Some preliminary statistics on the 2008 AIME I and 2008 AIME II:

                         AIME I   AIME II
Number taken   3830      2045
Median                    5         5
Mean                  4.93     4.95
Std Dev             2.39     2.77

There are still a few (200 or less) AIME II answer forms to come in yet and be scored, I do not expect these forms to affect these statistics appreciably.

Steve Dunbar
MAA Director, American Mathematics Competitions

I don't think the mean AIME score has been more than like, 3.1 since 1999.  Wow.

[quote="davidyko"](!) The means for both were almost 5? Doesn't that make this year's AIMEs significantly easier than past years' (2006 - 2.741, 2005 - 2.717, etc.)?[/quote]

yea but the floor was still 6 because Im willing to bet the AMC 12s this year were the hardest ever.  Thus, the main first selection index was still 204.0, and thus, this made the floor 6, and thats why there were 468!! kids <=10 grade that got this score, and thats why the index is so harsh for <=10th graders (202.5!!)

The indices were really similar, so 10th grade and below had almost no advantage:
the only way to qualify by the secondary index was to get an index of 202.5 - 203.5, and that is hard.  Only one of the following scores would give an index in that range:
6 / 142.5 = 202.5
9 / 112.5 = 202.5
8 / 123.0 = 203.0
7 / 133.5 = 203.5
and even then, 156 of them got one of those scores anyway.

[quote="annoyingpi"]The indices were really similar, so 10th grade and below had almost no advantage:[/quote]
Younger students taking the AMC 12 (thank you) had practically no advantage.  Students who took the AMC 10 had a significant advantage.

Of course, I did get the second one of those listed...  :P
Though I blame it on being sick during the AMC 12.

[quote="annoyingpi"]and even then, 156 of them got one of those scores anyway.[/quote]

There's also people with an index of above 204, but with an AMC 10 score.

Some unusual stats this year, IMO. For one, aren't the 6,000ish takers of the AIME significantly lower than, say, two years ago (18,000+...)? The 5 average is apparently quite high, and one might expect the index to skyrocket to accomodate this :maybe: . Especially if you compare this year to 2005, when the index was something like 233 (right?), yet the average was about 2.7.... :|

Understand that 2005 was an outlier year for having an easy AMC and many AIME qualifier. Having a minimal number of AIME qualifiers because of the hard AMC means that the people taking the AIME were, on average, stronger students. That might account for about 20% of the increase in the AIME mean. The rest of it was that this AIME was easier.

[quote="Zmastr"]If this AIME was so much easier, how did the floor only be 1 above the mean?  There must have been a lot of people with 5's.[/quote]

No thats because the AMC 12 was probalby the hardest AMC 12s ever (and a lot harder than pervious ones).

Thus that brought the AMC 12 index down, which brought the floor score down, which brought the secondary index up (more people with the lowre floor, so they have to be harsher)

[quote="the future"]No thats because the AMC 12 was probalby the hardest AMC 12s ever[/quote]
Ah - how short is memory. Go look up 2001.

[quote="annoyingpi"]6 / 142.5 = 202.5[/quote]142.5 is an impossible score.[quote="Boy Soprano II"]Younger students taking the AMC 10 had practically no advantage.  Students who took the AMC 10 had a significant advantage.[/quote]The first 10 should be a 12, I think.

Dude AIME i was easier than aime II. The reason why people scored so low on the former was because of stupid mistakes. Doing a problem by problem comparison would also prove my above statement.

[quote="the future"]It looked like the AIME II was [i]easier[/i] than the AIME I.  So much for all the complainign that the II was harder.[/quote]

The standard deviation was also significantly (significant not as in meaning large but as in important) larger meaning there were more high scores and low scores. High scores could simply be caused by the schools that took the AIME II or w/e.

Look: last year, more than 5% of AMC 12'ers qualified. Some of those might not be up to par with the higher scorers. That accounts for lower AIME scores overall.

Next: maybe all of the good people took AIME I?

Even if you add 2000 AIME takers to this year and say they got on average, a score of 0.5, you still get a mean about 1.1 higher than last year with a comparable amount of test-takers.

Well notice that the cutoff for the AIME in past years has just been 100.  Now that harder scoring methods are being put in place, cutoffs are going below 100, and we have far fewer (and thus probably more experienced) people taking the AIME.  Hence, the higher scores.

Just a theory: do you think that banning calculators on the AMC resulted in an AIME pool that was better at problem solving, rather than calculator brute-forcing?

AIME participants are more competitive this year (exactly 5% of AMC participants qualify, but before AMC scoring-change, 10% of AMC participants qualify), and I thought the first 4-5 problems on AIME I are really straight forward(a lot easier than the last few problems on this year's AMC12's). I think the AIME II average is high because many competitive schools had to offer AIME II (TJ, Exeter)

[quote="Carbon57"]Well, generally, that could be possible...but on this year's AMC, a calculator wouldn't have helped very much. The only thing it would really be useful is to prevent calculation errors...[/quote]That was on purpose.


[quote="4everwise"]In how many ways can a row of k seats be filled from a set of n people.[/quote]

[hide] eassy  P( n, k )=   n! /  (n-k)! [/hide]


[hide]

100^(1/4) = 10^(1/2) which is approximately 3.1

So the next number is most likely lower than 4 - if the next number is 2, then it's 10, which is way off.

Let's try 3.

100^(1/3)

5^3 = 125 and 3^3 = 27, and 100 is somewhere between 27 and 125, so 3 works.

3 and 4, so n =2.

[/hide]

[hide]If you raise each part of the inequality to the nth power, you get

3^n <100<5^n

if n=2 you get 9<100<25 which is untrue

if n=3 you get 27<100<125 which is true

if n=4 you get 81<100<625 which is true

if n=5 you get 243 <100<3125 which is untrue, so 3 and 4 are the only possibilities[/hide]

[quote=MCrawford]How many positive integers $n$ satisfy

\[ 3 < \sqrt[n]{100} < 5? \][/quote]

Solution with [b]stroller[/b], [b]mathscienceclass[/b], [b]mathleticguyyy[/b], [b]brianzjk[/b], [b]kevinmathz[/b], [b]awang11[/b], and [b]RadiantCheddar[/b].

Take $\log_{100}$ of both sides to get that
\[\log_{100}3 < \frac{1}{n} < \log_{100}5 .\]
We now estimate $\log_{100}3$ and $\log_{100}5$.

Note that $3^2<10$, so $\log_{100}3 < \frac{1}{4}$. We note the delicate bound of $10^2<3^5$, so $\log_{100}3 > \frac{1}{5}$. Thus, we have
\[\frac{1}{n} \ge \frac{1}{4}\implies n\le 4.\]

Note that $2^3<10$, so $\log_{10}2<\frac{1}{3}$. Note that $2^{10}>10^3$, so $\log_{10}2>\frac{3}{10}$. As
\[\log_{100}5 = \frac{1-\log_{10}2}{2},\]
we get the bound
\[\frac{1}{3}<\log_{100}5 < \frac{7}{20} < \frac{1}{2}.\]
Equivalently, we have $n\ge 3$, as $n$ is an integer.

Now, we have shown that $n\in \{3,4\}$. Since $4 < \sqrt[3]{100} < 5$ and $3<\sqrt[4]{100}<4$, both $n=3$ and $n=4$ work, so the solutions are $n=3,4$.

[quote=BrightMonkey]Uh.. why did we bump a post that was [b]19 years old[/b]. The rules clearly say that we aren't allowed to bump old threads[/quote]

You can bump a thread if you have anything to contribute.

[quote=mlgjeffdoge21]how to actually solve it ^[/quote]

That IS an actual solution. You may see [url]https://en.wikipedia.org/wiki/Taylor_series[/url] and the references there for the background theory on approximations of $\log$.


How many sets of ordered triples (a,b,c) exist such that a, b, and c are positive integers less than or equal to 10, and a+b+c=abc?

They can't all be greater than 1, so one must be 1 (WLOG a), so we have 1+b+c=bc, or (b-1)(c-1)=2. So c=2, b=3 and permutations of (1,2,3) are all the possibilities (6 in total).

Alternately, WLOG, we have $ a \leq b \leq c$. Thus, $ abc \equal{} a \plus{} b \plus{} c \leq 3c$, so $ ab \leq 3$. From here, we set a very low upper bound for $ a$ and $ b$, and we can find the desired solutions.

The restriction that they are all less than or equal to 10 can be removed.


i'm new here. :)

and... i have a problem that needs solving.

What is 1^3+2^3+3^3+4^3+5^3+6^3......+20^3  ???

and sorry, no calculators allowed.  T_T

Thank you in advance!

I can prove it for you.

[hide="Hint"]Use the identity $ (k \plus{} 1)^4 \minus{} k^4 \equal{} 4k^3 \plus{} 6k^2 \plus{} 4k \plus{} 1.$[/hide]

[hide="Proof"]Using the identity $ (k \plus{} 1)^4 \minus{} k^4 \equal{} 4k^3 \plus{} 6k^2 \plus{} 4k \plus{} 1,$

When $ k \equal{} 1, 2^4 \minus{} 1^4 \equal{} 4 \cdot 1^3 \plus{} 6 \cdot 1^2 \plus{} 4 \cdot 1 \plus{} 1$
When $ k \equal{} 2, 3^4 \minus{} 2^4 \equal{} 4 \cdot 2^3 \plus{} 6 \cdot 2^2 \plus{} 4 \cdot 2 \plus{} 1$
$ \vdots$
When $ k \equal{} n, (n \plus{} 1)^4 \minus{} n^4 \equal{} 4n^3 \plus{} 6n^2 \plus{} 4n \plus{} 1$.

Add both sides of these equations,

$ LHS \equal{} (2^4 \minus{} 1 ^4) \plus{} (3^4 \minus{} 2^4) \plus{} \dots [(n \plus{} 1)^4 \minus{} n^4].$

$ \text{A bunch of things cancel, leaving us with: }$
$ LHS \equal{} (n \plus{} 1)^4 \minus{} 1^4$

Now for the RHS.

$ RHS \equal{} 4 \sum_{k \equal{} 1} ^n k^3 \plus{} 6 \sum_{k \equal{} 1}^n k^2 \plus{} 4 \sum_{k \equal{} 1}^n k \plus{} n$
$ \equal{} 4 \sum_{k \equal{} 1} ^n k^3 \plus{} 6 \cdot \frac {n(n \plus{} 1)(2n \plus{} 1)}{6} \plus{} 4 \cdot \frac {n(n \plus{} 1)}{2} \plus{} n$
$ \text{Therefore,} 4 \sum_{k \equal{} 1} ^n k^3 \equal{} (n \plus{} 1)^4 \minus{} 1 \minus{} n(n \plus{} 1)(2n \plus{} 1) \minus{} 2n(n \plus{} 1) \minus{} n$
$ \equal{} (n \plus{} 1)[(n \plus{} 1)^3 \minus{} n(2n \plus{} 1) \minus{} 2n \minus{} 1$
$ \text{After a bit of expanding, adding, and subtracting, we get: }$
$ \equal{} (n \plus{} 1)(n^3 \plus{} n^2)$
$ n^2(n \plus{} 1)^2$
Now we know that $ 4 \sum_{k \equal{} 1} ^n k^3 \equal{} n^2(n \plus{} 1)^2$, but we need to divide by $ 4$ to get $ \sum_{k \equal{} 1} ^n k^3$.

Therefore, $ \boxed{\sum_{k \equal{} 1} ^n k^3 \equal{} \frac {n^2(n \plus{} 1)^2}{4} \implies [\frac {n(n \plus{} 1)}{2}]^2}$[/hide]

Hopefully that was conclusive enough - it was a pain to type that all in LaTeX.   :D

You could also prove it through induction.

Btw, FullMetalNaruto, welcome to the forum!  :D

Anyways, here's the induction method, in case anyone's wondering.
[hide]
So, we wish to prove that $ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} \dots \plus{} n^3 \equal{} \left[ \dfrac{n(n\plus{}1)}{2} \right]^2.$

We proceed by induction. The base case, $ n\equal{}1,$ clearly holds. We now assume that for some positive integer $ k$ we have \[ 1^3 \plus{} 2^3 \plus{} \dots \plus{} k^3 \equal{} \left[ \dfrac{k(k\plus{}1)}{2} \right]^2.\] For $ n\equal{}k\plus{}1,$ we have
\[ 1^3 \plus{} 2^3 \plus{} \dots \plus{} k^3 \plus{} (k\plus{}1)^3 \equal{} \left[ \dfrac{k(k\plus{}1)}{2} \right]^2 \plus{} (k\plus{}1)^3.\]
Simplifying that we get the desired result. [/hide]

[quote="Mewto55555"][quote="TachyonPulse"]Although induction can prove it, it is more enlightening to see how it's derived. This is because the same method [b]AdithyaGanesh[/b] can be applied for any sum of powers.[/quote]

The way I derive this stuff is calculate the first n+2 values and then some good ol' Lagrange interpolation.  :)[/quote]

Elaborate please?  Sounds interesting.  :)

It's fairly obvious that $ 1^m\plus{}2^m\plus{}...\plus{}n^m$ is a polynomial in terms of $ n$ with degree $ m\plus{}1$ (e.g. when m=3 this is the 4th degree polynomial that is the discussion of this thread).

It's also well-known that to calculate a polynomial of degree $ m\plus{}1$ you need $ m\plus{}2$ values.

Here, we have

$ f(0)\equal{}0$

$ f(1)\equal{}1$

$ f(2)\equal{}1\plus{}2^m$

etc.

and then just use [url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]Lagrange[/url] or [url=http://www.math-linux.com/spip.php?article72]Newtonian[/url] Interpolation. If $ m$ is small (as it usually is for contests) the whole procedure usually takes less time than Adithya's method.

[quote="randomguy64"]Could you explain why 1^3+...=(1+2+...+n)^2?[/quote]

Since $ (1^{3}\plus{}2^{3}\plus{}3^{3}\plus{}...\plus{}n^{3})\equal{}\left (\frac{ n(n\plus{}1)}{4} \right )^{2}\equal{}(1\plus{}2\plus{}3\plus{}...\plus{}n)^{2}$, it should be possible to rearrange one of the series to form the other. Can some show how this would be done?


A square is changed into a pentagon by building an equilateral triangle on top of it, as shown. By
what percent does the perimeter increase going
from the square to the pentagon?

Say the square has side length $ s$.  Then so does the triangle since it is equilateral and fits on top.  So the perimeter goes from $ 4s$ to $ 5s$.  So it increases by $ \25%$.

perimeter goes 4 to 5.

so increased by 1/4, which is 25%.

answer : 25

Yes, it is definitely 25.  I seem to have left out the 2 when I typed it in before. :oops:

[quote="isabella2296"]That's funny. You know how when the cursor hovers above $ \text{\LaTeX}$, a blue box-thingy says the code, minus the dollar signs? Like if you put your cursor on top of this: $ 2x - 3 = 5$, after a few seconds it will say in a small blue box-thingy: 2x - 3 = 5.

