MATHCOUNTS
modular arithmetic
geometry
area of a triangle
Heron\u0027s formula

2004 MC Target: number 8
A collection [i]S[/i] of integers is defined by the following 3 rules:  (I) 2 is in [i]s[/i]; (II) for every [i]x[/i] in [i]S[/i], 3[i]x[/i] and [i]x[/i]+7 are also in [i]S[/i];  (III) No integers except those defined by rules (I) and (II) are in S.  What is the smallest integer greater than 2004 which is NOT in [i]S[/i]?

1992 MC Team: Number 5
One side of a triangle is divided into segments of lengths 6cm and 8cm by the point of tangency of the inscribed circle.  If the radius of the circle is 4 cm, what is the length, in centimeters, of the shortest side of the triangle? [/i]

8
[hide]It was already posted earlier. The answer is 2009 I believe because you can never get a multiple of 7.[/hide]

Doesn't number two have something to do with the fact that the opposite lines of the triangle of 6 and 8 are also 6 and 8?

5
[hide]Two tangents of a circle that meet outside of the circle are equal. We can draw the triangle and label the lengths of the tangents $x, x, 6, 6, 8$, and $8$. Drawing the radii to the tangents we make a bunch of right triangles. We add up the areas of the triangles to get:

$4x + 24 + 32 = 56 + 4x$

We set this equal to Heron's formula and get:

$56 + 4x = \sqrt {(14 + x)(8)(6)(x)} \Rightarrow x = 7$

Comes out friendly. The shortest side is then $6 + 7 = 13$.[/hide]

[quote="NoSoupForYou"][hide="First Problem"]You take $2\cdot 3^a \pmod 7$, with increasing integer values of $a$. Eventually you get (in no particular order) $1,2,3,4,5,6 \pmod 7$, but no $0\pmod7$ Therefore the answer is the first multiple of $7$ greater than $2004$ : $\bold{2009}$. Hope that explains it.[/hide][/quote]

ok, but why did you choose seven to mod?

[quote="sirorange"][hide]The first one:

Clearly, 3x will never be divisible by 7 unless x is a multiple of 7, and any 7x + 2 Where x is a whole number will ever be divisible by 7 either, and the first multiple of 7 greater than 2004 is 2009, and a brief check will show that 2005-2008 will work. so 2009[/hide][/quote]

a couple of things: first, where did you get 7x+2?
Second, you don't give a clear proof that x will never be a multiple of 7 (you say 3x is not unless x is a multiple of 7, but you don't elaborate)

[quote="seungsoo"]In the problem, 2*3^n + 7x are terms of the sequence, where n and x are non-negative.[/quote]

Where did you get this?

I will attempt to explain it better.

We note that by multiplying 2 by 3 a sufficient number of times and then adding enough 7's, we can show that many numbers are in S.  Specifically, if we can construct any value mod 7 using $2\cdot 3^n$ while keeping the numbers reasonably small, we can add some number of 7's to reach any large desired number.  Now there are several ways of showing that $2\cdot 3^n$ can be any value mod 7 other than 0.  During the actual competition, guess and check should work out fine.  Then you can show that any relatively large number that is not a multiple of 7 is in S (make sure that $2\cdot 3^n$ does not need to be too large; in this case it doesn't).  Furthermore, it is obvious that $2\cdot 3^n$ is not a multiple of 7, so $2\cdot 3^n + 7x$ is also not a multiple of 7.

I understand why any number divisible by 7 is not in the set, and also why anything except 0 mod 7 is also in the set, but there obviously there are numbers not divisible by 7 that are in the set.

[quote="probablility1.01"] Then you can show that any relatively large number that is not a multiple of 7 is in S (make sure that  does not need to be too large; in this case it doesn't). [/quote]

How do you judge if 2004 is a relatively large number? Can you prove that no number less than 2009 is in S in a practical amount of time?

Oh, I remember the first one... I got it wrong. :( I think I ran out of time, because I wasn't solving it the optimal way...

You test $2\cdot 3^n$ for $1 \le n \le 6$ to check their mods.  A good mathcountser should know that $2\cdot 3^6 < 2004$.

[quote="Secret Asian"]2004 MC Target: number 8
A collection [i]S[/i] of integers is defined by the following 3 rules:  (I) 2 is in [i]s[/i]; (II) for every [i]x[/i] in [i]S[/i], 3[i]x[/i] and [i]x[/i]+7 are also in [i]S[/i];  (III) No integers except those defined by rules (I) and (II) are in S.  What is the smallest integer greater than 2004 which is NOT in [i]S[/i]?
 [/quote]

[hide="Solution"]
1) Mod 7, 3^n.

To get any number with a remainder of 2, use 2 + 7n.
To get any number with a remainder of 6, use 2(3) + 7n.
To get any number with a remainder of 4, use 2(3)(3) + 7n.
To get any number with a remainder of 5, use 2(3)(3)(3) + 7n.
To get any number with a remainder of 1, use 2(3)(3)(3)(3) + 7n.
To get any number with a remainder of 3, use 2(3)(3)(3)(3)(3) + 7n.

Notice how you can't get a multiple of 7. [b]2009[/b].

2)
[/hide]


Let $a,b,c,d$ be non-negative real numbers. Prove that
\[a(a-b)(a-c)(a-d)+b(b-a)(b-c)(b-d)+c(c-a)(c-b)(c-d)+d(d-a)(d-b)(d-c)+abcd \ge 0.\]

\[=\sum_{c}x^{4}-\sum_{s}x^{3}y+\sum_{s}x^{2}yz-3xyzw \]
follows from applying schur on vars taken three at a time

No, me@home, you are wrong.
It follows from the following reasoning:
let $a=\min\{a,b,c,d\}$ and $b=a+x,$ $c=a+y,$ $d=a+z.$
Hence, the hungkhtn's inequality equivalent to
$a^{4}+(x+y+z)a^{3}+(xy+xz+yz)a^{2}+$
$+\left(\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)\right)a+\sum_{cyc}x^{2}(x-y)(x-z)\geq0.$ :wink:

Yes, the solution of Arqady is true and nice (after expanding, we get two Schur inequality). My solution is to use EMV method, quite the same. Anyway, I really wait for another solution without expanding and also EMV.  :blush:


Let triangle ABC. Let X,Y,Z in segment BC,CA,AB respectively such that:
AX=BY=CZ and  angle BAX = angle CBY = angle ACZ.
Prove that triangle ABC is equilateral

I tried but I fail in finding the solution,some one can help me?


limit
real analysis
real analysis unsolved

Can you please confirm that first one is a big-O and second one is a small-o ?
For now with the given hypotheses all I can prove is that $f' = O(\frac{1}{x})$ (big-o).

Yes I confirm that the first one is a big $O$ and the second is a small $o$. That's what is given on the document I found on the net.

[quote="erdos"]let $f$ $\in C^{2}\left( \mathbb R,\mathbb R \right)$ and suppose that :
$i)$ $\underset{x\rightarrow +\infty }{\lim }f\left( x\right) =0$
$ii)$ $f^{\prime \prime }=O\left( \frac{1}{x^{2}}\right)$
Show that : $f^{\prime }=o\left( \frac{1}{x}\right)$[/quote]
Consider the Taylor expansion of $f$ at a fixed $x$, in the form $f(x+hx) = f(x) + hxf'(x) + \frac{h^2x^2}{2}f''(\xi)$ with $\xi$ between $x$ and $x+hx$ and $h > 0$.

After rearranging, one obtains $xf'(x) = \frac{f(x+hx) - f(x)}{h} - \frac{hx^2}{2}f''(\xi)$. Given $\epsilon$, choose $M$ sufficiently large such that $|f(x)| < \epsilon$ for $x > M$ and then choose $h = \sqrt{\epsilon}$. Then $|xf'(x)| \le 2 \sqrt{\epsilon} + \frac{\sqrt{\epsilon}}{2}C$, where $C$ is a bound for $x^2f''(x)$. This can be made as small as we please, so $xf'(x) = o(1)$ and $f'(x) = o\left(\frac{1}{x}\right)$.


geometry
complex numbers
projective geometry

For the longest time I thought 1/:inf:=0, is that true? (and I apologize if this was discussed before)

Also how is operations on :inf: defined? (For example for all real number A which does not contain ":inf:" in it can you say :inf:+A=:inf:-A=A:inf:=:inf:/A=:inf:?)

well for 1/:inf:=0 i would think it would be true because think about it, the graph 1/x has limit 0 for y so 1/:inf: is 0. Also, you can look at patterns... 1/2>1/3>1/4>1/5>...1/:inf:... it keeps getting smaller but still greater than 0 so if you went on forever you would get there.

Mmm, but there's also the opposing argument that y=0 is an assymptote in the graph y=1/x so y can never be 0...

Also here's this weird proof I came up with which say A/:inf:=1: Let's say A is all real number. So if A/:inf:=0 then this should work for A=:inf:-1. So by substitution you get (:inf:-1)/:inf:=:inf:/:inf:-1/:inf:=1-1/:inf:. Since 1/:inf: is defined as 0 that equals 1-0=0. By transitivity A/:inf:=1...???

1/ :inf: isn't anything. It doesn't make sense. lim<sub>x--> :inf: </sub> 1/x=0, but you can't say any more than that.

You can't really say any of those, can you? The most you can say is that limx->:inf: 1/x = 0. You can't evaluate infinity.

I think one needs to remember that infinity is a concept - and :inf: is a symbol for this concept. It's not a number. So none of those operations are defined. You can say that a number A plus a very large number is still a very large number, but not much else.

ya, like :inf: = "a very big number beyond representation by hand" so A + "a very big number beyond representation by hand" yields "another very big number beyond representation by hand" which is also :inf:

Uhm, not really in my opinion, since "a very big number beyond representation by hand" is a rather vague statement. I could say that 1050 fits your description, but 1050 is nowhere near :inf:; and 1050 + A doesn't come close to :inf: either. :)

On second thought - what JBL said.

The key point:  infinity is not a number.  It's a concept.  It describes what happens when things get bigger than any real number.  Now, it makes [i]intuitive[/i] sense to say that 1/ :inf:  = 0, 1/0 =  :inf:,
:inf: + A =  :inf: for any real number A, and many others.  However, it does not make [i]mathematical[/i] sense.

Note also that there are two entirely different sets of ideas about infinity:  one is this concept of "what happens as you take a series of numbers which is increacing and has no real limit."  The other is an idea about "how big sets are."  For example, the natural numbers and the real numbers are both clearly infinite sets -- that is, they have more than any finite number of elements.  However, there are also far more real numbers than natural numbers.  (In fact, if for every natural number, you took a set the size of the natural numbers, and then counted how many total numbers you had, it would still be smaller than the number of real numbers.)  The infinite cardinalities are very important, but they still aren't "numbers."

Well the whole infinity operation thing comes into play when I'm quickly trying to figure out the range of a graph and I usually just say infinity plus/minus/to the blank power/multiplied by/divided by/etc. is still infinity and when infinity is on the denominator the whole fraction's 0 just for the simplicity of it and I wanted to see if what I'm doing is correct. Is it? (both the method and the answer I get to as the range)

or, rather, an :inf: range, huh? sorry for the double post :)

[quote="Tare"]For the longest time I thought 1/:inf:=0, is that true? (and I apologize if this was discussed before)[/quote]

There is some discussion of this issue on the Math-Teach forum

http://mathforum.org/epigone/math-teach/choupryshax

where the short answer is no. The slightly longer answer is that infinity is not a number, as other participants here have said.

[quote="paladin8"]ok, how bout "a number not displayable by any means of technology to this date," that'd be much bigger yet would include a wide range[/quote]

No.  It's not a number, not even a big one.  It isn't "displayable" by [i]any[/i] technology, ever.  And Tare, what you are doing is probably not mathematically rigorous -- however, there is likely no reasonable situation in which it would give you the wrong answer.  That is, it's not rigorous, but it works.

Note of course that there is a problem when you deal with something like x :^2:/(x - 1) in that scenario, however.

You can't do "standard" arithmatic with infinity in that way because it doesn't fit the Ordered Field Axioms that you want regular numbers to satisfy. The ordered field axioms are:

[u]Addition:[/u]
1) closure under addition
2) commutivity
3) associativity
4) existence of a zero element
5) additive inverses for every element

[u]Multiplication:[/u]
1) closure under multiplication
2) commutivity
3) associativity
4) existence of a unit element
5) multiplicative inverses for every element
6) distributive law

[u]Order:[/u]
1) x + y < x + z if y < z
2) xy > 0 if x > 0 and y > 0

The rational numbers and real numbers are both ordered fields. Can you see why?

In addition, the real numbers have a very special property that you need in order to take limits and do things like calculus. This is called the least upper bound property. I won't go into detail about what that is, but it more or less says that the real numbers don't have any "gaps" like the rational numbers do.

