calculus
limit
modular arithmetic
induction
number theory proposed
number theory

A sequence $c_n$ is constructed after the following rule: We set $c_1 = 2$ and $c_{n+1}=\left[ \frac{3c_{n}}{2}\right] $ for every $n\geq 1$. [Hereby, [x] denotes the integer part of the number x.] Prove that

[b]a)[/b] This sequence $c_n$ contains infinitely many even numbers and infintely many odd numbers.

[b]b)[/b] The sequence $e_{n}=\left( -1\right) ^{c_{n}}$ is non-periodical.

[b]c)[/b] [Calculus fundamentals required.] There exists a real number G such that $c_{n}=\left[ \left( \frac{3}{2}\right) ^{n}G\right] +1$ for every natural n.

  Darij

Only a) and c) up until now:

a) Assume first that $c_n$ is even, $\forall n\ge n_0$. Then for all $n\ge n_0$ we have $c_{n+1}=\frac {3c_n}2$, and we will eventually run out of $2$s in the expansion of $c_n$, which gives a contradiction. Now assume $c_n$ is odd, $\forall n\ge n_0$. Then for $n\ge n_0$ we have $c_{n+1}=\frac{3c_n-1}2$, which is also odd, so $c_n$ must be $M_4+1$ for all $n\ge n_0$. We continue like this and we get $2^k|c_{n_0}-1,\ \forall k\ge 1$, which is absurd.

c) It's equivalent to showing that the intervals $I_n=[(\frac 23)^n(c_n-1),(\frac 23)^nc_n)=[a_n,b_n)$ have a non-void intersection. It is, however, pretty easy to show that $a_n$ is increasing and $b_n$ is decreasing (from the definition of $c_n$), and that $b_n-a_n\to 0$, so we can take $G=\lim_{n\to\infty}a_n$.

b) I'll use the following lemma: If $c_{m+1},c_{n+1}$ have the same residue $\pmod{2^x},\ x\ge 1$ and $c_m,c_n$ have the same parity, then $c_m,c_n$ have the same residue $\pmod{2^{x+1}}$. I won't prove it: just take separate cases when $c_m,c_n$ are odd/even etc.

Assume there is a $T>0$ s.t. $c_{n},c_{n+T}$ have the same parity $\forall n\ge 1$. Assume, furthermore, that we have shown $c_1\equiv c_{1+T}\pmod{2^k},\ c_2\equiv c_{2+T}\pmod{2^{k-1}},\ldots,c_{k-1}\equiv c_{k-1+T}\pmod {2^2}$. We know that $c_k\equiv c_{k+T}\pmod 2,c_{k+1}\equiv c_{k+1+T}\pmod 2$.

We apply the lemma for $m=k,n=k+T,x=1$ to find $c_k\equiv c_{k+T}\pmod {2^2}$. We apply the lemma again for $m=k-1,n=k-1+T,x=2$ to find $c_{k-1}\equiv c_{k-1+T}\pmod{2^3}$, and so on until we find $c_1\equiv c_{1+T}\pmod{2^{k+1}}$. We continue in this manner for $k\to k+1$ indefinitely. We have just obtained $2^m|c_{1+T}-c_1,\ \forall m\ge 1$, which is absurd.

This induction seems really nasty, so I wonder if it's correct. What do you people think?


[hide]For n^2 where n is even, we can see that n^2 = (n-4)^2 + (n/2 + 3)^2 - (n/2 - 5)^2
Thus, for n=100, we have 96^2 + 53^2 - 45^2. Adding these up we get 194.[/hide]

Edit: Correct answer now.

for c, don't you mean (n/2 - 5)? that would make your answer two less.

Here's another way to think of the problem:
The sum of b and c is always 2 less than the first number, so in that case b+c=98

a is always 4 less than the first number so a=96

96+98=194 [u]ans.[/u]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Hi,

there are many nontrivial groups $G$ isomorphic to their direct product $G \times G$, e.g. the free abelian groups of infinite rank, $(\mathbb{R},+), (\mathbb{Q}^*,*)$, ...
This leads to the following question:

Is there any finitely-generated nontrivial group $G$ s.t. $G \cong G \times G$?

I already know that such a group must be infinite, non-abelian, perfect, and that every proper normal subgroup must have an infinite index. But this doesn't really help :-)

And what's about the generalization: When $G$ is minimal spanned by $n>1$ elements, then of course $G \times G$ is spanned by $2n$ elements. But is this also the lower bound?

Just for the last one:
$G= S_n$ is generates by $(12), (123...n)$.
I think that $G \times G$ is generated by $( (12) , (12) ) , ((123...n) , (123...n))$ and $((1n),(1))$ then.

Thank you! :-)

But I have not made any progress with the first question ...

Hi

Can you explain your last assertion: "every proper normal subgroup must have infinite index"?

Vipul

$G \cong G \times G$ implies $G/N \cong G/N \times G/N$ for every normal subgroup $N$. When $N$ has finite index, $G/N$ is finite, s.t. $G/N = 1$, i.e. $G = N$.

Hi

I can't understand the statement that $G/N$ must be isomorphic to $G/N \times G/N$. This will hold only if the isomorphism takes $N$ to $N \times N$, and that isn't likely to hold.

Vipul


I have been looking over some calculus to try to stay warm (2 years since AP) and came across this problem that I don't quite understand.  It is probably pretty basic since its from a calculus review and not a physics one, though.

A tank full of water has a vertical side which is 5 feet long by 3 feet high. (The tank is rectangular in shape and its pipes are parallel to one another.) Find the fluid force on this side of the tank.

The pressure at depth $ x$ is $ \rho g x$, assuming that the tank is closed from above and so the atmospheric pressure doesn't influence anything. The force on the area $ \text dS = a \text dx$ is $ \text dF = a\rho g x\text dx$, and integrating from $ x=0$ to $ x=h$ gives $ F = a h^{2}\rho g/2$.


