Congruent equilateral triangles are used to tessellate the plane. How
many triangles share a common vertex?

[hide]
Each angle of an equilateral triangle is $ 60^{\circ}$. The maximum degree that the triangles can be is $ 360$. So there are $ \frac {360}{60} \equal{} \boxed{6}$ triangles.
[/hide]

Oh I get it now. I thought the answer was $ 4 $ because the score was $ 4 $ and I thought it was the answer :D .

[quote="quaksterjohny"]what does it mean to tesselate a plane?[/quote]

To tessellate a plane means to fill up a plane with the same shape without overlapping or gaps.

Or, as an alternative to hunter's solution, you know that 6 equilateral triangles make up a hexagon, which tessellates the plane.


In the finite sequence 7,8,6,9,5,10,...,0, the odd-numbered terms decrease by 1 and the even-numbered terms increase by 1. The last term in the sequence is 0. What is the sum of the terms?

[hide]120
[color=blue]Moderator kyyuanmathcount says: Uh-oh, here comes the reasons Nazi!!![/color][/hide]

[quote="ch1n353ch3s54a1l"]You're both incorrect  :(
I didn't get this problem at all. WHy does it say odd numbered terms decrease by 1 and vice versa? I don't see a decrease or increase by 1..[/quote]

Look at it. The 1st term is odd and starts at 7, so the next odd numbered term, the 3rd, will be decreased by one. That's why it's 6 and the 5th is 5 and so forth. The same is for even, except their increasing by one.

Try looking at every other number.

[hide]OH. I thought it was like 7,8,7,8,7,8, which made no sense. I have a solution to my own problem...
THe sequence is 7,8,6,9...and two terms are paired, summing to 15. Since the first term is 7, $15\times7=105$
[/hide]

ok then
[hide]
The sequence is thus
7,8,6,9,5,10,4,11,3,12,2,13,1,14,0

Note that all of the pairs sum to 15.

count-7 pairs + 0
$7\cdot15=105$[/hide]

writing it out is too much
[hide]
if you think about it then its just the sum of 1-14.
Using gauss's method you get 14*7.5=105[/hide]


geometry
circumcircle
geometric transformation
rotation
geometry unsolved

Assume that all the angles of a given triangle $ ABC$ are smaller than $ 120^{\circ}$. Equilateral triangles $ AFB, BDC$ and $ CEA$ are
constructed in the exterior of $ \triangle ABC$. Prove that the lines $ AD, BE$, and $ CF$ pass through one point $ S$ for which $ SD \plus{} SE \plus{} SF \equal{} 2(SA \plus{} SB \plus{} SC)$.

It is well known that the circumcircles of these triangles pass throw a point $ S$ such that $ \angle ASB\equal{}\angle BSC\equal{}\angle CSA\equal{}120^{\circ}$. I will give an easy proof

 of this fact: denote by $ S\equal{}AD \cap BE$. We deduce that $ BE$ is obtained from $ AD$ from a rotation of $ 60^{\circ}$ around $ C$. So $ \angle DSB\equal{}60^{\circ} \Rightarrow S$ is on the

circumcircle of $ BCE$ and similar on the circumcircle of $ ACE$. Finally $ \angle ASB\equal{}120^{\circ} \Rightarrow$ $ S$ is on the circumcircle of $ ABF$

For the second part it is enough to apply the Ptolemy's theorem

The point S of above proof (Torricelli's point - so we call it) has the property that his distance to the vertices of the triangle is minimum, namely it equals AD, since SD=SB+SC.

Best regards,
sunken rock


vector
probability
function
algebra
domain
topology
real analysis

I thought it was true that you can have a measure space that isn't even a topological space, but I can't think of an example. Does anyone know one? Or am I wrong?

Thanks.

Well, to start with, you can make any set into a topological space in any of several trivial ways, so you don't want to just ask for something that "isn't a topological space." But then there's a meta-principle in mathematics that when acknowledge two different structures on the same set, you require some sort of relationship. ([i]e.g.[/i], a "topological vector space" isn't just a set that is both a topological space and a vector space; you need some additional axioms, like the arithmetic operations being continuous.) The idea that links measure and topological spaces is the idea of a "regular Borel" measure.

But getting back to your original question: are there measure spaces for which we do not care whether or not there's a topological structure? I claim one broad family of such spaces is commonplace: they're called probability sample spaces. After all, the axioms of a probability space are just that it is a measure space of total mass $1.$

A random variable is a function whose domain is a probability sample space; we really should limit that to measurable functions. But look at the way we discuss random variables. We deal endlessly with their distributions, and whether the distribution is continuous (absolutely continuous), discrete, or something else, and so on. Those are all statements about the range of the random variable, not its domain. Even when we discuss more than one random variable on the same sample space, we ask how they are jointly distributed (including the special case of being independent) - but we never really look at them at the level of points in the sample space.

The way I like to phrase it: the probability sample space that is the domain of a random variable sits "behind the curtain." It's back there somewhere, but we almost never look. And if we almost never look back there, then we don't care whether it has a topology or not.

[quote="Kent Merryfield"]But getting back to your original question: are there measure spaces for which we do not care whether or not there's a topological structure? [/quote]
That is in fact what I meant. Thank you, your reply was very illuminating. :)


LaTeX
number theory
greatest common divisor
number theory unsolved

Solve the equation in positive integers $ x$,$ y$,$ z$:

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

This should be pretty straightforward. If I am not mistaken with my proof (simple, but long to be easly LaTeX-ed), I got the lone solution $ x\equal{}4,\, y\equal{}2,\,z\equal{}5$.

Look the equation modulo 3 and see that x is even.
So, factorize as (z+2^y_0)(z-2^y_0)=3^z

If you factorize, take the gcd!

Now, there are few cases :)

i'm going to assume that for 2^x: x <32 and 3^y < 2^32. given that this turns into a fairly easy coding problem. minutes to type up, and seconds to solve.


vector
trigonometry
analytic geometry
exterior angle
AoPS Books
trig identities
Law of Cosines

[quote]41. Trangle ABC is isosceles with base AC. Points P and Q lie on CB and AB respectively, such that AC=AP=PQ=QB. The number of degrees in angle B is:
A)  B)  C)  D)  E) None of the above
[/quote]

[hide]Answer: the correct answer to number 41 is A (25 and 5/7).  If you draw the triangle out, you get several isosceles triangles, isoseles triangle ABP, APC, QBP, and QPA.  Then using the exterior angle theorem (the some of two angles in a triangle is equal to the measure of the exterior angle of the third angle).  And using the fact that there are 180 degrees in a triangle.  Asumming angle B is x, then the sum of the interior angles in triangle ABC is 7x.  Therefore angle B is 25 and 5/7 of a degree[/hide]

[hide="41"]Let angle b=x.
Angle BPQ=x
Angle AQP=2x
Angle QAP=2x
Angle CPA=3x
Angle PCA=3x
Angle BAC=3x

3x+3x+x=7x=180
x=25  5/7. $\boxed{A}$[/hide]

[quote="frt"][hide="46"]
We can rewrite the two lines using vector form:
$\overrightarrow{u}=(3,4,3)+k(2,-1,-2)$ and $\overrightarrow{v}= (3,4,3)+k(-1,2,6)$

Let $\theta$ be the angle formed by these two vectors.
Then $\cos{\theta}=\frac{\overrightarrow{u}\cdot \overrightarrow {v}}{||\overrightarrow{u}||\cdot ||\overrightarrow{v}||}$.

(I don't have a calculator with me at this moment, so I can't get the actual answer.)
[/hide][/quote]And calculator says
[hide]146.4[/hide]
[EDIT]corrected a mistake in dot product :P

I'm still having issues with the triangle one, can someone put up a picture, please? I can't seem to get an image that makes sense.

Here's a rough picture. Hope this helps :)

[quote="bthings_2000"]
46. The two lines determined by the parametric equations give below intersect. To the nearest tenth of a degree, find the measure of the obtuse angle formed by these lines. (Free response)
Line 1: $(x,y,z)=(2t+3, 4-t, 3-2t)$
Line 2: $(x,y,z)=(3-t, 2t+4, 6t+3)$[/quote]

if you take 2 points on the line and subtract them, you will get a vector that is the direction of the line, that gets rid of the constants, and leaves it as a multiple of the coefficients of t, so using the dot product

$(2 , -1 , -2) \cdot (-1 , 2 , 6)=-16=3*\sqrt{41}*\cos\theta$

$\cos\theta=-\frac{16}{3*\sqrt{41}}\implies \theta \approx 146.40$

I still haven't managed to look over vectors yet in the AoPS books (darn US History is killing me :(), but a pretty similar solution can be done with the distance formula and law of cosines:
[hide]We want to find the intersection point, so let's take the x coordinates of the two lines and equate them: $2t + 3 = 3 - t \implies t = 0$. And plugging t = 0 into the coordinates for y and z pans out also, so the intersection point is (3, 4, 3). Now, we take any other point on line 1, let's just choose t = 1 for the heck of it and we get the point as (5, 3, 1). Now, we need another point on line 2, using t = 1 again, yields (2, 6, 9). So then, we can use distance formula $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}$ to find the distances between the points, or the lengths of the sides of a triangle connecting them. Now, we can use the cosine law with the distance between (5, 3, 1) and (2, 6, 9) as $c^2$ and we find:

$c^2 = 82, b^2 = 9, a^2 = 41$ (depending on which a and b you choose--it doesn't matter).

and we have:

$\cos C = \frac{82 - 9 - 41}{(-2)(3)(\sqrt{41})}$ and $C = 33.599^{\circ}$ which is not obtuse so the obtuse one must be supplementary and is equal to $146.4^{\circ}$.[/hide]
I'll have to check, but is the vector method somehow connected to cosine law? Gotta get started reading the books anyhow, good excuse as any :D

[quote="bthings_2000"]Source: Georgia Varsity Math Tournament 2002

41. Trangle ABC is isosceles with base AC. Points P and Q lie on CB and AB respectively, such that AC=AP=PQ=QB. The number of degrees in angle B is:
A) $25\frac{5}{7}$   B) $26\frac{1}{3}$   C) $30$   D) $40$   E) None of the above

46. The two lines determined by the parametric equations give below intersect. To the nearest tenth of a degree, find the measure of the obtuse angle formed by these lines. (Free response)
Line 1: $(x,y,z)=(2t+3, 4-t, 3-2t)$
Line 2: $(x,y,z)=(3-t, 2t+4, 6t+3)$[/quote]

Hey would you happen to have all the problems to this specific year? If so, can you please scan me a copy?


Putnam
algebra
polynomial
college contests

Show that for any positive integer $n$ there is an integer $N$ such that the product $x_1x_2\cdots x_n$ can be expressed identically in the form
\[x_1x_2\cdots x_n=\sum_{i=1}^Nc_i(a_{i1}x_1+a_{i2}x_2+\cdots +a_{in}x_n)^n\]
where the $c_i$ are rational numbers and each $a_{ij}$ is one of the numbers, $-1,0,1.$

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

Let $N=2^{n-1}$, $a_{i1}=1$ for all $i$, $a_{ij}\in\left\{-1,1\right\}$, $j>1$ and sets $\left(a_{i2},a_{i3},...,a_{in}\right)$ form all possible $2^{n-1}$ combinations, $c_i=1$ if number of negative $a_{ij}$ is even, and $c_i=-1$ if number of negative $a_{ij}$ is odd.
For example, $n=3$:

$\left(x+y+z\right)^3+\left(x-y-z\right)^3-\left(x+y-z\right)^3-\left(x-y+z\right)^3$.

It is clear $x_1...x_n$ has positive coefficient in each summand, moreover, it is clear if $x_i=0$ for some $i$ then sum equals 0 too, therefore sum is a nonzero polynomial, which is divisible by $x_1...x_n$.

[u][b]The author of this posting is : ComplexZeta[/b][/u]
____________________________________________________________________

Myth, is it so obvious that this works? (i.e. is it obvious enough for 10 points?) That is almost exactly what I did on the test. (I let $N=2^n$ instead of $2^{n-1}$.) Yesterday I wasn't expecting to get many points for it, but maybe I will.

[u][b]The author of this posting is : harazi[/b][/u]
____________________________________________________________________

Nice problem. It follows directly from an identity I posted (related to a problem with nilpotent matrices) not long time ago: $n! x_1 \cdot ... \cdot x_n = \left(\sum_{i=1,n}{x_i}\right)^n - \sum{\left(x_2+...+x_n\right)^n} + ...$. To prove that, just write in two different ways the product $\left(e^{xx_1}-1\right)\cdot ...\cdot\left(e^{xx_n}-1\right)$, develop into series and identify coefficients.

If we can show that a more constrained problem is true, then we will have shown problem A4 is true.

I can show that for any positive integer $n$ the product $x_1x_2\cdots x_n$ can be expressed identically in the form
\[x_1x_2\cdots x_n=\sum_{i=1}^{2^n-1} c_iT_i\],
where $T_i=(b_{i,1}x_1+b_{i,2}x_2+\cdots +b_{i,n}x_n)^n$,
the $c_i$ are rational numbers,
and each $b_{i,j}$ is one of the numbers, $0,1$.

For term $T_i$, we let $b_{i,1}b_{i,2}b_{i,3}\cdots b_{i,n}$ be the [i]n[/i]-bit binary representation of $i$.  Thus, each term of the summation contains a unique (non-empty) combination of $x_i$ terms, raised to the [i]n[/i]th power.

Define $f(i)$ as the number of $x_i$ terms in $T_i$.  I.e.:  $f(i)=\sum_{j=1}^n b_{i,j}$.

We let $S(m)$ be the set of terms, $T_i$, such that $f(i)=m$.

The number of terms in $S(m)$ is ${n\choose m}$, $1\le m \le n$.

Begin with the one element in $S(n)$.  Consider the full expansion of this element.  Fully expanded, there is one term of the form $n!\cdot x_1\cdot x_2\cdot x_3\cdots x_n$.  This is the one and only term containing the expression on the left hand side of our  problem statement.  We can see that $c_i$ for this term is $\frac{1}{n!}$.  Every remaining term in the full expansion of this element is also an [i]n[/i]th-order term.

Every expansion of every $T_i$ results in a set of $n$th-order terms.  These terms can be categorized by grouping them into partitions of $n$.  For example, if $n=5$, the partitions of $n$ are $5,~4+1,~3+2,~2+2+1,~2+1+1+1,~1+1+1+1+1$.  Partions of the form "$5$" are:  $s^5,~t^5,~u^5,~v^5,~w^5$.  Partitions of the form "$4+1$" are:  $s^4t,~ s^4u,~s^4v,~\ldots,~w^4v$.  [There are 20 distinct terms sharing this partition.]  Note that $s,~t,~u,~v,~w$ represent $x_1,~x_2,~x_3,~x_4,~x_5$.