When my cursor was on top of the $ 5$ that JRav wrote, it didn't come up as "5", but as "\25%", but it displays 5....[/quote]

thats because unless its an actual operation (fraction, square root, log), a \ in front of something nullifies the anteceding digit or word, so \25=5 in latex.

the blue box is just to view the latex code used to make it in case you wanna duplicate it for a response


Let $ x,y,z>0$,prove that:
$ \left( {\frac {x}{y}}\plus{}{\frac {y}{z}}\plus{}{\frac {z}{x}}\minus{}3 \right)
 \left( {\frac {x\plus{}y}{y\plus{}z}}\plus{}{\frac {y\plus{}z}{z\plus{}x}}\plus{}{\frac {z\plus{}x}{x\plus{}y}}\minus{}3
 \right) \leq  \left( {\frac {x}{y}}\plus{}{\frac {y}{z}} \right)  \left( {
\frac {y}{z}}\plus{}{\frac {z}{x}} \right)  \left( {\frac {z}{x}}\plus{}{\frac {x}
{y}} \right) \minus{}8.$
BQ


1. With the three digits of the integer 142, there are six two-digit positive integers that can be formed. When the sum of those two-digit integers is divided by the sum of the three digits of the original three-digit integer, the result is 22. When the same thing is done with the [i]k[/i] digits of a [i]k[/i]-digit integer, and each of the [i]k[/i] digits is different and non-zero, the result of dividing the sum of all of the possible two-digit integers by the sum of the [i]k[/i] digits of the original [i]k[/i]-digit integer is 66. What is the value of [i]k[/i]?

Let the digits be $ d_1,d_2,\dots,d_k$. Consider just the sum of the digits in the 1's place: The $ k$ digits will each repeat $ k \minus{} 1$ times, leading to a sum of $ (k \minus{} 1)\displaystyle\sum_{i \equal{} 1}^{k}{d_i}$. However, each digit will occur the same number of times in the 10's place as they do in the 1's place. This means that the sum overall is
$ 10(k \minus{} 1)\displaystyle\sum_{i \equal{} 1}^{k}{d_i} \plus{} \displaystyle\sum_{i \equal{} 1}^{k}{d_i} \equal{} 11(k \minus{} 1)\displaystyle\sum_{i \equal{} 1}^{k}{d_i}$.
Now, dividing by the sum of the digits, we have
$ \displaystyle\frac {11(k \minus{} 1)\displaystyle\sum_{i \equal{} 1}^{k}{d_i}}{\displaystyle\sum_{i \equal{} 1}^{k}{d_i}} \equal{} 11(k \minus{} 1) \equal{} 66$.

Therefore $ \boxed{k \equal{} 7}$.


[quote="Padmavathi"]I just love this problem...i spent hours...and when I finished, the answer seemed so obvious!

1
11
21
1211
111221
What is next in line?[/quote]
312211 :D
13112221
1113213211
31131211131221
13211311123113112211

[quote="bpms"][quote="i_like_pie"][quote="mgao"]here is an extension:

Can you know for sure whether the number of terms in the rows keep on either a) staying the same of b) getting larger?  If not, why is there a smaller row in some instance in the infinite number of rows than the preceding row?[/quote]
[hide]If you somehow get 111111111222222222333333333, or something like that, the next one would be 919293.[/hide][/quote]BUt you can't get [hide]111111111222222222333333333
because that is read as 1 one 1 one 1 one ... In which case it would be read differently. From this I conclude that the largest number you can get is 3... [/hide][/quote]
Look at mgao's solution.

[hide]After 111221 comes 312211 (three 1's, two 2's, one 1.)[/hide]

[quote="i_like_pie"][/quote][quote="bpms"][quote="i_like_pie"][quote="mgao"]here is an extension:

Can you know for sure whether the number of terms in the rows keep on either a) staying the same of b) getting larger?  If not, why is there a smaller row in some instance in the infinite number of rows than the preceding row?[/quote]
[hide]If you somehow get 111111111222222222333333333, or something like that, the next one would be 919293.[/hide][/quote]BUt you can't get [hide]111111111222222222333333333
because that is read as 1 one 1 one 1 one ... In which case it would be read differently. From this I conclude that the largest number you can get is 3... [/hide][/quote][quote="i_like_pie"]
Look at mgao's solution.

[hide]After 111221 comes 312211 (three 1's, two 2's, one 1.)[/hide][/quote]
What he means is that you can't have [b]4[/b] or more consecutive occurrences of a number. Let's look at
'3333'
as an example. This would mean that the sequence before it would have been
'333333'
because 3333 means "3 3's, 3 3's." However, that sequence would actually have been read as
'63,'
not 3333.

[quote="nebula42"][/quote][quote="i_like_pie"][/quote][quote="bpms"][quote="i_like_pie"][quote="mgao"]here is an extension:

Can you know for sure whether the number of terms in the rows keep on either a) staying the same of b) getting larger?  If not, why is there a smaller row in some instance in the infinite number of rows than the preceding row?[/quote]
[hide]If you somehow get 111111111222222222333333333, or something like that, the next one would be 919293.[/hide][/quote]BUt you can't get [hide]111111111222222222333333333
because that is read as 1 one 1 one 1 one ... In which case it would be read differently. From this I conclude that the largest number you can get is 3... [/hide][/quote][quote="i_like_pie"]
Look at mgao's solution.

[hide]After 111221 comes 312211 (three 1's, two 2's, one 1.)[/hide][/quote][quote="nebula42"]
What he means is that you can't have [b]4[/b] or more consecutive occurrences of a number. Let's look at
'3333'
as an example. This would mean that the sequence before it would have been
'333333'
because 3333 means "3 3's, 3 3's." However, that sequence would actually have been read as
'63,'
not 3333.[/quote]
Oh, I see. Then I'm not sure.

[quote="mgao"]I have seen this before...

[hide="My Solution"]

The arrangements go like this:

1
11 => because there is one one in the first row.
21 => because there are two ones in the second row.
1211 => because there is one two, and one one in the third row.
111221 => because there is one one, one two, then two ones in the preceding row...

So 312211 would be next.

Here is the whole thing:

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211

[/hide]

here is an extension:

Can you know for sure whether the number of terms in the rows keep on either a) staying the same of b) getting larger?  If not, why is there a smaller row in some instance in the infinite number of rows than the preceding row?[/quote]

Not quite right...it HAS to be in order which is the reason why it DOES keep either a) staying the same of b) getting larger
What i mean by In Order is...
1
11(One 1)
21(two 1s)
1211(One 2 and one 1)
111221(One 1, one two and two ones)
312211(Three 1s two 2s and two 1s)
13112221(One 3 One 1 Two 2s and 2 1s)
Ur not supposed to group all the ones and twos from all over and say  3 if there is a total of three 1s.
:) :D
I am not sure if the numbers decrease.....do they? CAN they?
This problem was given to engineers at Google and all of them solved it!!


Which of the fractions $\frac{13}{40}$, $\frac{17}{52}$, $\frac{21}{64}$, and $\frac{25}{76}$ is smallest?

surprised by the fact that no one showed their works

so instead of saying the answer

I'll show my works

[hide="works"]if you can't see right away the smallest fraction, try finding the least common multiple of each denominators[/hide]

A quicker method if you don't want to do a lot of calculating:
[hide]If a<b<c<d for any numbers a, b, c, and d, then 3a<3b<3c<3d, so we just need to find which of 39/40, 51/52, 63/64, and 75/76 is the smallest. Each of these fractions can be written as 1-(1/d) where d is the denominator of the fraction. Clearly the smaller 1/d is, the larger the fraction is, so the larger d is, the larger the fraction is. Thus we want the fraction with the smallest denominator, or 13/40.[/hide]
Note that this method doesn't work in general though, but chances are if you're asked to order horribly messy sets of fractions, there'll be some shortcut.

[hide]divide the numerator by the denominator, so u get decimals and see which one is the lowest. :) [/hide]

[hide="Don't read this--it's a hard way..."]I did 100/(denominator) then timesed that by the numerator.  It's one of the hard ways....[/hide]


geometry
geometric transformation
reflection

Sam is designing a triangular piece for a metal sculpture. He tells
Martha that two of the sides of the piece are 40 inches and 15 inches,and the angle opposite the 40-inch side measures 120°. Martha decidesto sketch the piece that Sam described. How many different triangles can she sketch that match Sam’s description?

(1) 1     (3) 3
(2) 2     (4) 0

[quote="peeta"]i dont think this is intermediate (probably not even getting started, because its regular school math...)
there is only one way, because two trangles with two sides equal and the angle opposite to the largest side equal are congruent, hence only one way to sketch/ construct...[/quote]
I think $2$
THINK why does $\text{ASS}$ congruency not work????

do u mean we got two sketches because of the reflection?

like u could sketch ABC as well as BAC ...

otherwise i dont get what u mean

In this case, there is only one solution. Draw a 120-degree angle, one of its sides being of length 15. Centered at the other endpoint of the 15 side, draw a radius 40 circle. This circle intersects the other side of the 120-degree angle once, so there is only one triangle.

There are situations in which two triangles are possible, just not this one.

yes I understand what you mean so there is only $1$ possible solution.


topology
real analysis
real analysis unsolved

Call a sequence $ x_n$ "sequentially dense" in $ X$ if the set $ \{x_n|n\geq N\}$ is dense in $ X$ for any $ N$. (Is there a real name for this?) One example of this would be $ r_1,r_2,...$, the usual enumeration of the rationals in $ [0,1]$.

Is it true that if $ x_n$ is sequentially dense in $ X$, then so is any rearrangement $ x_{p(n)}$ (where $ p: \mathbb{N} \to \mathbb{N}$ is any bijective rearrangement of the indexing integers)?

Yes.  Proof idea: Since $ p$ is bijective it cannot map infinite sets into finite ones, so we proceed by contradiction.

Proof: Since $ \lbrace x_n\mid n\geq N \rbrace$ is dense in $ X$, there are infinitely many elements of $ \lbrace x_n\rbrace_{n\in \mathbb{N}}$ in each member of the topology $ \mathcal{T}$ on $ X$ (for, otherwise, there is an open set $ U\in \mathcal{T}$ and an $ N = 1 + \max \lbrace k\mid x_k\in U\rbrace$ so that $ x_k\notin U$ for all $ k\geq N$).  Given any bijective rearrangement $ x_{p(n)}$, suppose $ x_{p(n)}$ is not dense in $ X$; then we can find an open set $ U\in \mathcal{T}$ so that for some finite $ M$ we have $ \lbrace x_{p(n)}\mid p(n)\geq M\rbrace\cap U = \emptyset$.  Since there are infinitely many $ k$ for which $ x_k\in U$, in particular there must be infinitely many $ x_{p(k)}\in U$ such that $ p(k) > M$ (for, otherwise, $ p$ must map the infinite set $ \lbrace k\mid x_k\in U\rbrace$ into the finite set $ [1,M]\cap \mathbb{N}$, and hence could not be bijective), so $ \lbrace x_{p(n)}\mid p(n)\geq M\rbrace\cap U \neq \emptyset$ and we have a contradiction.  Thus, $ \lbrace x_{p(n)}$ is sequentially dense in $ X$.


function
modular arithmetic
algebra unsolved
algebra

[quote="jin thynj"]Let Z be the integers. f : Z → Z is defined by f(n) = n - 10 for n > 100 and f(n) = f(f(n+11)) for n ≤ 100. Find the set of possible values of f.[/quote]

For $ 90 \le n \le 100$, $ f(n) \equal{} f(f(n \plus{} 11)) \equal{} f(n \plus{} 1)$

So $ f(90) \equal{} f(91) \equal{} ... \equal{} f(100) \equal{} f(101) \equal{} 91$

Now for $ n < 90$, let $ n \equal{} i \pmod{11}$ and $ 90 \le i \le 100$

we have  $ f(n) \equal{} f(f(...f(f(f(i)))...)) \equal{} f(f(...f(f(91))...))$
$ \equal{} f(f(...f(91)...)) \equal{} ... \equal{} f(91) \equal{} 91$

And clearly we have $ f(n \plus{} 10) \equal{} n$ for all $ n \ge 91$,
So the set of possible value of f $ \equal{} \{ x \in \mathbb{Z} | x \ge 91 \}$


algorithm
modular arithmetic
number theory proposed
number theory

It is easy to see that (r+1)/(r+5)=q/p. So r+1=q and r+5 = p. So r=2 , q=3 and p=7.

Actually $ r \plus{} 1 \equal{} q$ and $ r \plus{} 5 \equal{} p$ isn't necessary. Also, $ p \equal{} 3,q \equal{} 2,r \equal{} 7$ is another solution which doesn't satisfy your condition. :wink:

[hide="Hint"]Write it as $ \frac {r \plus{} 1}{q} \equal{} \frac {r \plus{} 5}{p} \equal{} k$, where $ k\in \mathbb N$. (why?)[/hide]

As suggested by Taidoda-X, note that $ \dfrac{p}{q}\equal{}\dfrac{r\plus{}5}{r\plus{}1}$. Euclid's algorithm immediately yields that $ \lambda(r)\equal{}(r\plus{}5,r\plus{}1)\equal{}(4,r\plus{}1)$. Hence, $ \lambda(r)\in\{1;2;4\}$. Note also that $ p>q$. The rest is little more than casework:
[list]
[*] $ \lambda(r)\equal{}1$: This readily implies $ r\plus{}1\equal{}q$, whence the solution $ (p,q,r)\equal{}(7,3,2)$
[*] $ \lambda(r)\equal{}2$: We get $ 2p\equal{}r\plus{}5$ and $ 2q\equal{}r\plus{}1$. Thus, additionning and subtracting the two equalities, $ p\plus{}q\equal{}r\plus{}3$ and $ p\minus{}q\equal{}2$. Suppose $ p,q,r\geq5$. Hence, $ p\plus{}q\equiv 0,2\mbox{ mod }6$ and $ r\plus{}3\equiv 2,4\mbox{ mod }6$, implying that $ p\equiv q\equiv 1\mbox{ mod }6$, contradicting the fact that $ p$ and $ q$ are twin primes. Checking the remaining cases, we find the solution $ (p,q,r)\equal{}(5,3,5)$.
[*] $ \lambda(r)\equal{}4$: Subtracting the equalities $ 4p\equal{}r\plus{}5$ and $ 4q\equal{}r\plus{}1$, we find that $ p\minus{}q\equal{}1$, which readility implies the solution $ (p,q,r)\equal{}(3,2,7)$.[/list]

I'am not sure is this correct....

We easily see that r=(5q-p)/(p-q)=-5+(4p)/(p-q)

As r is integer, we see that 4p=k(p-q), k is positive integer.
4p=2*2*p
p-q<p
We see that p-q=1 or p-q=2 or p-q=4
If p-q=1, then p=3, q=2, r=7

If p-q=2, then p=5,q=3, r=5 or p=6k+1, q=6k-1.
It follows that r=12k-3. Then r is not prime, contradiction.

