function
quadratics
inequalities unsolved
inequalities

The function $ f(x,y,z)\equal{}\frac{x^2\plus{}y^2\plus{}z^2}{x\plus{}y\plus{}z}$ is defined for every $ x,y,z \in R$ whose sum is not 0. Find a point $ (x_0,y_0,z_0)$ such that $ 0 < x_0^2\plus{}y_0^2\plus{}z_0^2 < \frac{1}{1999}$ and $ 1.999 < f(x_0,y_0,z_0) < 2$.

Let $ \delta \approx 0^ \plus{}$
Let $ \epsilon \approx 0^ \plus{}$
Let $ \kappa \equal{} \frac {2 \minus{} 4\delta \minus{} \sqrt {(4\delta \minus{} 2 \plus{} \epsilon )^2 \minus{} 4\cdot 6\delta ^2}}{2}$

We claim that $ \boxed{(x_0,y_0,z_0) \equal{} (2\delta \plus{} \kappa, \minus{} \delta \minus{} \delta)}$ satisfies:

$ x_0^2 \plus{} y_0^2 \plus{} z_0^2\approx 0^ \plus{}$ and $ f(x_0,y_0,z_0) \equal{} 2 \minus{} \epsilon\approx 2^ \minus{}$

Since $ \delta \approx 0^ \plus{}$ and $ \epsilon \approx 0^ \plus{}$:

$ \kappa \equal{} \frac {2 \minus{} 4\delta \minus{} \sqrt {(4\delta \minus{} 2 \plus{} \epsilon )^2 \minus{} 4\cdot 6\delta ^2}}{2}\approx$
$ \approx \frac {2 \minus{} 4\cdot 0 \minus{} \sqrt {(4\cdot 0 \minus{} 2 \plus{} 0)^2 \minus{} 4\cdot 6\cdot 0^2}}{2} \equal{}$
$ \equal{} \frac {2 \minus{} \sqrt {2^2}}{2} \equal{} 0$

or, $ \kappa\approx 0$

Thus $ x_0 \equal{} 2\delta \plus{} \kappa\approx 0 ,y_0 \equal{} z_0 \equal{} \minus{} \delta\approx 0^ \minus{}$
Hence $ x_0^2,y_0^2,z_0^2\approx 0^ \plus{}$

$ \kappa$ is a root of the quadratic $ x^2 \plus{} x(4\delta \minus{} 2 \plus{} \epsilon) \plus{} 6\delta ^2 \equal{} 0$.
Thus $ \kappa$ satisfies $ \kappa ^2 \plus{} \kappa (4\delta \minus{} 2 \plus{} \epsilon) \plus{} 6\delta ^2 \equal{} 0$
Which is equivalent to:
$ \frac {(2\delta \plus{} \kappa)^2 \plus{} 2\delta ^2}{\kappa} \equal{} 2 \minus{} \epsilon$
But the LHS is precisely $ f(x_0,y_0,z_0)$
Hence our claim is true.


kolan pishrafte khoobi bud, dasteshoon dard nakone. :)

Man ham tabrik migam. Faratar az entezar zaher shodand. (Albate tu Spania ye bar be khodeshun tabrik goftam.)


geometry
analytic geometry
graphing lines
slope
geometry unsolved

A, B, and C are the points (1,2), (0,0), and (-1, 3) respectively. What is the slope of the bisector of angle ABC?

I tried solving it using the half angle formula with the tan(22.5) and got the answer 2^(1/2) - 1, but I'm not sure that is correct. :maybe:

Thanks

I got $\frac{1}{9},$ but I'm not sure that's correct either. :huh:


ratio
modular arithmetic
algebra unsolved
algebra

I think you slightly mis-stated this obvious result: the last term in your list should probably read $ \frac{x_n\plus{}x_1}{2}$

Modulo $ 4$ each odd number $ x_i$ has remainder $ \minus{}1$ or $ \plus{}1$  and hence, any sum $ x_i\plus{}x_j$ has remainder $ 0$ or $ 2$. If the first $ n\minus{}1$ ratios in the list are even, we would have for $ i\leq n\minus{}2: x_i\minus{}x_{i\plus{}2}\equiv x_i\plus{}x_{i\plus{}1}\minus{}(x_{i\plus{}1}\plus{}x_{i\plus{}2})\equiv 0\bmod{4}$, in particular $ x_1\equiv x_3\equiv \ldots\equiv x_n\bmod{4}$, since $ n$ is odd. But then $ x_1\plus{}x_n\equiv 2x_1\equiv 2\bmod{4}$ and $ \frac{x_n\plus{}x_1}{2}$ is odd.

$ \frac {x_{k} \plus{} x_{k \plus{} 1}}{2} \equal{} 2y_k \plus{} r_k$, (with $ x_{n \plus{} 1} \equal{} x_1$) where $ r_k$ is $ 0$ or $ 1$ , so $ \sum_{k \equal{} 1}^n x_k \equal{} 2\sum_{k \equal{} 1}^n y_k \plus{} \sum_{k \equal{} 1}^n r_k$ .
But $ \sum_{k \equal{} 1}^n x_k$ is odd , so $ \sum_{k \equal{} 1}^n r_k$ is odd, so in the list $ ..,\frac {x_{k} \plus{} x_{k \plus{} 1}}{2},..$ the number of   "odd numbers" is "odd" .
:cool:

[quote="hjbrasch"]I think you slightly mis-stated this obvious result: the last term in your list should probably read $ \frac {x_n \plus{} x_1}{2}$

Modulo $ 4$ each odd number $ x_i$ has remainder $ \minus{} 1$ or $ \plus{} 1$  and hence, any sum $ x_i \plus{} x_j$ has remainder $ 0$ or $ 2$. If the first $ n \minus{} 1$ ratios in the list are even, we would have for $ i\leq n \minus{} 2: x_i \minus{} x_{i \plus{} 2}\equiv x_i \plus{} x_{i \plus{} 1} \minus{} (x_{i \plus{} 1} \plus{} x_{i \plus{} 2})\equiv 0\bmod{4}$, in particular $ x_1\equiv x_3\equiv \ldots\equiv x_n\bmod{4}$, since $ n$ is odd. But then $ x_1 \plus{} x_n\equiv 2x_1\equiv 2\bmod{4}$ and $ \frac {x_n \plus{} x_1}{2}$ is odd.[/quote]

Sorry but i don't think that "$ i\leq n \minus{} 2: x_i \minus{} x_{i \plus{} 2}\equiv x_i \plus{} x_{i \plus{} 1} \minus{} (x_{i \plus{} 1} \plus{} x_{i \plus{} 2})\equiv 0\bmod{4}$"is true cause you proved above that sum of any xi+xj has remainder 0 or 2! So what happens if x_(i)+x_(i+1)-(x_(i+1)+x_(i+2)) has 2 different remainders?? :maybe:

[hide]assume they're all even. then $ 4|x_i\plus{}x_{i\plus{}1}$ for each i from 1 to n-1, and $ 4|x_n\plus{}x_1$.

for any of these pairs, one of them must be $ equiv 1\pmod{4}$ and the other $ \equiv 3\pmod{4}$.

if $ x_1\equiv1$, then $ x_2\equiv 3$, $ x_3\equiv 1$, ... for every odd $ i$, $ x_i\equiv 1$. thus, $ x_n\equiv 1$. however, then $ x_1\equiv 3$, contradiction.

same if $ x_1\equiv 3$.[/hide]


modular arithmetic
number theory unsolved
number theory

Find how many distinct six-tuples of elements in $ Z_p$ there exist, such that:
$ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca \equal{} x^2 \plus{} y^2 \plus{} z^2 \minus{} xy \minus{} yz \minus{} xz$ (in $ Z_p$).
($ p$ is prime, $ p \geq 5$).

Reduces to $ (a \plus{} \frac{p\minus{}1}{2})^2 \plus{} (b \plus{} \frac{p\minus{}1}{2})^2 \plus{} (c \plus{} \frac{p\minus{}1}{2})^2 \equal{} (x \plus{} \frac{p\minus{}1}{2})^2 \plus{} (y \plus{} \frac{p\minus{}1}{2})^2 \plus{} (z \plus{} \frac{p\minus{}1}{2})^2$ (*). By using the fact that $ m^2 \minus{} n^2 \equiv k \pmod{p}$ has $ p\minus{}1$ solutions $ (m,n)$ for every $ k \neq 0$ and $ 2p\minus{}1$ for $ k \equal{} 0$, we can get that (*) has $ p^5 \plus{} p^3 \minus{} p^2$ solutions.

I'm confused by the first line: $ (a \plus{} \frac{p\minus{}1}{2})^2 \plus{} (b \plus{} \frac{p\minus{}1}{2})^2 \plus{} (c \plus{} \frac{p\minus{}1}{2})^2 \equal{} (x \plus{} \frac{p\minus{}1}{2})^2 \plus{} (y \plus{} \frac{p\minus{}1}{2})^2 \plus{} (z \plus{} \frac{p\minus{}1}{2})^2$, is equivalent to $ a^2\plus{}b^2\plus{}c^2\minus{}a\minus{}b\minus{}c\equal{}x^2\plus{}y^2\plus{}z^2\minus{}x\minus{}y\minus{}z$ - that isn't necessarily equivalent to $ a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca\equal{}x^2\plus{}y^2\plus{}z^2\minus{}xy\minus{}yz\minus{}xz$, is it?

Sorry about that. I guess what I meant to do was
$ (a\minus{}b)^2 \plus{} (b\minus{}c)^2 \plus{} (c\minus{}a)^2 \equal{} \ldots$
(works since $ 2$ is invertible mod $ p$). So now we have the additional constraint that those three squares have to add to $ 0$. There goes the original solution.


LaTeX
calculus
derivative
function
invariant
real analysis
real analysis unsolved

Hey.
I ran into the following problem and was unable to solve:
[code]
u = x + ay
v = x +by

I need to find value for a and b such that:
9 d^2f/dx^2 -9d^2f/dxdy + 2d^2f/dy^2 = 0
becomes:
d^2f/dudv = 0
[/code]

I did some work but i don't get how i am supposed how to show that...and what the hell that even means. Didn't have problems with mutivariable stuff right up until now.

Btw, how do you use the fancy math symbol on this board, is there a topic on that (ie. Sum notation etc...).

Thx in advance.

The fancy math symbol you are interested in is part of $ \text{\LaTeX}$.  To use it on these forums, no downloading is necessary.  There is a $ \text{\LaTeX}$ tab at the top left of the forum pages and a board dedicated to $ \text{\LaTeX}$ (and something else called Asymptote).  Here are some links that might help you:

http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideCommands.php
http://test.izyba.com/equationeditor/equationeditor.php

As for the problem, you need to express $ \frac{\partial^{2}f}{\partial u\partial v}$ in terms of the derivatives in $ x$ and $ y$. The coefficients of derivatives will involve $ a$ and $ b$. Choose $ a$ and $ b$ so that the coefficients are $ 9,-9,2$.

Here is a link for the multivariable [url=http://en.wikipedia.org/wiki/Chain_rule]Chain Rule[/url]. Use it twice: first for $ \frac{\partial f}{\partial v}$, and then to differentiate the result with respect to $ u$.

Hey, i 'solved' the problem however the solution is incorrect at some point, here is what i did:
$ \frac{\partial^{2}f}{\partial x^{2}}=f_{uu}+2f_{vu}+f_{vv}$
$ \frac{\partial^{2}f}{\partial y^{2}}=a^{2}f_{uu}+2abf_{vu}+b^{2}f_{vv}$
$ \frac{\partial^{2}f}{\partial y \partial x}=af_{uu}+(a+b)f_{vu}+bf_{vv}$
This i got from differentiation a function f wrt to x and y and using the defenitiont that :
$ u = x+ay$
$ v = x+by$
Now, that i differentiation this i put this in the original equation:
$ 9\frac{\partial^{2}f}{\partial x^{2}}-9\frac{\partial^{2}f}{\partial x \partial y}+2\frac{\partial^{2}f}/{\partial y^{2}}= 0$
This gave rise to the equation that needs to be equaled to 0 in three places (fuu,fvv,fvu) thus i get this:
$ 9-9a+2a^{2}= 0$
$ 9-9b+2b^{2}= 0$
$ 9-9a-9b+4ab = 0$

Which means that a or b could both be 3 and 3/2, however the answer specifically states that a =3 and b =3/2 .
Any ideas where i messed up?
Thx in advance.

You didn't mess up anything essential: both pairs $ a=3$, $ b=3/2$ and $ a=3/2$, $ b=3$ are OK ($ f_{uv}$ is the same as $ f_{vu}$, so the final equation is invariant with respect to the exchange of variables $ u$ and $ v$, which explains this symmetry). On the other hand, you cannot have $ a=b=3$, say, not because something is wrong with your differentiation but because it does not give you a change of variables in $ \mathbb R^{2}$.

The only thing that worries me a bit about your solution is your third equation $ 9-9a-9b+4ab=0$. You do not need to make the coefficient at $ f_{uv}$ equal to $ 0$ (Moreover, you need it to be different from $ 0$)! So, this equation should either not be there or it should read $ 9-9a-9b+4ab\ne 0$.

So is the answer a = 3 and b  = 3/2 only one of the possible two answers or do i still need to do something else?

Yes, it is one of two possible answers. Nothing remains to be done.


calculus integration trigonometry Putnam

Evaluate $\int_{0}^{\frac{\pi}{2}}{\frac{dx}{1+\tan^n{x}}}$, where n is any real greater than 0.  Evaluate it in terms of n.

The Putnam asked you to solve it for $n=\sqrt{2}$, by the way.

Because there's no actual integration involved?

Extension to a Putnam problem is considered a Pre-Olympiad problem?

But Putnam problems are not necessarily too difficult for Pre-Olympiad.

I assume you know the trick already, if you don't, once you see it, it's definitely on the level of a pre-olympiad problem (given that the pre-olympiad understand the concept of a definite integral)

You are absolutely right, blahblahblah!

