for any $x$, so $f_n$ is increasing in $n$ for any $n$ for which it is positive and any $x$ not equal to 0.


LaTeX
symmetry
trigonometry
geometry proposed
geometry

In ABC triangle,the bisectors of < ABC and <ACB meet AC and AB at D and E,respectively,and <ADE-<AED=60 degree.Find <ACB angle.

Posted before on http://www.mathlinks.ro/viewtopic.php?t=205308  :wink:

I'm surprised it is not solved by anyone...

Let <ADE=a
<AED=b
also.<B=2p & <C=2q
Thus, a-b=60
a+b-p=2q+p
and, a+b-q=2p+q
both of which gives, a+b=2(p+q)
thus, a=2(p+q)+60/2=p+q+30
b=p+q-30
Is there any more restriction on the problem?........cause otherwise p & q can be chosen of our own. What is the next step in the problem??

Please use $ \text{\LaTeX}$. Using it makes your posts more readable  :wink: .

Well, I'm pretty certain that $ \angle ACB=120^{\circ}$, after experimenting for a while. But that does not mean that all the angles of $ ABC$ are fixed!
[quote="Sunkern_sunflora"]
...
both of which gives, $ a+b=2(p+q)$
thus, $ a=2(p+q)+60/2=p+q+30$
[/quote]
I think this part is wrong. Can you explain?

[hide]
Let $ F$ be the point on $ [AE]$ such that $ DF \equal{} DE$, and $ D'\in DF$ and $ DE\perp ED'$
Let also $ \angle AED \equal{} \alpha$.

We have $ DF \equal{} DE$ $ \Rightarrow \angle FDE \equal{} \alpha$.
By a quick angle chase, we get $ \angle FD'E \equal{} 90^{\circ} \minus{} \angle D'FE \equal{} 90^{\circ} \minus{} \alpha \equal{} \angle D'EF$, so $ DF \equal{} FE \equal{} FD'$, and $ F$ is the center of the circle $ \Gamma$ circumscribed to $ \triangle DED'$. Since $ \angle CEF \equal{} 90^{\circ}$, $ CE$ is the tangent of $ \Gamma$ at $ E$.
This way, $ \angle DEC \equal{} \angle DD'E \equal{} \alpha$.
Concluding, $ \frac {\angle C}{2} \equal{} 180^{\circ} \minus{} \angle DEC \minus{} \angle CDE \equal{} 180^{\circ} \minus{} (180^{\circ} \minus{} \angle EDA) \minus{} \alpha \equal{} \angle EDA \minus{} \alpha \equal{} 60^{\circ}$, so that $ \angle ACB \equal{} 120^{\circ}$

$ QED$
[/hide]

EDIT: Wrong solution

Are you sure? There is something strange. I think if angle ACB=120 then ABC=120 because of symmetry ... but ... 120+120>180 ... impossible.

I think with that notation angle BAC should be calculated but I'm not sure.

Well, i got a mistake is my proof, and it is completely invalid... :(  the right angle just isn´t right. I'll try to correct it tomorrow( or get a new one :roll: )
However, i think the problem is right, because we can't use symmetry, since $ \angle ADE\minus{}\angle AED\equal{}60^{\circ}$, but $ \angle AED\minus{}\angle ADE\equal{}\minus{}60^{\circ}$. Also, i've draw the diagram in my computer, and it seems correct :P

I'm sorry! I'm wrong - there is no symetry ... I'm sleeping in Bulgaria it is night...

Thanks for the advise of using LATEX. But I failed to understand the diagram used by you. Could you please calrify?

solved
first i will alter the problem and prove an iff statement that the angle is 120 degrees
now the if statement is pretty straight forward
for only if we just contradict the if statement
(borislav_mirchev your problems are really good at times)

i have seen only 3 of them of which 2 were amazing the third was good{this is one of the amazing ones}

[quote="borislav_mirchev"] [color=darkred]In $ \triangle ABC$ the bisectors of $ \widehat {ABC}$ and $ \widehat {ACB}$ meet $ AC$ and $ AB$ at $ D$ and $ E$ respectively and $ m(\widehat {ADE}) \minus{} m(\widehat {AED}) \equal{} 60^{\circ}$ . Find $ m(\widehat {ACB})$ .[/color] [/quote]
[size=134][color=darkblue][b]Here is just a joke ![/b][/color][/size]

[color=darkblue][b][u]Trigonometrical Proof.[/u][/b] Denote $ I\in BD\cap CE$ and $ B \equal{} 2y$ . Since $ m(\widehat {ADE}) \minus{} m(\widehat {AED}) \equal{} 60^{\circ}$ $ \implies$ exists $ x\in (30,90)$

so that $ \left\|\begin{array}{c} m(\widehat {ADE}) \equal{} x \plus{} 30 \\
\ m(\widehat {AED}) \equal{} x \minus{} 30\end{array}\right\|$ obtain $ \left\|\begin{array}{c} A \equal{} 180 \minus{} 2x \\
 \\
\boxed {\ C \equal{} 2(x \minus{} y)\ }\end{array}\right\|$ . Show easily that $ \left\|\begin{array}{ccc} m(\widehat {IDE}) & \equal{} & x \minus{} y \minus{} 30 \\
 \\
m(\widehat {DEI}) & \equal{} & y \plus{} 30\end{array}\right\|$

and $ \left\|\begin{array}{ccc} m(\widehat {EIA}) & \equal{} & 90 \minus{} y \\
 \\
m(\widehat {AID}) & \equal{} & 90 \minus{} x \plus{} y\end{array}\right\|$ . Apply an well-known trigonometrc relation in the convex quadrilateral $ AEID$ :

$ \boxed {\ \begin{array}{c} \\
\ \sin\widehat {IAE}\sin\widehat {DEI}\sin\widehat {AID}\sin\widehat {EDA} \equal{} \sin\widehat {DAI}\sin\widehat {AED}\sin\widehat {EIA}\sin\widehat {IDE} \\
 \\
\sin (y \plus{} 30)\sin (90 \minus{} x \plus{} y)\sin (x \plus{} 30) \equal{} \sin (x \minus{} 30)\sin (90 \minus{} y)\sin(x \minus{} y \minus{} 30) \\
 \\
\sin (y \plus{} 30)\cos (x \minus{} y)\sin (x \plus{} 30) \equal{} \sin (x \minus{} 30)\cos y\sin(x \minus{} y \minus{} 30) \\
 \\
\sin (x \plus{} 30)[\sin (x \plus{} 30) \plus{} \sin (2y \plus{} 30 \minus{} x)] \equal{} \sin (x \minus{} y \minus{} 30)[\sin(x \plus{} y \minus{} 30) \plus{} \sin (x \minus{} y \minus{} 30)] \\
 \\
1 \minus{} \cos (2x \plus{} 60) \plus{} \cos (2y \minus{} 2x) \minus{} \cos (2y \plus{} 60) \equal{} 1 \minus{} \cos (2y \plus{} 60 \minus{} 2x) \plus{} \cos 2y \minus{} \cos (2x \minus{} 60) \\
 \\
\ [\cos (2x \minus{} 60) \minus{} \cos (2x \plus{} 60)] \plus{} [\cos (2y \minus{} 2x) \minus{} \cos 2y] \plus{} [\cos (2y \plus{} 60 \minus{} 2x) \minus{} \cos (2y \plus{} 60)] \equal{} 0 \\
 \\
\sqrt 3\sin 2x \plus{} 2\sin 2y \minus{} x)\sin x \plus{} 2\sin (2y \plus{} 60 \minus{} x)\sin x \equal{} 0 \\
 \\
\sqrt 3\cos x \plus{} [\sin (2y \minus{} x) \plus{} \sin (2y \plus{} 60 \minus{} x)] \equal{} 0 \\
 \\
\sqrt 3\cos x \plus{} \sqrt 3\sin (2y \minus{} x \plus{} 30) \equal{} 0 \\
 \\
\cos x \equal{} \sin (x \minus{} 2y \minus{} 30) \\
 \\
\cos x \equal{} \cos (120 \plus{} 2y \minus{} x) \\
 \\
x \equal{} 120 \plus{} 2y \minus{} x \\
 \\
x \minus{} y \equal{} 60 \\
 \\
\blacktriangleright\ \boxed {\ C \equal{} 120\ }\ \blacktriangleleft \\
\ \end{array}\ }$[/color]
[size=134][color=darkblue][b]

I am surely you didn't like it ![/b][/color][/size]

Why to not like?! It is hard work and application of knowledge, collected over the time so I like it. Thank you for the solution.

I know there may be synthetical proof and people like Leon, vittasko, Quatto Master, April, etc. may know or invent it but it is not possible every time to see such a proof at first glance.


geometry
trigonometry
AMC
USA(J)MO
USAJMO
perpendicular bisector
geometry proposed

[quote="pohoatza"][color=darkred]Let $ABC$ be a triangle such that $b>c$ and denote by $D,E,F$ the intersections of the its incircle with the sides $BC$, $AC$, $AB$ respectively.
Consider the circle $w=C(M)$ with the diameter $[BC]$. Denote the intersections $\{\begin{array}{c}X\in w\cap (NF\\\ Y\in w\cap (NE\\\ N \in AM \cap EF\end{array}$. Prove that $\frac{NX}{NY}=\frac{AB}{AC}$.[/color][/quote]
[color=darkblue][b][u]Proof I.[/u][/b] The well-known properties $\|\boxed{\begin{array}{c}IN\perp BC\\\\ X\in CI\ ,\ Y\in BI\end{array}}\|$ $\implies$ $\{\begin{array}{c}m(\widehat{YXI})=\frac{B}{2}\ ,\ m(\widehat{XYI})=\frac{C}{2}\\\\ m(\widehat{NIX})=90^{\circ}-\frac{C}{2}\ ,\ m(\widehat{NIY})=90^{\circ}-\frac{B}{2}\end{array}\|$ $\implies$

$\frac{NX}{NY}=\frac{IX}{IY}\cdot\frac{\sin \widehat{NIX}}{\sin \widehat{NIY}}=$ $\frac{\sin \widehat{XYI}}{\sin\widehat{YXI}}\cdot \frac{\sin \widehat{NIX}}{\sin \widehat{NIY}}=$ $\frac{\sin\frac{C}{2}}{\sin\frac{B}{2}}\cdot\frac{\cos\frac{C}{2}}{\cos\frac{B}{2}}$ $\implies$ $\frac{NX}{NY}=\frac{\sin C}{\sin B}$ $\implies$ $\boxed{\ \frac{NX}{NY}=\frac{AB}{AC}\ }$. [b][u]Remark. [/u][/b]$\frac{NF}{NE}=\frac{AC}{AB}$.

[b][u]Proof II.[/u][/b] Denote the intersection $Z\in BX\cap CY$. Observe that $Z\in ID$ and the point $I$ is the orthocenter of the triangle $BZC$, i.e. the triangle $XDY$ is the orthic triangle of $\triangle BZC$. From the well-known property "the ray $[DI$ is the bisector of the angle $\widehat{XDY}$" obtain $\frac{NX}{NY}=\frac{DX}{DY}$. Prove easily that $\{\begin{array}{c}m(\widehat{DXY})=B\\\ m(\widehat{DYX})=C\end{array}\|$ $\implies$ $\triangle XDY\sim\triangle BAC$ $\implies$ $\frac{DX}{DY}=\frac{AB}{AC}$. In conclusion, $\frac{NX}{NY}=\frac{AB}{AC}$.

[b][u]A general remark.[/u] If you know the above mentioned remarkable properties, then this very nice Pohoatza's problem is easily ![/b][/color]

[quote="Virgil Nicula"][size=117][color=darkred][b][u]A similar problem.[/u][/b] Let $C(I)$ be the incircle of $\triangle ABC$. Denote the second intersections $M$, $N$, $P$ of the circumcircle $C(O)$ of $\triangle ABC$ with the lines $AI$, $BI$, $CI$ respectively. Prove that $AM\perp NP$ and $R\in OM\cap NP\Longrightarrow$ $\frac{RN}{RP}=\frac{AB}{AC}$.[/color][/size][/quote]

[quote="Virgil Nicula"]
 $ X\in CI\ ,\ Y\in BI$

[/quote]

Please, Virgil Nicula, I have no idea how this comes!

[quote="cancer"][quote="Virgil Nicula"]
 $ X\in CI\ ,\ Y\in BI$[/quote]Please, Virgil Nicula, I have no idea how this comes![/quote]
See http://www.mathlinks.ro/Forum/viewtopic.php?t=155800

What? I can't see how it's related... are there any tangent circles?  :maybe:

[quote="Virgil Nicula"]See the definitions of the points $ X$, $ Y$ from the above enunciation !!.[/quote]

I already see! Unless the incircle and $ w$ are tangent??

[color=darkblue][b]NO ![/b]

1. $ N\in AM\cap EF$.
2. $ (NE$ and $ (NF$ are rays (semiline) !
3. The circle $ w$ has the center in the middle $ M$ of the side $ [BC]$ and the length of the its radius is equal to $ MB$ !

Sorry, this is the all ! Now please, ask you Pohoatza to explain away ...[/color]

[quote="pohoatza"]Let $ ABC$ be a triangle, let $ E, F$ be the tangency points of the incircle $ \Gamma(I)$ to the sides $ AC$, respectively $ AB$, and let $ M$ be the midpoint of the side $ BC$. Let $ N \equal{} AM \cap EF$, let $ \gamma(M)$ be the circle of diameter $ BC$, and let $ X, Y$ be the other (than $ B, C$) intersection points of $ BI$, respectively $ CI$, with $ \gamma$. Prove that
\[ \frac {NX} {NY} \equal{} \frac {AC} {AB}.
\]
[i]Author: Cosmin Pohoata[/i][/quote]
We have some well-known results in this configuration:
1. $ X$, $ Y$ lie on line $ EF$.
2. $ N$ lies on $ ID$, where $ D$ is the tangency point of $ \Gamma(I)$ to the side $ BC$.

Now, denote by $ P$ the intersection of $ AI$ with $ BC$. It's easily to see that $ \angle NIX\equal{}\angle PIC$, therefore triangles $ NIX$ and $ PIC$ are similar. It implies $ \frac{NX}{NI}\equal{}\frac{PC}{PI}$. Similarly, we conclude that $ \frac{NI}{NY}\equal{}\frac{PI}{PB}$.