It is not important to know the value of the coefficients for any given partition of $n$.  However, it is important that the coefficients for a given partition of $n$ are always the same (and always an integer), for every expansion of $T_i$.  I omit the proof due to time constraints.

Next, consider the full expansions of the $n$ elements in $S(n-1)$.  We will choose the same value of $c_i$ for all of these terms.  We will find a single value of $c_i$ is sufficent to cancel all of the terms containing exactly $n-1$ distinct values of $x_j$, from the previous expansion.  The elements of $S(n-1)$ will not introduce any new terms we have previously resolved.  The value we choose for $c_i$ is rational because the operations we use to choose its value are sums, differences, products, and quotients of integers and rational numbers.

Next, consider the full expansions of the ${n(n-1)/2}$ elements in $S(n-2)$.  We will choose the same value of $c_i$ for all of these terms.  Again, we will find a single value of $c_i$ is sufficient to cancel all of the terms containing exactly $n-2$ distinct values of $x_j$, from the previous expansions.  The elements of $S(n-2)$ will not introduce any new terms we have previously resolved.

We continue in this manner, considering the full expansions of the elements in $S(n-3)$, then $S(n-4)$, and so on to $S(1)$.  After applying the above technique to the elements in $S(1)$, all terms will have been eliminated, except the term of the form $x_1\cdot x_2\cdot x_3\cdots x_n$.

Example:  n=3
$T_7=(1x_1+1x_2+1x_3)^3$
$T_6=(1x_1+1x_2+0x_3)^3$
$T_5=(1x_1+0x_2+1x_3)^3$
$T_4=(1x_1+0x_2+0x_3)^3$
$T_3=(0x_1+1x_2+1x_3)^3$
$T_2=(0x_1+1x_2+0x_3)^3$
$T_1=(0x_1+0x_2+1x_3)^3$
$c_7=\frac{1}{6}~~~~~$ (3-distinct x-terms)
$c_3=c_5=c_6=-\frac{1}{6}~~~~~$  (2-distinct x-terms)
$c_1=c_2=c_4=\frac{1}{6}~~~~~$  (1-distinct x-terms)

[quote="Myth"]For those, who don't want to read whole rcv's reasoning (though I am not sure that rcv meant following):

Prove that
\[n!\cdot x_1...x_n=\sum_{k=1}^{n}(-1)^{n-k}\sum_{i_1<...<i_k}(x_{i_1}+...+x_{i_k})^n.\]

Believe me, it is very easy (like my very first solution ;) ).
 :D  :D  :D[/quote]

My proof was a constructive proof -- I.e., I showed that I could construct a solution, through an algorithmic procedure.  Your method is algebraic.

I deliberately made no claim as to what my coefficients would be, except to show that the first coefficient, $c_{2^n-1}=\frac{1}{n!}$, is a rational number.

If I am interpreting your notation correctly, your result is the algebraic equivalent of my procedural result.  Your method has the added benefit of proving the value of the coefficients.  My method merely showed that my $2^n-1$ terms had no more than $n$ distinct coefficients.

One more additional benefit: I invented it very very quickly after I have read your first phrase.
I like solutions, which come from an intuition.

My idea is to start with $(x_1 + x_2 + ... + x_n)^n$, eliminate all elements with $n-1$ distinct $x_i$ in the product by subtracting the appropriate $\sum c(x_{i_1}+ x_{i_2} + ... x_{i_{n-1}})^n$, then eliminate all elements with $n-2$ distinct $x_i$ in the product etc... This seems like a pain to write up though...

Not knowing of this appearance, I proposed essentially the same problem as Problem 4 in the Baltic Way 2017. See [url=https://artofproblemsolving.com/community/c6h1545114]here[/url].

[hide=Construction]
The idea is to take intuition from $n=2$, where $4xy = (x+y)^2-(x-y)^2$. For general $n$, we take the sum of all expressions of the form $(\pm x_1 \pm x_2 \pm ... \pm x_n)^n$. The catch is that we will choose the coefficient of each expression so that the $x_1x_2...x_n$ term is positive.

So then why must all of the other expressions cancel out? The reason is that for all other terms, at least one of the variables among $x_1, x_2, ... x_n$ is missing, and it can be positive or negative. However, the coefficient of the entire expression depending on all of the terms means that the coefficient of $x_j$ (let $j$ be this missing variable) is dependent on the signs of the rest of the variables as well as the coefficient of the general expression. Hence, we can take all $2^n$ combinations as we like. In particular, a phenomenon occurs where, for example, the term $x^2y$ depending on only $x$ and $y$ can be so that $(x+y+z)^3$ and $-(x+y-z)^3$ have the shared sign combination within but opposite sign outside, so that they cancel out. Letting the $c_i$ terms normalize the coefficient of $x_1x_2...x_n$, we complete the proof.
[/hide]


Is there some way to suppress the numbering of specific equations within an align environment (I want one of the equations numbered, but none of the others)?

Use \notag after the equation and before \\


vector
analytic geometry
trigonometry
calculus
calculus computations

Hi all.  It's been a long time since I've done much with math.   :blush:

Can someone please show me how the following vector (in cartesian):

ds = <dx,dy,dz>

transforms into the following vector (in spherical coordinates)

ds = <dr,r d\theta, r sin \theta d\phi>

???

Thanks!   :)

im not exactly sure what you want, but i think it has something to do with how you  transform cartesian points into spherical points.

The only proof I know isn't hard, but long and rather messy. I'll try to see if I can reproduce it.

Okay... We have the following:

$x=r\cos\theta\sin\phi$
$y=r\sin\theta\sin\phi$
$z=r\cos\phi$

Differentiating the above gives us:

$dx=dr\cos\theta\sin\phi - r d\theta\sin\theta\sin\phi + r d\phi\cos\theta\cos\phi$
$dy=dr\sin\theta\sin\phi + r d\theta\cos\theta\sin\phi + r d\phi\sin\theta\cos\phi$
$dz=dr\cos\phi-r d\phi\sin\phi$

We also have this:

$\hat{r}=\hat{x}\cos\theta\sin\phi+\hat{y}\sin\theta\sin\phi+\hat{z}\cos\phi$
$\hat{\theta}=-\hat{x}\sin\theta+\hat{y}\cos\theta$
$\hat{\phi}=\hat{x}\cos\theta\cos\phi+\hat{y}\sin\theta\cos\phi-\hat{z}\sin\phi$

Now if we solve the above three equations for $\hat{x},\hat{y},\hat{z}$ (you do the work!), we get:

$\hat{x}=\hat{r}\cos\theta\sin\phi-\hat{\theta}\sin\theta+\hat{\phi}\cos\theta\cos\phi$
$\hat{y}=\hat{r}\sin\theta\sin\phi+\hat{\theta}\cos\theta+\hat{\phi}\sin\theta\cos\phi$
$\hat{z}=\hat{r}\cos\phi-\hat{\phi}\sin\phi$

Then, since we have $d\vec{s}=\hat{x}dx+\hat{y}dy+\hat{z}dz$, we can insert the above six results for the unit vectors and the differentials, and solve away (a lot of things will cancel out by well-known trigonometrical identities) you ought to be able to do it. I won't! ;)


linear algebra
matrix
linear algebra unsolved

if we have a matrix A \in M{m,n}how can we find a matrix B in M{n,m} such that BA=I_n??

well Kuba this falls under college status (although not in Romania). I have split it :D

When such a matrix B exists, it is called a [i]left inverse[/i] of A. In the first place, in order for B to exist, the rank of A must be n. Now, let S be a non-singular nxn submatrix of A: then $ B \equal{} [S^{\minus{}1} \,\,\,\,0]$ is a left inverse of A.

Here necessarily $ m\geq{n}$ and $ rank(A)\equal{}n$; then you may choose $ B\equal{}(A^*A)^{\minus{}1}A^*$.


quadratics
number theory proposed
number theory

Find $ x,y\in Z$ such that $ x^3\minus{}y^3\equal{}3xy\plus{}1$

Solution:
$ x^3\minus{}y^3\minus{}3xy\minus{}1\equal{}(x\minus{}y\minus{}1)(x^2\plus{}y^2\plus{}x\minus{}y\plus{}xy\plus{}1)\equal{}0$.Thus either $ x\equal{}y\plus{}1$,or $ x^2\plus{}y^2\plus{}x\minus{}y\plus{}xy\plus{}1\equal{}0$.
Consider the second equation as a quadratic trinomial $ f(x)$.If it has an integer root,then its discriminant is square of an integer number,so $ (y\plus{}1)^2\minus{}4(y^2\minus{}y\plus{}1)\equal{}a^2$,hence $ \minus{}3y^2\plus{}6y\minus{}3\equal{}a^2$,which is impossible,if $ y\neq 1$.
If $ y\equal{}1$,then $ x\equal{}\minus{}1$.
Answer:$ x\equal{}y\plus{}1$,where $ x\in\mathbb{Z}$.And $ x\equal{}\minus{}1,y\equal{}1$.


inequalities trigonometry quadratics function

It looks terribly hard at a glance. It turns out to be an old problem of Vietnamese Mathematical Olympiad 1996. After simple substitution $a=\sqrt{\frac{vw}u}$..., we can write it as
Let $a,b,c$ be positive reals such that
\[ab+bc+ca+abc=4, \]
Prove that $a+b+c\geq ab+bc+ca.$

Is it an evidence that they are runing out of new inequalities for national Olympiads? If at all, what a pity!

It is a problem from VietnamMO :D

@shencaili: Can you post all problems from China TST 2007 to National Olympiad box?

I'm sorry,but we've just had two exams. :D

Otherwise, it is the intention of the proposer to remind his Chinese team members of Vietnam as a host country for this coming IMO. :P

All for a memorable IMO in Vietnam :wink:

is this proof correct:
:we have to prove $z+x+y\ge zx+xy+yz$ when $x+z+y+xyz=4$
let$x\ge y\ge z$ so that $z\le 1$
    \[x+z+y+xzy=4 \ \iff x=\frac{4-y-z}{1+yz}\]
substituting $x$ we have to prove
 \[\frac{4-y-z}{1+yz}+y+z\ge \frac{(4-y-z)}{1+yz}.(y+z)+yz\]
 \[\iff 4+yz(y+z)\ge 4(y+z)-(y+z)^{2}+yz+y^{2}z^{2}\]
it becomes to prove $f(y)=(1+z-z^{2})y^{2}+y(z^{2}-4+z)+(4-4z+z^{2})\ge 0$
which is equivalent to it having nonpositive discriminent
  \[D=(z^{2}+z-4)^{2}-4(4-4z+z^{2})(z+1-z^{2})\le 0\]
  \[\iff (z-1)^{2}(5z-8)\le 0\]
which is true as $z\le 1$ by assumption.

[quote="anuj kumar"]is this proof correct:
:we have to prove $z+x+y\ge zx+xy+yz$ when $x+z+y+xyz=4$
let$x\ge y\ge z$ so that $z\le 1$
\[x+z+y+xzy=4 \ \iff x=\frac{4-y-z}{1+yz}\]
substituting $x$ we have to prove
\[\frac{4-y-z}{1+yz}+y+z\ge \frac{(4-y-z)}{1+yz}.(y+z)+yz \]

\[\iff 4+yz(y+z)\ge 4(y+z)-(y+z)^{2}+yz+y^{2}z^{2}\]
it becomes to prove $f(y)=(1+z-z^{2})y^{2}+y(z^{2}-4+z)+(4-4z+z^{2})\ge 0$
which is equivalent to it having nonpositive discriminent
\[D=(z^{2}+z-4)^{2}-4(4-4z+z^{2})(z+1-z^{2})\le 0 \]

\[\iff (z-1)^{2}(5z-8)\le 0 \]
which is true as $z\le 1$ by assumption.[/quote]
but you made a mistake.Let  be positive reals such that

Prove that

$u,v,w>0$,such that $u+v+w+\sqrt{uvw}=4$
prove that $\sqrt{\frac{uv}{w}}+\sqrt{\frac{vw}{u}}+\sqrt{\frac{wu}{v}}\geq u+v+w$   ............(1)
  proof:Let
\[a^{2}= \frac{1}{4}u,b^{2}= \frac{1}{4}v,c^{2}= \frac{1}{4}w,{\rm{ }}{\rm{ }}a,b,c > 0 \]
we can write it as
Let  be positive reals such that
$a^{2}+b^{2}+c^{2}+abc = 1$...........(2)
Prove that
$\frac{{bc}}{a}+\frac{{ca}}{b}+\frac{{ab}}{c}\ge 2(a^{2}+b^{2}+c^{2})$  ........... (3)
by  (2),Let$a = \cos A,b = \cos B,c = \cos C$
thereinto $A,B,C$ are the  angles of acute angle$\Delta ABC$,
so that $(3) \Leftrightarrow \frac{{\cos B\cos C}}{{\cos A}}+\frac{{\cos C\cos A}}{{\cos B}}+\frac{{\cos A\cos B}}{{\cos C}}\ge 2(\cos^{2}A+\cos^{2}B+\cos^{2}C)$ ...... (4)
By"$x^{2}+y^{2}+z^{2}\ge 2yz\cos A+2zx\cos B+2xy\cos C$,thereinto $A,B,C$ are the  angles of $\Delta ABC$,$x,y,z \in R$“，we have
$\sum_{CycA,B,C}{\frac{{\cos B\cos C}}{{\cos A}}}\ge 2\sum_{CycA,B,C}{\sqrt{\frac{{\cos C\cos A}}{{\cos B}}}}\cdot \sqrt{\frac{{\cos A\cos B}}{{\cos C}}}\cdot \cos A = 2\sum_{CycA,B,C}{\cos^{2}A}$

another one:
$a,b,c \in (0,1], a^{2}+b^{2}+c^{2}=2$
prove that
$\frac{1-b^{2}}{a}+\frac{1-c^{2}}{b}+\frac{1-a^{2}}{c}\leq \frac{5}{4}$
i think it is much harder than the first one and the second one,
we can use schur inequality to solve the first one

[quote="anuj kumar"]is this proof correct:
:we have to prove $z+x+y\ge zx+xy+yz$ when $x+z+y+xyz=4$
let$x\ge y\ge z$ so that $z\le 1$
\[x+z+y+xzy=4 \ \iff x=\frac{4-y-z}{1+yz}\]
substituting $x$ we have to prove
\[\frac{4-y-z}{1+yz}+y+z\ge \frac{(4-y-z)}{1+yz}.(y+z)+yz \]

\[\iff 4+yz(y+z)\ge 4(y+z)-(y+z)^{2}+yz+y^{2}z^{2}\]
it becomes to prove $f(y)=(1+z-z^{2})y^{2}+y(z^{2}-4+z)+(4-4z+z^{2})\ge 0$
which is equivalent to it having nonpositive discriminent
\[D=(z^{2}+z-4)^{2}-4(4-4z+z^{2})(z+1-z^{2})\le 0 \]

\[\iff (z-1)^{2}(5z-8)\le 0 \]
which is true as $z\le 1$ by assumption.[/quote]
I think it is not right as you assume $x+z+y+xyz=4$ though it should be $xy+xz+yz+xyz=4$

@xuheiedgar: Can you please show how to prove the inequality with Schur?
Thank you.