Note that $\frac{1}{1+\tan^nx}+\frac{1}{1+\tan^n(\pi/2-x)}=1$. It follows that required integral is $\frac{\pi}{4}$ !!!


There is a collection of 2004 circular discs (not necessarily of the same radius) in the plane. The total area covered by the discs is 1 square meter. Show that there is a subcollection S of discs such that the discs in S are non-overlapping and the total area of the discs in S is at least 1/9 square meter.

Let's take $D_1$ disk with largest radius ($r_1$) of all disks.
if $D \cap D_1 \neq \emptyset$ then $D$ is covered by a disk of radius $3 \cdot r_1$ (clear by doing a drawing).

Let's take $D_2$ disk with next largest radius ($r_2$) of all disks $D$ such such $D \cap D_1 = \emptyset$.

.....

continue until there are no more disks to choose. Algorithm stops since there are a finite number of disks ($2004$ is a red herring here ;))

So we have $D_1, .., D_n$, with radius $r_1, ..., r_n$, non overlapping and such the disks with same center but with radius $3 \cdot r_1, ..., 3 \cdot r_n$ would cover all disks
Then $9 \times$ area of $D_i \geq 1$ : we can take those disks.


inequalities
function
integration
calculus
real analysis
real analysis unsolved

Let $f:[0,2]\to\mathbb{R}$ a function such that $|f(x)-f(y)|\leq|x-y|$ for any $x,y\in [0,2]$ and $f(1)=f(2)=1$.
 Prove that:
                           $1\leq\int_{0}^{1} f(x)dx\leq 3$ ;)

First I'll assume that he means $\displaystyle 1\leq\int_{0}^{2} f(x)dx\leq 3$ (otherwise $f(x)=|\frac{1}{2}-x|+\frac{1}{2}$ is a quick counter example).

Put $y=0$ and get $|f(x)-1|\leq x\Rightarrow 1-x\leq f(x) \leq 1+x$
Then put $y=2$ and get $|f(x)-1|\leq x\Rightarrow -1+x\leq f(x) \leq 3-x$

Putting the results together we se that $f(x)\geq\left\{ \begin{array}{ll}
1-x &x\in [0,1] \\
-1+x &x\in [1,2] \end{array} \right. = |1-x| \quad x\in [0,2]$

So the given integral $\displaystyle \int_{0}^{2} f(x)dx\geq\int_{0}^{2} |1-x| = 1$

The upper bound is proved similary.


prove that for every integer n
1) 4 does not divide n 2 + 2
2) 12 does not divide n 2 + 3n + 5

You don't need induction for the first one.  Consider the two cases: n is even or it is odd.  If n is even then n^2 would have a factor of 4.  When we add two this new number would be congruent to 2 mod 4 which is not divisible by 4.  If n is odd then odd*odd=odd ,odd+even=odd an odd number is never divisible by 4.

You can use a similar parity argument for the second one:

If n is even then it will look like

(even)^2+(even)(odd)+odd=  even+even+odd=odd

Anything that contains 12 as a factor contains 2 and thus must be even.

If n is odd then:

(odd)^2+(odd)(odd)+odd=odd+odd+odd=odd

Same thing.

The expression is always odd for the second question, and hence not divisible by 12.


inequalities
trigonometry
inequalities proposed

[quote="Michael Niland"]Triangle ABC is acute-angled

Prove that 2<(cosA+cosB)^2 +(cosB+cosC)^2 +(cosC+cosA)^2

This is also true if the triangle is right-angled[/quote]

Again, use the relation $\cos^2{A} + \cos^2{B}  + \cos^2{C} + 2\cos{A}\cos{B}\cos{C} = 1$, this is equivalent to:

(*) $2\cos{A}\cos{B}\cos{C} < \cos{A}\cos{B} +\cos{B}\cos{C} +\cos{C}\cos{A}$

For acute triangle, at least one angle (say $C$) is $\geq \pi/3$, then $2\cos{A}\cos{B}\cos{C} \leq \cos{A}\cos{B}$ and $0 < \cos{B}\cos{C} +\cos{C}\cos{A}$. Add together.

For right triangle, if $C=\pi/2$, then (*) becomes $0 < \cos{A}\cos{B}$ which is obviously true.

(*) $2\cos{A}\cos{B}\cos{C} < \cos{A}\cos{B} +\cos{B}\cos{C} +\cos{C}\cos{A}$

[quote="tranminhhoang"]I think this inequality is also true for all triangles not only acute triangles.[/quote]

You are right!

If $\Delta{ABC}$ is obtuse, WLOG, assume that $C > \pi/2$, then $0< A+B < \pi/2$ and $\cos(A+B) = -\cos{C} > 0$.

(*) is equivalent to:

$\displaystyle \frac{1}{\cos{A}} + \frac{1}{\cos{B}} + \frac{1}{\cos{C}} < 2$  or equivalent to:

$\displaystyle \frac{1}{\cos{A}} + \frac{1}{\cos{B}} < 2 + \frac{1}{\cos{(A+B)}}$.

Since $\displaystyle \min \left\{\frac{1}{\cos{A}},  \frac{1}{\cos{B}}\right\} < 2$ and $\displaystyle \max \left\{\frac{1}{\cos{A}},  \frac{1}{\cos{B}}\right\} <\frac{1}{\cos{(A+B)}}$, we have:

$\displaystyle \frac{1}{\cos{A}} + \frac{1}{\cos{B}}  = \min \left\{\frac{1}{\cos{A}},  \frac{1}{\cos{B}}\right\} + \max \left\{\frac{1}{\cos{A}},  \frac{1}{\cos{B}}\right\} < 2 + \frac{1}{\cos{(A+B)}}$.

we use the trivial identity $x+y = \min\{x,y\} + \max\{x,y\}$ in the above proof.

Ok  we  need  to  prove  that  $ cosAcosB+cosBcosC+cosCcosA>2cosAcosBcosC $. But  in any triangle we have  $cosAcosB+cosBcosC+cosCcosA=\frac{s^2+r^2-4R^2}{4R^2}  $  and  $ cosAcosBcosC=\frac{s^2-(2R+r)^2}{4R^2}  $.
After  some  boring  computations , we  need  to prove  that $ 4R^2+8Rr+3r^2>s^2 $  which is true  because
$ 4R^2+8Rr+3r^2-s^2> 4R^2+4Rr+3r^2-s^2=IH^2\geq0 $

Well Michael , i  think  this is  what  you need :
In  any triangle  ABC   show  that $ (cosA+cosB)^2+(cosB+cosC)^2+(cosC+cosA)^2 \geq2(2+\frac{r}{R}) $

Sorry ..make  the  RHS  $ 2(1+\frac{r}{R}) $


prove/explain why kinetic energy is not conserved in a completely inelastic collision and why it is conserved in an elastic collision.

Simply... K.E is conserved in an elastic collision because "elastic collision" is defined that way - Collision in which K.E is conserved.
It is not conserved in Inelastic collision just for the same reason.{They're defined so..)

Perfectly elastic and perfectly inelastic collisions are two extreme idealizations that serve as good approximations in some situations where tracking energy at the microscopic level is not of interest.

In reality, during a macroscopic collision, kinetic energy is converted into heat, sound pressure waves, deformation, etc. Rather than keeping track of where the macroscopic kinetic energy went, it is often more useful to consider this energy to be "lost", and to just model how much of the energy is "lost" during the collision.

This is precisely what the  [url=http://en.wikipedia.org/wiki/Coefficient_of_restitution]coefficient of restitution[/url] is -- the square of the COR equals the fraction of the relevant kinetic energy left after a collision (COR alone gives fraction of velocity). A perfectly elastic collision has COR = 1, and a perfectly inelastic collision has COR = 0.

It is easy to understand the role of the COR when you look in the center of mass frame. You know that, by conservation of momentum, the center of mass will not accelerate due to the collision, so it is an inertial frame. In a perfectly elastic collision, the two bodies will separate the same way they approached eachother. In a perfectly inelastic collision, the two bodies will immediately stop moving upon contact (that is, the loss of relevant kinetic energy is equal to the total kinetic energy before the collision in the center of mass frame). Intermediate situations occur when the COR is between 0 and 1.


Let $ a, b, c$ be positive real numbers. Prove that

$ (\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^{2} \ge (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})$.

The inequality is equivalent to
$ \frac {a}{c} \plus{} \frac {c}{b} \plus{} \frac {b}{a} \plus{} \frac {{a^2 }}{{b^2 }} \plus{} \frac {{b^2 }}{{c^2 }} \plus{} \frac {{c^2 }}{{a^2 }} \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 3$
Using the AM- GM inequality, we have
\[ LHS \ge \frac {{a^2 }}{{b^2 }} \plus{} \frac {{b^2 }}{{c^2 }} \plus{} \frac {{c^2 }}{{a^2 }} \plus{} 3 \ge \frac {{2a}}{b} \plus{} \frac {{2b}}{c} \plus{} \frac {{2c}}{a} \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 3\]
Thus the inequality is proved

Hi TT-P. Did you mean to post this as a new topic rather than replying to my topic?

[quote="AndrewTom"]Let $ a, b, c$ be positive real numbers. Prove that

$ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})^{2} \ge (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$.[/quote]
The inequality is equivalent to $ \sum_{cyc}x^2\plus{}\sum_{cyc}\frac{1}{x}\geq \sum_{cyc}x\plus{}3$. Multiply by $ 3$. Then we get that $ 3\sum_{cyc}x^2\plus{}3\sum_{cyc}\frac{1}{x}\geq 3\sum_{cyc}x\plus{}9$. But from AM-GM we have $ 2\sum_{cyc}x^2\plus{}\sum_{cyc}\frac{1}{x}\geq 3\sum_{cyc}x$. So it is enough to prove $ \sum_{cyc}x^2\plus{}2\sum_{cyc}\frac{1}{x}\geq 9$ which holds from AM-GM

[quote="TT-P"]Given non-negative numbers x, y, z such that x² + y² + z² = 1. Prove that
x³ + y³ + z³ + xyz ≤ 1.[/quote]

$ uvw$ easily kills it :wink:
don't post problems in foreign topics, just an advice


[quote="bryantacademy"][i]A regular, 2-D, 9*9 box is a magic square where the sum of all of the numbers in each diagonal, row, and column is the same. From left to right, the top row has the numbers (?, 1, ?), the middle row has (5, ?, 13), and the bottom row has (X, ?, 3). What does X represent? You will probably want to draw this. Rules: You may not enter anything but odd numbers in the empty spaces of the box and you may not enter any numbers that are greater than 17.[/i]

[b]Attention solafidefarms:[/b] you had an extraordinary formula to solve this problem, and I don't remember it. Could you please pm me it so as to not spoil the hunt for others working to solve this problem, and I will attatch it to this post as a thread when I see fit.[/quote]

How is it 9x9 if there are only three numbers for each row...? Am i misunderstanding something?

here a way to solve 3x3 and 5x5
         j
      a b c
   k d e f  l
      g h i
         m

(j should be above b, a should be above d, so everything should be centered)

say you are making a square with nubmers 1-9

put 1 where j is
2 where a is
3 where k is
4 at c
5 at e
6 and g
7 at l
8 at i
9 at m

then where j is put that number where h is (so it's 1)
then where m is put that number where b is (so it is 9)
then "switch the other two.
you have a square solved.

this works with any arithmetic sequence you just have to put them on diagonals
also, notice the numbers in the the diagonal are consecutive.

+edits+ didn't come out right

[hide]I got it one the first guess!

15         1           5
5           3           13
1(X)       17         3

X=1[/hide]

[hide]$5+13+?=1+2?$, simplify to $17+?=2?$ simplify that to $17=?$ and we have 5+13+17=17+17+1=35.  Now we plug it into the bottom row, 17+3=20, so we need 15 more to complete the box.  15=x[/hide]


how can you prove that there is at least one prime between x and 2*x ? (x is a positive integer)

This is Bertrand's postulate :)
Use the search function!

ex : http://www.mathlinks.ro/Forum/viewtopic.php?highlight=bertrand+postulate&t=40143

or : http://mathworld.wolfram.com/BertrandsPostulate.html


geometry
geometric transformation
rotation
reflection
real analysis
real analysis unsolved

There exits a $ Borel \mbox{} set$ in the plane such that , however move it in the plane, it contains only one $ gridpoint$?

What does "move" mean? Translations? Translations and rotations? Reflections?


geometry
circumcircle
trigonometry
geometry proposed

Let $ BB_1$ and $ CC_1$ be the altitudes of an acute-angled nonisosceles triangle ABC. The bisector of the acute angles between lines $ BB_1$ and $ CC_1$ intersects sides $ AB$ and $ AC$ at $ P$ and $ Q,$ respectively. Let $ H$ be the orthocenter of triangle ABC and let $ M$ be the midpoint of $ BC.$  Let the bisector of BAC intersect $ HM$ at $ R.$ Prove that quadrilateral $ PAQR$ is cyclic.

Let $ L$ be the midpoint of the arc $ BC$ of the circumcircle $ (O).$  Then $ \triangle ARH \sim \triangle LRM$

$ \frac {RM}{RH} \equal{} \frac {ML}{AH} \equal{} \frac {BC \tan \frac {A}{2}}{2BC \cot A} \equal{} \frac {1}{2} \tan A  \cdot \tan\frac {A}{2} \ \ (1)$

On the other hand, let $ K$ be the orthogonal projection of $ M$ onto $ AB.$

$ PK \equal{} \frac {_1}{^2}BC_1 \minus{} PC_1 \Longrightarrow \frac {PK}{PC_1} \equal{} \frac {BC_1}{2PC_1} \minus{} 1 \equal{} \frac {BC \cos B}{2HC_1\tan\frac {A}{2}} \minus{} 1$

$ \frac {PK}{PC_1} \equal{} \frac {BC \cos B}{2AH \cos B \tan \frac {A}{2}} \minus{} 1 \equal{} \frac {\tan A}{2\tan \frac {A}{2}} \minus{} 1 \equal{} \frac {1}{2}\tan A \cdot \tan \frac {A}{2} \  \ (2)$

From $ (1)$ and $ (2)$ we have $ \frac {RM}{RH} \equal{} \frac {PK}{PC_1} \Longrightarrow RP \perp AB.$ Analogously $ RQ \perp AC.$ Hence, $ P$ and $ Q$ lie on the circumference with diameter $\overline{ AR}.$


[quote="ABM"]幫幫解這題數學, 謝謝

Suppose f:[a,b]->R is an integrable function, and g:[0,a]->R by g(x)= f(a+b(b-a)x)

1. Show that g is integrable
2. Show that (b-a)$\int_{0}^{1}$g = $\int_{a}^{b}$f[/quote]

1.We can have a proof by using the definition of integrable,that is it has boundery and it is continuous,you can refer to the textbook.