A right square pyramid has a base with edges each measuring 3 cm and a height twice the perimeter of its base. What is the volume of the pyramid?

The area of a pyramid is 1/3bh.
The area of the base is 3^2=9.
The height equals 2*perimeter of base=2*(3*4)=2*12=24.
1/3*24*9=8*9=72.


Paul Dreyer just sent out a message with several pieces of information. One is that the teams won't be staying in Tonapah Hall - instead, we'll be in cluster of dorms out near the Tropicana Ave. side of campus - Dayton, Boyd, Rodman, Williams, with check-in at Dayton.

And he included a program.

[b]Friday, May 29th, 2009[/b]
3:00-3:30 P.M. New Coaches Meeting - CBC A106.
3:30-4:00 P.M. Coaches Meeting - CBC A106
4:00-4:30 P.M. Proctors Meeting - CBC A106
6:00-7:30 P.M. Team and Power Rounds - Assigned Rooms in CBC
8:00-9:30 P.M. Talk by Cynthia Vinzant, HCSSiM - Student Union Ballroom. (Refreshments provided by MathZoom)
11:00 P.M. Students must be in Residence Halls
12:00 Midnight Curfew - All Participants in Assigned Rooms

[b]Saturday, May 30th, 2009[/b]
7:00-8:30 A.M. Breakfast - Wilson Dining Commons
9:00-11:00 A.M. Individual and Relay Rounds - Student Union Ballroom
11:30 A.M. Tie Breaker - Student Union Ballroom
12:00-1:30 P.M. Lunch - Wilson Dining Commons
1:30-2:00 P.M. Super Relay and Awards Assembly - Student Union Ballroom

Hey, Paul - are we really going to get the tiebreakers in before lunch? Color me skeptical on that point.

[quote="Kent Merryfield"]
1:30 P.M. Tie Breaker - Student Union Ballroom
12:00-1:30 P.M. Lunch - Wilson Dining Commons

Hey, Paul - are we really going to get the tiebreakers in before lunch? Color me skeptical on that point.[/quote]

According to this, I think the tiebreakers are after lunch.

We definitely will if that schedule is right.  1:30am tiebreakers will be a cinch.  Just wake up the students who we think will ace the contest on Saturday.

</snark>

But seriously, I acknowledge the timing will be tight, but we cannot wait until after lunch for the tiebreakers.  Being a little late for lunch is okay.  Keeping the other three sites waiting for UNLV tiebreaker results will probably get me killed.

All right, all right. So I lost a digit typing it. I've edited.

SFBA stayed in Dayton in 2006, and I remember it being pretty nice.  It'll be a longer walk to everything, though.  We also won't all be in the same building, which is too bad.

Yeah, and it's a longer walk to the In-n-Out. Here, broadly, are where the teams are:

Macau: Rodman, 1st floor
Colorado: Rodman, 2nd floor (with a couple of coaches on 3rd floor)
Northern California: Rodman, 3rd floor
Desert Southwest: Boyd, 1st floor (with one coach on 2nd)
Roseburg HS: Boyd, 2nd floor
Oregon: Boyd, 2nd and 3rd floors
SFBA: Boyd, 3rd and 4th floors
San Diego: Williams, 1st and 2nd floors
Washington: Williams, 2nd and 3rd floors
Southern California: Williams, 3rd, 4th, and 5th floors
Utah: Dayton, 3rd floor north
Philippines: Dayton, 3rd floor north

At Iowa, we get clip boards!  you can snap them!  they make me abnoixious

What are you talking about? On the individual and relay rounds, we get clipboards too. Always have, even when we were in San Jose.

What's different about the West is that we do team and power on Friday evening - in a classroom building, one team to a room. Isn't that how you do those, albeit on Saturday morning?

[quote="dnkywin"]Argh I find it unfair that the UNLV people actually get tables; UGA has everyone in an auditorium using those little armrests as tables  :mad:[/quote]

i think we should protest, given that we are on the same team  :P

Yes, that's what we do at UGA. Rooms for team and power. Auditorium with the little desks for individual and relay. I assume that it will be mostly the same this year, but I've heard something about not being in the auditorium until awards. I don't know where the individual and relays will be.

Tom

[quote="worthawholebean"]At Penn State we get boards... without clips. That's obnoxious.[/quote]

Where can you even buy those things?

There is something to be said for how annoying the non-stop click-click-clicking of 500 lapboards gets after a few hours  :)

At Iowa we had desks, which worked nicely. (But that was well in the past... dunno how it is today.)

We definitely had clipboards as early as 2002 in Iowa.  I was at one time considering composing some music for several clipboard players, to be performed immediately prior to the individual round.  Never did, though.

UGA this year I've heard will have lapboards. Just like the good ole' days, when the Southeast went to Penn State. :)

Just to be clear: generating and Sly Si may be reminiscing about Iowa in years past, but both of them will be in Las Vegas this year. I see their names listed on the housing reservation list for the site, both listed as coaches (for San Diego and SFBA, respectively).

[quote="Paul Dreyer"]Hello, all!

Two quick things.

1.  Due to a scheduling snafu on UNLV's part, we'll have to delay lunch from 12:00 to 12:30.  With the added individual round, we might have been running a little long anyways.  Depending on how the day goes, we might do the super relay pre-lunch as well.  My goal is to have everything finished and folks on the road by the original schedule (or close to it).  This schedule change will be reflected in the programs you receive upon check-in.

2.  Cynthia Vinzant (from Berkeley and HCSSiM) sent me the title and abstract of her talk on Friday night.  It should be a fun one.

Hercules, Hydras, and Really Big Numbers

We'll witness an epic battle between Hercules and the Hydra, whose monstrous heads grow at an alarming rate. The longer Hercules takes, seemingly the worse his plight. A clever Hercules might be able to survive, but can a foolish one? He might just break arithmetic trying.

Paul
[/quote]

[quote="National Weather Service"]
Friday: Mostly sunny and hot, with a high near 97. Calm wind becoming northeast between 5 and 8 mph.