It's customary to sometimes add the two symbols -:inf: and +:inf: to the real numbers with the conventions: x + :inf: = +:inf: , x - :inf: = -:inf: , and x/ +:inf: = x/ -:inf: = 0, where x is any regular real number. And to leave the expressions  :inf: - :inf: ,  :inf: / :inf: , etc. as undefined. This system is called the extended real numbers, but it is not a field. Can you see why?

You can also do a similar thing with complex numbers (which are a field, but not an [i]ordered[/i] field) by adding one element called :inf: that makes the complex "plane" into a "sphere." You can make this idea more precise by using projective geometry - but again, this change makes the extended complex numbers no longer a field!!! [b]And the properties of a field are the key property that you need in order to do arithmatic[/b].

On a more philosophical level, the reason for the confusion about infinity comes mainly because we confuse the ordinality of a number with the actual number itself. What that means is, there are two different "meanings" to a number. One meaning of a number is as an element of a field (usually the ordered real numbers). This looks at the element as an abstract object that obeys the rules of arithmatic. The other meaning is the use of numbers to signify the SIZE of something. These are two different things! For example, the number "four" is not the same thing as the set {2, 3, 4, 5} which has a SIZE of four.

Often times when we think of infinity, we think of it as refering to the relative "size" of something. The technical definition of infinity is "something that is not finite." Then finite is definied as the size of something that can be up into a one to one correspondence with a set with a whole number of elements. So you see that even this technical definition of infinity refers to the relative "size" of something, not to it as a number in relation to the other real numbers.

There are, however, some theories in math that deal with arithmatic of infinite (and infinitessimal) ordinals. These are called non-standard analysis. The most famous such theory is the theory of Surreal Numbers - invented by John Conway while he was trying to understand the game of Go. Some good books you can use to learn about Surreal Numbers are:

Berlekamp, E. R.; Conway, J. H.; and Guy, R. K. [u]Winning Ways for Your Mathematical Plays, Vol. 1: Games in General.[/u]

Conway, J. H. [u]On Numbers and Games.[/u]

Conway, J. H. and Guy, R. K. [u]The Book of Numbers.[/u]

Conway, J. H. and Jackson, A. [u]"Budding Mathematician Wins Westinghouse Competition."[/u]

Gonshor, H. [u]An Introduction to Surreal Numbers.[/u]

Ebbinghause, et. al. [u]Numbers[/u]

Knuth, D. [u]Surreal Numbers: How Two Ex-Students Turned on to Pure Mathematics and Found Total Happiness.[/u]

The fourth book in that list is about a high school student, Jacob Lourie, who won the Westinghouse Competition with a project on Surreal Numbers. The last book in the list, I believe (but I haven't read it), is a fiction book that uses Surreal Numbers as a theme - sort of like [u]Flatlands[/u]. The book [u]Numbers[/u] is a great book that talks about all kinds of different number systems. Though it's somewhat advanced (probably appropriate for undergraduate level students or above).

Hi well just to through in something new to think about well we all know 0 is not negative and not positive and it goes on forever so wouldnt  :inf: be 0 since like :inf: 0 could also be seen as a theory since anything divided by 0 is not answerable one of my friends said that the answer to problems divided by 0 is :inf: so technically zero is :inf: right?

No, unfortunately that idea doesn't work because zero is defined as a number that when you add it to something the result is that something (i.e. 0 + a = a). But you can see that infinity + something wouldn't be the same something that you added.

To see in general why infinitity can't be treated at a number though, take a look at the field axioms that I listed above. Which ones would "infinity" not satisfy?

I hope nobody said this already, but theoretically :inf: + X= :inf:

That's what this entire post is about -- several people have explained why the equation you wrote is problematic.


I know that last year the floor value for the AIME was 7. How about previous years?

In the past, did everyone in 10th (or below) who earned the floor value go onto the USAMO? It looks like this year, it is limited to 80 students, and if there are more than 80 then they will factor in the AMC 10 scores, as they do with the AMC 12 participants. I am getting this information from the teachers manual on the AMC site.

Also, when they talk about the rules for 10th and below, are they talking about people who took the AMC 10, or are they including 10th (and below) who took the AMC 12?

Thanks!

10th and below includes 10th and below who took AMC12. I don't know about the other questions though.

By the way, here is last year's distribution of USAMO qualifiers by grade:
[code] 7th grade    1
 8th grade   15
 9th grade   25
10th grade   73
11th grade   61
12th grade   87
13th grade    1
     Total  263[/code]I think all the 10th graders and below who made the AIME floor value were invited to the USAMO.  I think the limit of 80 students was in the instructions from last year too, but I don't think it was enforced.

Hmm, a question, then.

If the 80-person limit is reached, then the USAMO index of each 10th grader is taken, and the 80 highest of these will reach the USAMO.  This part I understand. In the calculation of the USAMO index, however, do they only factor in the AMC 10 score, be it A, or B? Or do they factor in whichever test they did better on, be it 10A, 10B, 12A or 12B?

This news comes somewhat as a surprise to me, for if I had known that the USAMO index would be a factor for underclassmen, I would have taken the AMC10B rather than the 12, in order to maximize my chances. I really hope this "new" (is it new?) rule is enforced.

Does anyone know who the 7th grader that qualified is?

David Benjamin, Grade 7, William Henry Harrison High School, in West Lafayette, IN.

Taken from the list in official AMC site

He's also dnas on this website, but he doesn't post that often.


i think this is very strong,
i am not able to prove it  :(
Prove that for all $ x,y,z$ - positive reals -
$ \frac {x(x \plus{} y)}{z(y \plus{} z)} \plus{} \frac {y(y \plus{} z)}{x(x \plus{} z)} \plus{} \frac {z(z \plus{} x)}{y(x \plus{} y)} \plus{} \frac {8xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)} \ge 4$
or this one-
$ \sum_{cyclic} \frac {a^2}{b^2} \plus{} 8abc \ge 4$ (the two equivalent and the first is due to a nice transformation by nayel  :wink: )
the second one has the following conditions
1. $ 0 \le a,b,c \le 1$
2. $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$

Nayel's transformation is not very much helpful here atleast for me.....

Substitute $ a = cosA$, $ b = cosB$, $ c = cos C$ for an acute trinagle $ ABC$.

Lemma 1: $ 4 - 8\cos A\cos B\cos C = 4(\sum \cos^2 A )$.
Proof: This immediately follows from $ \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A\cos B\cos C$ which leads us to:
                                                      $ 4(\sum \cos^2 A ) = 4(1 - 2\cos A\cos B\cos C)$ which in turn implies
                                                      $ 4 - 8\cos A\cos B\cos C = 4(\sum \cos^2 A )$.
                                  -------------End Lemma -1-------------------

Lemma 2: $ \cos A\cos B\cos C \le \frac {1}{8}$ .
Proof:  This inequality is equivalent to proving that $ \sin (\frac {A}{2})\sin (\frac {B}{2})\sin (\frac {c}{2}) \le \frac {1}{8}$
. Now  the LHS of the above inequality can be written as $ \frac {1}{2} [ \cos({\frac {A - B}{2}) - cos(\frac {A + B}{2}) ] sin(\frac {C}{2}) = \frac {1}{2} [ \cos({\frac {A - B}{2}) - \sin(\frac {C}{2}) ] \sin(\frac {C}{2}) \le \frac {1}{2} [ 1 - \sin(\frac {C}{2})] \sin (\frac {C}{2}) = \frac {1}{2} [ \frac {1}{4} - (\frac {1}{2} - \sin\frac {C}{2})^2 ] \le \frac {1}{8}}}$. Equality holds good if $ ABC$ is equilateral.

Proof for Lemma 2 can also be given using Jensen's inequality which I am avoiding since any olympiad math problem can be solved without calculus{Pre calculus tools}. A proof using Cosine rule formulation and applying AM GM in the end is also possible.

----------------------ENd LEMMA 2-------------------

Now we are ready to solve the inequality.

By Lemma 1, it is enough to prove the inequality $ \sum (\frac {cosA}{cosB})^2 \ge 4(\sum cos^2A)$.

By the $ AM - GM$ inequality  we have

$ 2 (\frac {\cos A}{\cos B})^2 + (\frac {\cos B}{\cos C})^2 \ge \frac {3\cos ^2 A}{ (\cos ^2 A\cos ^2 B\cos ^2 C)^\frac {1}{3} } \ge 12 cos^2 A$. ( Using lemma 2 here )

Analogously writing the other two inequalties, we get

$ 3 \sum (\frac {cos A}{cos B})^2\ge 12(\sum cos^ 2 A)$and thus we get that$ \sum (\frac {cosA}{cosB})^2 \ge 4(\sum cos^2 A)$

Thus we end our proof for this inequality.


function
integration
real analysis
real analysis unsolved

Consider the sequence $f_n\in C^2([0,1])$ such that $f_n(0)=f'_n(0)=0$ and $\left|f''_n(0)\right|\leq 1$. Prove that there is a subsequence of the previos sequence which converges uniformly.

Are you sure all you have are conditions on the behavior of $f_n$ in $0$? Because I don't think that suffices. Take, for instance, $f_n$ to be any smooth function which vanishes on $\left[0,\frac 13\right]$ and takes the value $n$ on $\left[\frac 23,1\right]$. Clearly, no subsequence of such a sequence can converge uniformly.

I think that in the counterexample given by you $f_n\notin C^2([0,1])$ ...

I did say the functions were supposed to be smooth, as in infinitely differentiable, so it's even better than $C^2$. You can construct smooth functions with those properties.

Now I understand. I'll check the source of the problem.

[quote="gnosis"]I've made the correction.  :oops:

Consider the sequence $f_n\in C^2([0,1])$ such that $f_n(0)=f'_n(0)=0$ and $\left|f''_n(x)\right|\leq 1$. Prove that there is a subsequence of the previos sequence which converges uniformly.[/quote]

I think we can use there Arzela-Ascoli theorem.
First,since $f_n(x)=f_n(0)+f_n'(0)x+\frac{1}{2}f_n''(\psi)x^2=\frac{1}{2}f_n''(\psi)x^2$, so for all $n$ we have that $|f_n(x)| \leq \frac{1}{2}$ (*).

now, lets prove that the sequence $f_n$ is equicontinious. Since $f_n'(x)=\int_0^x f_n''(t)dt$, so $|f_n'(x)| \leq x$.
In the same manner $f_n(x)=\int_0^x f_n'(t)dt$, $f_n(y)=\int_0^y f_n'(t)dt$ which means that $|f_n(x)-f_n(y)|=|\int_y^x f'(t)dt| \leq \frac{1}{2}|x-y|^2$, which proves that the sequence $(f_n(x))_n$ is equicontinous and together with (*) it is bounded by $\frac{1}{2}$, now Arzela-Ascoli Theorem prove the problem assertion


The positive integers a_1, b_1, c_1, a_2, b_2, c_2 are such that
a_1+ a_2=31
b_1+b_2=32
c_1+c_2=1994

Prove that the products a_1* b_1* c_1 and a_2* b_2* c_2 are not equal.

*bump* can anyone offer some help/a solution/hints? I've spent a good amount of time thinking on the problem and Id like to know whether any of my ideas were leading in the right direction or whether all my failed approaches were nonsense.

Really a nice problem! You consider two cases: c_1=c_2=997 and c_1>c_2. The first case is trivial. In the second case gcd(c_1,c_2)<=2 because it must be a divisor of 1994. a_2b_2c_2 must contain all the prime divisors of c_1, so a_2b_2=>c_1/2 and with the same argument: a_1b_1=>c_2/2. Summing up these two inequalities you receive:
a_1b_1+a_2b_2=>(c_1+c_2)/2=997. Then you use a_2=31-a_1 and b_2=32-b_1 and get: a_1b_1+(31-a_1)(32-b_1)=>997. If I didn't make a calculation mistake it's the same as: (2a_1-31)(b_1-16)=>501. The left side is atmost (2*30-31)(31-16)=29*15=435<501 which is a contradiction.

Do you have an other solution(perhaps a shorter one), Iris?


inequalities
calculus
inequalities proposed

Given $ a, b, c \geq\ 0$. Prove that: $ \frac{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2} \geq\ \frac{ab(a\plus{}b)}{a^2\plus{}b^2} \plus{} \frac{bc(b\plus{}c)}{b^2\plus{}c^2} \plus{} \frac{ca(c\plus{}a)}{c^2\plus{}a^2}$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that: $ \frac {(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \geq\ \frac {ab(a \plus{} b)}{a^2 \plus{} b^2} \plus{} \frac {bc(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {ca(c \plus{} a)}{c^2 \plus{} a^2}$
 :)[/quote]
\[ \leftrightarrow 3 \ge \frac {ac \plus{} bc}{a^2 \plus{} b^2} \plus{} \frac {ab \plus{} bc}{a^2 \plus{} c^2} \plus{} \frac {ac \plus{} ba}{c^2 \plus{} b^2}
\]
I prove it by expanding and Schur inequality :(  (It is also reason that I don't post a complete proof
P/s:Sorry all, I also found a mistake in my calculus .Sorry!!!

