Somewhat standard result, I assume, but I am curious to see a fleshed-out proof.

Prove that if a category has products and equalizers, then it has all limits.

OK, let's see if I can guess how this would work....

Let $ \cal{C}$ be our category which is assumed to have products and equalizers. Let $ D$ be a diagram in $ \cal{C}$, that is a set of objects and certain morphisms between them. Specifically, suppose that the objects in $ D$ are $ \{X_i | i \in I\}$ for some index set $ I$, and the morphisms are $ \{\phi_j | j \in J\}$ for some index set $ J$, where $ \phi_j : X_{d(j)} \to X_{e(j)}$ for some maps $ d,e : J \to I$ which tell us what the domain and range of each morphism are.

We're supposed to construct a limit object for $ D$, which is an object $ L$ in $ \cal{C}$ and morphisms $ L \to X_i$ such that everything commutes and the appropriate universal property holds.

First, let's construct two product objects in $ C$ (using the assumption that products exist):

$ A \equal{} \Pi_{i \in I} X_i$ is the ordinary product of all the objects (disregarding any morphisms in $ D$)

$ B \equal{} \Pi_{j \in J} X_{e(j)}$ is the ordinary product, over all the morphisms, of the range of each morphism.

Notice that in order to define a mapping of anything into $ B$ you just need to define mappings, indexed by $ J$, into the individual objects $ X_{e(j)}$: by definition of a product, this determines uniquely a mapping into $ B$. Also observe that $ A$ comes equipped with projection maps $ \pi_i : A \to X_i$ for all $ i \in I$.

There are now two fairly natural maps we can construct from $ A$ to $ B$:

$ f: A \to B$ defined by $ f_j \equal{} \phi_j \circ \pi_{d(j)}$: first apply the projection from $ A$ to the domain of $ \phi_j$, and then apply $ \phi_j$ itself.

$ g: A \to B$ defined by $ g_j \equal{} \pi_{e(j)}$: just project into the range of $ \phi_j$.

As we noted, both $ f$ and $ g$ are defined as the product of the component maps $ f_j$ and $ g_j$ using the properties of $ B$ as a product.

Our limit object $ L$ is now defined as the equalizer of $ f$ and $ g$; more precisely, the equalizer $ L$ comes equipped with a morphism $ \psi: L \to A$, and we compose that with the projections $ \pi_i$ to obtain a set of morphisms $ \psi_i : L \to X_i$, as required for $ L$ to even be a candidate limit object.

Verifying that $ L$ and the $ \psi_i$ satisfy the requirements of a limit isn't too hard. First we need to show that everything commutes, which just says that for every $ j \in J$ we should have $ \psi_{e(j)} \equal{} \phi_j \circ \psi_{d(j)}$. This is indeed the case: since $ (L, \psi)$ is an equalizer, we have $ f \circ \psi \equal{} g \circ \psi$. The $ j$th component of this equality says that $ f_j \circ \psi \equal{} g_j \circ \psi$, namely $ \phi_j \circ \pi_{d((j)} \circ \psi \equal{} \pi_{e(j)} \circ \psi$ which is just what we need (recall that $ \psi_i \equal{} \pi_i \circ \psi)$.

Finally, if $ T$ is any test object equipped with a set of compatible morphisms $ \alpha_i : T \to X_i$ which commute with all the $ \phi_j$, we construct a mapping $ \alpha : T \to A$ as the product of all the $ \alpha_i$, and observe that $ \alpha$ equalizes $ f$ and $ g$ by the same calculation we just went through. Since $ L$ was the equalizer of $ f$ and $ g$, there must be a unique map $ T \to L$ as required by the universal property of the limit $ L$.

Hope this is detailed enough - it's a lot easier to follow with a diagram.


Let $ z\equal{}x\plus{}yi\in\mathbb C,$ where $ \{x,y\}\subset\mathbb R.$ Prove trhat
\[ |\sin z|\geq\frac{e^y\minus{}e^{\minus{}y}}{2}\]

Nice proof :)

I'm just confused about how you got to the third-to-last step from the one above it.

Thanks!

Factor into $ \left(\frac {e^y \plus{} e^{ \minus{} y}}2\right)^2(\sin^2x \plus{} \cos^2x) \equal{}\left(\frac {e^y \plus{} e^{ \minus{} y}}2\right)^2$.


linear algebra
matrix
inequalities
vector
linear algebra unsolved

Assume $ O \in M_{n}(\mathbb{R})$ , and $ O^{T}O\equal{}I_{n}$

and $ A\equal{} \left(\begin{array}{cccc}a_{1} &  &  & \\  & a_{2} &  & \\  & & ... & \\ &  &  & a_{n}\end{array}\right)$

if $ \lambda$ is eigenvalue of $ OA$

show that:

$ \min_{1 \leq j \leq n}|a_{j}| \leq |\lambda| \leq \max_{1 \leq j \leq n}|a_{j}|$

[quote="Carcul"]Is A a diagonal matrix whose elements are real?[/quote]

Yes :D

[hide="Right Inequality"]If $ \lambda$ is an eigenvalue of $ OA$ there is a non zero eigenvector $ x \equal{} (x_1,x_2, ..., x_n)$ such that $ \langle x,x \rangle \equal{} 1$ and $ OAx \equal{} \lambda x$. Therefore,

$ x^TOAx \equal{} \lambda x^Tx \equal{} \lambda$

$ \equal{} > |\lambda| \equal{} |x^TOAx| \equal{} |\langle O^Tx,Ax \rangle| \leq \sqrt {\langle O^Tx,O^Tx \rangle} \sqrt {\langle Ax,Ax \rangle}$

$ \equal{} \sqrt {x^TOO^Tx} \sqrt {\sum_i x_i^2a_i^2} \equal{} \sqrt {\sum_i x_i^2a_i^2} \leq \sqrt {\sum_i x_i^2 (max \,\,|a_i|)^2}$

$ \equal{} (max \,\,|a_i|) \sqrt {\sum_i x_i^2} \equal{} (max \,\,|a_i|) \sqrt {\langle x,x \rangle} \equal{} max \,\,|a_i|$.[/hide]

[hide="Left inequality"]If $ A \equal{} O_{n \times n}$ there is nothing to prove. For $ A \neq O_{n \times n}$ and $ \lambda$ an eigenvalue of A, there is a non-zero vector x such that $ \langle x,x \rangle \equal{} 1$ and $ OAx \equal{} \lambda x$ (1). Now multiplying both sides of (1) on the left by $ x^TAO^T$ we have

$ x^TAO^TOAx \equal{} \lambda x^TAO^T x \equal{}> x^TA^2x \equal{} \lambda \langle Ax, O^Tx \rangle$

$ \equal{}> | \lambda| \equal{} \frac{x^TA^2x}{|\langle Ax, O^Tx \rangle|} \geq \frac{x^TA^2x}{\sqrt{\langle Ax,Ax \rangle} \sqrt{\langle O^Tx,O^Tx \rangle}} \equal{} \frac{x^TA^2x}{\sqrt{x^TA^2x}} \equal{} \sqrt{x^TA^2x}$

$ \equal{} \sqrt{\sum_i x_i^2 a_i^2} \geq \sqrt{\sum_i x_i^2 (min \,|a_i|)^2} \equal{} min \, |a_i|$.[/hide]


Presuming you mean log to the base $ 10$, then taking logs of both sides to the base 10,

$ \log(x^{\log x}) > \log 10$

$ (\log x)^{2} > 1$

$ \log x > 1$ or $ \log x < \minus{}1$

$ x > 10$  or $ x < \frac{1}{10}$.


quadratics
number theory solved
number theory

Let $ n$ be a positive integer. Prove that the number $ 2^n\plus{}1$ does not have any prime divisor of the form $ 8k\minus{}1$, where $ k$ is a positive integer.

Very nice problem, was on the Shortlist for the Mathlinks Contest :D

Let q a prime factor of 2^n+1 such that q if 8k-1. Then n is odd ( since x^2+1 doesn not have prime factors of the form 4k+3). So, -2 is a quadratic residue modulo q. Since (-1/q)=(-1)^((q-1)/2) and (2/q)=(-1)^((q^2-1)/8) we find that (q-1)/2+(q^2-1)/8 is even, false.

(a/p) is the Legendre symbol. It is 1 if a is a quadratic residue modulo p and -1 in the other case.

oh, I understand !
Your solution is right.


let $ n$ be an even whole number, $ x_i \geq 1$ is real for $ i \equal{} 1,2,...,n$ and $ \prod_{i \equal{} 1}^{n} x_i \equal{} Q^n$
prove $ \sum_{i \equal{} 1}^{n} x_i x_{i \plus{} 1} \minus{} \sum_{i \equal{} 1}^{n} x_i \geq \frac {n}{2} (Q^2 \minus{} 1)$

($ x_{n \plus{} 1} \equal{} x_1$)

can this be solved by using the mixing variables method?
the case when $ n\equal{}2$ is very easy, but it gets hard when $ n\equal{}4$

Can you please tell what is the mixing variables mentod???


function
calculus
inequalities proposed
inequalities

let $x,y\ge0$ and $x+y=$ prove that:
$x^2y^2(x^2+y^2)\le2$

More general $(x > 0)\&(y>0)\&(n\geq 1) \Longrightarrow x^ny^n(x^2+y^2) \leq \frac{(x+y)^{2n+2}}{2^{2n+1}}$
One : $xy \leq  \frac{(x+y)^2}{4} $ and $x^ny^n(x^2+y^2)=x^ny^n((x+y)^2-2xy)$
Two : the function $f(t) = t^n(S^2-2t)$ , is increasing for $t\in[0,\frac{S^2}{4}]$, .. so here you are ..
 :cool:

$S=x^4y^2+x^2y^4=x^4(2-x)^2+(2-x)^4x^2$
Then we can use calculus to prove that $max(S)=2$


hello, i have computed the following system
$ x^2\equal{}36\plus{}y^2$
$ v^2\equal{}36\plus{}(18\minus{}y)^2$
$ u^2\plus{}100\equal{}v^2$
$ (u\plus{}x)^2\plus{}100\equal{}18^2$
solving these equations i have got $ x\equal{}\frac{54}{5}$.
Sonnhard.

Let the foot from $ C$ to $ \overline{AB}$ be $ D$ and from $ A$ to $ \overline{CB}$ be $ E$, so that $ \triangle BCD\sim\triangle BAE\implies\frac x6\equal{}\frac{18}{10}\equal{}\frac95\implies x\equal{}\boxed{\frac{54}5}$.

you guys are both wrong since they asked for a decimal.

sry for stupid revive


But it'll be proved by simple method..

For positive real numbers a,b,c,d, prove the inequality :

 \sum sym a^10 b^2 +  \sum sym a^4 b^3 c^3 d^2 >= 2 \sum sym a^7 b^3 c^2

( \sum sym a^p b^q c^r d^s means the sum of all a^p' b^q' c^r' d^s' as {p',q',r',s'} is all possible permutations of {p,q,r,s})

Thanks

Hehe the only nice thing that I see in this inequality is the fact that the degree of each term is the same.

OK I think I need help here.

Checking the possibility for a=b=c=d

If we count the permutations, i.e. the number of terms, for each 'symmetric' sum  \sum sym :

 \sum sym a 10 b 2  contains 4*3 = 12 terms
 \sum sym a 4 b 3 c 3 d 2  contains 4! = 24 terms
 \sum sym a 7 b 3 c 2  contains 4*3*2 = 24 terms
Since they are all of the same degree and after some algebraic simplification I quickly got 12a 12 +24a 12 =36a 12  for the LHS of the inequality but it is 48a 12  for the RHS of the inequality and that would rule out a lot of values, especially the case a=b=c=d. Is there something wrong or am I just missing something?

In symmetric sum for n variables, there are always n! terms, for example

For x, y, z, Sum_{sym} x^3 = 2(x^3+y^3+z^3)
                Sum_{sym} xyz = 6xyz

So Sum a^10*b^2 has 24 terms.

Namdung

OK so  \sum sym a 10 b 2 = \sum sym a 10 b 2 c 0 d 0

This should work out then. By the way, equality holds when a=b=c=d  :D

Now to simply matters if I can get to prove that

x 3 (y+z+t) + yzt( \sqrt (yz) +  \sqrt (yt) +  \sqrt (zt))  \geq  x \sqrt x ( yz( \sqrt y +  \sqrt z) + yt( \sqrt y +  \sqrt t) + zt( \sqrt z +  \sqrt t))

and that looks even more complex than the original :?


find the sum of an infinite sequence begining with 12/5 with a ratio of 4/5.

Use the fact that the sum of an infinite geometric series is $ \frac{a}{1\minus{}r}$, where $ a$ is first term and $ r$ is the common ratio. After that it is pure computation.

Also, here's a useful formula...

$ \sum_{n=1}^{infinity}\((\frac{x}{y})^n=\frac{x}{y-x}$

so in we rewrite the question as...

$ \frac{12}{5}+\frac{12}{5}(\frac{4}{5})+\frac{12}{5}(\frac{4}{5})^{2}+\frac{12}{5}(\frac{4}{5})^{3}+...=\frac{12}{5}[1+\frac{4}{5}+\frac{4}{5}^{2}+\frac{4}{5}^{3}+...]=\frac{12}{5}[1+\sum_{n=1}^{infinity}\((\frac{4}{5})^n]=\frac{12}{5}[1+\frac{4}{5-4}]=\frac{12}{5}[1+4]=\frac{12}{5}(5)=12$

[p.s. I'm not completely sure that the formula is absolutly correct, I spent 2 hours making it, It will take more time to verify it.]

Here's how to prove your formula:

$ \sum_{n \equal{} 1}^{\infty}\left(\frac xy\right)^n \equal{} \frac {\frac xy}{1 \minus{} \frac xy} \equal{} \frac {x}{y \minus{} x}$. This formula is a special case of the closed form of a geometric series (I wouldn't memorize it).