Friday Night: Partly cloudy, with a low around 76.

Saturday: Mostly sunny and hot, with a high near 99.
[/quote]


function
geometry
topology
algebra proposed
algebra

Well ,I came up with this problem in the last week ,but couldnt find a solution for it (This is probably as usuall ,Well_known )

[color=blue]Problem[/color]

Prove that there exist a bijective injective function from any closed shape in $R^2$ to any closed shape in $R^2$.

Well for defining ,CLOSED:

I mean a shape closed shape (By its usual meaning  :blush: sorry couldnt find a better explanation) with its internal area ,and it should be countinous too.

there shouldnt be any empty area inside the shape ,All of the internal area must be including the shape

I take it the shape has to be bounded. If so, pick a point X inside the shape and map it to the origin; map the boundary of the shape to the unit circle; and map each line segment with endpoints at X and on the boundary to a radius of the unit circle. Thus we can biject any such shape with the unit circle, and the conclusion follows.

That "shape" need not be convex, so your procedure of mapping rays to rays can fail. If I am not mistaken, there is a theorem due to Schoenflies stating that any homeomorphism from a closed simple curve in the plane to the unit circle can be extended to a homeomorphism of the entire plane onto itself, mapping the interior of the curve onto the interior of the circle (and the same for exteriors), but it's very difficult to prove, from what I hear.

It's in that category of results to which the Jordan Curve Theorem belongs: simple to state but which have only nasty, long known proofs.


induction
geometry
geometric transformation
homothety
combinatorics unsolved
combinatorics

There is a set of $ n$ coins with distinct integer weights $ w_1, w_2, \ldots , w_n$. It is known that if any coin with weight $ w_k$, where $ 1 \leq k \leq n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $ n$ for which such a set of coins exists.

[quote="orl"]There is a set of $ n$ coins with distinct integer weights $ w_1, w_2, \ldots , w_n$. It is known that if any coin with weight $ w_k$, $ 1 \leq k \leq n$, is removed from the set, the remaining coins can be split into two groups of the same weight. (The number of coins in the two groups can be different.) Find all $ n$ for which such a set of coins exists.[/quote]
All n is $ n\equiv 1 (\mod 2)$ and $ n > 5$
The first we will prove that n must be an odd number .
Exist a set such that $ \sum_{i = 1}^n w_i$ which is smallest.
From condition we have :
$ S_n\equiv w_i,\forall i = 1,..,n$ where $ S_n$ is sum of all $ w_i$
Therefore all element of set if parity .
If all of them is even then the set $ \{\frac {w_i}{2}\}$ also satisfy condition ,this is contradiction from the chose $ S_n$ is minimum .
It gives that all elements of set must be odd number .
Now suppose n is even then when we remove from the set a $ w_i$ then exactly one of two set after split has an odd numbers of element ,it gives that sum of all element is an odd number , but the other set has sum of all element is a even . Therefore thay can't equal .
So n must be an odd number .
The case $ n = 3,5$ it is easy to prove there is no set satisfy .
We will prove for all odd number great than 5 by induction .
Let :
$ f = 1*3*....*(2n - 1)$ ( n is odd number)
where * can take $ + , -$
The case n=7 is obvious by some carculation .
Suppose it true for $ 2n + 1$  we will prove for $ 2n + 3$
Easy to check that when we make a transform : chose  two odd consecutive and change them sign then we have $ |f' - f| = 2$
If we move a term from the set ${ \{1,...,2n - 1}$ then use the result $ |f' - f| = 2$ we can chose sign for $ 2n + 1,2n + 3$ for which $ f = 0$ (not contain the object which had been moved ) .
Take all terms which have sign + in a set ,the others in a set . Two sets have equal sum.
If we move $ 2n + 3$ or $ 2n + 1$ ,in two case it is quite easy to find the ways by consider two case $ n\equiv 1 (\mod 4)$ and $ n\equiv 3(\mod 4)$
So all all n odd and $ n > 5$ satisfy condition .

i think there no N to satisfy;
(**)it is clear that all Wi shoud have same parity!
consider W1<W2...<Wn,then if this set satisfy the condition then set of Wi-k or k.Wi do well,hence set of Wi-W1 can do that.
let Wi-W1=2^Mi.Xi  ,(Xi is odd).
let MIN(Mi)=p ,then set of Ri=(Wi-W1)/2^p can satisfy the condition,but we know that one of the Ri is even(R1=(W1-W1)/2^p=0) and  other Rj are odd BUT stand on (**)they shoud have same parity!unless Wi-W1=0,but it is appose to our data!!
is it true? :huh:

[quote="honav"]i think there no N to satisfy;
(**)it is clear that all Wi shoud have same parity!
consider W1<W2...<Wn,then if this set satisfy the condition then[color=#FF0000] set of Wi-k[/color] [color=#0000FF]or k.Wi [/color]do well,hence set of Wi-W1 can do that.
let Wi-W1=2^Mi.Xi  ,(Xi is odd).
let MIN(Mi)=p ,then set of Ri=(Wi-W1)/2^p can satisfy the condition,but we know that one of the Ri is even(R1=(W1-W1)/2^p=0) and  other Rj are odd BUT stand on (**)they shoud have same parity!unless Wi-W1=0,but it is appose to our data!!
is it true? :huh:[/quote]
Try to first fully read the conditions!
[quote="orl"][color=#004000]The number of coins in the two groups can be different[/color].[/quote]
Then carefully read the proposed solutions (as you are new to this forum, the number of stars of the poster is most often a good indication on the good quality of his work - shurely TTsphn is one of the most authoritative users, so chances he is wrong - and that noone pointed yet that till now - are very slim).

What you wrote is only true partially. Indeed, the set $kW_i$ (homothety) works, indifferently of the splitting into groups; however, the set $W_i - k$ (translation) works [b]only[/b] if the splitting is always made into groups of same cardinality. Indeed,[b] with this extra constraint[/b], the result is all weights must be equal (and it may even be extended to the case when the weights are real numbers, not necessarily integer!).