If p-q=4, then p=7,q=3,r=2 or p=6k+5, q=6k+1.
Then r=6k, r is not prime, again contradiction.

Solutions: (3,2,7), (5,3,5), (7,3,2).

My solution.....

[b]p/q-4/(r+1)=1
[p(r+1) - 4q] / q(r+1) = 1
p(r+1) - 4q = q(r+1)
pr+p-4q = qr+q
pr - qr = 4q + q - p
r(p-q) = 4q + q - p
r = [4q + (q - p)] / [p-q]
r = [4q - (p - q)] / p - q
r = 4q / (p-q) - 1
p-q|4q
p-q = 1 ; p-q = 2 ; p-q = 4[/b]
For the first and third case the solution [b] [7,3,2] [/b] ; but for second case [b] p,q,r [/b] isn't simple number.....

naye,will you post your full solution?I am curious.

Can you explain your solution more clearly?

$\# .$If $p-q=,$it has the only solution $p=3,q=2$ because then they are of different pairity,so the smaller one is even i.e. $q=2$.Then $r=7$
$\#2.$If $p-q=2$ then it is well known that every prime greater than $3$ is of the form $6n\pm 1$
Using this,$p\equiv 1\mod 6,q\equiv -1\mod 6$ but then $r+1=2q\implies r=2(6q-1)-1=3(2q-1)$ which is not a prime.So then the only possible value is $p=5,q=3$
$\#3.p-q=4,r+1=q$ so then $r=2,q=3$ is the only solution

Above is not true because $(r+1)(p-q)=4q$
[b]Solution:[/b]Let $r\equiv{1}(mod3)$ $\implies$ $q=3$$\implies$ $12=(p-q)(r+1)$ , done !
Let $r\equiv{2}(mod3)$ $\implies$ $p=3$ $\implies$ $3(r+1)=q(r+5)$ $\implies$ $q<3$ , done!
Let $r=3$ $\implies$ $p=2q$ , absurd.

[quote="individ"]Offer to find all the solutions of this equation, and not just simple numbers. Need to write a formula describing the solutions of this equation.[/quote]
In each case, 7 years old child can find solutions.

[quote="Ahiles"]Find all prime numbers $ p,q,r$, such that $ \frac{p}{q}\minus{}\frac{4}{r\plus{}1}\equal{}1$[/quote]
[hide="Solution"]We claim the solution set to the given equation is \[(p, q, r) \in \{(7, 3, 2); (5, 3, 5); (3, 2, 7)\}.\]
Firstly, observe that $\frac{p}{q} = 1 + \frac{4}{r + 1} > 1 \implies p > q.$
Now, notice that \[\frac{p}{q} - \frac{4}{r + 1} = 1 \implies pr + p - 4q = qr + q\]\[\implies 4q = (p - q)(r + 1).\]
Since $p > q$, each term in the above expression is positive.
Since $q$ is prime, the only divisors of the $\text{LHS}$ are $1, 2, 4, q, 2q, 4q.$ Therefore, we have six cases:
(1) $p - q = 4q; r + 1 = 1$
(2) $p - q = 2q; r + 1 = 2$
(3) $p - q = q; r + 1 = 4$
(4) $p - q = 4; r + 1 = q$
(5) $p - q = 2; r + 1 = 2q$
(6) $p - q = 1; r + 1 = 4q.$

The first three cases are absurd, as they imply $p = 5q, 3q, 2q$, respectively. Since $p$ must be prime, and $5q, 3q, 2q$ are obviously not prime, it follows that these cases are impossible.

In the fourth case, we have $r + 1 = q.$ This implies that $r$ and $q$ are consecutive primes. Therefore $r = 2, q = 3.$ This implies that $p = 4 + q = 7.$ Indeed this solution works, as \[\frac{p}{q} - \frac{4}{r + 1} = \frac{7}{3} - \frac{4}{2 + 1} = 1.\] This concludes the fourth case.

In the fifth case, we have $r + 1 = 2q \implies r = 2q - 1.$ In addition, we have $p - q = 2 \implies p = q + 2.$ Now, we consider our three primes in $\mod 3.$ Firstly, we consider the case when any of them is equal to $3$:
$p = 3 \implies q = 1$, contradiction, as $1$ is not prime;
$r = 3 \implies q = 2 \implies p = 4$, contradiction, as $4$ is not prime;
$q = 3 \implies p = 5, \implies r = 5.$ This is a valid solution, as \[\frac{p}{q} - \frac{4}{r + 1} = \frac{5}{3} - \frac{4}{5 + 1} = 1.\]

Now, when none of our primes are equal to $3$, we see that
$q \equiv q \pmod 3$
$p \equiv q + 2 \pmod 3$
$r \equiv 2q - 1 \pmod 3.$

Since neither $p$ nor $q$ can be divisible by $3$, the first two congruences imply that $q \equiv 2 \pmod 3.$ However, the third congruence then implies that $r \equiv 2 \cdot 2 - 1 \equiv 0 \pmod 3$, contradiction, as $r$ must be prime. This concludes the fifth case.

In the sixth case, we have $p - q = 1.$ This implies that $q$ and $p$ are consecutive primes. Therefore $q = 2, p = 3.$ This implies that $r = 4q - 1 = 7.$ This solution is valid, since \[\frac{p}{q} - \frac{4}{r + 1} = \frac{3}{2} - \frac{4}{7 + 1} = 1.\] This concludes the sixth case.

Since we have covered all possible cases, we deduce that the set of all solutions is \[(p, q, r) \in \{(7, 3, 2); (5, 3, 5); (3, 2, 7)\}.\][/hide]

[quote="Ahiles"]Find all prime numbers $ p,q,r$, such that $ \frac{p}{q}\minus{}\frac{4}{r\plus{}1}\equal{}1$[/quote]

Another way...

Obvious $p=q$ no solutions so...$(p-q,q)=1$

$\dfrac{p-q}{q}=\dfrac{4}{r+1}$  (1)

So $p-q=1$ or  $p-q=2$ or $p-q=4$

If $p-q=1$ then obvious $(p,q,r)=(3,2,7)$

If $p-q=4$  from (1) $q=r+1$ so obvious $(p,q,r)=(7,3,2)$

If $p-q=2$ from $(1)$ $r+1=2q$

Because $p-q=2$ if $q=3$ we have the solution $(p,q,r)=(5,3,5)$ but if $q>3$ then p>3$ and because $p-q=2$ must $p=1mod6$ , q=5mod6$ so from $r+1=2q$ we have $r=9mod6$ so only if $r=3$ then $q=2$ and $p=4$ no solution...

$(p,q,r)=(3,2,7),(7,3,2),(5,3,5)$

My solution:

Rearrange the condition as $\frac{p}{q}=\frac{r+5}{r+1}$. Observe that $\gcd{(r+5, r+1)}=\gcd{(4, r+1)}=x$.

Case 1: $x=1$. Then we must have $r$ even, so $r=2$. Then $p=7, q=3$ yielding the solution $(7, 3, 2)$.

Case 2: $x=2$. This is only true if $r=4a+1$ for some integer $a \ge 0$. Then we have $p=2a+3, q=2a+1$ because $\gcd{(2a+3, 2a+1)}=1$. Suppose we have a solution in the form $(2a+3, 2a+1, 4a+1)$. Then either $a \equiv 2 \pmod{3}$ or $a=1$. If $a \equiv 2 \pmod{3}$ then $4a+1$ is not prime, so $a=1$. This gives us the solution $(5, 3, 5)$.

Case 3: $x=4$. This is only true if $r=4a+3$ for some integer $a \ge 0$. Hence, $p=2a+2, q=2a+1$. The only primes that differ by $1$ are $2$ and $3$, so $a=1$ and we have the solution $(3, 2, 7)$.

So the solutions are $(p, q, r)=(3, 2, 7), (5, 3, 5), (7, 3, 2)$. We are done.

The given equation can be rearranged into the below form:

$4q = (p-q)(r+1)$

$Case 1: 4|(p-q)$

then we have

$q = ((p-q)/4)(r+1)$
$=> (p-q)/4 = 1$ and $q = r + 1$
$=> r = 2, q = 3$ and $p = 7$

$Case 2: 4|(r+1)$

then we have

$q = (p-q)((r+1)/4)$
$=> (p-q) = 1$ and $q = (r + 1)/4$
$=> p = q + 1 => q = 2, p = 3$ and $r = 7$

note that if $(r+1)/4 = 1$, then $q = (p-q) => p = 2q$ which is a contradiction.

$Case 3: 2|(p-q)$ and $2|(r+1)$

then we have

$q = ((p-q)/2)((r+1)/2)$
$=> (p-q)/2 = 1$ and $q = (r+1)/2$
$=> p = q + 2$ and $r = 2q - 1$
We have that exactly one of $q, q + 1, q + 2$ is a multiple of $3$.

$q + 1$ cannot be a multiple of $3$ since $q + 1 = 3k => q = 3k - 1$. Since $r = 2q - 1$ is prime, then we have $2(3k - 1) - 1 = 3(2k-1)$ is a prime.
$=> k = 1 => q = 2 => p = 4$ contradiction.

Also, $q + 2$ cannot be a multiple of $3$ since, $q + 2 = 3k => p = 3k => k = 1 => q = 1$ contradiction.

So, $q = 3k$
$=> k = 1 => q = 3, p = 5$ and $r = 5$

Thus we have the following solutions: $(7, 3, 2), (3, 2, 7), (5, 3, 5)$

Note that
\[\frac{p}{q} - \frac{4}{r+1} = 1 \iff (p-q)(r+1) =4q.\]
We must have $p>q$ and $p-q\mid 4q$. Because $\gcd (p-q,q)=1$, so we have $p-q\mid 4$.
If $p-q=1$, we have $(p,q,r) = (3,2,7)$.
If $p-q=2\iff p=q+2$, we also have $r=2q-1$. Note that $(p,q,r) = (5,3,5)$ is solution. Consider when $q>3$. Let $q= 6k+1\implies p= 6k+3$ or $q=6k+3\implies r=12k + 9$ for some positive integer $k$, but this contradiction $p,r$ is prime numbers.
If $p-q=4\iff p =q+4$, we alsp have $r = q-1$. Then, $(p,q,r) = (7,3,2)$.

Therefore, the solution : $(p,q,r) = (3,2,7),(5,3,5),(7,3,2)$.

Multiplying by $q(r+1)$ gives $p(r+1)-4q=q(r+1)$, or $pr-4q+p=qr+q$, equivalent to $pr+p=qr+5q$ after simplification. This means $p(r+1)=q(r+5)$, or $\frac{p}{q}=\frac{r+5}{r+1}$. Now we take cases. From Euclidean Algorithm, the maximum value of $\gcd(r+5,r+1)$ is $4$, attained with $r=4k+3$, giving $\frac{p}{q}=\frac{4k+8}{4k+4}=\frac{k+2}{k+1}$. This would mean $p=k+2, q=k+1$, only possible with $k=1 \implies (3,2,7)$. We cannot have $\gcd(r+5, r+1) = 3$, as $r+5 \equiv 0 \pmod 3$ means $r \equiv 1 \pmod 3$ and $r+1 \equiv 2 \pmod 3$. If $\gcd(r+5,r+1)=2$, then $r=4k+1$, so $\frac{p}{q}=\frac{2k+3}{2k+1}$. These two have GCD $1$, so $p=2k+3, q=2k+1, r=4k+1$. Assume $p, q, r \neq 3$. Then the second equation gives $k \equiv 2 \pmod 3$, but this is not possible due to the third equation. If $p=3$, then $k=0$, which does not work. If $q=3$, then $k=1$, which gives $(5,3,5)$. If $r=3$, then $k=\frac{1}{2}$, which does not work. Lastly, we could have $\gcd(r+5, r+1)=1$, which means $p=r+5, q=r+1$. Clearly $r$ then must be even, so $r=2$, which gives $(7,3,2)$. Our solutions are $\boxed{(3,2,7),(5,3,5),(7,3,2)}$.

[quote=daniel73]Solution below
[hide] Obviously, $ q(r \plus{} 5) \equal{} p(r \plus{} 1)$.  If $ p \equal{} q$, then $ r \plus{} 5 \equal{} r \plus{} 1$, absurd, so $ p,q$ are mutually prime since they are distinct primes.  Therefore, $ q$ divides $ r \plus{} 1$ and $ p$ divides $ r \plus{} 5$.
[/hide]

Not enough with this hint?  Then keep on reading...
[hide]
Now, consider $ d \equal{} \gcd(r \plus{} 1,r \plus{} 5)$.  Obviously $ d$ divides $ r \plus{} 5 \minus{} (r \plus{} 1) \equal{} 4$, and $ d \equal{} 1,2,4$, and furthermore $ q\frac {r \plus{} 5}{d} \equal{} p\frac {r \plus{} 1}{d}$.  Now, since $ \frac {r \plus{} 1}{d},\frac {r \plus{} 5}{d}$ are mutually prime, then now yes, $ p \equal{} \frac {r \plus{} 5}{d}$, $ q \equal{} \frac {r \plus{} 1}{d}$. [/hide]

If this is still not enough...
[hide]
Primes $ p$ and $ q$ have a difference equal to $ \frac {4}{d}$.  Obviously, $ d \equal{} 4$ iff $ p,q$ are consecutive integers, ie, iff $ p \equal{} 3$ and $ q \equal{} 2$, yielding $ r \equal{} 7$.  This is indeed one possible solution...

Primes $ p$ and $ q$ are twin primes (primes that differ by 2), and obviously both odd, iff $ d \equal{} 2$, in which case $ r \equal{} 2q \minus{} 1 \equal{} 2p \minus{} 5$.  If $ p,q$ are not 3, then they leave remainders 1 and 2 modulus 3 (their remainder modulus 3 cannot be equal since $ p,q$ differ by 2), and integer $ s$ exists such that $ p \equal{} 6s \plus{} 1$, $ q \equal{} 6s \minus{} 1$, or $ r \equal{} 12s \minus{} 3$, divisible by 3, absurd, so $ q \equal{} 3$, $ p \equal{} 5$, $ r \equal{} 5$.

Finally, primes $ p$ and $ q$ differ by 4, and obviously both are odd, iff $ d \equal{} 1$, in which case $ r \equal{} q \minus{} 1 \equal{} p \minus{} 5$, and $ r$ is even, or $ r \equal{} 2$, with $ q \equal{} 3$ and $ p \equal{} 7$.

[/hide][/quote]

Thanks! That second hint really helped me, then it was just casework. My sols were $(p, q, r) = (3, 5, 5), (3, 2, 7), and (2, 3, 7).$


trigonometry
calculus
linear algebra
matrix
function
integration
derivative

This is an exercise in using limited tools. You may assume the basic theorems of calculus and you may assume the existence-uniqueness theorem for differential equations. But do not assume any familiarity with trigonometric functions. Also, try to do this without using power series. (In other words, no matrix exponentials.)