You should know this formula:

$ \sum_{k \equal{} 0}^{n} r^k \equal{} \frac {r^{n \plus{} 1} \minus{} 1}{r \minus{} 1}$

Proof:

$ S \equal{} \sum_{k \equal{} 0}^{n}r^k$

$ rS \equal{} \sum_{k \equal{} 0}^{n}r^{k \plus{} 1}$

$ (r \minus{} 1)S \equal{} \sum_{k \equal{} 0}^{n}r^{k \plus{} 1} \minus{} r^{k} \equal{} r^{n \plus{} 1} \minus{} 1$ (Telescoping)

And clearly $ \lim_{n\to\infty}\frac {r^{n \plus{} 1} \minus{} 1}{r \minus{} 1} \equal{} \frac {1}{1 \minus{} r}$ if $ |r| < 1$.


quadratics
Vieta
inequalities
symmetry
conics
parabola
absolute value

Let $\alpha$ and $\beta$ be the roots of the equation $mx^2-x-2=0$. Find all real numbers $m$ such that $|\alpha|<1$ and $|\beta|<1$.

Then $-1<\alpha\beta<1$ and $-2<\alpha+\beta<2$.
From Vieta's Formulas, $-2<\frac{-(-1)}{m}<2$ and $-1<\frac{-2}{m}<1$.
The first inequality rules that $m<\frac{-1}{2}$ or $m>\frac{1}{2}$ and the second says $m<-2$ or $m>2$.
Thus, $m<-2$ or $m>2$.

[quote="towersfreak2006"]$m<-2$ or $m>2$[/quote]
That's not correct :(
In fact, try $m=\frac{5}{2}$. One of the roots is $\frac{1+\sqrt{21}}{5}$, which is greater than 1.
Or, try $m=-3$. It doesn't have real solutions!

Sorry, maybe my wording in the problem was unclear. So I edited it a little bit. The problem is asking the necessary and sufficient condition.

Okay, let's try another approach then.  :blush:
$x=\frac{-(-1)\pm\sqrt{(-1)^2-(4)(m)(-2)}}{2m}=\frac{1\pm\sqrt{1+8m}}{2m}$
The bottom limit for $m$ is $\frac{-1}{8}$, and the numerator is "closer" to $1$ than to $-1$ (by virtue of the $\frac{1}{2}$). Hence, we want the largest $m$ such that $1+\sqrt{1+8m}<2m$.
$1+8m<4m^2-4m+1$
$0<4m(m-3)$, so $m>3$ or $m<0$.

But $|m|>1$ or the absolute value of one solution will be greater than $1$. So then $m>3$. Is this correct? :?

Yup, that's correct! ;)

There's another way to solve this problem.
[hide]There are some conditions we have to check.
But before starting, we divide both sides to get $x^2-\frac{1}{m}x-\frac{2}{m}=0$.

$(1)$ First, we must have $D\geq0 \Longleftrightarrow 1+8m\geq0 \Longleftrightarrow m\geq -\frac{1}{8}$

$(2)$ Second, the axis of symmetry of the graph $f(x)=x^2-\frac{1}{m}x-\frac{2}{m}$ has to be between $x=-1$ and $x=1$, so $-1<\frac{1}{2m}<1$, which leads to $m<-\frac{1}{2}$ or $m>\frac{1}{2}$.

$(3),(4)$ Lastly, since the graph of $f(x)$ opens up, we must have $f(1)>0$ and $f(-1)>0$.
So, $f(1)>0 \Longleftrightarrow m<0$ or $m>3$
and $f(-1)>0 \Longleftrightarrow m<0$ or $m>1$

From $(1), (2), (3), (4)$, we obtain $m>3$ :)[/hide]

If $m=0$, then the root of the given equation is $x=-2$, so this case isn't suitable.

So, we are to consider for $m\neq 0$.Then we can rewrite the given equation as $x^2=\frac{1}{m}(x+2)$.

Therefore we should find the condition for which the parabola $y=x^2\ (-1<x<1)$ have two interception points with

a line $y=\frac{1}{m}(x+2) $ passing through a fixed point $(-2,0)$ regardless of the value of $m$.

Your solution look more difficult than frt's solution!


limit
function
calculus
calculus computations

prove that if $ f$ is differentiable at $ a$ ,then exist positive real $ \delta$ such that $ f$ is continuous on $ \left( a \minus{} \delta,a \plus{} \delta\right)$

[color=green]Unnecessary empty lines removed by moderator[/color]

No need to cry for help. The statement is wrong as stated.

Elaborating on what fedja said:

Let $ f(x)\equal{}\begin{cases}x^2,&x\text{ rational}\\0,&x\text{ irrational}\end{cases}$

Then $ f$ is differentiable at $ 0,$ but is discontinuous at every point other than $ 0.$

(I'll assume that fedja briefly thought of an example like this before he typed the message above.)

[quote="Timestopper_STG"][quote="Kent Merryfield"]Then $ f$ is differentiable at $ 0,$ but is discontinuous at every point other than $ 0.$
[/quote]
Can you show your proof ? Thanks :lol:[/quote]

Take a rational point $ a$, other than zero. Then $ \lim_{x\to a_ \minus{} } f(x) \equal{} 0$, $ \lim_{x\to a_ \plus{} } f(x) \equal{} 0$ over irrationals, but $ a^2$ over rationals. So the limit doesn't exist (as there is infinitely many rational and irrational points in any neighborhood of $ a$. Same holds for irrational $ a$. If the limit doesn't exist, function is discontinuous in the point.

(proof edited :) )

Hmm.. I realize that when looking at $ f$ as one function, values of $ f(x)$ jump around and thus, the limit does not exist as $ x\to{0}$.

But just out of curiosity - wouldn't its graph [i]look[/i] as if it had a limit? $ f(x)$ would look like it was two separate functions, $ y \equal{} x^2$ and $ y \equal{} 0$, though they'd have infinitely many holes. Those two curves, however, each approach $ 0$ as $ x\to{0}$, so thinking this way, why couldn't one say that $ \lim\limits_{x\to{0}}f(x)$ actually exists?

No one said that the limit as $ x$ approaches zero doesn't exist; it [i]does[/i] exist.  That's why Kent Merryfield specified that the function is discontinuous everywhere except $ 0$.  (Review the definition of a limit.)

Notice I wasn't talking about zero. The function IS continuous in zero (even better-differentiable! so it is the point $ a$ from the statement), but in every neighborhood of $ a$, it is discontinuous (Ah, I just noticed I've used letter $ a$ in my proof: not really good notation-it shouldn't be confused with $ a$ from the O.P.'s conjecture).

oh snap.. I was looking at some other topic dealing with limits and probably mixed it up with this one. Combined with not thinking clearly, I made a total fool of myself. Sorry! :oops:


Let $x_{i}$ be the number of times $i$ chosen as $a_{j}, 0 \leq i \leq n, j \in$ ${1, 2, \cdots, k}$.
Then we must find the number of solutions of this equation
$x_{1}+x_{2}+\cdots+x_{n}= k$
Answer : $\binom{n+k-1}{k}$

Let $b_{i}=a_{i}+i$
Implies  $1\leq{b_{1}}<{b_{2}}<...<{b_{k}\leq{n+k}}$
Thus ,the number of set$(a_{1},...,a_{k})$ equals the number of set $(b_{1},...,b_{k})$ and equals ${n+k}\choose{k}$
:)

tmbtw, u made a mistake
$a_{i}$ must be natural, then $2 \leq b_{1}$ and the result follows...
$\binom{n+k-1}{k}$


It is true and well-known indeed.  :P

$(I)$ is the incircle of $\triangle ABC.$ Line $ BC$ is the polar of $ D$ with respect to  $(I).$ Line $ EF$ is the polar of $ A$ with respect to $(I)$ and $ P$ is the pole of the tangent $\tau$ to $(I)$ through $ P.$ Since $ A,P,D$ are collinear $ \Longrightarrow$ $ EF,BC$ and $\tau$ concur.

[quote="greentreeroad"]I thought of a statement but can't determine whether it is true.

Let ABC be a scalane triangle. Let D, E, F be points that its incircle touches BC, AC, AB respectively. Let AD intersect the incircle again at P. Are the three lines, BC, EF and the tangent line to the incircle at P, concurrent?[/quote]
$ (FEPD)\equal{}\minus{}1$ then BC,EF and the tangent line to the incircle at P are concurrent.

PFDE is a harmonic quad (the tangents at E and F to its circumcircle concur at A - on PD), hence the tangents at D and P concur on EF.

Best regards,
sunken rock


[hide]The answer for the first problem is 10.

We have a square with side length x and 3 yards are added to it making the side lenght x+3. Then the area of the square is 69 yards more so...

$(x+3)^{2}$=$x^{2}+69$

$x^{2}+6x+9$=$x^{2}+69$

6x+9=69

6x=60
x=10[/hide]

How high up is the person in question 98?

For number 98 we need to know whats going on in 97.

sorry... yeah 97 is essential to 98.

so the question is..
If a person drops a water balloon off the rooftop of a "100 foot building", the height of the water baloon is given by equation H= -16T + 100, where T is the seconds.

So basically its 100 feet.

appreciate the feedback.

[quote="jt787"]These questions were asked for me to do for my homework, and im really struggling to translate these into formulas. Please help me figure out the following quadratic equations.

94. A square's side is increased by 3 yards, which corresponds to an increase in the area by 69 square yards. How many yards is the side of the intial square?

98. If the person in the exercise 97 throws the water balloon downward with a speed of 5 feet per second, the height of the water balloon is given by the equation
H= -16T^2-5T+100 where the T is in seconds. When will the water balloon hit the ground?[/quote]
[hide="Number 94"]$(x+3)^{2}=x^{2}+69$

$x^{2}+6x+9=x^{2}+69$

$6x+9=69$

$x=10$[/hide]

[quote="sonny"]Is the question considering gravity?[/quote]

yes.. the question regards gravity, and how much time it would take for a balloon to hit the ground from a 100 foot building.

Well if its a constant speed of 5 meters per second its just 20 seconds. However, if you need to include gravity you need to know how fast gravity pulls an object downwards.

ok i know whats wrong here. So yeah the answer for 97 is 2.5 seconds after it is dropped. Does that help?

[hide]Okay,this is what I get

If gravity is pulling the ball down at 5 feet per second then

its going to take the ball $2sqrt10$ seconds.

If the person already gives the ball a "push" of 5 feet per second then (considering gravity pulls an object down at 16 feet per second) the answer is approximately 3.23. I maybe wrong.[/hide]

[quote="jt787"]94. A square's side is increased by 3 yards, which corresponds to an increase in the area by 69 square yards. How many yards is the side of the intial square?

98. If the person in the exercise 97 throws the water balloon downward with a speed of 5 feet per second, the height of the water balloon is given by the equation
H= -16T^2-5T+100 where the T is in seconds. When will the water balloon hit the ground?[/quote]

[hide="94"]
$(x+3)^{2}-x^{2}=(3)(2x+3)=69 \Rightarrow x=10$[/hide]

[hide="98"]
We're given the equation for the balloon's distance, aren't we? Just put the height as $0$, then we have
$0=16t^{2}+5t-100 \stackrel{\text{Quad Formula}}{\Longrightarrow}t=\frac{-5 \pm \sqrt{6425}}{32}$. Negative root is extraneous (can't have negative time), so the time is $t=\frac{-5+\sqrt{6425}}{32}\approx 2.35$[/hide]

My previous answer is incorrect. The equation given is for that particular problem not a general equation. The answer can be derived using the formula

$-b+sqrtb^{2}-4ac/2a$

where a,b,c are : ax^2+bx+c

ans is  2.35


geometry
linear algebra
matrix
parallelogram
vector
trigonometry
linear algebra unsolved

Let we have $ \vec{a}$ and $ \vec{b}$ in $ \mathbb{R}^{2}$. The linear operator $ A$ from $ \mathbb{R}^{2}$ in $ \mathbb{R}^{2}$ works as:
\[ A\vec{x}\equal{}(\vec{a}\cdot\vec{x})\vec{a}\plus{}(\vec{b}\cdot\vec{x})\vec{b}\].
Find the needed and sufficient condition, which must hold for $ \vec{a}, \vec{b}$, such that the operator $ A$ will preserve the area of paralerograms.

Let $ P$ be the $ 2\times 2$ matrix whose columns are $ \vec{a}$ and $ \vec{b}.$ That is, $ P\equal{}\begin{bmatrix}a_1&b_1\\a_2&b_2\end{bmatrix}.$

The given linear transformation has the matrix $ A\equal{}PP^T.$

A linear transformation preserves the areas of quadrilaterals iff its matrix has determinant $ 1.$ (Or $ \minus{}1,$ but $ \det(PP^T)\ge 0.$)

Since $ \det(PP^T)\equal{}\det(P)\det(P^T)\equal{}(\det(P))^2,$ we must have $ \det(P)\equal{}\pm1.$ That is the condition you're looking for: $ |a_1b_2\minus{}a_2b_1|\equal{}1.$

Geometrically, this is the condition that $ \vec{a}$ and $ \vec{b}$ themselves span a parallelogram of area $ 1.$

Thank you, professor Kent, for answering on my three tasks.

I knew there must exist more straightforward solution, like yours, since I worked with vector product and directed angles for the condition $ |\vec{a} \times \vec{b}|\equal{}1$ to obtain.

WLOG we can put $ \mathbb{R}^{2}\hookrightarrow \mathbb{R}^{3}$ for the vector product to be defined.