Hence, we have $ \frac{NX}{NY}\equal{}\frac{NX}{NI}\cdot\frac{NI}{NY}\equal{}\frac{PC}{PI}\cdot\frac{PI}{PB}\equal{}\frac{PC}{PB}\equal{}\frac{AC}{AB}$. This completes our solutions.  :lol:

Dear April,
Your proof is Cosmin's. It was introduced in an aritcile of his.

Let $ D$ be the tangency point of $ BC$ and $ (I)$ and $ \mathcal {K}\equiv EF\cap BC$. Also, let $ X',Y'$ respectively be the intersections of $ BI,CI$ with $ EF$. It is well known that $ (\mathcal {K}DBC) \equal{} \minus{} 1$ $ \Longrightarrow$ $ (X'\mathcal {K},X'D,X'B,X'C) \equal{} \minus{} 1$. In the other hand, since $ X'$ lies on the perpendicular bisector of $ DF$, therefore, $ X'B$ is the bisector of $ \angle \mathcal {K}X'D$, thus, $ BX'\perp CX'$, which implies $ X'\equiv X$. With the same argument, we can also conclude that $ Y'\equiv Y$. As the consequence, $ E,F,X,Y$ are collinear.

Let $ A'$ be the intersection of $ ID$ and the line $ \triangle$ passes through $ A$ parallel to $ BC$. Let $ N'\equiv ID\cap EF$ and $ \mathcal {K'}$ be the intersection of $ EF$ with $ \triangle$. Since $ \angle AA'I \equal{} 90^{\circ}$ $ \Longrightarrow$ $ A'\in (AEF)$. Therefore, $ A'N'$ is the bisector of $ \angle EA'F$, but $ A'N'\perp A'\mathcal {K'}$ $ \Longrightarrow$ $ (A'E,A'F,A'N',A'\mathcal {K'}) \equal{} \minus{} 1$, which implies $ (EFN'\mathcal {K'}) \equal{} \minus{} 1$ $ \Longrightarrow$ $ A(CBN'\mathcal {K'}) \equal{} \minus{} 1$. But $ A\mathcal {K'}\|BC$, therefore, $ AN'$ passes through $ M$, which implies $ N'\equiv N$ $ \Longrightarrow$ $ I,D,N$ are collinear.

Now, it is easy to prove that $ \triangle YDX$ $ \cong$ $ \triangle BAC$ (a.a) and $ DI$ is the bisector of $ \angle YDX$, then $ \frac {NX}{NY} \equal{} \frac {DX}{DY} \equal{} \frac {AC}{AB}$. $ \square$

Well, the same trick ( $MR$, $BI$, $FE$ and $\gamma$ intersect in $X$) appears 7 years later at USAJMO. (http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3478583&sid=5997de18ba862fa0672276d599ecf7cb#p3478583). I first saw this trick in one of Cosmin's book.

($R$ is midpoint of $AC$)

USAJMO #6 is my problem too, so no copyright issues I hope :). By the way, the problem didn't mean to have a part (a) initially, but we figured the problem might be too difficult without isolating the known lemma (which unfortunately turned out not to be so well known, as there weren't more than 30 students who solved the first part).

Dear mathlinkers,
any ideas for the similar problem of Virgil Nicula?
Sincerely
Jean-Louis

Dear Virgil and Mathlinkers,
are you sure of your result (similar problem )?
Toying with the figure in question, I have a doubt?
Sincerely
Jean-Louis

[b]Lemma:[/b] $NI \perp BC$ (This is lemma 7 [url=http://yufeizhao.com/olympiad/geolemmas.pdf]here[/url])
[b]Proof:[/b] Let the line passing through $A$ parallel to $BC$ be $l$. Let $AN \cap \Gamma(I) \equiv P, Q$ such that $AP < AQ$. Let $FP \cap QE \equiv S$ and $FQ \cap PE \equiv T$.
We have that $PEQF$ is a harmonic quadrilateral. Applying Brokard to $PEQF$ we have $NI \perp TS$, so it is enough to show that $T, A, S, Z$ are collinear.
By Brokard, $ST$ is the polar of $N$, and by tangency $N$ is on the polar of $A$, thus La Hire yields $A$ is on the polar of $N$. It follows that $T, A, S$ collinear. Also, $PEQF$ is harmonic so $(Z, N; E, F)= -1$. Hence, $N$ is on the polar of $Z$, and La Hire again yields the desired.

Also, by the Right Angles on Intouch Chord Lemma $X, Y \in \gamma(M)$. Now we length chase a little.
By the Ratio Lemma, $\frac{NX}{NY} = \frac{XI}{YI} \frac{\sin \angle XIN}{\sin \angle YIN}$. By PoP or just similar triangles, this is $\frac{CI}{BI} \frac{\sin \angle BID}{\sin \angle CID}$. It's easy to see with lengths around the incircle and LoS that $\frac{CI}{BI} = \frac{\sin \frac{B}{2}}{\sin \frac{C}{2}}$ and $\frac{\sin \angle BID}{\sin \angle CID} = \frac{\cos \frac{B}{2}}{\cos \frac{C}{2}}$. Then by the double angle formula for sine we have $\frac{NX}{NY} = \frac{\sin B}{\sin C} = \frac{AC}{AB}$ as requested.

It is well-known that $X, Y$ lie on $FE$. It is well-known that $NI \perp BC$. We have $\angle XIN = 90^{\circ} - \frac{\angle B}{2}$ and $\angle NIY = 90^{\circ} - \frac{\angle C}{2}$. From Ratio Lemma, we have $$\frac{NX}{NF} = \frac{XI}{IY} \cdot \frac{\sin (90^{\circ} - \frac{\angle B}{2})}{\sin (90^{\circ} - \frac{\angle C}{2})} = \frac{XI}{IY} \cdot \frac{\cos \frac{\angle B}{2}}{\cos \frac{\angle C}{2}}$$

We have $BI = \frac{r}{\sin \frac{\angle B}{2}}$ and $CI = \frac{r}{\sin \frac{\angle C}{2}}$. So, $$\frac{NX}{NF} = \frac{\sin \frac{\angle B}{2} \cos \frac{\angle B}{2}}{\sin \frac{\angle C}{2} \cos \frac{\angle C}{2}} = \frac{\sin \angle B}{\sin \angle C} = \frac{b}{c} = \frac{AC}{AB}$$ as desired.

Let the midline $MI$ cut  $EF $ at $X'$  since $AB\parallel MX'$ then $\hat{B}= \widehat{MX'B}+\widehat{X'BM}, \widehat{X'BA}=\widehat{BX'M} $ so $MX'B$ is isoceles which implies $\widehat{BX'C}= \frac{\pi}{2}$ so $X=X'$ likewise we prove $ Y \in EF\cap MJ $ .
$\frac{NX}{NY} =\frac{\sin \widehat{XMN}\cdot MX}{\sin\widehat{YMN} \cdot MY}=\frac{\sin \widehat{MAB} }{\sin\widehat{MAC}}=\frac{AC}{AB}$.
[b]R HAS[/b]

[hide = solution]By well known Iranian Incenter, $X$ and $Y$ are on $EF$. Let $BY \cap CX = H$. Then, $I$ must be the orthocenter of $HBC$. Angle chasing with orthocenters gives$$\angle CHD = \angle CYX = \frac12\angle ABC$$and$$\angle BHD = \angle BXY = \frac12\angle ACB$$By Ratio Lemma we have$$\frac{NX}{NY} =  \frac{HX}{HY} \cdot \frac{\sin{\frac12\angle B}}{\sin{\frac12\angle C}} = \frac{HB}{HC} \cdot \frac{\sin{\frac12\angle B}}{\sin{\frac12\angle C}}= \frac{HB}{HC} \cdot \frac{CI}{BI}$$Notice that we also have$$\frac{HB}{HC} = \frac{\frac{BD}{\sin\angle BHD}}{\frac{CD}{\sin\angle CHD}}$$So we substitute this back into the former and get$$\frac{NX}{NY} = \frac{\frac{BD}{\sin\angle BHD}}{\frac{CD}{\sin\angle CHD}} \cdot \frac{CI}{BI}= \frac{\frac{DB}{BI}}{\frac{DC}{CI}} \cdot \frac{\sin{\frac12\angle B}}{\sin{\frac12\angle C}} = \frac{\cos{\frac12\angle B}}{\cos{\frac12\angle C}} \cdot \frac{\sin{\frac12\angle B}}{\sin{\frac12\angle C}}= \frac{\sin{\angle B}}{\sin{\angle C}} = \frac{AC}{AB}$$hence done. $\blacksquare$[/hide]

[hide=solution][quote=Romania TST 2007 Day 6 #2]Let $ ABC$ be a triangle, let $ E, F$ be the tangency points of the incircle $ \Gamma(I)$ to the sides $ AC$, respectively $ AB$, and let $ M$ be the midpoint of the side $ BC$. Let $ N \equal{} AM \cap EF$, let $ \gamma(M)$ be the circle of diameter $ BC$, and let $ X, Y$ be the other (than $ B, C$) intersection points of $ BI$, respectively $ CI$, with $ \gamma$. Prove that
\[ \frac {NX} {NY} \equal{} \frac {AC} {AB}.
\]
[i]Cosmin Pohoata[/i][/quote]
BY $\text{EGMO Lemma 1.45 ,} X,Y \in EF$ and by $\text{EGMO Lemma 4.17 ,} N \equiv DI \cap EF$.
Where  $D$ is the other in touch point.
Now say $BY \cap CX \cap IN=G$
Clearly $I$ is the orthocentre of $\triangle BGC$
Now the problem falls to trig.
Note that,
$$\angle NGX=\angle IGX=\angle IYX=\angle IBC=\frac{\angle B}{2}\rightarrow(1)$$
Similarly $\angle NGY=\frac{\angle C}{2}$ again note,
$$\angle GXN=\angle GXY=\angle GBC=90-\angle ICB=\frac{\angle A+\angle B}{2}=90-\frac{\angle C}{2}$$
and similarly see $\angle GYN=\frac{\angle A+\angle C}{2}=90-\frac{\angle B}{2}$
Now for the finish,observe
$$NX=GN\frac{\sin \frac{\angle B}{2}}{\sin \frac{\angle A+\angle B}{2}}=GN\frac{\sin \frac{\angle B}{2}}{\cos \frac{\angle C}{2}}$$
Obtaining similar ratio for $NY$ in terms of $GN$ and dividing gives,
$$\frac{NX}{NY}=\frac{\sin\frac{\angle B}{2}}{\sin\frac{\angle C}{2}}.\frac{\cos\frac{\angle B}{2}}{\cos\frac{\angle C}{2}}=\frac{\sin\angle B}{\sin\angle C}=\frac{AC}{AB}$$

Here we go again.
By Iran Lemma $X,Y \in EF$ also note that $\triangle IXY \sim \triangle ICB$ and bcs of meh config stuff $N = DI \cap \overline{EF}$ where $D$ is the other in touch point.
$$\dfrac{NX}{NY} = \dfrac{IX \sin \angle XIN}{IY \sin \angle YIN} = \dfrac{IC \sin \angle BID}{IB \sin \angle CID} = \dfrac{\tfrac{IC}{ID} \cos B/2}{\tfrac{IB}{ID} \cos C/2} = \dfrac{\sin B/2 \cos B/2}{\sin C/2 \cos C/ 2} = \dfrac{\sin B }{\sin C} = \dfrac{AC}{AB}$$

Let $P=AI\cap EF$, $Q=AI\cap BC$ and $D$ be the tangency point on $BC$. From a well known lemma from EGMO we know $DI$ passes through $N$. Now delete $AM$ and consider $N$ as the intersection of $DI$ and $EF$. Notice that $P, D$ are the corresponding point of the similar triangles $\triangle DIY$ and $\triangle CIB$. So $G, N$ are also corresponding points. Hence $NX/NY=CQ/QB=AC/AB$.

Let $DEF$ be the intouch triangle. We have $X=BI\cap EF$, $Y=CI\cap EF$, and $N=DI\cap EF$. Since $XYBC$ is cyclic, we get $\triangle IXY\sim\triangle ICB$ so if the isogonal of $ID$ with respect to $IBC$ intersects $BC$ at $Z$, then $\frac{NX}{NY}=\frac{ZC}{ZB}$. However, by Incenter-Excenter, the isogonal of $ID$ is $AI$, so $\frac{ZC}{ZB}=\frac{AC}{AB}$.

new favorite geo problem

[hide=Synthetic]
Let $D$ be the tangency point of the incircle $\Gamma(I)$ to side $BC$. It is well known that $D$, $I$, and $N$ are collinear. It is easy to see that $\angle CXI=\angle CXB=90^\circ$. Since $\angle CXI=\angle CEI=\angle CDI=90^\circ$, it follows that $CXEID$ is cyclic. Similarly, $BYFID$ is cyclic. We also have
\begin{align*}
\measuredangle XEF&=\measuredangle XEC+\measuredangle CEI+\measuredangle IEF\\
&=90^\circ+\measuredangle XIC+\measuredangle IAF\\
&=90^\circ+\measuredangle IBC+\measuredangle BCI+\measuredangle IAB\\
&=180^\circ
\end{align*}
so $X$, $E$, $F$ are collinear. Similarly, $Y$, $E$, $F$ are collinear. The radical axis of $(CXE)$ and $(BYF)$ is $\overleftrightarrow{DI}$, and $N$ lies on it. It follows that $NE\cdot NX=NF\cdot NY$. Additionally, we have
\[
\frac{NF}{NE}=\frac{\sin\angle FAN}{\sin\angle EAN}\cdot\frac{AF}{AE}=\frac{\sin\angle BAM}{\sin\angle CAM}\cdot\frac{CM}{BM}=\frac{AC}{AB}
\]
and the conclusion follows.
[/hide]

[hide=Barybash]
We are going to use barycentric coordinates on $\triangle ABC$. Let $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$ and $a=BC$, $b=CA$, $c=AB$. It follows that
\begin{align*}
E&=(s-c:0:s-a)\\
F&=(s-b:s-a:0)\\
M&=(0:1:1)
\end{align*}
so $N=(n:1:1)$ for some $n$. Since $E$, $F$, and $N$ are collinear,
\[
\begin{vmatrix}
n & 1 & 1\\
s-c & 0 & s-a\\
s-b & s-a & 0
\end{vmatrix}
=0
\]
so
\[
n=\frac{a}{s-a}.
\]
Observe that $X$ is the foot from $C$ to $\overleftrightarrow{BI}$. Let
\[
X=(a:x:c)=\left(\frac{a}{a+x+c},\frac{x}{a+x+c},\frac{c}{a+x+c}\right)
\]
so
\[
\overrightarrow{CX}=\left(\frac{a}{a+x+c},\frac{x}{a+x+c},-\frac{a+x}{a+x+c}\right).
\]
We also have
\[
\overrightarrow{BI}=\left(\frac{a}{a+b+c},-\frac{a+c}{a+b+c},\frac{c}{a+b+c}\right).
\]
Since $\overleftrightarrow{CX}\perp\overleftrightarrow{BI}$, it follows that
\[
a^2(cx+(a+c)(a+x))+b^2(ac-a(a+x))+c^2(ax-a(a+c))=0
\]
so $x=c-a$. Thus
\begin{align*}
\overrightarrow{NX}&=\left(\frac{a}{2c}-\frac{a}{2s-a},\frac{c-a}{2c}-\frac{s-a}{2s-a},\frac{c}{2c}-\frac{s-a}{2s-a}\right)\\
&=\left(\frac{a(b-c)}{2c(b+c)},-\frac{ab}{2c(b+c)},\frac{ac}{2c(b+c)}\right).
\end{align*}
Similarly, we have
\[
\overrightarrow{NY}=\left(\frac{a(c-b)}{2b(b+c)},\frac{ab}{2b(b+c)},-\frac{ac}{2b(b+c)}\right)
\]
so $c\cdot\overrightarrow{NX}=-b\cdot\overrightarrow{NY}$. The conclusion follows.
[/hide]


calculus
integration
geometry
rectangle
perimeter
number theory
prime numbers

ROUND 1
[b]Rules[/b]: [b]You have only 50 minutes to solve as many problems as you can. ABSOLUTELY NO CALCULATORS OR ANY OTHER RESOURCES ARE ALLOWED. This is based on the honor system only. Please PM me your answers. The answers are due by August 31st [/b] However, the most important rule is that those who participate in round 1 but not round 2 will be eaten. Seriously.