2.
Are you sure that g:[0,a]->R by g(x)= f(a+b(b-a)x), I think I can give a proof if g:[0,1]->R by g(x)= f(a+(b-a)x).

as g(x)= f(a+(b-a)x) and x belongs to[0,1]
  we get S[g*dx]=S[f*dx] x belongs to[0,1].
  f(a+(b-a)x), suppose t=a+(b-a)x, so,t belongs to[a,b] and dt=(b-a)dx.  So,
  we get S[g*dx]=S[f*dt] x belongs to[0,1] and t belongs to[a,b].
  that is (b-a)S[g]=S[f], S is integral here.


Here on AOPS, the latex compiles the way I want it to but in Miktex it complains when I try to use math code inside a tabular environment without using dollar signs. Of course I could just put dollars signs around all of the individual entries but that would take awhile for large tables. Is there another way to get all the elements of a tabular environment in math mode?

[code]\begin{tabular}{ l c r }
  y_1 & y_2 & y_3 \\
  y_4 & y_5 & y_6 \\
  y_7 & y_8 & y_9 \\
\end{tabular}[/code]

\begin{tabular}{ l c r } y_1 & y_2 & y_3 \\
y_4 & y_5 & y_6 \\
y_7 & y_8 & y_9 \\
\end{tabular}

You should not use a "tabular" inside math mode. Instead, use "array" inside a displayed equation, i.e.

[code]
\[
\begin{array}{ccc}
y_1 & y_2 & y_3 \\
y_4 & y_5 & y_6 \\
y_7 & y_8 & y_9 \\
\end{array}
\]
[/code]

$ \begin{array}{ccc}
y_1 & y_2 & y_3 \\
y_4 & y_5 & y_6 \\
y_7 & y_8 & y_9 \\
\end{array}$

Yeah, that is probably a better idea. Just for the record, I did figure out a way to do it with the "tabular" environment:

[code]\[\begin{tabular}{ >{$}c<{$} >{$}c<{$} >{$}c<{$} }
y_1 & y_2 & y_3 \\
y_4 & y_5 & y_6 \\
y_7 & y_8 & y_9 \\
\end{tabular}\][/code]

That doesn't work on AOPS but it works with MikTex.

Hi,

in that code \[ and \] are not necessary, the table can be set in text mode. Btw. the array package is needed to be able to use >{$} and <{$}.

Stefan


Sarah has $ \$112.72$ in her change purse. She may have quarters, dimes, nickels, pennies and paper money. What is the least number of coins she can have?

Obviously, she should have $ \$112$ in dollar papers to minimize her coins. This would leave us 72 cents. I should say 2 quarters, 1 dime, 1 nickel and 7 pennies.
$ \boxed{11}$?

How about 2 quarters, 2 dimes and 2 pennies?

That can be a better answer since my answer is based on the assumption that she has at least one of a kind, which is not clearly indicated in the prob.

I think we can assume there doesn't have to be at least one type of each coin, since it says: "She [b]may[/b] have quarters,...."

When I came across this question, I answered 6 and it was correct.

Wait for this problem is it assumed that there is a $112 dollar note?


calculus
integration
calculus computations

Show that $ \int_{0}^{1}{x^m(1\minus{}x)^n dx}\equal{}\frac{n}{m\plus{}1}\int_{0}^{1}{x^{m\plus{}1}(1\minus{}x)^{n\minus{}1}}dx$ for $ m>0,n>0$.
Hence, otherwise show that $ \int_{0}^{1}{x^m(1\minus{}x)^n dx}\equal{}\frac{m!n!}{(m\plus{}n\plus{}1)!}$

Similar problem from Japanese University entrance exam:

1989 Gakusyuin University entrance exam/Science

$ \int_0^1x^m(1\minus{}x)^n\,dx \equal{} B\left(m\plus{}1,n\plus{}1\right) \equal{} \frac{\Gamma(m\plus{}1) \Gamma(n\plus{}1)}{\Gamma(m\plus{}n\plus{}2)} \equal{} \frac{m! n!}{(m\plus{}n\plus{}1)!}$

Related to the problem

[color=blue]1972 Osaka University of Education entrance exam[/color]

Let $ B(p,\ q) \equal{} \int_0^1 x^p(1 \minus{} x)^qdx\ p > 0,\ q > 0$. Prove the following contexts.

(i) $ B(p,\ q) \equal{} B(q,\ p)$.

(ii) $ B(p,\ q) \equal{} B(p \plus{} 1,\ q) \plus{} B(p,\ q \plus{} 1)$.

(iii) $ B(p \plus{} 1,\ q) \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}B(p,\ q)$.

(iv) For positive integers $ p,\ q$,

$ B(p,\ q) \equal{} \frac {p!q!}{(p \plus{} q \plus{} 1)!}$.

Nice solution, Kouichi Nakagawa, tdl, carcul and kunny. My solution is the same Kouichi Nakagawa. And related problem of kunny can solve if use Integration by Parts, too.


[quote="M.A"]let a,b,c be positive real numbers prove that
 $ \frac{ab(a^3 \plus{} b^3)}{a^2 \plus{} b^2} \plus{} \frac{bc(b^3 \plus{} c^3)}{(b^2 \plus{} c^2} \plus{} \frac{ca(c^3 \plus{} a^3)}{c^2 \plus{} a^2}\geq (3abc(a^3 \plus{} b^3 \plus{} c^3))^{\frac{1}{2}}$ :)[/quote]

hint: (the cyclic sum x(b+c)/(y+z)) >=(3(ab+cb+ac))^1/2 :lol:


Hint[hide]
make them all equal denominators

$\frac{30}{15},\frac{30}{15},\frac{20}{15},\frac{10}{15},\frac{4}{15}$
ignoring all the denominators, we get $30,30,20,10,4$

I think you see it now :)[/hide]

Huh, I mean I know the solution, but I wonder how yours helps.

[hide]
i could plug it in in a graphing calculator to find the pattern (or the equation)

On the $x$ i put $1~5$ and the $y$ i put the values.
Then I found a linear regression equation ($y=ax+b$)
I got $-.48x+2.69333333$ now just plug in $6$ to the $x$
and i get $\frac{-14}{75}$ but is that the answer?
I'm not sure if this works.
the answer is not the same as above.
[/hide]

ya i know this is just a horrible attempt in solving normal sequence. I know this is not a linear sequence

[quote="8parks11"][hide]
i could plug it in in a graphing calculator to find the pattern (or the equation)

On the $x$ i put $1~5$ and the $y$ i put the values.
Then I found a linear regression equation ($y=ax+b$)
I got $-.48x+2.69333333$ now just plug in $6$ to the $x$
and i get $\frac{-14}{75}$ but is that the answer?
I'm not sure if this works.
the answer is not the same as above.
[/hide][/quote]

The problem is that a line must be one to one and my function isn't one to one, I will give you that.

Also, it will involve some thinking.

And your answer is wrong.

And do be cheap by saying that there will be a 4th degree polynomial passing through all the points.

[quote="mathgeniuse^ln(x)"][quote="8parks11"][hide]
i could plug it in in a graphing calculator to find the pattern (or the equation)

On the $x$ i put $1~5$ and the $y$ i put the values.
Then I found a linear regression equation ($y=ax+b$)
I got $-.48x+2.69333333$ now just plug in $6$ to the $x$
and i get $\frac{-14}{75}$ but is that the answer?
I'm not sure if this works.
the answer is not the same as above.
[/hide][/quote]

The problem is that a line must be one to one and my function isn't one to one, I will give you that.

Also, it will involve some thinking.

And your answer is wrong.

[b]And do be cheap [/b]by saying that there will be a 4th degree polynomial passing through all the points.[/quote]
I think that you made a typo

[hide]
well i thought it was just dividing each time by $.5$ higher.
Thought I could have solved it with a graph but i guess not.
anyways isn't it just dividing by $1,1.5,2,2.5,3..$ and so on?
so the next should be $\frac{4}{45}$

and this could be seen easier when it is in my form of
$30,30,20,10,4$

if this is wrong just tell me.
[/hide]

[hide="Hint:"]
Look at $\frac{a_{n}}{a_{n+1}}$. [/hide]

[hide="Challenge:"]
Find an explicit formula for this sequence. (The one intended, not fourth or higher-degree polynomial.) [/hide]

[quote="mathgeniuse^ln(x)"]I guess you can say, if you can't do it the first time, try try again.[/quote]

When I said that, I was hoping for you guys to realize that your answer was correct.


geometry
circumcircle
trigonometry
trig identities
Law of Sines

A triangle with sides $ a,b,c$ has its longest side at $ c$. Its circumradius is $ R$. Prove that the triangle is right angled if and only if $ a^2 \plus{} b^2 \equal{} 2Rc$

uh, $ 2R\equal{}c\Leftrightarrow \sin{C}\equal{}1\Leftrightarrow C\equal{}90^{\circ}$.

I'm not sure how your argument settles the problem.  What you've got shows that if a triangle is right, then $ a^2 \plus{} b^2 \equal{} 2Rc$, but I don't see the converse.  Maybe I am missing something obvious.

Well the triangle is right iff $ a^{2}\plus{}b^{2}\equal{}c^{2}$, so it suffices to prove that the triangle is right iff $ c^{2}\equal{}2Rc$, is that right?

I agree that the triangle is right iff $ a^2 \plus{} b^2 \equal{} c^2$.  I also agree that the triangle is right iff $ c^2 \equal{} 2Rc$.  I just don't see how either or both of those statements settle the original problem.

Oh, I see where you were coming from, I was for some reason assuming that if $ a^{2} \plus{} b^{2} \equal{} 2Rc$, they both had to be equal to $ c^{2}$.

[hide="Hint"]$ a^2\plus{}b^2\equal{}c^2\plus{}2ab \cos C \equal{} 2Rc \implies c \equal{} R \pm \sqrt{R^2 \minus{} 2R \cos C}$[/hide]

Apply the extended law of sines, $ \frac{c}{2R}\equal{}\sin\angle C$... So
$ \sin^2 A\plus{}\sin^2 B\equal{}\sin C$

Then we have $ C\equal{}\pi\minus{}A\minus{}B$, so

$ \iff \sin^2 A\plus{}\sin^2 B\equal{}\sin A\plus{}B$

Now I feel like it should be easier but I'm forgetting what identity would work here.

Sorry about bumping, but does someone have a good proof for this?

[hide="Big Hint"]Note that the longest side is opposite the largest angle in any triangle, and that the angle opposite the longest side creates an inscribed angle in the triangle's circumcircle. Use the fact that the central angle which opens the same arc is two times larger than said inscribed angle to prove that this inscribed angle will measure $ 90$ degrees iff $ c$ is the diameter of the circle (remember that $ 2R \equal{} D$).[/hide]

$ a^2 \plus{} b^2 \equal{} c^2 \implies c \equal{} 2R$. But any right angle is inscribed in a semicircle, in which case c, the diameter, would be 2R (what metafor said).

\[ b^2\plus{}c^2\equal{}2Ra\iff (b^2\plus{}c^2)^2\equal{}4R^2a^2\equal{}\frac{4a^4b^2c^2}{2a^2(b^2\plus{}c^2\minus{}a^2)\minus{}(b^2\minus{}c^2)^2\plus{}a^4}\]
\[ (b^2\plus{}c^2)^2(2a^2)(b^2\plus{}c^2\minus{}a^2)\minus{}(b^4\minus{}c^4)^2\plus{}a^4(b^2\plus{}c^2)^2\equal{}4a^4b^2c^2\]
\[ (b^2\plus{}c^2\minus{}a^2)((b^2\plus{}c^2)^2(2a^2)\minus{}(a^2\plus{}b^2\plus{}c^2)(b^2\minus{}c^2)^2)\equal{}0\]

If the first parenthesis is zero, we are done. If the second parenthesis is zero, then

$ (a^2)(b^4\plus{}6b^2c^2\plus{}c^4)\equal{}(b^2\minus{}c^2)^2(b^2\plus{}c^2)$. I'm not sure if there is a triangle that satisfies this with $ a>b,c$.