We have to find the conditions for:
\[ |v_{1} \times v_{2}|\equal{}|Av_{1} \times Av_{2}|\]
\[ |Av_{1} \times Av_{2}|\equal{}|\vec{a}\times \vec{b}|\left(\left\langle a,v_{1} \right\rangle \left\langle b,v_{2} \right\rangle \minus{} \left\langle a,v_{2} \right\rangle \left\langle b,v_{1} \right\rangle \right)\]
Now
\[ \left\langle a,v_{1} \right\rangle \left\langle b,v_{2} \right\rangle \minus{} \left\langle a,v_{2} \right\rangle \left\langle b,v_{1} \right\rangle\equal{} |a\parallel{}b\parallel{}v_{1}\parallel{}v_{2}|(\cos(\angle a,v_{1})\cos(\angle b,v_{2})\minus{}\cos(\angle a,v_{2})\cos(\angle b,v_{1}))\]
Further on
\[ \cos(\angle a,v_{1})\cos(\angle b,v_{2})\minus{}\cos(\angle a,v_{2})\cos(\angle b,v_{1}))\equal{}\frac{1}{2}\left(\cos(\angle a,v_{1}\plus{}\angle b,v_{2})\plus{}\cos(\angle a,v_{1}\minus{}\angle b,v_{2})\minus{}\cos(\angle a,v_{2}\plus{}\angle b,v_{1})\minus{}\cos(\angle a,v_{2}\minus{}\angle b,v_{1})\right)\equal{}
\frac{1}{2}\left(\cos(\angle a,b \plus{} \angle v_{1},v_{2})\minus{}\cos(\angle a,b \minus{} \angle v_{1},v_{2}) \right)
\equal{} \sin{(\angle v_{1}, v_{2})}\sin{(\angle a, b)}\]

Hence we get $ |v_{1} \times v_{2}|\equal{}|Av_{1} \times Av_{2}| \Leftrightarrow |a \times b|^{2}\equal{}1$


Let $ABCD$ be a convex quadrilateral such that $AC \perp BD$.

(a) Prove that $AB^2 + CD^2 = BC^2 + DA^2$.

(b) Let $PQRS$ be a convex quadrilateral such that $PQ = AB$, $QR = BC$, $RS = CD$ and $SP = DA$. Prove that $PR \perp QS$.

[hide="1)"]
Let $O=AC \cap BD$

Since $ABCD$ is convex, $O$ is in interior of it.

$AB^2 + CD^2 = (OA^2+OB^2) + (OC^2+OD^2)$
$BC^2 + DA^2 = (OB^2+OC^2) + (OD^2+OA^2)$

So, $AB^2 + CD^2 = BC^2 + DA^2$

[/hide]

[hide="2)"]
So we are asked to state the inverse: If $PQ^2+RS^2 = RQ^2+PS^2$ then $PR \perp QS$

We have $PQ^2-PS^2 = RQ^2-RS^2$

We construct a convex quadrilateral $PQRS$ with the sides of $ABCD$, as asked.
Of course, the diagonals intersect in the interior of $PQRS$.

In the triangle $PQS$ we bring the altitude $PK$. From the right triangles we get:

$\begin{array}{cc} PQ^2 = QK^2+KP^2 \\ PS^2 = SK^2+KP^2 \end{array}\} \Rightarrow PQ^2-PS^2 = QK^2 -SK^2\Rightarrow$

$PQ^2-PS^2 = (QK+SK)(QK-SK)$

Similarly, in the triangle $RQS$ we bring the altitude $RM$.

$\begin{array}{cc} RQ^2 = QM^2+MR^2\\ RS^2 = SM^2+MR^2 \end{array}\} \Rightarrow RQ^2-RS^2 = QM^2-SM^2\Rightarrow$

$RQ^2-RS^2 = (QM+SM)(QM-SM)$

If $PQ=PS$ then $PQRS$ is rhombus, so the diagonals intersect perpendicularly.

[img]http://tinypic.com/f39fdc.gif[/img][img]http://tinypic.com/f39bno.gif[/img][img]http://tinypic.com/f39cuq.gif[/img]

So, say that $PQ>PS$

$PQ^2-PS^2 = RQ^2-RS^2 \Rightarrow$

$(QK+SK)(QK-SK) = (QM+SM)(QM-SM) \ \ \ (1)$

Notice that the two sides of this equation are positive since $PQ^2-PS^2>0$

So, $QK>SK$ and $QM>SM$

Hence $K,M$ are on the ray $QS$ of course.

We'll show that  $K,M$  can't be on different sides of the point $S$, and if they are on the same side of $S$, then they coincide.

Suppose that $K$ is between $Q,S$.Then $QK+SK=QS$

$a)$ If $M$ is between $Q$ and $S$, then from $(1)$ we get $QS(QK-SK) = QS(QM-SM) \Rightarrow M \equiv S$

$b)$ If $M$ is on ray $QS$ such that $S$ is between $Q,M$

Then from $(1)$

$QS(QK-SK) = (QM+SM)(QM-SM)$
$QS(QK-SK) = (QM+SM)QS$
$QK-SK = QM+SM$

we get a false result because $QK-SK < QK < QS < QM < QM+SM$

So $M$ is between $Q,S$ hence $M \equiv K \Rightarrow PR \perp QS$

On the other hand, we could get the conclusion that $K,M$ are both out of segment $QS$, and they coincide, but then $PQRS$ whould be concave.[/hide]

I think , I have seen a beautyfull solution in << South Africa " s Olympiad >> , using areas. I do not remember the year of that olympiad. If someone has free time ,  can find this solution and write it down.

 [u]Babis[/u]

This proposed problem shows that

[b]"a quadrilateral [/b]$ABCD$ [b]is orthodiagonal iff[/b] $AB^2+CD^2=AD^2+BC^2$[b]".[/b]

Is it this one? :)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=south+africa&t=55896[/url]

Arne thanks. Yes !!! This is the solution. I had seen it in the << south Africa's olympiad >> site and not here.
Now I see that this forum is very ...rich !

 I have seen also a proof with vectors and inner product.

 [u]Babis[/u]

$AB^{2}+CD^{2}= AD^{2} +BC^{2} \Leftrightarrow A^{2} -2 \cdot A\cdot B+ B^{2} + C^{2} - 2\cdot C\cdot D+D^{2} = A^{2} -2 \cdot A\cdot D+ D^{2} + C^{2} - 2\cdot C\cdot B+B^{2} \Leftrightarrow (A-C)\cdot (B-D)= 0 \Leftrightarrow CA \bot DB$

[color=red][b]Lemma.[/b] Let $A,B$ be two fixed points. Then the geometrical locus of the point $L$ for which $LA^2-LB^2=k$ (a real constant) is a line $l$, where $l\perp AB$ .[/color]

[color=blue][u]Again you kill a butterfly with an atomic bomb ![/u][/color]

(a) Let $E$ be the intersection point of $AC$ and $BD$. It's in the interior of $ABCD$ since $ABCD$ is convex. Thus: $AB^2+CD^2=BE^2+AE^2+ED^2+EC^2=AD^2+BC^2$.

(b) Let $K$ be the intersection point of $PR$ and $SQ$. Again, it is in $PQRS$ since $PQRS$ is convex.

Now, lets call $\widehat{PKS}$ the angle $\alpha$. Then, if we use the cosine rule:

$SP^2+QR^2=PK^2+KS^2+KQ^2+KR^2-2 PK\cdot KS cos(\alpha)-2 KQ \cdot KR cos(\alpha)$
$=SR^2+PQ^2=PK^2+KS^2+KQ^2+KR^2+2 PK\cdot KQ cos(\alpha)-2 KS \cdot KR cos(\alpha)$.
From this it follows that $-cos(\alpha)(KS \cdot PK + KQ \cdot KR)=cos(\alpha)(KS \cdot RK + KQ \cdot KP)$. But, since distances are positive, if $cos(\alpha) \neq 0$, this means a negative number equals a positive number, contradiction! Thus $cos(\alpha)=0$ and hence $\alpha=\pi/2$. Q.E.D.


function
calculus
derivative
integration
real analysis
real analysis unsolved

A function $f:\mathbb R\to\mathbb R$ has the property that for every $x,y\in\mathbb R$ there is a $t\in(0,1)$ depending on $x$ and $y$ s.t. $f(tx+(1-t)y)=tf(x)+(1-t)f(y)$.

Is it true that $f\left(\frac{x+y}2\right)=\frac{f(x)+f(y)}2$ for all $x,y$?

May be I am saying something stupid but whatever.
Let $\psi$ be a function which is continious,but does not have derivative in any point(Van der Vaarden function for example).Then take $f(x)=\int_{0}^{x}{\psi}$.We have that $\psi$ is eveywhere nonmonotonic,so $f$ is everywhere nonconvex(nonconcave),so(since $f$ is continious) the condition holds.On the other hand the condition
$f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}$,
does not hold since otherwise we would have that $f$ is linear(since $f$ is continious).So the answer is no.
Sorry if I say something stupid.

[quote="LevonNurbekian"]We have that $\psi$ is eveywhere nonmonotonic,so $f$ is everywhere nonconvex(nonconcave),so(since $f$ is continious) the condition holds.[/quote] I don't understand this part. The function may be everywhere non-convex but its graph may lie completely under its secant line on some interval. Do I miss something? :?


For how many positive integers $ m$ as $ \frac{2004}{m^{2}\minus{}2}$ is a positive integer?

A) one
B) two
C) three
D) four
E) greater than 4

$ 2004\equal{}2^2\times3\times167$

So,
if $ m^2\minus{}2\equal{}1$, no solution,
if $ m^2\minus{}2\equal{}2$, $ m\equal{}\pm2$,
if $ m^2\minus{}2\equal{}3$, no solution,
if $ m^2\minus{}2\equal{}4$, no solution,
if $ m^2\minus{}2\equal{}6$, no solution,
if $ m^2\minus{}2\equal{}12$, no solution,
if $ m^2\minus{}2\equal{}167$, $ m\equal{}\pm13$,
if $ m^2\minus{}2\equal{}334$, no solution,
if $ m^2\minus{}2\equal{}501$, no solution,
if $ m^2\minus{}2\equal{}668$, no solution,
if $ m^2\minus{}2\equal{}1002$, no solution,
if $ m^2\minus{}2\equal{}2004$, no solution.

Therefore, $ m\equal{}\pm2, \pm13$.
Thus $ \boxed{B}$.

Well, very interested. But why $ m^{2}\minus{}2\equal{}1$ ?
Can you explain me ?

If $ m^2\minus{}2\equal{}1$, $ \frac{2004}{m^2\minus{}2}\equal{}\frac{2004}{1}\equal{}2004\in\text{positive integer}$.
But, $ m^2\minus{}2\equal{}1 \iff m\equal{}\pm\sqrt{3}\not\in\text{integer}$.

Ah thanks  :lol:
Very good. I understand now. Thank you very much.


Haber si vamos poniendo  en este topic ..
los eventos que se realizaran en este año...

quisieran que sigan este esquema:

[color=blue]1. [NOMBRE- DENOMINACION]

2. [LUGAR Y FECHA]

4. [NUM DE PARTICIPANTES, REQUISITOS DE CLASIFICACION,
NIVELES, CARACTERISTICAS DE  EXAMENES, ETC][/color]

-------------

Bueno, gracias

Carlos Bravo ;)
Lima - Peru

[color=blue] [size=150]XVIII Asian Pacific Mathematics Olympiad[/size][/color]

[color=blue]FECHA:
 - La tarde del lunes 13 de marzo[para America]
 - La mañana del Martes 14  de marzo[para la costa oeste del pacifico][/color]

Para mayor info: adjunto.

CARLOS BRAVO ;)

[quote="M4RI0"]Hay prueba de selección o asi nomas?[/quote]

No hay selección, tenés que ser seleccionado por tu país (por lo menos así es en Argentina)

[quote="DarthTenebrus"]En Peru solo tienes que ir a la Oficina de Matematicas de la PUCP e inscribirte ( nombre , edad , colegio , profesor asesor ) luego te acercas el lunes al lugar designado para el examen en la misma PUCP y listo .[/quote]

Mirá!!! En Argentina, en cambio, el comité olímpico te elige por tu actuación en la olimpíada nacional de ese año... van aproximadamente 40 o 50 personas.

[quote="DarthTenebrus"]En Peru solo tienes que ir a la Oficina de Matematicas de la PUCP e inscribirte ( nombre , edad , colegio , profesor asesor ) luego te acercas el lunes al lugar designado para el examen en la misma PUCP y listo .[/quote]

¿se paga algo :?:

En Perú no se paga ... ya sea una olimpiada (mayo, APMO) o examen de selección (IMO, Ibero ,Cono Sur)

$Tipe$

Bueno pues me falto decir que en Peru se puede inscribir tambien por fax pero ahorita no recuerdo el fax de la seccion de matematicas de la PUCP . Valga la aclaracion jeje .

una cosita mas pero esto que quede entre nos ( jaja ) :
Si uno se olvida de inscribirse ( ha pasado a veces ) y justo cuando se levanta se recuerda que ese dia era la prueba ( Mayo , APMO , Selectivos , etc )  entonces facil que se toma un taxi a la PUCP pero a la hora de entrar a la PUCP se va a tener problemas porque solo dejan pasan a los que se inscribieron pero por alli que uno le habla al encargado dle ingreso y entra . Ahora dentro de la PUCP uno se va corriendo hacia el aula en que se rinde el examen y si aun no ha pasado la media hora de iniciado el examen facil que uno entra a dar el examen
y durante el examen cuando te llega la lista de asistencia uno tiene que poner su nombre al final de la lista ( claro pues si no estan inscritos asi tien que ser )

Bueno asi ha pasado  a veces pero yo les recomiendo a todos los que rinden la prueba en Peru lo siguiente :

Hay que ser responsable con eso de la inscripcion porque eso de  hacerlo a ultima hora no es bueno . Por cierto las inscripciones empiezan aprox una semana antes del evento . y Sean puntuales

[quote="DarthTenebrus"]
Bueno asi ha pasado  a veces pero yo les recomiendo a todos los que rinden la prueba en Peru lo siguiente :

Hay que ser responsable con eso de la inscripcion porque eso de  hacerlo a ultima hora no es bueno . Por cierto las inscripciones empiezan aprox una semana antes del evento . y Sean puntuales[/quote]

Queda clara tu recomendacion .DarthTenebrus

Gracias ...