Some of these questions are proofs, so show your work on those questions to receive full credit. Since a lot of people are not advanced proof-writers, partial credit will be handed out generously.

The point value of each question is written after the question

[hide="CLICK WHEN READY (NO CALCULATORS)"]
1. How many positive integral values of $ a$ satisfy the equation $ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{8}$? (1 point)

2. Points $ X$ and $ Y$ lie on a circle of radius $ 2$ and center at $ O$ and the measure of $ \angle XOY$ is $ 30^{\circ}$.What is the area of the smaller region bounded by the circle and the chord joining $ X$ to $ Y$? Express your answer in terms of pi. (1 point)

3. What is the product of the real roots of the equation: $ 3x^2 \plus{} 10x \plus{} 61 \equal{} 2\sqrt {3x^2 \plus{} 10x \plus{} 76}$? (1 point)

4.  Deception Darts, a darts company, is holding a promotional sale for its newest dartboard, the Ruseboard. The Ruseboard has four possible scores that can be achieved by throwing only one dart: $ 5, 12, 14$, and $ 17$. The promotional manager tells David that if he scores $ X$ points on the Ruseboard given an unlimited amount of darts, David will walk home with a new Ruseboard. Being a math genius, David at once knows that there is no possible way to score the $ X$ points on the Ruseboard no matter how many darts he uses. What is the maximum possible value of $ X$? (1 point)

5. $ 8$ identical rooks are placed on a standard 8x8 chessboard such that it is impossible for any of the rooks to attack another. How many possible ways are there to make such an arrangement?  (In other words, no two rooks are in the same row or in the same column) (1 point)

6. One side of a rectangle is $ 7$ units. Express the area of the rectangle $ A$ in terms of its perimeter $ P$. (1 point)

7. If $ 2^W \equal{} 125$, $ 3^L \equal{} 32$, $ 4^O \equal{} 9$, and $ 5^G \equal{} 8$, what is the value of $ WLOG$? (1 point)

8.  A number $ a$ satisfies the equation $ a^2 \minus{} 7a \plus{} 2 \equal{} 0$. What is the sum of $ a$ and twice its reciprocal? (1 point)

[b]PROOFS[/b]
9.  Show that any positive integer raised to the third power can be expressed as the difference of two squares. (2 points)

10.  Show that $ \sqrt {1 \plus{} (x)(x \plus{} 1)(x \plus{} 2)(x \plus{} 3)}$ is always rational for all integers $ x$. (3 pts)

11. Determine, with proof, all prime numbers $ p$ such that:
a.) $ 2p \plus{} 1$ is a perfect square. (2 points)
b.) $ 2p \plus{} 1$ is a perfect cube. (3 points)

12. Prove that there cannot exist an infinite arithmetic sequence that consists of only square numbers. (2 pts)

 [/hide]

How many rounds and when is it due? Thanks!

EDIT: Oops, I missed that when I read through.  :blush:

[quote="Thunder365"]
[b]The answers are due by August 31st[/b][/quote]

[quote="Thunder365"]
However, the most important rule is that those who participate in round 1 but not round 2 will be eaten.[/quote]
so I assume there are 2 rounds.

[quote="Thunder365"]12. Prove that there cannot exist an infinite arithmetic sequence that consists of only square numbers. (2 pts) [/quote]

Error? You have to specify $ d\neq0,$ otherwise you could have the sequence $ 9,9,9,9,\dots.$

Also, are there any real roots on number 3?

Roflolcopt: Yeah sorry, $ d \neq 0$
PoP: Sorry, find the product of the roots of the equation, not necessarily real
athunder:lol, I remember what you told me at school about the ten round tournament. You can "start" the test on august 30th if you want :P

Yeah, and if you ask any questions and I dont answer it until like 3 hours later it [i]wont[/i] count toward your time for the test.

[quote="PowerOfPi"][quote="Thunder365"]3. What is the product of the [b]real roots[/b] of the equation: $ 3x^2 \plus{} 10x \plus{} 61 \equal{} 2\sqrt {3x^2 \plus{} 10x \plus{} 76}$? (1 point)[/quote]
So we ignore the bolded part?[/quote]

yes. The question reads:
3. What is the product of the roots of the equation: $ 3x^2 \plus{} 10x \plus{} 61 \equal{} 2\sqrt {3x^2 \plus{} 10x \plus{} 76}$? (1 point)

For number one, I'm guessing b has to be an integer too?

yep

The message is too small. Please make the message longer before submitting.

Tournament canceled due to lack of activity.

I will post the second round later for anyone who just wants to take it for fun.

[quote="Ihatepie"]I was going to send one in  :([/quote]

Exactly, you were going to send [i]one[/i] submission in. My [i]only[/i] submission :P .

You can still send it in. I will grade it and post round 2 just for you (and others who want to take it for fun)

[hide="1st Place"]ihatepie!!!! :winner_first:  :10:  :clap2: [/hide]

EDIT: You guys can take as long as you wanton this round, I dont really care. (if you even want to take it)   If you send me the answers, I will grade them.
[hide="Round 2"]1. For what real integer value(s) of $ a$ is $ ax\minus{}2a\plus{}5x$ always divisible by $ x\minus{}3$? (2 pts)

2. $ \triangle ABC$ has $ D$ as the midpoint of $ \overline {AB}$, $ E$ as the midpoint of $ \overline {AC}$, and $ F$ as the midpoint of $ \overline {BC}$. The medians of $ \triangle ABC$ intersect at point $ G$. If $ \overline {AD}\equal{}\overline {DG}\equal{} \overline {AE}\equal{}2$, find the length of the longest side of $ \triangle ABC$. (2 pts)

3. If $ f(a\plus{}b)\equal{}f(a)\plus{}f(b)\plus{}ab$ and $ f(5)\equal{}70$ find $ f(14)$.(2 pts)

4. For any given constant $ a$, what is the smallest value that $ f(x) \equal{} 4x^2\plus{}16x\plus{}a$ achieves? Express in terms of $ a$ if necessary.(2 pts)

5. Call a number [i] tehwin[/i] if it contains the digits $ 1,3,4,8$, and $ 9$ used exactly once each. One example of a [i] tehwin[/i] number is $ 19843$. What is the sum of all such [i]tehwin [/i] numbers? (2 pts)

6. Now call a set of numbers [i]hawt[/i] if it contains the digits $ 1,3,5,7,8$, and $ 9$, all used exactly once, with the property that each digit in the set is either greater than every digit that precedes it or each digit is less than every digit that precedes it. One such example of a set is $ 7, 5, 3, 8, 1, 9$. How many such [i]hawt[/i] sets are there? (2 pts)

7. Determine every possible positive integral ordered pair $ (a,b)$ when $ a^6 \equal{} b^2 \plus{} 127$. (2 pts)

8.  There are only $ 10$ ascending steps that stand in the way between Tanya and her free McFlurry at McDonalds. However, Tanya can only move up one step or two steps at a time. In how many ways can Tanya ascend the $ 10$ steps to get the free McFlurry? (2 pts)

[b]PROOFS[/b]

9. Find all positive primes $ p$ such that $ p$ is divisible by two consecutive terms of the sequence generated by $ a^2\plus{}3$. (4 pts)

10. Show that $ x\plus{}(x\minus{}1)(x)\plus{}(x\minus{}1)(x^2)\plus{}(x\minus{}1)(x^3)\plus{} \dots \plus{}(x\minus{}1)(x^n)\equal{}x^{n\plus{}1}$ for all positive integers $ x$ and $ n$.(4 pts)

11.  Prove that $ p^x\minus{}p^{x\minus{}2}$ is divisible by $ 24$ when $ p$ is any positive prime number, $ x$ is any positive integer, and $ p^x\minus{}p^{x\minus{}2} \ge 24$. (5 pts) [/hide]

Wow. First! This has never happened before. I am so happy!!!!!   :wink:

I gotta start looking at the problems XD


Prove that the remainder when we divide any prime number on $ 30$ will be $ 1$ or a prime number only.

 :D

[hide]If the remainder has a factor of 2, 3 or 5, then the original number would be divisible by 2, 3, or 5 as well (which ever applies, since 2, 3, and 5 are also factors of 30).

Thus, the remainder cannot have a factor of 2, 3, or 5.

The smallest composite number without a factor of 2, 3 or 5 is 7^2=49. However, this is already greater than 30, so it cannot be a remainder upon division by 30.

This means that the remainder when a prime is divided by 30 is non-composite; that is, it is either 1 or prime.[/hide]


With $ a,b,c > 0$. Prove that
 $ \frac {1}{\sqrt {81a^2 \plus{} (b \minus{} c)^2}} \plus{} \frac{1}{\sqrt {81b^2 \plus{} (c \minus{} a)^2}} \plus{} \frac {1}{\sqrt {81c^2 \plus{} (a \minus{} b)^2}} \geq \frac {1}{a \plus{} b \plus{} c}$

[quote="2424"]With $ a,b,c > 0$. Prove that
 $ \frac {1}{\sqrt {81a^2 \plus{} (b \minus{} c)^2}} \plus{} \frac {1}{\sqrt {81b^2 \plus{} (c \minus{} a)^2}} \plus{} \frac {1}{\sqrt {81c^2 \plus{} (a \minus{} b)^2}} \geq \frac {1}{a \plus{} b \plus{} c}$[/quote]
It's obvious after following Holder:
$ \left(\sum_{cyc}\frac {1}{\sqrt {81a^2 \plus{} (b \minus{} c)^2}} \right)^2\sum_{cyc}(b \plus{} c)^2(81a^2 \plus{} (b \minus{} c)^2)\sum_{cyc}(b \plus{} c)^2\geq\left(\sum_{cyc}(b \plus{} c)\right)^4$


Suppose $ BD$ is a median of a $ \triangle ABC$ ,and $ P ,Q$ are  two points on $ AB$ such that $ BP\equal{}PQ\equal{}QA$ , from the vertex $ C$ joining $ CP$ and $ CQ$ that meet the median $ BD$ at $ E$ and $ F$ repectively.

Find the ratio $ BE: EF: FD$

It's just Menelaus twice.
$ \frac{CD}{CA} \cdot \frac{AQ}{QB} \cdot \frac{BF}{FD}\equal{}1, so \frac{BF}{FD}\equal{}4$.
$ \frac{CD}{DA} \cdot \frac{AP}{PB} \cdot \frac{BE}{ED}\equal{}1, so \frac{BE}{ED}\equal{}1$.
So $ BE\equal{}x$, $ DE\equal{}x$, $ BF\plus{}FD\equal{}2x$, $ BF\equal{}\frac{4}{5} \cdot 2x\equal{} \frac{8}{5}x$, $ FD\equal{} \frac{2}{5} x$, $ EF\equal{}BF\minus{}BE\equal{}\frac{3}{5}x$. So the answer is $ 1: \frac{3}{5}: \frac{2}{5}\equal{} 5: 3: 2$

[quote="dgreenb801"]It's just Menelaus twice.
$ \frac {CD}{CA} \cdot \frac {AQ}{QB} \cdot \frac {BF}{FD} \equal{} 1, so \frac {BF}{FD} \equal{} 4$.
$ \frac {CD}{DA} \cdot \frac {AP}{PB} \cdot \frac {BE}{ED} \equal{} 1, so \frac {BE}{ED} \equal{} 1$.
So $ BE \equal{} x$, $ DE \equal{} x$, $ BF \plus{} FD \equal{} 2x$, $ BF \equal{} \frac {4}{5} \cdot 2x \equal{} \frac {8}{5}x$, $ FD \equal{} \frac {2}{5} x$, $ EF \equal{} BF \minus{} BE \equal{} \frac {3}{5}x$. So the answer is $ 1: \frac {3}{5}: \frac {2}{5} \equal{} 5: 3: 2$[/quote]

Would you please explain what Menelaus' thereom is and how to apply in this question?
Thanks.

Menelaus is exactly what I typed out, that's what the theorem is. It is always true.
[url]http://en.wikipedia.org/wiki/Menelaus_theorem[/url]

[quote="dgreenb801"]It's just Menelaus twice.
$ \frac {CD}{CA} \cdot \frac {AQ}{QB} \cdot \frac {BF}{FD} \equal{} 1, so \frac {BF}{FD} \equal{} 4$.
$ \frac {CD}{DA} \cdot \frac {AP}{PB} \cdot \frac {BE}{ED} \equal{} 1, so \frac {BE}{ED} \equal{} 1$.
So $ BE \equal{} x$, $ DE \equal{} x$, $ BF \plus{} FD \equal{} 2x$, $ BF \equal{} \frac {4}{5} \cdot 2x \equal{} \frac {8}{5}x$, $ FD \equal{} \frac {2}{5} x$, $ EF \equal{} BF \minus{} BE \equal{} \frac {3}{5}x$. So the answer is $ 1: \frac {3}{5}: \frac {2}{5} \equal{} 5: 3: 2$[/quote]

Thanks a lot for your instructions.
May be stupid ,  $ \boxed{\frac {CD}{DA} }\cdot \frac {AP}{PB} \cdot \frac {BE}{ED} \equal{} 1, so \frac {BE}{ED} \equal{} 1$.

is  a typo?

