A woman, while at the funeral of her grandfather, met this guy whom she did not know. She thought the guy was amazing, her dream man! She fell in love with him instantly, but never asked for his phone number and could not find him after the funeral. A few days later she killed her sister.

What is her motive in killing her sister?

No, she never knew the guy before, and she couldn't find the man afterwards. If her sister was the guy's wife, it contradicts.

do they have anything to do with eachother

The sister and the man? [b]Unknown[/b]. (The murderer more or less knew nothing about this man)

are they human?
is this about a love affair?
does the dead guy have anything to do with the murderer?

are they human? [b]Yes[/b]
is this about a love affair? [b]Yope[/b]
does the dead guy have anything to do with the murderer? [b]He's her grandfather...[/b]

[quote="Ihatepie"]oh, she thinks that he will come to her sisters funeral if he came to his grandfathers.
 :D  :first:[/quote]
Correct !  :first:

[quote="Ihatepie"]oh, she thinks that he will come to her sisters funeral if he came to his grandfathers.
 :D  :first:[/quote]

OMG that's mean though lol. Who does that now? The woman is too desperate

Love makes people to crazy things :D
Lol it's more like an obsession for this woman actually.

No reason to revive a 4.5 year old thread with a useless post. Locked.


topology
superior algebra
superior algebra unsolved

Hi guys,

I want to find a good book to study cohomology in Algebra ( NOT cohomology in topology)
any good book to recommend ?

thanks so much

Do you mean homological algebra¿ Or some special case like Galois cohomology¿

Rotman - An introduction to homological algebra


The solution of $ \sqrt{5x\minus{}1}\plus{}\sqrt{x\minus{}1}\equal{}2$ is:

$ \textbf{(A)}\ x\equal{}2,x\equal{}1 \qquad
\textbf{(B)}\ x\equal{}\frac{2}{3} \qquad
\textbf{(C)}\ x\equal{}2 \qquad
\textbf{(D)}\ x\equal{}1 \qquad
\textbf{(E)}\ x\equal{}0$

[quote="AIME15"]The solution of $ \sqrt {5x \minus{} 1} \plus{} \sqrt {x \minus{} 1} \equal{} 2$ is:

$ \textbf{(A)}\ x \equal{} 2,x \equal{} 1 \qquad \textbf{(B)}\ x \equal{} \frac {2}{3} \qquad \textbf{(C)}\ x \equal{} 2 \qquad \textbf{(D)}\ x \equal{} 1 \qquad \textbf{(E)}\ x \equal{} 0$[/quote]
[hide]Moving one radical to one side we now have:
$ \sqrt {5x \minus{} 1} \equal{} 2 \minus{} \sqrt {x \minus{} 1}$
$ 5x \minus{} 1 \equal{} 4 \minus{} 4\sqrt {x \minus{} 1} \plus{} x \minus{} 1$
$ 5x \minus{} 1 \equal{} x \plus{} 3 \minus{} 4\sqrt {x \minus{} 1}$
$ 4x \minus{} 4 \equal{} \minus{} 4\sqrt {x \minus{} 1}$
$ x \minus{} 1 \equal{} \minus{} \sqrt {x \minus{} 1}$
$ x^{2} \minus{} 2x \plus{} 1 \equal{} x \minus{} 1$
$ x^{2} \minus{} 3x \plus{} 2 \equal{} 0$
$ (x \minus{} 2)(x \minus{} 1) \equal{} 0$
But plugging both back in to see and find any extraneous roots, we see that 2 does not work, so the only answer is 1. D[/hide]
Thanks enndb0x, fixed! :P

I think the solution is $ x_1\equal{}1$ and there's no solution for $ x_2\equal{}2$

First, square both sides. This gives us

\[\sqrt{5x-1}^2+2\cdot\sqrt{5x-1}\cdot\sqrt{x-1}+\sqrt{x-1}^2=4 \Longrightarrow 5x-1+2\cdot\sqrt{(5x-1)\cdot(x-1)}+x-1=4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}+6x-2=4\] Then, adding $-6x$ to both sides gives us

\[2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}-2 =-6x+4\] After that, adding $2$ to both sides will give us

\[2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x+6\] Next, we divide both sides by 2 which gives us

\[\frac{2\cdot\sqrt{5x^2-6x+1}-2+2}{2}=\frac{-6x+4+2}{2} \Longrightarrow \sqrt{5x^2-6x+1}=-3x+3\] Finally, solving the equation, we get

\[5x^2-6x+1=(-3x+3)^2 \Longrightarrow 5x^2-6x+1=9x^2-18x+9\]\[\Longrightarrow 5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9)\]\[\Longrightarrow -4x^2+12x-8=0 \Longrightarrow -4(x-1)(x-2)=0\]\[\Longrightarrow x-1=0 \text{or}\ x-2=0 \Longrightarrow x=1 \text{or}\ x=2\] Plugging 1 and 2 into the original equation, $\sqrt{5x-1}+\sqrt{x-1}=2$, we see that when $x=1$

\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot1-1}+\sqrt{1-1}=2 \Longrightarrow \sqrt4+\sqrt0=2 \Longrightarrow 2+0=2 \Longrightarrow 2=2\] the equation is true. On the other hand, we note that when $x=2$

\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot2-1}+\sqrt{2-1}=2 \Longrightarrow \sqrt9+\sqrt1=2 \Longrightarrow 3+1=2 \Longrightarrow 4=0\] the equation is false. Therefore the answer is $\fbox{{\bf(D)} x=1}$

The shortest solution: the LHS is monotonous increasing function, the LHS is constant function, so if we have solution, we have maximum one solution, so $x=1$ is the only one solution. Done.

[hide=least bash algebra]
Let $x=y^2+1$. The whole equation becomes
$\sqrt{5y^2+4}=2-y\implies 5y^2+4=4-4y+y^2\implies 4y^2+4y=0\implies y(y+1)=0$
Hence,
$y=0\implies x=1$
$y=-1\implies x=2$
However, we can see that $x=2$ is extraneous, so the answer is $\boxed{x=1}$.

[b]A tricky solution:[/b] From $ \sqrt{5x\minus{}1}\plus{}\sqrt{x\minus{}1}\equal{}2$ if we multiply both side by the conjugated expression, we get:$2x=\sqrt{5x-1}-\sqrt{x-1}$, so after a substraction we get $2\sqrt{x-1}=2-2x\Rightarrow x=1$.

We can just plug in the possible answers and you realize 1 is the only answer.

Do you think that this is a proof? Can we probe all real numbers $x>=1$?  :maybe:

When x = 1, the function's value is 2; the function only exists for x >= 1; the function is monotone increasing on x >= 1; thus, 1 is the only solution

[quote=Aniv]i have the simplest solution
as a term $\sqrt{x-1}$ is given so $x\ge 1$
in qn the 2 values which are greater than or equal to 1 are 1 and 2
so put those values and find which one is satisfying the relation[/quote]

I also got $(D) x = 1$ as my answer, I used a similar method as you. I knew $x$ had to be a positive number, that left us with $\frac{2}{3}, 1,$ and $2$. After that I plugged in the values to see which one works.


number theory
least common multiple
modular arithmetic

One less than the number of coins must be a multiple of 3, 4, and 6. Any multiple of 12 (which is the LCM of 3, 4, and 6) will satisfy that condition. Thus, the number of coins is any multiple of 12 plus one. These are 13, 25, 37, 49, ...

[quote="Soccerblls2"]Makena has a certain amount of coins. When divided by 3,4,and 6 it has a remainder of 1. What is the number of coins? :D[/quote]

I think the question is asking for least amount maybe (?), in which case the answer is $ 13$ as explained by PoP.

Another way of writing this is:

$ k \equiv 1$  $ (\mod{12})$

Basically this is the expression talking about all $ k$ that have a remainder of $ 1$ when divided by $ 12$.  Here is more info on this, which is called modular arithmetic.
http://mathworld.wolfram.com/ModularArithmetic.html

Basically, all you do is take the LCM of the three numbers and add $ 1$ as PowerofPi explained.

Since the least common multiple is 12, then as Adithya said, k is congruent to 1(mod 12)

You can use a method called the "Chinese Remainder Theorem" to solve this problem. It is further explained here:
[url]http://www.cut-the-knot.org/blue/chinese.shtml[/url].

oh.
Well, you can just be like draw a table..and do it the long way..and you find out like 13 is the answer.

If the number of coins are divided by 3,4, or 6, and the reminder is 1, then the number of coins -1 is divisible by 3,4, or 6. So you have to find the least common multiple which is 12. Then, you have to add on the 1 that you subtracted. So any number that is a multiple of 12 and you add 1 could be the amount of coins she has.