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that: $ \frac {(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \geq\ \frac {ab(a \plus{} b)}{a^2 \plus{} b^2} \plus{} \frac {bc(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {ca(c \plus{} a)}{c^2 \plus{} a^2}$
 :)[/quote]
Try $ a\equal{}b\equal{}1, c\equal{}2$.

If you meant $ \frac {(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \le\ \frac {ab(a \plus{} b)}{a^2 \plus{} b^2} \plus{} \frac {bc(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {ca(c \plus{} a)}{c^2 \plus{} a^2}$
Try $ a\equal{}b\equal{}1,c\equal{}0.5$  :wink:

The following Inequalities are true for all $ a,b,c\ge0$:

[b]1/[/b]  $ \frac{(a\plus{}b\plus{}c)^3}{3(a^2\plus{}b^2\plus{}c^2)}\ge \frac {ab(a \plus{} b)}{a^2 \plus{} b^2} \plus{} \frac {bc(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {ca(c \plus{} a)}{c^2 \plus{} a^2}$

[b]2/[/b]  $ \frac {ab(a \plus{} b)}{c^2 \plus{} ab} \plus{} \frac {bc(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {ca(c \plus{} a)}{b^2 \plus{} ca}\ge a\plus{}b\plus{}c$

[quote="dduclam"]The following Inequalities are true for all $ a,b,c\ge0$:

[b]2/[/b]  $ \frac {ab(a \plus{} b)}{c^2 \plus{} ab} \plus{} \frac {bc(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {ca(c \plus{} a)}{b^2 \plus{} ca}\ge a \plus{} b \plus{} c$[/quote]
Yes, it's equivalent to $ \sum_{cyc}abc(a^2 \minus{} b^2)^2\geq0.$


1. What does mystery(35) return? What does mystery(101) return?
2. What's the complexity class of mystery()? In other words, what is $ f(x)$ if mystery(value) is $ O(f(x))$ where $ x \sim \ln(value)$?
3. Explain what the method mystery() does.
4. Explain why mystery() does what it appears to do.

[code]
long x, y, n, m, i, q, z;

public long mystery(long original)
{
		x = (int)Math.sqrt(original);
		y = (int)original / x;

		n = original - x * y;
		m = n;
		i = y - x + 1;

		if (n == 0)
		{
			return m;
		}
		for(q = 0; q <= x; q++)
		{
			if((x - q) % m == 0)
			{
				return m;
			}

			m += i;
			z = (int)((m - 1) / (x - q - 1));
			m = m - z * (x - q - 1);
			i += z + 2;
		}
}[/code]

[hide]I've figured out that mystery(num) returns an integer factor of num, but I have no idea why it does that, and I have no idea what the complexity class of mystery() is. If anyone can provide some mathematical analysis, I would greatly appreciate it. Thanks! :)[/hide]

EDIT> This should have gone into Computer Science and informatics... Oops. Sorry!

For 1., why don't you just run the program ?

Well, probably it makes sense to begin by writing down what the program is actually doing in a way that's not incomprehensible.  So, for example, we begin with input $ N$, then set $ x \equal{} \lfloor \sqrt{N}\rfloor$, $ n \equal{} \left\lfloor \frac{N}{x}\right\rfloor \equal{}  \left\lfloor \frac{N}{\lfloor \sqrt{N}\rfloor}\right\rfloor$, and then we compare $ xy$ with $ N$.  It seems to not be the case that this returns a factor of $ N$: for example, when you start with $ N \equal{} k^2$ a perfect square we end up with $ x \equal{} y \equal{} k$, $ m \equal{} n \equal{} 0$, and then it outputs $ 0$.


factorial
inequalities
number theory unsolved
number theory

Given a positive integer u, prove that the equation

n!=ua-ub have only finitely solutions (n,a,b) in positive integers.

[b]Edit: This solution is wrong. Please see the counterexample given by Myth below.[/b]

consider u>=4
let d=a-b
let p be a prime, p|(u-1)
let p^m || n!
since n!=u^b(u^d-1) and gcd(u^b,u^d-1)=1
therefore p^m || (u^d-1)
=> p^m | (u-1)d
=> d>p^m/u>=p^[n/p]/u (note:m=[n/p]+[n/p^2]+...>=[n/p])
=> n!=u^b(u^d-1)>u^d>u^{p^[n/p]/u}
The inequality above is true for finitely many values of integer n.

The proof for u=2 and u=3 is similar to above with slight modifications.

I think statement "pm | ud-1  ==> pm | (u-1)d" is false.

Myth pointed out an error in my proof. This is the corrected solution.

consider u>=3
let d=a-b
let p be a prime, p|(u-1)
let p^m || n!
since n!=u^b(u^d-1) and gcd(u^b,u^d-1)=1
therefore p^m || (u^d-1)
=> p^m | (u-1)(u+1)d
=> d>p^m/u^2>=p^[n/p]/u^2 (note:m=[n/p]+[n/p^2]+...>=[n/p])
=> n!=u^b(u^d-1)>u^d>u^{p^[n/p]/u^2}

The inequality above is true for finitely many values of integer n because the RHS is of the form x^{y^n},x>1,y>1, it grow very fast compared to LHS n!.

The proof for u=2 is similar to above with slight modifications and taking p=3.

[quote="Myth"]Will we ever see proof of your new statement
"p^m || (u^d-1) ==> p^m | (u-1)(u+1)d" ?[/quote]

This is the proof when p is odd prime. The proof for p=2 is similar with some modifications.

[b]lemma: p is odd prime, p^x||(u-1) ==> p^{x+y}||(u^{p^y}-1)

proof:[/b]
let u=kp^x+1, gcd(k,p)=1
u^{p^y}-1
=(kp^x+1)^{p^y}-1
=kp^yp^x+binom(p^y,2)(kp^x)^2+...
every term other than kp^xp^y is divisible by p^{x+y+1}
therefore p^{x+y}||(u^{p^y}-1)

[b]p^m||(u^d-1) ==> p^m||(u-1)d => p^m|(u-1)(u+1)d

proof:[/b]
let p^x||(u-1)
let p^y||d
let d=p^y.t, gcd(p,t)=1
let v=u^{p^y}

by lemma, p^{x+y}||(v-1)
v^{t-1}+...+v+1 is not divisible by p because v^{t-1}+...+v+1==1+...+1+1==t(mod p) and gcd(t,p)=1
therefore p^{x+y}||v^t-1=u^d-1
compare with p^m||(u^d-1), we have m=x+y

therefore p^x||(u-1),p^y||d ==> p^{x+y}||(u-1)d ==> p^m||(u-1)d ==> p^m|(u-1)(u+1)d

Are the Chinese TST 2004 published anywhere? If not, can anyone publish them?

recalling mihai manea's favorite fact:

If p is an odd prime and a and b are two different integers a=b(mod p) and n is a positive integer then the exponent of p in a^n-b^n is equal to the sum of the exponent of p in a-b and the exponent of p in n.

So the only case left is p=2 and i would like to see a proof of the lemma in this case, because for odd primes it is inmediately from mihai's favorite math fact!

[quote="pascual2004"]recalling mihai manea's favorite fact [...]
[/quote]
hey, how did you know that ?! :) do you know him? :P

There are two possible cases, n is odd and n is even..

[b]If a is odd integer and n>0 is even integer and 2^m||(a^n-1), then 2^m||(a-1)(a+1)(n/2).[/b]

There are more than one ways to prove this.

One way is to let a^2=k.2^x+1, the rest of the proof is like the proof I offered above (for odd prime p).

Another way is to factorize. For example, take n=16.
a^16-1=(a-1)(a+1)(a^2+1)(a^4+1)(a^8+1)
since 2||(a^2+1), 2||(a^4+1), 2||(a^8+1)
therefore 2^m||(a^16-1) ==> 2^m||(a-1)(a+1)(16/2)

[b]If a is odd integer and n>0 is odd integer and 2^m||(a^n-1), then 2^m||(a-1).[/b]

This case is easy.

Combine all three cases (two cases for p=2 and one for odd prime p), the following result is obtained. I use this result to prove the original problem.
p|(a-1),p^m||(a^n-1) ==> p^m|(a-1)(a+1)n

I don't know if the generalization is also true for p=2, you have to check it.

JEJE, No, i dont, but i would like to! at least i Have read he is a great problem solver....I found it in the imo 2001 site, were the contestants had a space to publish their info including favorite math fact....... :D  :D  :D

PD: Valentin, yours is empty, so you should check it out, i think it would be pretty cool to have that kind of stuff on every imo :cool:

[quote="pascual2004"]PD: Valentin, yours is empty, so you should check it out, i think it would be pretty cool to have that kind of stuff on every imo :cool:[/quote]
yeah, well at that time I was pretty busy buying nike basketball shoes :D ;)

I have found generalization also works for 2, that is:
if a and b are odd, and repalce u by a and 1 by b, so i think it is very interesting


There are a white, b black, and c red chips on a table. In one step, you may choose two chips of different colors and replace each one by a chip of the third color. Find conditions for all chips to become the same color.

The solution offered in the book says that a+b+c = 0 (mod 3) and a-b, b-c, a-c = 0 (mod 3) are the requirements, but I've come up with counterexamples to this so it doesn't appear to be correct.

What are your counterexamples?

Clearly $a+b+c \equiv 0 \mod 3$ is a necessary requirement, since the sum of the chips does not change under the transformation given, and if all chips are eventually of the same colour then $a=b=c$ and $a+b+c=3a \equiv 0 \mod 3$.

It can be shown that $a-b$ is invariant mod 3 under the transformation (and similarly $b-c$ and $a-c$) since either both $a$ and $b$ decrease by 1 or one decreases by 1 and the other increases by 2. Either way, the difference mod 3 does not change. If, in the end, $a=b=c$, $(a-b) \equiv 0 \mod 3$ then it must be to begin with as well.

edit:... oh, yes- I misread the question

I don't think that it is necessary for $a=b=c$. Take $a=2, b=1, c=4$. We trade a and c for 2 b's, so now we have $(a,b,c)=(1,3,3)$, and it's clear from here that we can just trade b's and c's for a's. An even more basic example: just take $b=c$, and we can let a be whatever we want.

The answer it gives is the answer to a different problem then the problem you wrote.  Go back and check the question: are you sure it says "for all chips to be the same color" and not "for each color to have the same number of chips"?

Literally, it says, "...Find conditions for all chips to become of the same color."

Yeah, [b]Phelpedo's [/b]statement of the problem is indeed correct.

[quote="JBL"]So, it is true that [i]one[/i] of the differences must be $\equiv 0 \pmod 3$, because of the second half of tim's argument (or else you'll never get two piles both congruent mod 3, which is certainly required to get two piles equal to 0).  (Note that he's solving the problem I suggested: getting each pile to the same size, rather than emptying two piles.  And he correctly shows that the condition you gave is the condition for that question.)

In fact, I believe this condition is good enough: the answer is $(a-b)(b-c)(c-a) \equiv 0 \pmod 3$.  All you have to do is show that condition is sufficient, which takes a teensy bit of casework: you just have to show how to shift 3 chips from one pile to another pile.  Then you can reduce the difference between the two piles which differ by a multiple of 3 to 0, and after that you're golden.[/quote]

For the problem I posted, the condition that one of the differences is 0 mod 3 is indeed both sufficient and necessary.


why doesnt LaTex appear in transcripts for me?

I don't know.  What do you see?  What browser are you using?

well LaTex appears in half and doesnt appear in the other half.  I use internet explorer

What version?  It works on IE 6 for Windows and IE 5.2 for Mac.


geometry
geometric transformation
reflection
trigonometry
geometry unsolved

Let $\ A_{1},B_{1},C_{1}$ be poins respectively on the sides $\ BC,CA,AB$ of a triangle $\ ABC$. Let $\ A_{2},B_{2},C_{2}$ be poins respectively reflection of $\ A,B,C$ in the midpoint of $\ B_{1}C_{1},C_{1}A_{1},A_{1}B_{1}$. Prove that:
1) $\ AA_{1},BB_{1},CC_{1}$ are cocurrent if only if $\ AA_{1},BB_{2},CC_{2}$ are cocurrent.
2) $\ S(A_{1}B_{1}C_{1})=S(A_{2}B_{2}C_{2})$ with $\ S(XYZ)$ is the area of the triangle $\ XYZ$.