I'm a friend of xuheiedgar. He and I use the same method to solve this problem in the exam (Schur). He's not at home now so I'll post the solution.

Take
$a=\sqrt{\frac{vw}{u}},\ b=\sqrt{\frac{wu}{v}},\ c=\sqrt{\frac{uv}{w}}$
we have
$ab+bc+ca+abc=4$
We just need to prove
$a+b+c\geqslant ab+bc+ca$
Actually if
$a+b+c<ab+bc+ca$
With Schur we have
$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\geqslant0\\ \Longleftrightarrow(a+b+c)^{3}+9abc\geqslant 4(a+b+c)(ab+bc+ca)$
So
$9abc\geqslant[4(ab+bc+ca)-(a+b+c)^{2}](a+b+c)\\ >[4(ab+bc+ca)-(ab+bc+ca)^{2}](a+b+c)\\ =abc(a+b+c)(ab+bc+ca)$
So we have
$\sqrt{3(ab+bc+ca)}\cdot(ab+bc+ca)\leqslant(a+b+c)(ab+bc+ca)<9$
That means
$ab+bc+ca<3$
Then it's easy that
$abc\leqslant\left(\frac{ab+bc+ca}{3}\right)^{3/2}<1$
Hence
$ab+bc+ca+abc<4$
Till now it's done.

thank you guys for pointing out my mistake .actually i had seen it wrongly .but same method works in correct condition too and to surprise gives the same quadratic function
so we have to prove  $x+y+z\ge zx+xy+yz$ when $xz+zy+yx+xyz=4$
 again from condition we have $x=\frac{4-yz}{z+y+yz}$ and which on substitution wants us to prove
 \[\frac{4-yz}{z+y+zy}+y+z\ge \frac{4-yz}{z+y+zy}.(y+z)+zy\]
  \[\iff y^{2}(1+z-z^{2})+y(z+z^{2}-4)+(4-4z+z^{2})\ge 0 \iff D=(z-1)^{2}.(5z-8)\le 0\]
 which is true.
nevertheless my mistake gave rise to newer inequality.

[quote="shencaili"]$u,v,w>0$,such that $u+v+w+\sqrt{uvw}=4$
prove that $\sqrt{\frac{uv}{w}}+\sqrt{\frac{vw}{u}}+\sqrt{\frac{wu}{v}}\geq u+v+w$[/quote]

$u+v+w+\sqrt{uvw}=4\Longleftrightarrow \sqrt{uvw}=4-(u+v+w)\ (u,v,w>0).$ Thus we can rewrite our inequality as follows.

$\sqrt{\frac{uv}{w}}+\sqrt{\frac{vw}{u}}+\sqrt{\frac{wu}{v}}\geq u+v+w$
$\Longleftrightarrow uv+vw+wu\geq (u+v+w)\sqrt{uvw}$
$\Longleftrightarrow uv+vw+wu\geq (u+v+w)\{4-(u+v+w)\}$
$\Longleftrightarrow(u+v+w)^{2}-4(u+v+w)+uv+vw+wu\geq 0$
$\Longleftrightarrow (u+v+w-2)^{2}+uv+vw+wu-4\geq 0.$
Under the condition $0<u+v+w<4,$ We are to prove that $uv+vw+wu\geq 4.$

[quote="shencaili"]$ u,v,w > 0$,such that $ u \plus{} v \plus{} w \plus{} \sqrt {uvw} \equal{} 4$
prove that $ \sqrt {\frac {uv}{w}} \plus{} \sqrt {\frac {vw}{u}} \plus{} \sqrt {\frac {wu}{v}}\geq u \plus{} v \plus{} w$[/quote]
Let $ u \equal{} \frac {4yz}{(x \plus{} y)(x \plus{} z)},$ $ v \equal{} \frac {4xz}{(x \plus{} y)(y \plus{} z)},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, $ w \equal{} \frac {4xy}{(x \plus{} z)(y \plus{} z)}$ and
$ \sqrt {\frac {uv}{w}} \plus{} \sqrt {\frac {vw}{u}} \plus{} \sqrt {\frac {wu}{v}}\geq u \plus{} v \plus{} w\Leftrightarrow uv \plus{} uw \plus{} vw\geq(u \plus{} v \plus{} w)\sqrt {uvw}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {16z^2xy}{(x \plus{} y)^2(x \plus{} z)(y \plus{} z)}\geq\sum_{cyc}\frac {4xy}{(x \plus{} z)(y \plus{} z)}\cdot\frac {8xyz}{(x \plus{} y)(x \plus{} z)(y \plus{} z)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3 \minus{} x^2y \minus{} x^2z \plus{} xyz)\geq0,$ which is Schur.

[hide="Solution"]
Let $ u \equal{} k^2$, $ v \equal{} l^2$, $ w \equal{} m^2$, so $ k^2 \plus{} l^2 \plus{} m^2 \plus{} klm \equal{} 4$, so let $ k \equal{} 2\cos A$, $ l \equal{} 2\cos B$, $ m \equal{} 2\cos C$ for an acute triangle $ ABC$.  Now, it suffices to have that $ \sum \frac{(2\cos A)(2\cos B)}{2\cos C} \ge \sum (2\cos C)^2$.  Let $ BC \equal{} a$, $ AC \equal{} b$, and $ AB \equal{} c$.  Then, $ 2\cos A \equal{} \frac{b^2 \plus{}  c^2 \minus{} a^2}{bc}$.  Finding similar relations for the rest, it suffices to have that \[ \sum \frac{(a^2 \plus{} c^2 \minus{} b^2)\cdot (b^2 \plus{} c^2 \minus{} a^2)}{abc^2\cdot (\frac{a^2 \plus{} b^2 \minus{} c^2}{a^2b^2})}\ge \sum \frac{(a^2 \plus{} b^2 \minus{} c^2)^2}{ab}\iff\] \[ \sum \frac{(a^2 \plus{} c^2 \minus{} b^2)^2(b^2 \plus{} c^2 \minus{} a^2)^2a^2b^2}{a^2b^2c^2\cdot \prod (a^2\plus{}b^2 \minus{} c^2)} \ge \sum \frac{c^2\cdot(a^2 \plus{} b^2 \minus{} c^2)^2}{a^2b^2c^2}\iff\] \[ \sum \frac{a^2b^2(a^2 \plus{} c^2 \minus{} b^2)^2(b^2 \plus{} c^2 \minus{} a^2)^2}{\prod (a^2\plus{}b^2 \minus{} c^2)} \ge \sum {c^2\cdot (a^2 \plus{} b^2 \minus{} c^2)^2}\] Let $ a^2 \plus{} b^2 \minus{} c^2 \equal{} z \equal{} 2ab\cos C > 0$, $ b^2 \plus{} c^2 \minus{} a^2 \equal{} x > 0$, $ a^2 \plus{} c^2 \minus{} b^2 \equal{} y > 0$.  Then, $ a^2 \equal{} \frac{y \plus{} z}{2}$ and similarly.  Now, it suffices to show that \[ \sum \frac{(y \plus{} z)(x \plus{} z)x^2y^2}{xyz} \ge 2\sum (x \plus{} y)z^2\iff\] \[ \sum x^3y^3 \plus{} 3x^2z^2y^2 \plus{} \sum x^3y^2z \ge 2\sum x^3y^2z\iff\] \[ \sum (xy)^3 \plus{} 3\cdot (xy)\cdot (yz) \cdot (xz) \ge \sum (xy)^2(xz)\] which is true by Schur. [/hide]

from titu andrescu 'problems from the books'
$ab+ac+bc+abc=4$. Denote m,n,k for which
$\frac{2}{m}=a,\frac{2}{n}=b,\frac{2}{k}=c => mnk=m+n+k+2 => \frac{1}{m+1}+\frac{1}{n+1}+\frac{1}{k+1}=1$ $x=\frac{1}{m+1},y=\frac{1}{n+1},z=\frac{1}{k+1}=>m=\frac{y+z}{x},n=\frac{x+z}{y},k=\frac{y+z}{x}=>a=\frac{2x}{y+z},b=\frac{2y}{x+z},c=\frac{2z}{x+y} $
and we want prove that
$a+b+c=\frac{2x}{y+z}+\frac{2y}{x+z}+\frac{2z}{x+y}\geq ab+ac+bc=\frac{2x}{y+z}.\frac{2y}{x+z}+\frac{2x}{y+z}.\frac{2z}{x+y}+\frac{2y}{x+z}.\frac{2z}{x+y}$
This inequality equivalent to $\sum_{cyc}x^{3}+\sum_{sym}x^{2}y+3xyz\geq 2\sum_{sym}x^{2}y$
and this is just schur

Here is another solution: We may rewrite the given condition as $\left(\sqrt{u}\right)^2 + \left(\sqrt{v}\right)^2 + \left(\sqrt{w}\right)^2 + \left(\sqrt{u}\right)\left(\sqrt{v}\right)\left(\sqrt{w}\right) = 4.$ It is then well-known that the triple $\left(\sqrt{u}, \sqrt{v}, \sqrt{w}\right)$ may be represented as $(2\cos A, 2\cos B, 2\cos C)$ where $A, B, C$ are the angles of an acute triangle. This follows from the well-known identity $\cos^2 A + \cos^2 B + \cos^2 C + 2\cos A\cos B\cos C = 1$; in particular, since $\max\{\sqrt{u}, \sqrt{v}, \sqrt{w}\} < 2$, we can set $\sqrt{u} = 2\cos A, \sqrt{v} = 2\cos B$ for angles $A, B \in \left(0, \frac{\pi}{2}\right)$, whence solving the resulting quadratic equation in $\sqrt{w}$ yields $\sqrt{w} = -2\cos(A + B) = 2\cos C.$

So, after the substitution, we are left to prove that \[\frac{2\cos A\cos B}{\cos C} + \frac{2\cos B\cos C}{\cos A} + \frac{2\cos C\cos A}{\cos B} \ge 4\left(\cos^2 A + \cos^2 B + \cos^2 C\right).\] Setting $x = \sqrt{\frac{\cos A\cos B}{\cos C}}, y = \sqrt{\frac{\cos B\cos C}{\cos A}}, x = \sqrt{\frac{\cos C\cos A}{\cos B}}$, the inequality is equivalent to \[x^2 + y^2 + z^2 \ge 2xy\cos B + 2yz\cos C + 2zx\cos A.\] This is just a well-known [hide=lemma][b][u]Lemma.[/u][/b]
If $x, y, z$ are real numbers, and $A, B, C$ are the angles of a triangle, then \[x^2 + y^2 + z^2 \ge 2xy\cos A + 2yz \cos B + 2zx \cos C.\]
[i][u]Proof.[/u][/i] (Note, this can also be proved using Jensen's Inequality, and I believe there is also a (much nicer) vector argument)
We rewrite the inequality as a quadratic in terms of $x$: \[x^2 - 2x(y\cos A + z\cos C) + y^2 + z^2 - 2yz \cos B \ge 0.\] By completing the square, we rewrite again, as \[(x - (y\cos A + z\cos C))^2 + y^2 + z^2 - 2yz \cos B \ge (y \cos A + z \cos C)^2.\] But clearly our first square term is nonnegative. Therefore it suffices to show that \[y^2 + z^2 - 2yx \cos B \ge (y \cos A + z \cos C)^2\]\[\iff y^2 + z^2 \ge y^2 \cos^2 A + z^2 \cos^2 C + 2yz \cos A \cos C + 2yz \cos B\]\[\iff y^2(1 - \cos^2 A) + z^2(1 - \cos^2 C) \ge\]\[\ge 2yz \cos A \cos C + 2yz \cos (180^{\circ} - A - C)\]\[\iff y^2 \sin^2 A + z^2 \sin^2 C \ge 2yz \cos A \cos C - 2yz \cos (A + C)\]\[\iff y^2 \sin^2 A + z^2 \sin^2 C \ge 2yz \sin A \sin C\] which is true, by AM-GM. This concludes the proof of lemma.[/hide].

[quote=shencaili]$ u,v,w > 0$,such that $ u \plus{} v \plus{} w \plus{} \sqrt {uvw} \equal{} 4$
prove that $ \sqrt {\frac {uv}{w}} \plus{} \sqrt {\frac {vw}{u}} \plus{} \sqrt {\frac {wu}{v}}\geq u \plus{} v \plus{} w$[/quote]

let $u=x^2, v=y^2,w=z^2 $ hence $x^2+y^2+z^2+xyz=4 $
$\iff x^2y^2+y^2z^2+z^2x^2\geq xyz(x^2+y^2+z^2) $
let $ x=2\sqrt{\frac{bc}{(a+b)(a+c)}} ,y=2\sqrt{\frac{ac}{(a+b)(b+c)}}, x=2\sqrt{\frac{ba}{(a+c)(b+c)}} $
$\iff \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq 2(\frac{ab}{(a+c)(b+c)}+\frac{bc}{(a+c)(b+a)}+\frac{ac}{(a+b)(b+c)}) $
$\iff a^3+b^3+c^3+3abc\geq a^2b+b^2a+c^2a+a^2c+b^2c+c^2b $
this is true.

$\Delta ABC$ is an acute triangle. [url=https://artofproblemsolving.com/community/c6h2637350p22808304]prove that:[/url]
$$\cos A\sin B+\cos B\sin C+\cos C\sin A<\dfrac{5\sqrt{2(\sin^2 A+\sin^2 B+\sin^2 C)}}{8}$$
(2007 China TST ?)