Carlos  Bravo ;)
Lima - Peru.


How many ordered pairs $ (x,\,y)$ with $ x$ and $ y$ rational numbers, are solutions to the equation $ x^2\plus{}y^2\equal{}4x\plus{}6y\minus{}15$?

hello, i have $ (x\minus{}2)^2\plus{}(y\minus{}3)^2\equal{}\minus{}2$? and here is no rational solution.
Sonnhard.

[hide]
$ x^2 \plus{} y^2 \equal{} 4x\plus{}6y\minus{}15$
Move the $ 4x\plus{}6y$ to the other side and complete the square.
$ x^2\minus{}4x\plus{}y^2\minus{}6y\equal{}\minus{}15$
$ (x^2\minus{}4x\plus{}4)\plus{}(y^2\minus{}6y\plus{}9)\equal{}\minus{}15\plus{}4\plus{}9$
$ (x\minus{}2)^2 \plus{} (y\minus{}3)^2\equal{}\minus{}2$
Since the square of a number is always positive, the answer is $ \boxed{0}$.
[/hide]

Are you allowed to post these problems?  Ask Mr. Sanders before you do so; it very clearly says "Copyright" at the bottom of the tests, and on the mathleague.org site they have these for [b]sale[/b]...

I didn't intend to post many problems; I just wanted to share some of the more interesting ones. In fact, the 4 that have already been posted are all that I wanted to post. However, since I'm not attempting to reproduce the entire contents of the competition, and definitely not for commercial purposes, I'm pretty sure that this doesn't infringe on the copyright. Still, since I'm no expert, I won't post any more problems from their competitions to avoid complications.


parameterization
limit
logarithms
calculus
derivative
function
calculus computations

find the following limit, if $ a,b >0$:
$ \lim_{n\to 0} \left (\frac{a^n\plus{}b^n}{2} \right )^{\frac{1}{n}}$

WLOG, $ a\ge b.$ Then

$ \frac{a^n}2\le\frac{a^n\plus{}b^n}2\le a^n$

$ \left(\frac{a^n}2\right)^{\frac1n}\le\left(\frac{a^n\plus{}b^n}2\right)^{\frac1n}
\le(a^n)^{1/n}$

$ \frac{a}{2^{1/n}}\le\left(\frac{a^n\plus{}b^n}2\right)^{\frac1n}\le a$

Since $ 2^{1/n}\to 1,$ the squeeze theorem is in operation, and the limit is $ a.$

Reexamining that "WLOG" gives us

$ \lim_{n\to\infty}\left(\frac{a^n\plus{}b^n}2\right)^{\frac1n}\equal{}\max(a,b).$

---

Let $ M_p(a,b)\equal{}\left(\frac{a^p\plus{}b^p}2\right)^{\frac1p}$ for any real $ p\ne0.$

We just showed that $ \lim_{p\to\infty}M_p(a,b)\equal{}\max(a,b).$ Now find two more limits:

$ \lim_{p\to0}M_p(a,b)\equal{}?$

$ \lim_{p\to\minus{}\infty}M_p(a,b)\equal{}?$

I tried using L'hospital rule. But I'm not sure of one step.

Lets take the "$ \ln$" of it. Then:
we get
$ \frac{1}{n}\ln\left ( \frac{a^n\plus{}b^n}{2} \right )$

Now if we take the limit of this we can apply the L'hospital rule, since the numerator and denominator both $ \to 0$ if $ n\to 0$:
$ \lim_{n\to 0} \frac{1}{n}\ln\left ( \frac{a^n\plus{}b^n}{2} \right ) \equal{} \lim_{n\to 0} \frac{2(a^n \ln a \plus{}b^n \ln b)}{2(a^n\plus{}b^n)}\equal{}\frac12(\ln a \plus{}\ln b) \equal{}\frac12\ln ab$

So we got :
$ \lim_{n\to 0}\ln\left ( \frac{a^n\plus{}b^n}{2} \right )^{\frac{1}{n}} \equal{}\frac12\ln ab$

Now here is the step i'm not sure about.

if $ \lim \ln f(x) \equal{} \ln g(x)$ then $ \lim f(x) \equal{} g(x)$ is this true?

if it is then:
$ \lim_{n\to 0}\left ( \frac{a^n\plus{}b^n}{2} \right )^{\frac{1}{n}}\equal{}\sqrt{ab}$

[quote="1.01math1.01"]I tried using L'hospital rule. But I'm not sure of one step.

Lets take the "$ \ln$" of it. Then:
we get
$ \frac {1}{n}\ln\left ( \frac {a^n \plus{} b^n}{2} \right )$

Now if we take the limit of this we can apply the L'hospital rule, since the numerator and denominator both $ \to 0$ if $ n\to 0$:
$ \lim_{n\to 0} \frac {1}{n}\ln\left ( \frac {a^n \plus{} b^n}{2} \right ) \equal{} \lim_{n\to 0} \frac {2(a^n \ln a \plus{} b^n \ln b)}{2(a^n \plus{} b^n)} \equal{} \frac12(\ln a \plus{} \ln b) \equal{} \frac12\ln ab$

So we got :
$ \lim_{n\to 0}\ln\left ( \frac {a^n \plus{} b^n}{2} \right )^{\frac {1}{n}} \equal{} \frac12\ln ab$

Now here is the step i'm not sure about.

if $ \lim \ln f(x) \equal{} \ln g(x)$ then $ \lim f(x) \equal{} g(x)$ is this true?[/quote]
It is wrong.
Here is an other solution:
Lemma :If $ \lim_{n\to\infty}\frac {x_{n \plus{} 1}}{x_n} \equal{} l$ then
$ \lim_{n\to\infty}\sqrt [n]{x_n} \equal{} l$
Let $ x_n \equal{} \frac {a^n \plus{} b^n}{2}$
$ \lim_{n\to\infty}\frac{x_{n \plus{} 1}}{x_n} \equal{} \max\{a,b\}$
So $ lim_{n\to\infty}(\frac {a^n \plus{} b^n}{2})^{\frac {1}{n}} \equal{} \max\{a,b\}$

[quote="1.01math1.01"]
if $ \lim \ln f(x) \equal{} \ln g(x)$ then $ \lim f(x) \equal{} g(x)$ is this true?[/quote]
I think you should write  " if $ \lim \ln f(x) \equal{} \ln g$ then $ \lim f(x) \equal{} g$ "  , and that's right because of $ f(x)\equal{}e^{\ln f(x)},\ g\equal{}e^{\ln g}$  :cool:

Take note, it's $ n\rightarrow 0$.
[hide]
The natural logarithm of $ \left( \frac {a^n + b^n}{2}\right)^\frac {1}{n}$ is
\[{ \frac {\ln \left( \frac {a^n + b^n}{2}\right)}{n}}
\]
Now letting $ f(n) = \ln \left( \frac {a^n + b^n}{2}\right)$, this limit as $ n\rightarrow 0$ of the above is $ f^\prime (0)$ by the definition of the derivative, since $ f(0) = 0$. Now
\[ f^\prime (n) = \frac {2}{a^n + b^n}\cdot \frac {a^n\ln a + b^n\ln b}{2}
\]
so that $ f^\prime (0) = \frac {\ln a + \ln b}{2}$.

Finally, the exponential function is continuous, so
\[ \lim_{n\rightarrow 0} \left( \frac {a^n + b^n}{2}\right)^\frac {1}{n} = e^\frac {\ln a + \ln b}{2} = \sqrt {ab}
\]
so your answer is correct.[/hide]

[quote="TTsphn"][quote="scorpius119"]

Now letting $ f(n) \equal{} \ln \left( \frac {a^n \plus{} b^n}{2}\right)$, this limit as $ n\rightarrow 0$ of the above is
so your answer is correct.[/hide][/quote]
Why $ n\to 0$ and it is not your function.
With your method you can solve as follow :
$ x_n \equal{} \frac {\ln(a^n \plus{} b^n)}{n}$
Apply Stolz lemma we have:
$ \lim_{n\to\infty}{x_n} \equal{} \lim_{n\to\infty}\ln(a^{n \plus{} 1} \plus{} b^{n \plus{} 1}) \minus{} \ln(a^n \plus{} b^n) \equal{} \lim_{n\to\infty}\ln(\frac {a^{n \plus{} 1} \plus{} b^{n \plus{} 1}}{a^n \plus{} b^n}$
$ \equal{} \ln(\max\{a,b\})$[/quote]

Please read the first post carefully, it is $ n \to 0$, not infinity.


What are the member rankings that go up as post count increases, and what are the post counts for each? (im saying rankings like Newbie, P versus NP, Riemannian Hypothesis, Navier-Strokes Equations, Poincare Conjecture, etc.)

0-19, New Member
20-49, P vs. NP
50-99, Hodge
100-249, Poincare
250-499, Rieman Hypothesis
500... I'm not sure

Does anyone search anymore?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159956

By the way, again: DON't taLK lIkE ThiS.


Find all functions such that $f: \mathbb{N}\rightarrow\mathbb{N}$ and
\[ \sum_{k=1}^nf(2k)=f(n)f(n+1)  \]
Is the identity function the only function satisfying this?  Prove it.

Masoud Zargar

No, the identity function is not. Here's a counter-example.

f(x) = 0

Then both sides are 0.

 :)

You know that by the identity function I mean $f(x)=x$, right?

He did manage to find a counter-example to the second part of your question :)

Also:  Are we working in $\mathbb{N}$ or what?

Edit:  Note that $f(n) \left( f(n+1) - f(n-1) \right) = f(2n)$.  I'm not totally sure where I can go with this, but every value of $f(2n)$ can be written this way, which puts some bounds on other possible solutions.

Yes, I understand that f(x) = x is the identity function.

However, f(x) = 0 also is a function that works, which anwers the second part of your question.

Now to answer the first part...

I always forget to mentin the properties of the numbers. :( Fixed.


Let $ n$ be a positive integer and $ S_n$ is the set of all positive divisors of n .
a) Prove that at most half of element in $ S_n$ with their last digits equal to 3 .
b) Prove that at most half of element in $ S_n$ with their last digit equal to 7 .

Seems that it's not that easy. :lol:

See a generalization here: http://mathnerds.com/mathnerds/best/DivisorExcess/solution.aspx

It is easy.
See also http://www.mathlinks.ro/viewtopic.php?t=113024 for a more general question.


algebra
polynomial
number theory solved
number theory

Does the equation: n!x^n+(n^2+n+1)x^(n-1)-(n!+1)x+(2n)!+3=0 n>=2 have integer solutions?

The answer is: nope :D
why? well the polinomial p(x)=n!x^n+(n^2+n+1)x^(n-1)-(n!+1)x+(2n)!+3 is from Z[x] and it has  (2n)!+3 as a "free term" which is odd, and we also have another two odd coeficiens:
n^2+n+1=n(n+1)+1 and n!+1 which is also odd cause n>=2.

cheers! hope it's correct!


inequalities
geometry
perimeter
inequalities solved

Let a,b,c be real numbers such that 0<a,b,c<1/2 and a+b+c=1.

Prove that

(3 \sqrt 3)abc \geq  \sqrt (1-2a) \sqrt (1-2b) \sqrt (1-2c) \geq  \sqrt 3/3(3-(a 2 +b 2 +c 2 ))

Upps, the left hand side is stronger than Schur. Doesn't matter, we will be brave!. Consider the traingle formed by a,b,c. Then sqrt((1-2a)(1-2b)(1-2c))=4S, where S is the area. So we have to prove that 4S<=3sqrt(3)abc, or R>=1/(3sqrt(3)), where R is the circumradius. But this comes down to sum sinA<=3sqrt(3)/2, which is true.
  Regarding the right side, I really do not understand the inequality. Is 3-(a^2+b^2+c^2) under the radical or not? Could you please write it more explicitely? Thank you.

Sorry, I am so inattentive.

The right side is  ( \sqrt 3/3)(3-8(a 2 +b 2 +c 2 ))

Sorry, again.

Right inequality is quite hard and interesting. It can be formulated as follow: Let a, b, c be sides of a triangle with perimeter 1 and area S. Show that
              4S + 8(a 2+b2+c2)/\sqrt 3  \geq  \sqrt 3

This inequality deserves observation.

Namdung

Yeah, I've also tried to prove it, but I failed.

Here is my solution:

Dehomogenusize, we have to show that
sqrt(3)sqrt((a+b-c)(b+c-a)(c+a-b)(a+b+c)) + 8(a^2+b^2+c^2) >= 3(a+b+c)^2 (1).

Put a = y^2 + z^2, b = z^2+x^2, c = x^2 + y^2, (1) becomes,
sqrt(3)sqrt(x^2+y^2+z^2)xyz >= (x+y+z)(x+y-z)(y+z-x)(z+x-y) which is triavial.

Note that the equality holds when a=b=c=1/3 and {a,b,c}={1/2,1/2,0}. That why the initial inequality is so hard to prove.

Namdung


calculus
integration
inequalities
logarithms
calculus computations

Prove without computing the integral that:

$ \int_{1}^{x} \frac{dt}{\sqrt{1\plus{}t^2}} \le ln(x)$

Does this work?