[quote="Altheman"]
\[ b^2 \plus{} c^2 \equal{} 2Ra\iff (b^2 \plus{} c^2)^2 \equal{} 4R^2a^2 \equal{} \frac {4a^4b^2c^2}{2a^2(b^2 \plus{} c^2 \minus{} a^2) \minus{} (b^2 \minus{} c^2)^2 \plus{} a^4}
\]

\[ (b^2 \plus{} c^2)^2(2a^2)(b^2 \plus{} c^2 \minus{} a^2) \minus{} (b^4 \minus{} c^4)^2 \plus{} a^4(b^2 \plus{} c^2)^2 \equal{} 4a^4b^2c^2
\]

\[ (b^2 \plus{} c^2 \minus{} a^2)((b^2 \plus{} c^2)^2(2a^2) \minus{} (a^2 \plus{} b^2 \plus{} c^2)(b^2 \minus{} c^2)^2) \equal{} 0
\]
If the first parenthesis is zero, we are done. If the second parenthesis is zero, then

$ (a^2)(b^4 \plus{} 6b^2c^2 \plus{} c^4) \equal{} (b^2 \minus{} c^2)^2(b^2 \plus{} c^2)$. I'm not sure if there is a triangle that satisfies this with $ a > b,c$.[/quote]
Nice soution!  Here is the proof for how no triangle exists with $ (a^2)(b^4 \plus{} 6b^2c^2 \plus{} c^4) \equal{} (b^2 \minus{} c^2)^2(b^2 \plus{} c^2)$ and $ 2Ra \equal{} b^2 \plus{} c^2$:
[hide="Proof"]
[b]CatalystOfNostalgia [/b] showed that either the triangle is right or the triangle is obtuse.  Considering the second case, we have that $ a^2 > b^2 \plus{} c^2$.  Yet, $ b^4 \plus{} c^4 \plus{} 6b^2c^2 > b^2 \plus{} c^2 \minus{} 2b^2c^2 \equal{} (b^2 \minus{} c^2)^2$, so $ a^2(b^4 \plus{} c^4 \plus{} 6b^2c^4) > (b^2 \minus{} c^2)^2(b^2 \plus{} c^2)$ and thus no equality can hold. [/hide]


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $F=\left(\begin{array}{cc}1&1\\ 1&0\\ \end{array}\right)$ be the so-called Fibonacci matrix. Try to find the elements of all  matrices  $B: =\sqrt{F}$     such that        $B^{2}=F .$

[quote="Kent Merryfield"]$B$ cannot be a real matrix, since $F$ is not positive definite. The eigenvalues of $B$ are the square roots of the eigenvalue of $F,$ meaning one of the eigenvalues of $B$ is real, namely $\pm\sqrt{\frac{1+\sqrt{5}}2},$ and the other eigenvalue is pure imaginary, namely $\pm\sqrt{\frac{1-\sqrt{5}}2}.$[/quote]OK !
However, in the problem it was'nt specified that $B$ must be a real matrix ! Perhaps $B$ belongs to a family of (complex) matrices.
 Thanks for interest,Alex

As written by KM B has, like F, 2 distinct eigenvalues (with 2 choices for each B's eigenvalue); thus B and F are diagonalizable; BF=FB therefore  if P¯1*A*P is a diagonal matrix, then P¯1*B*P is also a diagonal matrix.
Conclusion: there exist 4 complex  (non real) matrices B s.t. B²=F.
Moreover let spectrum(F)={a,b} and spectrum(B)={a',b'} where a'²=a and b'²=b; using the Lagrange's polynomials, we find explicitly the 4 solutions: B=1/(a'+b')*F+(a'b')/(a'+b')*Id.


A rectangular city block measures 400' by 500', including a 3-foot wide sidewalk which surrounds the block. After a blizzard uniformly dumped 18'' of snow, what is the total number of cubic feet of snow that needs to be cleared from the sidewalk surrounding this city block?

You have $ 4$ squares of size $ 3 \times 3$, $ 2$ rectangles of size $ 3 \times 400$, and $ 2$ rectangles of size $ 3 \times 500$.

This is $ 1.5(36 \plus{} 2400 \plus{} 3000)$, which is $ \boxed{8154}$.

Alternatively, the area of the sidewalk is 300x400-areaofcenter.

Area of center=(300-6)(400-6)=115836.

120000-115836=4164 sq. ft. MUltiplying by 1.5 feet (height of snow) we get 6246 cubic feet.

To ernie: The dimensions of the whole thing (including sidewalk) is 300x400, thats why you got wrong answer.

Actually, ernie, you assumed that the sidewalk was outside of the 400 by 500 square; it actually states that it is included in it.  Thus the volume should be

$ 1.5(2400\plus{}3000\minus{}36)$ rather than $ 1.5(2400\plus{}3000\plus{}36)$.  This yields an answer of...$ \boxed{8046}$

@mew: how ARE (not IS) the dimensions $ 300$ by $ 400$?


Show that for all positive integers $k,n$:
 $[((2k+1+$sqrt{4k^{2}+1}$)/2)^{n}]$  is divisible by $k$.


$x= \sqrt[4]{2}$

eh, i mean to write fourth root of two, dont know how to do it in latex...

[size=75][color=green]edited by mod[/color][/size]

[quote="arkmastermind"]$x= ^4\sqrt{2}$[/quote]

incorrect, i would not post a problem that trivial  ;) there is a mistake in your logic

[hide]Any possible solutions for $x$ will be given by the equation $x^{16}=16$, which has $\sqrt[4]{2}$ as the only positive real solution.

But suppose $a_0=1$ and $a_{n+1}=\sqrt[4]{2}^{a_n}$. Then $a_n<2$ which follows from induction; the base case clearly works and $a_k<2\Rightarrow a_{k+1}=\sqrt[4]{2}^{a_n}<\sqrt{2}<2$.

Since $a_n<2$, $\lim_{n\rightarrow \infty} a_n\leq 2$ so could not possibly be 16. Therefore, no solution.[/hide]

This problem is a take on a very famous problem, and it does point out something that a lot of people fail to think about... namely convergence. But that issue is probably too complicated for this forum.

We talked about one aspect of this once before [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39383]here[/url].

oh yeah marcus, you were telling us something about this on our way back from arml..

Yes, you can show that $x^{x^{x^{...}}}$ can never converge to anything larger than $e$ or smaller than $1/e$ for any real value of x. Studying the convergence and behavior of things like this would make a good research project for someone.

well anything greater than 1 means it increases to infinity and everything 0 to 1 would mean it approaches 0... likewise for the negative numbers.

what does $e$ play in this question? sorry i'm new at this

[quote="arkmastermind"]oh, yeah  :blush:

lol, i wonder why what i did was wrong though. i took log base 2 of both sides..[/quote]

as discussed, there is no solution, however, more explicitly, your mistake was that you assumed there was a solution

It can be shown through induction that an infinite power series in the format $n^{n^{n^{...}}}$ converges iff. $n$ can be written in the form $x^{1/x}$.

The maximum value of this function is $n=e^{\frac{1}{e}}$, which occurs at $x=e$.

Related thread: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77942]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77942[/url]


trigonometry
inequalities
inequalities proposed

[quote="Xixas"]Let $\alpha, \beta, \gamma$ be the angles of some acute triangle. Prove the inequality:
$\cos(\alpha-\beta)\cos(\beta-\gamma)\cos(\gamma-\alpha) \ge 8 \cos{\alpha}\cos{\beta}\cos{\gamma}$[/quote]
It's true for all real $\alpha,$ $\beta$ and $\gamma$ such that $\alpha+\beta+\gamma=\pi.$ :wink:


If:  a + 2(x + 1) = s, what is x + 1, in terms of s and a?

I am sitting an acumen test and need to brush up on these.
Could someone please explain the method to tackling these questions.
Thank you

[hide] We have that
\[a+2(x+1)=s \]
Thus, we can subtract $a$ from both sides to get that
\[2(x+1)=s-a \]
Now, we can divide both sides by two to get that
\[\boxed{x+1=\frac{s-a}{2}}\] [/hide]


harcki mitoone lotfan javab in soal ro be man bege!

noghteyi mesle M, ro tooye mosalas ABC, biyabid betori ke
MA * MB * MC, maximum she.

nemidoonam, chera hich kodoometoon reply nemikonid!  :(  :!:

valli in soal rah hale tahlilish badihiye, moadelehaye faseleha ro benevisid
va moshtagh begirid, valli vaghean kesi nemitoone in soal ro hendesi hal
kone?

Age manzoret [b]min[/b] be jaye [b]max[/b].

Injaro negah bekon:

http://en.wikipedia.org/wiki/Fermat_point
http://mathworld.wolfram.com/FirstFermatPoint.html
http://www.cut-the-knot.org/Generalization/fermat_point.shtml

na manzooram min nabood, khob age min ro bekhaym noghtaro be samte
yecki az rasa meil midim.

man max, ro mikham ...  :huh:

zemnan, noghteye ferma, noghteyiye ke hasele jame faselash kamtareene
vali man hasele zarb neveshtam!  :P

PS. be nazaretoon in soal rah hendesi nadare?  :blush:

jenab aghaye hatami, mishe begid in soal rah hale hendesi dare ya na!?
be fahamam jenabe eslahpazir maro sare kar gozashte, ya man dochare
kamboode IQ shodam? :mad:

motmaEn bash rahe hendesi dare,chon too moghadamati ham bood ,dooste man hal karde bood,farda azash miporsam.

khanoom / aghaye good girl, dar eshtebahi!
soal moghadamati ye chize dige boode... tavahom zadi...

PS. shart mibandam! :lol:  :P

mr masalan creative,be jaye shartbandi boro ye kuchulu englishet ro taghviat kon, :P
ye tahghighi ham dar morede mafhoome tavahom anjam bede!!!
hala man ye eshtebi kardam,....... :lol:


I'm going to be writing a mock AMC 8. Is anyone interested?

Please refrain from participating if you are over eighth grade

can i help, i have already written an AMC 10 and i made someone who is not that smart take it, and they got a reasonably high score, so i guess it was too easy. Maybe some problems may be good for AMC 8. can i send you the questions by pm and if you like any of them, you can put them on your test?

Anyone can participate. Just post here.

Once I post it, then just submit answers.

I'll Join! I'm in 5th grade so I can do this for a while yet...

alright, here is a preliminary list of people who are participating.

1. 12markkram34
2. BOGTRO
3. tinytim
4. FanstasyLover
5. Urc
6. neurosurgeon1118

wait 1=2, are you in or not?

I'm willing to check it over if you want...

Yeah once I'm done I'll send it to Altheman and 1=2

Ive never taken the amc 8, so i have a question. Is it like the amc 10 where if u guess you get no points if wrong?

Seriously, who needs AMC 8 practice? it all depends on how careless you are...

As I said, its for teh fifth graders, and those pretty young. When people are in fifth grade, it is in my assessment that they might need some practice :P

Its become more like an AMC 10 but whatever.

umm i hope its not too late to sign up...

The test site has been up for a while, but it looks like nobody has taken the test yet.  Is it still being checked?

How is this test going to work? Are you going to pm it, or is it posted on here?

another question: are you going to be using that website for submitting answers?

I've been on vacation, so I didn't have backend access to my server...there are a couple bugs to sort out.

I still have yet to even finish the mock test. Its probably going to be a mock amc 10. A link will be posted. You must go through a couple links and you have 45 min to do the whole test.

All questions should be in PDF form on the test so no one gets more than 45 min.

yeah, since #H34N1 isn't done with the test.

Everyone may still join. It'll probably take another week. :(

Ok, since it's going to probably be a mock AMC 10 (you said it yourself), may I join (I will be a freshman this year)?

Yeah. The harder problems (#20-25) are around AMC-12/AIME level

If it is going to be more like a mock AMC 10/12, then can you extend the time limit?

are any of the problems from that document that i sent you, if so i guess i can't take it.

yeah, but if you promise not to get a perfect. just kidding.

so here is the updated list:

1. 12markkram34
2. BOGTRO
3. tinytim
4. FanstasyLover
5. Urc
6. neurosurgeon1118
7. maththemonster
8. suma_milli
9. abacadaea
10. AIME15
11. MathAndKnowledge
12. shentang
13. pythag011

and #H34N1, can you send me a copy of the test before you post it? and are you allowing 9th grades and higher, if not Mathand is out. lol.

@cognos: its okay xD

Up through 10th grade.

I made it a mock AMC 10 since an AMC 8 would be an understatement

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212730

There it is


The diameter of each wheel of a bicycle is 36 inches.  If you are traveling at a speed of 45 mph, through how many revolutions per minute are the wheels turning?

This should be in Classroom Math
[hide]First convert mph to miles per minute to feet per minute to inches per minute:
$ 45 \div 60\cdot5280\cdot12=47520$ inches per minute
Now we find the circumference of the tire, $ 36 \cdot \pi = 36\pi$ inches per revolution
Then we divide: $ \frac{47520 in./min}{36\pi in./r}$= approx. $ 4147 r/m$. [/hide]


http://www.artofproblemsolving.com/Forum/resources.php?c=182
Here it says that there are 412 AIME problems which clearly isn't true.

I think there was one AIME where both Miths and I were adding problems at the same time...so I had to go and delete the extras I posted.  If I had posted 7 problems and then deleted that would explain the discrepancy...which is quite plausible.


|z_1|=1, |z_2|=k , |z_3|=2-k, z_1=cos a+i sin a , z_2=k(cos b+i sin b),
z_3=(2-k)(cos c+i sin c) and z_1+z_2+z_3=0,

(1) fint the minimum of  cos(b-c)=? and  k=?

(2) fint the maximum of   cos(b-c)=? and k=?

Can anybody answer me ?  :maybe:

I really don't think this should be in the highschool basics board


Prove that the Cartesian product AxB={(a,b): a in A, b in B} of two finite sets is again finite, with |AxB|=|A| |B|.

All I know is that if A is finite, then there are only finitely many elements in A, and same for B. So then if they are both finite sets, we cannot take their union and obtain a larger order since the union only consists of elements from A and from B.

How do I prove this though?  Do I have to say something like |AuB| < or = |A| |B|?

First try proving that if $ A,B$ are finite and $ B \equal{} B_1\cup B_2$ with $ B_1,B_2$ disjoint then $ A\times B \equal{} \left(A \times B_1\right)\cup \left(A \times B_2\right)$.  Use this to conclude $ A\times B \equal{} \bigcup_{b\in B}\left( A\times \{b\}\right)$.  This is a union of $ |B|$ disjoint sets each having cardinality $ |A|$, which establishes $ |A\times B|\equal{}|A\parallel{}B|$.


[b]Problem 1.[/b] Prove that
$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}-2(a^{2}+b^{2}+c^{2}) \ge \sqrt{3}-2$
where $a,b,c > 0$ and $ab+bc+ca=1$.