The first part is easy! Just note that if we let $C_{3}=CC_{2}\cap A_{1}B_{1}$ then from the sine law applied to the triangles $\triangle A_{1}CC_{3}$ and $\triangle B_{1}CC_{3}$ we obtain $\frac{\sin\angle C_{2}CB_{1}}{\sin\angle C_{2}CA_{1}}=\frac{\sin\angle C_{3}CB_{1}}{\sin\angle C_{3}CA_{1}}=\frac{CB_{1}}{CA_{1}}$. The product of this equality with two its analogues for another vertices of $ABC$ gives the desired equivalence since the left hand side would be a trig-Ceva product for the lines $AA_{2}, BB_{2}, CC_{2}$ and the right hand side would be a usual Ceva product for the lines $AA_{1}, BB_{1}, CC_{1}$.


algebra
polynomial
number theory proposed
number theory

Let $ n$  be an integer  greater than 2, and let $ p$ be a prime . Prove that polynomial
$ P(x)\equal{} x^n \plus{} p^2 x^{n\minus{}1}\plus{}...\plus{} p^2 x\plus{}p^2$ cannot be written as the product of two polynomials with integer coefficients.

[quote="Rust"]Eisenshtein theorem.[/quote]
Do you mean Eisenstein criterion?If so,then you are wrong, Eisenstein criterion works only if $ p^2$ does not divide $ a_0$.

For Eisenshtein theorem $ p|a_1,a_2,...,a_{n\minus{}1}, \ p^2\not a_0, p\not |a_n$.
But prove for these case $ p^2|a_1,...,a_{n\minus{}1}, p^3\not|a_0, p\not |a_n$ is same.

[quote="Rust"]For Eisenshtein theorem $ p|a_1,a_2,...,a_{n \minus{} 1}, \ p^2\not a_0, p\not |a_n$.
But prove for these case $ p^2|a_1,...,a_{n \minus{} 1}, p^3\not|a_0, p\not |a_n$ is same.[/quote]
Eisenstein criterion:
If $ p\in\mathbb{P}$ and $ p$ divides $ a_0,a_1,a_2,\dots,a_n$,but $ p\not |a_n$ and $ p^2\not | a_0$ then polynomial
$ P(x) \equal{} a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \dots \plus{} a_1x \plus{} a_0$ is irreducible over $ \mathbb{Z}[x]$.
But in your case $ p^2$ is not a prime.


function
inequalities
real analysis
real analysis unsolved

Let $ f: \mathbb{R} \to \mathbb{R}$ a function with the property of Darboux. We know also that $ f$ is increasing on $ \mathbb{R} \setminus \mathbb{Q}$. Prove that $ f$ is continous.

This is rather inelegant but I'm rusty  :blush:

Suppose by contradiction (ie. that f is not monotonic) there exists x < y for which f(x) > f(y), where x is rational (and y is therefore irrational).

Both hypotheses on f along with the inequality f(x) > f(y) implies that f maps the rationals in ]x, y[ onto the open interval ]f(x) < f(y)[. Wrong cardinality for both sets.

Suppose that there are $ a \in \mathbb{Q}$ and $ b \in \mathbb{R} \setminus \mathbb{Q}$ such that $ a<b$ but $ f(a)>f(b)$, since $ f$ has the Darboux property, if follows that there is $ \epsilon >0$ such that $ f(x)>f(b)$ , $ \forall x \in (a\minus{}\epsilon, a\plus{}\epsilon)$.But $ \mathbb{R} \setminus \mathbb{Q}$ is dense in $ \mathbb{R}$, therefore we can always find $ c \in \mathbb{R} \setminus \mathbb{Q}$ such that $ c \in (a\minus{}\epsilon, a)$ (which implies $ c <b$). So $ f(c)>f(b)$ contradiction.

When you deal with the case $ a,b$ are both in $ \mathbb{Q}$, you can always find $ m \in \mathbb{R} \setminus \mathbb{Q}$ such that $ a<m<b$, and work the same as above.

Language note: it's a property. "Propriety" means something entirely different.


I have the following problem: "List the 16 different relations on the set {0, 1}."
Does anyone have a clue what they mean?

It doesn't make sense for me... :?
What is the context for this problem?

No it doesn't make sence for me eighter. Its all what is written in the problem. I will findout what they mean and post it here, if anyone does not find  it out before me.

It must mean binary relations: subsets of $S\times S$.

[hide]You find the number of relations on a set A by taking 2^n^2, where n is the number of elements in the set (thus giving you 16.) The n^2 represents the #(AXA).

A set R is a relation on A iff R is a subset of AXA.

 The relations are:

the empty set;
(0,0);
(0,1);
(1,0);
(1,1);
(0,0), (0,1);
(0,0),(1,0);
(0,0), (1,1);
(0,1), (1,0);
(0,1), (1,1);
(1,0), (1,1);
(0,0), (0,1), (1,0);
(0,0), (0,1), (1,1);
(0,0), (1,0), (1,1);
(0,1), (1,0), (1,1);
(00), (0,1), (1,0), (1,1).

Sorry this is so long. It's a lot easier to read when everything is on a different line.

Hope this helps![/hide]

[quote="amin_b"]This problem is asking for for following relations:

EmptySet
{00},{01},{10},{11}
{00,01},{00,10},{00,11},{01,01},{01,10},{01,11}
{01,10,11},{00,10,11}...

and so on. You will end up with 16 relations.

Hope that helps!![/quote]
i think you are right amin_b. :D


just a doubt................
How do you prepare DMSO and where do u use it????????

as a biproduct ozonolysis using dimethyl sulphide instead of zinc,as a polar aprotic solvent

Or by oxidizing dimethyl sulphide with other more readily available oxidizing agent. Dimethyl sulphide can in turn be prepared (if necessary) with methyl iodide and sulphide ion.


algebra
polynomial
inequalities
Putnam
search
algorithm
superior algebra

Determine all polynomials $\sum_{k=0}^{n}a_kx^k$ with $a_k=\pm 2$ or $a_k=\pm 1$ ($0\leq k \leq n , 1\leq n <\infty $ ) such that each has only real zeros

I proved that $n\leq 8$ for any such polynomial, so the problem is reduced to a finite number of cases (not too big) and I really think the computer should help us.

I considered they are too simple to be presented.  :P . They are just trivial remarks: let $x_1,...,x_n$ be the roots of the polynomial $ a_0+...+a_n x^n$ with coefficients $1,-1,2,-2$. We have $ \sum_{i=1,n}{x_i^2}=\frac{a_{n-1}^2-2a_{n-2}a_n}{a_n^2}$. We have two cases:
  Case 1: if $ |a_n|=2$, then we obtain $\sum_{i=1,n}{x_i^2}\leq 3$ and also $\sum_{i=1,n}{x_i^2}\geq n\cdot\sqrt[n]{(x_1...x_n)^2}$, from where we find that $ (3/n)^n\geq 1/4$ which does not hold for n greater than 7.
  Case 2: $|a_n|=1$. Then we find that $\sum_{i=1,n}{x_i^2}\leq 8$ and from the AM-GM inequality we also have $\sum_{i=1,n}{x_i^2}\geq n$ from where the conclusion.

haha, "the computer should help us"...

it's a putnam problem though; you get no computers there.  and they especially had no *good* computers back in 1968.

I think you did not understand. This is not a Putnam problem, it is an open question lanced by Moubi. The problem from Putnam I can solve without help of computer.  :P

Moubi, I really have more interesting things to do than to make a program to find these polynomials.  ;)  :P

so impolite...

I think we can discuss here how to check polynomials for real roots only.
Any ideas? Maybe there are nice theorems devoted to it?

My initial idea was to use the well known inequality $\sqrt[k]{\frac{\sigma_k}{C_n^k}}\geq\sqrt[k+1]{\frac{\sigma_{k+1}}{C_n^{k+1}}}$. Indeed, we need to "shift" all roots firstly in order they became positive.
I tried it in this problem to establish estimate on $n$, but I failed. That's why I didn't see the simpliest harazi's idea. :(
Anyway, it can be a simple test for real roots to check it with a computer.

No, Myth, it is not impolite to say that you have more interesting things to do than to write a program and check 1500 polynomials. But anyway, I will not launch another useless discussion on this theme.

[quote="Moubinool"][quote="harazi"] $n\leq 8$ I really think the computer should help us.[/quote]
Here you can use computer. Did you find  the list of polynomials ?[/quote]
I gave a task to my students to write a computer program which search for all such polynomials (note that I read a course on advanced computer algorithms in my University. So I made it as an extra task for my students ;) ). The program found such polynomials only for $n\leq 4$. No other powers are suitable.
I hope the program is correct.

[quote="Myth"]
I gave a task to my students to write a computer program which search for all such polynomials. The program found such polynomials only for $n\leq 4$. No other powers are suitable.
[/quote]

Could we have the list for polynomial degree less 4 ?

BTW, I sent books 1 month ago. Have you received them? :?

Polynomials with positive leader coefficients of degree 3 and 4.

------------ DEGREE 4 -----------------
x^4+2*x^3-2*x^2-2*x+1=0
Корень: -2.414, кратность: 1
Корень: -1.000, кратность: 1
Корень: 0.414, кратность: 1
Корень: 1.000, кратность: 1

x^4+x^3-2*x^2-x+1=0
Корень: -1.618, кратность: 1
Корень: -1.000, кратность: 1
Корень: 0.618, кратность: 1
Корень: 1.000, кратность: 1

x^4-x^3-2*x^2+x+1=0
Корень: -1.000, кратность: 1
Корень: -0.618, кратность: 1
Корень: 1.000, кратность: 1
Корень: 1.618, кратность: 1

x^4-2*x^3-2*x^2+2*x+1=0
Корень: -1.000, кратность: 1
Корень: -0.414, кратность: 1
Корень: 1.000, кратность: 1
Корень: 2.414, кратность: 1

---------- DEGREE 3 ----------------

2*x^3+2*x^2-x-1=0
Корень: -1.000, кратность: 1
Корень: -0.707, кратность: 1
Корень: 0.707, кратность: 1

2*x^3+2*x^2-2*x-1=0
Корень: -1.452, кратность: 1
Корень: -0.403, кратность: 1
Корень: 0.855, кратность: 1

2*x^3+2*x^2-2*x-2=0
Корень: -1.000, кратность: 2
Корень: 1.000, кратность: 1

2*x^3+x^2-2*x-1=0
Корень: -1.000, кратность: 1
Корень: -0.500, кратность: 1
Корень: 1.000, кратность: 1

2*x^3-x^2-2*x+1=0
Корень: -1.000, кратность: 1
Корень: 0.500, кратность: 1
Корень: 1.000, кратность: 1

2*x^3-2*x^2-x+1=0
Корень: -0.707, кратность: 1
Корень: 0.707, кратность: 1
Корень: 1.000, кратность: 1

2*x^3-2*x^2-2*x+2=0
Корень: -1.000, кратность: 1
Корень: 1.000, кратность: 2

2*x^3-2*x^2-2*x+1=0
Корень: -0.855, кратность: 1
Корень: 0.403, кратность: 1
Корень: 1.452, кратность: 1

x^3+2*x^2-x-1=0
Корень: -2.247, кратность: 1
Корень: -0.555, кратность: 1
Корень: 0.802, кратность: 1

x^3+2*x^2-x-2=0
Корень: -2.000, кратность: 1
Корень: -1.000, кратность: 1
Корень: 1.000, кратность: 1

x^3+2*x^2-2*x-1=0
Корень: -2.618, кратность: 1
Корень: -0.382, кратность: 1
Корень: 1.000, кратность: 1

x^3+2*x^2-2*x-2=0
Корень: -2.481, кратность: 1
Корень: -0.689, кратность: 1
Корень: 1.170, кратность: 1

x^3+x^2-x-1=0
Корень: -1.000, кратность: 2
Корень: 1.000, кратность: 1

x^3+x^2-2*x-1=0
Корень: -1.802, кратность: 1
Корень: -0.445, кратность: 1
Корень: 1.247, кратность: 1

x^3+x^2-2*x-2=0
Корень: -1.414, кратность: 1
Корень: -1.000, кратность: 1
Корень: 1.414, кратность: 1

x^3-x^2-x+1=0
Корень: -1.000, кратность: 1
Корень: 1.000, кратность: 2

x^3-x^2-2*x+2=0
Корень: -1.414, кратность: 1
Корень: 1.000, кратность: 1
Корень: 1.414, кратность: 1

x^3-x^2-2*x+1=0
Корень: -1.247, кратность: 1
Корень: 0.445, кратность: 1
Корень: 1.802, кратность: 1

x^3-2*x^2-x+2=0
Корень: -1.000, кратность: 1
Корень: 1.000, кратность: 1
Корень: 2.000, кратность: 1

x^3-2*x^2-x+1=0
Корень: -0.802, кратность: 1
Корень: 0.555, кратность: 1
Корень: 2.247, кратность: 1

x^3-2*x^2-2*x+2=0
Корень: -1.170, кратность: 1
Корень: 0.689, кратность: 1
Корень: 2.481, кратность: 1

x^3-2*x^2-2*x+1=0
Корень: -1.000, кратность: 1
Корень: 0.382, кратность: 1
Корень: 2.618, кратность: 1

Thank's your students  for this list, Myth

Hmm... The list seems incomplete: how about $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2$ for example?  :?