[quote="TTsphn"]
Let :
$ f \equal{} 1*3*....*(2n \minus{} 1)$ ( n is odd number)
where * can take $ \plus{} , \minus{}$
Easy to check that when we make a transform : chose  two odd consecutive and change them sign then we have $ |f' \minus{} f| \equal{} 2$
[/quote]

I think chose  two odd consecutive and change them sign then we have $ |f' \minus{} f| \equal{} 4$.
Since $(2n+3)+(2n+7)=(2n+1)+(2n+5)+4$, suppose it true for odd number $n$, we can prove for $n+2$.

We will prove more general result: For any odd $n \geqslant 7$, there exit a partition $\left\{ {1,3,...2n - 1} \right\} = A \cup B$, $\sum\limits_{x \in A} x  = \sum\limits_{x \in B} x $, and $\left\{ {1,3,...2n - 1} \right\} = C \cup D$, $\sum\limits_{x \in C} x  =2+ \sum\limits_{x \in D} x $.

A)For $n = 7$: if cut $1$, then $13 + 11 = 3 + 5 + 7 + 9$; and $\left( {5 + 9 + 11} \right) - \left( {3 + 7 + 13} \right) = 2$.
if cut $3$, then $13 + 9 + 1 = 5 + 7 + 11$; and $\left( {11 + 13} \right) - \left( {1 + 5 + 7 + 9} \right) = 2$;
if cut $5$, then $13 + 9 = 1 + 3 + 7 + 11$; and $\left( {3 + 7 + 13} \right) - \left( {1 + 9 + 11} \right) = 2$;
if cut $7$, then $13 + 3 + 5 = 1 + 9 + 11$; and $\left( {9 + 13} \right) - \left( {1 + 3 + 5 + 11} \right) = 2$;
if cut $9$, then $13 + 7 = 1 + 3 + 5 + 11$; and $\left( {13 + 7 + 1} \right) - \left( {3 + 5 + 11} \right) = 2$;
if cut $11$, then $13 + 5 + 1 = 3 + 7 + 9$; and $\left( {1 + 3 + 7 + 9} \right) - \left( {13 + 5} \right) = 2$;
if cut $13$, then $11 + 7 = 1 + 3 + 5 + 9$; and $\left( {11 + 7 + 1} \right) - \left( {3 + 5 + 9} \right) = 2$.

B)If the result is true for a odd $n \geqslant 7$, then for the $n + 2$, we plus $2n + 1,2n + 3$.
If cut a $1\leqslant {w_i} \leqslant 2n - 1$, then there exit a partition $\left\{ {1,3,...2n - 1} \right\} = A \cup B$, $\sum\limits_{x \in A} x  = \sum\limits_{x \in B} x $, and $\left\{ {1,3,...2n - 1} \right\} = C \cup D$, $\sum\limits_{x \in C} x  =2+ \sum\limits_{x \in D} x $.
Let $A' = D \cup \left\{ {2n + 3} \right\}$, $B' = C \cup \left\{ {2n + 1} \right\}$, then $\sum\limits_{x \in A'} x  = \sum\limits_{x \in B'} x $. Let $C' = A \cup \left\{ {2n + 3} \right\}$, $D' = B \cup \left\{ {2n + 1} \right\}$, then $\sum\limits_{x \in C'} x  - \sum\limits_{x \in D'} x  = 2$.
Next we suppose cut $2n + 1$ or $2n + 3$.
C)For $n = 4k - 1$, if cut $2n + 3$, we can make $2k$ pairs, and the sum of each pair is $2n + 2$, take $k$’s pairs to be $A$, and $\left\{ {1,2n + 1} \right\} \subseteq A$, the rest to $B$.Then $\sum\limits_{x \in A} x  = \sum\limits_{x \in B} x $; Let $C = B \cup \left\{ 1 \right\}$, $D = A\backslash \left\{ 1 \right\}$, we have $\sum\limits_{x \in C} x  - \sum\limits_{x \in D} x  = 2$.
If cut $2n + 1$, let $C' = A \cup \left\{ {2n + 3} \right\}\backslash \left\{ {2n + 1} \right\}, D' = B$, then $\sum\limits_{x \in C'} x - \sum\limits_{x \in D'} x  = 2$；$A' = D \cup \left\{ {2n + 3} \right\}\backslash \left\{ {2n + 1} \right\},B' = C$, so $\sum\limits_{x \in A'} x  = \sum\limits_{x \in B'} x $.

D)For $n = 4k + 1$, if cut $2n + 3$, without $n,n + 2$, we can make $2k$ pairs, and the sum of each pair is $2n + 2$, take $k$’s pairs(include $\left\{ {1,2n + 1} \right\}$) and $\left\{ {n + 2} \right\}$ to $C$, rest to $D$, then $\sum\limits_{x \in C} x  - \sum\limits_{x \in D} x  = 2$. Let $C\backslash \left\{ 1 \right\} = A$, and $D \cup \left\{ 1 \right\} = B$, then $\sum\limits_{x \in A} x  = \sum\limits_{x \in B} x $.
If cut $2n + 1$, without $1,3,2n - 5,2n - 3,2n - 1,2n + 3$, we can make $2k - 2$ pairs, and the sum of each pair is $2n - 2$, take $k - 1$’s pairs$ \cup \left\{ {2n + 3,2n - 5,1} \right\}$ to $A$, rest to $B$, then $\sum\limits_{x \in A} x  = \sum\limits_{x \in B} x $.
Let $C = B \cup \left\{ 1 \right\}$, and $D = A\backslash \left\{ 1 \right\}$, then $\sum\limits_{x \in C} x  - \sum\limits_{x \in D} x  = 2$.
So we done.

I haven't looked at any of the solutions here, but I would like to solve this problem. Can someone please clarify the following: (1) can the weights be negative integers or 0? (2) when we split the remaining coins into 2 groups, can one of the groups be empty?