I think it is $ \frac {CD}{CA} \cdot \frac {AP}{PB} \cdot \frac {BE}{ED} \equal{} 1, so \frac {BE}{ED} \equal{} 1$.

Agree? :)

[quote="dgreenb801"]Yes, it is a typo.[/quote]

Thanks for your nice reply.

I understood the proof . :trampoline:


I do not really understand the difference between if, only if, and if and only if.

$\boxed{ex: }$
a number is even if it is divisible by 2.
a number is even only if it is divisible by 2.
a number is even if and only if it is divisible by 2.

difference?  :maybe:

thanks!

First statement follows:
If p then q.

(If it is divisible by two, then it is even. Does not exclude any other possibilities.)

Third statement follows:
If p then q, If q then p.

(If and only if it is divisible by two, then it it even. States that an even number HAS to be divisible by two. No other possibilities.)

I think the second statement is similar to the third statement.

There, the correct one is the 3º.

One thing is if (or "implicancia simple"):

e.g.: $x=y \Longrightarrow x(x-1)=y(y-1)$

If $x=y$, it's obvious that $x(x-1)=y(y-1)$, but if $x(x-1)=y(y-1)$, it's not necessary true that $x=y$ (try $(x;y)=(0,25;0,75)$)

If and only if (or "doble implicancia")

$a \Longleftrightarrow b$ means that, if $a$ is true, then $b$ is true, and if $a$ is false, then $b$ is false. On the same way, if $b$ is true, $a$ is true, and, if $b$ is false, $a$ is false.

e.g.: $k+3=7 \Longleftrightarrow k=7-3=4$, but, if $k+3 \ne 7 \Longleftrightarrow k \ne 4$

the statement "if" is a one way statement.

for example, if we have if [i]p[/i] then [i]q[/i] then we can only conclude what is says. if [i]p[/i] is true, then [i]q[/i] is true.

however, "if and only if" is a two way statement

if instead of the above, we have [i]q[/i] is true iff [i]p[/i] is true, then it means if [i]q[/i] is true, then [i]p[/i] is true. also, if [i]p[/i] is true, then [i]q[/i] is true.

okay i think that sounds kind of confusing, but sorry i don't know how to explain otherwise[/i]

I've reworded the statements.

a number is even if it is divisible by 2.
A number divisible by 2 is even.

a number is even only if it is divisible by 2.
It doesn't matter if the number is divisible by any other number the number must be divisible by 2 to be even.

a number is even if and only if it is divisible by 2.
A number divisible by 2 is even and it doesn't matter if the number is divisible by any other number the number must be divisible by 2 to be even.

Examples:

Mathnerd314 lives in the United States [b]if[/b] he lives in Missouri.

Mathnerd314 is using Windows XP [b]only if[/b] he is using a computer.

A number is prime [b]if and only if[/b] it has only two factors, one and itself.

thanks guys its getting clearer now.

 :D


geometry
inequalities
cyclic quadrilateral
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral,inscirbed in a circle centered at $ I$.Prove the following inequality:
$ (AI\plus{}DI)^2\plus{}(BI\plus{}CI)^2 \leq (AB\plus{}CD)^2$

Maybe you want to said:  circunscribed around a circle with center I

[quote="pro_math"]Let $ ABCD$ be a cyclic quadrilateral,inscirbed in a circle centered at $ I$.Prove the following inequality:
$ (AI \plus{} DI)^2 \plus{} (BI \plus{} CI)^2 \leq (AB \plus{} CD)^2$[/quote]
This ineq is true,please slove it!

[quote="pro_math"]Let $ ABCD$ be a cyclic quadrilateral,inscirbed in a circle centered at $ I$.Prove the following inequality:
$ (AI \plus{} DI)^2 \plus{} (BI \plus{} CI)^2 \leq (AB \plus{} CD)^2$[/quote]
Please stop posting the problems in [b]Math & Youth Magazine[/b] ! Keep your honesty, you should not post all unsolved problems in M&Y.


Let $u,v,w$ be vectors.

Prove:
$u$ dot ($v$ cross $w$) = ($u$ cross $v$) dot $w$

That's the definition of the scalar triple product.  :)

$u\cdot (v\times w)$

Let $v = v_{1}i+v_{2}j+v_{3}k$, and $w = w_{1}i+w_{2}j+w_{3}k$. Now take their cross product and dot it with $u=u_{1}i+u_{2}j+u_{3}k$. Show that this big mess is equal to $u\times (v\cdot w)$

can anyone tell me what is "cross" what is "dot"?

vectors is not my strong point.  :(

http://en.wikipedia.org/wiki/Cross_product

http://en.wikipedia.org/wiki/Dot_product

Wikipedia is your friend.

[quote="vishalarul"]That's the definition of the scalar triple product.  :) [/quote]

No.

[quote="vishalarul"]Show that this big mess is equal to $u\times (v\cdot w)$[/quote]

That last quantity isn't even defined.

Anyways, there are a couple of ways to proceed here.  Since the cross product is a three dimensional object, write out $u=(u_{1},u_{2},u_{3})$, etc, and expand the whole thing.  This is easy.

Second:  Interpret $u\cdot(v\times w)$ as the volume of a certain geometric object.  Interpret $(u\times v)\cdot w$ as the volume of the same object.

Third:  $u\cdot(v\times w)$ can be easily expressed as a determinant.  Same with $(u\times v)\cdot w$.  Use the fact that a row swap in a determinant just changes the sign of the determinant.

[quote="blahblahblah"]

Third:  $u\cdot(v\times w)$ can be easily expressed as a determinant.  Same with $(u\times v)\cdot w$.  Use the fact that a row swap in a determinant just changes the sign of the determinant.[/quote]

how to do this? i don't know determinant very much.

[quote="paladin8"]Wikipedia is your friend.[/quote]


Let $a,$ $b$ and $c$ are positive numbers.
Prove that $\sqrt\frac{a}{b+2c}+\sqrt\frac{b}{c+2a}+\sqrt\frac{c}{a+2b}\geq\sqrt[4]8$

It's a particular case of the inequality (reality, a conjecture) I post here

http://www.mathlinks.ro/Forum/viewtopic.php?t=103052

So you are in process of completing this conjecture, Arqady? I think it's very hard.

[quote="arqady"][quote="hungkhtn"]It's a particular case of the inequality (reality, a conjecture) I post here

http://www.mathlinks.ro/Forum/viewtopic.php?t=103052

So you are in process of completing this conjecture, Arqady?  [/quote]
Not precisely. I posted it earlier:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75688  :P :lol:[/quote]

Yes, I see  :P . But I also want to ask you if you can solve the problem with the general case of $k$? I don't know about the link you show, of course.

We have a stronger inequality :
$\sum_{cyclic}\sqrt{\frac{a}{b+c}}\ge 2$
(because $\sum_{cyclic}\sqrt{\frac{a}{2b+2c}}\le \sum_{cyclic}\sqrt{\frac{a}{b+2c}}$)
But the inequality given is very famous  :lol:

[quote="ehku"]We have a stronger inequality :
$\sum_{cyclic}\sqrt{\frac{a}{b+c}}\ge 2$
(because $\sum_{cyclic}\sqrt{\frac{a}{2b+2c}}\le \sum_{cyclic}\sqrt{\frac{a}{b+2c}}$)
[/quote]
Id est, $\sqrt2\leq\sum_{cyc}\sqrt{\frac{a}{2b+2c}}\le \sum_{cyc}\sqrt{\frac{a}{b+2c}}.$
But $\sqrt[4]8>\sqrt2.$ :wink:


number theory
greatest common divisor
logarithms
induction
inequalities
number theory proposed

Obviously, not induction! On the other hand, I don't know if Vess has a simple solution and I really hope he has one and he will not use the very deep estimations in analytic number theory that were used by Pomerance, Adleman and Rumely. If this is not the case, I won't even look on this problem anymore, since this is not a competition to post the most impossible problems using high-tech results. But I am sure he has an "easy" solution.

[quote="vess"]Prove that there exist infinitely many $n$ for which $2^{n}-1$ and $3^{n}-1$ have a common divisor bigger than $\exp(\exp(\sqrt{\log{n}}))$. (This estimate is rough).[/quote]
May be $exp(exp(\sqrt{ ln(ln(n))}))$, because $exp(exp(\sqrt{ ln(n)}))>3^{n}$ for big n.

for any a,b exist N, suth that $exp(x)>bx^{a}$. Let $N=exp(N^{2}(a,b)=exp(N(2,2)^{2})$. If $x>N$, then $exp(\sqrt{lnx })>2x$, therefore $exp(exp(\sqrt{lnx}))>exp(2x)>3^{x}$.
May be $exp(exp(\sqrt{ln(ln(n))}))$ or $exp(\sqrt{\ln(n)})$.

If you did not hurry so much, you would really be great! Your inequality $e^{\sqrt{\ln(x)}}>2x$ is obviously wrong for large $x$, simply because it would imply $\sqrt{\ln(x)}>\ln(x)$, and I have some doubts that this holds for $x>e$ for instance.

I am sorry. You are right. $exp(\sqrt{ln x})>b(ln x )^{a}$.

we can prove a much stronger result. Let $l$, $k$ an integers and $(l,k)=1$. Let $ \pi(x, k, l)$ denote the number of primes $p \leq x$ with $p \equiv l [k]$. We will use a classical theorem in the advanced number theory:

[b]Theorem $[1]$ : There exist $c_0> 0$  such that  $ \pi(x, k, l)=\frac{x}{\phi(k) \ln x}(1+O(\frac 1{\ln x})) $   hold uniformly for all $k \leq \exp (c_0 \frac {\ln x}{\ln \ln x}) $ except possibly for these $k$ which are multiples of a certain integers $k_0=k_0(x) \geq 3$.[/b]

[b]$[1]$[/b] : T.TATUZAWA ' On the number of primes in an arithmetic progression' Jap J.Math 21, 93-111 (1951).

Let $k=\prod _{p \leq c_1 \frac {\ln x}{\ln \ln x}, p \#q}{p}$, $0 < c_1 < c_0 $ where $q$ is prime, $q \mid k_0(x)$ and $q \leq c_1 \frac {\ln x}{\ln \ln x}$.

We have, for sufficiently large $x$ :

           \[k \leq \exp (\sum _{p \leq c_1 \frac {\ln x}{\ln \ln x}}{\ln p})=\exp (\theta(c_1 \frac {\ln x}{\ln \ln x}) )< \exp (c_0 \frac {\ln x}{\ln \ln x})\]

where ${\theta(x)=\sum _{p \leq  x}}{\ln p} \sim x$

Since $\ln q=O( \ln \ln x)$
\[k= \exp (\sum _{p \leq c_1 \frac {\ln x}{\ln \ln x}}{\ln p}- \ln q) > \exp (c_2 \frac {\ln x}{\ln \ln x})\]   where $0 < c_2 < c_1$.
Let $S$ denote the number of solutions of $m(p-1) \equiv  0  [k]$  $(*) $ where $m \leq x$ and $p \leq x$ is prime.
For each $d \mid k$, let $S_d$ denote te number of solution of $(*)$ with $d \mid p-1$, $ \frac{k}{d} \mid m$ and $(m,d)=1$.
We have \[ \sum_{p \leq x, d \mid p-1}{1} > c_3 \frac {x} {\phi (d) \ln x}\]  where $0 < c_3 < 1$
and \[\sum_{m \leq x, \frac{k}{d}\mid m, (m,d)=1}{1} > [ \frac {x} {k}]\phi (d)\]
So
\[S_d > c_3 \frac {x} {\ln x}[ \frac {x} {k}] > c_4 \frac{x^2}{k \ln x} \] where $ 0 < c_4 < c_3$
hence  \[S=\sum_{d%Error. "midk" is a bad command.
}{S_d} > c_4 d(k) \frac{x^2}{k \ln x}\]
here $d(k)$ is the number of divisors of $k$.
We have $d(k)=2^ {\sum_{p \leq c_1 \frac {\ln x}{\ln \ln x}, p \#q}{1}} $  and  $\sum_{p \leq c_1 \frac {\ln x}{\ln \ln x}, p \#q}{1} \sim c_1 \frac {\ln x}{(\ln \ln x)^2}$
Now the number of integers $n \leq x^2$ for which $k\mid n$ is $[\frac {x^2}{k}]$.
Also for each solution $m, p$ of $(*)$, $m(p-1)$ is such an $n$. Thus there is some $n \leq x^2$ with $k\mid n$ and $n$ has at least
\[ \frac{S}{[\frac {x^2}{k}]} > c_4 \frac {d(k)}{\ln x} > \exp (c_5 \frac {\ln x}{(\ln \ln x)^2}) > \exp (c_6 \frac {\ln n}{(\ln \ln n)^2})\]
representations as $m(p-1)$.
[b]We have prove that if $\lambda(n)$ is the number of primes $p$ such that $p-1 \mid n$ then there exists $c > 0$, such that
\[\lambda(n) > \exp (c \frac {\ln n}{(\ln \ln n)^2})\][/b].
holds for infinitely positive integers $n$.
Now, by Fermat little theorem
\[(a^n-1, b^n-1) \geq \prod_{p-1\mid n}{p} > \exp (c_7  \lambda(n))\]
where $a, b$ are positive integers $> 1$.
Finally
[b]\[(a^n-1, b^n-1) > \exp (\exp (c_8\frac {\ln n}{(\ln \ln n)^2})) \]
holds for infinitely positive integers $n$.[/b]


When 22.957 g of an irregularly shaped sample of cesium was placed in a container and filled to the 25.00 mL mark, the mass of the combined mass of the cesium and ethyl alcohol was 37.1892 g. What is the density of this element, if the density of the ethyl alcohol is 1.114 g/mL? Assume the volumes are additive.

Isn't there any reaction between Cesium and ethyl alcohol?


calculus
integration
calculus computations

can anyone please tell me something about
APPLICATION OF INTEGRALS IN STUDENT ENGINEERING

What I can tell you is that Google will be able to tell you way more than the users of the forum.