[quote=shencaili]$ u,v,w > 0$,such that $ u \plus{} v \plus{} w \plus{} \sqrt {uvw} \equal{} 4$
prove that $ \sqrt {\frac {uv}{w}} \plus{} \sqrt {\frac {vw}{u}} \plus{} \sqrt {\frac {wu}{v}}\geq u \plus{} v \plus{} w$[/quote]

Beautiful

[quote=shencaili]$ u,v,w > 0$,such that $ u + v + w + \sqrt {uvw} = 4$  prove that $$ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w$$[/quote]
Let $ u, v, w$ be positive real numbers such that $ u\sqrt {vw} + v\sqrt {wu} + w\sqrt {uv} \geq 1$. [url=https://artofproblemsolving.com/community/c6h247578p1358629]Prove that[/url]$$ u + v + w\geq \sqrt 3$$

[quote=Scrutiny][quote=shencaili]$ u,v,w > 0$,such that $ u \plus{} v \plus{} w \plus{} \sqrt {uvw} \equal{} 4$
prove that $ \sqrt {\frac {uv}{w}} \plus{} \sqrt {\frac {vw}{u}} \plus{} \sqrt {\frac {wu}{v}}\geq u \plus{} v \plus{} w$[/quote]

Let $z=\sqrt{\frac{uv}{w}},x=\sqrt{\frac{vw}{u}},y=\sqrt{\frac{wu}{v}}.$ We need to prove that if $xy+yz+zx+xyz=4$, then $x+y+z\ge xy+yz+zx$.
It's equivalent to prove that if $x+y+z=xy+yz+zx$, then $xy+yz+zx+xyz\ge 4.$
Let $p=x+y+z,q=xy+yz+zx,r=xyz.$ Note that $p=q$, so $q+r\ge 4 \iff q\frac{q}{p}+r\ge 4\frac{q^3}{p^3}\iff p^2q^2+p^3r\ge 4q^3.$
By Schur we have $p^3-4pq+9r\ge 0\implies p^2q^2+\frac{9q^2r}{p}\ge 4q^3.$ We just need to prove $p^3r\ge \frac{9q^2r}{p}$, which is trivial since $p^2\ge 3q.\blacksquare$[/quote]

inventive

Let $ u,v,w > 0$ and  $u + v + w + \sqrt {uvw}=4 .$ Prove that $$ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geq u + v + w \geq  \sqrt {u} + \sqrt {v} + \sqrt {w}$$

[quote=AMN300]Substitute $x^2=u, y^2=v, z^2=w$. So we have $x^2+y^2+z^2+xyz=4$. Thus by the identity $\cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1$, we can write $x=2\cos A, y = 2 \cos B, z = 2 \cos C$ where $A, B, C$ are three angles of an acute triangle. So it remains to prove that \[ \sum_{cyc} \frac{\cos A \cos B}{\cos C} \ge 2 \cos^2 A + 2 \cos^2 B + 2 \cos^2 C\] Applying the lemma \[ x^2+y^2+z^2 \ge 2xy \cos B + 2yz \cos C + 2zx \cos A\] with $x^2=\frac{\cos A \cos B}{\cos C}, y^2= \frac{\cos B \cos C}{\cos A}, z^2 = \frac{\cos C \cos A}{\cos B}$ yields the desired.

By the way, I was trying to prove the inequality $\sum_{cyc} \frac{\cos A \cos B}{\cos C} \ge 2 \cos^2 A + 2 \cos^2 B + 2 \cos^2 C$ with the quadratic formula after the identity $\cos^2 A +\cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C=1$ (treating as quadratic in $\cos A \cos B \cos C$) or using Cauchy. However I couldn't do so - can anyone find a solution with the two aforementioned methods?[/quote]

Super nice

Let $u=4\cos ^2A,v=4\cos ^2B,w=4\cos ^2C.$ Then
$$u+v+w+\sqrt{uvw}=4(\cos ^2+\cos ^2B+\cos ^2C+2\cos A\cos B\cos C)=4.$$
$$ \sqrt {\frac {uv}{w}} + \sqrt {\frac {vw}{u}} + \sqrt {\frac {wu}{v}}\geqslant u + v + w$$
$$\Leftrightarrow\sum_{cyc}\frac{2\cos A\cos B}{\cos C}\geqslant 4\sum_{cyc}\cos ^2A.$$
$$\Leftrightarrow\sum_{cyc}\left(\sqrt{\frac{\cos A\cos B}{\cos C}}\right)^2\geqslant 2\sum_{cyc}\sqrt{\frac{\cos B\cos C}{\cos A}}\cdot\sqrt{\frac{\cos C\cos A}{\cos B}}\cdot\cos A.$$
This is the 嵌入不等式.$\blacksquare$


A set of positive integers $a_1, \ldots, a_n$ satisfies $a_1+2a_2+\cdots +na_n=1979$. When $n$ is even, we say this set is even, and odd otherwise. Prove that the number of even sets and the number of odd sets are equal.


inequalities
conics
parabola
inequalities proposed

[quote="kam"]I think it must be $x_n_+_1=x_1$[/quote]
Take care, $ x_{n+1}:= x_n$  as proposed.

AHHH i keep finding softer inequality! :huh:

My idea for lhs, (rhs is just cauchy!) was to consider the parabola $x^2-(a+b)\sqrt2+2ab$, it is negative for all numbers in $(a\sqrt{2},b\sqrt{2})$ and all numbers $\sqrt{x_k^2+x_{k+1}^2}$ are in that interval, that gived an upper bound for the sum of the roots, however, i dont know what else to do.


linear algebra
matrix
number theory
greatest common divisor
linear algebra unsolved

Let A be an nXn matrix with integer entries. There are B, C nXn matrices with integer entries and det(B) = det(C) = 1 such that BAC is upper triangular. What is the proof of this??

joao1,

Here is one way for you to start:
For $i \not= j$, denote by $T_{i,j}\in \mathrm{SL}_{n}(\mathbb{Z})$ the matrix with $1$ on the diagonal, $1$ at row $i$ column $j$ and $0$ elsewhere.
If $X=(x_{1}\quad x_{2}\cdots x_{n}) \in \mathbb{Z}^{n}$, (right)multiplying $X$ by some $T_{i,j}$ you obtain $=(0 \quad 0 \cdots y)$ where $y$ is the gcd of the $x_{k}$.
Start with $n=2$ and use Euclide algorithm.


Find the largest two-digit number that is equal to three times the product of its digits.

If we let the integer be $ \overline{ab} \equal{} 10a \plus{} b$, then

$ 10a \plus{} b \equal{} 3ab\implies ab \minus{} \frac {10}3a \minus{} \frac b3 \equal{} 0$

$ \implies\left(a \minus{} \frac13\right)\left(b \minus{} \frac {10}3\right) \equal{} \frac {10}9\implies(3a \minus{} 1)(3b \minus{} 10) \equal{} 10$,

where our goal is to maximize $ a$.

Testing $ 3a \minus{} 1 \equal{} 10$, $ 3b \minus{} 10 \equal{} 1$, there are no solutions, but $ 3a \minus{} 1 \equal{} 5$, $ 3b \minus{} 10 \equal{} 2$ does indeed yield a solution: $ (a,b) \equal{} (\boxed2,4)$. The only other solution is $ (a,b) \equal{} (1,5)$.


AMC
USA(J)MO
USAMO
modular arithmetic
inequalities
trigonometry
induction

I personally thought it was a little harder than last year...tell us what you think.
The hardest question has got to be number 3, and the easiest is number 4. I thought it was interesting that they put a graph theory problem for number 6,... anyway.

I thought it was really difficult, but it was my first time. 1 and 4 were definitely the most approachable, though I thought 5 wasn't too bad.

Hmm...yeah I thought it was maybe a little harder than usual.

Can somebody please look at my solution for #1?

[hide]Consider any prime in the form p=6c+1.

By FLT, $ \frac{2^{6c}\minus{}1}{2^{2c}\minus{}1} \equiv 0\pmod p$.

Also, that number is in the form $ a^2\plus{}a\plus{}1$ for integer a.  Therefore, for all primes that are one mod 6, there exists an integer $ a$ with $ a^2\plus{}a\plus{}1\equiv 0\pmod p$.  Furthermore, all $ b\equiv a\pmod p$ also have $ b^2\plus{}b\plus{}1\equiv 0\pmod p$.

For the general statement, consider n+1 primes in the form 6c+1 (there are infinite such primes, so n+1 will exist.  Let them be $ p_i$, $ i\equal{}0,1,\cdots n$, and for each $ p_i$, $ a_i^2\plus{}a_i\plus{}1\equiv 0\pmod p_i$.

By Chinese Remainder Theorem, there exists an integer $ d \equiv a_i \pmod p_i$ for all the $ p_i$.

Hence, $ d^2\plus{}d\plus{}1$ is divisible by at least n+1 primes and so we are done.  [/hide]

I thought:

#1 - got this one, extremely easy
#2 - got somewhere probably 1-2. I used inversion, which came out extremely nicely. I heard that this one was harder than #3. Apparently you had to use symmedians. I did hear that Delong Meng used inversion.
#3 - apparently, this wasn't too hard

#4 - I got this one, but the proof took forever to write up (after which I didn't have too much time for the others)
#5 - I'm pretty experienced with problems like this, but I didn't have too much time to work on it - I got somewhere, so like 1-2
#6 - Did not work on it.

I'm looking at a score of 15-19. Day 1 was hard because of #2, but possibly easy if you worked on #3 and managed to get it. Day 2 was doable to get #4,5, BUT #4 took so long to write up. I heard of a few people who got #1,4,5.

The other person at my school had a very similar experience to me, he got #1,4, some of #2,5, and also took forever to write up #4.

I guess my opinions are a little different.  I thought it was definitely a step down in difficulty from last year.  The style of the problems was a little different. (more the kind I like, I suppose)

IMO, the order of difficulty of the problems, from easiest to hardest, was

1, 4, 5, 2, 3, 6

I'm somewhat surprised (and relieved) that there was no inequality, and somewhat happy that the hard ones were combinatorics.  Good, clean problems though!

EDIT: Actually, among 5, 2, and 3, I don't know which was the hardest.  I mean, everyone will find different problems to be at different relative difficulties based on individual strengths and weaknesses.

Oh, and the problems I got:

#1: CRT, quick and easy
#2: Pretty much an angle chase, I had to use law of sines at one step but nice problem
#3: Surprisingly this didn't take long.  Induction, after proving an interesting lemma.

#4: Easy to solve, took a long time to write up.
#5: Cool problem, you had to find how to make sum|a_i| a monovariant.  Glad I got it.
#6: Interesting.  Using group theory I was able to prove that if there exists a good configuration, then the number of them is a power of two.  I then tried to prove the existence part using induction on the number of vertices, showing that from any configuration, there exists a way to "fix" it and make all of the mathematicians happy.  Too bad I didn't finish putting my ideas down (and I don't think they worked on their own anyways).

[quote="FeastOrFamine46"]Haha, #4 was ridiculous. My response took 14 pages.[/quote]

Lol I thought I was pretty long with 10 pages.

For #1, you basically have to show that you can get $ m^2 \plus{} m \plus{} 1$ to have indefinitely many prime factors. You can write it as $ \frac {(2m \plus{} 1)^2 \plus{} 3}{4}$, so if $ \minus{} 3$ is a quadratic residue modulo an odd prime $ p$, then you can get $ m^2 \plus{} m \plus{} 1$ to be a multiple of $ p$. Then just show that there are infinitely many primes with $ \minus{} 3$ as a quadratic residue, and use Chinese Remainder Theorem to finish it.

Just so people know, the answer to #4 was anything that has no more than two 1s in its base 2 representation; equivalently, anything that can be expressed as the sum of two powers of 2. I checked this with numerous people.

How did you use group theory, K81o7?

Woah, I was the longest at my school with a 6-pager for #4.

What do you think they'll give for #5 if I mentioned that the problem works when two of the numbers are in rational ratio (with proof) and that we can scale all the numbers by a factor without changing the problem? (I also included some junk about Euclidean Algorithm, but it looks like that won't be part of a solution).  0? 1?

m^4+m^2+1=(m^2-m+1)(m^2+m+1)
m^2-m+1 is relatively prime to m^2+m+1
then we know that for the number with the format m^2+m+1, the number of the prime factors does not have up boundary
this is my solution of number one
is this right? :D

I also thought it was a lot easier than usual. 1, 3, 4, 5. For easy problems I suppose 1 and 4 were a bit tough. Someone I know who probably could have gotten 2007/1 and 2006/2 was unable to get either one completely.

I really thought 3 was one of the easier problems. I got that problem in somewhere around 20 minutes and it didn't take long to write up. Krishanu, did your solution involve taking a configuration and altering the paths without changing how many there are?

I'm not too good at graph theory, but 6 did seem like it was really quite difficult. Probably not as hard as the hardest problems from other years. It's surprising to hear that the hard part was proving there is at least one configuration. I suspected that might not be trivial, but I never managed to prove that an existing configuration implied they came in a power of two.

What's the verdict on 2? I hope I'm not the only one who struggled and failed to get it (2.5 hours and I almost didn't get around to 3). My angle chasing must suck (I already knew that of course) if that's all it took, because that's what I spent almost all of the time trying.

kyyuan: You may as well face it that score estimates from people on this forum will almost certainly suck. Just don't worry about it.

15-19
1,4 were easy
I think it is much easier to get score than last year.

[quote="krakola45"]By the way, they do have solutions up: [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008USAMOsoln.pdf[/url][/quote]

Okay, after looking at this, apparently, I was insanely close to finishing the second solution to #2. I was basically at the point where if I said clearly, this follows, then I would have gotten 5-6. I got the parallelogram, did all the inversion steps, I just didn't notice that the paralellogram automatically means that F' is on P'N'. Might I get more than 2?

for number 5 we can prove that $ r_i$ are all rational or irrational

case 1: all rational
scale so they become integers.  then use the euclidean algorithm on 2 numbers...

case 2: all irrational
we can show that the ratios are all rational by dividing the $ \sum a_ir_i\equal{}0$ by $ r_1$.  then scale everything to integers and its the same as number 1

does this work?

Hm, my score is probably >=3. I thought it was a harder test. #2 was ehk, I spent 3 hours on it and got nowhere (except a :roll: from my grader)

[quote="Sean Gu"][quote="darktreb"][quote="K81o7"]Wow.

With the number of high scores being claimed, I'm suddenly getting worried that 5 problems solved plus some partial credit on the sixth may not even be enough for black.  The competition for the team is going to be FIERCE this year...[/quote]

I'm willing to bet both that you didn't get more than 35 points and that the cutoff for black will not be 35 or higher.

In fact I'd bet almost anything that the cutoff for black won't be 35 or higher. Any takers?[/quote]
35points?....a new record if that becomes true :D[/quote]

35 point cutoff happened in 2002.

[quote="darktreb"][color=green]Please stop the pyramid quoting, this thread is getting spammy. -mod[/color]

35 point cutoff happened in 2002.[/quote]
I didn't even know what was Olympiad back then :oops:

[quote="darktreb"]I'm willing to bet both that you didn't get more than 35 points and that the cutoff for black will not be 35 or higher.