[hide]Let the number be $ x$. Consider $ x\minus{}1$. It is divisible by $ 3,4, \text{and} 6$.
Thus, it must be a multiple of $ \text{lcm}(3,4,6) \equal{} 12$ so we say $ x\minus{}1 \equiv 0 \pmod{12}$.
Then $ x \equiv 1 \pmod{12}$, so $ x$ can be $ 1, 13, 25, ...$.[/hide]

with numbers that small it isnt hard to guess and check, although if the numbers were to get larger then guessing and checking would get tougher.

theres a whole bunch of numbers, the list starts with 13 then keep adding 12

13,25,37........................

Let $ x$ be the number of coins Makena has.
$ x \mod 3 \equiv 1$
$ x \mod 4 \equiv 1$
$ x \mod 6 \equiv 1$
Obviously, since $ 3$ is a factor of $ 6$, we can remove that statement. Now to find the minimum:
$ x \minus{} 1 \mod 4 \equiv 0$
$ x \minus{} 1 \mod 6 \equiv 0$
The LCM of 4 and 6 is $ \frac{4}{2} \times \frac{6}{2} \times 2 \equal{} 12$, therefore:
$ x \minus{} 1 \equal{} 12$
$ x \equal{} 13$
Checking, we have:
$ 13 \mod 3 \equiv 1$
$ 13 \mod 4 \equiv 1$
$ 13 \mod 6 \equiv 1$
Done.


sorry i've done this before, but i'm looking for the solution. is this from the Bulgarian National Olympiad?

i dunno. we'd done it in a camp for a national olympiad and our guide mentioned taiwan. i have no idea myseelf.

i got a solution too by constructing a trinomial with a,b,c as roots.. it leads to a solution but its miserably messy.. i was looking for elegance..

The inequality is symmetric so It is always easy to prove  :D
Put $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$. The inequality is equivalent to:
\[ (3q\minus{}5)p^2\plus{}(p\minus{}1)q^2\plus{}9pq\ge 24p\plus{}3q\plus{}27\]
\[ LHS\ge 4p^2\plus{}2q^2\plus{}6pq\ge 12p\plus{}6(q\minus{}1)p\plus{}6p\plus{}2q^2\ge RHS\]

I think all of symmetric inequailty can be solved by "p,q,r technique" easily  :wink:


radius of circle O = $ 6$

Shaded area = $ 2 * 9\pi \equal{} \boxed{18\pi}$

Basicly move the shaded areas into where you see that they fit in so that the shaded area is just the area of 2 circles of radius 3.


function
inequalities unsolved
inequalities

Find the minimum value of the expression $ \frac{a\plus{}1}{a(a\plus{}2)}\plus{}\frac{b\plus{}1}{b(b\plus{}2)}\plus{}\frac{c\plus{}1}{c(c\plus{}2)}$, where $ a,b,c$ are positive real numbers with $ a\plus{}b\plus{}c \le 3$.

It is straightforward when you write it :
find the minimum of $ \frac{x}{x^2\minus{}1}\plus{}\frac{y}{y^2\minus{}1}\plus{}\frac{z}{z^2\minus{}1}$ for $ x,y,z\geq 1$ and $ x\plus{}y\plus{}z\leq 6.$

we have $ \frac{x}{x^2\minus{}1}\geq\frac{16\minus{}5x}{9}\Leftrightarrow (x\minus{}2)^2(5x\plus{}4)\geq 0,Right$
$ \Rightarrow LHS\geq\frac{48\minus{}5\sum x}{9}\geq\frac{48\minus{}5.6}{9}\equal{}2$

[quote="karis"]we have $ \frac {x}{x^2 \minus{} 1}\geq\frac {16 \minus{} 5x}{9}\Leftrightarrow (x \minus{} 2)^2(5x \plus{} 4)\geq 0,Right$
$ \Rightarrow LHS\geq\frac {48 \minus{} 5\sum x}{9}\geq\frac {48 \minus{} 5.6}{9} \equal{} 2$[/quote]
What about $ x\equal{}1/2$. Then $ x^2\minus{}1<0$ and hence $ LHS \le RHS$.

hello, i think we have $ \frac{a\plus{}1}{a(a\plus{}2)}\plus{}\frac{b\plus{}1}{b(b\plus{}2)}\plus{}\frac{c\plus{}1}{c(c\plus{}2)}\geq2$ and the equal sign holds for $ a\equal{}b\equal{}c\equal{}1$.
Sonnhard.

[quote="FelixD"][quote="karis"]we have $ \frac {x}{x^2 \minus{} 1}\geq\frac {16 \minus{} 5x}{9}\Leftrightarrow (x \minus{} 2)^2(5x \plus{} 4)\geq 0,Right$
$ \Rightarrow LHS\geq\frac {48 \minus{} 5\sum x}{9}\geq\frac {48 \minus{} 5.6}{9} \equal{} 2$[/quote]
What about $ x \equal{} 1/2$. Then $ x^2 \minus{} 1 < 0$ and hence $ LHS \le RHS$.[/quote]
Felix....... Remember that $ x \equal{} a \plus{} 1$ so $ x \ge 1$ ! ;) :P

Btw. the function $ f(a) \equal{} \frac{a\plus{}1}{a(a\plus{}2)}$ is decreasing and convex. So $ LHS \equal{} f(a) \plus{} f(b) \plus{} f(c) \ge 3f \left ( \frac{a\plus{}b\plus{}c}{3} \right ) \ge 3f(1) \equal{} 2$.

[quote="FelixD"]Find the minimum value of the expression $ \frac {a \plus{} 1}{a(a \plus{} 2)} \plus{} \frac {b \plus{} 1}{b(b \plus{} 2)} \plus{} \frac {c \plus{} 1}{c(c \plus{} 2)}$, where $ a,b,c$ are positive real numbers with $ a \plus{} b \plus{} c \le 3$.[/quote]
It's obviouly trues by Am-Gm because:
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \geq\ 3$
And $ \frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2} \geq\ 1$
The general problem:
Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \leq\ 3$. ($ m \geq\ k \geq\ 0$)
Prove that: $ \sum\ \frac {a \plus{} k}{a(a \plus{} m)} \geq\ \frac {3(1 \plus{} k)}{1 \plus{} m}$
 :)

[quote="nguoivn"][quote="FelixD"]Find the minimum value of the expression $ \frac {a \plus{} 1}{a(a \plus{} 2)} \plus{} \frac {b \plus{} 1}{b(b \plus{} 2)} \plus{} \frac {c \plus{} 1}{c(c \plus{} 2)}$, where $ a,b,c$ are positive real numbers with $ a \plus{} b \plus{} c \le 3$.[/quote]
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \geq\ 3$
And $ \frac {1}{a \plus{} 2} \plus{} \frac {1}{b \plus{} 2} \plus{} \frac {1}{c \plus{} 2} \geq\ 1$[/quote]
Could you explain how to use AM-GM?

[quote="FelixD"]I think that I can explain^^:

$ \frac {a \plus{} 1}{a(a \plus{} 2)} \equal{} \frac {1}{2}\left( \frac {1}{a} \plus{} \frac {1}{a \plus{} 2} \right)$.

Then AmGm.  :wink:

edit: I'm not sure if I've answered your question.  :roll:[/quote]
Thanks :) I'm pretty sure I would use AM-HM to the ineqs nguoivn posted, but now I understood the solution.


geometry
geometric transformation
reflection

A beam of monochromatic light falls under a small angle on one of the paralel surfaces of a glass plate(the coef of refraction of glass n) with thickness d. The length of the plate is much bigger than its thickness. The plate is in air, The coeficient of reflection is [b]R[/b]. We neglect the loss of energy by absorbtion. Light intensity in the beam is I[b][size=59]0[/size].
     a) Explain the obtaining of interference figures, their type, and conditions of maximal vizibility.
     b) Find the expressions for the repartition of intensity in the interference figures.
     c) Verify the law of energy conservation.[/b]

Assume that the angle which light makes with the normal is ${\alpha}$.Then we have two rays that are interfering,one that is reflected from the surface and the another that is first refracted into the medium and then undegoes a reflection at the second interface and is again refracted back to the original medium at the first interface.

from simple geometry it can be proved that the path difference is $2nd,cos{\alpha}$.We also have an additiional phase difference of ${\pi}$ due to reflection at the second interface.Hence the total path difference is
 $P$ =  $2nd, cos{\alpha}$ + $\frac{\lambda}{2}$

for max intensity we have $P$ = $n{\lambda}$

OK, thanks :lol: , and what  about the part b) and c) :|

a)Actually there will be a set of rays which won't interfere at all,unless there is a way to make them interfere,for example a lens,but in this case there will not be interference figures ,just a spot situated in the focal plane.Now visibility is defined as V=Imax-Imin/Imax+Imin where Imin =0 so the maximal visibility is obtained for the path difference showned by ashwinkrishnan.

b)all i can say is that the rays that go back in the air (if R is the same in each case of reflection) with the following intensities:R*I0,R*(1-R)^2*I0,R^3*(1-R)^2*I0,R^5*(1-R)^2*I0,R^7*(1-R)^2*I0,...