As i have been 'in engineering' for a moment, I'd say :

heat transfers in thermodynamics

moments of inertia in mechanics

just one of the many  examples...


Region $ R$ contains all the pointsw $ (x,y)$ such that $ x^2\plus{}y^2 \le 100$ and $ sin(x\plus{}y) \ge 0$. Find the area of region $ R$.

hm is the measurement degrees or radian?

no idea

tats why i'm asking this problem

It's probably in degrees.

And it's easy. First, notice that the first equation describes a circle of radius 10. Then we have $ \sin (x + y) \ge 0$, or $ - 180 \le x + y \le 180$. This means that $ x + y$ has to be between $ - 180$ and $ 180$, which is always true since the maximum and minimum of $ x + y$ in the circle are $ \pm 10\sqrt 2 \approx \pm 14.14$. Therefore the area is $ 100\pi$.

If it is in radians, then we have $ - \pi \le x + y \le \pi$, so the desired area is in between bounded by the following three inequalities:
\begin{align*}x^2 + y^2 & \le 100 \\
x + y & \ge - \pi \\
x + y & \le \pi\end{align*}
So we are finding $ 100\pi - \zeta$, where $ \zeta$ is the area of the two "hamburger buns" inside the first inequality, but outside the last two, which should be trivial with basic trigonometry.

[quote="Yongyi781"]

 Then we have $ \sin (x \plus{} y) \ge 0$, or $ \minus{} 180 \le x \plus{} y \le 180$. This means that $ x \plus{} y$ has to be between $ \minus{} 180$ and $ 180$,
[/quote]

uh...are you sure?

$ 0\le x\plus{}y\le 180$

$ \sin (\minus{}90)\equal{}\sin 270 \equal{} \minus{}1$

Ahh... wrong am I again. I thought too quickly.

It should be $ \minus{}360 \le x \plus{} y \le \minus{}180$ or $ 0 \le x \plus{} y \le 180$.

Clearly, $ x \plus{} y$ never reaches $ \minus{}180$, so the desired area is simply one half of the circle, or $ 50\pi$.

[quote="Yongyi781"]Ahh... wrong am I again. I thought too quickly.

It should be $ \minus{} 360 \le x \plus{} y \le \minus{} 180$ or $ 0 \le x \plus{} y \le 180$.

Clearly, $ x \plus{} y$ never reaches $ \minus{} 180$, so the desired area is simply one half of the circle, or $ 50\pi$.[/quote]

assuming the problem is in degrees

but its probably in radians

@(^_^) No, it's in degrees. I asked.  :wink:


calculus
derivative
limit
trigonometry
LaTeX
calculus computations

These are probably very easy to do, but I missed a few days of class a few weeks ago so I didn't hear the teacher's lecture on the subject. If someone could show me these problems worked out in detail that would be great.

1. Using the limit definition of the derivative, find f'(x) for f(x)=(x+1)/(x-1)

2. Find the derivative of f(x)=(sinx-1)/(cosx+1)

2.
$\displaystyle \frac{(\sin(x)-1)(\cos(x)-1)}{(\cos(x)+1)(cos(x)-1)}$

denominator is $\displaystyle -\sin(x)^2$
then seperate and differntiate each term..

orrrr you can do the quotient rule on both of them!

ah whoops, you needed the limit definition for #1. pretty simple set up:

$\lim_{\Delta{x} \to {\infty}} \left(\frac{\frac{x+\Delta{x} + 1}{x+\Delta{x}-1} - \frac{x+1}{x-1}}{\Delta{x}}\right)$

Also can be done by simplifying denominator first, then using the definition as seen above by Tr

I assume adidasty meant $\Delta x \rightarrow 0$ in his response ;)  For part 2, assuming you don't need to use the limit definition, you can use the quotient rule to differentiate, as follows:

\begin{eqnarray*}
\ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] & = & \frac{g(x) \cdot \frac{df}{dx} - f(x) \cdot \frac{dg}{dx}}{[g(x)]^2} \\
 & = & \frac{(\cos x + 1)(\cos x) - (\sin x - 1)(- \sin x)}{(\cos x + 1)^2} \\
& = & \frac{\cos ^2 x + \cos x + \sin ^2 x - \sin x}{(\cos x + 1)^2} \\
& = & \frac{1 + \cos x - \sin x}{(\cos x + 1)^2}
\end{eqnarray*}

lol yeah going to zero. latex mistake. sorry  :?


Find the maximum value of:
[size=200]L[/size]   [size=150]=[/size]  $ \frac{a^2\plus{}1}{b^2\plus{}1}\plus{}\frac{b^2\plus{}1}{c^2\plus{}1}\plus{}\frac{c^2\plus{}1}{a^2\plus{}1}$
which $ a,b,c\ge0$    and       $ a\plus{}b\plus{}c\equal{}1$

Even I don't have way to solve it, if you guys can't solve it, just give me suggestions.
I need you help. :help:  :help:  :help:

[quote="Potla"][quote="Unidranix"]Find the maximum value of:
[size=200]L[/size]   [size=150]=[/size]  $ \frac {a^2 + 1}{b^2 + 1} + \frac {b^2 + 1}{c^2 + 1} + \frac {c^2 + 1}{a^2 + 1}$
which $ a,b,c\ge0$    and       $ a + b + c = 1$[/quote]
Er, I am quite doubtful about the rightness of the ineq:
$ \frac {a^2 + 1}{b^2 + 1} + \frac {b^2 + 1}{c^2 + 1} + \frac {c^2 + 1}{a^2 + 1}\geq 3\sqrt [3]{\frac {a^2 + 1}{b^2 + 1} \frac {b^2 + 1}{c^2 + 1} \frac {c^2 + 1}{a^2 + 1}} = 3$ from AGM inequality; but your conditions are not needed.
Equality holds iff $ a = b = c( = \frac {1}{3}$ in this case$ )$.
 :huh:[/quote]
We want the maximum :P

Oh Potla, you're so funny, this inequality is not that easy.
My teacher gave it to me suggested me that the maximum value is $ \frac {7}{2}$ when $ a \equal{} 1,b \equal{} c \equal{} 0$ or permutation so we can't use AM-GM-HM, Cauchy, ....

Probably L achieves it maximum when a=1, b=0, c=0 and their permutations. I suggest a way to prove that ... calculate L's value for that permutation and then try to use pqr method to prove the statement.

Let $ (p,q,r)$ be a permutation of $ (a,b,c)$, such that $ p \ge q \ge r$ (Clearly $ p \plus{} q \plus{} r \equal{} 1$). The sequences $ (p^2 \plus{} 1, q^2 \plus{} 1, r^2 \plus{} 1)$ and $ \left( \frac {1}{r^2 \plus{} 1}, \frac {1}{q^2 \plus{} 1}, \frac {1}{p^2 \plus{} 1} \right)$ are similarly sorted, thus, by rearrangement

$ \frac {a^2 \plus{} 1}{b^2 \plus{} 1} \plus{} \frac {b^2 \plus{} 1}{c^2 \plus{} 1} \plus{} \frac {c^2 \plus{} 1}{a^2 \plus{} 1} \le \frac {p^2 \plus{} 1}{r^2 \plus{} 1} \plus{} \frac {q^2 \plus{} 1}{q^2 \plus{} 1} \plus{} \frac {r^2 \plus{} 1}{p^2 \plus{} 1} \equal{} \frac {p^4 \plus{} r^4 \minus{} 2p^2r^2}{(p^2 \plus{} 1)(r^2 \plus{} 1)} \plus{} 3\equal{}\frac {(p^2 \minus{} r^2)^2}{(p^2 \plus{} 1)(r^2 \plus{} 1)} \plus{} 3.$

Consider $ f(r) \equal{} \frac {(p^2 \minus{} r)^2}{(p^2 \plus{} 1)(r \plus{} 1)}$. Clearly $ f(r)$ is decreasing, thus, it takes it maximum at $ r \equal{} 0$. Therefore, $ \frac {(p^2 \minus{} r^2)^2}{(p^2 \plus{} 1)(r^2 \plus{} 1)} \plus{} 3 \le \frac {p^4}{p^2 \plus{} 1} \plus{} 3 \le \frac {p^3}{2} \plus{} 3 \le \frac {7}{2}.$

Sorry because of my bad English.
 :maybe: FelixD, I found some mistake in your solution.
Cause $ (a,b,c)$ are permutation in[size=150] L[/size] so you can set:$ a\equal{}max(a,b,c)$ or $ a\ge b$  and $ a \ge c$
But you can't arrange: $ a\ge b\ge c$
It's true: when you set $ a\equal{}max(a,b,c)$, the roles of $ b$ and $ c$ are not the same.
Don't change $ a$, change $ b$ by $ c$ and $ c$ by $ b$, you have another expression:
$ \frac{a^2\plus{}1}{c^2\plus{}1}\plus{}\frac{c^2\plus{}1}{b^2\plus{}1}\plus{}\frac{b^2\plus{}1}{a^2\plus{}1}$
different from [size=150]L[/size] :(

@Unidranix: I've wether assumed that $ a$ is the maximum of $ (a,b,c)$ nor that $ a \ge b \ge c$.
So my solution is perfectly right  :P

[quote="Unidranix"]Sorry, maybe I'm stupid but:
Let  $ (p,q,r)$  be a permutation of $ (a,b,c)$ , such that  $ p\ge q\ge r$
Means you assumed the roles between $ p,q,r$ meaned you assumed the roles between $ a,b,c$ but you can't.
Am I right? :maybe:
Somebody please explain to me. :surrender:[/quote]
A permutation changes the order of the elements in a set.
E.g. $ (3,2,1)$ is a permutation of $ (1,2,3)$, and $ (2,1,1)$ is a permutation of $ (1,2,1)$. In this problem $ (p,q,r)$ is a permutation of $ (a,b,c)$, such that the elements of $ (p,q,r)$ are the same as of $ (a,b,c)$, but they are sorted differently, i.e. $ p \ge q \ge r$

It's an old problem (for about 3 years ago). Sorry because I can't find the link of it. So, I post my proof again:
 :)

In general: Let $ a_1$, $ \dots$, $ a_n$ be non-neagtive real numbers with sum $ 1$. Prove that
$ \sum_{cyc}\frac {a_1^2 \plus{} 1}{a_2^2 \plus{} 1}\le n \plus{} \frac {1}{2}$.

@nguoivn: I have to say thank you very much.
Finally, I found the proof.
And thanks all of the participants, too

It's nice that you finally found a prove.
But do you accept mine now?^^


fie $p$ prim
aratati ca $C_{2p}^{p}-2$ este divizibil cu $p^{3}$ ($C_{2p}^{p}={2p \choose p}$)

Problema este si in cartea Pobleme de aritmetica si teoria numerelor de Laurentiu Panaitopol si Alexandru Gica.Nu este 3p in loc de 2p?

o fi si cu 3p
problema pe care o stiu eu este asa. (cu 2p) . am zis ca e cunoscuta :) totusi, e interesanta.

Avem $\binom{2p}{p}-2 = \frac{2}{(p-1)!}\left[ \prod_{i=1}^{p-1}(p+i)-\prod_{i=1}^{p-1}i \right]$. Prima parte $\left( \frac{2}{(p-1)!}\right)$ are sens in $\mathbb Z \slash p^{3}\mathbb Z$, deci e suficient sa demonstram ca \[\prod_{i=1}^{p-1}(p+i)-\prod_{i=1}^{p-1}i \equiv 0 \left( \bmod \, p^{3}\right)\]  \[\, \Longleftrightarrow \, \, p^{p-1}+p^{p-2}\sum_{i}i+\ldots+p^{2}\sum_{i<j}\frac{(p-1)!}{ij}+p \sum_{i}\frac{(p-1)!}{i}\equiv 0 \left( \bmod \, p^{3}\right)\] Teorema lui Wolstenholme ne zice ca $p^{2}\left| \sum_{i}\frac1{i}\right.$.

Fie $Q \in \mathbb Z \slash p \mathbb Z [X]$, $Q(X) = X^{p-1}-1$. Din teorema mica a lui Fermat, $Q$ are radacinile distincte $1,2,\ldots,p-1$. Cum $Z \slash p \mathbb Z$ este corp si $Q$ are gradul $p-1$, inseamna ca $Q(X) = (X-1) \ldots (X-(p-1))$. Din aceasta deducem cu usurinta ca polinoamele simetrice elementare (mai putin produsul) sunt divizibile cu $p$, in particular, $p \left| \sum_{i<j}\frac{(p-1)!}{ij}\right.$.

E bine? Dacă am făcut vreo greşeală, dau vina pe ora înaintată.

Cât despre cealaltă problemă, aia chiar îmi pare neabordabilă.

exact asa am facut`o si eu
nu e insa singura demonstratie...sigur exista una mai brute force care nu foloseste wolstenholme..


vector
geometry
linear algebra
linear algebra theorems

What's the difference between vector space and affine space? I read the Mathworld and Wikipedia entries, but I still don't really get the idea. (I have studied vector spaces a fair amount but I never heard of affine spaces except on this forum and what I just read.)

As a math student from one of top universities  :) when it comes to incidence geometry, I would like to answer this..
An affine space is an incidence structure.  I can provide a very strict definition if you like(ask me for it then) but the bottomline is that the "elements" of an affine space are all COSETS of subspaces (I think you are supposed to omit the trivial and total subspace).

A very fundamental difference is that the zero vector is of no importance at all in an affine space while it is in a vector space, or a bit more mathematical if you like : "the zero vector is stabilized by all automorphisms of the vector space, but in an affine space it isn't by stabilized by all automorphisms of the affine space".

Ok thanks, I think I vaguely see what you mean...

Here is an example. Consider $\mathbb R^{3}$. The set $\{(x,y,z): x=1\}$ is an affine subspace of $\mathbb R^{3}$, but it is not a linear (vector) subspace.

Ok I understand what it is now. But what do you do with it? (I am sure there are a ton of things, but the only one I'm aware of is the game SET and I don't really know hte mathematics of that anyway.)

Every vector space is an affine space. Every affine space with a fixed point (origin) is a vector space. So, in a sense, there is no much difference between them. In particular, many geometric theorems about ${\mathbb R}^{d}$ are really theorems about the affine space $\mathbb R^{d}$.

There are two cases when the distinction between affine spaces and vector spaces becomes important.
1) An affine subspace $\neq$ a linear subspace.
2) An affine transformation $\neq$ a linear transformation.