In fact I'd bet almost anything that the cutoff for black won't be 35 or higher. Any takers?[/quote]

Do you not think it was possible for anyone to get over 35 points?  I'm not necessarily expecting over 35 points, but I'm hoping for somewhere in that area.

And there are a significant number of good people who claim to have solved 5 problems, as we know...

I think you can trust past moppers can at least somewhat accurately predict their own usamo scores...

EDIT: Also, I don't think the cutoff for black will be 35 or higher, but I feel it will be around 28 or so, possibly a little higher, possibly a little lower.

As far as we know, problem 6 was solved only by Eric Larson.

[quote="greentreeroad"][quote="K81o7"]
And there are a significant number of good people who claim to have solved 5 problems, as we know...[/quote]

How many have you actually heard including yourself. Also, is Zhai going to get perfect score? :D[/quote]

The other people who have told me or said they solved 5 are Zhai, Shaunak, and Sam Elder (SamE) at least.  I would guess Lars (Eric Larson) solved at least 5, maybe 6.  Then maybe there are a few others on here I'm forgetting.

[quote="101101101"]I solved #6 with only about ten minutes remaining; my solution was the intended one (although presented somewhat more disgustingly), but it is pretty unclear in places and occasionally underspecified. I honestly don't know how much credit they'll give me--I could see getting 1-2, but I could also imagine getting 6-7. Realistically I'm hoping for 5.

Altogether I answered 5 questions plus the outline of the coordinate bash for #2 (same as Mellon). I think I have enough sloppiness and inaccuracy that I'm going to get closer to 28 than 35, but for all I know both < 28 and > 35 are possible. I'm betting the cutoff is somewhere in 30-32.[/quote]

What's your name? Are you Eric Larson?
P.S if you don't mind :D

I'm Paul Christiano. I was at '06 mop and failed spectacularly last year. Hopefully this year will be somewhat better (I've never overestimated my score in the past, although I've never had so many horrible proofs either).

Just for the record...

The later the problem appears, the more difficult it is to obtain partials.  Graders don't mind giving out 2 points on, say, #4 because it wasn't that hard anyway, but , unless you have NON-TRIVIAL progress, graders will not give out any partials on, say, #3.

So don't count on substantial partials for any problems except #1 or #4 -- and it seems that most people posting here have solved those.  My guess is that the generic partial-hopeful solution will earn 2 partials on #2, 1 on #3, 3 on #5, and 0 on #6.

:D hope I have some real improvement next year///

^^ Now the above post is really what I was referring to when I said useless posts in the other thread. Does not pertain to the previous post, has no content whatsoever, and just takes up space on my screen.

I know I'm kind of whining, but this is seriously irking me.

This thread is getting sort of spammy, so I've deleted some meaningless posts. Please stay on topic and don't post if you have nothing to say.


modular arithmetic
algebra unsolved
algebra

i don't understand. u said that it's easy but you post it in unsolved problem  :huh:
http://www.mathlinks.ro/viewtopic.php?p=508187#508187

[quote="cnyd"]$ x_{1}\in\mathbb{Z^{ + }}$  ,$ x_{n + 1} = \sum\limits_{k = 1}^{n}{{x_{k}^{2}}}$,
$ 2006|x_{2006}$  $ \implies$  find minimum $ x_{1}$[/quote]

We have $ x_{n + 1} = x_n + x_n^2 = x_n(x_n + 1)$

$ 2006 = 2\cdot 17 \cdot 59$
$ x(x + 1)$ is always even and so $ 2|x_{2006}$ $ \forall x_1 > 0$
Looking at the expression $ x(x + 1)$ modulo $ 17$, it's immediate to see that this expression can be $ 0$ or $ - 1$ if and only if $ x$ is $ 0$ or $ - 1$ and so we need $ x_1 = 0$ or $ x_1 = - 1$ modulo $ 17$
Looking at the expression $ x(x + 1)$ modulo $ 59$, it's immediate to see that this expression can be $ 0$ or $ - 1$ if and only if $ x$ is $ 0$ or $ - 1$ and so we need $ x_1 = 0$ or $ x_1 = - 1$ modulo $ 59$

So $ x_1$ is the smallest positive integer such that ($ x_1 = 0\pmod{17}$ or $ x_1 = - 1\pmod{17}$) and ($ x_1 = 0\pmod{59}$ or $ x_1 = - 1\pmod{59}$)

$ x_1 = 0\pmod{17}$ and $ x_1 = 0\pmod{59}$ $ \implies$ smallest $ x_1 = 17\cdot 59 = 1003$
$ x_1 = 0\pmod{17}$ and $ x_1 = - 1\pmod{59}$ $ \implies$ $ x_1 = 17x_2$ and $ x_2 = 52\pmod{59}$ and so smallest $ x_1 = 17\cdot 52 = 884$
$ x_1 = - 1\pmod{17}$ and $ x_1 = 0\pmod{59}$ $ \implies$ $ x_1 = 59x_2$ and $ x_2 = 2\pmod{17}$ and so smallest $ x_1 = 59\cdot 2 = 118$
$ x_1 = - 1\pmod{17}$ and $ x_1 = - 1\pmod{59}$ $ \implies$ $ x_1 = 17x_2 + 16$ and $ x_2 = 58\pmod{59}$ and so smallest $ x_1 = 1002$

And the smallest is $ x_1 = 118$ (and it's easy to see that $ x_2 = 7\cdot 2006$ and so $ x_n = 0\pmod{2006}$ $ \forall n > 1$)


A square with side length 5 inches and a 3-by-6-inch rectangle are each drawn on the coordinate plane such that there are 12 square inches of overlap. What is the area, in square inches, of the portion of the square that is not overlapping the rectangle?

The square has area $ 5 \times 5 \equal{} 25$, and $ 12$ of these square inches are overlapping with the rectangle. Therefore, $ 25 \minus{} 12 \equal{}13$ square inches of the square are not overlapped.


How do you write piecewise functions out in latext? I tried using the \substack command but that makes the functions too small.

Use the cases environment which requires the amsmath package.
[code]f(x)=
\begin{cases}x^n\sin\left(\frac{1}{x}\right) & x\neq 0 \\
0 & x=0
\end{cases}gives[/code]
$ f(x)= \begin{cases}x^{n}\sin\left(\frac{1}{x}\right) & x\neq 0 \\ 0 & x=0 \end{cases}$
The [b]&[/b] lines up the conditions and [b]\\[/b] separates the 2 lines

[quote="NeverOddOrEven"]How do you write piecewise functions out in latext?[/quote]That should LaTeX or even $ \text{\LaTeX}$ without a [b]t[/b] on the end and is pronounced [i]laytek[/i] or [i]laytech[/i]


How do you experimentally find the value of k (spring constant) for a spring?  It seems like you need a known mass and use Hooke's law.

Here's one type of experiment:

Assume small objects with different masses $m_{1}$,$m_{2}$........,and attach them one by one to the ceiling by the spring with unknown spring constant $k$.Then set them in motion with a small amplitude.Take timer and count $N$ number oscilations.Then stop the timer.We need the following equation:

$t=NT$
T-is period of oscilations
$T=2\pi\sqrt{\frac{m}{k}}$

$t=2N\pi\sqrt{\frac{m}{k}}$

According to the last equation we change the masses and plot a graph time($t$) versus $\sqrt{m}$.From the tangent of the graph we get the numerical value for the following constant:

$tan\alpha=\frac{2\pi N}{\sqrt{k}}$

It gives the result for $k$:

$k=\frac{2\pi N}{tg\alpha}$

There's an easier way to do this, but may not be as accurate.

Get some objects of various masses. Measure those masses, and use $F=ma$ to calculate the force of gravity on them (use $a=-9.8 m/s^{2}$, but that's just an assumption). Then use $F=-kx$ (Hooke's Law). Since the force exerted on the spring is equal to the force of gravity, $ma=-kx$. Since you know m, a, and can measure x (the distance the spring has stretched from its equilibrium state), you can find k. To make your measurement more accurate, plot a graph of F vs. -x, using different objects, and then k would be the slope of that line.


geometry
incenter
circumcircle
perpendicular bisector
geometry unsolved

Let $ ABC$ be a triangle and $ (O)$ be its circumcircle. The internal angle bisectors of the angles $ BAC$, $ ABC$, $ BCA$ intersect circle $ (O)$ again at $ A_1$, $ B_1$, $ C_1$, respectively. Call $ A_2$, $ A_3$ the intersections of $ A_1B_1$, $ A_1C_1$ with $ BC$ and $ A_0$ midpoint of $ A_2A_3$. Points $ B_0$, $ C_0$ are defined analogously. Prove that $ AA_0$, $ BB_0$, $ CC_0$ concur if and only if triangle $ ABC$ is equilateral.

[b]EDIT.[/b] My apologizes to all. Thank you dear [b]K81o7[/b], and [b]¬[ƒ(Gabriel)³²¹º]¼[/b] for pointed out my mistake.

Let $ I$ be the incenter of $ \triangle{ABC}$.  It's well known and easy to prove that $ A_1$ is the circumcenter of $ \triangle{BIC}$, and similar statements for $ B_1$ and $ C_1$.  Therefore, $ A_1I\equal{}A_1B$, $ C_1I\equal{}C_1B$, and so $ A_1C_1$ is the perpendicular bisector of $ BI$.  Then $ A_3I\equal{}A_3B$, $ C_2I\equal{}C_2B$.  But because $ BI$ is the bisector of angle $ A_3BC_2$, we have $ BA_3\equal{}BC_2$, and so $ BA_3IC_2$ is a rhombus.

We then get a whole bunch of parallelisms, and thus get that $ A_3B_2, B_3C_2,$ and $ C_3A_2$ are the lines through $ I$ parallel to $ AB, BC,$ and $ CA$, respectively.

Now notice $ BA_3\equal{}IA_3$, and $ CA_2\equal{}IA_2$.  Thus, \[ \frac{BA_0}{CA_0}\equal{}\frac{IA_3\plus{}A_3A_0}{IA_2\plus{}A_2A_0}\equal{}\frac{AB\plus{}BD}{AC\plus{}CD}\]
where $ D$ is the midpoint of $ BC$.  So, by Ceva's theorem, it reduces to showing $ (2a\plus{}b)(2b\plus{}c)(2c\plus{}a)\equal{}(2b\plus{}a)(2c\plus{}b)(2a\plus{}c)$, which isn't true.  What did I mess up here?

ehm...April are you sure that the question is true?  :huh:

Most likely you still mean "isosceles" rather than "equilateral".

  darij

[quote="April"]I'm sorry dear all my friends. I've edited my message. Now it's correct.  :D[/quote]

Hm so the condition reduces to $ a^2b\plus{}b^2c\plus{}c^2a\equal{}b^2a\plus{}c^2b\plus{}a^2c$.  This becomes $ a(c\minus{}b)(b\plus{}c\minus{}a)\equal{}bc(c\minus{}b)$, which then becomes $ 0\equal{}(a\minus{}b)(a\minus{}c)(b\minus{}c)$, so the thing is true iff there are two sides of $ \triangle{ABC}$ of equal length.


analytic geometry
LaTeX
Asymptote
AoPSwiki

I'm doing a school project, and I'd like to be able to create coordinate grids for use as diagrams in a LaTeX document.
Can anyone tell me a program, etc. that I can use to create coordinate diagrams?
Thanks.

If you mean something like this then you don't need another program as LaTeX itself can draw it:
$\setlength{\unitlength}{2mm}\begin{picture}(30, 20) \linethickness{0.075mm}\multiput(0, 0)(1, 0){31}{\line(0, 1){20}}\multiput(0, 0)(0, 1){21}{\line(1, 0){30}}\linethickness{0.15mm}\multiput(0, 0)(5, 0){7}{\line(0, 1){20}}\multiput(0, 0)(0, 5){5}{\line(1, 0){30}}\linethickness{0.3mm}\multiput(5, 0)(10, 0){3}{\line(0, 1){20}}\multiput(0, 5)(0, 10){2}{\line(1, 0){30}}\end{picture}$
You can copy this code and adjust it as you wish.
[code]\setlength{\unitlength}{2mm}[
\begin{picture}(30, 20)
 \linethickness{0.075mm}
 \multiput(0, 0)(1, 0){31}{\line(0, 1){20}}
 \multiput(0, 0)(0, 1){21}{\line(1, 0){30}}
 \linethickness{0.15mm}
 \multiput(0, 0)(5, 0){7}{\line(0, 1){20}}
 \multiput(0, 0)(0, 5){5}{\line(1, 0){30}}
 \linethickness{0.3mm}
 \multiput(5, 0)(10, 0){3}{\line(0, 1){20}}
 \multiput(0, 5)(0, 10){2}{\line(1, 0){30}}
\end{picture}[/code]You'll find more about the picture environment in [url=http://tug.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url] in section 5.2

You might also want to have a look at Asymptote which has a new [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_%28Vector_Graphics_Language%29]AoPSWiki guide[/url]


Find all integers $x,y,z$ satisfying $x^{2}+y^{2}+z^{2}-2xyz=0$.

It's clear that there are 0 or 2 odd numbers in the set $\{x, y, z\}$. If there were 2 the left side wouldn't be divisible by $4$ while the right is. So $x, y, z$ are all even. After making the substitution $x=2x_{1}$ etc. and dividing by 4 the equation rewrites as:
$x_{1}^{2}+y_{1}^{2}+z_{1}^{2}-4x_{1}y_{1}z_{1}= 0$
By exactly the same argument as before $x_{1}, y_{1}, z_{1}$ all are even. We can continue this process infinitely long, which means that $x=y=z=0$.


Put $ n$ numbers 0 and $ n$ numbers 1 in a row, we want to to cut the row into two parts such that in each part, the number of 0 is equal to the number of 1. Denote A be the number of cases which we can not do as the above demand, B be the number of cases which we can do as the above demand in only one way. Prove that B=2A.

Catalan number.

when $ a_n\equal{}\dfrac{1}{n\plus{}1}\dbinom{2n}{n}$, $ a_n\equal{}a_0a_{n\minus{}1}\plus{}a_1a_{n\minus{}2}\plus{}...\plus{}a_{n\minus{}1}a_0$ is true.

and which is $ 2A\equal{}2*RHS\equal{}2*LHS\equal{}B$


find all the solutions of the following equation
\[ z^{\frac{4}{3}} \equal{} 1 \plus{} i
\]
(it has 3 solutions).

Close, but I believe you have to take $ 2^{3/8}$ out, not just $ \sqrt2$.


Solve the equation $ x^4 \minus{} 4x^3 \minus{} 10x^2 \plus{} 37x \minus{} 14 \equal{} 0$.

Well, thank you!