Common sense says that weights are never negative, and most likely not zero either. The solutions in this thread use that weights are positive. (When weights can be negative, I find exactly one additional $n$ that works.)

If the weights are all positive, naturally neither group can be empty (except the trivial case $n=1$). Turns out that with weights can be negative, that still doesn't matter.

The solutions written above are not understandable; they are written in broken english and not clearly explained. Can someone pls put the above solutions into a readable form, or better explain the solution. Thank you.

[b]Lemma 1.[/b] [i]The number of weights $n$ must be an odd number.[/i]

[i]Proof.[/i] Assume that $S_n = \sum_{i=1}^n w_i$ is minimal. If any coin is removed the remaining sets sum must be a multiple of $2$. Thus, \[w_2+w_3+\cdots+w_n \equiv 0 \mod 2\] \[w_1+w_3+\cdots+w_n \equiv 0 \mod 2\] \[\vdots\] \[w_1+w_2+\cdots+w_{n-1} \equiv 0 \mod 2.\] Summing these up we get $(n-1)(w_1+w_2+\cdots+w_n) \equiv 0 \mod 2$. Thus, we have two cases: $n \equiv 0 \mod 2$ and $\sum_{i=1}^n w_i \equiv 0 \mod 2$ and $n \equiv 1 \mod 2$ and $\sum_{i=1}^n w_i \equiv 1,0 \mod 2$. In the first case, $w_i \equiv 0 \mod 2, \forall i = 1,\ldots,n$. This is impossible as the new set $\left \{\dfrac{w_i}{2}\right \}$ also works but is smaller contradicting the minimality of $S_n$. Thus, $n$ is odd.

[b]Lemma 2.[/b] [i]The number of weights $n>5$.[/i]

[i]Proof.[/i] For $n = 3$, no such set exists since we would be left with two numbers $a,b$, which can't be equal since they are distinct.

For $n=5$, we will have $4$ numbers left which must be broken into two groups that are equal. Let the five weights be $w_1 < w_2 < w_3 < w_4 < w_5$. Suppose we remove $w_1$. Then we must have $w_2+w_5 = w_3+w_4$ and/or $w_5 = w_2+w_3+w_4$. The second case is not possible as in this case we would have $w_5 = w_2+w_3+w_4 > w_1+w_3+w_4$, which means the coins cannot be split when we remove $w_2$. Thus we must have $w_2+w_5 = w_3+w_4$. This shows that $w_1+w_5 < w_3+w_4$. But then no splitting is possible when we remove $w_2$ since $w_5 < w_3+w_4$ tells us that the group with $w_5$ must contain exactly $2$ coins. The second coin cannot be $w_1$, and as $w_5 > w_4 > w_1$ the second coin is $> w_1$ we have no other choice.

[b]Lemma 3.[/b] [i]Any odd n > 5 works.[/i]

Proof. Let $f = 1 \ast 3 \ast \cdots \ast (2n-1)$ for some odd number n and the $\ast$ symbol can take either $+,-$ where the weights of the coins are $1,3,\ldots,2n-1$. Now suppose that we can satisfy the condition of the problem for some $2n + 1$ and then we need to show that $2n + 3$ is possible. Then define a transformation on $f$ call it $f'$ where $|f'-f| = 2$ which we form by taking two odd consecutive integers and changing their sign to form $f'$. We then move a term from the set $\{1,\ldots,2n-1\}$ then by using the result $|f'-f|$ we change the sign on $2n+1$ or $2n+3$ using $|f'-f| = 2$ to get $f = 0$. Thus two sets have equal sum if we move $2n+3$ or $2n+1$ which holds for $n \quad 1 \pmod4$ or $n \equiv 3 \pmod{4}$.

The answer is all odd $n\ge 7$. Without loss of generality, assume that $w_1< w_2 <\dots < w_n$ and that $\gcd(w_1,w_2,\dots,w_n)=1.$ Now, note that if $S=w_1+w_2+\dots+w_n$ then $S-w_1,S-w_2,\dots,S-w_n$ are all even. Therefore, $w_k$ have a fixed parity for $1\le k\le n.$ Since they are not all even, they are all odd. This implies that $S$ is odd. Since $S$ is odd, $n$ must be odd.

This is not possible for $n=3$ because removing $w_1$ would imply $w_2=w_3$.

For $n=5$, removing $w_5$, we see that $A_5=(w_2+w_3,w_1+w_4)$ or $B_5=(w_1+w_2+w_3,w_4)$ are the only possible ways to split them into two groups, for size reasons. Similarly, removing $w_4$, we see that $A_4=(w_2+w_3,w_1+w_5)$ or $B_4=(w_1+w_2+w_3,w_5)$ are the only possible splitting ways.

If we use $A_5$ and $A_4$ or $B_5$ and $B_4$ then we get $w_4=w_5$, absurd. If we use $A_4$ and $B_5$ then $2w_1+w_5=w_4$, absurd. Thus, we must use $A_5$ and $B_4$. Now, if when we remove $w_3$ we use $w_1+w_2+w_4=w_5$ then from $B_4$ we know $w_3=w_4$. Therefore, $w_1+w_5=w_2+w_4$ so $2w_1+w_3=w_4.$ However, $w_4-w_3=w_2-w_1$ so $2w_1=w_2-w_1$ so $3w_1=w_2.$ If when we remove $w_2$, we use $w_1+w_5=w_3+w_4$ then this contradicts $w_1+w_5=w_2+w_4.$ Thus, $w_1+w_3+w_4=w_5.$ This implies that $w_2=w_4$ by $B_4$, so that is also impossible. Therefore, $n=5$ is impossible.

For $n\ge 7$ we claim that $1,3,5,\dots,2n-1$ will work. All we need to prove, in fact, is that there is some way to place $+$'s or $-$'s between the numbers $1,3,5,\dots,2n-1$ and before $1$ so that the result can be any number $1,3,5$, to $2n-1$.