Actually I obtained another expression for the path difference:
P=2d*sqrt(n^2 - sin(a)^2) - lambda/2.
I think yours is also correct but your angle alfa must be the angle which the refracted ray makes with the normal.

:idea:
For the intensity distribution in the interference figures i obtained something like this (notify me if I'm wrong)
I=I(o)cos(t)^2.
where I(o) is the intensity in maximums and angle t=pi*P/lambda
I(o)=4A^2, where A- amplitude of the light

something tells me you considered a difractional system but this cannot be applied here.we have a set of paralel rays,a infinity of them with a certain direction which is different from a diffractional system.THERE WILL NOT BE INTERFERENCE FIGURES.The best way you can consider is that all the rays sum up in a point .

I=K*A^2 for each ray.but if we have 2 rays that interfere with different intensities then I=[sqrt(I1)+sqrt(I2)]^2 so.....u're wrong


1. Let $ D$ be the foot of the bisector of the angle $ \angle BAC$ in the triangle $ ABC$ with $ \angle BAC \equal{} 60^{\circ}$ Consider the reﬂections $ M$ and $ N$ of the point $ D$ across the sides $ AB$ and $ AC$. Let $ MN \cap AD \equal{} \{P\}$.
a) Find the angles of the triangles $ AMN$ and $ ADN$.
b) Prove that $ P$ is the midpoint of $ AD$.

2. Let $ ABC$ be a triangle with $ AB \equal{} AC$. Lines $ l_1$ and $ l_2$ pass through $ B$ and $ C$ respectively, such that $ l_1 \perp AB$ and $ l_2 \perp AC$.
Prove that the lines $ l_1$ , $ l_2$ and the bisector of angle $ \angle BAC$ are concurrent.

3. Let $ ABC$ be a triangle with $ AB \equal{} AC$. Consider an interior point $ M$ such that $ \widehat{MBA} \equal{} \widehat{MCA}$. Prove that $ AM \perp BC$.

4. Let $ ABC$ be a triangle with $ AB \equal{} AC$. Isoscelles triangles $ ABM$ and $ ACN$ with bases $ AB$ and $ AC$ are errected outside triangle $ ABC$.
Prove that the altitudes $ MP \perp AB$, $ NQ \perp AC$ and $ AA' \perp BC$ are concurrent.

̂̂5. In the triangle $ ABC$ the bisector of the angle $ \widehat{BAC}$ touches $ BC$ at $ D$. Consider $ E$ on $ AC$ such that $ DE \| AB$. Let $ F$ be a point on $ AB$ such that $ EF \perp AD$. Prove that $ DN \| AC$.

Note: Diagrams are always appreciated  :D

[hide="1"]Since $ < BAD \equal{} < CAD \equal{} 30^o$ then the reflected lines $ AN$ and $ AM$ about $ AB$ and $ AC$ complete the equilateral triangles $ ADM$ and $ ADN$ thus $ ANDM$ is a rhombus , its diagonals $ AD,MN$ cut at their middle $ P$ , $ \Delta AMN$ is isosceles with $ < MAN \equal{} 120^o$[/hide]

[hide="2"]If $ l_1,l_2$ cut in $ E$ just observe that the triangles $ ABE$ and $ ACE$ are congruent --> $ EB \equal{} EC$ --> E lies on the bisector of $ < BAC$.[/hide]

[hide="3"]$ < MBA \equal{} < MCA$ with $ < ABC \equal{} < ACB$ implies $ < MBC \equal{} < MCB$ ---> $ \Delta MBC$ is isosceles ---> $ MB \equal{} MC$---> $ M$ lies on the midperpendicular of $ BC$----> $ AM \perp BC$[/hide]

[hide="4"]It is easy to see that the concurrency point is the circumcenter of $ \Delta ABC$[/hide]

[hide="5"]$ N$ is not defined in the problem[/hide]


Let $ f: \mathbb{N*} \to \mathbb{N*}$.  Given:
1) $ f(a)f(b)\equal{}f(ab)$ for all coprime $ a$ and $ b$
2) $ f(p\plus{}q) \equal{} f(p) \plus{} f(q)$ for all primes $ p$ & $ q$.
Find $ f(2002)$.

[quote="stephencheng"]...
$ f(2)f(2) \equal{} f(2 \plus{} 2) \equal{} 2f(2)$
...[/quote]
Where does that come from? :huh:
Here's my solution. It has some parts similar to stephencheng's though.

Let $ f(2)\equal{}x,$ $ f(3)\equal{}y$

From 2), we get:
i)$ f(4)\equal{}2f(2)\equal{}2x$
ii)$ f(5)\equal{}f(2)\plus{}f(3)\equal{}x\plus{}y$
iii)$ f(7)\equal{}f(2)\plus{}f(5)\equal{}2x\plus{}y$
iv)$ f(9)\equal{}f(2)\plus{}f(7)\equal{}3x\plus{}y$

From 1), $ 3x\plus{}2y\equal{}f(5)\plus{}f(7)\equal{}f(12)\equal{}f(4)f(3)\equal{}2xy$
It is clear that $ x\not \equal{}1$
So $ x>1$ and $ y\equal{}\frac{3x}{2x\minus{}2}$

Suppose that $ x>4$
From $ 2x\minus{}2<3x<4x\minus{}4$, we get $ 1<\frac{3x}{2x\minus{}2}\equal{}y<2$ which is impossible since $ y\in\mathbb{N}$
So $ x\equal{}2,3$ or $ 4$
Plugging the values we get $ x\equal{}2, y\equal{}3$ or $ x\equal{}4,y\equal{}2$
From $ x(x\plus{}y)\equal{}f(2)f(5)\equal{}f(10)\equal{}f(3)\plus{}f(7)\equal{}2x\plus{}2y$
Suppose $ x\equal{}4, y\equal{}2$, we get $ 24\equal{}12$ which is impossible.
So $ x\equal{}2,y\equal{}3$

So we know now that $ f(2)\equal{}2,f(3)\equal{}3,f(5)\equal{}5,f(7)\equal{}7$ and $ f(9)\equal{}9$

From $ f(11)\plus{}7\equal{}f(11)\plus{}f(7)\equal{}f(18)\equal{}f(2)f(9)\equal{}18$ So $ f(11)\equal{}11$
And from $ f(13)\equal{}f(2)\plus{}f(11)\equal{}13$
$ \therefore f(2002)\equal{}f(2)f(7)f(11)f(13)\equal{}2002$

[quote="beginner01"][quote="stephencheng"]...
$ f(2)f(2) \equal{} f(2 \plus{} 2) \equal{} 2f(2)$
...[/quote]
Where does that come from? :huh:
[/quote]

It seems that the question does not say $ p, q$ must be distinct .  :maybe:

@stephencheng: Yep, I think you got something wrong. $ p$ and $ q$ must be primes to satisfy $ f(p+q)=f(p)+f(q)$. Then you can't say $ f(9) = f(3) + f(6)$. Oh, $ p$ and $ q$ don't have to be distinct.
And $ f(4)=f(2)f(2)$ isn't right either.
I've come up with this solution:

$ f(2002)=f(2)f(7)f(11)f(13)$
$ f(2002)= f(14)f(11)f(13) = 2f(7)f(11)f(13)$ as $ f(2p) = 2f(p)$.
Then $ f(2) = 2$.

Right, $ f(12) = f(5) + f(7) = (f(2) + f(3)) + (f(2) + f(5)) = \vdots = 3f(2) + 2f(3)$
And $ f(12) = f(3)f(4) = 4f(3)$. From these two, we have $ f(3)=3$.  :maybe:

$ f(7) = f(2) + f(5) = 2f(2) + f(3) = 7,$ and $ f(11) = f(14)-f(3)=f(2)f(7)-f(3)=14-3=11.$
Finally $ f(13) = f(11) + f(2) = 13$  :D
But can we show that $ f(x) = x$, at least for all primes $ x$?


[quote="tuantam1lan"]let a,b,c be non-negative number such that $ a \plus{} b \plus{} c \equal{} 2$,no two of which are zero.prove that:
$ \frac {bc}{2a \plus{} \sqrt {4a^2 \plus{} bc}} \plus{} \frac {ca}{2b \plus{} \sqrt {4b^2 \plus{} ac}} \plus{} \frac {ab}{2c \plus{} \sqrt {4c^2 \plus{} ab}} \leq 1$[/quote]
It is very old, too.
$ \frac{5}{2}(a \plus{} b \plus{} c) \ge \sum \sqrt {4a^2 \plus{} bc}.$
See here: http://www.mathlinks.ro/viewtopic.php?t=197672


quadratics
algebra
polynomial
Vieta
quadratic formula

This problem is proof. This may seems hard but it's basically simple manipulation.