[i]Let be m and n intengers > 0. Proof that[/i]

$\frac{1}{\sqrt[m]{n+1}}+\frac{1}{\sqrt[n]{m+1}}\geq1$

[hide] We start by this lemma : [b]Bernoulli's inequality[/b]

If x > - 1 and n an intenger > 0, then

$(1+x)^{n}\geq1+nx$

Come back to the problem. By this lemma, we can write :

$(1+\frac{m}{n})^{n}\geq1+m$

and

$(1+\frac{n}{m})^{m}\geq1+n$

Then,

$\frac{1}{\sqrt[m]{n+1}}+\frac{1}{\sqrt[n]{m+1}}\geq\frac{1}{1+\frac{n}{m}}+\frac{1}{1+\frac{m}{n}}=1$[/hide]

By AM-GM we have $\sqrt[m]{n+1}\leq \frac{n+m}{m}$ and $\sqrt[n]{m+1}\leq \frac{n+m}{n}$. Done!

[quote="HTA"]oh , its remind me of France's inequalities (i dont remember exacly the year)
let x;y be positve reals number , prove that
$x^{y}+y^{x}\geq 1$[/quote]
It's France 1996 :lol:
[hide]If $a,b \ge 1$, the inequality is obviously true
If $0<a,b<1$:
We have two solutions:
[u]Solution 1:[/u]
We will prove that: $a^{b}\ge \frac{a}{a+b-ab}$
But it is clearly true because:
$a^{1-b}=(1+a-1)^{1-b}\le a+b-ab$
Similarly, we get the desired result
[u]Solution 2:[/u]
$\frac{1}{x^{y}}=(1+\frac{1}{x}-1)^{y}\leq 1+\frac{y}{x}-y<1+\frac{y}{x}=\frac{x}{x+y}$
$\Rightarrow x^{y}\geq \frac{x}{x+y}$
Similarly, we get the desired result
[/hide]

Just use Am-Gm
We only need to prove in case $x,y<1$
So, exist $m,n \in \mathbb{N}$ such that
$\frac{1}{n}\ge x\ge \frac{1}{n+1}$
$\frac{1}{m}\ge y\ge \frac{1}{m+1}$
So we only need to prove that
$\frac{1}{n+1}^{\frac{1}{m}}+\frac{1}{m+1}^{\frac{1}{n}}\ge 1$
We have
$\sqrt[n]{m+1}\le \frac{m+n}{n}$
So $\sqrt[n]{\frac{1}{m+1}}\ge \frac{n}{m+n}$
Similarly
 $\sqrt[m]{\frac{1}{n+1}}\ge \frac{m}{m+n}$
QED


Prove that you can find a polynomial $p(x) = Ax^2 + Bx + C$ that passes through any three points $(x_1,y_1),(x_2,y_2), and (x_3,y_3)$, where all $x_i$ are distinct.

Well the fact that all x's are different shows that it is a function. Frome there, can't you just say $y_n = Ax_n^2 + Bx_n + C$ where $n=1,2,3$ and then go on to solve for each coefficient?

We have the system in $A$, $B$ and $C$:

$f(x_1) = Ax_1^2 + Bx_1 + C = y_1$
$f(x_2) = Ax_2^2 + Bx_2 + C = y_2$
$f(x_3) = Ax_3^2 + Bx_3 + C = y_3$

Thus we get the determinant

$\triangle = \left| \begin{array}{ccc} x_1^2 & x_1 & 1 \\ x_2^2 & x_2 & 1 \\ x_3^2 & x_3 & 1 \end{array}\right| = (x_3 - x_2)(x_3 - x_1)(x_2 - x_1)$

Since all $x_i$ are different, $\triangle \not= 0$, and system above is a Cramer's System. Therefore $A = \frac{\triangle_A}{\triangle}$. If those three points aren't collinear (seems that condition wasn't given), we have $\triangle_A = - \left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right| \not= 0.$

Hence $A \not= 0$ and we are donne. That shows we have only one polynomial $p(x)=Ax^2+Bx+C$ satisfying conditions.

If anyone is interested in more information about the idea LoboSolitario has introduced you should be able to find it by searching for information on Lagrange Interpolating Polynomials


geometry
circumcircle
inequalities
absolute value
geometry proposed

Let T be interior point for acute triangle ABC. Points M, N, P, Q, S and L on the sides of the triangle are such that the line segments MN, PQ and SL pass through T and are parallel to the sides of the triangle. Prove that

                $MT \cdot TN+PT \cdot TQ+LT \cdot TS \le R^2$

where R is the circumradius of the triangle ABC.

 It is my problem and a Bulgarian proposal for IMO'06.

Nice problem!  :D
Actually we can prove that
$MT.TN+PT.TQ+LT.TS=-P_{T/(O)}=R^{2}-OT^{2}\le R^{2}$
general:
$\overline{MT}.\overline{TN}+\overline{PT}.\overline{TQ}+\overline{LT}.\overline{TS}=P_{T/(O)}\forall T$ on plane!
I will post solution letter!

Let $KL \parallel BC,\ MN \parallel CA,\ PQ \parallel AB$ with $K \in CA,\ M \in AB,\ P \in BC.$ Let X, Y, Z be the feet of normals from T to BC, CA, AB and let A', B', C' be the midpoints of BC, CA, AB. The right angle triangles $\triangle MTZ \sim \triangle OBA'$ are similar, having equal angles $\angle MTZ = \angle OBA' = 90^\circ-\angle CAB,$ hence $\frac{A'B}{OB}= \frac{MT}{ZT},$ $MT = \frac{OB \cdot ZT}{A'B}= \frac{2R \cdot ZT}{AB}.$ Similarly, $NT = \frac{2R \cdot XT}{BC}.$ Since $\angle ZTX + \angle ABC= 180^\circ$ $\Longrightarrow$

$MT \cdot NT = 4R^{2}\ \frac{ZT \cdot XT}{AB \cdot BC}= 4R^{2}\ \frac{[\triangle ZTX]}{[\triangle ABC]}.$

In exactly the same way,

$KT \cdot LT = 4R^{2}\ \frac{[\triangle YTZ]}{[\triangle ABC]},\ \ \ PT \cdot QT = 4R^{2}\ \frac{[\triangle XTY]}{[\triangle ABC]}.$

Summing up,

$KT \cdot LT+MT \cdot NT+PT \cdot QT = 4R^{2}\ \frac{[\triangle XYZ]}{[\triangle ABC]}$

Absolute value of the power of a triangle interior point T to its circumcircle (O) is $R^{2}-TO^{2} > 0$ and area of the pedal triangle $\triangle XYZ$ of the point T is $[\triangle XYZ] = [\triangle ABC] \cdot \frac{R^{2}-TO^{2}}{4R^{2}}$ (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38164]Area of pedal triangle[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5658]Beautiful problem[/url]). As a result,

$KT \cdot LT+MT \cdot NT+PT \cdot QT = R^{2}-TO^{2}\le R^{2}$

[quote="Megus"][quote="estoyanovvd"]
It is my problem and a Bulgarian proposal for IMO'06.[/quote]

You mean it's on shortlist 2006?  :o  :huh:  :?: Then it shouldn't be posted....[/quote]
 No,it was two times only proposal on 2005 and 2006!
They don't like her!!!!Why?!

@Yetti: Fine!!! It's exactly like my solution!!!

[b][u]Very nice and easy problem ![/u][/b]

[color=darkred][b][u]Lemma (well-known).[/u][/b] Let $ABC$ be a triangle. Then

$\boxed{\ x+y+z=1\Longrightarrow yza^{2}+zxb^{2}+xyc^{2}\le R^{2}\ }\ .$[/color]

[color=darkblue][b][u]Proof of the proposed problem.[/u][/b]

$\|\begin{array}{c}\{\begin{array}{c}B\prec C_{2}\prec B_{1}\prec C\ ,\ B_{1}C_{2}=xa\\\\ C\prec A_{2}\prec C_{1}\prec A\ ,\ C_{1}A_{2}=yb\\\\ A\prec B_{2}\prec A_{1}\prec B\ ,\ A_{1}B_{2}=zc\end{array}\\\\ T\in A_{1}A_{2}\cap B_{1}B_{2}\cap C_{1}C_{2}\\\\ \{\begin{array}{c}A_{1}A_{2}\parallel BC\\\\ B_{1}B_{2}\parallel CA\\\\ C_{1}C_{2}\parallel AB\end{array}\end{array}\|\Longrightarrow\|\begin{array}{c}ABC\sim TC_{2}B_{1}\sim C_{1}TA_{2}\sim B_{2}A_{1}T\\\\ \{\begin{array}{c}A_{1}T=BC_{2}\ ,\ TA_{2}=B_{1}C\\\\ B_{1}T=CA_{2}\ ,\ TB_{2}=C_{1}A\\\\ C_{1}T=AB_{2}\ ,\ TC_{2}=A_{1}B\end{array}\end{array}\|\Longrightarrow$

$\|\begin{array}{c}\boxed{\ x+y+z=1\ }\\\\ TA_{1}=za\ ,\ TA_{2}=ya\\\\ TB_{1}=xb\ ,\ TB_{2}=zb\\\\ TC_{1}=yc\ ,\ TC_{2}=xc\end{array}\|\Longrightarrow \sum TA_{1}\cdot TA_{2}=\boxed{yza^{2}+zxb^{2}+xyc^{2}\le R^{2}}\ .$

[b]Remark.[/b] The proposed inequality is truly generally, i.e. for any point $T\ \ ,\ \ \sum \overrightarrow{TA_{1}}\cdot \overrightarrow{TA_{2}}\le R^{2}\ .$[/color]

[quote="estoyanovvd"]
[b]Problem:[/b] Let $M$ be interior point for acute triangle $ABC$. Points $X, T\in AB; Y, U\in CA; Z, V\in BC$ so that: $XY\parallel BC; ZT\parallel CA; UV\parallel AB$.  Prove that: $\overline{MU}.\overline{MV}+\overline{MT}.\overline{MZ}+\overline{MX}.\overline{MY}\le R^{2}$
where $R$ is the circumradius of the triangle $ABC$.[/quote]
[b][color=blue]gemath[/color][/b], I can help you. :D
[b]Lemma:[/b] Let a circle $(O;R)$ and three points $A, B, C$ lie on a line. Then we have:
$P_{A/(O)}.\overline{BC}+P_{B/(O)}.\overline{CA}+P_{C/(O)}.\overline{AB}+\overline{BC}.\overline{CA}.\overline{AB}=0$
[b]Proof:[/b]
$P_{A/(O)}.\overline{BC}+P_{B/(O)}.\overline{CA}+P_{C/(O)}.\overline{AB}+\overline{BC}.\overline{CA}.\overline{AB}$
$=(OA^{2}-R^{2}).\overline{BC}+(OB^{2}-R^{2}).\overline{CA}+(OC^{2}-R^{2}).\overline{AB}+\overline{BC}.\overline{CA}.\overline{AB}$
$=OA^{2}.\overline{BC}+OB^{2}.\overline{CA}+OC^{2}.\overline{AB}+\overline{BC}.\overline{CA}.\overline{AB}-R^{2}.(\overline{BC}+\overline{CA}+\overline{AB})$
$=OA^{2}.\overline{BC}+OB^{2}.\overline{CA}+OC^{2}.\overline{AB}+\overline{BC}.\overline{CA}.\overline{AB}=0$ (by Stewart's theorem)
[b]Solution for the problem:[/b]
$P_{M/(O)}.\overline{XY}+P_{X/(O)}.\overline{YM}+P_{Y/(O)}.\overline{MX}+\overline{XY}.\overline{YM}.\overline{MX}=0$
$\Longrightarrow P_{M/(O)}=P_{X/(O)}.\frac{\overline{MY}}{\overline{XY}}-P_{Y/(O)}.\frac{\overline{MX}}{\overline{XY}}+\overline{MX}.\overline{MY}$
$=\frac{\overline{XA}.\overline{XB}.\overline{MY}}{\overline{XY}}-\frac{\overline{YA}.\overline{YC}.\overline{MX}}{\overline{XY}}+\overline{MX}.\overline{MY}$
$=\frac{\overline{MU}.\overline{XB}.\overline{MY}}{\overline{MY}}+\frac{\overline{MT}\overline{YC}.\overline{MX}}{\overline{MX}}+\overline{MX}.\overline{MY}$
$=\overline{MU}.\overline{MV}+\overline{MT}.\overline{MZ}+\overline{MX}.\overline{MY}$
$\Longrightarrow\overline{MU}.\overline{MV}+\overline{MT}.\overline{MZ}+\overline{MX}.\overline{MY}=R^{2}-OM^{2}\le R^{2}$

yes that's right April, it likes me too, but I think it is in the book of your teacher  :D


advanced fields
advanced fields unsolved

Could you help me with this problem.
with method of images
∇$ ^2u$(ξ,η)=h(ξ,η) ,ξ>0 ,η>0
u(0,η)=f(η)
$ u$η(ξ,0)=g(ξ)

Triple post: lol :
http://www.mathlinks.ro/viewtopic.php?p=1400829#1400829
http://www.mathlinks.ro/viewtopic.php?p=1400830#1400830


If I have time, I will make a mock MathCounts test. I want to know what level to make it, so [b]please choose your preference in the poll![/b]

[b]To sign up, add your name to this list[/b] (and keep other people on it too, copying the list):

[hide="Participants"]
1.
2.
3.
4.
5.

*Note that it isn't limited to 5 people. Add numbers as necessary.
[/hide]

[b]I will probably not need problem-makers or graders, but if you wish to help, add your name to this list:[/b]

[hide="Help"]
1.
2.
3.
4.
5.

*Note that it isn't limited to 5 people. Add numbers as necessary. Also, I may not need any helpers, so nothing is guaranteed.
[/hide]

You can sign up as a participant and helper, and if I need helpers, I will go down the list and take your name off the participants list. You can sign up for just helping too.

I will most likely do a [b]Sprint Round[/b], and if I have time I will add a Target Round and Team Round.

It will probably be in [b]mid- to late-October[/b], and I will post rules then (they should
[color=red][size=9]PowerOfPi is not responsible for any physical or mental harm caused by this mock test.[/size][/color]

Good Luck!

[hide="Participants"]
1.Cliu
2.Thunder365
3.
4.
5.[/hide]

Sounds cool. Im in

I'm in. /think Why am I even joining?
LOLMFG ya heard that?
/facehammer...