For $n=7$, we use
\begin{align*}
1-3-5-7-9+11+13 &= 1\\
1+3-5-7+9-11+13 &= 3\\
1+3+5+7-9+11-13 &= 5\\
1-3-5+7+9+11-13 &= 7\\
1+3+5-7+9+11-13 &= 9\\
1-3+5-7-9+11+13 &= 11\\
1+3+5-7+9-11+13 &= 13
\end{align*}
Now, suppose it is possible with $(1,3,\dots,2n-1)$ Then we'll prove it's possible with $(1,3,\dots,2n+3).$ If we are trying to make some number $1\le k\le 2n-1$, then take any way to make $k$ with $(1,3,\dots,2n-1)$. Then, we subtract $2n+1$ and add $2n+3$ to make $k+2.$ Now, add a negative sign before the $1$ to make $k$. If the sign before $1$ is already negative, then we add $2n+1$, subtract $2n+3$, and then make the sign before $1$ positive.

If we are trying to make $2n+1$, then you take the way to make $2n-1$, add $2n+3$ then subtract $2n-1$ and then leave it alone. Now, to make $2n+3$, if the $1$ is negative, then do what we did for $2n+1$ but change that negative before the $1$ to positive. Otherwise, we can always find two consectuvie numbers $2i-1,2i+1$ such that $2i-1$ is positive and $2i+1$ is negative. Now, if we flip those two signs then we can make $2n+3$ with $(1,3,\dots,2n-1)$. However, using what we did for $1\le k\le 2n-1$ we can also make $2n+3$ using $(1,3,\dots,2n+3)$ as desired.


ratio
trigonometry
geometry
geometric transformation
dilation
inequalities
analytic geometry

Let $ (O)$ is a circle, let $ AB, CD$ are two diameters and $ AB\perp CD$, let $ E$ is a point on $ AD$ arc. Let $ EC$ meet $ OA$ at $ M$, $ EB$ meet $ OD$ at $ N$. Prove that: $ \frac {OM}{AM} \plus{} \frac {ON}{DN}\ge \sqrt{2}$.

set a co ordiate system and radius of circle 1
use small amount ofhomothesy to calculate the ratio
simpify the trignometric inqualitiy
this gives us the obvious trignometric inqualitiy

[quote="thanhnam2902"] [color=darkred]Let $ C(O,r)$ be a circle with diameters $ AB$ , $ CD$ so that $ AB\perp CD$ . For a point  $ E$ which

belongs to the arc $ AD$ denote $ M\in EC\cap OA$ , $ N\in EB\cap OD$ . Prove that $ \frac {OM}{AM} + \frac {ON}{DN}\ge \sqrt {2}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Denote $ \|\begin{array}{c}
\tan\widehat {ECD}=x\\\\
\tan\widehat {EBA}=y\end{array}$ . Observe that $ \{x,y\}\subset [0,1]$ and $ x+y+xy=1$ because $ m(\widehat {ECD})+m(\widehat {EBA})=45^{\circ}$ . Since

$ \|\begin{array}{ccc}
MO=Rx & \implies & \frac {MO}{MA}=\frac {x}{1-x}=\frac {x}{y(1+x)}\\\\
ON=Ry & \implies & \frac {NO}{ND}=\frac {y}{1-y}=\frac {y}{x(1+y)}\end{array}$ obtain $ \frac {OM}{AM} + \frac {ON}{DN}=$ $ \frac {x}{y(1+x)}+\frac {y}{x(1+y)}\ge$ $ \frac {2}{\sqrt {(1+x)(1+y)}}=$ $ \frac {2}{\sqrt {1+xy+1-xy}}=\sqrt 2$ .[/color]

Thank Virgil Nicula very much. Your solution is very nice.  :P  :maybe:

[quote="gauravpatil"]this solution is really teriffic[/quote] Then post yours.

i meant Virgil Nicula's solution was great
[size=200]generalities[/size]
observe that dilations with $ O$ as the centre keeps the ratios in the inequality for a given point $ E$ constant
so WLOG take the radius of the circle equal to $ 1 unit$
i set up a coordinate system $ O$ as the origin $ AB$ along $ x$ axis and $ CD$ as $ y$ axis
again Wlog $ arcAD$ lies in the first quadrant
let $ angleDOE \equal{} \theta$
[size=200]solution[/size]
now we drop a perpendicular from $ E$ to $ AB$ and $ CD$ with feet say $ K$ and $ L$ respectively
observe that $ EK \equal{} OL \equal{} \cos{\theta }$ and also $ EL \equal{} KO \equal{} \sin{\theta }$
so $ \frac {EK}{ON} \equal{} \frac {BK}{BO}$ (kind of homothesy)
but $ BO \equal{} 1 \equal{}$radius of circle and $ BK \equal{} BO \plus{} OK \equal{} 1 \plus{} \sin{\theta}$
and $ EK \equal{} \cos{\theta }$ so $ ON \equal{} \frac {\cos{\theta}}{1 \plus{} \sin{\theta }}$
similarly we get $ OM \equal{} \frac {\sin{\theta}}{1 \plus{} \cos{\theta }}$
this implies $ \frac {ON}{ND} \equal{} \frac {\sin{\theta}}{1 \plus{} \sin{\theta} \minus{} \cos{\theta}}$
and $ \frac {OM}{MA} \equal{} \frac {\cos{\theta}}{1 \plus{} \cos{\theta} \minus{} \sin{\theta}}$
so now we just have to prove that
$ \frac {\sin{\theta}}{1 \plus{} \cos{\theta }} \plus{} \frac {\cos{\theta}}{1 \plus{} \cos{\theta} \minus{} \sin{\theta}} \ge \sqrt {2}$
simplifying this we get that
$ (\sin{\theta} \minus{} \frac {1}{\sqrt {2}})(\cos{\theta} \minus{} \frac {1}{\sqrt {2}}) \le 0$
this is pretty obvious