But actually I was interested in the standard way for solving it (without Cardano's formulas), not in the effective solutions (I could put the equation in wolframalpha for just finding the roots!)  :)

Not sure if this helps but with a little trial and error

$ x^{4} \minus{} 4x^{3} \minus{} 10x^{2} \plus{} 37x \minus{} 14 \equal{} (x^{2} \plus{} ax \plus{} 2)(x^{2} \plus{} bx \minus{} 7)$

$ \equal{} x^{4} \plus{} (a \plus{} b)x^{3} \plus{} (ab \minus{} 5)x^{2} \plus{} (2b \minus{} 7a)x \minus{} 14$.

Cofficients of $ x^{3}$ and $ x$ give $ a \equal{} \minus{} 5, b \equal{} 1$.

Therefore $ x^{4} \minus{} 4x^{3} \minus{} 10x^{2} \plus{} 37x \minus{} 14 \equal{} (x^{2} \minus{} 5x \plus{} 2)(x^{2} \plus{} x \minus{} 7)$.

This gives two quadratic equations whose solutions (I hope) are those given by Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we get
$ x_{1,2} \equal{} \frac {1}{2}(5\pm\sqrt {17})$
$ x_{3,4} \equal{} \frac {1}{2}( \minus{} 1\pm\sqrt {29})$
Sonnhard.[/quote]

That doesn't help anyone who needs the solution, not the answers. For god's sake, to obtain that result he could stuck this into wolfram alpha.

Thank you very much, AndrewTom!  :D

Now it's clear...  :)

hello, that's correct, we have
$ x^4\minus{}4x^3\minus{}10x^2\plus{}37x\minus{}14\equal{}(x^2\minus{}5x\plus{}2)(x^2\plus{}x\minus{}7)$
Sonnhard.

[quote="AndrewTom"]Not sure if this helps but with a little trial and error

$ x^{4} \minus{} 4x^{3} \minus{} 10x^{2} \plus{} 37x \minus{} 14 \equal{} (x^{2} \plus{} ax \plus{} 2)(x^{2} \plus{} bx \minus{} 7)$

[/quote]
I like this way, as an option, but how did you come to

$ (x^2 \plus{} ax \plus{} 2)(x^2 \plus{} bx \minus{} 7)$ as the form$ ?$

How did you eliminate these as possibilities:

$ (x^2 \plus{} ax \minus{} 2)(x^2 \plus{} bx \plus{} 7)$
$ (x^2 \plus{} ax \minus{} 1)(x^2 \plus{} bx \plus{} 14)$
$ (x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \minus{}14)?$

E.g. for the first one

$ a\plus{}b\equal{}\minus{}4$,
$ 7a\minus{}2b\equal{}37$

gives $ a\equal{}29/9$ and $ b\equal{}\minus{}65/5$ (Sorry CTRL key gone again, can't LaTeX),
contradicting $ ab\plus{}5\equal{} 10$.


My little brother was pouring me some milk for dinner. He was filling my 10 ounce glass with 2% milk when he ran out of milk at exactly 6 ounces. He got some more milk out of the fridge, but it was 1% milk. He filled the glass.
My question is, what '%' milk did I have for dinner last night? Because I don't really know what the '%' means on milk, let's just say the '%' means % of the total milk which is milkfat. If so, what % milkfat did my milk have?

Bonus question:
Let's say my brother had used skim milk (milkfat=0%). He wants to get me the same '%' milk as in the first question, in a 10 ounce glass. How much of each 0% and 2% milk should he mix together?

[hide]
For the first part, there is 6ounces of 2% milk.
That is $6\times.02=.12$ ounces of fat.
Then he adds 4 ounces of 1%.
That is $4\times.01=.04$ ounces of fat.
Total is $.16=16\%
$ fat.
There is 10 ounces so $1.6/%
$
Ill do bonus later[/hide]

biffanddoc, I think tz was right. how do I have 16% milk when I mixed 2% and 1%? I should get between 1 and 2 percent milk.
I think you forgot that it was out of 10 ounces total, not just 1 ounce.

[hide="Bonus"]Since it should be at 1.6, there should be $\frac{16}{4}=4$ times as much 2%. Therefore, 8 oz. of 2% and 2 of 0%.[/hide]

[quote="biffanddoc"][hide="bonus"]
He needs .16 ounces of fat still.
so 8 ounces of $2\%$ and 2 of $0\%$[/hide][/quote]

That is so much simpler because its [i]0[/i]%.  :blush:

Yes, whole milk is completely fat :roll:

Actually its 3%

[quote="k8reindeer"]I realize that now
I made it bonus because it didn't actually happen and I came up with it on the spot.
How about the same problem but we have to change it to...
whole milk?
4% milk?
50% milk?
n% milk?
where the % is still the % milkfat.
does whole milk have 100% milkfat?[/quote]

You can't get 1.6% with 2% and 4/50%


saludos a todos  he visto un problema muy interesante es  de una  nacional de ucrania  en el 92
 y dice asi:
sean a,b,c,d en el intervalo   [1,2] y piden demostrar

     $( \frac{a+b} {b+c}) + (\frac{c+d }{d+a})\leq  4( \frac{a+c}{b+d})$
[quote] vamos  aver que tal  esta no?[/quote][/quote]

Bueno, por la condicion $a,b,c,d \in [1,2]$, tenemos que $b+d \leq 4$, entonces, solo necesita mostrar :
$\frac{a+b}{b+c} + \frac{c+d}{d+a} \leq a+c$.

Y... tengo un pochito tiempo para mostrarlo  :rotfl:

Entonces solo habría que demostrar que:

$a(b+c) \geq a+b$
$b(a-1)+a(c-1) \geq 0$

$c(a+d) \geq c+d$
$c(a-1)+d(c-1) \geq 0$

Mm.. podemos suponer que : $(1 \leq) a \leq c$, entonces $c = a+k$, $k \geq 0$.
Y lo que queriamos es :
$\frac{a+b}{b+c} + \frac{c+d}{d+a} \leq 2a + k$ ...

Pero si demuestras que:

$\frac{a+b}{b+c} \leq a$

$\frac{c+d}{d+a} \leq c$

queda demostrada la desigualdad

Buenos dias...

he posteado esta pregunta ...
en la Advanced Section ...

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85459]Aki[/url]

Bueno esperemos alguna sugerencia ....

Revisar..

Carlos Bravo ;)
Lima _ PERU

Este es mi primer aporte, aún tengo problemas con el LATEX pero ya me estoy acostumbrando. Profe(AYALA), creo que este es el problema que me queria decir la otra vez. Nos vemos.
LEMA: Sean $x,y,z,w\in\mathbf{R}^+$. Se cumple que: $\frac{x}{y}\le\frac{z}{w}\longrightarrow\frac{x+z}{y+w}\le\frac{z}{w}$. Lo cual es obvio.

En el problema considerare dos casos:

CASO 1: $a\le c$. Veamos: $a\le c\longrightarrow\frac{a}{c}\le\frac{b}{b}\longrightarrow\frac{a+b}{b+c}\le 1$. Además: $a\le c\longrightarrow\frac{d}{d}\le\frac{c}{a}\longrightarrow\frac{c+d}{d+a}\le\frac{c}{a}$. Sumando ambos resultados tenemos: $\frac{a+b}{b+c}+\frac{c+d}{d+a}\le\frac{a+c}{a}\le a+c\le 4\frac{a+c}{4}\le 4\frac{a+c}{b+d}$. Por transitividad obtenemos lo pedido.

CASO 2: $c\le a$. Veamos: $c\le a\longrightarrow\frac{c}{a}\le\frac{d}{d}\longrightarrow\frac{c+d}{d+a}\le 1$. Además: $c\le a\longrightarrow\frac{b}{b}\le\frac{a}{c}\longrightarrow\frac{a+b}{b+c}\le\frac{a}{c}$. Sumando ambos resultados tenemos: $\frac{a+b}{b+c}+\frac{c+d}{d+a}\le\frac{a+c}{c}\le a+c\le 4\frac{a+c}{4}\le 4\frac{a+c}{b+d}$. Por transitividad obtenemos lo pedido.

Se observa que en ambos casos la igualdad se da sólo si $b+d=4$. Esto eso $b=d=2$. Reemplazando, estos valores para b y d implican que $a=c=1$. Esto es la igualdad se da solamente cuando $a=c=1$y$b=d=2$.


In acute triangle ABC, let CE and BD be the heights of sides AB and AC respectively, CE intersects BD at H, the circle with diameter DE intersects AB, AC at F, G, respectively, FG intersects AH at K. If |BC| = 25, |BD| = 20, |BE| = 7, find |AK|.

GO!

Using Pyphagorean theorem for triangles $BEC$ and $BDC$ we obtain
$CE=24$
$DC=15$
Using Ptolemei for inscribed quadrilateral $BCDE$ we obtain $BD\cdot CE=DE\cdot BC+DC\cdot EB$, so
$DE=15$
We have triangle $DCE$ is isosceles $\Rightarrow$ $\angle DEC=\angle DCE$, but  in inscribed quadrilateral $BCDE$  $\angle DCE=\angle DBE$, so
$\angle DEC=\angle DBE$
It follows that triangles $BDE$ and $HDE$ are similar. It gives us following equalities:
$HE=21/4$
$DH=45/4$
$BH=35/4$
Since $DE$ is a diametr we conclude $\angle DFE=\pi/2$ $\Rightarrow$ $DF||CE$ and triangles $DBF$ and $BEH$ are similar. It gives us
$FB=16$
$DF=12$
$FE=9$
We see that $DF=CE/2$ and $DF||CE$ $\Rightarrow$ $DF$ is medial line in $ACE$ and, hence
$AF=9$
Now we can calculate $AB$
$AB=25$
We obtain that $ABC$ is isosceles, so $CE=AP$ ($AP$ is third altitude)
$AP=24$
Since $DE FG$ is inscribed we conclude $\angle AFG=\angle ADE$; and since $BCDE$ is inscribed we conclude $\angle ADE=\angle ABC$, so
$\angle AFG=\angle ABC$
It implies that $AKF$ and $APB$ are similar; it yelds that
$AK=9\cdot 24/25$

I know.
Moreover, I know bit shorter solution.


floor function
modular arithmetic
AMC
AIME
function
number theory

[quote="paladin8"][quote="bpms"][quote="4everwise"]How many positive integer multiples of 1001 can be expressed in the form $10^{j}-10^{i}$, where $i$ and $j$ are integers and $0\leq i < j \leq 99$?[/quote]
[hide]1001=7*11*13 so the number must be divisible by all three of these. Looking at the equation mod 11 gives us that
$(-1)^{j}=(-1)^{i}$ so they just need to have the same parity, that doesn't help us much.
mod 13 does though.
$10^{j}=10^{i}$, and with fermats little theorem we determine that $10^{12}$ is 1 mod 13 which helps a lot.
mod 7
$3^{j}=3^{i}$ with Fermats little theorem we get that $3^{6}$ is 1 mod 7 which is a subset of the last one. So we know that the numbers must differ by 12.
This is equivolent to $\sum^{99}_{n=0}\lfloor \frac{n}{12}\rfloor$ which is $32+84+72+60+48+36+24+12+0=368$[/hide][/quote]

You'll notice, however, that Fermat's Little Theorem does not say that $p-1$ is the [i]smallest[/i] power for which it is $1$. In fact, we have $10^{6}\equiv 1 \pmod{13}$ (which we can easily see from the fact that $10^{3}= 1000 \equiv-1 \pmod{13}$).[/quote]
Okay, never mind what I said, I was grossly wrong.

$16\cdot49=\boxed{784}$ :P
Which is the correct answer.

Ugh, why does Carmichael number appear on the aime?  :roll:  :huh:
Could someone please suggest some ways of memorizing those properties of the carmichael function used in evaluating carmichael numbers? I'm not very good at memorizing stuff (sometimes I cannot even be sure if I'm memorizing some formulae correctly, which is really awful especially on tests, so I have to just take the risk...)... thanks for any suggestions...

$ 10^j \minus{} 10^i \equal{} 10^i(10^{j \minus{} i} \minus{} 1)$, and clearly 1001 doesn't have any factors in common with $ 10^i$. So we need
$ 10^{j \minus{} i} \equiv 1 \equiv \minus{} 1000 \pmod {1001}$, or $ 10^{j \minus{} i \minus{} 3} \equiv \minus{} 1 \pmod {1001}$. It is easy now to see that $ j \minus{} i \minus{} 3 \equal{} 3$ is the first possible exponent satisfying this. So for every three numbers the exponent is raised the right side will be multiplied by -1, so it needs to be raised by 6 to bring it back to -1, so $ j \minus{} i \minus{} 3 \equal{} 6k \plus{} 3$ for $ k \ge 0$. So j-i can equal $ 6,12,18....96$.
If $ j \minus{} i \equal{} n$ there are $ 100 \minus{} n$ possible values for i and a corresponding value for j. So the answer is
$ (100 \minus{} 6) \plus{} (100 \minus{} 12) \plus{} ....(100 \minus{} 96) \equal{} 100 \cdot 16 \minus{} 6 \cdot (1 \plus{} 2 \plus{} ...16) \equal{} 784$.

[quote=Calculus123]I have a question. We know that J - I has to equal 6, because 10^j - i = 1(mod 7), and by the toient function of 7, j - i has to equal 6. But using the same logic, it is true that j - i has to equal 10. Why is this?[/quote]

Ok, so we have that $10^{10} \equiv 1 \pmod {11}$. This is correct. However, this does [b]not[/b] imply that $10^{10} \equiv 1 \pmod {1001}$. To show this, we can use the CRT, and evaluate $10^{10}$ for the prime factorization of $1001$, namely 7,11,13. Thus, we see that
$$\begin{aligned}
&10^{10} \equiv 4 \pmod {7} \\
&10^{10} \equiv 1 \pmod {11} \\
&10^{10} \equiv 3 \pmod {13} \\
&\implies 10^{10} \equiv 991 \pmod {1001}. \\
\end{aligned}$$
Which is obviously not $1 \pmod{1001}$.

Additionally, modulo 6 works so well in this problem because
$$\begin{aligned}
&10^{6} \equiv 1 \pmod {7} \\
&10^{6} \equiv 1 \pmod {11} \\
&10^{6} \equiv 1 \pmod {13} \\
&\implies 10^{6} \equiv 1 \pmod {1001}. \\
\end{aligned}$$

Standard NT [i]again[/i].

It suffices to find all $0\leq i<j\leq 99$ such that $10^{j-i}\equiv 1\pmod{1001}$. Note that $10^6\equiv (10^3)^2\equiv (-1)^2\equiv 1\pmod{1001}$ so the order of $10^n\pmod{1001}$ is $6$. Hence, we have $j-i=6, 12, \dots, 96$. Summing the cases, up, we the total amount of $i,j$ is $99\cdot 16-6\left(\frac{16\cdot 17}{2}\right)+16=784$.