Consider the differential equation $y''+y=0$ with two different initial conditions:

Define $c(t)$ to be the unique function such that $c''+c=0, c(0)=1,c'(0)=0.$

Define $s(t)$ to be the unique function such that $s''+s=0,s(0)=0,s'(0)=1.$

1. Prove that $c'=-s$ and $s'=c.$

2. Prove that $c(-t)=c(t)$ and $s(-t)=-s(t).$

3. Prove that $c^{2}(t)+s^{2}(t)=1$ for all $t.$

4. Prove that $c(x+y)=c(x)c(y)-s(x)s(y).$

5. Prove that $s(x+y)=s(x)c(y)+c(x)s(y).$

[With 3, 4, and 5 in hand you ought to be able to prove any other trig identity you can name.]

One way is to define $\arcsin x = \int_{0}^{x}\frac{dt}{\sqrt{1-t^{2}}}$ (for $x\in (-1,1)$). Define $s$ suitably based on this, and define $c=s'$. Then just check that $s$ and $c$ solve the differential equations.

Probably not what you were looking for but in the sense of formally defining $\sin$ and $\cos$ it seems like a sensible way to do it if you want to avoid power series.

You can also get away without using the existence-uniqueness theorem for differential equations to solve your problem since you can easily show from the properties you've listed that $s$ must have an inverse on the interval $[-\pi/2,\pi/2]$ with derivative $\frac{1}{\sqrt{1-t^{2}}}$, where you define $\pi$ as $2$ times the first positive zero of $s$. We actually went through all of this stuff in the first course in mathematics at Lund University, but I don't know if it was worth the time.

I think I made a mistake b/c I only got number 2, but I'll post incase I did some of it right and people can build off it.

I started out with an assumtion that y is in the form of $e^{rt}$, which turned out correct but maybe a different solution is the one I need.  so $y''=r^{2}e^{rt}$.  Then I plugged in and divided by $e^{rt}$ and got $r^{2}+1=0$. so $r={i,-i}$ and:
$y=k_{1}e^{it}+k_{2}e^{-it}$
$y''=-k_{1}e^{it}+-k_{2}e^{-it}$
added equals zero, and
$y'=k_{1}ie^{it}-k_{2}ie^{-it}$
I used the initial conditions given for c and s to find k1 and k2 for the equations and got:
$c=\frac{e^{it}}{2}+\frac{e^{-it}}{2}$
$s=\frac{e^{it}}{2}-\frac{e^{-it}}{2}$

number 2:
$c(-t)=\frac{e^{i(-t)}}{2}+\frac{e^{-i(-t)}}{2}=\frac{e^{-it}}{2}+\frac{e^{it}}{2}=c(t)$
$s(-t)=\frac{e^{i(-t)}}{2}-\frac{e^{-i(-t)}}{2}=\frac{e^{-it}}{2}-\frac{e^{it}}{2}=-s(t)$

That's outside the intent of the problem. Assuming the solution is an exponential [b]is[/b] assuming familiarity with the trigonometric functions and/or power series. The idea is not to find formulas for $c$ and $s$ but simply to assume that $c$ and $s$ are uniquely defined by the differential equation and then to build their properties from that differential equation.

[u][b]Proof that c(-t) = c(t) and s(-t) = -s(t):[/b][/u]

let y(t) = c(-t)

=> y'(t) = - c'(-t)      and         y''(t) = c''(-t)

since c(t) satisfies c'' + c = 0 for all t =>    c(-t) also does

and y(0) = c(-0) = c(0) = 1,  y'(0) = - c'(-0) = 0

=>  y(t) must equal c(t) by uniqueness of the solution
=> c(-t) = c(t)

similarly by assuming y(t) = -s(-t)
it can be shown that s(t) = y(t) = -s(-t)

[u][b]Proof of the 4th and 5th parts[/b][/u]

Since c(t) is a solution to the DE for all t => so is c(t+d)

=> X = c(t+d)  and  X(0) = c(d) ,  X'(0) = -s(0+d) = -s(d)   is a unique solution...

Define f = c(t) c(d) - s(t) s(d)

f can be trivially proved to be a solution to the DE as it is a linear combination of two solutions...

f(0) = c(0) c(d) - s(0) s(d) = c(d)

f'(0) = -s(0) c(d) - c(0) s(d) = -s(d)

=> by uniqueness of solution: f = X  FOR ALL d and t

Using a very similar argument... s(t+d) = s(t)c(d) + c(t)s(d)

[u][b]A question that is currently really bothering me...[/b][/u]

Can it be shown using the same tools that s(t+2 pi) = s(t) and c(t+2 pi) = c(t) ????

Anyways...
Thanks Dr. Merryfield for the nice exercise...  :)

[quote="abdo88"]
Can it be shown using the same tools that s(t+2 pi) = s(t) and c(t+2 pi) = c(t) ????
[/quote]
It can be shown that $s$ and $c$ are periodic. Indeed, $s^{2}+c^{2}=1$ is fixed, so once we have 3 equal values for $s$, we have two points where both $c$ and $s$ are the same and we can apply the uniqueness theorem then to show periodicity. Now let us show that $s=0$ for infinitely many points. Indeed, suppose that $s$ preserves sign on $[a,+\infty)$. Then $c$ is a monotone function there, so it must preserve sign as well, maybe on another ray $[b,+\infty)$. Thus, we can choose a ray on which both $s$ and $c$ are monotone and preserve sign. Now, since $s'=c$, the integral $\int^\infty |c|$ must converge. Similarly, $\int^\infty |s|$ must converge. So, $\int^\infty (|s|+|c|)$ must converge but the integrand is never less than 1, so we get a contradiction. Now, we just [b]define[/b] $2\pi$ to be the least period of $s$.

[quote="fedja"]Now, we just [b]define[/b] $2\pi$ to be the least period of $s.$[/quote]
I would have said that if fedja hadn't. In fact. I should have included a part of the original problem that pointed in this direction.


trigonometry
symmetry
geometry
trapezoid
angle bisector
geometry unsolved

As the picture shows, let $ BCDE$and$ ACFG$ be squares,and $ AC$≠$ BC$.

If $ CM=CN$,prove that $ \Delta ABC$ is a right-angled triangle!

It does not require a lot of effort to solve this problem.

Firstly, you might want to simplify the question so that people can approach it easily:

Given a triangle $ \Delta ABC$ ($ AB\ne AC$). To the outside of the triangle, draw two isosceles right triangles $ \Delta ABM$ and $ \Delta ACN$ $ BA=BM, CA=CN, \angle ABM =\angle ACN = 90^{o}$
$ BN, CM$ intersect $ AC, AB$ at $ P, Q$ respectively. Suppose that: $ AP= AQ$.
Prove that: $ \angle BAC = 90^{o}$

It's much easier now.

OK, here is the solution outline:
1. The attitude $ AH$ of the triangle $ ABC$, $ BN$, $ CM$ concurrent at $ I$ and $ \angle PHA=\angle QHA$
2. According to Ceva's theorem,
\[ \frac{QA}{QB}.\frac{HB}{HC}.\frac{PC}{PA}=1\]
\[ \longrightarrow \frac{HB}{HC}.\frac{PC}{QB}=1\]
\[ \longrightarrow \frac{HB}{QB}=\frac{HC}{PC}\]
\[ \longrightarrow \sin \angle BQH : \sin \angle QHB = \sin \angle CPH : \sin \angle PHC\]
\[ \longrightarrow \sin \angle BQH=\sin \angle CPH\]
Therefore, $ \angle BQH= \angle CPH$ or $ \angle BQH+\angle CPH = 180^{o}$
If $ \angle BQH = \angle CPH$, it turns out that $ \Delta ABC$ is isosceles which contradicts the given statement.
If $ \angle BQH+\angle CPH = 180^{o}$, $ \angle BAC = 90^{o}$

Let $ B'$ be, the point on $ AC$ $ ($ [b][size=100]creatorvn’s[/size][/b] wording $ ),$such that $ AB' = AB.$  Also let $ C',$ $ N'$ be, the points on the extensions of $ AB,$ $ AM$ $ ($ towards $ B,$ $ M$ $ )$ respectively, such that $ AC' = AC$ and $ AN' = AN.$

Because of the symmetry of the configuration, with respect to the angle bisector of $ \angle A$ and because of $ AP = AQ,$ it is easy to show that the segment line $ B'N',$ passes through the point $ Q.$

Because of now, $ \frac{AB'}{B'C}= \frac{AB}{BC'}= \frac{AM}{MN'},$ we have that $ MB'\parallel CN'$ and so, from the trapezium $ CN'MB',$ we conclude that

the sideline $ AB$ of $ \bigtriangleup ABC,$ bisects its bases $ CN'$ and $ MB'.$

That is we have $ KM = KB',$ where $ K\equiv MB'\cap AB.$

We denote as $ A'$ the orthogonal projection of $ B'$ on $ AB$ and so, it is easy to show that $ A'B' = BM$

$ ($ from the congruent right triangles $ \bigtriangleup BMK,$ $ \bigtriangleup A'KB'$ $ ).$

So, we have $ A'B' = MB = AB = AB'$ $ ,(1)$

From $ (1)$ $ \Longrightarrow$ $ A'B'\equiv AB'$ $ \Longrightarrow$ $ A'\equiv A$ $ \Longrightarrow$ $ AB'\perp AB$ and the proof is completed.

Kostas Vittas.


geometry
incenter
circumcircle
collinearity
IMO Shortlist

[quote="orl"]Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i = 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i + 1}$ and $ A_iA_{i + 2}$ (the addition of indices being mod 3). Let $ B_i, i = 1, 2, 3,$ be the second point of intersection of $ C_{i + 1}$ and $ C_{i + 2}.$ Prove that the circumcentres of the triangles $ A1B_1I,A_2B_2I,A_3B_3I$ are collinear.[/quote]
[hide="Solution"]
Invert around the incircle and let the incircle meet $ A_1A_2$, $ A_2A_3$, and $ A_1A_3$ at $ F$, $ D$, and $ E$ respectively.  Let the midpoints of $ EF$, $ DF$, and $ ED$ be $ M$, $ N$, and $ P$ respectively.  Since both the incircle and $ C_1$ are tangent to $ A_1A_2$ and $ C_1$ passes through $ I$, we see that $ C_1$ and the incircle are tangent at $ F$.  Thus, the image of $ C_1$ is the line tangent to the incircle at $ F$, which is $ A_1A_2$.  Similarly, $ C_2$ is mapped to $ A_2A_3$ and $ C_3$ is mapped to $ A_1A_3$.  Since $ B_1$ is the intersection between $ C_1$ and $ C_3$, we see that it is mapped to the intersection between $ A_1A_3$ and $ A_1A_2$, which is $ A_1$.  Furthermore, it is well-known that $ A_1$ is mapped to the midpoint of $ EF$, which is $ M$.  Thus, the circumcircle of $ \triangle A_1B_1I$ is mapped to $ A_1M$.  Similarly, the circumcircles of $ A_2B_2I$ and $ A_3B_3I$ are mapped to $ A_2N$ and $ A_3P$ respectively.  Notice that $ A_2N$, $ A_3P$, and $ A_1M$ concur at $ I$ because they are angle bisectors of $ \triangle A_1FE$, $ \triangle A_2FD$, and $ \triangle A_3DE$ since triangles $ A_1FE$, $ A_2FD$, and $ A_3DE$ are isosceles.  Thus, the circumcircles of $ \triangle A_1B_1I$, $ \triangle A_2B_2I$, and $ \triangle A_3B_3I$ all meet at another point, say $ X$.  The circles then have common radical axis $ XI$ and hence all lie on the perpendicular bisector of $ XI$, and thus are collinear.

[b]Remark[/b]: In fact, because the inverted images of the circles intersect at $ I$, and we are inverting around $ I$, we see that the three circles are actually all tangent to each other at $ I$.  The end still works since the three centers lie on the line perpendicular to the common tangent to the three circles at $ I$.  [/hide]

Denote $ L_i (i\equal{}1,2,3)$ the intersections of $ A_iI$ and $ (O)$. We have $ L_i$ is the circumcenter of triangle $ IA_{i\plus{}1}A_{i\plus{}2}$ so $ L_iL_{i\plus{}1}$ is the perpendicular bisector of $ [IA_{i\plus{}2}]$
Since $ L_1O_1, L_2O_2, L_3O_3$ concur at $ I$ so $ L_1L_2L_3$ and $ O_1O_2O_3$ are perspective. By [b]Desargues[/b]'s theorem, we get $ (L_1L_2\cap O_1O_2), (L_2L_3\cap O_2O_3), (L_1L_3\cap O_1O_3)$ are collinear. They are circumcenters of triangles $ IA_1B_1, IA_2B_2, IA_3B_3$
[b]Remark:[/b] The end still works when $ O_i$ lies on $ IA_i$ and we don't need $ (O_i)$ tangent to $ A_{i\plus{}1}A_{i\plus{}2}$

The centers of three circles passing through the same point $ I$ and not touching each other are
collinear if and only if they have another common point. Hence it is enough to show that the
circles $ A_iB_iI$ have a common point other than $ I$. Now apply inversion at center $ I$ and with
an arbitrary power. We shall denote by $ X`$ the image of $ X$ under this inversion. In our case,
the image of the circle $ C_i$ is the line $ B`_{i \plus{} 1}B` _{i \plus{} 2}$
while the image of the line $ A_{i \plus{} 1}A_{i \plus{} 2}$ is the
circle $ IA`_{i \plus{} 1}A`_{i \plus{} 2}$ that is tangent to$ B`_{i}B`_{i \plus{} 2}$, and $ B`_{i}B`_{i \plus{} 2}$. These three circles have equal radii, so their centers$ P_1,P_2,P_3$ form a triangle also homothetic to$ B`_{1}B`_{2}B`_{3}$. Consequently, points
$ A`_1,A`_2,A`_3$
 that are the reflections of $ I$ across the sides of $ P_1P_2P_3$, are vertices of a triangle alsohomothetic to$ B`_1B`_2B`_3$. It follows that$ A`_1B`_1,A`_2B`,A`3B`3$are concurrent at some point $ J`$,
that the circles $ A_iB_iI$ all pass through $ J$.
Probem 10.

Original diagram:

Let the incircle be tangent to $A_1A_2A_3$ at points D, E, F on sides $A_2A_3, A_3A_1, A_1A_2$ respectively. Also let L, M, N be the intersection of lines $A_1I, A_2I, A_3I$ with the incircle respectively. Let points $B_1, B_2, B_3$ be defined as stated in the problem.

The condition of the circumcenters of $(A_1B_1I), (A_2B_2I), (A_3B_3I)$ being collinear combined with the fact that I lies on all three circles implies that the three circles are coaxial. Hence, there exists another point, J, that lies on all three circles.