Let be the point $ K\equiv AD\cap BE$ and it is easy to show that $ MK\parallel CD,$ because of $ \angle MKA \equal{} \angle MEA,$
from the cyclic quadrilateral $ MAEK$ $ ($ $ \angle MAK \equal{} \angle MEK \equal{} 45^{o}$ $ )$  and $ \angle MEA \equal{} \angle CDA \equal{} 45^{o}.$

So, we conclude that $ \frac {OM}{MA} \equal{} \frac {KD}{KA}$ $ ,(1)$

Applying the [b][size=100]Menelaus theorem[/size][/b], in the triangle $ \bigtriangleup OAD$ with transversal $ BNK,$

we conclude that $ \frac {KD}{KA}\cdot \frac {BA}{BO}\cdot \frac {NO}{ND} \equal{} 1$ $ \Longrightarrow$ $ \frac {KD}{KA} \equal{} \frac {1}{2}\cdot \frac {ND}{NO}$ $ ,(2)$

Hence, from $ (1),$ $ (2),$ it is enough to prove that $ \frac {OM}{MA} \plus{} \frac {ON}{ND} \equal{} \frac {ON}{ND} \plus{} \frac {1}{2}\cdot \frac {ND}{ON} \geq \sqrt {2},$

which is true and the proof is completed.

Kostas Vittas.

[quote="vittasko"]Let be the point $ K\equiv AD\cap BE$ and it is easy to show that $ MK\parallel CD,$ because of $ \angle MKA \equal{} \angle MEA,$
from the cyclic quadrilateral $ MAKE$ $ ($ $ \angle MAK \equal{} \angle MEK \equal{} 45^{o}$ $ )$  and $ \angle MEA \equal{} \angle CDA \equal{} 45^{o}.$

So, we conclude that $ \frac {OM}{MA} \equal{} \frac {KD}{KA}$ $ ,(1)$

Applying the [b][size=100]Menelaus theorem[/size][/b], in the triangle $ \bigtriangleup OAD$ with transversal $ BNK,$

we conclude that $ \frac {KD}{KA}\cdot \frac {BA}{BO}\cdot \frac {NO}{ND} \equal{} 1$ $ \Longrightarrow$ $ \frac {KD}{KA} \equal{} \frac {1}{2}\cdot \frac {ND}{NO}$ $ ,(2)$

Hence, from $ (1),$ $ (2),$ it is enough to prove that $ \frac {OM}{MA} \plus{} \frac {ON}{ND} \equal{} \frac {ON}{ND} \plus{} \frac {1}{2}\cdot \frac {ND}{ON} \geq \sqrt {2},$

which is true and the proof is completed.

Kostas Vittas.[/quote]
magnificent just great!!

How often on this forum a simple problem is mishandled
   by being solved by some method that blows it out of all
   reasonable proportions!

   Fortunately, it did not happen to this lucky problem, yet.

   This one is another candidate for PWW, a nice illustration
   of Van-Aubel's theorem  in a right triangle.

   OM/AM + ON/DN = OX/XY  where X is an intersection of AN and DM

   on OY.  Locus of X is clearly a reflection of arc AD in chord AD,

   which is also an arc of circle centered at the 4th vertex W of

   the square  AODW  radius WA. It makes it obvious when the MIN

   of the given sum of ratios takes place and at what value.

   Thank you.
   M.T.

[quote="armpist"] [color=darkred]This is a nice illustration of [b]Van-Aubel's theorem[/b]  in a right triangle.[/color] [/quote]
[color=darkblue][b]Very cool the [u]Armpist[/u]'s proof ! It is the best. Thank you, [u]Armpist[/u] ![/b][/color]

[quote="thanhnam2902"] [color=darkred]Let $ C(O,r)$ be a circle with diameters $ AB$ , $ CD$ so that $ AB\perp CD$ . For a mobile point  $ E$ which

belongs to the arc $ AD$ denote $ M\in EC\cap (OA)$ , $ N\in EB\cap (OD)$ . Prove that $ \frac {OM}{AM} + \frac {ON}{DN}\ge \sqrt {2}$ .[/color] [/quote]

[color=darkblue][b][u]Armpist's proof.[/u][/b] Denote $ \|\begin{array}{c}
X\in AN\cap DM\\\
Y\in AD\cap OX\end{array}$ . Observe that $ m(\widehat {AXD})=135^{\circ}$ (costant) . The geometrical locus $ \Lambda$ of the point $ X$

is symmetrically with the arc $ AED$ w.r.t. the line $ AD$ and the incenter $ I$ of $ \triangle AOD$ belongs to $ \Lambda$ . Apply the [b]Aubel's relation[/b]

in the $ O$ - right triangle $ AOD\ \ : \ \frac {XO}{XY}=\frac {MO}{MA}+\frac {NO}{ND}$ . Therefore,  the ratio $ \frac {XO}{XY}$ is minimum iff $ X: =I$ and in this case

$ \frac {IO}{IY}=\frac {OA+OD}{AD}=\sqrt 2$ . In conclusion, $ \frac {OM}{AM} + \frac {ON}{DN}\ge \sqrt {2}$ for any point $ E$ which belongs to the arc $ AD$ .[/color]


I cannot find the forum for the class I am in.  What should I do?


[hide]
The discriminant must be greater than zero.

$b^{2}-4ac>0\Rightarrow 4-4k>0\Rightarrow k<1$.

$\boxed{k<1,k\neq 0}$.
[/hide]

[quote="Interval"]Find k such that the equation kx^2 + 2x + 1 = 0 has TWO distinct real solutions.

What exactly is the question asking?[/quote]
To have 2 real solutions, the discriminant, $b^{2}-4ac$, in the quadratic formula, $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, needs to be positive.
[hide]The discriminant is $4-4k$ so $4-4k>0$
$1-k>0$
$k<1$ but $k\ne0$ because then the quadratic formula would be $\frac{-b\pm\sqrt{b^{2}-0}}{0}$ and it would be undefined.
$\therefore k<1, k\ne0$[/hide]

So basically, we plug and chug using the discriminant, right?

The question is not worded in simple terms.