Prove that sum of the roots of quadratic equation is $\frac {-b}{a}$ and product of the roots is $\frac {c}{a}$.

I'll post both hint and the answers to help the people who are going to try on this.

[hide="hint"]What are two roots that are defined in the Quadratic Formula?[/hide]

[hide="close_hint"]There is a sign $\pm$. How does this helps us on the product and sum?[/hide]

[hide="Answer"]
For product:

$\frac {-b + \sqrt{b^2-4ac} }{2a} *\frac {-b - \sqrt{b^2-4ac}}{2a} $
$\frac {b^2-(b^2-4ac)}{4a^2}$
$\frac {4ac}{4a^2}$
$\frac {c}{a}$

For sum:

$\frac {-b+ \sqrt{b^2-4ac}}{2a} +\frac {-b - \sqrt{b^2-4ac}}{2a}$
$\frac {-2b}{2a}$
$\frac {-b}{a}$

Done.[/hide]

Extra challenge: See if you can find an even slicker way than the proposed solution!  Then think about polynomials with more than 2 roots. . .

I'm going to try a different way than Silverfalcon's.

[hide]The quadratic equation is ${ax^2}+{bx}+{c} = 0$.

I'm going to make $a = a_1\cdot a_2$ and $b = c_1\cdot a_2+c_2\cdot c_1$ and $c = c_1\cdot c_2$.

We can factor the above polynomial into $(a_1x+c_1)(a_2x+c_2)$.  The roots are $\frac{-c_2}{a_2}$ and $\frac{-c_1}{a_1}$.

The sum of the roots is $\frac{-c_2\cdot a_1 + -c_1\cdot a_2}{a_1\cdot a_2}$.  Since the numberator is equal to $-b$, and the denominator is equal to $a$, we can just plug them in to get $\frac{-b}{a}$.

The product of the roots is $\frac{-c_1\cdot -c_2}{a_1\cdot a_2}$.  Since the numberator is equal to $c$, and the denominator is equal to $a$, we can plug them in to get $\frac{c}{a}$[/hide]

[hide]
I used a direct approach . . .

Let $p$ and $q$ be the roots of: $ax^2 + bx + c = 0\;\;\Rightarrow\;\;[1]\;x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

Then the equation is of the form: $(x - p)(x - q) = 0\;\;\Rightarrow\;\;[2]\;x^2 - (p+q)x + pq = 0$

Since [1] and [2] are the same equation, their corresponding coefficients are equal.
Therefore: $p+q = -\frac{b}{a},\;\;pq = \frac{c}{a}.$[/hide]

This theorem was proved by Vieta in the mid 1600's.

[hide]

Let $s_i$ be the sum of the product of distinct polynomial roots $r_j$ of the polynomial equation of degree $n$

\[a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=0\]

where the roots are taken $i$ at a time.  For example,

\[s_1=r_1+r_2+...\]
\[s_2=r_1r_2+r_1r_3+r_1r_4+r_2r_3+...\]

and so on, Vieta's Formula states that

\[s_i=(-1)^i\left(\frac{a_{n-i}}{a_n}\right)\]

Proof for a Quadratic:

\[a_2x^2+a_1x+a_0=a_2(x-r_1)(x-r_2)=a_2[x^2-(r_1+r_2)x+r_1r_2]\]

Thus

\[s_1=r_1+r_2=\frac{-a_1}{a_2}\]
\[s_2=r_1r_2=\frac{a_0}{a_2}\]

Using similar reasoning this can be shown for a cubic polynomial and extended to a polynomial of degree $n$.

Much information on Vieta's Formulas can be found at http://www.mathworld.com

[/hide]

See if you can come up with the proof for the polynomial

\[a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0=0\]

using the proof of the quadratic.

Can you explain more about Vieta's formula. It seems very interesting and useful.  :)

Basically, Vieta's Formula allows you to calculate the sum product of the roots taken $n$ at a time. If $n=1$ then basically you are calculating the sum, since you are taking them one at a time. On the other hand, if $n=n$ then you are finding the product of the roots since you are taking all of them together (or $n$ at a time). Thus, you can calculate the sum of the roots taken two at a time, three at a time, four at a time,..., $n$ at a time. Here is an example, from AoPS:

If three roots of $x^2+Ax^2+Bx+c=0$ are $-1$, $2$, and $3$ then what is the value of $2C-AB$?


geometry
3D geometry
Pythagorean Theorem

Attached

My message is too short...now it isn't

[hide="answer and solution"]
72

Use the Pythagorean Theorem and you will find that the slant height of the cone is 15, and the slant height of the cone is the radius of the circle. The lateral surface area of the cone is 180pi (pi*r*slant height), and the area of the full circle is 225pi. 180pi/225pi = 4/5, which means that there is 1/5 of the circle cut out. 1/5 times 360 = [b]72[/b].
[/hide]


Let $a1;...an$ be a positive real sequence.Prove that
$\overline{\mbox{lim}}\left(\frac{a_1+a_{n+1}}{a_n}\right)^n\geq e$.

No, consider the sequence defined by \[ a_n=\frac{1}{2^n} \]
As $n$ grows large, $\frac{1+\frac{1}{2^{n+1}}}{\frac{1}{2^n}}=\frac{1}{2^n}+\frac{1}{2(4^n)}$ which approaches 0.

Simply apply the definition of $e=\lim_{r\to \infty} (1+r^{-1})^r$


number theory
greatest common divisor
induction

Let $a,b$ be natural numbers and $(a,b)$ denote the gcd of $a$ and $b$
[b][color=blue](1)[/color][/b] Prove that if $(a,b) = 1$ then $(a^k,b^n) = 1$ where $k,n$ are any natural numbers.
[color=blue][b](2)[/b][/color] Is the converse of [b][color=blue](1)[/color][/b] true?

[quote="varun"]Can we use Principle of Mathematical Induction to prove the (1)[/quote]

You show it then  :)

(1) if $ (a,b) \equal{} 1$ then $ (a^k ,b) \equal{} 1$ for any natural number $ k \ge 0$.
Proof: For $ k \equal{} 0$, the result is obvious.  Let $ (a,b) \equal{} 1$ and $ k > 0$.  By Bézout's identity $ ax \plus{} by \equal{} 1$ for some integers$ x,y$. Then

$ 1 \equal{} (ax \plus{} by)^k \equal{} a^k x^k \plus{} b\sum\limits_{i \equal{} 1}^k {\left( \begin{array}{l} k \\
i \\
\end{array} \right)} a^{k \minus{} i} b^{i \minus{} 1} \equal{} a^k x_k \plus{} by_1$

And therefore $ (a^k ,b) \equal{} 1$.

Now if $ (a,b) \equal{} 1$ and $ k,n \ge 0$ then by (1) $ (a^k ,b) \equal{} 1$.  set $ a^k \equal{} A$ then $ (b,A) \equal{} 1$ and by (1) again $ (b^n ,A) \equal{} 1$.  So $ (b^n ,a^k ) \equal{} 1$.

(2) Yes, since if $ p$ is prime such that $ p|a,b$ then $ p|a^k ,b^n$.


[b]Step 1)[/b] Draw a 3x3 box on a piece of paper.
[b]Step 2)[/b] Put an "n" in the first box of the first row, nothing in the second box, and a 2 in the third box. This completes row 1.
[b]Step 3)[/b] Put nothing in the first or second boxes of row 2 and a 9 in the third box. This completes row 3.

[u]Nothing goes in the last three boxes.[/u]

In this magic square, each number from 1 to 9 will appear once, with the result that each row, column, and diagonal has the same sum. What number must appear in the space marked "n"?

A) 3
B) 4
C) 5
D) 6
E) 7

I have seen solafidefarms solve these multiple times, but I'm not able to solve them myself.

[hide="Hint"]
One of the diagonals must be 4,5,6.
[/hide]

First we want to find out what the sum of numbers in each row is. We know the sum of numbers from $1$ to $9$ is $45$, so there is a sum of $15$ in each row, column. A $4$ goes in the bottom right-hand corner.

See the diagram for additional notation.

Going to the middle column, $t+2=u+4$, $u+2=v+4$. Thus $v<u<t$ with each difference equal to $2$. Therefore $3u=15$ (remembering $t+u+v=15$) and so $u=5$. Hence $n=6$

n - t - 2
m - u - 9
p - v - 4


There are N(1<=N<=50) towns and M routes between them. Recently, some new travel agency was founded for servicing tourists in these cities. We know cost which tourist has to pay, when traveling from town i to town j which equals to Cij<=100 and time needed for this travel - Tij<=100. There are many tourists who want to use this agency because of very low rates for travels, but agency still has only one bus. Head of agency decided to organize one big travel route to gain maximal possible amount of money. Scientists of the company offer to find such a cyclic path G, when greedy function f(G) will be maximum. Greedy function for some path is calculated as total cost of the path (sum of Cij for all (i,j) - routes used in path) divided by total time of path (similar to Cij). But nobody can find this path, and Head of the company asks you to help him in solving this problem.