Inverted diagram:

Let us now invert the diagram about the incircle. It follows that circles $C_1, C_2, C_3$ invert to triangle $B_1^*B_2^*B_3^*$. Sides $A_2A_3, A_3A_1, A_1A_2$ invert to circles of diameter $DI, EI, FI$ respectively. Let the centers of these circles be $O_1, O_2, O_3$ respectively. Since L, M, N lie on the incircle, $L=L^*$, $M=M^*$ and $N=N^*$.

Note that $(A_1B_1I)$ inverts to the line formed by points $A_1^*$ and $B_1^*$. Hence it suffices to show that lines $A_1^*B_1^*$, $A_2^*B_2^*$, and $A_3^*B_3^*$ are concurrent.

This statement is equivalent to proving triangle $A_1^*A_2^*A_3^*$ is similar to triangle $B_1^*B_2^*B_3^*$. Doing so would imply the existence of a homothecy mapping $A_1^* \rightarrow B_1^*$ and likewise for the other 4 points. This would imply collinearity.

To prove this, we note that $O_1O_2 \parallel B_1^*B_2^*$ since IM is perpendicular to both lines. (The first perpendicularity follows from radical axis and the second from the fact that IM is the perpendicular bisector of the segment formed by the inverted tangency points on $B_1^*B_2^*$.) Hence, $O_1O_2O_3$ is similar to $B_1^*B_2^*B_3^*$.

However, the medial triangle of $O_1O_2O_3$ is similar to $A_1^*A_2^*A_3^*$ by a scale factor of $\frac{1}{2}$. Hence,  $O_1O_2O_3$ is congruent to $A_1^*A_2^*A_3^*$. From this, it follows that $B_1^*B_2^*B_3^*$ is similar to $A_1^*A_2^*A_3^*$, as desired.

Let $D_1,D_2,D_3$ be the contact points of the incircle. Under incircle inversion (we'll be lazy with the stars), we see that $B_1^*B_2^*B_3^*$ is a triangle externally tangent to the three circles $(ID_1),(ID_2),(ID_3)$, which all have equal radius since $ID_1=ID_2=ID_3$. Since the radii are all equal, we see that $B_iB_{i+1}\parallel D_iD_{i+1}$, so $B_1B_2B_3$ is homothetic to $D_1D_2D_3$, say $B_i=(1-k)J+kD_i$ for some fixed point $J$ and number $k$. We also see that $A_i^*$ is the midpoint of $D_{i+1}D_{i+2}$, so it's not hard to check that
\[\frac{2k}{2k+1}A_i+\frac{1}{2k+1}B_i=\frac{k}{2k+1}(D_i+D_{i+1}+D_{i+2})+\frac{1-k}{2k+1}J,\]
which is constant. Thus, all the $A_iB_i$ concur, so pre-inversion, we have that the circles $(A_iB_iI)$ are coaxial, as desired.

[quote=IMO Shortlist 1997, Q9]Let $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i \equal{} 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i\plus{}1}$ and $ A_iA_{i\plus{}2}$ (the addition of indices being mod 3). Let $ B_i, i \equal{} 1, 2, 3,$ be the second point of intersection of $ C_{i\plus{}1}$ and $ C_{i\plus{}2}.$ Prove that the circumcentres of the triangles $ A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear.[/quote]

Let $DEF$ be the contact triangle of $\triangle ABC$, let the midpoints of $DE,EF,FD$ be $A_3^*,A_1^*,A_2^*$ respectively. Let the circumcenters of $\triangle A_1IB_1,A_2IB_2,A_3IB_3$ be $O_1,O_2,O_3$ respectively. Now we will perform an inversion($\Psi$) around the incircle $$\begin{cases}\Psi:A_1\mapsto A_1^* \\ \Psi:A_2\mapsto A_2^* \\ \Psi:A_3\mapsto A_3^* \\  \Psi:B_1\mapsto \text{Radical Axis} (C_2,\odot(I))\cap \text{Radical Axis} (C_3,\odot(I))=B_1^* \\ \Psi:B_2\mapsto \text{Radical Axis}  (C_1,\odot(I)) \cap \text{Radical Axis} (C_3,\odot(I))=B_2^* \\ \Psi:B_3\mapsto \text{Radical Axis} (C_2,\odot(I))\cap \text{Radical Axis} (C_1,\odot(I))=B_3^*\end{cases}$$

Since, $\triangle B_1^*B_2^*B_3^*$ is homothetic to $\triangle DEF$ which in turn is homothetic to its medial triangle of $\triangle A_1^*A_2^*A_3^*$. So, $\triangle B_1^*B_2^*B_3^*$ is homothetic to $\triangle A_1^*A_2^*A_3^*$. This implies that $A_1^*B_1^*,B_1^*C_1^*,C_1^*A_1^*$ are concurrent. Now the images of their circumcenters will be reflection of $I$ over $A_1^*B_1^*,A_2^*B_2^*$ and $A_3^*B_3^*$ respectively. So, by a Homothety with a scale factor of $2$ we get that $I,O_1^*O_2^*B_2^*$ are concyclic. Hence, inverting back we get that $O_1,O_2,O_3$ are collinear.

Notice that $C_1=(IB_2B_3)$. Clearly the center of $(IB_2B_3)$ lies on $A_1I$. By Fact 5, the center of $(IA_2A_3)$ lies on $A_1I$, so $(IB_2B_3)$ is tangent to $(IA_2A_3)$. We now invert around $I$ with arbitrary radius.

We know that $A_2^*A_3^*\parallel B_2^*B_3^*$, and similarly for the other pairs of sides, and thus $\triangle A_1^*A_2^*A_3^*$ is homothetic to $\triangle B_1^*B_2^*B_3^*$. Therefore, $A_i^*B_i^*$ for $i=1,2,3$ concur, which means $(A_iB_iI)$ are coaxial, and we are done.

The center of $(IA_2A_3)$ is the midpoint of $\widehat{BC}$ and the center of $(IB_2B_3)$ also lies on $\overline{A_1I}$ so $\overline{A_2'A_3'}\parallel\overline{B_2'B_3'}$ where we are inverting about $I.$ Hence, $\overline{A_iB_i}$ concur and so $(A_iB_iI)$ are coaxial. $\square$

Apply the inversion with the respect to $I$. We leave to the reader to draw the inverse picture. Notice that the condition that $I$ is the incentar now reads that the circumcircles $A^*_i A^*_{i+1}I$ are of the same radii. Indeed if $R$ is the radius of the circle of inversion and $r$ the distance between $I$ and $XY$ then the radius of the circumcircle of $\triangle IX^*Y^*$ is equal to $R^2/r$. Now we use the following statement that is very easy to prove: "Let $k_1, k_2, k_3$ be three circles such that all pass through the same point $I$, but no two of them are mutually tangent. Then the centers of these circles are colinear if and only if there exists another common point $J \neq I$ of these three circles."

In the inverse picture this transforms into proving that the lines $A^*_1B^*_1, A^*_2B^*_2,$ and $A^*_3B^*_3$ intersect at a point.

In order to prove this it is enough to show that the corresponding sides of the triangles $A^*_1A^*_2A^*_3$ and $B^*_1B^*_2B^*_3$ are parallel (then these triangles would be perspective with respect to the infinitely far line). Afterwards the Desargue's theorem would imply that the triangles are perspective with respect to a center. Let $P^*_i$ be the incenter of $A^*_{i+1}A^*_{i+2}I$, and let $Q^*_i$ be the foot of the perpendicular from $I$ to $P^*_{i+1}P^*_{i+2}$. It is easy to prove that $$\overrightarrow{A_1^*A_2^*}=
2\overrightarrow{Q_1^*Q_2^*}=
-\overrightarrow{P_1^*P_2^*}.$$ Also since the circles
$A^*_iA^*_{i+1}I$ are of the same radii, we have, $P^*_1P^*_2 || B^*_1B^*_2$, hence $A^*_1A^*_2||B^*_1B^*_2$. As desired!

Let $D_i$ be touch point of incircle with $A_{i+1}A_{i+2}$ and $\gamma$ is inversion wrt this circle. $\gamma (A_i)$ is midpoint of $D_{i+1}D_{i+2}$ and all circles $\odot (I\gamma(A_i)\gamma(D_{i+2})\gamma(A_{i+1}))$ are congruent. Therefore $\gamma(C_i)\parallel D_{i+1}D_{i+2}$ and since $\overline{\gamma(A_i)\gamma(B_i)}$ concur (by homothety) $\odot (IA_iB_i)$ are coaxial as desired.

It suffices to prove the circles are coaxial. Let $DEF$ be the intouch triangle. Note thatof $\Gamma_1^\ast$ is exactly the circle with diameter $\overline{ID}$, etc.
We proceed by inversion around $I$.
[b][color=red]Claim:[/color][/b] The triangle $A_1^\ast A_2^\ast A_3^\ast$ is the medial triangle of $DEF$.
[i]Proof.[/i] Circles $\Gamma_2$ and $\Gamma_3$ are mapped to the circles with diameter $\overline{IE}$ and $\overline{IF}$, hence their second intersection $A_1^\ast$ is exactly the midpoint of $\overline{EF}$. $\blacksquare$

[b][color=red]Claim:[/color][/b] The triangle $B_1^\ast B_2^\ast B_3^\ast$ is homothetic to triangle $DEF$.
[i]Proof.[/i] This is the triangle determined by the lines $\Gamma_1^\ast$, $\Gamma_2^\ast$, $\Gamma_3^\ast$. Since $\Gamma_1^\ast$ is clearly perpendicular to $\overline{A_1 I}$, it is parallel to $\overline{EF}$, and similarly. $\blacksquare$
This means $A_1^\ast B_1^\ast$, $A_2^\ast B_2^\ast$, $A_3^\ast B_3^\ast$ are indeed concurrent as needed.

It suffices to show that the three circles are coaxial, i.e. intersect twice. Let $DEF$ be the intouch triangle. Invert around the incircle, denoting images with $\bullet'$. Then $\overline{A_2A_3}$ gets sent to the circle $\omega_1$ with diameter $\overline{ID}$, etc., and the circle $C_i$ gets sent to the common external tangent to $\omega_{i+1}$ and $\omega_{i+2}$ which is closer to $A_1'$ than $I$ for $1 \leq i \leq 3$ (note that $\triangle DEF$ is acute). We wish to prove that $\overline{B_1'A_1'},\overline{B_2'A_2'},\overline{B_3'A_3'}$ concur. Observe that $A_1'$ is the midpoint of $\overline{EF}$, etc., hence $\triangle A_1'A_2'A_3'$ is the medial triangle of $\triangle DEF$. Furthermore, it is clear that $\overline{DE} \parallel \overline{B_1'B_2'}$ since $\omega_1$ and $\omega_2$ have equal radius, so triangles $DEF$ and $B_1'B_2'B_3'$ are homothetic, hence there is a homothety sending $\triangle A_1'A_2'A_3'$ to $\triangle B_1'B_2'B_3'$ which implies the desired concurrence. $\blacksquare$

who made these point labels :sob:

solved with [b]popop614[/b]

We relabel the diagram one time. Let the triangle be $ABC$, still with incenter $I$. The $\Gamma_i$ closest to each $A,B,C$ will be $\omega_A, \omega_B,\omega_C$. The circles will intersect at $X_A, X_B, X_C$ instead of $B_1, B_2, B_3$. Let $DEF$ be the contact triangle.
\\
\\
We invert around the incircle. $A'B'C'$ becomes the medial triangle of $DEF$, and the sides of triangle $X_A'X_B'X_C'$ are the common tangents of the three pairs of circles from $(A'IB'F), (A'IC'E), (B'IC'D)$ that lie strictly outside of $DEF$, as these three circles are the inverses of $AB, AC, BC$. Let the circles have centers $O_D$, $O_E$, and $O_F$. Note that these circles have the same diameter, which is also the circumradius of $DEF$. This implies that the common tangent of $(A'IB'F)$ and $(A'IC'E)$ must be parallel to $O_EO_F$, and similar for the other two. As a result, $O_DO_EO_F$ is similar to $X_A'X_B'X_C'$. However, a homothety of factor $2$ at $I$ sends $O_DO_EO_F$ to $DEF$, and another homothety of factor $-\frac12$ at the centroid of $DEF$ sends $DEF$ to $A'B'C'$. Thus, $A'B'C'$ is similar to $X_A'X_B'X_C'$, implying that there must be a center of homothety. As a result, $A'X_A', B'X_B', C'X_C'$ concur. Therefore, $(AX_AI),(BX_BI),(CX_CI)$ are coaxial, so their centers are collinear.

Let the contact triangle be $\triangle DEF$. Invert about the incircle:
[list]
[*] The sides of $\triangle A_1A_2A_3$ invert to congruent circles with diameters $ID$, $IE$, and $IF$.
[*] Circles $C_i$ map to pairwise external tangents of the above circles.
[*] The desired circles map to $A_i^*B_i^*$.
[/list]

Thus the condition is equivalent the desired circles being coaxial, so we need to show the three lines $A_i^*B_i^*$ being concurrent. This simply results from $\triangle A_1^*A_2^*A_3^*$ and $\triangle B_1^*B_2^*B_3^*$ being homothetic with parallel sides. $\blacksquare$

All indices are taken modulo $3$. Also rename $C_i$ to $\Gamma_i$.
Let the projection of $I$ onto $A_iA_{i+1}$ be $D_{i+2}$. Then we do inversion wrt the incircle. Note that $\overline{A_iA_{i+1}}^* = (ID_{i+2})$ and $\Gamma_i^*$ is the common external tangent of $(ID_{i+1})$ and $(ID_{i+2})$. $(A_iB_iI)^* = \overline{A_iB_i}$.
Our condition is equivalent to $(A_iB_iI)$ being coaxial which is then equivalent to $A_iB_i^*$ concurring. So then $A_iB_i$ must also concur. We can take an affine transformation sending $A_1A_2A_3$ to an equilateral triangle and by symmetry, $A_iB_i$ must concur.
the affine transfomation is a bit dubious sorry

Invert about the incircle, let $P_i$ be the inverse of $A_i.$ By Fact 5, $A_iI$ is a diameter of $(A_{i+1}A_{i+2}I),$ so $I$ is the orthocenter of $P_1P_2P_3.$ Now let $Q_i$ be the inverse of $B_i.$ Since the $(IP_iP_{i+1})$ have the same radius we see $Q_{i+1}Q_{i+2}\perp IP_i\perp P_{i+1}P_{i+2}$ so $P_1P_2P_3,Q_1Q_2Q_3$ are homothetic so $P_1Q_1,P_2Q_2,P_3Q_3$ concur. Thus inverting back the circumcircles of $A_1B_1I,A_2B_2I,A_3B_3I$ are coaxial which finishes.


find all soluation of the simultaneaus equation
$y =x+\sqrt{x+\sqrt{x+\sqrt{x+.....+\sqrt{x+\sqrt{y}}}}}$
,$x+y =6$
where there are 1975 square roots in first equation

It is obviosly, that $x=2,y=4$ is solution. If y>4, then x=6-y<2 and $x+\sqrt y <y$, because $y-(6-y+\sqrt y )=(2\sqrt y+3)(\sqrt y-2)>0$. Therefore $\sqrt{x+\sqrt y }<\sqrt y$, $\sqrt{x+\sqrt{x+\sqrt y }}<\sqrt y$,... give $x+\sqrt{x+\sqrt{x+\sqrt{x+...}}}=z<y$ (can not be $z=y$). If $y<4$ we have $z>y$ (can not be $z=y$). It proof, that these solution is unique.