How is a student to know that the use of the discriminant is needed here?

Thanks

The discriminant is what determines if the solutions are real, and how many there are.

Let the solutions be $r=\frac{a\pm\sqrt{b}}{c}$.  If $b<0$, then $\sqrt{b}\not\in\mathbb{R}$.  If $b=0$, then the roots are the same, because $\frac{a-0}{c}=\frac{a+0}{c}$.  If $b>0$, then nothing is strange.


[quote="TUYEN_LOVER"]prove that:
with a,b,c>1
$ \sqrt {log_ba} \plus{} \sqrt {log_ac} \plus{} \sqrt {log_cb} \ge 3$
thank for Help ^^[/quote]

$ \sqrt {log_ba} \plus{} \sqrt {log_ac} \plus{} \sqrt {log_cb} \geq3\sqrt[3]{\sqrt {log_ba}*\sqrt {log_ac}*\sqrt {log_cb}}\equal{}3\sqrt[6]{log_ba*log_ac*log_cb}\equal{}3\sqrt[6]{\frac{log_na}{log_nb}*\frac{log_nc}{log_na}*\frac{log_nb}{log_nc}}\equal{}3$

$ a\equal{}b\equal{}c$


1) If $ \alpha,\beta,\gamma$ are the roots of the equation $ x^{3}\plus{}qx\plus{}r\equal{}0$ then evaluate $ \sum \frac{\alpha}{\beta\plus{}\gamma}$.

2) If $ x\equal{}2\plus{}2^{\frac{2}{3}}\plus{}2^{\frac{1}{3}}$ then evaluate $ x^{3}\minus{}6x^{2}\plus{}6x$

[hide="Solution #1"]

Let $ S_1 \equal{} \alpha \plus{} \beta \plus{} \gamma$, $ S_2 \equal{} \alpha \gamma \plus{} \alpha \beta \plus{} \beta \gamma$ and $ S_3 \equal{} \alpha \beta \gamma$.

And by Viete's relations, $ S_1\equal{}0$, $ S_2\equal{}q$, $ S_3\equal{}\minus{}r$.

Note that $ \sum\frac {\alpha}{\beta \plus{} \gamma} \equal{} \dfrac{S_1\cdot(\alpha^2 \plus{} \beta^2 \plus{} \gamma^2) \plus{} 3S_3}{S_1\cdot S_2 \minus{} S_3}$

Also we should see that since $ S_1$ is negative the coefficient of $ x^2$ in the polynomial, then $ S_1 \equal{} 0\implies \sum\frac {\alpha}{\beta \plus{} \gamma} \equal{} \dfrac{3S_3}{ \minus{} S_3}$

Which, if I substituted everything correctly, should make,
\[ \sum\frac {\alpha}{\beta \plus{} \gamma} \equal{} \boxed{ \minus{} 3}\]
[/hide]


conics
ellipse
geometry
geometric transformation
reflection
angle bisector

Consider an ellipse with major and minor axes of length $2a$ and $2b$ respectively. chose a point $Q$ on the outside of the ellipse, in the direction of the major axes. Draw one of the tangent lines to the ellipse. At the tangent point, draw the perpendicular to the tangent line. The point $P$ is the intersection with the major axis of length 2a.

What is $|OP|\cdot|OQ|$?

($O$ is the center of the ellipse)

[hide]Let $F_{1},F_{2}$ be the focuses of the ellipse ($F_{2}$ closer to $Q$)
and $A$ the tangent point

It is well known that the line $AP$ reflects the line segment $F_{1}A$ into $F_{2}A$.
So $AP$ is the internal angle bisector of $\angle F_{1}AF_{2}$

$AQ\perp AP \Rightarrow AQ$ is the external angle bisector of $\angle F_{1}AF_{2}$

So the division $(F_{1},P,F_{2},Q)$ is harmonic $\Rightarrow OP \cdot OQ ={OF_{1}}^{2}$[/hide]


hey guys

i am trying to learn cauchy and apply it to inequalities, however i need help.

how would you apply cauchy to this problem:

If $a, b, c, d$ are positive reals, prove that:

$\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}\geq \frac{64}{a+b+c+d}$

Cauchy can be applied in this way: $(a+b+c+d)\left( \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d} \right)\geq (1+1+2+4)^2$ because $\sqrt{4}=2$ and $\sqrt{16}=4$.

Dividing by $a+b+c+d$ arrives at the result.

In order to use Cauchy, you need a product of two sums, which are usually in terms of variables but can be formed from constants (such as writing $n$ as a sum of $n$ $1^2$'s).

Products of sums or sums of products should ring the Cauchy bell.  Most of the time, you need to find a tricky substitution or some transformation to the inequality to Cauchy it.

[quote="sen"]how would you prove the trivial, $a^2+b^2+c^2\geq ab+bc+ac$ using cauchy?[/quote]

You could be like $(a^2+b^2+c^2)(b^2+c^2+a^2) \ge (ab+bc+ca)^2 \Rightarrow a^2+b^2+c^2 \ge ab+bc+ca$.

oh yepp i see. thanks man

If a, b, c, d are positive reals such that $(a^2+b^2)^3=c^2+d^2$

prove that $\frac{a^3}{c} + \frac{b^3}{d} \geq 1$

Here's a nice extension of Cauchy (known as Holder):

$\left(\sum a_i^p\right)^{\frac{1}{p}}\left(\sum b_i^q\right)^{\frac{1}{q}} \ge \sum a_ib_i$.

Then we have

$(c^2+d^2)^{\frac{1}{3}}\left(\frac{a^3}{c}+\frac{b^3}{d}\right)^{\frac{2}{3}} \ge a^2+b^2$

Cube both sides and rearrange to get

$\left(\frac{a^3}{c}+\frac{b^3}{d}\right)^2 \ge \frac{(a^2+b^2)^3}{c^2+d^2} = 1$

so

$\frac{a^3}{c}+\frac{b^3}{d} \ge 1$ as desired.