You are right. I don't know why the program didn't catch it :? Though it search for roots of high orders.
BTW, it is not a first mistake in this program. I checked 8 variants of it. And everytime there was some mistakes. So my students will have to do an additional work to correct it finally.


Hey, who here is doing the ECOO on saterday? Alex and I will be there again this year.

Its a team programing competition for schools in ontario.

Teams of four, one computer, four questions, three hours. points for question answered and points for early finishes.

Its a good deal of fun, becuase of the group aspect. There are three rounds, place top 9 or 10 in each round, proceed to the next.

Oh, and they give free pizza and pop at every level!


Find the exact values of sin(u/2), cos(u/2), and tan(u/2) using the half angle formulas and given that:

$ \tan{u} \equal{} \frac{5}{8}$
and
$ \pi < u < \frac{3\pi}{2}$ (u is in 3rd quadrant)

[hide="HINT"]
Find the hypotenuse of the triangle.  Use that triagle to find the values of the trig-functions.  Substitute those values into the formulæ for half angles.
[/hide]


Given parallelogram $ABCD$, we choose $P$ on $AD$ and $Q$ on $BC$ randomly. $AQ$ and $BP$ meet at $X$ and $PC$ and $QD$ meet at $Y$. Prove that $\overleftrightarrow{XY}$ divides parallelogram $ABCD$ into two equal areas (regardless of the locations of $P$ and $Q$).

(The result is obvious from [url=http://mathworld.wolfram.com/PappussHexagonTheorem.html]Pappus's Hexagon Theorem[/url], but is there any way to prove this without using the theorem?)

basically you are asking for us to prove that theorem...

i suggest vectors...the parallel sides really do not matter...(how do you show collinearity with vectors?  ;) )


geometry
combinatorics solved
combinatorics

Suppose D_1, D_2, ..., D_n are closed disks lying in a plane, such that every point in the plane is covered by at most 2003 disks. Prove that there exists a disk which intersects no more than 14020 other disks.

If a disk D_i intersects 14021 disks or more such that no more than 2003 meet at a point, one of those disks is smaller than D_i.
Proof:
    Lemma 1: If 7n disks larger than D_i intersect D_i, there is a point covered by n+1 of these 7n disks.

14021 = 7*2003, so if D_i touches 14021 disks larger than it, there is a point covered by 2004 disks, and since only 2003 disks can cover a given point there must be a smaller disk touching each disk. This is a contradiction because n is finite so there must be a smallest disk.

I'll prove the lemma soon.

OK this was eluding me for a while but now I've got it.
Proof of Lemma 1:
If we have 7n disks no smaller than D_i overlapping D_i, we want to find the the maximum amound that cover soome point, and show this is greater than n. By shrinking the disks, we can only decrease this maximum. If all of the disks are the same size as D_i, we still have the maximum > n because we must choose centers for these disks: either inside D_i or between D_i and a concentric circle with twice D_i's radius. We can divide that area into 6 parts, each a section of the circle with angle 60, and D_i is a seventh part. If we split up the 7n disks into these 7 sections, one section must have at least n disks, and the midpoint of the 60 degree arc of the larger circle is covered by all ther circle centered in that section. However, it is impossible for each section to have exactly n disks since the disks in the outer sections would be tangent to the disk in D_i. Therefore, there must be a point with n+1 of these larger disks covering it.


[quote="babyloverain"]Given a,b,c>0 and abc=1.Prove that:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \ge a \plus{} b \plus{} c \plus{} \frac {(a^2c \plus{} b^2a \plus{} c^2b \minus{} 3)^2}{ab \plus{} bc \plus{} ca}$[/quote]
It is equivalent to $ \sum_{cyc}(a^3\minus{}3a^2b\plus{}6a^2c\minus{}4abc)\geq0,$ which is obvious.

It is also equivalent to the following:

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

which is Andrescu lemma (it's no matter whether $ \frac{a\minus{}b}{b}$, etc. are negative or positive.) :lol:


One hundred people were surveyed. Of these, $ 87$ indicated they liked Mozart and $ 70$ indicated they liked Bach. What is the minimum number of people surveyed who could have said they liked both Mozart and Bach?

The minimum occurs when everyone likes at least one of them, so we have $ 87\plus{}70\minus{}x\equal{}100 \implies x\equal{}\boxed{57}$.


Prove or disprove:

$a,b,c,t>0$

$\frac{a^2+tb^2}{ta+b}+\frac{b^2+tc^2}{tb+c}+\frac{c^2+ta^2}{tc+a}\ge a+b+c$

Use Cauchy's inequality, since $a,b,c,t>0$

$\frac{a^2}{ta+b}+\frac{b^2}{tb+c}+\frac{c^2}{tc+a}\geq \frac{(a+b+c)^2}{(ta+b)+(tb+c)+(tc+a)}=\frac{(a+b+c)^2}{(t+1)(a+b+c)}$

$=\frac{a+b+c}{t+1}\ \cdots [1].$

Similarly we have $\frac{tb^2}{ta+b}+\frac{tc^2}{tb+c}+\frac{ta^2}{tc+a}\geq \frac{t}{t+1}\cdot (a+b+c)\ \cdots [2].$ Adding the equations $[1],[2]$

yielding $\frac{a^2+tb^2}{ta+b}+\frac{b^2+tc^2}{tb+c}+\frac{c^2+ta^2}{tc+a}\geq a+b+c.\ Q.E.D.$

[quote="conceive"]That's amazing!! Good work kunny. Anyway we usually call it Cauchy-Schwarz  ;) not Caushy-Schwalz's .. :|[/quote]

I made a typo. Thank you!

kunny

$\frac{a^2}{ta+b}+\frac{b^2}{tb+c}+\frac{c^2}{tc+a}\geq \frac{(a+b+c)^2}{(ta+b)+(tb+c)+(tc+a)}$
sorry, can you explain this part? cause im not familiar with cauchy schwarz and i've only seen the application of two variables.. i mean can you explain what you used in your $a_k$, $b_k$ values for the inequality:
$\left(\sum_{k=1}^{n}{a_kb_k}\right)^2\le \left(\sum_{k=1}^{n}{a_k^2}\right)\left(\sum_{k=1}^{n}{b_k^2}\right)$

All right, minsoens. :)

For[b] positive numbers [/b]$a,b,c,x,y,z$ use Cauchy-Schwalz's inequality for three variables $x,y,z,$
we have $\{(\sqrt{x})^2+{(\sqrt{y})^2+\{(\sqrt{z})^2\}\left\{\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2+\left(\frac{c}{\sqrt{z}}\right)^2\right\}\geq \left (\sqrt{x}\cdot \frac{a}{\sqrt{x}}+\sqrt{y}\cdot \frac{b}{\sqrt{y}}+\sqrt{z}\cdot \frac{c}{\sqrt{z}}\right)^2,}$ yielding

\[ \boxed{\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\geq \frac{(a+b+c)^2}{x+y+z}} \]

oh.. :blush: thanks
i guess it would be possible to get the similar result for any number of variables?

[quote="kunny"]Use Cauchy-Schwalz's inequality, since $a,b,c,t>0$

$\frac{a^2}{ta+b}+\frac{b^2}{tb+c}+\frac{c^2}{tc+a}\geq \frac{(a+b+c)^2}{(ta+b)+(tb+c)+(tc+a)}=\frac{(a+b+c)^2}{(t+1)(a+b+c)}$

$=\frac{a+b+c}{t+1}\ \cdots [1].$

Similarly we have $\frac{tb^2}{ta+b}+\frac{tc^2}{tb+c}+\frac{ta^2}{tc+a}\geq \frac{t}{t+1}\cdot (a+b+c)\ \cdots [2].$ Adding the equations $[1],[2]$

yielding $\frac{a^2+tb^2}{ta+b}+\frac{b^2+tc^2}{tb+c}+\frac{c^2+ta^2}{tc+a}\geq a+b+c.\ Q.E.D.$[/quote]
In fact, you've used the Inequality of Bergstrom, but it's the same.


[This is a very interesting problem，I hope you can enjoy it.] In triangle ABC (a=BC；b=AC；c=AB); M  is the midpoint of side AC. Prove that inscribed circle of triangle divides segment BM into there segments that are the same length if and only if : a/5=b/10=c/13 :roll:

$ CT < CM \Longleftrightarrow \frac{_{a \plus{} b \minus{} c}}{^2} < \frac{_b}{^2} \Longleftrightarrow a < c.$ Then $ BP \equal{} MQ \Longleftrightarrow a \equal{} BC \equal{} CM \equal{}\frac{_b}{^2}.$
$ MQ \equal{} QP \equal{} \frac{_1}{^3}MB \Longleftrightarrow$ $ MT^2   \equal{} MQ \cdot MP \equal{} \frac{_2}{^9} MB^2 \Longleftrightarrow$ $ \frac{_1}{^4} (c \minus{} a)^2 \equal{} \frac{_2}{^9} \cdot \frac{_1}{^4}(2c^2 \plus{} 2a^2 \minus{} b^2) \equal{} \frac{_1}{^{9}}(c^2 \minus{} a^2) \Longleftrightarrow$
$ 9(c\minus{}a) \equal{} 4(c\plus{} a) \Longleftrightarrow \frac{_c}{^{13}} \equal{}  \frac{_a}{^5} \equal{}  \frac{_b}{^{10}}$

it seems the indians have generated a problem from another olympiad to pose it in there national olympiad. is this valid?

I am sorry :blush: . I saw this problem on a book , maybe the author of the book has made a mistake.


geometry
circumcircle
symmetry
real analysis
geometry solved

Let A'B'C'D'E' be a convex pentagon and suppose B'C', D'E' meet at A; C'D', E'A' meet at B; etc, so we have a convex pentagon ABCDE obtained from extending the sides of A'B'C'D'E'.

Prove that the second intersections of the neighboring pairs of circumcircles of AC'D' , BD'E', CE'A', DA'B', EB'C' intersect at 5 concyclic points.

This is probably one of my favorites :)

Solution:

Let the circumcircles of $ DA'B'$ and $ CA'E'$ meet at $ P$.  Similarly, let the pairs of circumcircles which meet at $ B',C',D',E'$ have second intersections $ Q,R,S,T$ respectively.

First, let's establish the fact that the second intersections of the circumcircles are outside $ A'B'C'D'E'$.  Let $ X$ and $ Y$ be the circumcenters of $ \triangle DA'B'$ and $ \triangle CA'E'$.  It will suffice to show that

$ \angle XA'B' \plus{} \angle B'A'E' \plus{} \angle YA'E' > \pi$
$ \iff \frac {1}{2}(\pi \minus{} \angle B'XA') \plus{} \angle B'A'E' \plus{} \frac {1}{2}(\pi \minus{} \angle A'YE') > \pi$
$ \iff \angle B'A'E' > \frac {1}{2}(\angle B'XA' \plus{} \angle A'YE') \equal{} \angle B'DA' \plus{} \angle A'CE'$

which follows from (using exterior angles)

$ \angle B'A'E' \equal{} \angle B'DA' \plus{} \angle DB'C$
$ \equal{} \angle B'DA' \plus{} (\angle B'AC \plus{} \angle A'CE') > \angle B'DA' \plus{} \angle A'CE'$

Now, we show that the points $ A',Q,T,B,E$ are concyclic.  Note that

$ \angle EQA' \equal{} \angle EQB' \plus{} \angle B'QA'$
$ \equal{} (\pi \minus{} \angle EC'B') \plus{} \angle B'DA'$
$ \equal{} \pi \minus{} \angle EBA'$

because $ \angle EC'B' \equal{} \angle B'DA' \plus{} \angle EBA'$ again by exterior angles.  Thus, $ Q$ lies on the circumcircle of $ A'BE$ and by a similar argument, $ T$ lies on the same circle.

Again, by analogy, we have that the points $ A,B',C,R,P$ are also concyclic.

Finally, we will show that $ \angle QRP \equal{} \angle QTP$, which will show that $ P,Q,R,T$ are concyclic, and by symmetry, $ Q,P,T,S$ will also be concyclic and the result will follow.  We have

$ \angle QRP \equal{} \angle QRB' \plus{} \angle B'RP$
$ \equal{} \angle QEB' \plus{} \angle PCA'$
$ \equal{} \angle QTA' \plus{} \angle PTA' \equal{} \angle QTP$

and we're DONE!