Is this solution correct:
 Let $g(G)=\frac{1}{f(G)}$. So we should find the cycle $G$ such that $g(G)$ is minimal. We use modification of Floyd-Warchall algorithm and let minimal cost for the path from i to j be $\frac{a[i, j].t}{a[i, j].c}$. After we apply Floyd-Warshall algorithm and the solution for $g$ is $min\{ \frac{a[i, i].t}{a[i, i].c} \}$

say path a->b, t1, c1. where g() min.

path b->c (adjacent say) then t2,c2.

note it is not necessary to be i=1 where ti/ci+t2/c2 will b min.

am i wrong?


calculus
integration
trigonometry
calculus computations

Evaluate:

$\int \frac{\cos^{5}2 \theta}{\sin^{2}\theta}.d \theta$

[quote="N.T.TUAN"][hide="hint"]Put $y=\cot{\theta}$.  :wink: [/hide][/quote]

Sorry I still don't get it. Mind elaborating?

$d\cot{\theta}=-\frac{1}{\sin^{2}{\theta}}\cdot d \theta$ and $\cos{2\theta}=\frac{\cot^{2}{\theta}-1}{\cot^{2}{\theta}+1}$.

[quote="BanishedTraitor"]I tried and I ended up with $\int (\frac{1-y^{2}}{1+y^{2}})^{5}.dy$ is that right?

I still don't know how to integrate it.[/quote]

well , a more directly way (but quite tedious) might be expanding out everything ....

$\frac{\cos^{5}2x}{\sin^{2}x}\\=\frac{(1-2\sin x)^{5}}{\sin^{2}x}\\=\frac{1}{\sin^{2}x}-\frac{10}{\sin x}+40-20\sin x+80\sin^{2}x-32\sin^{3}x$

Then you integrate one by one seperately

So did I, shyong. This integral doesn't need to make special thing all. :D

That's also what I did but I was hoping for a less tedious method. But I still don't how the suggested substitution could work.

How about this:
let $sin\theta = \frac{z^{2}-1}{2iz}$, $cos\theta = \frac{z^{2}+1}{2z}$, $d\theta = \frac{dz}{iz}$.


Does anyone know what's wrong with the website?

[url]http://ace.delos.com[/url] works fine for me, but not [url]http://ace.delos.com/usacogate[/url].

it seems none of them are available now. so it is their internal problem may be.


geometry
inequalities
parallelogram
inequalities solved
Geometric Inequalities

Let $ABCD$ be a convex quadrilateral and $K,L,M,N$ be points on $[AB],[BC],[CD],[DA]$, respectively. Show that,
\[
\sqrt[3]{s_{1}}+\sqrt[3]{s_{2}}+\sqrt[3]{s_{3}}+\sqrt[3]{s_{4}}\leq 2\sqrt[3]{s}
\]
where $s_1=\text{Area}(AKN)$, $s_2=\text{Area}(BKL)$, $s_3=\text{Area}(CLM)$, $s_4=\text{Area}(DMN)$ and $s=\text{Area}(ABCD)$.

Alternatively, draw in the diagonals $\overline{AC}$ and $\overline{BD}$. It is clear that $\overline{KL}, \overline{LM}, \overline{MN},$ and $\overline{NK}$ are midlines of $\triangle{ABC}, \triangle{BCD}, \triangle{CDA},$ and $\triangle{DAB}$ respectively; whence, $S_1 + S_2 + S_3 + S_4 = \frac{S}{2}$. Therefore the result follows by Jensen's inequality. In fact, we find the follow generalization

$S_1^r + S_2^r + S_3^r + S_4^r \le 4 \cdot \left(\frac{S}{8}\right)^r$ for all $0 \le r \le 1$ and the corollary result

$S_1^r + S_2^r + S_3^r + S_4^r \ge 4 \cdot \left(\frac{S}{8}\right)^r$ for all $r \le 0$ and $r \ge 1.$

$ (\frac{S_1}{S})^{\frac{1}{3}}\equal{}(\frac{A(ABD)}{S}\cdot\frac{AK}{AB}\cdot\frac{AN}{AD})^\frac{1}{3}\leq\frac{1}{3}(\frac{A(ABD)}{S}\plus{}\frac{AK}{AB}\plus{}\frac{AN}{AD})$ (where $ A(XYZ)$ equals to the area of $ XYZ$) Summing these 4 inequalities gives the desired result.

As a plus, the equality holds when $ABCD$ is a parallelogram and $K,L,M,N$ are midpoints.


OK, i [b][u]finally[/u] [/b]managed to solve this integral last night..... LOL !  :D

show

$\int_{0}^{\frac{\pi}{2}}\;\cos^{-1}{\left( \frac{\cos x}{1+2\cos x}\right)}\;dx\;=\;\boxed{\frac{5}{4}\;\zeta{(2)}}$


[quote="236factorial"]There are 12 ordered pairs of integers (x,y) such that $x^2+y^2=25$. What is the greatest possible sum of x+y?[/quote]

[hide]There are actually 6 different ordered pairs of $(x,y)$ because you can just switch the values of x and y and there would be no difference. Those 6 possible ordered pairs are:

$(3,4)$
$(-3,-4)$
$(-3,4)$
$(-4,3,)$
$(5,0)$
$(-5,0)$

Of those, the highest possible value of $x+y$ would be when they are both positive, leaving us with $(3,4)$ and $(5,0)$, of which $(3,4)$ is greater. So the greatest possible value would be $7$.[/hide]

[hide]
the positve possible solutions are:
(3,4)
and
(5,0)
so 7 is our answer[/hide]


In a triangle $ ABC$ the midpoint of $ BC$ is $ M$. Let $ L$ and $ N$ be the intersections of $ AM$ with the incircle (with center $ I$) of $ ABC$.

Prove that if  $ CA\equal{}AB\plus{}AM$ then $ \angle{NIL}\equal{}\frac{2\pi}{3}$

 :)

[hide="hint"]
Let $ P$ be the projection of $ I$ over LN. Examine somthing about triangle $ PIN$
[/hide]

 :)

Assume that $AC > AB$ and let $P$ be the projection of $I$ onto $\overline{AM}.$  If $r$ denotes the inradius of $\triangle ABC,$ we have

$[\triangle ABM] \equal{} \frac {_1}{^2}[\triangle ABC] \equal{} \frac {_1}{^2}IP \cdot AM \plus{} \frac {_1}{^2}r \cdot BM \plus{} \frac {_1}{^2}r \cdot AB$

$[\triangle ABM] \equal{} \frac {_1}{^2}(IP \cdot AM \plus{} \frac {_1}{^2}r \cdot BC \plus{} r \cdot AB)$

$\frac {_1}{^2}r(AB \plus{} AC \plus{} BC) \equal{} IP \cdot AM \plus{} \frac {_1}{^2}r \cdot BC \plus{} r \cdot AB$

$\Longrightarrow \frac {_1}{^2}r(AC \minus{} AB) \equal{} IP \cdot AM  \Longrightarrow \ IP \equal{} \frac {_1}{^2}r \equal{} \frac {_1}{^2}IL$

Therefore, in the right $\triangle IPL,$ we have $\angle ILP \equal{} 30^{\circ}$ $\Longrightarrow$ $\angle NIL \equal{} 120^{\circ}.$


floor function
combinatorics proposed
combinatorics

Given $ n$ points in a line so that any distance occurs at most twice, show that the number of distance occurring exactly once is at least $ \lfloor n/2 \rfloor$.

[i]V. T. Sos, L. Szekely[/i]

See the solution here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=958002&sid=32fc9a34fd472782015c2a1de7984516#p958002


Let $(a_n)$ be a sequence of positive reals such that $a_n\le a_{2n}+a_{2n+1}$. Prove that $\sum_0^\infty a_i$ diverges.

I need only $a_1 > 0$.  Let $\displaystyle \sum_{i=0}^{\infty} a_i$ suppose a convergent serie : $\displaystyle \sum_{i=1}^{\infty} a_i = L$ (finite).
Then I denote $S_n= \sum_{i=0}^n a_i$. It's easy to see that $S_n \leq S_{2n+1}-a_1$ , so $L \leq L - a_1$, so $a_1 \leq 0$ (manifestly false)
 :cool:  Finally the serie doesn't converge , and that's easy to generalize for $a_n \leq a_{mn}+a_{mn+1}+...+a_{mn+m-1}$:cool:

My solution:

Take $b_n=\sum_{i=2^n}^{2^{n+1}}a_i$. We have from the problem that $s_n$ is a increasing sequence. And $\sum_{i=1}^\infty a_i=\sum_{i=1}^\infty s_n$ which diverges at infinity since $s_n \nto 0$


Starting on January 1, what month will it be when exactly one-fifth of the year has passed?

there are 12 months in a year which is 100% of the year.
one-fifth is equal to 20% of the year.
20% of 12 is 2.4
2.4 months will have passed making the month [b]March[/b]

i get feb.
u get 2.4 which is like nearly half the way through feb. how can it be in march?