[b]Problem 2.[/b] Prove that
$1 \le \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\le \frac{3\sqrt{2}}{2}$
where $a,b,c \ge 0$

[quote="ehku"]
[b]Problem 2.[/b] Prove that
$1 \le \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\le \frac{3\sqrt{2}}{2}$
where $a,b,c \ge 0$[/quote]
The left inequality:
Let $a+b+c=1.$ Hence, $\sum_{cyc}\sqrt{\frac{a}{a+b}}=\sum_{cyc}\frac{a}{\sqrt{a^{2}+ab}}\geq\frac{1}{\sqrt{a^{3}+b^{3}+c^{3}+a^{2}b+b^{2}c+c^{2}a}}.$
Thus, remain to prove that $(a+b+c)^{3}\geq a^{3}+b^{3}+c^{3}+a^{2}b+b^{2}c+c^{2}a,$ which true.
The right inequality see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549

The first is mine, i have post it along time before on diendantoanhoc forum, you can prove it by using SOS method :)

No, I think it had been posted for a long time, longer than you! :D

Can you post your full solution for Problem 1, toanhocmuonmau ?

[b]Problem 3. (ehku)[/b] Prove that
$\sqrt[7]{\frac{a^{4}}{(a+b)^{4}}}+\sqrt[7]{\frac{b^{4}}{(b+c)^{4}}}+\sqrt[7]{\frac{c^{4}}{(c+a)^{4}}}\le \frac{3}{\sqrt[7]{16}}$
where $a,b,c \ge 0$


Express the following sum as a fraction:
$1/2+1/2^{2}+1/2^{3}...1/2^{10}$

[hide]
1023/1024, because for this pattern it is 2^x-1/2^x
[/hide]

and we have a winner!!! :showoff:  :first:

:10:
How do you  get a fraction in LaTeX?

you first need to put a dollar sign then you need to do \frac{whatever you want your numerator to be}{whatever you want your denominator to be} then another dollar sign.
 try it on the texer that is how i learned.


geometry
rectangle
ratio
trigonometry
complex numbers
geometry proposed

Let $ABCD$ be a convex quadrilateral. Rectangles $ABC'D\"$, $BCD'A\"$, $CDA'B\"$ and $DAB'C\"$ with centroids $P$, $Q$, $R$ and $S$ are constructed on the outside of $ABCD$ so that all four rectangles are similar, but $AB$ and $CD$ are long sides, whereas $BC$ and $AD$ are short sides. Show that $PR$ and $QS$ are perpendicular.

It's a nice problem. Let the oriented segments AB, BC, CD,  DA, PR, QS be interpreted as complex numbers a, b, c, d, p, s. Denote k the ratio of sides of the rectangle. We have
$p=\frac{b-d} 2+k\frac {i(a-c)} 2$,
$s=\frac{c-a} 2+\frac 1 k \frac{i(b-d)} 2$.
Showing that PR and QS are perpendicular is equivalent to showing that $\frac p s$ is purely imaginary.  Because
$\frac p s=-ik$, we are done.

It is known that the diagonals of a quadrilateral are mutually perpendicular iff the sums of squares of the opposite sides are equal. Therefore, it suffices to prove that
\[PQ^2+RS^2=PS^2+QR^2 \quad (*)\]
To compute $PQ^2$ use the cosine law in the triangle. Denoting $\angle {PBA}=\alpha$ we obtain
\[PQ^2=\frac{AB^2}{4\cos^2{\alpha}}+\frac{BC^2}{4\sin^2{\alpha}}+\frac{AB\cdot BC\cdot \sin{\angle{ABC}}}{2\cos{\alpha}\sin{\alpha}}
\]Writing the other three equalities and replacing in $(*)$ the requested equality reduces to
\[AB\cdot BC\cdot \sin {\angle {ABC}}+CD\cdot DA\cdot \sin {\angle {CDA}}=BC\cdot CD\cdot \sin {\angle {BCD}}+DA\cdot AB\cdot \sin {\angle {DAB}},
\]which is obvious (both sides equal twice the area of $ABCD$)

[quote="Arne"]It is nice indeed, but the only solution I have uses complex numbers as well.

Anyone with a synthetic solution?[/quote]
Well, if a complex number solution does not involve any "brute-force computation",  then it is always possible to "convert" it into a synthetic one.

Anyway, this problem is actually extended Van Aubel's Theorem, which states one more fact $PQ=RS$ for the case of square.


Let $a_1=1, a_2=3$, and $a_{n+2}=(n+3)a_{n+1}-(n+2)a_n$, $\forall n\geq1$.
(a) Determine all $n\in\mathbb{N}$, such that $9\nmid a_n$.
(b) Determine all $n\in\mathbb{N}$, such that $11\mid a_n$.


Proved that:
2^(p-1) -1  not divisible p^2       with  p is prime number

It is first prime, suth that $p^{2}|2^{p-1}-1$, second $p=3511$, third big (more 6 000 000 000). But I think, that there are infinetely many primes, suth that $p^{2}|2^{p-1}-1$. I don't know exist prime p, suth that $p^{3}|2^{p-1}-1$.


Vanessa takes a test worth 85 points.  If the minimum percent required to earn an A is $ 90\%$, and no rounding can be done, what is the least number of whole points she can earn on the test to receive an A?

To earn a 90%, she must get at least $ 90\%\cdot85\equal{}76.5$ right. But the answer is an integer, so is $ \boxed{77}$


analytic geometry
graphing lines
slope
rotation
trigonometry
calculus
derivative

I'm not sure whether this problem goes in the Intro from or not, but here it is

A cannon is on the peak of a hill that slopes uniformly downwards at an angle phi.  At what angle theta should the cannon shoot at to achieve maximum range?
Answer Clue: If phi = 60 degrees, then theta = 15 degrees.

Ah that problem from K&K.

Slope your axes and it should be easy from there

(Define a new angle alpha = theta + phi which is the angle aboev the new horizontal in the new axis orientation and find R(alpha) and then find Rmax

Ahhh..interesting idea.
I have  solved it out, but cannot find the maximum angle theta

When you have the rotate the x and y axis clockwise such that the x axis falls along the sloped hill, then there is acceleration in both the new y and x directions right?

Would the x acceleration be g sin (phi) and the y acceleration be - g cos (phi)?

If so, then the following are my equations.  I can't figure out how to solve it

Let alpha = theta + phi, v = initial velocity, h = height of the ball in terms of the new rotated axis, and R = range

h = v sin (alpha) t - 1/2 g cos (phi) t^2
When the ball hits the ground, h will be 0, so
1/2 g cos (phi) t = v cos (alpha)
t = 2 * (v cos (alpha))/ (g cos (phi))

The kinematics equation for the x - direction is
R = v cos (alpha) t + 1/2 g sin (phi) t^2

If I plug in t into this equation, I get a huge messy equation and I do not know how to maximize R.

Any ideas?

I haven't checked through your intial work, but from $ R\equal{}v\cos (\theta \plus{} \phi) t \plus{}C$, [phi is constant I think], you can just find $ dR/d\theta$ and set it equal to zero do first derivative test blah

[quote="#H34N1"]I haven't checked through your intial work, but from $ R \equal{} v\cos (\theta \plus{} \phi) t \plus{} C$, [phi is constant I think], you can just find $ dR/d\theta$ and set it equal to zero do first derivative test blah[/quote]

R = v cos (angle) + C is only when there's no acceleration in the x - direction though

When you tilt your axes, gravity is no longer pointing perpendicular to the x-axis, it's pointing a bit off to the right, so gravity will cause both y-decceleration and x-acceleration, I think?

First thing:  your equation for x should be $ x\equal{}v*sin\alpha...$ and $ 1/2 *g*cos(\phi)t^2$

as to your statement in the most recent post, true. or well, it's less than what would have been.

That's what I got, that there would be some acceleration in the new x-axis.

When I id this, it came out pretty ugly, but it could have been much worse. (the first three times I tried, I came out with something different...   :oops: )
But I got $ cot(\phi)\equal{}tan(2\alpha)$ and I was unable to simplify further.  It works though.

shouldn't it be x = v * cos (alpha) t + 1/2  g * sin (alpha) t ^2

what you have is x = v * sin (alpha) t + 1/2 g * cos (alpha) t^2

I'm positive that the x component of the velocity vector V is Vcos(alpha)

whoops yeah, i take that back, my bad.  :oops:  lol

EDIT:  1337 posts!

EDIT 2:  The acceleration is dependent on phi (the angle we turn the axes by) so it should be 1/2g*sin(phi)t^2 instead of alpha.
You had it wrong before [quote]R = v cos (alpha) t + 1/2 g cos (phi) t^2 [/quote]
as well, but you seem to have fixed something and gotten something else wrong.  :roll:

[quote="zapi2007"]
But I got $ cot(\phi) \equal{} tan(2\alpha)$ and I was unable to simplify further.  It works though.[/quote]
$ cot(\phi) \equal{} tan(2\alpha)$
$ arctan(cot(\phi)) \equal{} 2\theta \plus{} 2\phi$
$ (arctan(cot(\phi)) \minus{} 2\phi)/2 \equal{} theta$
Plugging in $ \phi \equal{} 60$ gives $ \theta \equal{} \minus{}45$ though where $ \theta should \equal{} 15$
oddly i found that
$ cos^2(\phi) \equal{} tan(2\alpha)$
This doesn't work either given the answerclue

hmmm...if cot($ \phi$)=1/$ \sqrt3$ tan(2*$ \alpha$)...maybe it should be cot($ \phi$)=-tan(2*$ \alpha$)...

Either way, if cot($ \phi$)=1/$ \sqrt3$=-tan(2$ \alpha$), 2*$ \alpha$=150 or 330.  then $ \alpha$=75 or 165.  But the parameters are $ \alpha$>$ \phi$ and 165 would go backwards... so...$ \theta$=15.

For a solution without calculus see here:

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=291839450&t=241782


Let $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}6$
Prove that
$ \sum \sqrt{a^2\plus{}6} \minus{} a \minus{} b \minus{} c \geq 3\sqrt{2}$
don't use "trigonometry"

[quote="nguyensilva"]Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} 6$
Prove that
$ \sum \sqrt {a^2 \plus{} 6} \minus{} a \minus{} b \minus{} c \geq 3\sqrt {2}$
don't use "trigonometry"[/quote]
The desired inequality can be written as follows
$ \sum \sqrt {(a\plus{}b)(a\plus{}c)}\ge a\plus{}b\plus{}c\plus{}\sqrt{3(ab\plus{}bc\plus{}ca)}$
See here:
http://www.mathlinks.ro/viewtopic.php?p=1632827#1632827
or
http://www.mathlinks.ro/viewtopic.php?t=297110


Through the vertices of the triangle $ ABC$ we draw three lines wich intersect the opposed sides in $ A',B',C'.$ If through a point $ M$, interior to $ ABC$, we draw parallels to $ AA',BB',CC'$ then these lines intersect $ (BC),(CA),(AB)$ in $ A'',B'',C''$. Prove that:
$ 1)$ $ \frac{MA''}{AA'}\plus{}\frac{MB''}{BB'}\plus{}\frac{MC''}{CC'}\equal{}1.$
$ 2)$ If $ AA',BB',CC'$ are concurrent then: $ \frac{AA'}{AM}\plus{}\frac{BB'}{BM}\plus{}\frac{CC'}{CM} \ge \frac{9}{2}.$

[quote="moldovan"]Through the vertices of the triangle $ ABC$ we draw three lines wich intersect the opposed sides in $ A',B',C'.$ If through a point $ M$, interior to $ ABC$, we draw parallels to $ AA',BB',CC'$ then these lines intersect $ (BC),(CA),(AB)$ in $ A'',B'',C''$. Prove that:
$ 1)$ $ \frac {MA''}{AA'} \plus{} \frac {MB''}{BB'} \plus{} \frac {MC''}{CC'} \equal{} 1.$
$ 2)$ If $ AA',BB',CC'$ are concurrent then: $ \frac {AA'}{AM} \plus{} \frac {BB'}{BM} \plus{} \frac {CC'}{CM} \ge \frac {9}{2}.$[/quote]
I can't slove problem 2 :">
with prob 1 , By drawing AH and MA''' to BC, we have $ \frac{MA'''}{AH}\equal{}\frac{MA''}{AA'}$
similarly, we have QED


Is the following possible?

Compute $\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\cdots}}}}$.

or

Compute $\sqrt[1]{1+\sqrt[2]{2+\sqrt[3]{3+\sqrt[4]{4+\cdots}}}}$.

I'm not exactly sure, but after looking at diff values of n, won't it diverge?

I tell you what. That sum looks UGLY, as in it probably involves $\pi$ or something.

[quote="miyomiyo"][hide="For number 2"]We have $a_{1}= 1$ and $a_{n}= \sqrt[n]{n+a_{n+1}}$, so solving for $a_{n+1}$ gives us $a_{n+1}={a_{n}}^{n}-n$.  This gives us $a_{2}=0$ which makes no sense.  What did I do wrong?[/hide][hide="Similarly 1"]We have $a_{1}= 1$ and $a_{n}= \sqrt{n+a_{n+1}}$.  Solving for $a_{n+1}$, we get $a_{n+1}={a_{n}}^{2}-n$.  This gives us $a_{2}= 0$ which makes no sense.  What did I do wrong?[/hide]What did I do wrong?  Can somebody help?[/quote]
You're using the wrong series. What you're doing would be the series for \[\sqrt{ \ldots+\sqrt{3+\sqrt{2+\sqrt{1}}}}\] which is much different.