(Wow that write-up was longer than I expeced... on second thought, maybe I should've posted this in the Geometry forum  :blush: )

[quote="ThAzN1"]
(Wow that write-up was longer than I expeced... on second thought, maybe I should've posted this in the Geometry forum  :blush: )[/quote]

I'll move it for you ;)

Here is my solution (which is similar to ThAnz1 's, but as I use some interesting theorems, so I decide to post it)
  [b]Theorem 1[/b] (Miquel 's)
  Suppose $a, b, c, d$ are four lines on a plane. Denote by $abc$ the triangle formed by $a, b, c$. Analogously, we define $bcd, cda, abd$. Then the circumcircles of $abc, bcd, cda, abd$ are concurrent.
  [b]Theorem 2[/b] (Lebesgue 's)
  Let $ABCD$ be a convex cyclic quadrilateral. Let $(a), (b), (c), (d)$ be four circles such that $A$ and $B$, $B$ and $C$, $C$ and $D$, $D$ and $A$ respectively lie on $(a), (b), (c), (d)$. If $A', B', C', D'$ are the second intersections of $(a)$ and $(b)$, $(a)$ and $(b)$, $(c)$ and $(d)$, $(d)$ and $(a)$, then $A'B'C'D'$ is cyclic.
   Back to the problem, let $A_1, B_1, C_1, D_1, E_1$ be the second intersections of $(A'E'C)$ and $(A'B'D)$, $(A'B'D)$ and $(B'C'E)$, $(B'C'E)$ and $(C'D'B)$, $(C'D'B)$ and $(D'E'B)$, $(D'E'B)$ and $(E'A'C)$. By applying theorem 1 to four lines $B'A', A'E', E'D', D'C'$, we obtain that $C, E, A', E_1$ lie on a circle. Similarly, we obtain $C, E, A', C_1$ lie on a circle. Hence $CE_1C_1E$ is cyclic.
   Consider 4 points $C, A', B', E$ which are collinear in that order (notice that here we can consider $CA'B'E$ is a cyclic quadrilateral). By applying theorem 2 with circumcircles $(CA'E_1A_1), (A'B'A_1B_1), (B'EB_1C_1), (ECC_1E_1)$, we obtain $E_1A_1B_1C_1$ is cyclic. In the same manner, we have $A_1B_1C_1D_1$ is cyclic. Hence 5 points $A_1, B_1, C_1, D_1, E_1$ are on a circle.

[url=http://www.math.ust.hk/excalibur/v6_n1.pdf]Here (Mathematical Exaclibur 2001 (vol. 6), no. 1, p. 1-2)[/url] you will find a discussion of this problem and a nice example of political misuse of mathematics...

  Darij

In "Excalibur 2001 (vol 6)" say this problem was posed by the President Jiang Zemin to the student in a meeting. It is a charming action !


What is the least positive three-digit multiple of 7?

you don't need to divide it, the guess of the solution is possible.

$ 100\minus{}70\equal{}30$, $ 28<30<35$
So $ 70\plus{}35\equal{}\fbox{105}$

[quote="isabella2296"]The smallest 3-digit integer is 100. When we divide this by 7, we get about 14. Thus our answer is $ 15 \times 7 \equal{} \boxed{105}$.[/quote]

What is did (in equation) is this:

$ \lceil(\frac {100}{7})\rceil \times 7$

This works for everything, so the equation is...

$ \lceil(\frac {\text{smallest possible number}}{\text{divisor}})\rceil \times \text{ divisor}$

[size=75][color=red]Fixed LaTeX. -Isabella[/color][/size]

Note that $ \lceil$ and $ \rceil$ means the smallest integer less than or equal to the number inside those symbols.

It's smallest integer [b]greater[/b] than or equal to the number in the equation, izzy.


Two people play a game in which they take turns rolling a single six sided die. If someone rolls the number, their opponent just rolled or any multiple of that number, then they win; otherwise the game continues and their opponent takes a turn. What is the probability that the first player wins the game on the second turn?

Uh from what I can tell of the problem, the second player goes on the second turn, so 0?

Perhaps what he meant is that the first player wins the game on the second turn of the first player?
Hmm...I'm not sure either.

The question, I believe, is asking what is the probability that the first player wins on the second round of rolling the dice. Thats how the problem was worded. I know its pretty confusing  :(

[hide="Solution"]There are 6 multiples of 1, 3 multiples of 2, 2 multiples of 3, 1 multiple of 4, 1 multiple of 5, and 1 multiple of 6 among the numbers 1 to 6.

Hence, the probability of a player winning on any round is $ \frac{1}{6}\left(1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{6}\plus{}\frac{1}{6}\plus{}\frac{1}{6}\right)\equal{}\frac{7}{18}.$

In order for the first player to win on the second round, the second player must lose on the first round and the first player must win on the second round.

The probability of such event occurring is $ \frac{11}{18}\cdot\frac{7}{18}\equal{}\frac{77}{324}.$

Answer: $ \frac{77}{324}$[/hide]

Thats the answer that I got. The answer key claims that the answer is 59/216 however.

The reason that doesn't work is because the first roll of the second player isn't equally likely to be any number. If player 2 rolls a 1, then they only win if the first player rolled was a 1, but if they roll a 6, they win if the first player rolled a 1,2,3, or 6.

Cases after player one and player two roll once
$ 7/18$ chance player two won.

$ 11/18$ chance player two hasn't won; subcases:

$ 5/36$ chance player two rolled a 1
$ 1/9$ chance player two rolled a 2
$ 1/9$ chance player two rolled a 3
$ 1/12$ chance player two rolled a 4
$ 1/9$ chance player two rolled a 5
$ 1/18$ chance player two rolled a 6

Calculating the probability that player 1 wins yields:

$ (\frac {5}{36}) (1) \plus{} (\frac {1}{9}) (\frac {1}{2}) \plus{} (\frac {1}{9}) (\frac {1}{3}) \plus{} (\frac {1}{12}) (\frac {1}{6}) \plus{} (\frac {1}{9}) (\frac {1}{6}) \plus{} (\frac {1}{18}) (\frac {1}{6})$

$ \frac {5}{36} \plus{} \frac {1}{18} \plus{} \frac {1}{27} \plus{} \frac {1}{72} \plus{} \frac {1}{54} \plus{} \frac {1}{108}$

$ \frac {30}{216} \plus{} \frac {12}{216} \plus{} \frac {8}{216} \plus{} \frac {3}{216} \plus{} \frac {4}{216} \plus{} \frac {2}{216}$

$ \frac {59}{216}$


$a_{1}, a_{2}, a_{3}$ are the first three terms in an arithmetic series of reals. Find all (a_1, a_2, a_3) such that

$1) a_{2}^{2}=a_{1}a_{3}+9$
$2) a_{1}^{4}+a_{2}^{4}+a_{3}^{4}=4737$

hint:

[hide]let the middle term be x and denote the other terms using the difference of the sequence[/hide]

(2,5,8) and (-8,-5,-2) with the first and third terms switched accounts for all possible comboos.

oh yea, the sign on the middle term doesn't matter

a better substitution:

$a_{1}=x-d$
$a_{2}=x$
$a_{3}=x+d$

working in equation one first, finding d, then continuing is the idea you should use.

How would you prove that (2,5,8), (8,5,2), (-8,-5,-2), (-2,-5,-8) are the only solutions?

uh.. after you do the actual work i suggested, you see that:
$a_{2}=\pm5, \pm\sqrt{-61}$

so by the initial conditions, those four soultuions are clearly the only possiblities.

[quote="vishalarul"]Since it is an arithmetic sequence, $a_{2}= \frac{a_{1}+a_{3}}{2}$

${a_{2}}^{2}=a_{1}a_{3}+9$
$\frac{{a_{1}}^{2}+2a_{1}a_{3}+{a_{3}}^{2}}{4}=a_{1}a_{3}+9$
$\frac{{a_{1}}^{2}-2a_{1}a_{3}+{a_{3}}^{2}}{4}=9$
$\frac{{a_{1}}^{2}-2a_{1}a_{3}+{a_{3}}^{2}}=36$
$(a_{1}-a_{3})=\pm6$
$a_{1}=a_{3}\pm6$
$a_{2}=a_{3}\pm3$
I'm not sure what to do next... Anybody have any ideas?[/quote]

After doing that you have basically found out that the terms of the sequence have a common difference of 3.  Now you can let the first three terms be $x-3,x,x+3$ (or $x,x+3,x+6$, etc.) and solve in the second equation.

um, it's obviously a biquartic? and an easily factorable one at that

and considering you didn't even USE the second equation, I'd think it'd be worth trying to deal with this 'messy quartic' before asking us for help

[quote="vishalarul"]I guess you are right... Wouldn't it be easier to use $x-3$, $x$, and $x+3$? However, I was worried about taking everything to the 4th power and adding. Isn't it a messy quartic?[/quote]

absolutely not.  the odd powers would cancel and the expression could be easily expanded w/ binomial theorem.  in this case, it factors nicely into the form (x^2+n)(x^2-m)=0


prove that one is not straight by using contradiction

At least this is in the right forum for a change. Seeing as this is topic is probably spam, might as well make it worth a laugh.

[youtube]N5UkFfcGQrs[/youtube]

show him one of those inapporopriate magazines


if ax^2+bx+c has no real roots and a+b+c<0 then prove that c<0

It is more beautiful to write in Latex  :wink:

If $ ax^2\plus{}bx\plus{}c$ has no real roots and $ a\plus{}b\plus{}c<0$ then prove that $ c<0$

how can i do that????
plz help i m new here

[hide]Let $ f(x) \equal{} ax^2\plus{}bx\plus{}c$, $ f(1)\equal{}a\plus{}b\plus{}c<0$ so $ f(x) <0$, so $ f(0) \equal{} c < 0$ as required[/hide]

See here. [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]  :lol:


geometry
circumcircle
area of a triangle
Heron\u0027s formula

Quadrilateral ABCD is inscribed in a circle. AB=6, BC=7, CD=4, and AC=8. What is the area of this circle?

[hide]Well, anyways, it is clear that $ \bigtriangleup ABC$ is also inscribed in the circle. By Heron's formula, the area of the triangle is $ \sqrt {s(s - a)(s - b)(s - c}$, where $ s = \frac {8 + 6 + 7}{2} = \frac {21}{2}$. The area is then $ \sqrt {\frac {21}{2}\cdot \frac {5}{2} \cdot \frac {9}{2} \cdot \frac {7}{2}} = \sqrt {\frac {3^3 \cdot 5\cdot 7^2}{2^4}} = \frac {21\sqrt {15}}{4} = K$.
Then, the circumradius is given by $ \frac {abc}{4K} = \frac {8\cdot 6\cdot 7}{21\sqrt {15}} = \frac {16}{\sqrt {15}}$.
The area of this circle is therefore $ \pi {\left( \frac {16}{\sqrt {15} \right)}^2 = \boxed{\frac {256 \pi}{15}}}$[/hide]


Anyone here interested in astronomy  :D  ? I guess there should be lots of you.

I propose making a kind of marathon for astronomy, if there are people that enjoy watching the sky and so on :).
Anyhow, anybody here going at the astronomy olympiad? Do you know sites with training problems or anything like?

Cheers

yeah
I'm interested in it
I study astronomy by myself
and I have expriences in astronomy olympiad
it's great to have  astronomy marathon  :rotfl:

It would be great to have an astronomy olympiad.
I have bben to the international astronomy olympiad twice(2005,2006 both golds).
Search for The Paradoxical Universe on google.It is a problem book with 250 questions on astronomy but it's in Russian>use a good translator and u can figure out the questions pretty easily.

if any one can guide about the type of questions for 2nd round of astro olympiad?
where can we have sample questions?


geometry
number theory unsolved
number theory

1) if 1 + 2^n + 4^n = prime number so prove that n=3^k
2)x^2=a       so find x,y,z if a-b=8c-1
   y^3=b
   z^3=c
3)1000!=x(mod 2003) x=?
And one geometry
4)we have a circumscribed quadrilateral.And the tangent points are connected.Prove that the diagonals and this two lines are intersecting at same point

Please post different problems in different topics
For number 1 check
http://www.mathlinks.ro/Forum/viewtopic.php?t=150412

Daniel


Let $a,b>0$ and $E(a,b)=\sqrt{a\sqrt{b\sqrt{a\sqrt{b...}}}}$. Compute $E^3(a,b)$.

Well, in the book says $E^3(a,b)$. I think the they meant $[E(a,b)]^3$

I think you're right, because it makes the problem easier.