[hide=Solution]We wish to find integers $0\le i<j\le 99$ such that $1001\mid 10^j-10^i,$ which means that$$10^j\equiv 10^i\pmod{1001}.$$Since $j>i,$ let $j=i+k$ such that $k>0.$ Substituting, we have$$10^{i+k}\equiv 10^i\pmod{1001}.$$We can cancel $10^i$ since $\gcd(10^i,1001)=1.$ Our equation is then$$10^k\equiv 1\pmod{1001}.$$
Note that $10^3\equiv 1000\equiv -1\pmod{1001}.$ Thus,$$10^6\equiv \left(10^3\right)^2\equiv (-1)^2\equiv 1\pmod{1001}.$$Since $10^6$ is the smallest power of $10$ greater than $1$ that is congruent to $1\pmod{1001},$ we know $6\mid k.$

Therefore, $k=6,12,\ldots,96$ since we must have $k<100.$ If $k=6c,$ then there are $99-6c+1$ values of $j$ (from $6c$ to $99$). Thus, the total number of $i,j$ pairs is$$\sum_{c=1}^{16}{99-6c+1}=\sum_{c=1}^{16}{100-6c}=\sum_{c=1}^{16}{100}-6\sum_{c=1}^{16}{c}.$$This is equal to$$1600-6\cdot \frac{16\cdot 17}{2}=1600-3\cdot 16\cdot 17=\boxed{784}.$$[/hide]

I... don't get why this is in HSM, are older aime problems supposed to be here?

Anyways since 1001|10^(j-i)-1 and order of 10 wrt 1001 modulo is 6, 6|j-i, now casework (17 numbers are 0-3 mod 6, hence 4*17choose2, plus 16 numbers that are 4 or 5 mod 6 gives another 2*16 choose 2) for a total of 784 ways.[quote=peelybonehead][color=#00f]Key finding:[/color]
$$10^j - 10^i \equiv 0 \pmod{1001} \implies j \equiv i \pmod 6.$$
Then proceed by casework to get $\boxed{784}$ number of positive multiples of $1001.$[/quote]

i think the main idea WAS to use orders and show that, you literally skipped over the main part


floor function
inequalities proposed
inequalities

$n$ is a positive integer. Prove that:

$\left| \left\{\frac{n}{1}\right\}-\left\{\frac{n}{2}\right\}+\left\{\frac{n}{3}\right\}-\cdots-(-1)^{n}\left\{\frac{n}{n}\right\}\right| < \sqrt{2n}$

( note: if $t \in R$, $\left\{t\right\}= t-\left\lfloor t\right\rfloor$ ) :D

:mad:  what this words .  respect other if you are real Iranian
where is religion? :?:

:huh: if i said anything bad i'm really sorry, would you plz tell me why are you angry?!!!

ok everybody this is one of the tough and cute problems of Mathematics Magazine, i've seen its solution in magazine but i was wondering if anyother other solution would be obtained... :)


greatest common divisor
number theory
MATHCOUNTS
relatively prime

Put the 4-digit numbers into two groups based on whether

they are the products of two 2-digit numbers or not.

Which group is greater?

[quote="anirudh"]What do you mean by "products of 2 digit numbers or not". Do you mean that the number of 2 digit factors. I don't clearly understand this problem.  :maybe:[/quote]
If the number can be expressed as the product of two 2-digit numbers, then it fits into that group. Example: $50\cdot50=2500.$ 2500 is in that group.

[quote="bpms"][quote="i_like_pie"][quote="anirudh"]wouldn't this mean whether they are relatively prime to all digits from 10-99?

and therefore I would say that the products gorup would be greater.

I'm not sure if i am right though[/quote]
Can you please explain what relatively prime means? Then I might be able to answer your question. :)[/quote]
Two numbers are relatively prime if their gcf (sometimes known as gcd) is 1.[/quote]
Thank you, but I still don't know how to do the problem.

[quote="i_like_pie"][/quote][quote="bpms"][quote="i_like_pie"][quote="anirudh"]wouldn't this mean whether they are relatively prime to all digits from 10-99?

and therefore I would say that the products gorup would be greater.

I'm not sure if i am right though[/quote]
Can you please explain what relatively prime means? Then I might be able to answer your question. :)[/quote]
Two numbers are relatively prime if their gcf (sometimes known as gcd) is 1.[/quote][quote="i_like_pie"]
Thank you, but I still don't know how to do the problem.[/quote]This problem seems very complicated but it probably has an elegant solution. Intuition is telling me that the answer is the group of numbers that can be written as a product of 2 2digit numbers. Intuition can be wrong though (cantor's diagnolization, summation of an infinite series...)

Moved to Mathcounts for being too hard for MOEMS. And perhaps too hard for Mathcounts too...

Well i cant answer the problem, but i just have to say that just because a 4-digit number is not relatively prime to 2-digit numbers, it doesnt mean that it can be made up of 2 2-digit numbers. for example, 2197 is not relatively prime to 13, but it cannot be formed from 2 two-digit numbers. So it might not have to do with factoring that much.

[quote="Ignite168"]I don't think this problem is too hard for MathCounts but I may be doing it wrong.

[hide]You have 90 choices for each number you want to multiply together. That gives you $8100$ possible numbers. Then, divide by two because each product can be expressed two different ways. That gives you $4050$ possible numbers. Now squares cannot be expressible so add 90 back on to get $4140$. Now obviously, not all of these will even be 4 digit, in fact 32 is the first number that multiplied to itself will give you a 4 digit number. However, there is $9000$ four digit numbers so even if you included ALL numbers expressable as a product of two two digit numbers, you would still not get a greater group than the four digit numbers.

The four digit numbers not expressable as a product of two two digit numbers is greater [/hide][/quote]
That method is good and not to hard for mathcounts at all. It is very elegant.

Setting an upper limit is a very good technique on a vague problem like this (true or false).

The exact number of elements in each set would be extremely difficult to calculate (I think  :maybe: ).

[quote="ProtestanT"][quote="Ignite168"]I don't think this problem is too hard for MathCounts but I may be doing it wrong.

[hide]You have 90 choices for each number you want to multiply together. That gives you $8100$ possible numbers. Then, divide by two because each product can be expressed two different ways. That gives you $4050$ possible numbers. Now squares cannot be expressible so add 90 back on to get $4140$. Now obviously, not all of these will even be 4 digit, in fact 32 is the first number that multiplied to itself will give you a 4 digit number. However, there is $9000$ four digit numbers so even if you included ALL numbers expressable as a product of two two digit numbers, you would still not get a greater group than the four digit numbers.

The four digit numbers not expressable as a product of two two digit numbers is greater [/hide][/quote]

You still don't know which  numbers the 4140 numbers are, So you cannot add up the numbers to be sure the sum is smaller than the other group's sum...[/quote]
I'm pretty sure that moldlee meant the cardinality. If he/she meant sum then this problem is way too hard


2^n.a+b is square forevery n=0,1,2,3(positive integer)a and b are  integers.
2^n means 2 of power n.

It has been solved on this forum already...I can't find the link.

The solution is so short (and beautiful) that we can write it again. But I am sick a little bit of this problem, I've seen it at least 6 times (I posted it first, at least 1 year ago). If $a>0$ then if we put $ x_n=\sqrt{a\cdot 2^n+b}$ then show that $ 2x_n-x_{n+2}$ is a sequence of integers which converges to 0, thus becomes 0 from a certain point. Done.

another way is to look at $4(a2^{n}+b)-(a2^{n+2}+b)$ which becomes too small to be a difference of squares.  It's a pretty open-ended problem


superior algebra
superior algebra solved

Let $(G,\cdot)$ be a group and let $F$ be the set of elements in the group $G$ of finite order. Prove that if $F$ is finite, then there exists a positive integer $n$ such that for all $x\in G$ and for all $y\in F$, we have
\[ x^n y = yx^n. \]

For each $x\in G$, the conjugation by $x$ induces a permutation of $F$. This means that each such permutation, when raised to the power $|F|!$, yields the identity map, so it suffices to choose $n=|F|!$.


number theory
relatively prime
calculus
calculus computations

[url=http://bildr.no/view/488201][img]http://bildr.no/thumb/488201.jpeg[/img][/url]

1. in (i) why is $ a_{n_k}\equal{}2(2^{3^k}\plus{}1)$ the subsequence ?
2. in (ii) how to prove the relatively prime of $ b_n$?

Thanks

Ps. the forum can not be uploaded pic directly . I feel http://bildr.no/ does not keep that pic forever.

How to insert pic directly ?

[quote="water2011"][url=http://bildr.no/view/488201][img]http://bildr.no/thumb/488201.jpeg[/img][/url]

1. in (i) why is $ a_{n_k} \equal{} 2(2^{3^k} \plus{} 1)$ the subsequence ?
2. in (ii) how to prove the relatively prime of $ b_n$?

Thanks

Ps. the forum can not be uploaded pic directly . I feel http://bildr.no/ does not keep that pic forever.

How to insert pic directly ?[/quote]
(i) $ a_{n_k} \equal{} (n_k)^2 \plus{} 2 \equal{} (2^{(3^k\plus{}1)/2})^2\plus{}2 \equal{} 2^{3^k\plus{}1} \plus{} 2 \equal{} 2(3^k \plus{} 1)$
(ii) First $ b_i \equiv 1 \bmod 2$. Second assume $ j > i$: $ \gcd(b_i, b_j) \equal{} \gcd(b_i,(b_1b_2...b_{j\minus{}1})^2\plus{}2) \equal{} \gcd(b_i,2) \equal{} 1$.


In a reality we should choose $5$  of the $15$  initial  players.
In the $15$  people there are $3$  couples(a man-a woman) for which we know that the couples are not divided, i.e either both of two will select or both of them will disqualify .
The selection of the $5$ winners will become in two stages :
In the first, will be selected $10$ of the $15$  players and in the second   will be selected $5$  of the $10$  winners of the first stage.
In how many ways could be choosed  $5$ ?

You find it difficult or not understandable ?


geometry
circumcircle
rectangle
trigonometry
geometry proposed

We denote as $M,$ the intersection point of $BD,$ $EF$ and so, we have that the quadrilateral $BENM$ is cyclic, because of $BD\perp EF.$

Hence, we conclude that $\angle NEF = \angle NBD$ $,(1)$

From the similar right triangles $\bigtriangleup BNE,$ $\bigtriangleup BNC,$ we have that

$\frac{EN}{BN}= \frac{BE}{BC}=\frac{EF}{AC}= \frac{EF}{BD}$ $\Longrightarrow$ $\frac{EN}{BN}= \frac{EF}{BD}$ $,(2)$

From $(1),$ $(2)$ we conclude that the triangles $\bigtriangleup ENF,$ $\bigtriangleup BND,$ are similar and so, we have that $\angle NDB = \angle NFE$ $,(3)$

From $(3),$ $\Longrightarrow$ the quadrilateral $DNMF,$ is cyclic.

Hence, we have that $\angle DNF = \angle DMF = 90^{o}$ and the pproof is completed.

Kostas vittas.

[quote="sinajackson"]$E$ and $F$ are points on sides $AB$ and $BC$ of square $ABCD$ respectively such that: $BE=BF$. If $BN$ is the altitude of triangle $BEC$, Prove that $DNF$ is right angle.[/quote]

[color=red] It is a nice problem from Leningrad or All Russian olympiad.[/color].

[i] After Vitta's nice (and [color=red]new[/color] to me !) solution , I send you one more solution, not metric.[/i]

 If $BN$ meets  $AD$ at $Z$ , then triangles $ABZ, BEC$ are equal (since $\angle ABZ= \angle BCE$ . Thus $AZ=BE=BF$ and $DZ=CF$. If $DF$ meets $CZ$ at $O$ , then :
  $ON= \frac{ZC}{2}= \frac{DF}{2}$, because in right triangle $ZNC$ segment $NO$ is median and $ZC=DF$.

 This means that triangle $DNF$ is right , since $NO$ is a median in this triangle.

A nice and easy problem ![hide="A short proof."][color=darkblue]Denote $G\in BN\cap AD$. From $\triangle ABG\equiv\triangle BCE$ obtain $AG=BE$, i.e. $AG=BF$. Thus, the point $N$ belong to the circumcircle of the rectangle $CDGF$. But the segment $[DF]$ is another diameter of this circle. Therefore, $ND\perp NF$.[/color][/hide][b][u]Stergiu[/u][/b], are only four minutes between the our messages !

[quote="Virgil Nicula"]
Stergiu[/u][/b], are only four minutes between the our messages ![/quote]

It seems ,  we solve problems about the same time !!!

    Greetings.-

[color=darkblue][b]A metrical proof[/b][/color] [hide="with the following simple remark applied to the right triangle NBC :"][color=darkred] In the $A$- right triangle $ABC$, $AC>AB$, denote $D\in BC$, $AD\perp BC$
and  $M\in (AC)$ for which $AM=AB$. Then $\tan\widehat{CDM}=\frac{CM}{CA}$.[/color][/hide][color=darkblue]Indeed, $\tan\widehat{CNF}=\frac{CF}{CB}=\frac{CF}{CD}=\tan\widehat{CDF}$, i.e. $\widehat{CNF}\equiv\widehat{CDF}$ a.s.o.[/color]

 [b][u]Stergiu[/u][/b], (again) now is only one minute between our messages ! Greetings !


Start with an isoceles triangle with base angles of 72 degrees and equal sides with length 1.  Draw an angle bisector for one of the base angles.  This will result in a (smaller) isoceles triangle with base angles 72 degrees.  Continuing at the terminus of the first angle bisector, draw an angle bisector of the smaller triangle, getting an (even smaller) isoceles triangle with base angles 72 degrees.  Continue this process ad infinitum.

(a) What is the total length of all the angle bisectors that are drawn?

(b) Do the angle bisectors have approach a limit point?  Can you describe it in terms of the original triangle?

The length of the first angle bisector is $\frac{\sin36}{\sin108}=\frac{\sqrt5-1}2$ by the Law of Sines.
The second one:  $\frac{x}{\sin36}=\frac{\frac{\sqrt5-1}2}{\sin108}$    which happens to be $\left(\frac{\sqrt5-1}2\right)^2$
This pattern continues.

The total length of the angle bisectors is thus $\displaystyle\sum_n{\left(\frac{\sqrt5-1}2\right)^n}$    which is $\frac{\sqrt5+1}2$


A star-polygon is drawn on a clock face by drawing a chord from each number to the firth number counted clockwise from that number. That is, chords are drawn from $ 12$ to $ 5,$ from $ 5$ to $ 10,$ from $ 10$ to $ 3,$ and so on, ending back at $ 12.$ What is the degree measure of the angle at each vertex in the star-polygon?