?that last part was kinda spammy
btw, in!

I sign up. btw can you send a reminder message when you put the rounds up? I'm probably going to forget

I'm in. Actually, I sort of want a between STate-Nat test :)

Please copy the lists. It saves me some time.  :wink:

[hide="Participants"]
1. Cliu0301
2. Thunder365
3. alexautin
4. cgyao15
5. jerry125
6. AwsomeToad
7. ernie ??? (you said you will help, but never specified if you will participate)

*Note that it isn't limited to 5 people. Add numbers as necessary.
[/hide]

[hide="Helpers"]
1. ernie
2.
3.
4.
5.

*Note that it isn't limited to 5 people. Add numbers as necessary. Also, I may not need any helpers, so nothing is guaranteed.
[/hide]

Also, there is a high chance that the other mock MathCounts administrators might help with this and me with the other one.

How many people are interested in this:

A School, Chapter, State, and National round full competition with Sprint and Target for each? This may or may not happen, so no guarantees. You will have to qualify for the next round each time, and after it is over, a full answer key will be posted, so you can compare your answers (if you aren't in the competition) to the correct answers.

IF this happens, it will be a combination of this and the other mock test.

[hide="Participants"]
1. Cliu0301
2. Thunder365
3. alexautin
4. cgyao15
5. jerry125
6. AwsomeToad
[/hide]

[hide="Helpers"]
1. ernie
2.
3.
4.
5.
[/hide]

Come on, only six people?

I'll join.  Your "qualifying for the next round" idea sounds good to me.  I voted for the nationals test in the poll if that helps with anything  :)

[hide="Participants"]
1. Cliu0301
2. Thunder365
3. alexautin
4. cgyao15
5. jerry125
6. AwsomeToad
7. jxl28
[/hide]

[hide="Helpers"]
1. ernie
2.
3.
4.
5.
[/hide]

[hide="Participants"]
1. Cliu0301
2. Thunder365
3. alexautin
4. cgyao15
5. jerry125
6. AwsomeToad
7. jxl28
8. king!
[/hide]

[hide="Helpers"]
1. ernie
2.
3.
4.
5.
[/hide][/quote]

I'm in! I voted for a nationals test.

[hide="Participants"]

1. Cliu0301
2. Thunder365
3. alexautin
4. cgyao15
5. jerry125
6. AwsomeToad
7. jxl28
8. king!
9. SonyWii
[/hide]

[hide="Helpers"]
1. ernie
2.
3.
4.
5.
[/hide]

I'm joining. I wonder if I can do it..

By the law of AoPS, any test meant to be at level $ x$ will end up being at least $ x\plus{}1$. I want a National level test, thus, I voted for state.

Participants:
1. Cliu0301
2. Thunder365
3. alexautin
4. cgyao15
5. jerry125
6. AwsomeToad
7. jxl28
8. king!
9. SonyWii
10. isabella2296

Helpers:
1. ernie
2.
3.
4.
5.

Okay, I think I will start work on the School Sprint (still working with vallon and pytheagle). I am hoping we can start this in a couple weeks, provided we get at least 20 signups.

Also, helpers are welcomed.  :)  And because it will have a variety of levels, ignore the poll.

OK I'll join.

I only have 3 school tests(2007-2009) so I want one to mirror that level(I'm aiming for a 45, but at my school I really just need a 30)

[hide="Participants"]
jxl28
athunder
Thunder365
mathboy101
AIME15USAMO
cgyao15
the cliu
TheMan1998
jerry125
SuperNerd123
\$LaTeX\$
Canton
algebra1337
AwesomeToad
bulldog23
ButteredButNotEaten
KingJamesMvP
varunrocks
Maybach
gogators
yaofan
[/hide]
by pytheagle and vallon22

$ \plus{}$

[hide="Participants"]
1. Cliu0301
2. Thunder365
3. alexautin
4. cgyao15
5. jerry125
6. AwsomeToad
7. jxl28
8. king!
9. SonyWii
10. isabella2296
11. Maybach
12. mathboy101
[/hide]
[hide="Helpers"]
1. ernie
2.
3.
4.
5.
[/hide]
by PowerOfPi

$ \equal{}$

[hide="Participants"]
jxl28
athunder
Thunder365
mathboy101
AIME15USAMO
cgyao15
the cliu
TheMan1998
jerry125
SuperNerd123
\$LaTeX\$
Canton
algebra1337
AwesomeToad
bulldog23
ButteredButNotEaten
KingJamesMvP
varunrocks
Maybach
gogators
yaofan
alexautin
King!
SonyWii
isabella2296
[/hide]
[hide="run by"]
pytheagle
vallon22
ernie
PowerOfPi
[/hide]

[b][size=150][color=red]From now on, all conversation will be continued in the other thread![/color][/size][/b]


counting
IMO Shortlist
combinatorics
bijection

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="Severius"]first number the $1$s in the sequence from right to left, then the $2$s from right to left, and so on (for example, the full sequence $213244$ becomes $314265$).[/quote]

I don't really understand your construction, HOW exactly are you assigning a permutation to a full sequence?

Well... you take the full sequence. First, you look at just its $1$s, and number them from right to left; that is, you assign $1$ to the rightmost $1$, $2$ to the following $1$ to the left, ..., $k$ to the leftmost $1$, where there are $k$ $1$s in the sequence. Then, you number the $2$s in the same fashion, but the first number you assign is $k+1$ (which goes to the rightmost $2$), then $k+2$, and so on; the leftmost $2$ is assigned $k+m$, where there are $m$ $2$s. We proceed like this until the whole sequence is numbered. Then, for example, $2211124334$ becomes $653214(10)879$, if I've done it right.

[quote="pavel25"]A full sequence have to start with $1$
Because for all $k\geq2$ the first occurrence of $k-1$ comes before the last occurrence of $k$.
The next term in the sequence can be $1$ or $2$.
And if our term is $k$ the next term can be $k$ or $k+1$ - $2$ options each time we adding a term
So for each $n$ the number of full sequences equal to ${2^{n-1}}$[/quote]

Aha, I see the problem with your logic.  Notice that the third term can be $1, 2$, or $3$ because the problem only stipulates the first and last occurrences of $k-1, k$ respectively, so there are $p$ possibilities for the $p^{th}$ term and the answer is $n!$ as stated above.

[quote="orl"]Let $n$ be a positive integer. A sequence of $n$ positive integers (not necessarily distinct) is called [b]full[/b] if it satisfies the following condition: for each positive integer $k\geq2$, if the number $k$ appears in the sequence then so does the number $k-1$, and moreover the first occurrence of $k-1$ comes before the last occurrence of $k$. For each $n$, how many full sequences are there ?[/quote]
we want to determaine the $F_n$
we circle the place of the greatest number
then we choose $i$ places from $n$ places and then we put the $n-i$ number4s left in $F{}_n{}_-{}_i$ ways
and now the $i$ places:
$i)$one of them be the greatest one
$ii)$two of them be the greatest one
.
.
.
$\Rightarrow F_n=\sum\limits_{i=1}^n{{{n}\choose{i}}F{}_n{}_-{}_i}$
;)  :D

Here's a boring solution.

Let $ a_{n}$ be the desired number of full sequences of length $ n$. We show that $ a_{n} \equal{} na_{n \minus{} 1}$. It will follow easily by induction that $ a_{n} \equal{} n!$ for all $ n$. Let our sequence be $ x_{1},x_{2},\ldots,x_{n}$. Clearly, some term in the sequence must be 1. Let $ x_{i} \equal{} 1$, such that $ i$ is maximal. We claim that the number of ways to fill in the other $ n \minus{} 1$ values of the sequence is $ a_{n \minus{} 1}$. This will prove the claim, as we just sum this from $ i \equal{} 1$ to $ i \equal{} n$ to get $ na_{n \minus{} 1}$. We will show a bijection between $ (x_{1},x_{2},\ldots,x_{i \minus{} 1},x_{i \plus{} 1},x_{i \plus{} 2},\ldots,x_{n})$ and full sequences of length $ n \minus{} 1$.

If we take our full sequence of length $ n$, note that $ (x_{1},x_{2},\ldots,x_{i \minus{} 1},x_{i \plus{} 1},x_{i \plus{} 2},\ldots,x_{n})$ satisfies the full sequence conditions for $ k\ge3$, and if one of $ x_{1},x_{2},\ldots,x_{i \minus{} 1}$ is equal to 1 (since we took $ i$ maximal, none of the other terms are), then it satisfies the full sequence conditions for $ k\ge2$, and is a full sequence by itself. On the other hand, if $ x_{i}$ is the only 1 in our original full sequence, the minimum value of the other terms is 2 (note that there must be a 2, or else every term would be a 1), and we can subtract 1 from every term to get a full sequence of length $ n \minus{} 1$. Thus, each full sequence of length $ n$ with $ x_{i} \equal{} 1$ maps to a full sequence of length $ n \minus{} 1$. We see that no two full sequences of length $ n$ map to the same full sequence of length $ n \minus{} 1$: if they're the same, then either the full sequences of length $ n$ are the same, or if they are translations, we have a full sequence that is the result of adding 1 to every non $ x_{i}$ term of another full sequence, which is not possible because we need a 2 in $ (x_{i \plus{} 1},\ldots,x_{n})$, meaning we had a 1 in this subsequence in the original full sequence, contradicting the maximality of $ i$ (WOW that was run-on).

For the other direction, we show that every full sequence of length $ n \minus{} 1$ maps to a unique full sequence of length $ n$. Things basically work backwards the same way, if there is a 1 in the last $ n \minus{} i$ terms of our full sequence of length $ n \minus{} 1$, add 1 to every term, and stick in a 1 in the appropriate place to get a full sequence of length $ n$. If not, stick in the 1 without doing anything. Note that the resulting full sequences are unique because if two sequences of length $ n \minus{} 1$ map to the same sequence of length $ n$, they must be translations of each other, and one doesn't contain a 1 and is not a full sequence. So we're done.

Here is another induction solution :

Let $\chi_k ^n (i,j)$ denote the number  of full sequences of length $n$, with  maximal element $k$, first appearance of $k$ is at $i$, and last appearance is at $j$. Its easy to that  $i=j$, then $k \ge 2$ ans if $k$ appears more than once in any of the full sequences in $\chi_k ^n (i,j)$, then $1 \le k \le n-1$.

We want to find $T_n =  \sum _{1 \le i \le j \le n} \sum_{k=1}^{n}\chi_k ^n (i,j)  =  \sum _{2 \le i =  j \le n} \sum_{k=1}^{n}\chi_k ^n (i,j)  +  \sum _{1 < i \le j \le n} \sum_{k=1}^{n}\chi_k ^n (i,j)$          (*)

[b]Lemma 1:[/b]

$\sum _{2 \le i =  j \le n} \sum_{k=1}^{n}\chi_k ^n (i,j)  = \sum _{2 \le i =  j \le n} \sum_{k=2}^{n}\chi_k ^n (i,j)= \sum _{1 \le p \le q \le n-1}\sum_{k=2}^{n}(n-p) \chi_{k-1}^{n-1}(p,q)$  $ = \sum _{1 \le p \le q \le n-1}\sum_{k=1}^{n-1}(n-p) \chi_{k}^{n-1}(p,q) $

Proof: If we remove $k$, by a bijection we see the following :

$\chi_k ^n (i,i) = \sum_{1 \le p \le i-1} \chi_{k-1}^{n-1}(p,q) $

$ \sum_{1 \le i \le n}\chi_k ^n (i,i) = \sum_{a=2}^{n}\sum_{p=1}^{a-1} \chi_{k-1}^{n-1}(p,q)   = \sum _{1 \le p \le q \le n-1}(n-p) \chi_{k-1}^{n-1}(p,q)$

So indeed $ \sum _{2 \le i =  j \le n} \sum_{k=2}^{n}\chi_k ^n (i,j)= \sum _{1 \le p \le q \le n-1}\sum_{k=2}^{n}(n-p) \chi_{k-1}^{n-1}(p,q)$

[b]Lemma 2:[/b] $\sum_{k=1}^{n-1} \sum _{1 \le i < j \le n-1} \chi_k^n (i,j) = \sum _{1 \le p \le q \le n-1 } \sum_{k=1}^{n-1} p \; \chi_{k}^{n-1} (p,q)$

Proof : If we remove the first $k$, by a bijection we see the following :

${\chi_k ^n (i,j)}_{i<j} = \sum_{p=a}^{b-1}  \chi_{k}^{n-1} (p, b-1) $

So $\sum _{1 \le i < j \le n-1 }\chi_k ^n (i,j) = \sum_{b=2}^{n} \sum_{a=1}^{b-1} \sum_{p=a}^{b-1}  \chi_{k}^{n-1} (p, b-1)  $  $= \sum_{b=2}^{n} \sum_{p=1}^{b-1}  p \;    \chi_{k}^{n-1} (p, b-1)  $ $= \sum_{1 \le p \le q \le n-1} p \: \chi_k ^{n-1} (p,q)$

So summing over all $k$, we get $\sum_{k=1}^{n-1} \sum _{1 \le i < j \le n-1} \chi_k^n (i,j) = \sum _{1 \le p \le q \le n-1 } \sum_{k=1}^{n-1} p \; \chi_{k}^{n-1} (p,q)$

Using both the lemmas in (*) we get

$T_n =  \sum _{1 \le i \le j \le n} \sum_{k=1}^{n}\chi_k ^n (i,j) =   n\;  \sum _{1 \le p \le q \le n-1} \sum_{k=1}^{n-1}\chi_k ^{n-1} (p,q) = n T_{n-1}  $

So $T_n = n!$

There is a quick bijection between these sequences and the permutations of $1,2,...,n$!!

For such a sequence with numbers from $1$ to $k$, we write:

the position of the first $k$, then the position of the second $k$, ..., then the position of the last $k$, then the position of the first $k-1$, ..., then the position of the last $k-1$, ..., then the position of the first $1$, ..., then the position of the last $1$.

For any permutation, look at its first number and write a $k$ in that position. If the next number is bigger, write a second $k$ in that position, otherwise write a $k-1$ in that position. Continue like this. Everytime the number increases, rewrite the number you just wrote in that position. Otherwise, write that number minus $1$, in that position. at the end you can deduce value of $k$

This is thus a bijection so $\boxed{n!}$ is answer!

[quote=Severius]The answer is $n!$.

To prove it we'll find a bijection between the permutations of $1, 2, \ldots, n$ and the full sequences of length $n$.