[hide]$E(a,b)=\sqrt{a\sqrt{b\sqrt{a\sqrt{b...}}}}$
Let $x = E(a,b)$:
$x=\sqrt{a\sqrt{bx}}$
$x^2=a\sqrt{bx}$
$x^4=a^2bx$
$x^3=a^2b$
$[E(a,b)]^3=a^2b$[/hide]


group theory
abstract algebra
linear algebra
matrix
quadratics
superior algebra
superior algebra unsolved

Let $p,q$ be odd primes such that $q$ is a divisor of $|SL_{2}(Z/pZ)|$. Then any $q$-Sylow of $SL_{2}(Z/pZ)$ is cyclic.

The order of $SL_{2}(\mathbb{Z}/p\mathbb{Z})$ is $(p-1)p(p+1)$. Since each odd prime divides only one of these factors, it suffices to construct cyclic subgroups of order $p-1,p,p+1$ respectively.

The matrices $\begin{pmatrix}1&a\\0&1\end{pmatrix}$ are a subgroup naturally isomorphic to $\mathbb{Z}/p\mathbb{Z}$, which is cyclic of order $p$.
The matrices $\begin{pmatrix}a&0\\0&\frac1a\end{pmatrix}$ for $a\neq0$ are a subgroup naturally isomorphic to $\left(\mathbb{Z}/p\mathbb{Z}\right)^{*}$, which is cyclic of order $p-1$.

For order $p+1$, we step back slightly. Let $r$ be any quadratic nonresidue mod $p$. The field with $p^{2}$ elements can be written as $\{a+b\sqrt{r}\}$, with $a,b\in \mathbb{Z}/p\mathbb{Z}$. This field is naturally isomorphic to the field of matrices of the form $\begin{pmatrix}a&rb\\b&a\end{pmatrix}$. The nonzero matrices of this form form a cyclic group (the multiplicative group of the field) with order $(p-1)(p+1)$. The determinant is a surjective homomorphism from this group to $\left(\mathbb{Z}/p\mathbb{Z}\right)^{*}$. Its kernel (the intersection with $SL_{2}$) therefore has order $p+1$. As a quotient of a cyclic group, it is cyclic.
Putting the pieces together, our cyclic subgroup of order $p+1$ is $\left\{\begin{pmatrix}a&rb\\b&a\end{pmatrix}: a^{2}-rb^{2}=1\right\}.$


[color=red]Here's a problem that we have for homework, I figured out a way to do them, but I was wondering if there was easeir ways, so please reply.

Augustus De Morgan said in 1864, "At some time in my life the square of my age was the same as the year in which I was that age." In what year was De Morgan born? Show how you arrived at your answer.

Please help.
[/color]
[color=darkred]
[i]This thread has been hijacked by Treething to make it more legible =o[/i]
[/color]

Ok, here's another problem:

People used toothpicks to make squares, and on the first design, there was 4 tooth picks and the area was one, and it goes on like this
                   Toothpick   Area
                         4              1
                        10             3
                        18             6
                        28             10
and it goes on like that.. and the question is what is the number of tooth picks and the area of the 15th term

Thanx :)

LuCy4EvA wrote:Ok, here's another problem:

People used toothpicks to make squares, and on the first design, there was 4 tooth picks and the area was one, and it goes on like this
                   Toothpick   Area
                         4              1
                        10             3
                        18             6
                        28             10
and it goes on like that.. and the question is what is the number of tooth picks and the area of the 15th term

Thanx

Here's a hint. Look for a pattern (duh). One of the most useful patterns are [hide]triangle numbers[/hide] and they're useful in this case.

[quote=ghu2024][quote=mira74]
[b][color=#f00]Lemma:[/color][/b] $0 \leq \text{any age} \leq 130$
[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love:
-----[/quote]

[url]https://en.wikipedia.org/wiki/Li_Ching-Yuen[/url] This dude claims to be 256...[/quote]

sorry ghu2024, but Li Ching Yuen looks to be at most 116, unlike my homie braidon, whose mom looks 127 :love:

Sar, I have an issue with your steamy first lemma. You said
[b][color=#f00]Lemma:[/color][/b] $0 \leq \text{any age} \leq 130$
but I myself do not believe this. The lemma should be this:
[b][color=#f00]Lemma:[/color][/b] $-1 \leq \text{any age} \leq 130$
[i]Proof:[/i] none of us know anyone over 130. However, it is important to note that fetuses are human lives, a well established fact by thousands of mathematicians. This can be supported by work done by the mathematician National Right to Life in their paper [url]http://www.nrlc.org/abortion/wdlb/[/url], and topologist Donald J. Trump [url]https://twitter.com/realdonaldtrump/status/1129954110747422720?lang=en[/url], who builds on work done by Ronald Reagan, who was actually not in fact a mathematician but is well known for his role of being a president. Therefore, it is possible for an age to be negative.
However, I have not met any fetuses under the age of -1. Therefore, the bounds $-1 \leq \text{any age} \leq 130$ are sufficient.
I am assuming this problem was tackled by someone with a notable lack of olympiad experience, as this is an easy mistake to miss if you have not practice USAMO geometry questions, which typically have similar style proofs. I invite Skrublord420 to proofread my work due to his notable achievements concerning Alphatrion's prime number conjectures.

What roles does $f'(a)=-2a+1$ play @5above

[quote=Treething]Well, he could be 100 or -100, right ;) ?[/quote]

How can someone be -100?

[quote=m1377]Lmao the girl who originally posted this hasn't visited AOPS in 11 years. Shes probably in her 30s right now[/quote]

Nah, if she was in middle school, then she was at most 13 yrs old. So, 13+11= 24.

[quote=Euleramanujan]What roles does $f'(a)=-2a+1$ play in mira74's post[/quote]

Hello,

Through evaluating the derivative of $f$ with respect to $a$, we are able to conclude that the function $f$ is decreasing, allowing us to utilize the inequalities $f(a) \geq f(42)$ for all $a \leq 42$ (in the valid age range) and $f(a) \leq f(44)$ for all $a \geq 44$.

Best,

mira74

[quote=BrightMonkey]the real question is : why did we bump a 16 year old post??[/quote]

to solve the problem!

Thank you for clearing up what role  $f'(a)=-2a+1$ plays.

[quote=mira74]After 18 years of nonstop work...

Solution with [b]  Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee,  Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b],  [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].

[size=150][b]Problem from post 1[/b][/size]:

We proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \in \mathbb{Z}$. Aquí es el idea clave:
-----
[b][color=#f00]Lemma:[/color][/b] $0 \leq \text{any age} \leq 130$
[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love:
-----

Define $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:

$a \leq 42$: His age in $1864$ is $$f(a) \geq f(42) = 142 > 130,$$ which is absurd.
$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\boxed{1806}.$$
$a \geq 44$: His age in $1864$ is $$f(a) \leq f(44) = -28 < 0,$$ which is absurd.

Thus, we have exhausted all cases, and are done. $\blacksquare$

[size=150][b]Problem from post 3[/b][/size]:

We proceed with engineering. See the attached diagram.

Denote by $\kappa(n)$ the number of toothpicks in the $n^{\text{th}}$ term, and denote by $\tau(n)$ the area of the $n^{\text{th}}$ term. Note that upon taking the discrete derivative $\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:

\begin{tabular}{|c|c|c|c|c|}\hline
$n$ & $1$ & $2$ & $3$ \\\hline
$\Delta_\tau(n)$ &  $2$ & $3$ & $4$\\\hline
$\Delta_\kappa(n)$ & $6$ & $8$ & $10$\\\hline
\end{tabular}
الآن هنا هو المطالبة الرئيسية:
-----
[color=#f00][b]Lemma[/b][/color]: $\Delta_\tau(n)$ and $\Delta_\kappa(n)$ are linear functions.
[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\boxed{}$
-----
Channeling SPIRIT of Ramanujan implies $\Delta_{\tau}(n) \equiv n+1$ and $\Delta_{\kappa}(n) \equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine
$$\tau(n) = \tau(1) +\sum_{k=1}^{n-1} \Delta_{\tau}(k)= 1 + \sum_{k=1}^{n-1} \left(k+1\right) = 1+\left( -1 + \frac{n(n+1)}{2} \right)= \frac{n(n+1)}{2}$$
and $$\kappa(n) = \kappa(1) + \sum_{k=1}^{n-1} \Delta_{\kappa}(n) = 4 + \sum_{k=1}^{n-1}\left( 2k+4\right) = 4 + 2 \cdot \frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$
and evaluating at $n=15$ gives a final answer of $\tau(15)=120$ area and $\kappa(15)=270$ toothpicks, so we are done. $\blacksquare$[/quote]

mira74, did you really work for 18 years just to find the answer?

@above

No, I only found part of the solution. The rest was done by [b]  Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee,  Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b],  [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].

Was this problem seriously this complicated? wow

[quote=mira74]After 18 years of nonstop work...

Solution with [b]  Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee,  Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b],  [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].

[size=150][b]Problem from post 1[/b][/size]:

We proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \in \mathbb{Z}$. Aquí es el idea clave:
-----
[b][color=#f00]Lemma:[/color][/b] $0 \leq \text{any age} \leq 130$
[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love:
-----

Define $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:

$a \leq 42$: His age in $1864$ is $$f(a) \geq f(42) = 142 > 130,$$ which is absurd.
$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\boxed{1806}.$$
$a \geq 44$: His age in $1864$ is $$f(a) \leq f(44) = -28 < 0,$$ which is absurd.

Thus, we have exhausted all cases, and are done. $\blacksquare$

[size=150][b]Problem from post 3[/b][/size]:

We proceed with engineering. See the attached diagram.

Denote by $\kappa(n)$ the number of toothpicks in the $n^{\text{th}}$ term, and denote by $\tau(n)$ the area of the $n^{\text{th}}$ term. Note that upon taking the discrete derivative $\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:

\begin{tabular}{|c|c|c|c|c|}\hline
$n$ & $1$ & $2$ & $3$ \\\hline
$\Delta_\tau(n)$ &  $2$ & $3$ & $4$\\\hline
$\Delta_\kappa(n)$ & $6$ & $8$ & $10$\\\hline
\end{tabular}
الآن هنا هو المطالبة الرئيسية:
-----
[color=#f00][b]Lemma[/b][/color]: $\Delta_\tau(n)$ and $\Delta_\kappa(n)$ are linear functions.
[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\boxed{}$
-----
Channeling SPIRIT of Ramanujan implies $\Delta_{\tau}(n) \equiv n+1$ and $\Delta_{\kappa}(n) \equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine
$$\tau(n) = \tau(1) +\sum_{k=1}^{n-1} \Delta_{\tau}(k)= 1 + \sum_{k=1}^{n-1} \left(k+1\right) = 1+\left( -1 + \frac{n(n+1)}{2} \right)= \frac{n(n+1)}{2}$$
and $$\kappa(n) = \kappa(1) + \sum_{k=1}^{n-1} \Delta_{\kappa}(n) = 4 + \sum_{k=1}^{n-1}\left( 2k+4\right) = 4 + 2 \cdot \frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$
and evaluating at $n=15$ gives a final answer of $\tau(15)=120$ area and $\kappa(15)=270$ toothpicks, so we are done. $\blacksquare$[/quote]

18 years.  :sleeping:

[quote=BrightMonkey]1974@above[/quote]

really? How did the computers even function?

yeah I searched it up on google.The actual AOPS was actyllu the books

[quote=mira74]After 18 years of nonstop work...

Solution with [b]  Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee,  Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b],  [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].

[size=150][b]Problem from post 1[/b][/size]:

We proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \in \mathbb{Z}$. Aquí es el idea clave:
-----
[b][color=#f00]Lemma:[/color][/b] $0 \leq \text{any age} \leq 130$
[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love:
-----

Define $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:

$a \leq 42$: His age in $1864$ is $$f(a) \geq f(42) = 142 > 130,$$ which is absurd.
$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\boxed{1806}.$$
$a \geq 44$: His age in $1864$ is $$f(a) \leq f(44) = -28 < 0,$$ which is absurd.

Thus, we have exhausted all cases, and are done. $\blacksquare$

[size=150][b]Problem from post 3[/b][/size]:

We proceed with engineering. See the attached diagram.