We know that for $ t > 0$,

$ \frac{1}{\sqrt{1\plus{}t^2}} \leq \frac{1}{t}$

Integrate both sides (I'm sure that some sort of statement would need to be used to justify this),

$ \int_1^x \frac{dt}{\sqrt{1\plus{}t^2}} \leq \int_1^x \frac{dt}{t}$

And this gives us the statement of:

$ \int_1^x \frac{dt}{\sqrt{1\plus{}t^2}} \leq \ln x$


Prove that if a triangle has sides $ a, b ,$ and $ c$ such that they satisfy $ a\plus{}b\minus{}c\equal{}2$ and $ 2ab\minus{}c^2 \equal{} 4$, then the triangle must be equilateral.

This has been posted before:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216691]Link[/url]


let $ \lambda \geq 8$ and $ a$ , $ b$ , $ c$ positives reals numbers , prove that :

$ \frac{(a\plus{}b\plus{}c)^3}{a^3\plus{}b^3\plus{}c^3\plus{}3\lambda abc } \geq \frac{9}{1\plus{}\lambda}$

Inequality equivalents to $ (\lambda \minus{} 8)(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3(1 \plus{} \lambda)(a \plus{} b)(b \plus{} c)(a \plus{} c)\geq27 \lambda abc$. Because of $ \lambda \geq8$ and $ a^3 \plus{} b^3 \plus{} c^3\geq3abc$ , $ (\lambda \minus{} 8)(a^3 \plus{} b^3 \plus{} c^3) \plus{} 3(1 \plus{} \lambda)(a \plus{} b)(b \plus{} c)(a \plus{} c)\geq (\lambda \minus{} 8)3abc \plus{} 3(1 \plus{} \lambda)8abc \equal{} 27\lambda abc$ and the proof is ended.


Prove that there are only two real numbers $x$ such that \[(x-1)(x-2)(x-3)(x-4)(x-5)(x-6) = 720\]

[quote="Jan"]Just note that $6! = 720$, so conclusion $x=0$ or $x=7$ follows immediately.  :)[/quote]

obviously that gives two solutions, but i don't see how it shows that there are only two solutions...a graphing exercise works i guess...but that is not particularly rigirous

They are indeed the only solutions, it's only a simple application of AM-GM...
If you really want to see this proved, here it is:

Assume there is another solution $a$.

The case $0 < a < 7$ is the only nontrivial part:

In order to have the product positive, we must have $4 < a < 5$ or $6 < a < 7$ (the other cases follow by symmetry-axis $x=3.5$)

So first case: \[(a-1)...(a-4)(5-a)(6-a) \leq \left(\frac{2a+1}{6}\right)^{6}< \left(\frac{2a+1}{6}\right)^{6}< 2^{6}< 720\]

Second case is trivial. Take the derivate of the product using the extended product rule. You get a sum of $6$ positive terms for $a > 6$, so you are done since $6$ is a zero of the product.


AMC
USA(J)MO
USAMO
summer program
Mathcamp
PROMYS
Ross Mathematics Program

Mr. Dunbar, are the actual USAMO scores given our or just score intervals, and will these be posted on the internet soon?

Yan Zhang

For the past two years they've posted your score; I can't imagine they won't do it again.

If Mr. Dunbar does read this- will there be anyway to find out what the MOP index cutoff was? I'd like to know how close I was...

However, a number of people who received calls from Zuming were only given an interval, and before the past two years they only gave out score intervals.

[quote="Eugevir"]For the past two years they've posted your score; I can't imagine they won't do it again.
[/quote]

Oh, that's good.  Hopefully I won't be too disappointed when they show up.

[quote="Setofallsets"]Zuming told me (or at least I think he did) that the exact scores would be posted when he talked to me.  Maybe they'll post them tomorrow.[/quote]

Individual Scoring is out. The Average is 7.8 pts.

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-results.html

Last week was a bad week for me.  I did a terrible job on USAMO and I messed up a state competition.  I don't know what happened.  Maybe I should take a rest.

I got a 17, 1 point away from honourable mention!  I wish I didn't forget to put the equality conditions for #1.  Also, there are 12 HMs, and usually there are 17;  why couldn't they just include the people who got a score of 17 in HM? that would make 19 of them, which is not much higher than 17...

HMs sure were clustered very tightly.

How come TJ doesn't have any winners/ HMs this year? Odd...

Yes, I'd noticed that too...it seems very odd that TJ all of a sudden doesnt have anyone...

didn't Brian Rice go to TJ? he's a winner

Brian most certainly does not go to TJ. In fact, I think he hates TJ.

You're probably confusing him w/ eric price...

oh wait, you're right...., i looked at the big roster wrong... :?

I don't [i]hate[/i] TJ, I'm just rather envious of those who get to go there (I live in southwestern VA) and am a little antagonistic towards it.

[quote="GreenTwilight"]I seem to have gotten a lot lower than I expected.  I got got 2 whole questions right, plus partial credit on 3 other questions, but I got 11.  I don't know why, but it seems that the graders screwed me.  Oh well...[/quote]

I read on some USAMO thing or another that it is common for students to overestimate their scores, then be disappointed when they receive their actual results.  I believe this is the case here.  The graders do not ``screw'' anyone.  Plus, partial credit is hard to quantify, and it's possible that what you thought was partial credit material was actually worth no points.  I know that was the case when I took the USAMO.

[quote="GreenTwilight"]I seem to have gotten a lot lower than I expected.  I got got 2 whole questions right, plus partial credit on 3 other questions, but I got 11.  I don't know why, but it seems that the graders screwed me.  Oh well...[/quote]

Do you mind saying which questions you thought u had right, outlines of proofs for them? I wouldn't mind taking a look and seeing if/where you might have some errors. (You can PM me if you dont want to post)

Yes it is common for students to overestimate their score, but I guess I'm the exception. I underestimated my score by 3 points!

My brain literally shut down for 9 hours, so I got a 3! I was expecting a 0. About 20 minutes after I handed them in, though, I figured out how to do 2 of them! What a pity.

Better luck next year.

I figured out how to do #1 about 15 mins afterward, too. Oh well. When the pressure if off, you think a lot more clearly.

[quote="texas137"][quote]Individual Scoring is out. The Average is 7.8 pts. [/quote]

This is one of those cases where it may be better to look at the median rather than the mean to guage how you did. Roughly one-half made 3 or less.[/quote]

That's because there was a 42, a 41, and a 40 at the top this year. Whereas last year, there were two perfect scores, then the next up was a 36.

I got a 10. I think that's odd, because I don't think I got 3 points on #2 (that's pretty rare, isn't it?) and I didn't think I got credit on Problem 1. Oh well, I'm not complaining. :D

I'm definitely one of those people who overestimate their score.  I was expecting 5 or 6 on #4, and around 1 each for #2 and #3, but somehow ended up with 3 total.  In retrospect, I probably got 0 for #3 (I had the common mistake answer), 0 for #2 (probably didn't have enough there for partial), and 3 for #4 (not the most elegant proof, and badly written on top of that).

Zableman, I sympathize with you.

I also got a pathetic 3...

For #1, my proof was sorta shaky, and I wrote down the wrong equality condition. I am assuming that's where all 3 points came from.

For #2, I tried to prove that SOME linear combination was possible, but I used multiples of 2 rather than multiples of powers of 2. I am assuming they gave me a big fat 0 for my stupidity.

#3, Haha. I was never going to get that one.

#4, GRAHHHH! Game problem! Lesson learned: NEVER look at number 6 before number 4. I rejected Andreescu's gift of 7 points. (I solved this in five minutes shortly after the exam)

#5, I missed the key insight. Still kicking myself for it, but I'm sure that I got 0 points.

#6, LOL this was a good one. I made all these stupid assumptions and made some really dumb argument. I would have gotten more points had I drawn a picture of Trogdor the Burninator.

Well, that's my USAMO story. I hope I'll go to mathcamp instead then.

Is Mathcamp, PROMYS, or ROSS better?

[quote="masamune2387"]

For #2, I tried to prove that SOME linear combination was possible, but I used multiples of 2 rather than multiples of powers of 2. I am assuming they gave me a big fat 0 for my stupidity.

[/quote]

I used multiples of two... it worked.

Sorry I took so long to respond.  I was 99% sure that I had gotten questions 1 and 4 right.  (1, 2, 3 and 4 before posting on this site).  But after I realized that I got 2 and 3 wrong, and 1 and 4 right, I assumed a worst case scenario - one in which I would get NO partial credit for 2 and 3, but that I got 1 and 4 right.  That would lead to 14, but I got 11.  (I actually walked out of the test thinking I would get at least a 20 or so.)  I'm sure the graders were right, I was just a little bit angry.  But I am 100% sure that I proved to MYSELF the answers to 1 and 4.  I might have been a little unclear in writing something in my proof or something like that (I'm sure something I've written in this post is confusing already)  Saying that the graders "screwed me" was a little bit of anger as soon as I saw my score.

Anyway, I'm fine now.  Although I will always believe that if I was at my best while taking the test, I would absolutely have gotten Honorable Mention.

As for posting my solutions, I really have no idea exactly what I wrote.  I may have just written something incorrectly or not proved something that I knew was true or something along those lines.  With the exception of ARML I have no real experience writing proofs.  This is my first (and unfortunately last) year of USAMO.  And proving stuff on ARML isn't really that hard...

Oh well, I'm alright now.

Wow, the Winner's list is full of seniors. I also know that Janos Kramar is graduating this year. That makes at most two non-graduating winners.


geometry calculus quadratics trigonometry function AMC AIME

cuz imma take algebra 2 (10th grade class), and im only goin 2 8th grade

btw, did i mention honors?  :D  :D  :D

That really isn't huge on a math forum...

Anyway, geometry next year.

yeah i kinda know that

oh well

i was jw

I'm going to be in 7th grade next year and I'll be going to 9th grade Geometry!!  :D

I'm in 8th grade and I am taking calculus. :| Next year I'm ... doing something.

I'll be in 8th grade, and will take Geometry Honors and Precalculus Honors.

*could be taking BC Calc, but can't due to an unfortunate scheduling conflict a few years ago...

How are you all so smart?

Does AoPS prepare you for higher school math or vice versa?

Next year I'll be in 9th grade and I'll be taking both Algebra 2 and Geometry.

It's not really so much that they are really smart. Anyone can progress up to calculus in middle school, as long as they study or take summer classes. Me, I prefer to just go farther into the topics that I already know, instead of going up to a new topic. I find my current class easy (I have a perfect average of 101 right now) and I just do things like mc and what I said before instead of going higher up, leaving me with no math choices in high school.

[quote="Ihatepie"] I find my current class easy (I have a perfect average of 101 right now) and I just do things like mc and what I said before instead of going higher up, leaving me with no math choices in high school.[/quote]

Wow. That's such a coincidence!! I have an average of 102 or something like that(but my Algebra class is so easy...and all the kids there are so dumb...they hate math and school in general and the girls wear like, 5 tons of make-up).

[quote="Urc"]How are you all so smart?[/quote]

Me? Smart? I wish. Probably the only reason I started skipping two years of math last year in 5th grade is because 6th grade math(which I was taking at that time) was too easy(my god, they were still on adding fractions with like denominators) so I moved on to pre-alg, so this year I'm in alg and next year I'll be taking geo(blah! hate geo).

My school is stupid.  They don't let you take math higher than what you should be prepared for in your grade. -_-  Basically I'm surrounded by people who don't know how to do x /2 = 35/7.  Luckily I have mathcounts. Thank god!

Yeah... in China, the "average" 6th grader knows how to solve quadratic equations. In America, the "average" 6th grader knows how to add, subtract, multiply, divide, and add fractions with like denominators (not making fun of anybody here...).

The "average" AoPSer, however, knows much more. :D

Yeah, my parents are always saying how in India they did so much more and got 30 pages of homework a day.  Here I get no homework from school.

Next year in 10th I'm taking Pre-Calc but I think I'm going to do some indepedent calculus studies (assuming my 3 AP's will give me some free time).

@suma-mili

yeah, well im indian, and my parents tell me how much harder it wuz in india

like they were doing like biology and geometry in like 5th grade (btw indian school starts a year before American school)

whil were learning that suking our thumbs is a bad habit  :rotfl:  :rotfl:  :rotfl:

My new school is sooo much better than my local high school! I'm going to 10th Grade and taking advanced precalculus IV (with lots of proofs) and BC Calculus I. :D  If I had stayed in my old school as a 9th grader, they would only have let me take Algebra II with Trig, which was what I did in 8th grade. :mad:

at my school, i'm taking one called "math three" ...which is...uh. idk how to describe it.
http://www.charterschool.org/downloads/courseOutlines/mathematics/MathIII.pdf

i know most of the stuff here, but if i advanced, i'd be forced to take the amc12 which i'd fail. heh. plus it's good to review(:

[quote="1=2"]I'm going into 9th and I understand that Mathematics cannot be grouped into categories without confusion. :)[/quote]

LOL.

Blah, I discovered AoPS in just the last half of this past year. That's when I decided to stop being fail and do some math.

School math is too trivial. Pre-calculus fails too much, so I took it over this summer and now I'll take BC Calc next year as a sophomore.

Idk much about calculus though, so I'll (hopefully!??) enjoy it. (oh please, please, be a challenge).

Oh yes, and to add to that story before...I wasn't all too good at math either before seeing AoPS. But, after working through some harder stuff, I couldn't help but fall asleep in my math class.

Im going to 9th grade next year and taking Algebra ii honors (10th grade). But in my school, they only let me skip one grade!! The principal says that we're not that smart.. lol anyways do u guysa go to magnet school, public schools , private ?

Public, but all the smart people are put into one team, and a team does basically everything together, so its kinda like a private school for me in that Im always surrounded by smart people.