$ y \equal{} Ae^{k(x \minus{} vt)}$

Bounded refers to y. Y refers to the disp of an individual particle while x is the medium and t the time. Obviously the medium can stretch till $ \infty$ and so can time.

I presume, he means that the above function does not exhibit wave like properties because the value of y can even be $ \infty$ which seems improbable for a wave. Or basically a wave should exist between finite values. Like Sin or Cos.

Condition for a  function $ f(x,t)$ to be a progressive wave:
1. It should be continous everywhere and always.

2. $ \frac{\d^{2} f }{\d t^{2}} = v^{2} \frac{\d^{2}f }{\del x^{2}}$

Here i am refering to second order partial derivative.
Oh sm1 temme how i must write 'dough square / dough t square.' etc

[quote="achu_182"]in the book, they say that in addition to this, the function must be bounded. i think what chronoz is trying to say is that the range should be bounded b/w 2 finite values.[/quote]

Exactly, the disp should only between 2 FINITE numbers. I think, if you logically analyze, the infinite disp, is clearly not a wave.

[quote="Conjurer"]For any value of x or t , how can y be infinity?[/quote]

Thats an exponential function and obviously y can just keep on increasing indefinitely.

IN HC Verma there are various wave equations of the form y = Ae^@.


least common multiple
number theory proposed
number theory

[quote="pohoatza"]Find all integer positive numbers $n$ such that:
$n=[a,b]+[b,c]+[c,a]$, where $a,b,c$ are integer positive numbers and $[p,q]$ represents the least common multiple of numbers $p,q$.[/quote]

Let $a=2^{p}q$, $b=2^{p}$ and $c=1$
Then $n=2^{p}(2q+1)$

And, since any positive integer n can be written as $n=2^{p}(2q+1)$, any positive integer may be written as $n=[a,b]+[b,c]+[c,a]$, where $a,b,c$ are integer positive numbers.

Sorry (edited later) : This demo does not work with $n=2^{p}$ since this would imply $a=0$.
So it remains to study the case $n=2^{p}$

End (edited later than later) : It does not exist solution for $2^{p}$ :
Let $a=2^{u}a'$, $b=2^{v}b'$ and $c=2^{w}c'$ with a', b' and c' odd and $u\geq v\geq w$
Then $n=2^{u}([a',b']+[a',c'])+2^{v}[b',c']=2^{v}(2^{u-v}([a',b']+[a',c'])+[b',c'])$
and, since $2^{u-v}([a',b']+[a',c'])+[b',c']$ is odd and $>1$, n can't be $2^{p}$

And the response is : "any $n>2$ different from $2^{p}$, $p>1$

--
Patrick

[quote="pohoatza"]Find all integer positive numbers $ n$ such that:
$ n \equal{} [a,b] \plus{} [b,c] \plus{} [c,a]$, where $ a,b,c$ are integer positive numbers and $ [p,q]$ represents the least common multiple of numbers $ p,q$.[/quote]
[b]Lemma:[/b]
$ n$ can be written on the form iff $ 2n$ can be written on that form.
[b]Proof:[/b]
If $ n \equal{} [a,b] \plus{} [b,c] \plus{} [c,a]$ then $ 2n \equal{} [2a,2b] \plus{} [2b,2c] \plus{} [2c,2a]$.
If $ 2n \equal{} [a,b] \plus{} [b,c] \plus{} [c,a]$ we have to cases:
$ 2 \mid a,b,c$ then $ n \equal{} \left [ \frac {a}{2}, \frac {b}{2} \right ] \plus{} \left [ \frac {b}{2}, \frac {c}{2} \right ] \plus{} \left [ \frac {c}{2}, \frac {a}{2} \right ]$.
$ 2 \mid a,b, 2 \nmid c$: $ n \equal{} \left [ \frac {a}{2}, \frac {b}{2} \right ] \plus{} \left [ \frac {b}{2}, c \right ] \plus{} \left [ c, \frac {a}{2} \right ]$. $ \square{}$

Now if $ n \equal{} 2m \plus{} 1$ then $ n \equal{} [m,1] \plus{} [1,1] \plus{} [1,m]$. But $ [a,b] \plus{} [b,c] \plus{} [c,a] \ge 3 > 1$ so $ n \equal{} 1$ has no solution. Using the lemma we obtain that $ n$ can be written on the form iff it is not a power of $ 2$. $ \square{}$


algebra
polynomial
search
real analysis
real analysis unsolved

Show that $ f\left(x\right) \equal{} 1 \plus{} \frac {x}{1!} \plus{} \frac {x^2}{2!} \plus{} ...\plus{}\frac {x^{2k}}{\left(2k\right)!}$ is positive for all real $ x$ and all integers $ k\geq 0$.

For $ x\ge 0,$ this is a sum of nonnegative terms, hence clearly positive. Hence for the rest of the problem, I will assume that $ x<0.$

By Taylor's Theorem, $ e^x\equal{}f(x)\plus{}\frac{e^{\xi}x^{2k\plus{}1}}{(2k\plus{}1)!}$ for some $ \xi\in(\minus{}x,0).$ We can rearrange this to

$ f(x)\equal{} e^x\minus{}\frac{e^{\xi}x^{2k\plus{}1}}{(2k\plus{}1)!}.$

But $ x^{2k\plus{}1}<0$ for negative $ x$ and $ e^{\xi}>0$ no matter what $ \xi$ is. Hence we are adding a positive term to $ e^x,$ and $ e^x$ is positive. Hence $ f(x)$ is positive.

You can also use the fact that a real polynomial of even degree is positive everywhere if and only if it's positive in every real root of its derivative. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1553761007&t=121981]here[/url] or [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=636349#636349]here[/url].


group theory
abstract algebra
superior algebra
superior algebra unsolved

The fallowing statement is true or not :
$G$ is a group and $H$,$K$are subgroups of $G$.
Core($H$)=Core($K$) if and only if $H$=$K$.
if it's not tru make and example.

[quote="jmerry"]What is "Core"? I don't think that's standard terminology in English.[/quote]
Core($H$)=$\bigcap_{x \in G}xHx^{-1}$

OK, I don't know a standard term for that.

It's blatantly false- that intersection is the largest normal subgroup of $H$. If your group has non-normal subgroups, there will be counterexamples. For one, consider $A_{n}$ for some $n>5$; that intersection will include only the identity if $H$ is any proper subgroup, and there are lots of proper subgroups.


function
inequalities proposed
inequalities

$\mathrm{Suppose\;that\;}m\in\mathbb{N}\mathrm{.\;For\;each\;}$ $p\in\mathbb{N}$ $\mathrm{def\mbox{i}ne\;}$ $f_{m}(p)=k$ $\mathrm{where\;}k\mathrm{\;is\;the\;largest\;real\;number\;for\;which}$
\[a^{4m+1}+b^{4m+1}+c^{4m+1}-3(abc)^{p}\geq k(1-(abc)^{p}) \]
$\mathrm{for\;all\;positive\;real\;}a,b,c\mathrm{\;such\;that}$
\[a^{2m+1}+b^{2m+1}+c^{2m+1}=3. \]
$\mathrm{Find\;}$
\[\mathfrak{S}=\sum_{j=1}^{2m}f_{m}(j). \]

[hide="Hint"]The function $f(p)=\frac{a^{2m+1}+b^{2m+1}+c^{2m+1}-3(abc)^{p}}{1-(abc)^{p}}$ is decreasing in $p.$[/hide]


i think this goes here

In Victorian times three women Misses A,B and C are sitting in a railway carriage. The train
goes through a tunnel and a large quantity of smoke is blown into the carriage. When the train
emerges from the tunnel Miss A sees that the faces of Misses B and C have been blackened by the
smoke and she starts laughing at them. However, Miss B and Miss C are also laughing (obviously
at seeing a smoke-blackened face and not realizing that they are in the same state.)
Explain why Miss A should realize that her own face has also been blackened.

[hide]When Misses A laughs, she get's a hankerchief and dries the tears, with some black coming off onto the hankerchief.[/hide]

[hide]Um... this riddle has many flaws in it. I won't go into them, but the basis of the riddle is that if A didn't have a black face, and B starts laughing, then C would know that B was laughing at her, not A, and then not be in a laughable position. [/hide]


number theory
prime numbers
number theory proposed

[quote="manjil"]Prove that no infinite arithmetic sequence contains only perfect powers.[/quote]

Let the sequence be $ u_n\equal{}a(b\plus{}nc)$ with $ c>0$ and $ (b,c)\equal{}1$

Dirichlet theorem says that there are infinitely many prime numbers which may be written $ b\plus{}nc$ and, since we have a finite number of prime factors in $ a$, infinitely many $ u_n$ have a prime factor whose order is $ 1$.
Hence the result.

I guess this sequence is not constant . pco solution is good but  we can solve by an elementary solution . Let the sequence be $ a_n\equal{}a_0\plus{}nd$ . Take p be a prime which is relative prime to d . Because $ \{a_0\plus{}nd\}_{n\equal{}0}^{p^2\minus{}1}$  is a complete residue mod $ p^2$ so exist $ n_0$ such that $ a_n\equiv p(\mod p^2)$ ,this fact shows that $ v_p(a_{n_0})\equal{}1$ so it can't be a power .


So it looks like the math season is about over. Bummer. :(

What is everyone going to do in the summer or in the next couple of months besides school since it will end soon?

Sorry, I don't know how to set up a poll.

:|
Hao's gonna go on AoPS and stalk the MC forum.
Alan, me, and Andrew are gonna go to MP.
Other people, I don't know.

'Math season' isn't over. Math isn't seasonal. Keep doing math.

And no, I don't mean re-do the MathCounts problems that you've already done. I don't mean do new ones either. There will be plenty of time for that next year.

Meh, after MP I have no idea what I'll do...
Well, I would only have like 3 weeks or so anyways...

I meant like math contests, excluding you really smart people who are doing all these stuff I don't know. Of course you should do math all the time, but I don't you won't do it as much. Anyways, I am doing the most exciting thing over summer. First I stay home. Then I stay home. Then I stay home. Then I go to Cape Cod and Canada in a trip. Then I stay home. Then I stay home. Then I stay home. Then I stay home. Get the picture? Yep, I stay home for the whole summer, probably playing stuff, going to people's houses, go play tennis and swimming and doing math. Yeah I have no life. :o . Oh, then I start my 8th grade carrier.

Eh, none of us are taking it.
(None of us in MS)

girl scout camp, arguing with people who tell me to go to math camp, trip to NH with family, swim, run, leave summer reading to last second, avoid all thinking possible, and no math for me over summer break.


I'm not such helpless with this but I think a good explanation of this would be really helpful :) .

It says on my book:
Let f (x) = a_n x^n + a_n-1 x^n-1..... a_2 x^2 + a_1 x + a_0 be a polynomial with real coefficients and a_0 not equal to zero.

1. The number of positive real zeros of f is either equal to the number of variations in sign of f (x) or less than that number by an even integer.
2. The number of negative real zeros of f is either equal to the number of variations in sign of f (-x) or less than that number by an even integer.

Let $Z$ be the positive zeroes of $f(x)$ and $C$ the number of changes of sign of the sequence of its coefficients. We then have $C-Z\geq 0$, furthermore, $C-Z$ is an even number.

The proof below is from [b]'Problems and Theorems in Analysis II'[/b] by George Plya and Gabor Szeg.

Let $\alpha_1,\dots,\alpha_Z$ are the positive zeroes of $f(x)$. thus $f(x) = g(x) (x-\alpha_1)\cdots(x-\alpha_Z)$, where $g(x)$ is a polynomial of degree $(n-Z)$ with real coefficients. The number of changes of sign of $g(x)$ is $\geq 0$, that of $g(x)(x-\alpha_1)$ is $\geq 1$, that of $g(x)(x-\alpha_1)(x-\alpha_2)$ is $\geq 2$, ...., that of $f(x) = g(x) (x-\alpha_1)\cdots(x-\alpha_Z)$ is $\geq Z$.

Let $a_\alpha$ and $a_\omega$ be the first and the last non-vanishing coefficient of $f(x)$, $\alpha\leq\omega$, only $\alpha < \omega$ is interesting. $0<\xi<\alpha_1\leq \alpha_2\leq \cdots \leq \alpha_Z<X<\infty$. if $\xi$ is sufficiently close to 0 and $X$ to $\infty$, it can be deduced that
\[\hbox{sgn }f(\xi) = \hbox{sgn }a_\alpha,\quad  \hbox{sgn }f(X) = \hbox{sgn  }a_\omega\]
From the aid of above, if $C=1$ then $Z=1$

yes, i don't think this problem is a beginner either.

The following is some complement of the prove above:

[b]Proposition 1[/b] Let the values of $f(a)$ and $f(b)$ be different 0. The interval $a<x<b$ contains an even or an odd number of zeros of $f(x)$ according as $f(a)$ and $f(b)$ have the same or opposite sign.

[b]Proposition 2[/b] Let the values of $a_j$ and $a_k$ be different 0. The subsequence $a_j,a_{j+1},\dots,a_{k-1},a_k$ contains an even or an odd number of changes of sign according as $a_j$ and $a_k$ have the same or opposite sign.

[b]Proposition 3[/b] If the finite number sequence $a_0,a_1,\dots,a_n$ has $C$ changes of sign, then the sequence
\[a_0, a_1 - a_0, a_2 - a_1, \dots, a_n - a_{n-1}, -a_n\]
formed from it has not fewer than $C+1$ changes of sign.

[b]Proposition 4[/b] The polynomials $P(x)$ and $P(ax)$ has the same number of changes of sign if $a$ is positive.

[b]Proposition 5[/b] Let $\alpha>0$. Passing from the polynomial \[a_0+a_1x+a_2x^2 + \cdots + a_n x^n\]
to the polynomial
\[(\alpha-x)(a_0+a_1x+a_2x^2 + \cdots + a_n x^n) = \alpha a_0 + (\alpha a_1-a_0)x + (\alpha a_2-a_1)x^2 + \cdots -a_n x^{n+1}\]
the number of changes of sign increases : in fact, it increases by an odd number.

Proof of Proposition 5 :

Replace $x$ by $\alpha_x$ (Proposition 4) then apply Proposition 3.