[hide]
12/5=2.4
March. :blush:  Sorry about the confusion.
[/hide]

Woah

Lets not get confused here. The answer is actually [b]March.[/b]
2.4 means that 2 months have passed not we're on the second month. If the months are confusing you, use days instead.

[hide=" Days"]1/5 * 365= 73

You can clearly tell that after 73 days its actually in [b]March[/b][/hide]

math92...You need to learn to argue and stick with your answer. Good Job though

o i m sorry about that!
i was just confused
 :blush:


factorize this such that the FACTORS DO NOT HAVE ANY ROOTS IN THEM .

$a^3+1/a^3 + 2$

no roots in them?????So u cant write them as squares and then do (a+b)^2?

I am sooo dumb!!!Its....
[hide]a^3+1/a^3-1^3+3[/hide]
I hope this gives the ans or are ur conditions something else?

The most i can go is

$\frac{(a+1)^2(a^2-a+1)^2}{a^3}$ .

What you want actually ?  :?

[quote="shadysaysurspammed"]I am sooo dumb!!!Its....
[hide]a^3+1/a^3+1^3-3[/hide]
I hope this gives the ans or are ur conditions something else?[/quote]
Wrong a bit I think I agree with shyong,
And this is that sum in 9th isnt it.....Sachin? Did it by Milan sir  Way...

hey shady , shyong and riddler all 3 of u are correct , u guys did by different methods,  ;)

riddler, shady 's method is quite correct, in fact he used the shortest method ,
but yours and shyong's method is corrrect too ( the milind sir way i think)

the only reason i posted this cos to get aNOTHER  answer than the mugging (a+b)^2 method in which the factors have roots :lol:

[quote="riddler"][quote="shadysaysurspammed"]I am sooo dumb!!!Its....
[hide]a^3+1/a^3+1^3-3[/hide]
I hope this gives the ans or are ur conditions something else?[/quote]
Wrong a bit I think I agree with shyong[/quote]

In ur face riddler :lol:  :rotfl: I think u are wrong.

[quote="shadysaysurspammed"]What's wrong in mine????I'm getting the formula $a^3+b^3+c^3-3abc$ directly with what i've done....the hidden thing is just the expansion to come to the form $a^3+b^3+c^3-3abc$,its not the factorisation answer....[/quote]
shut up! tell me ur sol..

shady,

what you did wrong is that

you wrote as

$2 = 1^3$

which isn't true

Ah, the sad side of moderation has revealed itself.  :|

[quote="shadysaysurspammed"]Sorry silverfalcon,i promise it wont happen again.riddler's a good friens,very,very good friend and lives abroad....just teasin him :rotfl: Please dont ban me,i love this place.....

Yeah now for shinwoo.....I wrote $2$ as -1+3...made a little typo mistake.[/quote]

Don't worry. I know how you and riddler are dedicated to this forum and banning will happen iff admins agree anyway (and I never did it so don't worry about it).

[quote="Silverfalcon"]riddler and shadysaysurspammed watch your language in this forum (I know shadysayurspammed didn't swear but he offended riddler).

I don't think this is the first time I see you guys having trouble in this forum concerning the language. If this goes out more, I'm going to contact administrator by myself AND block you two from posting. Trust me, I can make it happen.  ;)

I don't want and like doing that so guys, juet get along with each other well, okay?[/quote]
Sorry :blush:

[quote="G-UNIT"]I personally see no problem in the conduct of their behaviour.  But what I find extremely disturbing is how they constantly communicate through some form of uncivilized chat-language.  Shady, eminem, riddler I understand you guys are friends, but please remember this is a forum, viewed by other people, its already f***ing annoying having to read through all the spam on the forums, you guys are just making it worse.  Please be considerate to other individuals, there are instant-messaging services for such things, thanks.[/quote]

Dude,what spams are u talking about....and isnt ur language a bit too offensive for the other readers?I think u should be considerate enough to understand that,u could've told us that by messaging us :roll:  :D


Suppose I have 6 shirts and 5 ties.  How many shirt-and-tie outfits can I make?

[hide=Solution]There are six choices for a shirt and six choices for a tie, so there are $6\cdot5=\boxed{30}$ different outfits I can make.[/hide]

[quote=math154]This is simply $ 6\cdot5\equal{}\boxed{30}$.[/quote]

pls hide wsolutions


inequalities
geometry
circumcircle
trigonometry
geometry proposed

Given triangle $ABC$ circumradius $R$ and othorcenter $H$ prove that
$3R+OH\ge HA+HB+HC\ge 3R-2OH$

$HA = 2R \cos{A}$, then use $\sum{\cos{A}}=\frac{R+r}{R}$, so $\sum{HA}=2R+2r$.

and now only put $OH$ in one side and square the relation, using $OH=3OG \Longleftrightarrow OH^{2}= 9R^{2}-(a^{2}+b^{2}+c^{2})=9R^{2}-(2p^{2}-2r^{2}-8Rr)$ and use gerretsen.

[quote="pohoatza"]$HA = 2R \cos{A}$, [/quote]
Are you sure for all triangle ?, I think it is true for acute triangle only!

Actually if triangle is acute then $3R\ge HA+HB+HC$ but when it isn't acute it needs my inequality, hope there are nice solutions from everybody.


geometry
perimeter
incenter
angle bisector
geometry solved

Let $ABC$ be a triangle such that $AB>BC>CA$. Let $m_1,m_2, m_3$ be the midpoints of broken lines $ABC, BCA, CAB$ . Show that lines $Cm_1, Am_2, Bm_3$ have a common point.
($m_1$ midpoint of broken line $ABC$ means : if $AB>BC$ then $Am_1=m_1B+BC$.)

as payman_pm remind defenitly non of $m_i$s are gona be on the $AC$ so the problem is not correct. i will check the problem again and edit the post

[quote="Sailor"]Well , an interesting problem is the following:

Let $l_1$ be the parallel from $m_1$ to the bisector of respective angle(B), and the others. Prove that $l_1,l_2,l_3$ are concurent .[/quote]
yes,that's realy intersting (i'll post a solution later),
Sailor,do you know what is the source of this problem?

Here is my solution:
if my proof is'nt obvious please look at the shape.
let $l_b$ meet $AB$ ,$AC$at $T_1$,$M$ extend $l_b$ to meet $BC$ at $T_1'$ we have :
$\angle T_1T_1'B=\angle T_1'T_1B=\angle\frac{B}{2}$ thus $\triangle T_1T_1'B$
is isoceles.
let $d$ be a line such that passes $A$ and parallel by $T_1T_1'$ ,let $D$ meet $BC$at $Q$
$BQ=BA$ and $T_1'B=T_1B$ -----> $T'1Q=T_1A$
and
$T_1A=T_1'c$ that means $T_1'$ is the midpoint of $QC$ thus
$M$ is the midpoint of $AC$ .
similary we can show that $l_c$ and $l_a$ passes midpoint of $AB$ and $BC$ so $l_a,l_b,l_c$ are
are the bisectors of PEDAl triangle of $ABC$ that means they are concurent.

Nice, it's almost the same as my solution!
About the source, I belive it's: Russia '96 or '97.

Let's not be ridicolous about the source of the problem, it is a really classic fact from triangle geometry. Maybe the best formulation is the following:

[color=blue]Let ABC be a triangle and A', B', C' be the midpoints of its sides BC, CA, AB. Then, the angle bisectors of triangle A'B'C' bisect the perimeter of triangle ABC. In other words, the angle bisector of the angle C'A'B' passes through the midpoint of the broken line CAB, the angle bisector of the angle A'B'C' passes through the midpoint of the broken line ABC, and the angle bisector of the angle B'C'A' passes through the midpoint of the broken line BCA. These three angle bisectors are called the [i]cleavers[/i] of triangle ABC. Of course, being the three angle bisectors of triangle A'B'C', these cleavers concur at one point, namely at the incenter of triangle A'B'C'. This incenter is also called the [i]Spieker point[/i] of triangle ABC.[/color]

I won't give another solution of this problem, since we have already enough ones posted on the internet. Just look at http://www.mathlinks.ro/Forum/viewtopic.php?t=20400 or at http://www.mathlinks.ro/Forum/viewtopic.php?t=34612 .

  Darij

definetly no one is ridicolous about the source of the problem.
we have it in our training courses. you can ask omid hatami (he is in our team this year and he can make you asure)(it is for 1998). and it can be for russia too .why not?

i think lomos_lupin is right.(i have seen this problem before),and there is no reason for ridicolos about the problem. :!:  :?