[quote="modularmarc101"]im in a private school but not that many people we consider smart, mathematically anyways. that's why it kinda ruins it for us but its mainly our principal.  :mad:[/quote]

Like I've wanted to say to a lot people here:

Screw the principal, screw the school, screw the whole system! If you've truly got the drive, you're going to yell in that retard's face and say

"Yo. What the fail? I know I'm smart enough to do more math. And you know I'm going to end up doing it anyways. So why don't you save me the trouble and just agree right now?"

or something to that extent. I've had my fair share with retarded schools (though maybe not to the same degree as some people here). If it ends up that the school won't let you take advanced classes (after a really good fight), then go ahead and do some independent study. They can't stop you from that.

Edit: I go to a public school.

We already had a big fight, two years ago so I could skip Pre-algebra to go to Algebra 1 in 7th grade, and it was a big fight. My mom and dad had to go and they had so many meeting with her, the math department and the headmaster, and I ended up being changed to algebra after like a trimester of prealgebra! And this also happened with my older brother for taking other classes like physics , etc. This year we have a new headmaster so maybe he'll be on our side this time..

I'm WAY behind everybody else in AoPS.  I'm going in to 7th grade, taking Algebra I over the summer, and am going to take Algebra I in school this year too.  Anyways, here are my goals I want to achieve(likely never going to happen, but I'll give it a shot :lol: -
                          Make it to Mathcounts Nationals
                          Get a perfect score on the AMC8
                          Qualify for the AIME through the AMC10
                           Get a 40/40 on the multiple choice "Math-League" competition
                           and, I know it has nothing to do with math, but I'd also like to make it to the State Geographical Bee

                       Currently, the only ones that look remotely possible are the AmC8 and the "Math League" competition, but to me, the most important is the first one(albeit the one least likely to happen)

Eh I gave up on Mathleague this year. After dropping from a perfect in 6th to a 39 in 7th, I realized that Mathleague is only a contest of who can make the least stupid mistakes, which I fail at.

but you get a cool plaque  :P

Agreed.  Anyways, the Math League Competition is relatively easy, since you get to use a calculator.  :lol: Unfortunately, the time factor is a REAL problem.  Also, the tests now are harder than the samples they have on the website because there are now 5 choices instead of four.

Just to tell you, modularmarc101, my situation was almost exactly like yours going into 7th grade.  I had it a little easier; after a lot of discussion (arguing) and looking at my ACT scores (from 6th grade!!!) they decided that I would be a good fit for 8th grade algebra I honors.  I was nervous, but it ended up being an easy class, of course...  Argh!  The state of the American school system! :!:

P.S.  And then wanting to skip math classes became a trend with some others at my school (not to brag, but they had a noticeably lower understanding, and, most of all, love and enthusiasm for math than I did!).

Yeah, it's just that they don't look at math like the way we do...They got no idea what we're capable of, cuz they just think about math inside the curriculum and not at what we do in competitions.

[quote="lowtopology"]And then wanting to skip math classes became a trend with some others at my school (not to brag, but they had a noticeably lower understanding, and, most of all, love and enthusiasm for math than I did!).[/quote]

Hm yep, same exact thing in my school. Skipping math classes is so ridiculously easy that people just skip to look smart or be with their friends. They dont actually care about math at all.


If I have one circle and one conic, I can find infinit homographies to map the circle onto the conic.

What if I have two circles and two conics? Could I find one or more homographies that maps the two circles onto the conics?

three circles and three conics ?

Martin


function
calculus
derivative
Support
real analysis
real analysis theorems

Is there an intuitive way to understand why so many analysis theorems require that a function be $C^{1}$? Why is $C^{1}$ so common whereas something like $C^{2}$ or $C^{\infty}$ is seen less often?

There's nothing particularly interesting going on; it's just that a large class of theorems are true for "nice enough" functions, and exactly what "nice enough" means comes from the proof. Needing to take one derivative is a lot more common than needing to take two derivatives.

When something really strong is invoked as a hypothesis - $C^{\infty},$ or Schwartz class, or compact support - that's often overkill. You only really need some of those derivatives, or some rate of decay. The overkill hypothesis is just a matter of getting all of that out of the way first and not worrying about it.


How many positive integers $ n$ satisfy $ 3<\sqrt[n]{100}<5$?

We must have $ 3^n<100<5^n$, which is true for $ \boxed{2}$ values of $ n$, $ 3$ and $ 4$.


Prove the identity

$ \csc 20^{\circ} \minus{} \cot 40^{\circ} \equal{} \tan 60^{\circ}$


calculus
integration
calculus computations

Prove that:

$\int^a_0 \frac{f(x)}{f(x) + f(a-x)}dx$

equals $\frac{a}{2}$

If $I=\int_0^a \frac{f(x)}{f(x)+f(a-x)} \, dx$
then subtitution $a-x=u$ give $I=\int_0^a \frac{f(a-u)}{f(a-u)+f(u)} \, du$
then compute $2I$.


Count them all.  You are lucky that any circel which goes through 3 (or more points)  does not have a radius = 3. (Your diagonal points  have to be considered though becuase the diagonal is less than 6) .

i believe there are a total of 10 circles.  I drew it out on a piece of paper and got 10.  Remember that  2 unique circles and intersect the sets of 2 points.

One way to approach it is to draw circles of radius 3 around each of your four points. You know that the centers of the circles you're looking for must lie on these drawn circles. Naturally, the centers must be where exactly 2 of the drawn circles intersect. As Gyan said, no circle with radius 3 will contain 3 of your points, so you know that no 3 of your drawn circles will intersect at the same place. But, also as Gyan said, since the distance between points on opposite ends of the square is less than 6, the circles around the 2 opposite points will intersect twice. So you can see that the circles around any pair of points intersect in 2 places and thus each pair of points defines 2 circles of the type you want. Since there are 4C2 = 6 possible pairs, there are (I think) 2*6 = 12 circles with radius 3 containing exactly two of the listed points.


Are u allowed to, as a means of proving a given statement, allowed to manipulate that statement u are trying to prove so u show that it is indeed true. for example, if i was asked to prove the pythagorean theorem, could i manipulate the pythagorean theorem into a statement that i know to be true, and would that be considered a valid proof? ( i know there are much simpler ways of proving the pythagorean theorem, but this is just a general example)

I think the answer is yes, working backward is fine [b]as long as all your steps are reversible.[/b]
For example, if  you manipulate a statement that you want to prove, it's not a good idea to multiply by zero or an expression that could equal zero.  (Because then when going forward, you'd have to divide by zero.)

To avoid such mistakes (i.e. multiplying by 0), you can reverse your entire proof. Start from the statement you know to be true, and perform each step normally. Here you don't have to worry about reversibility, and it will become very clear if you erroneously assumed all of your steps reversible in your original, backwards proof.

Richard Rusczyk's lengthy article on how to write a solution covers this topic: http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteForward.php

I'm new to the whole thing (relatively).  One thing I've noticed is

If you cant create a logical explanation of an intuitive thought, and the idea is unlikely to be seconded by the proof's viewer, then I would stay away from that argument, as it is now technically 'unfounded'.

Good terminology.

Good order in the presentation of the proof often makes or breaks it.

Edit $ t%Error. "neqw" is a bad command.
$


[hide]-4. With 3 consecutive even integers, if one of them was positive, you would either get a positive number, or if you began with 2, you would have -2 as the smallest number and they would cancel each other out.

However, if you start with 0, and then have -2 and -4, you get -6 and the smallest number is -4. [/hide]

[quote="MCrawford"]The sum of three consecutive even integers is -6.  What is the least of the three numbers?[/quote]

[hide]
I get:
$-4\\
-4+-2+0=-6$
[/hide]

[quote="MCrawford"]The sum of three consecutive even integers is -6.  What is the least of the three numbers?[/quote]

[hide]
Let
x = 1st consecutive even integer
x+2 = 2st consecutive even integer
x+4 = 3st consecutive even integer

x+x+2+x+4 = -6
3x+6 = -6
3x = -12
x = -4

The least of the three numbers is -4.
[/hide]

[hide]-4 is correct. SoccerBriany40 has the best idea to solve these in my opinion.[/hide]

[hide] Turn it into an algebraic equation.

x  +  ( x + 2 )  +  ( x + 4) = 3x + 6 = -6

3x + 12= 0

3x= -12

x=-4

The smallest number is x so the answer is [b]-4[/b] [/hide]

[quote="MCrawford"]The sum of three consecutive even integers is -6.  What is the least of the three numbers?[/quote]

[hide]$x+x+1+x+2=-6$

$3x=-9$

$x=-3$[/hide]

the problem says "smallest even integer" your answer is odd.
[hide]
instead of
x+x+1+x+3=-6 it should be x+x+2+x+4=-6
so solve that and you get x=-4[/hide]


[quote="eunice914"]a, b, c, d ≥ 0
prove that    a^2 + b^2 + c^2 + d^2 + abcd + 1≥ ab + ac + ad + bc + bd + cd[/quote]
Use $ abcd\plus{}1\geq2\sqrt{abcd}.$ :wink:

I tried with your hint, arqady.
but I can't solved this problem.

Please  post your solution, arqady.


Dear Friend:
                 With the advent of "String Theory," "Higgs Boson,"
it seems that modern "Mathematical Theoretical,Experimental
Physics" are on the thresthold of mathematical "Brink of Equa-
tion." Any confirmation regarding my mathematical "hypothesis?"

                                               Respectfully,
                                               Mikhail

Well... it's either not the best place to ask, or... time will show.


We have $ a^5\plus{}1 \geq a^3\plus{}a^2 \iff (a^3\minus{}1)(a^2\minus{}1)$ is true for all a>0. Therefore $ 2a^5\minus{}2a^2\plus{}7\geq 2a^3 \plus{}5$,and similar results apply for $ b$ and $ c$.

Therefore $ (2a^{5}\minus{}2a^{2}\plus{}7)(2b^{5}\minus{}2b^{2}\plus{}7)(2c^{5}\minus{}2c^{2}\plus{}7) \geq (2a^3\plus{}5)(2b^3\plus{}5)(2c^3\plus{}5)$
$ \equal{}(a^3\plus{}a^3\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1)(1\plus{}1\plus{}b^3\plus{}b^3\plus{}1\plus{}1\plus{}1)(1\plus{}1\plus{}1\plus{}1\plus{}c^3\plus{}c^3\plus{}1) \geq(a\plus{}a\plus{}b\plus{}b\plus{}c\plus{}c\plus{}1)^3$
$ >(2a\plus{}2b\plus{}2c)^3 \equal{} 8(a\plus{}b\plus{}c)^3$


Here is a fun triangle problem I did today.

In $\triangle ABC$, $AC = 6$ and $BC = 5$. Point $D$ on $AB$ divides it into segments of length $AD = 1$ and $DB = 3$. What is the length $CD$?

Anyway, if it is,
[hide]
Using the triangle inequality:

CD+1>6 so CD>5
1+6>CD so 7>CD

7>CD>5

so CD=6[/hide]

well actually , if you use cosine rules ...

[hide]
you will get $\cos B=\frac{1}{8}$ . And hence you will get $CD=5.5$
[/hide]


trigonometry algebra function domain calculus derivative integration

consider the taylor series for $ \sin{\theta}\equal{}\frac{2}{\pi}\minus{}\frac{4}{\pi}\sum_{1}^{\infty}\frac{\cos{2n\theta}}{4n^2\minus{}1}$ for $ 0\le\theta\le\pi$.

differentiating this we have:  $ \cos\theta\equal{}\frac{4}{\pi}\sum_{1}^{\infty}\frac{2n\sin{2n\theta}}{4n^2\minus{}1}$

but integrating we have:
$ \cos{\theta}\equal{}\minus{}\frac{2}{\pi}\theta\plus{}\frac{4}{\pi}\sum_{1}^{\infty}\frac{\sin{2n\theta}}{(4n^2\minus{}1)2n}$

how are the two expressions above the same?!

(we know that the fourier series for $ \theta$ is $ 2\sum_1^{\infty}\frac{(\minus{}1)^{n\plus{}1}}{n}\sin{n\theta}$ ).

A Fourier series gives you a periodic function. Even though you're talking about half-domain series, you can think of your Fourier cosine series as giving you an even function. Thus, while you called the function in your first expansion $ \sin\theta,$ it might be better thought of as $ |\sin\theta|.$ That makes it an even function.

Then your derivative is really the odd function that corresponds to $ \cos\theta$ on $ (0,\pi).$ I can write that as

$ \begin{cases}\cos\theta&0<\theta<\pi\\\minus{}\cos\theta&\minus{}\pi<\theta<0\end{cases}$ , extended to be periodic of period $ 2\pi$

Notice now that this is a discontinuous function. It has jump discontinuitities at $ 0$ and $ \pi.$ (Given the jump discontinuities, we expect that the coefficients should be $ O(n^{\minus{}1})$ as $ n\to\infty,$ which they are.)

That now means that the second derivative of this isn't even a function. There is a way to express that second derivative, but it would involve the Dirac delta function.

Now, what about the integral? That's not even properly defined. The problem is that the only way a periodic function can have a periodic antiderivative would be if its average value over a period is zero. And the average value of $ |\sin\theta|$ isn't zero; it is $ \frac{2}{\pi}.$ (Notice that that average value shows up as the constant term of the Fourier series.) Hence it doesn't have a periodic antiderivative, and hence your "Fourier series" isn't a Fourier series. There's actually no contradiction here.

---

As a general word of warning: when you are working with Fourier series, you must be constantly aware of the domain on which your functions are defined, and you must be constantly aware that your functions may be piecewise defined.