Proof of Proposition 3 :

We denote the indices of change in the question by $\nu_1+1, \nu_2+1, \dots, \nu_C+1$, $0\leq nu_1<nu_2<\cdots<\nu_C\leq n-1$. The following two subsequences
\[a_0,a_1-a_0,\dots,a_{\nu_1+1}-a_{\nu_1} \]
\[a_{\nu_C+1}-a_{\nu_C},a_{\nu_C+2}-a_{\nu_C+1},\dots,a_n-a_{n-1},-a_n \]
each have a change of sign. For if $a_\alpha$ is the first non-vanishing term of the first sequence, then we have $0\leq \alpha \leq \nu_1$ and
\[ \hbox{sgn }(a_n-a_{n-1}) = \hbox{sgn }a_\alpha = -\hbox{sgn }a_{\nu_1+1} = -\hbox{sgn }(a_{\nu_1+1} - a_{\nu_1})
\]
now apply Proposition 2. Similarly in the case of the second subsequence above.

[quote="confuted"][quote="liyi"]something long and complex[/quote]
That's not going to be understood whatsoever by someone posting questions in the "Getting Started" forum. That's the sort of thing that goes in the "advanced" forum. So, silverfalcon, don't be worried if you didn't understand a word of liyi's post.[/quote]

Well, don't write off what liyi wrote, either.  Descartes' Rule of Signs is certainly a non-trivial result in analysis, and it is instructive to see a proof of such a theorem.


Not yet. There's still $ 2\frac{1}{2}$ days left. :D

Lawlz for me there are 10 days left. Yay for my school having 3-week long winter breaks. :D

[quote="k00lperson"]sigh
Don't you have another week of interval too?
LUCKY. Anyway, i only have like 1 day left....:([/quote]

Yeah. That's Intersession week, which is like technically no actual classes. :)


The equation $ \sqrt {x \plus{} 4} \minus{} \sqrt {x \minus{} 3} \plus{} 1 \equal{} 0$ has:

$ \textbf{(A)}\ \text{no root} \qquad\textbf{(B)}\ \text{one real root}$
$ \textbf{(C)}\ \text{one real root and one imaginary root}$
$ \textbf{(D)}\ \text{two imaginary roots} \qquad\textbf{(E)}\ \text{two real roots}$

hello, rewriting your equation in the form $ \sqrt{x\plus{}4}\plus{}1\equal{}\sqrt{x\plus{}3}$,squaring both sides we have
$ \sqrt{x\plus{}4}\equal{}\minus{}1$, its obviously there's no real solution. Squaring agian we have $ x\equal{}\minus{}3$, but $ x\equal{}\minus{}3$ fulfills not our equation, so there's no root. The answer is $ \text{A}$.
Sonnhard.

[hide="HIDE, different solution"]Note that $ x\plus{}4>x\minus{}3$, so $ \sqrt{x\plus{}4}\minus{}\sqrt{x\minus{}3}$ is positive. You add 1, so the result is positive. Therefore there are no roots.[/hide]


Let $K$ be a field of characteristic $0$. Prove that for any finitely generated subgroup $G$ of $GL(n,K)$, there is a torsion free subgroup of finite index in $G$.

This can't be true in general. An example:

Let $K$ be the algebraic closure of the field $F_{2}$ with 2 elements. $K$ can be interpreted as the union of fields of order $2^{n}$ for each positive integer $n$, and all elements of $K^{*}$ are torsion.
If $n=1$, $GL_{n}(K)=K^{*}$. The only torsion-free subgroup is the trivial subgroup, which has infinite index.

Actually, it fails for all $n$ if $K$ is any field with infinitely many roots of unity. Every subgroup of finite index must intersect the subgroup of matrices of the form $\lambda I$ with $\lambda$ a root of unity nontrivially, since this subgroup is infinite.

The problem should read: Prove that for any finitely generated subgroup $G$ of $GL(n,K)$, there is a torsion free subgroup of finite index in $G$.

The problem has been discussed before (although no proof has been posted) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=cassels&t=77343]here[/url].


calculus
function
algebra
domain
trigonometry
irrational number
calculus computations

$ \text{1.}$ Give some examples for functions that is continuous only in two points and its domain is the real numbers.
For example $ f(x)$ defined below is continuous only in points $ x\equal{}1$ and $ x\equal{}\minus{}1$ and its domain is the real numbers:
$ f(x)\equal{}x^{2}\minus{}1\;\;\;\; x\in\mathbb{Q}$
$ f(x)\equal{}0\;\;\;\;\;\;\;\; x\in\mathbb{R}\minus{}\mathbb{Q}$
$ \text{2.}$ Give some examples for functions $ \mathbb{R}\rightarrow\mathbb{R}$ such that they're continuous only in integer points.
For example:
$ f(x)\equal{}\sin\pi x\;\;\;\; x\in\mathbb{Q}$
$ f(x)\equal{}0\;\;\;\;\;\;\;\; x\in\mathbb{R}\minus{}\mathbb{Q}$

For number one :
$ f(x) \equal{} x^2$, for rational number $ x$, and $ f(x) \equal{} x$, for irrational number $ x$, it is only continuous in the point $ x\equal{}0$ and $ x\equal{}1$.


Prove that if  $xyz = 1$ then we have  $3 + \frac{x}{z} + \frac{y}{x} + \frac{z}{y} > x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$

After the substitution $x=a/b, y=b/c, z=c/a$ and little transformations you get the well-known inequality:

$a^3+b^3+c^3+3abc>a^2b+ab^2+a^2c+ac^2+b^2c+bc^2 $

Misha

[quote="Misha123"]After the substitution $x=a/b, y=b/c, z=c/a$ and little transformations you get the well-known inequality:

$a^3+b^3+c^3+3abc>a^2b+ab^2+a^2c+ac^2+b^2c+bc^2 $[/quote]

This actually should be

$a^3+b^3+c^3+3abc\geq a^2b+ab^2+a^2c+ac^2+b^2c+bc^2$

and it is the most famous special case of the Schur inequality. In the original inequality of Metru, the > sign must be replaced by an $\geq$ sign; equality holds if and only if x = y = z = 1.

  Darij


How many integers between 100 and 300 have both 11 and 8
as factors?

This means we're looking for the integers that are divisible by $ 11 \times 8 \equal{} 88$. There are only $ \boxed{2}$ of them between 100 and 300.


group theory
abstract algebra
function
Galois Theory
superior algebra
superior algebra unsolved

Let $ K \equal{} \mathbb{Q}(\sqrt [n]{a})$, where $ a \in \mathbb{Q}, a>0$.
Suppose $ [K: \mathbb{Q}]\equal{}n$.
Prove that if $ n$ is odd then $ K$ has no nontrivial subfield which are Galois over $ \mathbb{Q}$ and if $ n$ is even then the only nontrivial subfield of $ K$ which is Galois over $ \mathbb{Q}$ is $ \mathbb{Q}(\sqrt {a})$.

[hide="Solution"]Let $ L \equal{} K(\zeta^n)$ be the splitting field of $ x^n \minus{} a \equal{} 0$. Let $ H \equal{} Gal(L/K)$ (as a subgroup of $ G \equal{} Gal(L/\mathbb{Q}))$. By the fundamental theorem of Galois theory, the result above amounts to showing that no normal subgroup of $ G$ contains $ H$, except possibly for a subgroup of index $ 2$ if $ n$ is even.

We have that $ G$ is the group of permutations on $ \mathbb{Z}/(n)$ given by addition of an element of $ \mathbb{Z}/(n)$ or multiplication by an element of $ U_n$, also a semi-direct product of $ C_n$ and $ U_n$. Then $ H \equal{} U_n$, and suppose $ H \le N \triangleleft G$ with $ N \neq G$. Let $ K \equal{} N \cap C_n$, and let $ a$ be a generator of $ K$. For any $ b \in U_n$ and $ c \in C_n$, then, by the normality of $ N$, the function on $ \mathbb{Z}/(n)$ defined by $ f(x) \equal{} b(x \plus{} c) \plus{} a \minus{} c \equal{} bx \plus{} bc \minus{} c \plus{} a$ must be in $ N$, so that $ bc \minus{} c \in K$. We can let $ c \equal{} a \minus{} 1$ so that $ (a,c) \equal{} 1$, since $ a \neq 1$. Then $ (b \minus{} 1) \in (a)$ for all $ b \in U_n$. This is only true if there is at most one nonzero coset of $ (a)$ in $ C_n$, so $ a \equal{} 2$, and $ n$ is even. From this it follows that $ [G: N] \equal{} 2$, and we are done.

Note: $ U_n$ denotes the group $ \mathbb{Z}/(n)^{\times}$[/hide]


[quote="M.A"]Let $ a,b,c > 0$ prove that
\[ \frac {a}{a^2 \plus{} bc} \plus{} \frac {b}{b^2 \plus{} ac} \plus{} \frac {c}{c^2 \plus{} ba}\ge\frac {3}{a \plus{} b \plus{} c}
\]
[/quote]
It's equivalent to the following:
$ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\plus{}\sum_{cyc}(a^3b^3\plus{}a^3b^2c\plus{}a^3c^2b\plus{}a^2b^2c^2)\geq0.$

[quote=M.A]Let $ a,b,c>0$ prove that
\[ \frac{a}{a^2\plus{}bc}\plus{}\frac{b}{b^2\plus{}ac}\plus{}\frac{c}{c^2\plus{}ba}\ge\frac{3}{a\plus{}b\plus{}c}\][/quote]
Let $ a,b,c>0 .$ Prove that
\[ \frac{a}{a^2\plus{}bc}\plus{}\frac{b}{b^2\plus{}ac}\plus{}\frac{c}{c^2\plus{}ba}\ge\frac{9}{2(a\plus{}b\plus{}c)}.\]


inequalities
search
inequalities proposed
SRMC 2002

I tried to search SRMC problems,but i didn't find them(I found only SRMC 2006).Maybe someone know where on this site i could find SRMC problems?I have all SRMC problems,if someone want i could post them, :wink:
Here is one of them,this is one nice inequality from first SRMC:
Let $ n$ be an integer with $ n>2$ and $ a_{1},a_{2},\dots,a_{n}\in R^{\plus{}}$.Given any positive integers $ t,k,p$ with $ 1<t<n$,set $ m\equal{}k\plus{}p$,prove the following inequalities:
a)
$ \frac{a_{1}^{p}}{a_{2}^{k}\plus{}a_{3}^{k}\plus{}\dots\plus{}a_{t}^{k}}\plus{}\frac{a_{2}^{p}}{a_{3}^{k}\plus{}a_{4}^{k}\plus{}\dots\plus{}a_{t\plus{}1}^{k}}\plus{}\dots\plus{}\frac{a_{n\minus{}1}^{p}}{a_{n}^{k}\plus{}a_{1}^{k}\plus{}\dots\plus{}a_{t\minus{}2}^{k}}\plus{}\frac{a_{n}^{p}}{a_{1}^{k}\plus{}a_{2}^{k}\plus{}\dots\plus{}a_{t\minus{}1}^{k}}\geq\frac{(a_{1}^{p}\plus{}a_{2}^{p}\dots\plus{}a_{n}^{p})^{2}}{(t\minus{}1) ( a_{1}^{m}\plus{}a_{2}^{m}\plus{}\dots\plus{}a_{n}^{m})}$
b)$ \frac{a_{2}^{k}\plus{}a_{3}^{k}\dots\plus{}a_{t}^{k}}{a_{1}^{p}}\plus{}\frac{a_{3}^{k}\plus{}a_{4}^{k}\dots\plus{}a_{t\plus{}1}^{k}}{a_{2}^{p}}\plus{}\dots\plus{}\frac{a_{1}^{k}\plus{}a_{2}^{k}\dots\plus{}a_{t\minus{}1}^{k}}{a_{n}^{p}}\geq\frac{(t\minus{}1)(a_{1}^{k}\plus{}a_{2}^{k}\dots\plus{}a_{n}^{k})^{2}}{( a_{1}^{m}\plus{}a_{2}^{m}\plus{}\dots\plus{}a_{n}^{m})}$ :wink:

[quote="Erken"][quote="galois01"]Erken what is SRMC ? :?:[/quote]
It is Silk Road Mathematics Competiton :wink: .It is very nice olympiad and quite hard olympiad. :)[/quote]
Thank's a lot for your explanation it's the first time I hear about this olympiad  :D

   $[a]$ For the first part, denote the left-hand-side by $S$. Using Cauchy-Schwarz inequality, we have,
    $$
    S\left(
    \sum_{i=1}^p a_i^p(a_{i+1}^k+\dots+a_{i+t-1}^k)\right)\geq (a_1^p+\cdots+a_n^p)^2,
    $$
    where, the indices above are taken modulo $n$. In particular, the inequality boils down to proving, $$
    (t-1)(a_1^m+\cdots+a_n^m)\geq (a_1^pa_2^k+a_1^pa_3^k+\cdots+a_1^pa_t^k+a_2^pa_3^k+\cdots+a_n^pa_1^k+\cdots+a_n^pa_{t-1}^k).
    $$
    Observe that, both sides have, upon opening, $n(t-1)$ terms. Now, keeping in mind that, $m=k+p$, consider the vectors,
$$
v_1=(\underbrace{a_1^k,a_1^k,\dots,a_1^k}_{n-1},\underbrace{a_2^k,\dots,a_2^k}_{n-1},\dots,\underbrace{a_n^k,\dots,a_n^k}_{n-1}),
$$
and,
$$
v_2=(\underbrace{a_1^p,a_1^p,\dots,a_1^p}_{n-1},\underbrace{a_2^p,\dots,a_2^p}_{n-1},\dots,\underbrace{a_n^p,\dots,a_n^p}_{n-1}).
$$
The inequality above, is simply a result of rearrangement inequality, applied on $v_1$ and $v_2$.

$[b]$ For the second part, break the each terms into sum of individual terms, that is, for instance,
$$
\frac{{a_2^k + a_3^k +  \cdots  + a_t^k}}{{a_1^p}}=\frac{a_2^k}{a_1^p}+\frac{a_3^k}{a_1^p}+\cdots+\frac{a_t^k}{a_1^p}.
$$
Now, apply Cauchy-Schwarz on left-hand-side with weights $(a_1^pa_2^k+\cdots+a_1^pa_t^k+\cdots+a_n^pa_1^k+\cdots+a_n^pa_{t-1}^k)$, to get,
$$
{\rm LHS}\cdot (a_1^pa_2^k+\cdots+a_1^pa_t^k+\cdots+a_n^pa_1^k+\cdots+a_n^pa_{t-1}^k)\geq ((t-1)(a_1^k+\cdots+a_n^k))^2.
$$
Now, using the fact that,
$$
(a_1^pa_2^k+\cdots+a_1^pa_t^k+\cdots+a_n^pa_1^k+\cdots+a_n^pa_{t-1}^k)\leq (t-1)(a_1^m+\cdots+a_n^m),
$$
we arrive at the result.