If $x,y,z$ are positive numbers, prove that $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}}\ge\sqrt{\frac{3}{2}(x+y+z)}$.

we may assume:
$x+y+z=1$,
then we should to prove:
$\frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}+\frac{z}{\sqrt{1-z}}\ge\sqrt{\frac{3}{2}}$
you can prove it by convex function.
but there is another way.
use arrangement inequality or :
easily we have:$\frac{x}{\sqrt{1-x}}+\frac{y}{\sqrt{1-y}}+\frac{z}{\sqrt{1-z}} \ge \frac 1 3\left(\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1-y}}+\frac{1}{\sqrt{1-z}}\right)$
then we should prove:
$\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1-y}}+\frac{1}{\sqrt{1-z}} \ge 3\sqrt{\frac{3}{2}}$
use Caughy-Schwarz's inequality:
we have:
$\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1-y}}+\frac{1}{\sqrt{1-z}} \ge \frac9{\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}}$
the rest is for you ;)

I remember that I had proposed this nice inequality for RNMO 2005. It even appeared in the shortlist. Anyway, i'll prove a stronger one. Here it goes:

  [color=blue][b] Problem.[/b]

    Let $a, b, c$ three positive real numbers. Prove the inequality:

                          \[ \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq (a+b+c)\sqrt{\frac{a+b+c}{2(ab+bc+ca)}}.  \]

    [/color]

   [i] Solution.[/i]

      Using Cauchy-Buniakovski-Schwarz, we have
\[ \sqrt{a}\sqrt{a(b+c)}+\sqrt{b}\sqrt{b(c+a)}+\sqrt{c}\sqrt{c(a+b)}\leq\sqrt{a+b+c}\sqrt{a(b+c)+b(c+a)+c(a+b)}  \]
 which is equivalent with \[ a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\leq\sqrt{2(a+b+c)(ab+bc+ca)}.  \] (*)
 Again Cauchy-Schwarz, proves
\[ (a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b})\cdot(\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}})\geq (a+b+c)^{2}  \]
de unde \[ \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq\frac{(a+b+c)^{2}}{a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}},  \]
But , from (*)  \[ \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq\frac{(a+b+c)^{2}}{\sqrt{2(a+b+c)(ab+bc+ca)}}  \]
The conclusion follows. :)  ;)

This problem can be solved easily by Holder's inequality.
Use H\"older's inequality, we have \[\left( \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\right)^{2}\left( a(b+c)+b(c+a)+c(a+b) \right) \geq (a+b+c)^{3}\]  \[\sum\frac{a}{\sqrt{b+c}}\geq \sqrt{\frac{\left( \sum a \right)^{3}}{2\sum ab}}\geq \sqrt{\frac{3\left( \sum a \right)^{3}}{2\left( \sum a \right)^{2}}}=\sqrt{\frac{3}{2}\left( a+b+c \right)}\]

Jensen's.

$f(a)=\frac{1}{\sqrt{a}}$ \[\implies \left(\frac{1}{x+y+z}\right)\left(\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}}\right)=\frac{xf(y+z)+yf(z+x)+zf(x+y)}{x+y+z}\]  \[\ge f\left(\frac{x[y+z]+y[z+x]+z[x+y]}{x+y+z}\right)=f\left(\frac{2[xy+yz+zx]}{x+y+z}\right)\]  \[=\sqrt{\frac{1}{2}\left(\frac{x+y+z}{xy+yz+zx}\right)}\ge \sqrt{\frac{3}{2}\left(\frac{x+y+z}{(x+y+z)^{2}}\right)}=\sqrt{\frac{3}{2}\left(\frac{1}{x+y+z}\right)}\] Whichever way is easiest to you...but what exactly is Holder's inequality?

I can't remember how manu times it has been posted.
http://www.mathlinks.ro/Forum/viewtopic.php?t=87487


Is it OK if I let the HMMT Problems be added to the AoPSWiki? Because that will make it a lot easier to get to there from AoPS.

[hide="These people"]
Tournament Directors
    * Rishi Gupta, Harvard, transferring to MIT
    * Yi-Hsin Lin, MIT
    * Eric Price, MIT

Problem Czars
    * Yi Sun, Harvard
    * Yufei Zhao, MIT

Webmaster
    * Beth Schaffer, MIT

Treasurer
    * Keone Hon, MIT

Registration Director
    * Maria Monks, MIT

Graphic Designers
    * Anders Kaseorg, MIT
    * Amali Vaz, MIT
[/hide] might be able to answer your question. You can look up their contact information [url=http://web.mit.edu/bin/cgicso?options=general&query=]here[/url].  :)


function
number theory proposed
number theory

By observing that any natural number can be written as $ sr^{2}$ with $ s$ squarefree, show that $ \sqrt{x}\leq 2^{\pi (x)}$. Where $ \pi (x)$ be the number of prime less than or equal to $ x$.

$ p_{n}=n^{th}\ prime\ number$

if $ x\in \{0,1,2\}$ it's very easy to chek that $ x\le 2^{2\pi(x)}$

if $ x\ge 3$
then $ \pi(x)\ge 2$
$ \forall n\in\mathbb{N}: \ \exists \ prime\ p\in[2^{n},2^{n+1}[$
then $ \forall n\in\mathbb{N}: \ p_{n}\le 2^{n+1}$
so if $ n\ge 3$: $ \sqrt{p_{n}-1}\le 2^{\frac{n+1}{2}}\le 2^{n-1}$
then we have $ \sqrt{x}\le \sqrt{p_{\pi(x)+1}-1}\le 2^{\pi(x)}$ (because $ \pi(x)+1\ge 3$)

My solution is similary. This is other solution, not of mine :D

For any set $ S$ of primes, denote $ f_{S}(x)=|\{1\leq n\leq x| \gamma (n)\subseteq S\}|$, here $ \gamma (n)$ is set of prime divisors of $ n$.

Now assume that $ |S|=t$, write such an $ n$ as $ n=r^{2}s$, $ s$ is squarefree. Since $ 1\leq r^{2}s\leq x$, we have $ r\leq \sqrt{x}$ and there are at most $ 2^{t}$ choices of $ s$, so $ f_{S}(x)\leq 2^{t}\sqrt{x}$. Put $ \pi (x)=m$ , choose $ S=\{p_{1},...,p_{m}\}$ then $ f_{S}(x)=x$ and $ x\leq 2^{m}\sqrt{x}=2^{\pi (x)}\sqrt{x}$ and we're done.


find all $m,n \in Z^{+}$  such that :
              $mn + 1$ divisibled by $m^2 +n^2$

This problem is well-know and difficult too :) .
You can see the problem(with solution) from IMO 1988 like it this [url=http://www.kalva.demon.co.uk/imo/isoln/isoln886.html]www.kalva.demon.co.uk/imo/isoln/isoln886.html[/url]
Also the problem from Canadian National olimpiad 1998[url=http://www.kalva.demon.co.uk/canada/casoln/csol985.html]www.kalva.demon.co.uk/canada/casoln/csol985.html[/url]
In the lest site ther is the solution of this problem ;)


The adjacent map is part of a city: the small rectangles are rocks, and the paths in between are streets. Each morning, a student walks from intersection A to intersection B, always walking along streets shown, and always going east or south. For variety, at each intersection where he has a choice, he chooses with probability $\frac{1}{2}$ whether to go east or south. Find the probability that through any given morning, he goes through $C$.

[asy]
defaultpen(linewidth(0.7)+fontsize(8));
size(250);
path p=origin--(5,0)--(5,3)--(0,3)--cycle;
path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle;
int i,j;
for(i=0; i<5; i=i+1) {
for(j=0; j<6; j=j+1) {
draw(shift(6*i, 4*j)*p);
}}
clip((4,2)--(25,2)--(25,21)--(4,21)--cycle);
fill(q^^shift(18,-16)*q^^shift(18,-12)*q, black);
label("A", (6,19), SE);
label("B", (23,4), NW);
label("C", (23,8), NW);
draw((26,11.5)--(30,11.5), Arrows(5));
draw((28,9.5)--(28,13.5), Arrows(5));
label("N", (28,13.5), N);
label("W", (26,11.5), W);
label("E", (30,11.5), E);
label("S", (28,9.5), S);[/asy]

$\textbf {(A) } \frac{11}{32} \qquad \textbf {(B) } \frac 12 \qquad \textbf {(C) } \frac 47 \qquad \textbf {(D) } \frac{21}{32} \qquad \textbf {(E) } \frac 34$

Am I good at providing diagrams or what?  :D

[hide="hint"]Since he goes from left to right, he must always pass through the column to the left of B at some point. [/hide]

[hide="other hint"]Brute foce it with a tree diagram :yup:[/hide]

[hide]$D, {21\over 32}$

What I did was draw it on paper then from the top left, wrote down the probability of reaching each intersection, so A was 1, the two adjacent intersections were $\frac 12$ each, and so on.[/hide]

[hide]

From $A$ to $C$ there are $6C3=20$ ways .
From $A$ to $B$ there are $7C3=35$ ways .

So the probability are $\frac{20}{35}=\frac{4}{7}$

$\boxed{C}$
[/hide]

[hide="Whats wrong with"] He has to go through either the point above B or the point beside B. Therefore both are equally probably, for 1/2. Where did I go wrong?[/hide]

Simple, there are more ways to get to C following the stipulations set up in the problem than there are to get to the point to the left of B.  (6C3)>(6C2).

I thought that at first too.  Unfortunately that will only turn out to be true if there are the same number of streets going across as there are streets going down.  If they were the same then it would be 1/2.  You can use pascals triangle sideways to see this.

I agree, it looks like, due to symmetry, there are the same number of ways from A to C as from A to the point left of D, and thus each are equally likely, so B) 1/2.

It may look symmetrical, but for it to be 1/2, the rectangle formed using A and B in the city must be a square.

Simple combinatorics show that it is 4/7

Ah, whoops. I didn't look at the picture closely enough. I thought it was 5x5.
Can you explain how you use choose in this problem and what it means?

Note that if you get to C, the only way you can go is directly to B.  So if we count the ways to get to C, then divide by the ways to get to B, that's our answer.

To get to C, you have to go east 3 times and south 3 times.  Consider the arrangement EEESSS, and there are $\frac{6!}{3!3!} = 20$ ways to arrange this, so there are 20 ways to get to C.

To get to B, you have to go east 3 times and south 4 times.  Similarly, we use the arrangement EEESSSS, and there are $\frac{7!}{3!4!} = 35$ ways to arrange this, so there are 35 ways to get to B.

Then the answer is $\frac{20}{35} = \frac 4 7 \implies \boxed C$

every trip, he must go down (D) 4 times, and right (R) 3 times, clearly, the ways to arrange the word
DDDDRRR will be all the different paths, and all the combinations of that word where the last letter is D will have him pass through point C, so
$\frac{ \binom63}{\binom74}=\frac{4}{7}$

I don't know of any tutorials, but I've heard Mathematics of Choice is an excellent book on Intermediate Combinatorics.

[quote="Hokkage"]Ah, I understand now. Thanks guys. A question: how many ways can you arrange AAABBBCCCDDD? Anyone know of any good combinatorics tutorials on the web? I know the basics, but a better grounding would be good.[/quote]

$\frac{12!}{3!3!3!3!}=369600$, this is just based on if you have all the permuatations if you consider each A different, there will be a total of 3! counts for the same string if you consider the A's to be the same, that logic is the same for B,C,D

a similar idea to this is the multinomial theorem, and generally different "words" with some letters is a very standard problem that is just following the above process


[quote="kunny"]The real numbers $x,y$ varies being satisfied with $x^2+y^2\leqq 1$.
Let $m\geqq 0$ be a constant number,express the maximum and minimum value of $xy+m(x+y) $ in terms of $m$.[/quote]

By $AM-GM$ twice, $xy\leq x^2+y^2$, $(x+y)^2\leq 2(x^2+y^2)$.  The result follows immediately.

Oh, I wonder it is easy to solve in such a way,blahblahblah. :?

When $x^2+y^2=1$, we get that the minimum value is $\frac {-(m+1)}{2}$.  It shouldn't be hard (read: if I had a pencil and paper handy) to show that this gives us the minimum.

Regrettably you are wrong, blahblahblah. :?


I would like to know if it would be more advantagous to choose my AMC12 score which is a 132.5 for the index or just try for the floor value?

I don't think you can choose. If your index is high enough to place you in the top however-many-they-choose, you'll get in by index; if not, you'll have a chance at the floor.


geometry
perimeter
inradius
angle bisector

im not sure if there are middle schoolers that learned some basic geometry...

an equilateral triangle with side length of 3 contains an inscribed circle. what is the radius of that circle?

For a general triangle, $A=s\cdot r$, where $A$ is the triangle's area, $s$ is the semiperimeter (perimeter divided by two), and $r$ is the inradius (length of the radius of the inscribed circle).

[quote="nat mc"]Draw from the center of the circle to a side and to a vertex.  This forms a 30-60-90 traingle, with long leg $\frac32$, making the short leg, or radius this multiplied by $\frac{\sqrt3}3$, which gives us $\frac{\sqrt3}2$[/quote]

thats a better and quicker way than how i do it. I form a 30-60-90 triangle by drawing a angle bisector from one of the angles of the triangle. The length of the bisector is 1.5 *sqrt 3. Multiply that by 1/3 to get the radius

join the centre of the cicle  to the vertices and then draw perpendiculars from the radius to each side.
area of the tri=$root$3/4*3^2=1/2r(3+3+3)
r=3^1/2/2 ;) [/b]


function
induction
trigonometry
algebra
functional equation
calculus
calculus computations

Find all continuous real valued functions $f(x)$ such that $f\left(\sqrt{x^{2}+y^{2}}\right)=f(x)+f(y)$ for all $x$ and $y$.

[hide="what i have so far"]
$f'(x)=\frac{df}{dx}=\frac{2x}{\sqrt{x^{2}+y^{2}}}f'(\sqrt{x^{2}+y^{2}})$
and
$f'(y)=\frac{df}{dy}=\frac{2y}{\sqrt{x^{2}+y^{2}}}f'(\sqrt{x^{2}+y^{2}})$

from this I got $\frac{dy}{dx}=\frac{x}{y}\Longrightarrow ydy=xdx\Longrightarrow y=\pm\sqrt{x^{2}+C}$
its obvious that $f(0)=0$ so I have
$y=\pm\sqrt{x^{2}}\Longrightarrow y=\pm|x|$
but i have the $f(x)$ must be even and $\pm|x|$ is an odd function. where did i make a mistake?
or is it that there doesn't exist a function that satisfies the conditions?[/hide]

Let's tackle this algebraically. Consider for the moment the case in which $x$ and $y$ are both positive. Define $g(x)=f\left(\sqrt{x}\right).$ The functional equation now becomes (for $x,y>0$)
\[g(x+y)=g(x)+g(y)\]
It's fairly well known that the only continuous functions that satisfy this are of the form $g(x)=ax$ for some $a\in\mathbb{R}.$ To establish that, let $a=g(1),$ show by induction that $g(n)=an,$ show by another induction that $g\left(\frac{p}{q}\right)=a\frac{p}{q},$ then extend to the reals by continuity.