I still dont see how the following holds:

$ \sum_1^{\infty}\frac {\sin{2n\theta}}{(4n^2 \minus{} 1)2n}\minus{}\frac {( \minus{} 1)^{n \plus{} 1}}{n}\sin{n\theta}\equal{}\sum_{1}^{\infty}\frac {2n\sin{2n\theta}}{4n^2 \minus{} 1}$

 :maybe:

It doesn't. Did you read a word of what I wrote?

@henry23 First of all, you should integrate correctly. If you want the antiderivative of $ \sin\theta$ that is $ 0$ at $ 0$, it is $ 1\minus{}\cos\theta$, not $ \minus{}\cos\theta$. This missing $ 1$ whose sine Fourier series on $ [0,\pi]$ I leave to you to find accounts for the difference.

@Kent What you said is right, of course, but I think you missed the point here. The sine or cosine Fourier series on $ [0,\pi]$ should be unique, so if one gets two identities valid on $ [0,\pi]$ whose left hand side is some common function and whose right hand sides are combinations of sines, the coefficients at sines should coincide. I think this is what gave henry23 a headache. ;)


function
calculus
integration
geometry
calculus computations

Hi
I got this question that I don't quite understand!

Question:
Plot a graph of g(t) where:

g(t)=intergral of y(x) dx           (range is 0 to t)

y(x)=6x^2-7x-3*exp(-1/5*x)

You should be able to see that g(t) is the area under the curve y(x) in the interval 0<x<t .(you might like to plot y(t) on the same plot). Use this insight to interpret the observed shape of g(t).

Note: My question is what is y(t)?? how does this relate to the graph of g(t)??
And also what is there to interpret??

Thanks
Chris

Using this [url]http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus[/url]

since $g(t)= \int_{0}^{t}y(x) \, dx$ you have that $g'(t)=y(t)$ .

Some informations you can get from $g'(t)$ are the critical points $(g'(t)=0 )$

that give you the possible local max or min of $g(t)$ and

where $g(t)$ is increasing $(g'(t)=y(t) >0)$ or decreasing $(g'(t) <0 )$.

For example $g'(t)=y(t)<0$ for $t \in [0,1.42947]$ therefore $g(t)$ is

decreasing in that interval (the root of $g'(t)=0$ is aprox. $1.42947$) .


Well $a \geq b \geq c$ and we have LHS $\geq \frac{3}{2c}\geq \frac{3}{1+c^{2}}\geq$ RHS... should be strict?

Yes I also found that it is strict but I posted in the form that prososed at the contest . I am waiting for nice solutions  :wink:

Could someone verify that the stonger version posted in the post above is true ? If yes , does there exist a nice proof ?
Thank you

I solved the second problem in case that
$a\geq b\geq c$ and $x\geq y\geq z$ .
If you want I will post my proof but it is quite easy , just Cauchy and Chebychev .
Could someone prove the other cases ?Please help.

[quote="silouan"]What about the following stronger
 :wink:
Let $a,b,c$ nonzero numbers and $x,y,z>0$ with $x+y+z=3$
Prove that
$\frac{\sqrt{3}}{2}\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}\geq\frac{x}{a^{2}+1}+\frac{y}{b^{2}+1}+\frac{z}{c^{2}+1}$[/quote]
This inequality is not true, let $x=3,y=z=0$, then the inequality becomes
\[\frac{\sqrt{3}}{2}\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}\geq\frac{3}{a^{2}+1}\]
Now, let $b=c\rightarrow \infty$, it becomes
\[\frac{\sqrt{3}}{2}\sqrt{\frac{1}{a^{2}}}\ge \frac{3}{a^{2}+1}\]
Let $a=1$, it's wrong. :)

The best constant for the inequality
\[k\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}\ge \frac{x}{a^{2}+1}+\frac{y}{b^{2}+1}+\frac{z}{c^{2}+1}\]
Is $k=\frac{3}{2}$, but with this value, it is obvious. :)

Thank you very much , nice informations .


This is not a homework assignment, just something I am having trouble figuring out and thought someone may be able to help.

How many different combinations of six numbers are there using all of the numbers between 1 and 45  :?:

i.e. 1,2,3,4,5,6 - 1,2,3,4,5,7 - 1,2,3,4,5,45 - - - 40,41,42,43,44,45

I have tried to figure this out a few times but constantly get confused or think i have left some out.

note: The combinations do not include those that already have the same numbers i.e. If you had 1,2,3,4,5,6 you would not need 6,5,4,3,2,1 as a combination as all six numbers are already counted as a combination.

Please leave your answers to this problem if you choose to try to figure it out as it will be interesting to see what everyone comes up with.

Cheers
 :lol:

Is this a serious topic?
$^{45}C_{6}=8145060$

are combinations of this sort allowed:

2.3.4.1.4

is it only me or does jgreer seem distinctly thales-ish to others as well?

why would people use more than one account on the forum?

they need an account for each personality they have?
coz i have 2 personalities, me and my alter ego Benny

but that doesnt mean i need 2 accounts...

They may create new accounts to get rid of bad reputation.

Kind of obvious?!?
45C6?!?
Like... duh. This [i]is[/i] a math forum, by the way.

to answer lena's question, no, you can't have the same number twice within the sequence. And to those who think I was under another name on this site, YOUR ON DRUGS, this is my first time on the site and yeah it's a math site, and I'm not convinced that 45C6 is the correct answer. But hey, thats why I came to a math site, because my question was to do with Numbers. If I had a question about cars I would go there but my question wasn't about cars.. fair enough??

I'm a photographer not a math genius.

it is by definition $^{45}C_{6}$, read 45 choose 6 (see rd's hint above), or number of combinations of 45 distinct things taking 6 at a time without replacement. for the formula and derivation see any standard intermediate algebra text book like Hall & Knight or Bernard & Child. Ok?

[quote="kyyuanmathcount"][quote="randomdragoon"]I thought Koda was also an alternate personality.

I think you made me want to have an alternate personality. But because both rarely surface at the same time, I think I'll just leave it like this.

[b]I choose U[/b], Pikachu. That's a hint.[/quote]

Jesusfreak?  You haven't used that one for a while then.

Edit: NVM, then Daermon?[/quote]
kyuanmcount, what are you asking me?  :maybe:
hee hee my sig is funny


Problem
Find the one hundredth smallest positive integer that can be written using no digits other than digits 0 and 1 in base 3.

[hide]Solution
The goal is to count in base 3 using only binary digits. The $ 100^{\text{th}}$ smallest positive binary integer is $ 100 \equal{} 1100100_2$, so the $ 100^{\text{th}}$ smallest positive integer that can be written with only the binary digits is $ 1100100_3 \equal{} \boxed{981}$. [/hide] What does 100 have to do with the number given in the problem? There is a condition that the number can be written with no other digits than 0 and 1 in base 3, not two.

ok this kinda relates to the question, how do you take a number in a base and convert it to another base e.g. how would you convert this $ 1303_4$ to base 3??? or $ 768_9$ to base 2.  Is there a formula that you could use???

[b]aufha:[/b]  The keyword here is [i]bijection.[/i]  If you write out the sequence of numbers that can be written with only zeroes and ones in base $ 3$, you should notice a pattern:

$ a_1 \equal{} a_{1_2} \equal{} 1_3$
$ a_2 \equal{} a_{10_2} \equal{} 10_3$
$ a_3 \equal{} a_{11_2} \equal{} 11_3$
$ a_4 \equal{} a_{100_2} \equal{} 100_3$
$ a_5 \equal{} a_{101_2} \equal{} 101_3$

The point is that the "counting" we do to get from one index is the next is the same kind of "counting" that we do to find the next number that can be written using zeroes and ones in base $ 3$, except that the first count is in base $ 2$ and the second count is in base $ 3$.  In other words, we have a very natural description of the function $ f(n) \equal{} a_n$; we take the binary expansion of $ n$ and interpret it in base $ 3$.

@zolojetto:  I always convert the numbers from the given base into base 10, and then convert them into the desired base.  This may not be the fastest method, but it works.

[quote="t0rajir0u"]
The point is that the "counting" we do to get from one index
[/quote]

What does the "one index" refer to?

[quote="t0rajir0u"]
except that the first count is in base 2 and the second count is in base 3
[/quote]

What is the first and second count? Do you mean the first two columns you wrote?

[quote="t0rajir0u"]
We have a very natural description of the function f(n) = a_n; we take the binary expansion of n and interpret it in base 3.
[/quote]

What is so natural about the function? I'm really missing something here...

[quote="aufha"][quote="t0rajir0u"]
The point is that the "counting" we do to get from one index
[/quote]

What does the "one index" refer to?[/quote]

I was referring to the subscript of $ a_n$, which was written in base $ 2$ to make the connection to the value of $ a_n$ clearer.  To get from one index to the next, we just add $ 1$.  The important pattern here is how this affects the binary representation of $ n$.

[quote="aufha"]

[quote="t0rajir0u"]
except that the first count is in base 2 and the second count is in base 3
[/quote]

What is the first and second count? Do you mean the first two columns you wrote? [/quote]

The "first count" refers to the index (the subscripts).  The "second count" refers to the values of $ a_n$ (the last column in base $ 3$).

[quote="aufha"][quote="t0rajir0u"]
We have a very natural description of the function f(n) = a_n; we take the binary expansion of n and interpret it in base 3.
[/quote]

What is so natural about the function? I'm really missing something here...[/quote]

"Natural" in the sense that it's easy to describe, and in the sense that it follows naturally from the pattern that you should hopefully have noticed by now.

I suppose I should make more explicit what's really going on here.

[b]Underlying lemma:[/b]  Given two sequences $ a_n a_{n\minus{}1} ... a_0$ and $ b_m b_{n\minus{}1} ... b_0$ of binary digits (that is, ones and zeroes),

$ (a_n a_{n\minus{}1} ... a_0)_2 < (b_m b_{m\minus{}1} ... b_0)_2$

if and only if

$ (a_n a_{n\minus{}1} ... a_0)_3 < (b_m b_{m\minus{}1} ... b_0)_3$.

In other words, the set of numbers that can be written with only ones and zeroes in base $ 3$, when sorted in increasing order, has base-$ 3$ representations that are precisely the representation of all of the positive integers in base $ 2$, and in that precise order.  Can you prove this lemma?  And after proving it, can you see why it implies the result?  :)

I'm sorry. I am still missing something..  :(

"To get from one index to the next, we just add 1."
$ a_2 \equal{} a_{10_2} \equal{} 10_3$

I don't understand. The subscript of $ a_2$ is $ 2$ and the subscript of $ a_{10_2}$ is $ 10_2\equal{}2$. Where do we notice that we add one?

I don't.. see the pattern although I know it must be an obvious one.

"The next" refers to the next row, not the next column.  To get from $ a_2 \equal{} a_{10_2}$ to $ a_3 \equal{} a_{11_2}$, we just add $ 1$ (the first time in base $ 10$ and the second time in base $ 2$).  What I'm saying is that the carrying required to add $ 1$ to the index in base $ 2$ is the same carrying required to find the next number expressible in base $ 3$ using only zeroes and ones.

[quote="t0rajir0u"][quote="aufha"][quote="t0rajir0u"]
The point is that the "counting" we do to get from one index
[/quote][/quote][/quote]

[b]Underlying lemma:[/b]  Given two sequences $ a_n a_{n \minus{} 1} ... a_0$ and $ b_m b_{n \minus{} 1} ... b_0$ of binary digits (that is, ones and zeroes),

$ (a_n a_{n \minus{} 1} ... a_0)_2 < (b_m b_{m \minus{} 1} ... b_0)_2$

if and only if

$ (a_n a_{n \minus{} 1} ... a_0)_3 < (b_m b_{m \minus{} 1} ... b_0)_3$.

/quote]

If our two numbers are $ (a_n a_{n \minus{} 1} ... a_0)_2$ and $ (b_m b_{m \minus{} 1} ... b_0)_2$ are $ 10_2$ and $ 11_2$ respectively, we see that, plugging into the second line of inequalities in the above,

$ 10_3<11_2$
$ 3<3$

Where did I miscalculate?

[quote="aufha"]
$ 10_3 < 11_2$
$ 3 < 3$
Where did I miscalculate?
[/quote]
Well, it should be
$ 10_3 < 11_3$, which gives $ 3<4$.

This problem is extremely similar to a 1986 aime problem.
You can think of them as on/off, for 0 and 1, and so you can translate it into binary.


I noticed just now that AoPS doesn't have the little RSS icon in the URL bar to link to http://www.artofproblemsolving.com/Forum/feed.php, which seems to be a perfectly functional feed.

Shouldn't it be put there with the following code:

[code]
<link rel="alternate title="RSS" href="feed.php" type="application/rss+xml" />[/code]

Also: the blog RSS feeds don't show up in the proper layout for me. Attached are screenshots of the blog RSS and main RSS. the diference is obvious. I don't know how this could be fixed so could it be?

I will see what's wrong with the blog feed when I'm back from the IMO (late next week probably).

Again, I can't really do anything regarding this from Vietnam. You'll have to wait till I get back (mid-late next week).


how do you find the circumference of an ellipse, or rather what is the formula for the circumference of an ellipse?

There isn't a "nice" formula.  The [url=http://en.wikipedia.org/wiki/Elliptic_integral]integrals[/url] used to compute this circumference do lead to some rather deep mathematics, but nowhere will you find an expression you can work with on paper.

so......how can i "nicely" find the circumference of an ellipse? (because it was a ciphering question at a math tourny)

My guess is that a special ellipse was given as the problem or maybe there were certain conditions in the problem you missed that would've made it much simpler. I'm pretty sure a normal math contest for students below college doesn't ask for such a thing.

Either a) it wasn't an ellipse or b) they didn't ask for the circumference.  (It might've ended up being degenerate, in which case the answer would've been $ 0$.)


Of the five points (3, 10), (6, 20), (12, 35), (18, 40) and
(20, 50), what is the sum of the $ x$-coordinates of the points that
lie in the region above the line $ y \equal{} 2x \plus{} 7$ in the coordinate
plane?