$ \textbf{(A)}\ 20 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 30 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 60$

[hide="Solution"]Each angle intercepts 2 hours worth of clock (draw the star to see this).

2/12 = 1/6 of 360.

Because the angles are situated with a vertex along the side of the circle(clock), instead of having a vertex in the center, the angle is really 1/2 of the arc it subtends due to some random geo. thm. I'm forgetting the name of.

1/6 of 360 is 60 degrees,

and the acute angles are half of that, for an answer of $\boxed{30}$ degrees, or choice $\boxed{C}$
[/hide]

[quote="4everwise"]A star-polygon is drawn on a clock face by drawing a chord from each number to the firth number counted clockwise from that number. That is, chords are drawn from $12$ to $5,$ from $5$ to $10,$ from $10$ to $3,$ and so on, ending back at $12.$ What is the degree measure of the angle at each vertex in the star-polygon?

$\text{(A)}\ 20 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 60$[/quote]

[hide="solution"]There are 6 points so they divide the arc into 60 degree sectors.  The angle intercepting the sector is half of 60 or [b]30[/b].[/hide]

Pretty simple, if you know that on a clock that between two numbers is 30 degrees the this is between 2 sets of 2number--- such a 5 and 7.
so if you use the origin as alfa the the angle is alfa/2
so 60/2
Answer [b]30[/b][/b]


solve the inequality:
\[ \frac{x\plus{}3}{x\minus{}1}\leq0\]

show answer in interval notation

[quote="DaReaper"]solve the inequality:
\[ \frac {x \plus{} 3}{x \minus{} 1}\leq0
\]
show answer in interval notation[/quote]
[hide]Take the first case where  $ x > 1$. Multiplying the bottom doesn't change the sign so we simply have $ x\leq \minus{} 3$, which contradicts our first assumption so throw this out.
Second case is where $ x < 1$. Multiplying the bottom does change the sign and now we have $ x\geq \minus{} 3$.
So our following interval is $ \minus{} 3\leq x < 1$ So in interval notation this would be $ [\minus{}3,1)$[/hide]


[quote="Allan Z"]ABC has $ a \equal{} 13$, $ b \equal{} 14$, and $ c \equal{} 15$. X, Y, Z are points on sides a, b, c respectively, such that $ \frac{BX}{XC}\equal{} 2$, $ \frac{CY}{YA}\equal{} 3$, and $ \frac{AZ}{ZB}\equal{} 5$. The segments AX, BY, CZ intersect in pairs at three noncollinear points P, Q, R. Compute the area of PQR.

EDIT: i got a weird answer
[hide]$ \frac{5887}{228}$[/hide]

can someone confirm that? or if i am wrong, can someone show me the solution?[/quote]

Could you show us your solution, and then we can tell if it is right or wrong?

I got a weird answer as well, that is similar, but maybe I screwed up.

use of coordinate geometry, but run into a huge mess

This would be a good place to use barycentric coordinates.

Yeah, that is what I used to.  I thought that you could only use it if all concurrent, but I don't think that is true.

[quote="vishalarul"]I don't understand Barycentric coordinates. I tried looking it up at Wolfram Mathworld ([url=http://mathworld.wolfram.com/BarycentricCoordinates.html]here[/url]), and I don't get what they mean by "masses".

[asy]import cse5; pair B=MP("B",D((0,0))), C=MP("C",D((13,0))), A=MP("A",D(IP(CR(B,15),CR(C,14))),N), X=MP("X",D((B+2C)/3)), Y=MP("Y",D((3A+C)/4),N), Z=MP("Z",D((4B + A)/5),N), P=MP("P",D(IP(A--X, B--Y))), Q=MP("Q",D(IP(B--Y, C--Z))), R = MP("R", D(IP(C--Z, A--X)),N); D(A--B--C--cycle); D(A--X); D(B--Y); D(C--Z);[/asy]

Now, if it's possible to find the barycentric coordinates of $ P,Q,R$, then it's possible to find the area, just do this:

Let $ P(x_{1}, y_{1}, z_{1})$, $ Q(x_{2}, y_{2}, z_{2})$, $ R(x_{3}, y_{3}, z_{3})$ be the barycentric coordinates. Then, from the link from Wolfram, the area of $ \triangle PQR$ is given by:

$ [PQR] \equal{}\left |\begin{array}{ccc}x_{1}& y_{1}& z_{1}\\ x_{2}& y_{2}& z_{2}\\ x_{3}& y_{3}& z_{3}\end{array}\right |\cdot [ABC]$

Since $ [ABC] \equal{} 84$, we get:

$ [PQR] \equal{} 84\cdot\left |\begin{array}{ccc}x_{1}& y_{1}& z_{1}\\ x_{2}& y_{2}& z_{2}\\ x_{3}& y_{3}& z_{3}\end{array}\right |$[/quote]

Mass points is basically a way of simplifying a problem for example if given a line of lenght 5x that is divided into two smaller lines 2x and 3x , we can give the end points "values" so that if you multiply the endpoint with the smaller line closest to it on both sides, the products are equal.


Let $x,y$ be real numbers. Show that if the set ${{cos(n\pi x)+cos(n\pi y)}}$, $n$ is a positive integer, is finite, then $x,y$ are rational


calculus
integration
logarithms
calculus computations

if ${ \int_{x}^{xy}\{ f(t) }\, dt$  is independent of x and f ( 2 ) = 2  ,find the value of  ${ \int_{1}^{x}\{ f(t) }\, dt$

Given $ F(y) \equal{} \int_{x}^{xy} f(t)dt$ with $ f(2) \equal{} 2$.
Find $ \int_{1}^{x} f(t)dt$.

$ F(y) \equal{} \int_{x}^{xy} f(t)dt \implies F'(y) \equal{} f(xy)x \implies \frac{F'(y)}{x} \equal{} f(xy)$.
$ f(2) \equal{} 2 \implies F'(y)\equal{} \frac{4}{y} \implies F(y) \equal{} 4 \log y$.
Hence, $ F(\frac{1}{x}) \equal{} \minus{}4 \log x \implies \int_{1}^{x} f(t)dt \equal{} 4 \log x$


In $\triangle ABC, AB = 4, BC = 5$, and $AC = 7$. Point $X$ is in the interior of the triangle
such that $AX^2 +BX^2 +CX^2$ is a minimum. What is $X$, and what is the value of this
expression?

[hide="an approach"]put the triangle on a cartesian coordinate system, and let X be (x,y), then do the distance forumla, and it will work out okay because squaring the distance formula makes the sum of 2 squares, and those are easy to minimize, just complete the square, and make sure your point is in the triangle[/hide]

[hide="solution"]

We begin by placing each point on the coordinate plane.  Let $A$ be at $(0,0)$, $B$ at $(0,4)$.  Using equations for circles, we have:

$x^2 + y^2 = 49$
$x^2 + (y-4)^2 = 25$

The second equation is equivilant to:

$x^2 + y^2 - 8y = 9$

Subtracting this from the original equation:

$8y = 40$, $y = 5$, and by substituting, $x = \sqrt{24} = 2\sqrt{6}$

Now we find $AX^2 + BX^2 + CX^2$:

$AX^2 + BX^2 + CX^2 =\& x^2 + y^2 + x^2 + (4-y)^2 + (2\sqrt{6} - x)^2 + (5 - y)^2$
$= 3x^2 + 3y^2 - 4\sqrt{6}x -18y +65$
$= 3(x - \frac{2\sqrt{6}}3)^2 + 3(y-3)^2 + 30$

The minimum value of this last expression is when the squared terms are zero, or when $X = (\frac{2\sqrt{6}}3, 3)$ (which is inside of the triangle).  30 is also the minimum value of $AX^2 + BX^2 + CX^2$ when the squared values are zero.

Looking back on alltheman's advice, this solution used the same approach.
[/hide]

[hide="Solution using well-known formula"]By Leibniz's formula, $AX^2+BX^2+CX^2=3GX^2+\frac{1}{3}(AB^2+BC^2+CA^2)$ where $G$ is the centroid of $\triangle ABC$. Therefore, the minimum of $AX^2+BX^2+CX^2$ is attained when $X$ is the centroid, whereas the required value is $\frac{1}{3}(4^2+5^2+7^2)=30$.[/hide]


\[ \frac{(x^2+2)}{\sqrt(x^2+3x+2)}  \]

write in interval notation

I assume I factor the denominator first, but I get confused from there  :(

I really need help, an entire assignment of mine depends on this

As stated the problem makes more sense.  You probably want to write something like

'express the solution set to $f(x)>0$ in interval notation', where $f$ is your above function.

Interval notation's really easy and simlifies writing solution sets greatly.  Here:

If the problem's asking for the domain of x, we have
 \frac{(x^2+2)}{\sqrt((x+1)(x+2))}

Looking for x s.t. the denominator is real, we find that x is greater than -1 and less than -2.  In interval notation, this becomes
(-infinity,-2) U (-1, infinity).

What does "looking for x s.t." mean?

so, in another problem of mine
\[ \frac{\sqrt(x-7)(x-8)(x-9)}{\sqrt(x-10)}  \]
would it be (10, infinity) ?

Are you sure aren't you searching for max & m in? In this case, max will be $\infty$ puttind $x\to 2$ V 1. for min, i've minimizet $\frac{x^2+2}{x^2+3x+2}$ for AM-GM, but looking for the original expression, i'd try a variable chanching

sorry about how my post above turned out, i'm just learning latex

[quote="mathmanman"]What is google?  :roll:
http://www.google.com/search?q=interval+notation[/quote]

Fine... what does that have to do with the given fraction? Or in other words, what are we supposed to do?

I guess you're supposed to find the domain of definition of this function... :)

Anyway, I don't think it is the good section to post this.


Let $ A_n$={$ (x,y)|x \in \mathbb{R}, \minus{}\frac 1 {n} < y < \frac 1 {n}$} be a subset of $ \mathbb{R}^2.$
Then for any $ n \in \mathbb{N}$,
Lebesgue measure of $ A_n$ = $ \infty$, right??

yes (for every $ n\in\mathbb{N}$ and every $ m\in\mathbb{R}_{\geq 0}$ the box $ [\minus{}mn/2,mn/2]\times [\minus{}1/(2n),1/(2n)]$ lies strictly inside $ A_n$ and has measure $ m$, so letting $ m\to\infty$ shows that $ m(A_n)\equal{}\infty$)


show that $ 2Arc sinx \equal{} Arc cos (1\minus{}2x^2)$

by the way..i am a complete beginner to trigonometry..so i would appreciate it if you explain it more in detail.

good day

Um, if you are a complete beginner to trig, why did you get this question? just asking

all I can think of is to take $ \tan$ of both sides and simplify, but there might be an easier way.

EDIT: actually, that won't turn out very pretty

EDIT2: oh wait, it said to show that...

For the equation to have sense we must have $ x\in [\minus{}1,1]$. So let's set $ x\equal{}\sin(\theta)$(note that this is always possible), remeber that $ 1\minus{}2x^2\equal{}1\minus{}2\sin^2(\theta)\equal{}\cos(2\theta)$

$ 2arc\sin(x)\equal{}2arc\sin(\sin(\theta))\equal{}2\theta\equal{}arc\cos(\cos(2\theta))\equal{}arc\cos(1\minus{}2\sin^2(\theta))\equal{}arc\cos(1\minus{}2x^2)$

 :)


limit
geometry
3D geometry
calculus
calculus computations

Find $\lim_{x \rightarrow 1} \frac{\sqrt x - x^2}{1 - \sqrt x}$.

My solution.

Rewrite the expression as $\lim_{x \rightarrow 1} - \frac{(\sqrt{x})^4 - \sqrt x}{1 - \sqrt{x}}$. Factoring out $\sqrt{x}$ in numeartor, we get $\lim_{x \rightarrow 1} - \frac{(\sqrt x)((\sqrt{x})^3 - 1)}{1 - \sqrt x}$. Note the difference of cube in there so we can rewrite it as:

$\frac{(\sqrt x)(1 - \sqrt{x})(x + \sqrt{x} + 1)}{1 - \sqrt x}$. Bottom cancels leaving $\lim_{x \rightarrow 1} \sqrt{x}(x+\sqrt x+ 1) = \lim_{x \rightarrow 1} 1(1+1+1) = 3$.

Yes it is true. Another solution is:

Using L'Hospital we have: ${L=\lim_{x \to 1} \frac{\frac{1}{2\sqrt{x}}-2x}{-\frac{1}{2\sqrt{x}}}=-2(\frac{1}{2} - 2})=3$

That is correct.

Here's a little tip, and the way I would do it: don't be afraid to make substitutions and rename things. In this particular case, I would let $u=\sqrt{x}.$ Note that $x\to1$ iff $u\to1.$ That makes our limit

$\lim_{u\to1}\frac{u-u^4}{1-u}=\lim_{u\to0}\frac{u(u-1)(u^2+u+1)}{u-1}$

And we can see that the limit will be 3.

Of course, this is exactly the algebra that you did. I haven't really added anything to your method. It's just that the substitution makes the whole thing so much easier to see.

Ah - a L'Hôpital post, while I was working on this. I figured that was coming. Stay with your algebra, Silverfalcon - you'll have a better understanding of what is going on. But even if you were going to use L'Hôpital, doing the above substitution first will make that easier as well.

Thanks for the response. The substitution is indeed "clever" even though it's not anything new.

My class didn't learn that "L law" yet even though I think I know what that is (I probably learned independently during summer but I can't remember). So yes, I try to solve these limits problems without using of "L law."


Is there a way to create conditional comments in LaTex?

For example, I would like to create a single file for problem sets and solutions.  Instead of using the % to comment out solutions, I would like to set a global switch at the beginning file to turn on and off the solutions.

You can use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=optional]optional package[/url] - the documentation is in ../tex/latex/ltxmisc/optional.sty.
For example:
[code]\documentclass{article}
\usepackage[solution]{optional}
\begin{document}

Question\\
What is $1 + 1$?

\opt{solution}{
Solution\\
2
}
\end{document}[/code] prints everything but if you change
\usepackage[solution]{optional}
to
\usepackage[nosolution]{optional}
then none of the text in \opt{solution}{...} is printed.
You can use words other than [i]solution[/i]  and [i]nosolution[/i] if you wish.
For very long optional text you put the text in an external file and use \opt{solution}{\input{[i]filename[/i]}}

[size=75]Edited to give a better solution.[/size]

The probsoln package is another possibility.