It was certainly given in Russian MO '97 (but probably not in the final round)!


algebra
polynomial
inequalities
algebra proposed

Consider a polynomial with coefficients of real numbers $ \phi(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ with three positive real roots. Assume that $ \phi(0)<0$, prove that \[ 2b^3\plus{}9a^2d\minus{}7abc \le 0.\]
[i]Hery Susanto, Malang[/i]

[quote="Raja Oktovin"]Consider a polynomial with coefficients of real numbers $ \phi(x) \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ with three positive real roots. Assume that $ \phi(0) < 0$, prove that
\[ 2b^3 \plus{} 9a^2d \minus{} 7abc \le 0.\]
[i]Hery Susanto, Malang[/i][/quote]

Let $ x_1,x_2$ and $ x_3$ be the roots of $ \phi(x)$,then :

$ \phi(x) \equal{} a(x \minus{} x_1)(x \minus{} x_2)(x \minus{} x_3) \equal{} ax^3 \minus{} a(x_1 \plus{} x_2 \plus{} x_3)x^2 \plus{}$ $ a(x_1.x_2 \plus{} x_2.x_3 \plus{} x_3.x_1)x \minus{} a(x_1.x_2.x_3)$

Therefore :

$ b \equal{} \minus{} a(x_1 \plus{} x_2 \plus{} x_3)$ , $ c \equal{} a(x_1.x_2 \plus{} x_2.x_3 \plus{} x_3.x_1)$ and $ d \equal{} \minus{} a(x_1.x_2.x_3)$,

Since $ \phi(0) < 0$,If follows that $ a(x_1.x_2.x_3) > 0$ which lead to $ a > 0$,

We have to prove that :

$ 2(x_1 \plus{} x_2 \plus{} x_3)^3 \plus{} 9(x_1.x_2.x_3) \geq 7(x_1 \plus{} x_2 \plus{} x_3)(x_1.x_2 \plus{} x_2.x_3 \plus{} x_3.x_1)$

From Schur's inequality ( 3rd degree ) we have :

$ x_1.x_2.x_3 \geq \frac { 4(x_1 \plus{} x_2 \plus{} x_3)(x_1.x_2 \plus{} x_2.x_3 \plus{} x_3.x_1) \minus{} (x_1 \plus{} x_2 \plus{} x_3)^3 }{9}$,

Hence,It suffices to prove that :

$ \left(x_1 \plus{} x_2 \plus{} x_3\right) \left( (x_1 \plus{} x_2 \plus{} x_3)^2 \minus{} 3(x_1.x_2 \plus{} x_2.x_3 \plus{} x_3.x_1) \right) \geq 0$,

Which is true because :

$ x_1 \plus{} x_2 \plus{} x_3 \geq 0$ and $ (x_1 \plus{} x_2 \plus{} x_3)^2 \minus{} 3(x_1.x_2 \plus{} x_2.x_3 \plus{} x_3.x_1)$ $ \equal{} \frac {(x_1 \minus{} x_2)^2 \plus{} (x_2 \minus{} x_3)^2 \plus{} (x_3 \minus{} x_1)^2}{2} \geq 0$

Assuming the roots of the equation are $ u,v,w$ and by viete's theorem

$ b \equal{} \minus{} a(u \plus{} v \plus{} w) \ , \ c \equal{} a(uv \plus{} vw \plus{} wu) \ , \ d \equal{} \minus{} a(uvw)$

$ \phi(0) \equal{} d \equal{} \minus{} a(uvw) < 0$ , since $ u,v,w > 0$ , one can easily conclude that $ a > 0$

So it equivalent if we prove

$ \minus{} 2a^3(u \plus{} v \plus{} w)^3 \plus{} 9a^2. \minus{} a(uvw)\le7a. \minus{} a(u \plus{} v \plus{} w).a(uv \plus{} vw \plus{} wu)$

$ 2(u \plus{} v \plus{} w)^3 \plus{} 9uvw\ge7(u \plus{} v \plus{} w)(yv \plus{} vw \plus{} wu)$

$ \left(2(u^3 \plus{} v^3 \plus{} w^3) \plus{} 6\sum u^2v \plus{} 12uvw\right) \plus{} 9uvw\ge7\sum u^2v \plus{} 21uvw$

$ 2(u^3 \plus{} v^3 \plus{} w^3)\ge\sum u^2v$

it is true since by AM-GM

$ \sum (u^3 \plus{} u^3 \plus{} v^3) \ge 3\sum u^2v$

done

If $u,v,w$ are the roots then inequality is equivalent by Viete formulas to
$2(u^3+w^3+v^3)\ge uv(u+v)+vw(v+w)+(u+w)uw$ which is true by AM GM


MATHCOUNTS probability geometry rectangle ratio inequalities perimeter

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

[hide="1"]The area of the board is 16pi, the area of the largest circle. The area of the innermost circle is 4pi. The area of a circle with radius three is 9 pi, so the area of the area of the annulus is 5pi. So the answer is (5pi)/(16pi)=5/16[/hide]
[hide="2"]If the common radius of the balls is 4, then the diameter is of the can is 8. The height is 24. These dimensions are big enough to hold all three balls and small enough to make sure they don't move. The volume of the can is then 384 cm. The volume of a ball is 256pi/3. The volume of all three balls is 256pi. The answer is 384-256pi cm 3[/hide]
[hide="3"]The narrow end of the winter is 3.5. The radii of the semicircular sheet of glass is then 1.75 and the perimeter minus the straight edge 1.53125pi. So the answer is 11.5+1.53125pi which is around 16 ft.[/hide]
[hide="4"]The sets would be {1,5,6,7} and {2,3,4,8}, which gives a result of 210. The square root of 8! is around 200 so we can have some confidence in our answer.[/hide]
[hide="5"]9 is 3/4 of 12, so the roses would cost 20*3/4=15.
6 is 1/2 of 12, so the daises would cost 5*1/2=2.5.
18 is 3/2 of 12, so the chrysanthemums would cost 10*3/2=15.
15+2.5+15=32.5 dollars.[/hide]
[hide="6"]1/x+1/y=(x+y)/xy
1/12+1/20=2/15
12+20=32[/hide]
[hide="7"]x^2=(x-4)^2+(x/2+3)^2-(x/2-5)^2 for even x.
Plugging in 100 for x, we get 96^2+53^2-45^2.
96+53+45=194.[/hide]
[hide="8"]$\frac{l}{w}=\frac{d}{l}$
$l^2=2d$
$l^2+4=d^2$
$l^2=d^2-4=2d$
$d^2-2d-4=0$
$\frac{2\pm\sqrt{4+16}}{2}$
$1+\sqrt{5}$[/hide]
[hide="9"]Using the quadratic formula: $\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}$. Taking either of the roots gives an answer of four.[/hide]
[hide="10"]The inequalities simplify to $x\le 6$ and $x\ge 2.5$. The maximum is 6 and the minimum 2.5. Their difference is 7/2.[/hide]

Commenting on your solution for #9, you don't need the quadratic formula

We want $x - \dfrac{1}{x} = \dfrac{x^2 - 1}{x}$

We are given $x^2 - 4x - 1 = 0$ or $x^2 - 1 = 4x$

therefore, $\dfrac{x^2-1}{x} = \dfrac{4x}{x} = 4$

$w^5$ QED


Let $ ABC$ is a triangle with $ (O)$ is its circumcircle.A Circle $ (I)$ is tangent internally to $ (O)$at point $ T$ and both $ AB,AC$,A Circle $ (I_1)$ is tangent internally to $ O)$ and both $ AB,AT$,A Circle $ (I_2)$ is tangent internally to $ O)$ and both $ AT,AC$,
Show that circle $ (I_1)$ and $ (I_2)$tangent to each other
I want  to say that ,a proof with inversion is always expected

Denote W1,W2 are point that ⊙O2,⊙O3 tangent to AT
From 2003 Chinese National Team Test ⊙I2I3T is tangent to ⊙O at point T
(T1 is a point moves on arc BC, then ⊙I2I3T is always intersect ⊙O at point T )
Form mixtilinear property ,T,M,P; T,N,Q are collinear   so does T3,W2,Y ; T3,V,Q ; T2,U,P ; T2,W1,X
and also I is the midpoint of MN.I2 is the midpoint of UW1 , I3 is the midpoint of VW2
So I2I3∥PQ∥MN
We have known that UW1∥PX , VW2∥YQ
We are required to show W1=W2
So if PX,YQ,AT are concurrent then we can complete the proof
〈=〉AP/PY*YT/TX*XQ/QA=1  (*)
AP/PY*YT/TX*XQ/QA=sin(∠ACT/2)/sin(∠ABT/2)* sin(∠TAC/2)/ sin(∠B/2)* sin(∠C/2)/ sin(∠BAT/2)
Notice in any triangle there exists r=4R*sin(∠A/2)* sin(∠B/2)* sin(∠C/2)
So AP/PY*YT/TX*XQ/QA=r3* (sin(C/2))^2 /  r2* (sin(B/2))^2
=> we have to show r3/r2=(sin(B/2)* sin(B/2))/ (sin(C/2)* sin(C/2))
But r3/r2=(TI3* sin(B/2))/(TI2* sin(C/2)=TQ/TP *( sin(B/2))/ (sin(C/2))=(sin∠QBT/ sin∠PCT)* (sin(B/2))/ (sin(C/2))
Form mixtilinear property we gain that I,T,B.M are concyclic I,T,C,N are concyclic
 sin∠QBT/ sin∠PCT= sin∠IMT/ sin∠INT
sin∠IMT/ sin∠ITM=IT/IM=IT/IN= sin∠INT/ sin∠ITN
=> r3/r2=(sin(C/2))^2 /  (sin(B/2))^2
=> (*) is true
THUS WE ARE DONE!