That gives us $f(x)=ax^{2},$ at least for $x>0.$

Suppose $x\le0.$ Chose $y>|x|,$ and note that $f(x)+f(y)=f\left(\sqrt{x^{2}+y^{2}}\right)=f(-x)+f(y),$ implying that $f(-x)=f(x)$ and $f$ is an even function. Our conclusion is that all such $f$ have the form $f(x)=ax^{2}.$

o ok. thanks.
question, what made you think of making the substitution $g(x)=f\left(\sqrt{x}\right)$?

@mao kid

you are only given that f is continuous, how do you know that f is differentiable? ( in post 1)

@  dr kent- nice solution.

thanks for the help once again, and good point about the fact that i wasnt told that the function was differentiable.

if you dont mind could someone please point me in the correct direction for the following problem.

determine all differentiable functions $f(x)$ such that $f(x)+f(y)=f\left(\frac{x+y}{1-xy}\right)$ for all real $x$ an $y$ with $xy\neq{1}$

also could you give me some general pointer on how to attack function problems like these.
thanks.

I recognize that as the functional equation for the tangent; that suggests substituting $g(t)=f(\tan t)$.


Can everybody post questions of basic geometrical questions here?
Thanxx :lol:

Here's an interesting one that I posted before:

Take $\triangle ABC$, such that $\angle C=90^\circ$. Take two points $D$ and $E$ on line $AB$ such that $AD=DE=EB$. Draw line segments $CD$ and $CE$. That should divide $\triangle ABC$ into three smaller triangles. If the area of $\triangle CAD$ is 5, then what is the area of $\triangle ABC$?

 :) Enjoy!

I suggest you order the AoPS textbook.  They have very good geometric questions.
Heres one:
The line joining the midpoints of the diagonals of a trapezoid has length 3.   If the longer base is 97, what is the shorter base?


Proof that:

$ 2^{43112609} \minus{}1$ is prime numer!

Trivial casework. You better start checking now as it will take you some time.

somebody at UCLA got 100,000 USD for [url=http://www.guardian.co.uk/science/2008/sep/29/maths]that...[/url]


geometry
circumcircle
cyclic quadrilateral

Let $p, q$ be 2 lines intersecting at a point $X$ and $E$ a point not lying on either line. Let $A_iB_iC_iD_i$ be all cyclic quadrilaterals with the same diagonal intersection $E$, such that the points $A_i, B_i$ lie on the same ray $XP \in p$ of the line $p$ and the points $C_i, D_i$ on the same ray $XQ \in q$ of the line $q$. Show that

[b](a)[/b] the circumcircles $(O_i)$ of these cyclic quadrilaterals belong to the same pencil of non-intersecting circles,
[b](b)[/b] their common diagonal intersection $E$ lies on the center line of the pencil, being one of its base points,
[b](c)[/b] the radical axis of this pencil cuts the the segment $XE$ at its midpoint $M$.

Comment: For the propositions (a), (b), it is not necessary for the lines $p, q$ to intersect, the degenerate case of parallel lines with their intersection $X$ at infinity is perfectly valid. However, with the point $X$ at infinity, the midpoint $M$ of the segment $XE$ is undefined.

A very interesting result, good job Yetti! Though I can't believe you couldn't prove it, since the proof mostly uses your favorite results from inversive/projective geometry.

Anyway, let ABCD be a cyclic quadrilateral with its vertices A and B on the line p and its vertices C and D on the line q, and with its diagonals AC and BD intersecting at the point E. Then, we have to prove that the circumcircles of all such quadrilaterals ABCD are coaxal - and, more precisely, they form a non-intersecting coaxal family of circles with the point-circle E (this is the circle with center E and radius 0) being a member of this family -, that their common radical axis passes through the midpoint M of the segment XE and that their common center line passes through the point E.

In addition to the points $E=AC\cap BD$ and $X=AB\cap CD$, let's define a third point $Y=BC\cap DA$. Since the quadrilateral ABCD is cyclic, we can say much about the triangle EXY: This triangle is autopolar with respect to the circumcircle of the quadrilateral ABCD, and the center O of this circumcircle is the orthocenter of triangle EXY. The points E and X are conjugate points with respect to the circumcircle of the quadrilateral ABCD. (This was all discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=5326 .)

Now, consider the complete quadrilateral formed by the lines p = AB, BC, q = CD and DA. The diagonals of this quadrilateral are AC, BD and XY. By the complete quadrilateral theorem from projective geometry, the lines XE and XY divide the angle between the lines p = AB and q = CD harmonically. In other words, the line XY is the harmonic conjugate of the line XE with respect to the angle between the lines p and q. Or, equivalently, if we denote by y the harmonic conjugate of the line XE with respect to the angle between the lines p and q, then the line XY coincides with the line y. This is an important result, as the line y depends only on the points X and E and the lines p and q and not on the actual cyclic quadrilateral ABCD.

Denote the perpendicular to the line y through the point E by r.

Now, since O is the orthocenter of triangle EXY, we have $OE\perp XY$, what, in view of XY = y, becomes $OE\perp y$. Thus, the point O lies on the perpendicular to the line y through the point E. In other words, the circumcenter of the quadrilateral ABCD lies on the line r. Translating into the language of inversive geometry, where the line r is a degenerate circle and where a circle is orthogonal to a line if and only if its center lies on the line, we can restate this as follows: The circumcircle of the quadrilateral ABCD is orthogonal to the line r. This is the first key.

For the next, we remember that the points E and X are conjugate points with respect to the circumcircle of the quadrilateral ABCD. Consequently, the circle with diameter XE is orthogonal to the circumcircle of the quadrilateral ABCD.

Now we are ready to combine the puzzle. What we have obtained is: The circumcircle of quadrilateral ABCD is orthogonal to both the line r and to the circle with diameter XE. Now, all circles orthogonal to two fixed circles form a coaxal family of circles. Thus, the circumcircle of quadrilateral ABCD belongs to the coaxal family of all circles which are orthogonal to both the line r and to the circle with diameter XE.

Now, neither the line r nor the circle with diameter XE depend on the actual quadrilateral ABCD; they only depend on the points X and E and the lines p and q. Thus, the circumcircles of all quadrilaterals ABCD belong to one coaxal family, namely to the coaxal family of all circles which are orthogonal to both the line r and to the circle with diameter XE. The point-circle E also belongs to this family, since it is orthogonal to the line r (in fact, the point E lies on the line r) and to the circle with diameter XE (in fact, the point E lies on the circle with diameter XE). Thus, the common center line of this coaxal family of circles must pass through the point E. Also, it follows that the coaxal family is a non-intersecting one (in fact, only non-intersecting coaxal families of circles contain point-circles). Finally, the common radical axis of this coaxal family of circles must pass through the point M, since all circles of our coaxal family are orthogonal to the circle with diameter XE, and this circle is obviously centered at the midpoint M of the segment XE. This proves all of your assertions.

A side-note:

[quote="yetti"]Comment: For the propositions (a), (b), it is not necessary for the lines $p, q$ to intersect, the degenerate case of parallel lines with their intersection $X$ at infinity is perfectly valid. However, with the point $X$ at infinity, the midpoint $M$ of the segment $XE$ is undefined.[/quote]

It is not undefined. The midpoint between an Euclidean point and an infinite point is the latter infinite point (the Euclidean point is "insignificant"). Only the midpoint between two infinite points is undefined. And yes, assertion [b](c)[/b] very well holds with the point X being infinite.

  Darij

Thanks a lot for the solution. I did not even try to solve the problem, I just noticed these facts while solving some other geometry problems, the solution of which turned out to be elementary. In addition, I was asked, why do I not post any problems, so I posted this one as soon as I was reasonably sure about it. Judging by your solution, I would have hard time at best. Thanks again,

Yetti

See the attached pdf for a solution.

--Alex


Let ABC be a triangle ,andd L  line through C,parallel to AB. The internal bisector of A intersects side BC at D and L at E.The internal bisector of B intersects side AC at F and L at G.If GF=DE, prove that AC=BC.

Suppose, on the contrary that $ AC\ne BC$. WLOG we may assume that $ BC>AC$. Also $ A>B$ which implies $ AE<BG$. Now
\[ GF\equal{}\frac{BG\cdot CF}{AC}\equal{}\frac{BG\cdot CF}{CF\plus{}FA}\equal{}\frac{BG}{1\plus{}\frac{FA}{CF}}\equal{}\frac{BG}{1\plus{}\frac{AB}{BC}}>\frac{AE}{1\plus{}\frac{AB}{AC}}\equal{}\frac{AE}{1\plus{}\frac{BD}{DC}}\equal{}\frac{AE\cdot CD}{BC}\equal{}DE.\]
A contradiction.


Prove: that its impossible for three consecutive integers to sum to another perfect square.

I let the three consecutive integers be $ (x \minus{} 1)^2$, $ (x)^2$, and $ (x \plus{} 1)^2$

Adding these we get: $ (x \minus{} 1)^2$ $ \plus{}$ $ (x)^2$ $ \plus{}$ $ (x \plus{} 1)^2$ $ \equal{}$ $ 3x^2 \plus{} 2$

Now we have to prove that $ 3x^2 \plus{} 2$ $ \neq$ $ y^2$ where $ y$ is a integer.

$ 3x^2 \plus{} 2$ isn't a perfect square so it can't equal $ y^2$ but I don't know how to prove that.

So if anybody could show me how to finish this off that would be great!

[quote="Evolved"]$ 3x^2 \plus{} 2$ isn't a perfect square so it can't equal $ y^2$ but I don't know how to prove that.

So if anybody could show me how to finish this off that would be great![/quote]

suppose that $ y \equal{} 3k \plus{} a$ with $ a \in \{0,1,2\}$

then $ y^2 \equal{} (3k \plus{} a)^2 \equal{} 9k^2 \plus{} 6ak \plus{} a^2$
then $ y^2 \equal{} 3(3k^2 \plus{} 2ak) \plus{} a^2$ and notice that for each possibility of $ a$, $ a^2 \in \{0,1,4\}$
then possible quadratic residues of $ y^2$ mod $ 3$ are $ 0$ or $ 1$ never a $ 2$

we see that all perfect squares are either 0 or 1 mod 4. so if x^2 is 0 mod 4. the 3x^2+2 becomes 2mod4 which can never be a perfect square. But I am not able to prove it when x^2 is 1 mod 4.

Or consider number 0, 1, and 2 (mod 3). Squaring them each leaves 0, 1, and 1 as a remainder when divided by 3, so no perfect square can leavea  remainder of 2 when divided by 3. But in $ 3x^2\plus{}2$, you know that $ 3x^2$ is divisible by 3, so adding 2 makes $ 3x^2\plus{}2$ automatically equiv to 2 (mod 3). Therefore, this is not a perfect square.

The qsn seems incomplete....

[quote] "prove it is impossible for 3 consecutive integers to sum to a perfect square"...[/quote]


geometry
perpendicular bisector
geometric transformation
geometry proposed

Let $ ABC$ be a triangle. Right triangles $ ADB$ and $ AEC$ are constructed outwardly such that $ \angle ADB\equal{}\angle AEC$ are right and $ \angle DAB\equal{}\angle EAC$. Let $ M$ be the midpoint of $ BC$. Prove that $ \angle MDE\equal{}\angle DAB$.

Let $ N,L$ be the midpoints of $ AB,AC.$ Denote $ \angle DAB \equal{} \angle EAC \equal{} \varphi.$ It's easy to see that $\triangle  NMD$ and $\triangle LEM$ are congruent, because of $ \angle MLE \equal{} \angle MND \equal{} \angle A \plus{} 2\varphi$ and $ NM \equal{} LE \equal{} \frac {_1}{^2}b,$ $LM \equal{} ND \equal{} \frac {_1}{^2}c.$ Hence, $ MD \equal{} ME$ $\Longrightarrow$ $\triangle  MED$  is isosceles with apex $M.$ Therefore

$\angle DME \equal{} \angle A \plus{} \angle LME \plus{} \angle NMD \equal{} \angle A \plus{} 180^{\circ} \minus{} \angle MLE \equal{} 180^{\circ} \plus{} 2\varphi$

$ \angle DEM\equal{}\angle EDM \equal{} 90^{\circ} \minus{} \frac {_1}{^2}(180^{\circ} \minus{} 2\varphi) \equal{} \varphi$

Take P on the perpendicular bisector of BC such that angle MBP = angle DBA, with P and A separated by BC. There is a spiral similarity around B mapping D to M and A to P and another one around C mapping E to M and A to P, both similarities having the same arguments; we got DM = EM, the angle between them 2* angle ABD, the conclusion follows.

Best regards,
sunken rock


Quadraliteral $ A, B, C, D$ is inscribed into circle $ o$. Bisectors of angles $ \angle DAB$ and $ \angle ABC$ intersect at point $ P$. Bistectors $ \angle BCD$ and $ \angle CDA$ intersect at point $ Q$. Point $ M$ is midpoint of arc $ BC$ (the arc without points $ D$ and $ A$). Point $ N$ is midpoint of arc $ DA$ (the arc without points $ B$ and $ C$). Prove that line $ PQ$ is perpendicular to $ MN$.

Let $ X \equal{} AD\cap BC$. Wlog, suppose $ A\in(XD),B\in (XC)$ . Then $ P$ is excenter opposite of to the point $ X$ of $ \triangle XAB$ and $ Q$ is incenter of $ \triangle XCD$. Hence $ PQ$ is internal bisector of $ \angle AXB$. So it is suffice to show $ NM$ is perpendicular to internal bisector of $ \angle AXB$. This is hold because $ ON \equal{} OM,ON\perp XA,OM\perp XB$.


Because Latin in the mother of all the romance languages and a great part of english, I'v decided to make a game. I'll post a latin word and give it's meaning. The next person posts a derivitive of the latin word (preferably english) and post another latin word and it's meaning.