[quote=Erken][quote="galois01"]Erken what is SRMC ? :?:[/quote]
It is Silk Road Mathematics Competiton :wink: .It is very nice olympiad and quite hard olympiad. :)[/quote]

Thanks. Are there any official websites?

the aops contest collections are complete for SRMC
Path: Contests \ International Contests \ [url=https://artofproblemsolving.com/community/c804124_silk_road]Silk Road[/url]

as a source for adding the missing problems in aops I used [url=http://matol.kz/nodes/93]this[/url] page, I do not think that it is an official page but at least it has all the problems


Does anyone have any reccommendations about math olympiad challenges by Andresscu/Gelca?

I hope to qualify for USAMO this year, and I am wondering if that book will directly help me with the aime/amc or if I should work on other things like HMMT etc.

i think its slightly above your target level, if you're working on AIMEs

Math Olympiad Challenges is probably more for someone who's actually trying to do well on USAMO, not qualify for it. If you want to qualify for USAMO, I highly recommend AoPS Volume 2.

[quote="aahmadi"][quote="Altheman"]is that the blue or the yellow/orange/green book? the blue one is pretty diffiuclt at first, the motivation for the solutions are unclear, but part of the work is figuring that out on your own[/quote]
Which one is more difficult? I think I have the "yellow/orange/green " book (Andreescu/Feng) and I felt like that was an IMO prep book.[/quote]

I think Challenges (blue) is harder than Treasures (multi-colored).


These are two similar problems from what seems to be the Swedish equivalent to Mathcounts, from two different qualification rounds (2001 and 2006, I think). One point per correct answer for a maximum of three per question.

1. Find three different ways of forming the number $ 123$ by using the numbers $ 5, 6, 7, 7, 13$ and $ 18$ and the operations $ \plus{}, \minus{}, *, /$. Usage of parenthesis is allowed. (Each number may only be used once, but there are no limitations of the usage of operations.)

2. By using four $ 2$'s one can express the number $ 21$, by $ 21 \equal{} 22 \minus{} \frac2{2}$.
(a) Find a way to express $ 37$ by using five $ 3$'s.
(b) Find a way to express $ 20$ by using four $ 9$'s.
(c) Find a way to express $ 100$ by using six identical numbers.

What strategy do you use when solving problems of this type? Plain brute force or something else? I didn't bother much with the first question, but I answered the second one completely. The first two questions I came up with the answer quite quickly by just looking at the numbers, but the third took me some time.

[hide](a) $ 33\plus{}3\plus{}\frac3{3}$

(b) $ \frac{99}{9}\plus{}9$

(c) $ 5!\minus{}5*5\plus{}5\minus{}5\plus{}5$[/hide]

for the third one, a simple method.
[hide]25+25+25+25+25-25[/hide]

For the number 2 (c),

couldn't you have just put

[hide] 55+55-5-5[/hide][/code]


geometry
3D geometry
polynomial
parameterization
number theory proposed
number theory

find all the integers $ {x, y, z}$ such that:

$ {x^2 \plus{} xy \plus{} y^2} \equal{} {z^3}$

What i have done (nothing much of course):

if $ {x\minus{}y}$ is a perfect cube, then the equation has no non-trivial solutions:

$ x \minus{} y \equal{} k^3$
$ (x \minus{} y)(x^2 \plus{} xy \plus{} y^2) \equal{} (x \minus{} y)(z^3)$
$ x^3 \minus{} y^3 \equal{} (k^3)(z^3)$
( let $ t \equal{} ({kz})^3$ )
$ x^3 \equal{} y^3 \plus{} t^3$
which by Fermat's Last Theorem, has no solutions where $ xyt \neq 0$ .

(and else: the equation Has a solution: $ (x, y, z) \equal{} (3 ,3 ,3)$ )

$ x^2 \plus{} xy \plus{} y^2 \equal{} N(x \minus{} y \omega)$ is the norm of an [url=http://mathworld.wolfram.com/EisensteinInteger.html]Eisenstein integer[/url]; it follows that $ z^3$ and therefore $ z$ must also be such an integer.  Write

$ x \minus{} y \omega \equal{} (a \minus{} b \omega)^3 \equal{} a^3 \minus{} 3a^2 b \omega \plus{} 3ab^2 ( \minus{} 1 \minus{} \omega) \minus{} b^3 \implies$
$ x \minus{} y \omega \equal{} (a^3 \minus{} 3ab^2 \minus{} b^3) \plus{} ( \minus{} 3a^2 b \minus{} 3ab^2) \omega$.

An infinite family of solutions is therefore given by

$ (x, y, z) \equal{} (a^3 \minus{} 3ab^2 \minus{} b^3, 3a^2 b \plus{} 3ab^2, a^2 \plus{} ab \plus{} b^2)$.

The complete set of solutions can be deduced by noting that $ x \minus{} y \omega$ can be replaced with (five? eleven?) other expressions that give the same norm (such as $ x \minus{} y \omega^2$ or $ x \omega \minus{} y \omega^2$).  For example, $ (3, 3, 3)$
 is not generated by the above parameterization, but choosing $ a \equal{} b \equal{} 1$ does get you $ ( \minus{} 3, 6, 3)$, and then note that $ \minus{} 3 \minus{} 6 \omega \equal{} \minus{} 3 \omega \plus{} 3 \omega^2$.  These solutions are equivalent up to multiplication by units and conjugation.

A complete characterization of the integers of the form $ x^2 \plus{} xy \plus{} y^2$ can be deduced along similar lines as the argument [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url] for sums of two squares.


[hide="ans"]
$\boxed{A}$
[/hide]
 :blush:  :blush:  :blush:  :blush:

[quote="riddler"][hide="ans"]
$\boxed{B}$
[/hide]
Fermat-Euler- $252-248=4$.....[/quote]

the ansewr is $\boxed{C}$


algorithm
invariant
combinatorics unsolved
combinatorics

An arbitrary finite graph is given. Prove that vertices can be colored in black and white so that, for each vertex, at least half of its neighbours are the opposite color from the vertex itself.

This same question appeared in our Midterm test! (I am a computer science student)

Here is the solution:

Give an arbitrary colouring to the graph. Call the colouring $C$. For each colouring $C$, define $S(C)$ to be the number of edges with different coloured end points.

Suppose your colouring $C$, does not satisfy the 'at least 1/2 neighbours of other colour', then there is some (say white) vertex $v$ which has more than 1/2 its neighbours also white. Now keeping all the other vertices the same colour, obtain a new colouring $C'$ by colouring $v$ black. Observe that $S(C) < S(C')$. Since $S(C) \leq E$, the number of edges in the graph, this algorithm of picking an arbitrary colouring and moving 'bad' vertices around will ultimately give a colouring which satisfies the constraints of the problem.

I found that it can be solved by semi-invariant:(sorry if I did not use English  exactly!)
Call a vertice 'special' if the number of its same color neighbour is greater than the number of opposite color ones.We do according the following rules:In each step,re-color one 'special' vertice and after finite step we have the desired graph.
  Indeed, consider the number of vertex which have the same color then  after each step this quantity is decreasing but it obtain finite values so we are done!!!
   Sorry,I did not see the above solution,may it's similar to my one!!!


Let $ABC$ be a triangle such that $min \{\measuredangle{A}, \measuredangle{B}, \measuredangle{C}\}\geq 45^{0}$. Let $A'B'C'$ be the medial triangle of $ABC$. Show that for any point $P$ lying inside triangle $A'B'C'$ we have the following inequality
     $PA+PB+PC > \sqrt{2}(PA'+PB'+PC')$.

oh like Erdos inequality!
we have $PA'^{2}=\vec{PA'}.\vec{PA'}=\frac{1}{2}(\vec{PB}+\vec{PC}).\vec{PA'}\rightarrow PA'=\frac{1}{2PA'}\vec{PA'}(\vec{PB}+\vec{PC})$ then
$\sum PA'=\sum\frac{1}{2}\vec{PA}(\frac{\vec{PB'}}{PB'}+\frac{\vec{PC'}}{PC'})\le \frac{PA}{2}|\frac{\vec{PB'}}{PB'}+\frac{\vec{PC'}}{PC'}|(1)$
$\alpha=\measuredangle{(\vec{PB'},\vec{PC'})}$ then $|\frac{\vec{PB'}}{PB'}+\frac{\vec{PC'}}{PC'}|=2cos\frac{\alpha}{2}(2)$ then
$max\{cos\frac{\alpha}{2},cos\frac{\beta}{2},cos\frac{\gamma}{2}\}(PA+PB+PC)\ge \sum cos\frac{\alpha}{2}PA(3)$
and from supposition $\rightarrow min\{\measuredangle{A'},\measuredangle{B'},\measuredangle{C'}\}\ge 45^{0}\rightarrow max\{\alpha,\beta,\gamma\}\ge 90^{0}\rightarrow max\{cos\frac{\alpha}{2},cos\frac{\beta}{2},cos\frac{\gamma}{2}\}\le \frac{1}{\sqrt{2}}$
from $(1),(2),(3),(4)$ this problem is the fact! :)

As soon as I submitted this problem, I realized the proof with Erdos' inequality :oops: . This inequality is indeed a corollary of Erdos.  :wink:

I have a different proof (not complicating and not using Erdos). By the way, your proof is very nice.

I have similar problem , I keep intact all your suppositions but change supposition "$P$ inside medial triangle" by "$P$ inside $\triangle ABC$" prove that:
$PA'+PB'+PC'+\frac{m_{a}+m_{b}+m_{c}}{\sqrt{2}}> \sqrt{2}(PA+PB+PC)$
hope you are interested in,treegoner! :)


function
inequalities
algebra unsolved
algebra

We will use $(F_{m},F_{n})=F_{(m,n)}$.
Case 1)$(n,s)=1$ or $(m,s)=1$ or $(n,r)=1$ or $(m,r)=1$
Then, $n=r$ or $m=r$ or $n=s$ or $m=s$. Therefore, there are solutions which satisfy $\{m,n\}=\{r,s\}$.
Case 2)Else
Then, $(n,s)!=1, !=n$ and $(n,r)!=1, !=n$. Therefore, $(n,s), (n,r)\<\frac{n}{2}$. By expanding the general term of the sequence, we can show that ${F_{[\frac{n}{2}]}}^{2}\le F_{n}$. However,
\[F_{n}=(F_{r}F_{s},F_{n})=(F_{(r,n)}F_{(s,n)},F_{n})\le(F_{(r,n)},F_{n})(F_{(s,n)},F_{n})=F_{(r,n)}F_{(s,n)}\le{F_{[\frac{n}{2}]}}^{2}<F_{n}\]
, which is contradiction.

[quote="Hawk Tiger"]I am sorry that I didn't read you solution carefully.
But, I think there is another solution you haven't consider
$n=1,r=2$,$m=s$
Note that $F_{1}=F_{2}=1$ :P[/quote]
Oh, sorry.  :blush: I'd just ignored the fact that when $n=1$ or $2$, the inequality doesn't hold.  :blush:


If $M\in M_{n\times n}(F)$ can be expressed as: \[
M=\left(
\begin{array}{cc}
A & B \\
0 & C \\\end{array}
\right),\]then $\text{det}(M)=\text{det}(A)\text{det}(C)$. ($A$ and $C$ are square matrices.)

Um...can you please elaborate? I have no idea what you mean.  :blush:


probability
ratio
quadratics
geometric sequence
geometric series
algebra
quadratic formula

hi i got a question for this assignment that i just cant seem to solve.

Lord Byng is playing Kits in the senior girls soccer provincials championship. The game is tied after regulation and is sent to a shootout. The rules are both teams take one shot: if one team scores and the other dosent the scoring team wins. if both score or neither score both teams take another shot. This continues until one team finally wins. If Lord Byng has a shooting % of 3/4 and Kits has a shooting % of 3/5 what is the probability Lord Byng wins?

ok so what i figured it would be like is

P(of always scoring)= (O.75 x 0.6) + (0.45)^2+ .....

since first term is 0.45 and ratio is also 0.45 sum of infinity would be (0.45/(1-0.45)) = 9/11 ? (this seems a bit too high...)

and P( of always missing)= (0.25 x 0.4) + (0.1)^2 + .....
                         = 1/9
i hope thats right... but then how do u solve for the P( of Kits winning) because there are so many paths

i was hoping to do
 P( of Lord Byng winning)= 1 - (probability of kits winning or always scoring and missing)

actually i have one more question its not on probability though

if the product of the first 3 terms of a geometric sequence is -8 and their sum is 14/3 what is the common ratio?

so i got
let a= 1st term and r= ratio
 a(ar)(ar[squared])=-8
 a + ar + ar[squared]= 14/3

i dunt really know how to solve from there

[hide="Problem 1"]To answer your first question, note that

P(Byng wins) = P(Byng wins on first round) + P(Byng wins on second round) + P(Byng wins on third round) + ...

                      = .75*.4 + (.75*.6 + .25*.4)(.75*.4) + (.75*.6 + .25*.4)^2(.75*.4) + ...

here a_1 = .75*.4 = .3 and r = (.75*.6 + .25*.4) = .55

thus, the sum of this infinite geometric series is (.3)/(1-.55) = [u][b]2/3[/b][/u][/hide]

[hide="Hint for problem 2"]
As you wrote, we have two equations:
$a\cdot ar\cdot ar^2=-8$
$a+ar+ar^2=\frac{14}{3}$

If you simplify the 1st equation, we get $a^3r^3=-8\Longleftrightarrow ar=-2$.

Then, see what happens if you divide both sides of the second equation by $ar$.
You should be able to solve for $r$ from here.
[/hide]

[hide="problem1"]
The probability of only Lord Bryng scoring is (3/4)(1-3/5)=6/20
The probability of only Kits scoring is (1-3/4)(3/5)=3/20

Since on any given round Lord Bryng is twice as likely to win, the probability of each winning is in a 2:1 ratio

So Lord Bryng wins 2/3 of the time and
Kits wins 1/3 of the time.
[/hide]

For probabilty problems in which the game continues indefinitly untill someone wins, it is usually easier to figure out the ratio of each outcome rather than coming up with infinite geometric sequences.

for problem 2
[hide]By the last hint, $ar=-2$. We substitute that into $a+ar+ar^2=14/3$ to get $a-2r=20/3$. We seem stuck, until we remember that $ar=-2$. We divide both sides by $a$ to get $a=-2/r$. We substitute that into $a-2r=20/3$ to get $-2-2r=20/3$. We multiply by $r$ to get $-2-2r^2=20r/3$. If you can't figure it out from there, read on. [hide]You get every term to the left-hand-side and multiply both sides by $-1$ to get $2r^2+20r/3+2=0$. We use the quadratic formula to get $r=(-8/3,-2/3)$[/hide][/hide] Please correct me if I made any mistakes.[/hide]

I think you made a mistake in the last step (quadratic formula).
You can factor the LHS. Try multiplying it by $\frac{3}{2}$ first to get rid of the fractional coefficient.