The point must satisfy $ y>2x\plus{}7$. Plug this in for all five points. (6,20), (12,35), and (20,50) satisfy this. Add up their x-coordinates to get $ 6\plus{}12\plus{}20\equal{}38$.

i used to always put the other two coordinates...

ah-ha!  :rotfl:


AMC USA(J)MO USAMO AIME inequalities induction geometry

Hello,
I'm in 10th grade and I'm preparing for AIME and USAMO.  (I'm not sure whether I'll qualify; I got a 126 on AMC 10A and 7 on AIME, so I'm on the edge.)  Once I saw there was a possibility I might qualify for USAMO, I looked at past competitions, and I could do hardly anything on them, and since learning math is fun and useful anyway, I started to study for USAMO.  I had begun working through [i]Art and Craft of Problem Solving[/i] a while before AIME, and I understood most of it.  A math teacher suggested [i]Math Olympiad Treasures[/i] next, but it was somewhat too difficult - I understood the solutions presented in the book after I read them, but I could never imagine coming up with one myself.  Does anyone know of any intermediate books, somewhere between [i]Art and Craft[/i] and [i]Treasures[/i] that will help with AIME/USAMO and I can use as a "stepping stone" between the two?

Thanks for any suggestions!

[quote="Temperal"]USAMO doesn't really have any advanced concepts. According to some book (Olson's?), those with "natural" problem-solving creativity are designed to get into MOP, where advanced concepts for IMO are taught to them.

126 AMC 10 and 7 AIME probably isn't a high enough index to qualify.[/quote]

I personally think 126+70 is (just barely,maybe exactly on)

yea most of the newer USAMO problem you see dont really follow strict concepts and are just based on creativity/abstract thinking.

You could start off with some basic proofs (easier proofs than USAMO) to get you a feel for proofs.  Like prove some famous ones like AM-GM (inequalities not tested, butit uses good induction) or Cauchy-Schwartz inequalify, or jsut general easier proofs (like stewarts theorm)

A lot of times, there are many ways to prove a probelm

Every year, there is atleast one geometry problem and for that look for SIMILAR TRIANGLES (Power  of the point), concyclicty, area ratios, and sometimes, construction of special triangles are needed.  Also draw a good picture

For everything else, I cant really give you any advice since Ive really never solved a single NT problem on the USAMO

I don't have [i]Treasures[/i], so I don't know its level, but I recommend [i]Problem-Solving Strategies[/i] by Arthur Engel as the next step beyond Zeitz.  It is pretty comprehensive and should have problems that are approachable for you.

That said, if you can do "hardly anything" on past USAMOs, I have doubts about whether you have actually mastered Zeitz - don't just read it, do the problems and make sure you can apply the strategies he is presenting.

[quote="linj"]I am sorry for I do not know much about US university.
But could anyone tell me is harder to attend MIT or Harvard? :roll:[/quote]MIT is more concentrated on maths/sciences/engineering.

Thanks for your help.
[list]
I know I might not qualify for USAMO, but it's not as if learning math is a bad thing!  Besides, I know that Art and Craft will also help with next year's AIME if this year doesn't work out.

I'll see if I can find Engel in a bookstore and take a look at it (or if one of my teachers has it).

Geometry is probably something I should work on in particular.

I've been doing the problems and understanding the concepts involved for the most part, but there are 2 or 3 chapters in Art and Craft I'll want to go over (like geometry).  Maybe another reason why I didn't get very far in the USAMOs I looked at was time - I only spent about half an hour looking at each of the past 2 years.  I don't know (for those who have taken USAMO in previous years) how long it takes to see how you'll generally approach the problems, but in 30 minutes or so, I had general strategies for maybe 2 of them, and idea about a third.  Anyway, that was during the week and I didn't have much time; I'll take another look this weekend.[/list]

[quote="linj"]I am sorry for I do not know much about US university.
But could anyone tell me is harder to attend MIT or Harvard? :roll:[/quote]

fyi this is the "USAMO Preparation" thread not the "MIT vs. Harvard" thread

i have Engel, Problem Solving Through Problems, Treasures, Challenges. which is the easiest because they all make me want to pull my hair out!


inequalities
complex numbers
algebra unsolved
algebra

Let $ a,b,c$  be 3 complex numbers such that

                                      $ a|bc|\plus{}b|ac|\plus{}c|ab|\equal{}0$

 Prove that $ |(a\minus{}b)(b\minus{}c)(c\minus{}a)| \geq 3\sqrt 3 |abc|.$

Why don't you put it in the Inequalities Section?

http://www.mathlinks.ro/Forum/index.php?f=51

The case $ abc \equal{} 0$ is trivial. We have

$ 0 \equal{} \frac {\sum a|bc|}{|abc|} \equal{} \sum \frac {a}{|a|}$

$ \frac {a}{|a|}$, $ \frac {b}{|b|}$ and $ \frac {c}{|c|}$ are unitary complex numbers, therefore the above relation implies that the angles between them are all $ \frac {\pi}{3}$. The cosin's law now give us

$ |a \minus{} b| \equal{} \sqrt {|a|^2 \plus{} |ab| \plus{} |b|^2}$

and we want to prove that

$ \prod \sqrt {|a|^2 \plus{} |ab| \plus{} |b|^2} \equal{} \prod |a \minus{} b| \equal{} |\prod (a \minus{} b)| \ge 3\sqrt3|abc|$.

It suffices to prove that

$ \forall x, y \in \mathbb{R}_{ \plus{} } \sqrt {x^2 \plus{} xy \plus{} y^2} \ge \sqrt {3xy}$

and multiply the equations obtained when $ \{x, y\} \equal{} \{|a|, |b|\}$, $ \{|b|, |c|\}$ and $ \{|c|, |a|\}$. But

$ \sqrt {x^2 \plus{} xy \plus{} y^2} \ge \sqrt {3xy} \Longleftrightarrow x^2 \plus{} xy \plus{} y^2 \ge 3xy \Longleftrightarrow (x \minus{} y)^2 \ge 0$.

The equality occurs when $ |a| \equal{} |b| \equal{} |c|$ or $ a \equal{} 0$ and $ b \equal{} c$ or $ b \equal{} c \equal{} 0$ or in a permutation of the last two cases.


Hi
My first post in the olympiad section
This might be easy for u guys,but I wish to know how to solve these type of sums,I am a begginer after all :D
Prove
8(abcd + 1) > (a+1)(b+1)(c+1)(d+1)
where a,b,c,d are real nos greater than 1
Help plzzzzzzzzz :oops:

[hide="Hint"]
$ (a\minus{}1)(b\minus{}1) > 0$

$ \Rightarrow ab \minus{} a \minus{} b \plus{} 1 > 0$

$ \Rightarrow 2ab \plus{} 2 > ab \plus{} a \plus{} b \plus{}1$

$ \Rightarrow 2(ab\plus{}1) > (a\plus{}1)(b\plus{}1)$
[/hide]

I think you mean that $ \sum_{cyc}^{a \rightarrow b \rightarrow c \rightarrow d} \frac{1}{a\plus{}b\plus{}c} \ge \frac{16}{3(a\plus{}b\plus{}c\plus{}d)}$.

If so, then it follows directly from Titu's Lemma, the Cauchy-Schwarz Inequality or Power Mean Inequality.

I really didnt get what u said :( ,it's an AM-GM problem,that's what was given in challenge and thrill(no 11,exercise 11.3).Better see for urself.

It's not an AM-GM problem.It's a AM-HM problem. :lol:

Serious?
Never heard of that,may be u can invent one :(
[hide]if any such thing really exists,forgive me for my ignorance :P [/hide]
Yet,I'll give a try.

Well the inequality is obviously wrong,For example take a=b=c=d=1 then the inequality doesnt work.You can do the modified version of antiparticle even without using Titu's Lemma(A useful corolarry of Cauchy's Inequality),by using AM-HM Inequality.HM=Harmonic Mean,AM-HM Inequality says $ AM\geq HM$

Wait a sec,
AM>=HM doesnot hold true when the terms are proper fractions.
[hide]AM-HM technique is not given as any formal inequality solving method in Challenge & thrill,with any proof.U better post a proof of it to make it all clearer :read: [/hide]

Well the inequality $ RMS\geq AM \geq GM \geq HM$ is the most standard inequality.It is clearly explained in http://www.artofproblemsolving.com/Wiki/index.php/RMS-AM-GM-HM .You can also see the inequality section of Arthur ENGEL.

Yah,knew that kind of,but not the proof and all.
[hide]never occured to me that I need to apply that :P [/hide]

Well Here is the proof by

$ \sum_{cyc}\frac{1}{a\plus{}b\plus{}c}\ge\frac{16}{3(a\plus{}b\plus{}c\plus{}d)}
\Leftrightarrow \sum_{cyc}\frac{a\plus{}b\plus{}c}{4}\ge \frac{4}{\sum_{cyc}\frac{1}{a\plus{}b\plus{}c}}$ Which is AM-HM Inequality.The Equality Occurs iff $ a\plus{}b\plus{}c\equal{}b\plus{}c\plus{}d\equal{}c\plus{}d\plus{}a\equal{}d\plus{}a\plus{}b \Leftrightarrow a\equal{}b\equal{}c\equal{}d$

[quote="Scary math"]
Prove
8(abcd + 1) > (a+1)(b+1)(c+1)(d+1)
where a,b,c,d are real nos greater than 1
[/quote]

You can repeatedly apply tchebycheff's inequality
$ (abcd \plus{} 1)\ge\frac {(ab \plus{} 1)(cd \plus{} 1)}{2}$
$ (ab \plus{} 1)\ge\frac {(a \plus{} 1)(b \plus{} 1)}{2}$ and so on..


[quote="jhredsox"]is it just easiest to find a number that when you add 6 to it, that number would have 4 times a bigger square than the original? (which is obviously 6)[/quote]
[hide]
The formula for the volume of a cylinder is $V_{\text{cylinder}}=\pi r^{2}h.$  If you notice that quadrupling the height quadruples the volume, your proportionality sense might tell you you'll have to double the radius.  I think that would be the easiest solution.

But, if your proportionality sense doesn't tell you that, then you can solve the problem with algebra:

$V_{\text{bigger cylinder}}=V_{\text{bigger cylinder}}$

$\pi r^{2}(h+6)=\pi (r+6)^{2}\times 2$

$r^{2}\times 8= (r+6)^{2}\times 2$

$4= \frac{(r+6)^{2}}{r^{2}}$

$2= \frac{(r+6)}{r}$

$2r= r+6$

$r= 6$
[/hide]

[quote="jhredsox"]Would the easiest way be:

                original                  after adding 6
radius(^2):  r1^2                          r2^2
height:        2                               8

note the second height is 4x bigger than the original

8(r1^2)  is the same as   2(r2^2)

is it just easiest to find a number that when you add 6 to it, that number would have 4 times a bigger square than the original? (which is obviously 6)

-jorian

i might post some others later[/quote]

I did it your way, and personally I thought I got the answer pretty fast


Prove that the equation $x^n + y^n = z^n$, where $n > 1$ is an integer, has no
solution in integers $x, y, z$ with $0 < x, y\le n$.

It's called Fermat's Last Theorem.
 i'm not enclosing the proof because it's too long.
 check it out for yourslef at : [url=http://mathworld.wolfram.com/FermatsLastTheorem.html]Link for Fermat's Last Theorem[/url]
i don't think that $x, y \leq n$ is required

Well, I don't think he wants a proof of that type, but one for this special case ;)
But it was posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=45697


The fact that for absolutely no reason whatsoever, you replace $i$ with $-1$?

[quote="orangesurf024"](ixi +3)/2 = (  -1 times  -1 +3)/2[/quote]As blahblahblah said, this is your problem.

\[\frac{i\cdot{i}+3}{2}\ne\frac{(-1)\cdot(-1)+3}{2}\]

since $i$ is not $-1$ but the square root of $-1$:

\[i=\sqrt{-1}.\]

oops!
those -1s are supposed to be square roots of neg 1

I was recently at the mall adn came across a "two for the price of one" sale.  I only wanted one item so I asked the store clerk how much a single item would be.
      "Same as teh price for two" came the reply.
      "so one is the same as two?" I checked.
       "Yep, sure is!" vouched the clerk.

The problem with the other one is that $\sqrt{ab}\not=\sqrt{a}\cdot\sqrt{b}$ in general when we're talking about imaginary numbers.

I see that you have also posted this in the Getting Started forum.  In general, don't place the same topic in two different boards, because people will read it anyway.  But it looks like you're new so its okay, just don't do it in the future.  :)

there was, a similar topic in the "other problem solving topics" forum... check it out.

Go look at the topic in Other Problem Solving Topics.  I'm locking this one.


This is very easy AHSME problem that I want to let beginners to try it. Not the advancers, but the beginners.

A rectangular field is 300 feet wide and 400 feet long. Random sampling indicates that there are, on the average, three ants per square inch throughout the field [12 inches = 1 foot.] Of the following, the number that most closely approximates the number of ants in the field is:

A) 500 thousand
B) 5 million
C) 50 million
D) 500 million
E) 5 billion

[hide]The key here is that in 1 square foot there are 12^2 square inches.[/hide]

[hide]120000*144*3 = 51,840,000 C. 50 million

I know I should multiply it out and should just estimate, but it wasn't that hard to multiply so I did it neways  :D[/hide]

I guess it's an easy problem

the field is  3600inches wide and 4800long

which makes the area of it : 17280000^2"

since there are average 3 ants per square inch, we multiply them by 3

which we get about 50 millions!!

the actual number is 51,840,000

god it would look really ugly if u saw 50 million ants in your garden

happy new year!