Verbs should be in first person singular present
Nouns in nominative singular
Make any other parts of speech easy to understand.

please play only if you have a latin word that you know a derivitive for.

Heres one to start off.
Audio~I hear

Audition
durus - hard.

Incidentally, it is not very helpful to give only the first person singular of the verb, since most derivations stem from the supine (ie the fourth bit), eg frango, frangere, fregi, fractus - I break, from which the derivations are 'fraction', 'fracture', 'fractious' etc. all of which clearly stem from fractus.

-_- fine. How it shows up in the dictionary for verbs.

Durable?

Canis = dog

Edifice

(adj.) alius, alia, aliud: other, another.

[quote="InsanelyBored"]alias

Currio -  run[/quote]

Curry

jk  :lol: , current

Nauta - sailor

sector

flamma- fire

could the word also be Greek roots? I know more Greek roots than Latin. :lol:

vituperate: to attack one verbally

teneo tenere tenui tentus: hold, keep

tenent

wow, a lot of people take latin one.

This is an easy one:

Ego (accusative me) (daitive mihi) (i think genetive mea)-I, me, to me, mine

Please correct if wrong

its ego, mei, mihi, me, me

ego

obtrunco-assassinate

doesn't that mean I kill? obstuctionate. I think that is a word

Pastor=shepard

um pastor? lol because in the bible context pastors are considered the "herders of sheep".

anyways:
duco, ducere- lead

INDUCE (IT WAS NOT AN ELEMENT)

PROCAX, PROCACIS -- ARROGANT (I THINK)

[quote="fredbel6"][quote="futuremontrealian"]is that lead as in Pb or lead as in "to lead"?
sorry for not knowing  :blush: i'll look it up or something.

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

edit: hehe i guess it wouldn't be as in "Pb" since that is derived from latin's plumbum...  i still haven't looked it up though[/quote]
Pb is plumbum, meaning lead.

"ducere" is the Latin verb meaning "to lead".   :wink: The first person singular of the present perfect tense is "duxi" and the past participle is "ductum".

Mussolini, who ruled Italy during the Second World War, was often referred to as "Il Duce"
Another related word in English is "aquaduct" (it "leads water") and among the many other examples, we can find "conduct", "to reduce", "to abduct".

Edit : Indeed, Latin is the mother of the Romance languages (French, Italian, Spain, Portuguese, Romanian,..). English. Angles, Saxons and Jutes, who spoke Germanic languages and lived in Europe, settled in England in about the fifth century AD. Their languages combined to become Old English, it's only after the Norman conquest in 1066 that a lot of Romance elements slipped into it (like the ones discussed here)
So that means English is Germanic, but a lot more romanized than Germanic languages like Dutch and German.  In fact, in Dutch we often have a Romance and a Germanic word for the same thing.

Perhaps it might be interesting to add the German/Dutch equivalent as well? :wink:[/quote]

Fredbel you do not seem to know what you are talking about.
First of all, there is no such thing as "present perfect". There are only present, imperfect, future, perfect, pluperfect, and future perfect.
Also, duco's "first person singular present" is not duxi (which is perfect) but it is duco, then ducere, duxi, and ductum.
Perhaps your best course of action might be to actually look up the words/matters at hand before you correct someone, as you do not seem to have a reliable knowledge base.

anyways, disregarding fredbel's rant about the romance languages, let us go on with the game.

so: precocious

moneo, monere, monui, monitum- warn

[quote="InsanelyBored"]

Fredbel you do not seem to know what you are talking about.[/quote]
Well that is nice.

[quote]First of all, there is no such thing as "present perfect". There are only present, imperfect, future, perfect, pluperfect, and future perfect. [/quote]
Well yeah, it does correspond with that tense in English. :huh:

[quote]Also, duco's "first person singular present" is not duxi (which is perfect) but it is duco, then ducere, duxi, and ductum. [/quote]
I didn't claim anything that.
[quote]Perhaps your best course of action might be to actually look up the words/matters at hand before you correct someone, as you do not seem to have a reliable knowledge base. [/quote]Well that is nice.

[quote]Fredbel you do not seem to know what you are talking about.
First of all, there is no such thing as "present perfect". There are only present, imperfect, future, perfect, [b]pluperfect[/b], and future perfect.

Perhaps your best course of action might be to actually [b]look up the words/matters at hand before you correct someone, as you do not seem to have a reliable knowledge base. [/b]
anyways, disregarding fredbel's rant about the romance languages, let us go on with the game.
[/quote]

 :roll:  you missed something?

people make mistakes. that's normal.

Back to the game. Moniter

Cubiculum: bedroom

hey fredbel i just did a report on belgium!  i could've asked you a few questions :) lol.

is it just me or is that the last post was completely irelevent?

only in your imagination, ashpotter.

hmm... how about ponere, to put/place (i think)...

ponder?

ok.
another easy one generated by me

Candidatus- (i'm not even going to follow my own rules becuase this is so easy. Just look at the word and stare :ninja:  :w00t:

Circumnavigate

WHATEVER YOU KNOW THE VERB IN. GEEZ!!!

pluit: rain (check spelling)

[quote="Dave Vather"]Fenster (German for Window :D)
this one should be easy
Circa - around

BTW:
[quote] Verbs should be in first person singular present  [/quote]
Actually, in Spanish and German, verbs are found in the dictionary in infinitive.[/quote]

Ooo... I was thinking of "defenestrate" because thats english for "to throw out of a window" (Defenestration of Prague).  I vote "defenestrate" as the coolest word in the dictionary!!!!


algebra
function
domain
algebra unsolved

Suppose that two given functions $f$ and $g$ are defined on the interval $[0,2k]$ for some $k>0$ . Can we always find a pair $(x,y)$ of real numbers such that $|xy+f(x)+g(y)|\geq k^{2}$?


geometry
3D geometry
frustum
ratio
number theory
relatively prime

A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum-shaped solid $F$, in such a way that the ratio between the areas of the painted surfaces of $C$ and $F$ and the ratio between the volumes of $C$ and $F$ are both equal to $k$. Given that $k=m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

EDIT: whitehorseking88 has the correct answer. :blush:

Just wanted to provide a solution with the correct answer.

Let the radius of the cross-section be $k$.

So using $[SA] = \frac{1}{2}rs$ to express the lateral surface area and using $[A] = \pi r^2$ for the circles, we have a ratio.

We know the ratio of the volumes of C to F is:

$\frac{k^3}{27 - k^3}$

And that the surface areas that are painted of C to F is in the ratio:

$\frac{\frac{5 \pi k^2}{3}}{(15\pi - \frac{5 \pi k^2}{3}) + 9\pi}$.

Setting them equal, lots cancel.

So we end up with $\frac{k}{27 - k^3} = \frac{\frac{5}{3}}{24 - \frac{5}{3}k^2}$,
which if you multiple out gives you $k = \frac{15}{8}$.

SO plugging $k$ back into $\frac{C}{F} = \frac{k^3}{27 - k^3}$, gives us $\frac{125}{387}$.

The sum is $512$.

[hide=Solution]Let $C$ have base-radius $3x$, height $4x$, and slant height $5x$. We know that the ratio of paint areas of $C$ and the original cone is equal to the ratio of volumes of $C$ and the original cone. Using the formulas $A=\pi r^2$ and $LSA=\pi rl$, we get:
$$\frac{15x^2\pi}{24\pi} = x^3$$
$$x=\frac{15}{24} = \frac{5}{8}$$
Thus, $x^3 = \frac{5^3}{8^3}$, so $k=\frac{5^3}{8^3-5^3} \implies 5^3 + (8^3-5^3) = \boxed{512}$.


Let $\lambda$ be a real number s.t.
$0<\sqrt{2002}-\frac ab<\frac\lambda{ab}$ is satisfied for an infinity of integer pairs $(a,b)$. Prove that $\lambda\ge 5$.

[quote="Xell"]First observation: $2002=2.7.11.13$
We find that $a^{2}\equiv 0,1,2,4\ [7]$,$a^{2}\equiv 0,2,6,7,8,10\ [11]$, $a^{2}\equiv 0,1,3,4,9,10,12\ [13]$. So, if $a,b$ are integers such that $2002b^{2}-a^{2}>0$, then $2002b^{2}-a^{2}\geq 10$.
[/quote]

From here, because $2002b^2-a^2<\left(\frac {\lambda}{a}\right)^2+2\lambda$, assuming $\lambda<5$, gives us that $2002b^2-a^2<10-\epsilon$ for $a$ big enough. But if there are infinitely many such $(a,b)$ it's easy to show that $a-s$ cannot be bounded. From this we derive the contradiction.


this is similiar to beat the mod, but you cannot die. At the end, you will be given a rank, which will affect the events of the next thread (no, the next thread is not going to start until much later, probably one or two months later. Sign up here, 20 people can join. The topics are a surprise.

[quote="jjx1"]Why do we have to wait one or two months?

I sign up[/quote]
Because the next thread starts when this one ends, and this one will take one or two months.

i will join

if the game is still up and running

aka if u didnt abandone it.

hmmm,m

mathlete, when will we start?

And please keep signing up!!!

hmmmmm

sorry for advertising, but

JOIN BOMBS AWAY!!!

It is  (supposed) to be fun!

Kuz i created it!!!

i guess mathlete owes ME a favor, for keeping this thread alive......

I will tell you in advance that the first task is called "trust"

This video is supposedly me creating a level called "final test", then doing it. However, I might be lying. I might have just used animation rather than playing it through. You PM me whether you think I actually did this level or I just animated it.

It has something to do with this attachment.

I join!

The message is too small. Please make the message longer before submitting.

Scoreboard:

lotsofmath - 1
Garyzx - 1
hedgehog12 - 1
spiderboy95 - 0
Joe10112 - 0
nt811581147 - 0
Dojo - 0
jjx1 - 0

The answer is that I actually did it.
round 2 coming up soon

[size=200]Round 2 - Skill[/size]

You shall solve the problem on my blog. Pm me the answers.

My signature should have a link to my blog.

can you tell me the problem? that would be faster


Find the next term in the sequence:  $ 3, 9, 18 \cdots $

  Moreover, make a generalization for the $nth$ term.

[quote="JesusFreak197"][hide="Answer"]The next term in the sequence is $\boxed{30}$. The $n^{th}$ term in the sequence is given by $\displaystyle\sum_{n=1}^\infty3n$.[/hide][/quote]

That is correct!

[quote="math92"]is there an easier answer, that i would understand?[/quote]
[hide]The $nth$ term is $\frac{3n(n+1)}{2}$[/hide]

thanks nat mc your answer makes more sense to me. :)

[hide]And the sum of the first $n$ terms is: $\frac{(n)(n+1)(n+2)}{2}$[/hide]

Like this sequence:

3, 9, 18, 1000000004, 1000000004, 1000000004, 1000000004, 1000000004...

Which is what a math professor gave as the answer to a sequence problem.

[quote="JesusFreak197"]Actually, it isn't. If you plug in $n=3$, you get $20$, which is not right.

If it was given that $3$ and $9$ were the first two terms, you could come up with a number of different formulas, like $a_n=a_{n-2}+a_{n-1}+\frac{a_{n-2}+ a_{n-1}}{2}$. :lol:[/quote]
If oyu plug in 3 you get 30 which is 3+9+18....

Oh, whoops, I made an arithmetic mistake. :blush:


$ABCD$ is a parallelogram. $E$ is on $AB$ such that $AE = 4EB$. $F$ is on $BC$ such that $3BF = 2FC$. $DB$ cuts $EF$ at $K$. Find $DK: KB$

Since 3 points $B,\ K,\ D$ are colliniear, we can set $BK: KD=1-t: t.$
thus we have $\overrightarrow{AK}=t\overrightarrow{AB}+(1-t)\overrightarrow{AD}.$ On the other hand we can set $\overrightarrow{AK}=k\overrightarrow{AF}=k\left(\overrightarrow{AB}+\frac{3}{5}\overrightarrow{AD}\right).$

edit: I made mistake to misread the text.

kunny

[quote="kunny"]We can set $\overrightarrow{AK}=k\overrightarrow{AF}$[/quote]

We can't since $A,K,F$ aren't collinear.

[hide="But this is a tribute to your idea :)"]Put $\overrightarrow{AB}=\vec a, \overrightarrow{BC}=\vec b$.

If $\overrightarrow{AK}=k\overrightarrow{AD}+(1-k)\overrightarrow{AB}=(1-k)\vec a+k\vec b$, and

$\overrightarrow{BK}=t\overrightarrow{BE}+(1-t)\overrightarrow{BF}=t\left(-{1\over 5}\vec a\right)+(1-t){2\over 5}\vec b$,

then we have

$\overrightarrow{AK}-\overrightarrow{BK}=\overrightarrow{AB}=\vec a$, giving

$\left(1-k+{t\over 5}\right)\vec a+\left(k-{{2-2t}\over 5}\right)\vec b=\vec a$

hence

$1-k+{t\over 5}=1$ and $k-{{2-2t}\over 5}=0$

Solving the system we get $k={2\over 15}, t={2\over 3}$

Therefore $\overrightarrow{AK}={13\over 15}\vec a+{2\over 15}\vec b$ and $\overrightarrow{DK}=\overrightarrow{DA}+\overrightarrow{AK}=-\vec b+{13\over 15}\vec a+{2\over 15}\vec b={13\over 15}(\vec a-\vec b)={13\over 15}\overrightarrow{DB}$.

Therefore $DK: KB=13: 2$[/hide]

You are right, Farenhajt. :blush:

kunny


function
complex analysis
complex analysis unsolved

Let $ \{f_n\}$ be a sequence of analytic injective functions which converges to an analytic function $ f$. Show that $ f$ is either injective or $ f$ is a constant function.

Mmm, I just realized this follows from [url=http://en.wikipedia.org/wiki/Hurwitz%27s_theorem]Hurwitz's Theorem[/url] applied to the sequence $ \{f_n(z)\minus{}z\}_{n \in \bf{N}}$.