Denote by $\kappa(n)$ the number of toothpicks in the $n^{\text{th}}$ term, and denote by $\tau(n)$ the area of the $n^{\text{th}}$ term. Note that upon taking the discrete derivative $\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:

\begin{tabular}{|c|c|c|c|c|}\hline
$n$ & $1$ & $2$ & $3$ \\\hline
$\Delta_\tau(n)$ &  $2$ & $3$ & $4$\\\hline
$\Delta_\kappa(n)$ & $6$ & $8$ & $10$\\\hline
\end{tabular}
الآن هنا هو المطالبة الرئيسية:
-----
[color=#f00][b]Lemma[/b][/color]: $\Delta_\tau(n)$ and $\Delta_\kappa(n)$ are linear functions.
[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\boxed{}$
-----
Channeling SPIRIT of Ramanujan implies $\Delta_{\tau}(n) \equiv n+1$ and $\Delta_{\kappa}(n) \equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine
$$\tau(n) = \tau(1) +\sum_{k=1}^{n-1} \Delta_{\tau}(k)= 1 + \sum_{k=1}^{n-1} \left(k+1\right) = 1+\left( -1 + \frac{n(n+1)}{2} \right)= \frac{n(n+1)}{2}$$
and $$\kappa(n) = \kappa(1) + \sum_{k=1}^{n-1} \Delta_{\kappa}(n) = 4 + \sum_{k=1}^{n-1}\left( 2k+4\right) = 4 + 2 \cdot \frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$
and evaluating at $n=15$ gives a final answer of $\tau(15)=120$ area and $\kappa(15)=270$ toothpicks, so we are done. $\blacksquare$[/quote]

Wouldn't it have been 16 years...

https://artofproblemsolving.com/community/q1h43826p277736 How AoPS looked in the early years... o.O

[quote=franzliszt][quote=BrightMonkey]the real question is : why did we bump a 16 year old post??[/quote]

So that (s)he could get 44 upvotes! :P[/quote]

???????

[quote=GoogleNebula]franzliszt, congratulations on 2020 posts :).[/quote]

uh. Congratulations on 58 posts?

[quote=m1377]Lmao the girl who originally posted this hasn't visited AOPS in 11 years. Shes probably in her 30s right now[/quote]

lmao I am not even 11...

[quote=mira74]After 18 years of nonstop work...

Solution with [b]  Anish Kashyap, Evan Ren, Derek Zhu, Kyle Lee,  Luke Choi, Richard Wu, Shreyas Swaminathan, Sidhart Krishnan, Sriram Ananthakrishnan, Suhas Kotha, Sumith Nalabolu[/b],  [b]Vedant Kumud[/b] and [b]Vittal Thirumulai[/b].

[size=150][b]Problem from post 1[/b][/size]:

We proceed with mathematics. Let us assume that in the year that he mentioned, his age was $a$, where $a \in \mathbb{Z}$. Aquí es el idea clave:
-----
[b][color=#f00]Lemma:[/color][/b] $0 \leq \text{any age} \leq 130$
[i]Proof:[/i] none of us know any negative year olds or anybody over 130. our homie braidon says his mom is 200 years old but i saw her and she doesnt look a day above 127 :love:
-----

Define $f(a)=1864-a^2+a$ to be his age in 1864. Since $f'(a)=-2a+1$, it is decreasing for $a>1$. Now, we obtain the following cases:

$a \leq 42$: His age in $1864$ is $$f(a) \geq f(42) = 142 > 130,$$ which is absurd.
$a = 43$: His age in $1864$ is $f(43) = 58$, which is valid, giving a birth year of $$43^2-43=\boxed{1806}.$$
$a \geq 44$: His age in $1864$ is $$f(a) \leq f(44) = -28 < 0,$$ which is absurd.

Thus, we have exhausted all cases, and are done. $\blacksquare$

[size=150][b]Problem from post 3[/b][/size]:

We proceed with engineering. See the attached diagram.

Denote by $\kappa(n)$ the number of toothpicks in the $n^{\text{th}}$ term, and denote by $\tau(n)$ the area of the $n^{\text{th}}$ term. Note that upon taking the discrete derivative $\Delta_f(n)=f(n+1)-f(n)$ of the desired functions, we obtain the following table of values:

\begin{tabular}{|c|c|c|c|c|}\hline
$n$ & $1$ & $2$ & $3$ \\\hline
$\Delta_\tau(n)$ &  $2$ & $3$ & $4$\\\hline
$\Delta_\kappa(n)$ & $6$ & $8$ & $10$\\\hline
\end{tabular}
الآن هنا هو المطالبة الرئيسية:
-----
[color=#f00][b]Lemma[/b][/color]: $\Delta_\tau(n)$ and $\Delta_\kappa(n)$ are linear functions.
[i]Proof:[/i] Simple application of the Fundamental Theorem of Engineering. $\boxed{}$
-----
Channeling SPIRIT of Ramanujan implies $\Delta_{\tau}(n) \equiv n+1$ and $\Delta_{\kappa}(n) \equiv 2n+4$. Indeed, you may check that these functions satisfy the conditions. Taking the discrete integral (known to pedestrians as a [i]summation[/i] allows us to determine
$$\tau(n) = \tau(1) +\sum_{k=1}^{n-1} \Delta_{\tau}(k)= 1 + \sum_{k=1}^{n-1} \left(k+1\right) = 1+\left( -1 + \frac{n(n+1)}{2} \right)= \frac{n(n+1)}{2}$$
and $$\kappa(n) = \kappa(1) + \sum_{k=1}^{n-1} \Delta_{\kappa}(n) = 4 + \sum_{k=1}^{n-1}\left( 2k+4\right) = 4 + 2 \cdot \frac{(n-1)n}{2} + 4(n-1) = n(n+3),$$
and evaluating at $n=15$ gives a final answer of $\tau(15)=120$ area and $\kappa(15)=270$ toothpicks, so we are done. $\blacksquare$[/quote]

18 years indeed...

[quote=huayd695]Why is her age relevant?[/quote]

idk lol

[quote=BrightMonkey]1974@above[/quote]

 :o

I wish these people were still on AoPS :/


Find all solutions to the system:

\begin{eqnarray*}
x^2+y^2+x+y &=& 8\\
xy + x + y &=& 5
\end{eqnarray*}

If you find Intermediate problems easy, please leave this challenge problem for less experienced students to solve.

Source: Alabama State Contest

We have:
$\left\{ \begin{array}{l}
 x^2  + y^2  + x + y = 8 \\
 xy + x + y = 5 \\
 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 \left( {x + y} \right)^2  + \left( {x + y} \right) - 2xy = 8 \\
 xy + \left( {x + y} \right) = 5 \\
 \end{array} \right.$

We denote: $s = x + y,p = xy$ and obtain:
$\left\{ \begin{array}{l}
 s^2  + s - 2p = 8 \\
 s + p = 5 \\
 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 p = 5 - s \\
 s^2  + 3s - 18 = 0 \\
 \end{array} \right. \Leftrightarrow $

$ \Leftrightarrow \left\{ \begin{array}{l}
 p = 5 - s \\
 s = 3 \\
 \end{array} \right. \vee \left\{ \begin{array}{l}
 p = 5 - s \\
 s =  - 6 \\
 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 s = 3 \\
 p = 2 \\
 \end{array} \right. \vee \left\{ \begin{array}{l}
 s =  - 6 \\
 p = 11 \\
 \end{array} \right.$

For this reason
$\left\{ \begin{array}{l}
 x^2  + y^2  + x + y = 8 \\
 xy + x + y = 5 \\
 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
 x + y = 3 \\
 xy = 2 \\
 \end{array} \right. \vee \left\{ \begin{array}{l}
 x + y =  - 6 \\
 xy = 11 \\
 \end{array} \right. \Leftrightarrow $

$ \Leftrightarrow \left\{ \begin{array}{l}
 x = 1 \\
 y = 2 \\
 \end{array} \right. \vee \left\{ \begin{array}{l}
 x = 2 \\
 y = 1 \\
 \end{array} \right. \vee \left\{ \begin{array}{l}
 x =  - 3 - i\sqrt 2  \\
 y =  - 3 + i\sqrt 2  \\
 \end{array} \right. \vee \left\{ \begin{array}{l}
 x =  - 3 + i\sqrt 2  \\
 y =  - 3 - i\sqrt 2  \\
 \end{array} \right.$


i saw the following lemma in a soln:

Let $A,B,C,x,y,z$ be six positive reals s.t. $A+B+C=\pi$. Prove that \[ \sum xy \sin^2 C \leq \dfrac{(x+y+z)^2}{4} .\]

in that soln there was no proof and i can't do it.

ps: sorry if a proof was posted before, but i can't find it.

Hello Perfect radio!
The inequality in your post is essentially this inequality:

X^2 +Y^2 +Z^2 >= 2YZCosP +2ZXCosQ +2XYCosR (* )  where P, Q, R are the angles of a triangle
Adding 2YZ +2ZX +2XY to both sides gives:

(X+Y+Z)^2 >= 2YZ(1+CosP) +2ZX(1+CosQ) +2XY(1+cosR) =4YZ(CosP/2)^2 +4ZX(CosQ/2)^2 +4XY(CosR/2)^2     (**)
(Using the identity CosP =2(CosP/2)^2 -1 etc)

Now note that if P, Q and R are the angles of a triangle then A=pi/2 -P/2, B=pi/2-Q/2 and C=pi/2-R/2 are the angles of another triangle and sinA=cosP/2, sinB=CosQ/2 and sinC =CosR/2
Substituting for CosP/2... in the expression on the right of (**) leads to the inequality we want to prove.

To prove (*),This reduces to (X -ZCosQ -YcosR)^2 +(YSinR-ZSinQ)^2 which is positive.


I took the January SAT IIs.. and got 800 on both Math IIC and Physics.
Which is great:), BUT my percentiles aren't fabulous ... 90 and 92.
Should I retake?

No, you have received the best possible score.  The 90th percentile just means that 9 or 10 percent of the takers received an 800 too.  Taking it again won't help.

Don't worry about percentiles.

The only way to raise your percentile when you have already reached the top standard score (800) is to convince all of the tens of thousands of other people who want to take the next test to do badly. In other words, don't worry about your percentile score.

But now you see why 800 scores on SAT II tests alone are not impressive enough to ensure getting into a top college. Lots of other people have those scores too.

Good luck in your application process.

Thanks to both of you for the advice. I'm not going to retake. :)


[quote="ayush_2008"]a man has infinite amount of money. he distribute his money in n people in a multiple of 4,such that each person does not get more than 4n-1 rupees..in how many ways he can do that.[/quote]
[color=blue]
Let us consider $ \ 4$ as an unit or a share $ \ \implies$ each person can get at the most $ \ n \minus{} 1$ units.
Every person can be allotted $ \ 1$ or $ \ 2$ or $ \ 3\ \cdots$ or $ \ n \minus{} 1$ units $ \ \implies$ total ways for a person is $ \ n \minus{} 1\ .$
$ \implies$ Total number of ways is $ \ (n \minus{} 1)^n$

Alternatively:

Number of ways = coefficient of $ \ x^{n(n \minus{} 1)}$ in  $ \ \left(x \plus{} x^2 \plus{} x^3 \plus{} \cdots \plus{} x^{n \minus{} 1}\right)^n\left(1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} \cdots\right)$
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[quote="ayush_2008"]sir you said that Let us consider  4 as an unit or a share implies each person can get at the most n - 1 units. n-1 units implies that a person will get 4(n-1)>4n-1 Rs.also how can we handle that case in which one person can get 1,2 ,3 4, or any  Rs .[/quote]

See
$ 4(n\minus{}1)\equal{}4n\minus{}4<4n\minus{}1$

For your second query, I understood that each person is receiving money in a multiple of $ \ 4$

Is it your problem or not?

[quote="makar"][quote="ayush_2008"]sir you said that Let us consider  4 as an unit or a share implies each person can get at the most n - 1 units. n-1 units implies that a person will get 4(n-1)>4n-1 Rs.also how can we handle that case in which one person can get 1,2 ,3 4, or any  Rs .[/quote][/quote]

See
$ 4(n \minus{} 1) \equal{} 4n \minus{} 4 < 4n \minus{} 1$

For your second query, I understood that each person is receiving money in a multiple of $ \ 4$

Is it your problem or not?[/quote
yes sir it is my problem

[quote="ayush_2008"][quote="makar"][quote="ayush_2008"]sir you said that Let us consider  4 as an unit or a share implies each person can get at the most n - 1 units. n-1 units implies that a person will get 4(n-1)>4n-1 Rs.also how can we handle that case in which one person can get 1,2 ,3 4, or any  Rs .[/quote]

See
$ 4(n \minus{} 1) \equal{} 4n \minus{} 4 < 4n \minus{} 1$

For your second query, I understood that each person is receiving money in a multiple of $ \ 4$

Is it your problem or not?[/quote
yes sir it is my problem[/quote][/quote]

Then how can a person will get $ \ 1,\ 2,\ 3,$ or any amount  Rs . as every time he is getting in a multiple of $ \ 4$ .

In my calculation suppose person (1) got $ \ 3$ units or share, then he got Rs. $ \ 12$.

Hello sir \
 sir in this question the total amount of money to be distributed is multiple of 4.not necessary the money given to each person is multiple of 4. for ex if there are three relatives then one can get any amount of  money(<4n-1)but total sum should be multiple of four.