Take an arbitrary full sequence of length $n$ and assign it a permutation of $1, 2, \ldots, n$ as follows: first number the $1$s in the sequence from right to left, then the $2$s from right to left, and so on (for example, the full sequence $213244$ becomes $314265$).

Now, it's easy to see that the transformation is reversible: take a permutation of $1, 2, \ldots, n$ and label the $1$ with a $1$. From then on, if the last number we've looked at is $k$ (suppose we've labeled it $m$), we look at $k+1$; if $k+1$ is to the right of $k$ in the permutation we label it $m+1$, otherwise we label it $m$. It is easy to check that this procedure gives a full sequence, and that if we apply the first transformation to it, we get back the original permutation. So this gives a bijection between the two sets and we're done.[/quote]

What is the motivation for this bijection? Thank you very much!

Also, does the following solution work?
By removing the largest integer that first appears from a full sequence of length $n$, we get a full sequence with length $n-1$. For each full sequence with length $n-1$, there are $n$ ways to make it a full sequence with length $n$, so $a_n=na_{n-1}$, where $a_n$ is the number of full sequences, so $a_n=n!$

[hide=Solution]
The answer is $n!$.

We construct a bijection between full sequences and permutations of $1,2,\ldots,n$. Let $a_1,a_2,\ldots,a_n$ be a full sequence, and suppose that its largest term is $k$. Now, for each $i=k,k-1,\ldots,2,1$ in that order, write out all the indices $j$ in increasing order for which $a_j=i$. This gives a permutation of $1,2,\ldots,n$.

For the reverse, let $\sigma$ be a partition of $1,2,\ldots,n$. Break $\sigma$ into maximal increasing subsequences; suppose there are $k$ of them. Then, for $i=1,2,\ldots,k-1,k$, the indices $j$ for which $a_j=i$ in our full sequence are exactly the numbers in the $i$th maximal increasing subsequence.

It is easy to see that these operations are reverses of each other, so the answer is the number of permutations of $1,2,\ldots,n$, which is $n!$.
[/hide]

We will show that if there are $s_n$ full sequences, then $s_{n+1}=(n+1)s_n$, which will show $s_n=n!$. Consider the $s_n$ sequences, and consider any arbitrary full sequence $a_1\ldots a_n$. For any $0\le i\le n$, consider the new sequence $a_1\ldots a_iba_{i+1}\ldots a_{n}$, where $b=\max(1+\max(a_1,\ldots, a_i),\max(a_{i+1},\ldots, a_{n}))$ (Here, let the maximum of $0$ numbers to be $0$). We will show that this new sequence is a full sequence of length $n+1$. The first condition is fulfilled because $\max(a_1,\ldots, a_n,b)\le \max(a_1,\ldots, a_n)+1$. Also, the first occurrence of any number $k\le \max(a_1,\ldots, a_n)$, the second condition is fulfilled. We only need to check if $b=\max(a_1,\ldots, a_n)+1$ and $k=b$, but then we must have $b=1+\max(a_1,\ldots, a_i)$ so then our second condition is filled.

Hence, each full sequence of length $n$, there are $n+1$ corresponding full sequences of length $n+1$. It suffices to show that for any full sequence of length $n+1$, say $c_1\ldots c_{n+1}$ there is a unique number in the $c_{j+1}$, $1\le j\le n+1$, such that $c_{j+1}=\max(1+\max(c_1,\ldots, c_j), \max(c_{j+2},\ldots, c_{n+1}))$ (Once again, the maximum of $0$ numbers is $0$), and that this sequence $c_1\ldots c_jc_{j+2}\ldots c_{n+1}$ is still full. We will show that $c_{j+1}$ is in fact the first occurrence of $\max(c_1,\ldots c_{n+1})$. If $c_{j+1}$ is not the maximum, then there is an occurrence of $c_{j+1}+1$ after it by assumption, which contradicts that $c_{j+1}< \max(c_{j+2},\ldots, c_{n+1})$. If $c_{j+1}$ is not the first occurrence of the maximum, then clearly $c_{j+1}<1+\max(c_1,\ldots, c_j)$. Now, we show that if $c_{j+1}$ is the first occurrence of the maximum, then $c_{j+1}=\max(1+\max(c_1,\ldots, c_j), \max(c_{j+2},\ldots, c_{n+1}))$. In fact, if there is only one occurrence of the maximum, then, $c_{j+1}=1+\max(c_1,\ldots, c_j)$ by assumption, and if there is more than one occurrence, then $c_{j+1}=\max(c_{j+2},\ldots, c_{n+1})$. Furthermore, it's clear that the sequence $c_1\ldots c_jc_{j+2}\ldots c_{n+1}$ is a full sequence of length $n$, so we are done.

The answer is $n!$. We will establish a bijection between permutations of $1,2,\ldots,n$ and full sequences of length $n$.

Take a full sequence, and suppose that $i$ appears $a_i$ times. Replace the $j$th occurrence of $i$ in the full sequence with $a_1+a_2+\ldots+a_i+j$. The resulting sequence is clearly a permutation of $1,2,\ldots,n$ since every number we write is different and ranges from 1 to n. Also, this process is reversible and this shows that this is a bijection. Just take a permutation and let $1,2,\ldots,a_1$ be a subsequence, with $a_1$ as large as possible and replace all of those numbers with 1. Then, let $a_1+1,a_1+2,\ldots,a_1+a_2$ be a subsequence, with $a_2$ as large as possible and replace all of those numbers with 2, and so on. Clearly, the last $k-1$ must occur after the first $k$, or else the first $k$ would be a $k-1$ instead, so this is a full sequence. Hence, we have the answer is $n!$.

[hide=Solution]We claim the answer is $n!$.

Let $S$ be the set of all permutations of  $(1, 2, 3, …, n)$, where $s = (s_1, s_2, …, s_n)$ denote a permutation of $S$. Let $A$ be the set of all full sequences, where $a = (a_1, a_2, …, a_n)$ denotes a full sequence.

Now, we define a function $f: A \to S$ as follows:
At each step $i$: Let $j$ denote the index of the last occurrence of $1$. Then, let $s_j = n+1-i$ in $s$. Now, if the remaining elements of $a$ are all at least $2$, then we subtract all the elements by $1$.

We claim this function is bijective.

Indeed, we consider the function $f^{-1}: S \to A$ as follows:

At each step $i$: Let $j$ denote the index such that $s_j = i$. If $i = n$, then let $a_j = 1$. Otherwise, let $k$ denote the index such that $s_k = i+1$. Now, if $j<k$, then we let $a_j = a_k$. Otherwise, we let $a_j = a_k + 1$.

First, we claim that this always gives a valid full sequence. For each number $x>1$, let $y$ denote the index of $x$ corresponding first addition of $x$ in the sequence in the algorithm above. Now, let $z$ denote the index corresponding to the addition 1 step before the first addition of $x$. Now, note that $a_z = x-1$, because clearly, at each step, the number added either stays the same or increases by $1$. Furthermore, we have $z < y$ by design of the algorithm, so we have an occurrence of $x-1$ before the occurrence of $x$ for all $x>1$, and thus this must be a valid full sequence.

Next, we claim that $f^{-1}$ is indeed the reverse function of $f$. Now, note that $n$ in the permutation has the same index last occurrence of $1$ in the sequence in both functions. Furthermore, note that removing the last occurrence of $1$ either
a)Keeps it at a full sequence if there are additional $1$s, because the location of the first occurrence of $x$ and the last occurrence of $x+1$ doesn’t change for all $x$

b)If there are no additional $1$s, then it’s still a full sequence if we subtract $1$ from all other remaining elements, since the location of the first occurrence of $x$ and the last occurrence of $x+1$ doesn’t change for all $x \geq 2$. Furthermore, the location of $n$ is before the location of $n-1$ in the permutation.

Now, we let $a’$ denote the full sequence corresponding to the one obtained in either a) or b). Then, we can induct down, since we now apply this algorithm between $a’$ and all the permutations of $(1, 2, …, n-1)$. Clearly, this function is bijective for $n=1$, since the only full sequence is $(1)$ and the only permutation is $(1)$, proving the claim.[/hide]

Wait, $n!$ sequences? Am I misreading the problem or what? Somebody, please tell me which sequences am I missing.

When $n=3,$ there are $5$ [hide=sequences]\begin{align*}
123 \\
212 \\
122 \\
112 \\
111
\end{align*}[/hide].

When $n=4,$ there are $16$ [hide=sequences]
\begin{align*}
1234 \\
1233 \\
1323 \\
3123 \\
1223 \\
2123 \\
1223 \\
1123 \\
1112 \\
1122 \\
1212 \\
2112 \\
1222 \\
2122 \\
2212 \\
1111
\end{align*}
[/hide]

@below Thanks

[quote=Wizard_32]Wait, $n!$ sequences? Am I misreading the problem or what? Somebody, please tell me which sequences am I missing.

When $n=3,$ there are $5$ [hide=sequences]\begin{align*}
123 \\
212 \\
122 \\
112 \\
111
\end{align*}[/hide].

When $n=4,$ there are $16$ [hide=sequences]
\begin{align*}
1234 \\
1233 \\
1323 \\
3123 \\
1223 \\
2123 \\
1223 \\
1123 \\
1112 \\
1122 \\
1212 \\
2112 \\
1222 \\
2122 \\
2212 \\
1111
\end{align*}
[/hide].[/quote]
You are missing a lot of them.
1121,1211,1221,2121,1213,1231,1232,2312,2132
And for the other one you didnt write 121

I claim there are $n!$ full sequences. To do this, we establish a bijection between the set of full sequences of length $n$ and the set of permutations of $\{1,2,\ldots,n\}$. Begin with a permutation $P$ of $\{1,2,\ldots,n\}$ and partition it into decreasing subsequences of consecutive integers. For example, the permutation $13542$ would be partitioned into $[1], [3,2], [5,4]$. Number these subsequences $s_1,s_2,\ldots,s_k$ by increasing order of maximal element. Now, each $1 \leq i \leq n$ is in exactly one $s_j$. For each $i$ in the permutation, replace $i$ with $j$ where $i \in s_j$. In the above example, $13542$ would map to $12332$.

We now prove that this sequence is full. Take some permutation $P$ and two subsequences $s_k$ and $s_{k-1}$ in its partition. We must show that the first element of $s_{k-1}$ comes before the last element of $s_k$. However, the last element of $s_k$ is one greater than the first element of $s_{k-1}$. This means that if the first element of $s_{k-1}$ came after the last element of $s_k$, the last element of $s_k$ would be in $s_{k-1}$, contradiction. Thus, the produced sequence is full.

This process is easily reversible: begin with a full sequence of length $n$ and say there are $a_i$ occurrences of $i$ for each $i$. Then, create a permutation of $\{1,2,\ldots,n\}$ by, for each $i$, first numbering each occurrence of $i$ in descending order then adding $\sum_{j=1}^{i-1} a_j$ to the result. For example, the sequence $12332$ gives $a_1=1,a_2=2,a_3=3$, so it corresponds to the permutation $1(1+2)(3+2)(3+1)(1+1)=13542$, as desired.

[quote=tenplusten]I really liked CantonMathGuy's solution.But can someone check this solution?(Because recursive solutions for this problem dont seem easy .)
[quote=MathPanda1]

Also, does the following solution work?
By removing the largest integer that first appears from a full sequence of length $n$, we get a full sequence with length $n-1$. For each full sequence with length $n-1$, there are $n$ ways to make it a full sequence with length $n$, so $a_n=na_{n-1}$, where $a_n$ is the number of full sequences, so $a_n=n!$[/quote][/quote]
This is exactly what i did! To elaborate more, one takes a full sequence of length $n$ and remove the first occurrence of the largest number in the sequence giving us a $n-1$ number sequence which is obviously also full. Then notice that for a $n-1$ number full sequence, there are $n$ places where we can put a new number, however this number is fixed with respect to this position since it has to be the first occurrence of the largest number in the $n$ number sequence it is going to be a part of.

We solve this question by recurrence. Take any full sequence of length $n-1.$ Now, consider any one of the integers in the sequence. Now, we want to place a number to the right of it so that it is the first time the largest number appears in the new sequence. For example, in $212332$ we can place a $4$ between the two $3$'s to get $2123432.$ Now, we claim that in each of these integers, there is exactly one integer you can insert such that the resulting sequence is full.
\\$~$\\
Obviously, this number you are inserting, it must be at most one more than the largest integer currently on the list, and it must be at least the largest integer currently on the list. Suppose $k$ is the current largest integer. If there are any $k$'s to the left of where you are inserting the new number, then inserting a $k$ won't work. On the other hand, if there aren't any $k$'s to the left and you insert a $k+1$ then the sequence won't be full.
\\$~$\\
Now, note that the left-most space cannot have a number inserted in this way, so for each full sequence of length $n-1$ you can make $n$ full sequences of length $n.$ Also note that each full sequence of length $n$ there is only one full sequence of length $n-1$ that can be made into this sequence, because you must remove the first time the large integer appears. Now, since there is $1$ full sequence of length $1$ there are $n!$ full sequences of length $n.$

@above why does the newly inserted integer need to be at least the largest integer in the old sequence?

DAM.

The answer is $n!$.
We construct a bijection between full sequences with largest term k and terms $a_1,a_2,\dots,a_n$ in those spots, with permutations of $1,2,\dots,n$.

Break the permutation of 1->n into maximal increasing subsequences; suppose there are $k$ of them. Then, for $1\le i\le k$, the indices $j$ for which $a_j=i$ in our full sequence are exactly the numbers in the $i$th maximal increasing subsequence, where increasing subsequences are read left to right but the order of each subsequence block is read right to left. As an example, (6)(4)(237)(5)(1) goes to 1334253 ((1) goes to 1, (5) goes to 5th place in the second sequence with value 2 since it's the second subsequence, etc.), and (56{10})(7)(24)(39)(8)(1) goes to 1434665236; it's obvious that this works and that it's reversible, since the first occurrence of k-1 in the full sequence is the k-1th subsequence, which indeed occurs before the kth subsequence.
The problem is now evident, since there are n! ways for the permutations.

also when awesomeming just gets nodiffed by a user who has 1 post :skull:
jk dont block me like calvin is threatening to its a joke you probably dont check feed cuz u have too many isls that they would all be unread and you cant read all them


请教各位：一些数学符号如：根号，分数线等等该怎么打呀？

http://www.mathlinks.ro/viewtopic.php?t=250428

分数：\frac{}{}, 根号: \sqrt