If you invert around $ A$, this problem becomes the following known result: In triangle $ ABC$, let $ D$ be the point where the $ A$-excircle is tangent to side $ BC$.  Then the $ A$-excircle of triangle $ ABD$ and $ A$-excircle of triangle $ ACD$ are externally tangent (at the same point on $ AD$).

Dear Mathlinkers,
I saw this problem in
Yiu Paul, Notes on Euclidean Geometry (1999) 99 ; http://www.math.fau.edu/yiu/EuclideanGeometryNotes.pdf.
Perhaps did some one have another reference?
Sincerely
Jean-Louis


geometry
geometric transformation
quadratics
rotation
analytic geometry
algebra

In triangle ABC, AB = square root of 7, BC = sqrt.13, CA =sqrt. 19. Draw circles centered at A, B, C respectively, and let its radius be 1/3, 2/3, and 1 respectively. Prove: On the three circles there exsits a point A', B', C' (A' on circle A, B' on circle B, C' on circle C), so that triangle ABC and triangle A'B'C' are congruent triangles.

See [url]http://www.mathlinks.ro/viewtopic.php?t=42559[/url].

The same error again. It should be

(1) either $ AB \equal{} \sqrt {7}, BC \equal{} \sqrt {13}, CA \equal{} \sqrt {19}$ as stated, but $ r_A : r_B : r_C \equal{} 2 : 1 : 3,$

(2) or and $ r_A : r_B : r_C \equal{} 1 : 2 : 3$ as stated, but $ AB \equal{} \sqrt {7}, BC \equal{} \sqrt {19}, CA \equal{} \sqrt {13}.$
___________________________________________________

Actually, I now realize that the side lengths are to a large extent irrelevant. It is just necessary to show that for the given triangle lengths $ a, b, c,$ there is a point $ F,$ such that $ AF : BF : CF \equal{} 1 : 2 : 3,$ i.e., that the quadratic equation corresponding to the construction of $ F$ has real roots. The circles $ (A), (B), (C)$ with radii $ r_A : r_B : r_C \equal{} 1 : 2 : 3$ are then seen from $ F$ under the same angle and the $ \triangle ABC$ can be rotated around $ F$ into a congruent $ \triangle A'B'C'$ with $ A' \in (A), B' \in (B), C' \in (C).$
___________________________________________________

Assume that $ c \equal{} AB \equal{} \sqrt {7}, a \equal{} BC \equal{} \sqrt {13}, b \equal{} CA \equal{} \sqrt {19}$ and $ r_A : r_B : r_C \equal{} 1 : 2 : 3$ are both correct. One way to construct the point $ F$: Draw 3 concentric circles $ \mathcal F_A, \mathcal F_B, \mathcal F_C$ centered at $ F$ with radii $ \frac {_1}{^2}b, b, \frac {_3}{^2}b.$ A line $ l$ through $ F$ cuts $ \mathcal F_A, \mathcal F_C$ at $ C', A, C,$ such that $ C', F, A, C$ follow on the line in this order. Then $ AC' \equal{} 2b$ and $ AC \equal{} b.$ Construct $ \triangle ABC$ with the given sides $ a, b, c$ on the segment $ AC \equal{} b.$ Construct $ \triangle AB'C'$ with double the given sides $ 2a, 2b, 2c$ on the segment $ AC' \equal{} 2b,$ such that the $ \triangle AB'C' \sim \triangle ABC$ are centrally similar with the center $ A$ and similarity coefficient $ \minus{} 2$ (the points $ B', A, B$ are collinear). Draw a circle $ \mathcal O$ with diameter $ BB'.$ Assume that $ \mathcal O$ intersects $ \mathcal F_B$ at 2 points $ B_1, B_2.$ Construct points $ C_1, C_2 \in \mathcal F_C,$ such that $ \angle C_1AC \equal{} \angle B_1AB$ and $ \angle C_2AC \equal{} \angle B_2AB.$ Then the $ \triangle AB_1B_2 \sim \triangle AB_2C_2 \sim \triangle ABC$ are similar and $ AF : B_1F : C_1F \equal{} AF : B_2F : C_2F \equal{} 1 : 2 : 3.$ It remains to show that for the given values of $ a, b, c,$ the circles $ \mathcal O, \mathcal F_B$ intersect.

Put the coordinate origin at $ F.$ $ B$ is the intersection of 2 circles with equations $ (x \minus{} \frac {_1}{^2}b)^2 \plus{} y^2 \equal{} c^2$ and $ (x \minus{} \frac {_3}{^2}b)^2 \plus{} y^2 \equal{} a^2.$ Solving for $ x, y$ yields the coordinates of $ B$:

$ B \equal{} (x_B, y_B) \equal{} \left(\frac {2b^2 \plus{} c^2 \minus{} a^2}{2b},\ \frac {\sqrt {4b^2c^2 \minus{} (b^2 \plus{} c^2 \minus{} a^2)^2}}{2b}\right).$

Therefore (substituting the given values), $ x_B^2 \plus{} y_B^2 \equal{} \frac {3b^2 \plus{} 6c^2 \minus{} 2a^2}{4} \equal{} \frac {73}{4} < 19 \equal{} b^2,$ which means that $ B$ is inside of $ \mathcal F_B.$ Since the $ \triangle AB'C' \sim \triangle ABC$ are centrally similar with center $ A \equal{} (\frac {_1}{^2}b, 0)$ and similarity coefficient -2, the coordinates of $ B'$ are $ B' \equal{} (x_{B'}, y_{B'}) \equal{} ( \minus{} 2x_B \plus{} \frac {_3}{^2}b, \minus{} 2y_B).$ Therefore (substituting the given values),

$ x_{B'}^2 \plus{} y_{B'}^2 \equal{} 4(x_B^2 \plus{} y_B^2) \minus{} 6bx_B \plus{} \frac {_9}{^4}b^2 \equal{} 3c^2 \plus{} a^2 \minus{} \frac {_3}{^4}b^2 \equal{} \frac {79}{4} > 19 \equal{} b^2,$

which means that $ B'$ is outside of $ \mathcal F_B.$ Since $ BB'$ is a diameter of $ \mathcal O,$ $ B$ is inside and $ B'$ outside of $ \mathcal F_B,$ the circles $ \mathcal O, \mathcal F_B$ intersect at 2 points $ B_1, B_2.$


inequalities
AMC
USA(J)MO
USAMO
inequalities proposed

[quote="pvthuan"]Let $a,b,c,d$ be positive real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=4.\] Prove that \[a+b+c+d\geq ab+bc+cd+da.\] That's not very hard inequality.[/quote]
This problem was posted a couple months ago, but no one was able to come up with a solution. Then, pvthuan posted his solution. Recently, I was thinking about this problem and came up with another way to solve it.
[hide="Here is his solution:"][quote="pvthuan"]Here is the solution:

We rewrite the inequality as \[a+b+c+d\geq (a+c)(b+d),\] or \[\frac1{a+c}+\frac1{b+d}\geq1.\] It remains to prove that \[a+b+c+d\leq4,\] since $1/(a+c)+1/(b+d)\geq4/(a+b+c+d)$.
I think now you can do the rest.[/quote][/hide]
I'm probably doing something stupid because I don't see that last step.
Anyway I will post my solution soon... let's see if anyone can get it before I post :wink:

OK here is is. I will post the whole solution including the last step(so it's easier to understand):
We have $8=(a^{2}+1)+(b^{2}+1)+(c^{2}+1)+(d^{2}+1)\ge 2(a+b+c+d)$
So $4\ge a+b+c+d$, so $4/(a+b+c+d)\ge 1$
Also $1/(a+c)+1/(b+d)\geq 4/(a+b+c+d)\ge 1$
So $1/(a+c)+1/(b+d)\geq 1$
So $a+b+c+d\ge ab+ad+bc+cd$ QED

There is another solution in post 7 of http://www.mathlinks.ro/Forum/viewtopic.php?t=97950

Let $s = a+b+c+d$, then the ineq is equivalent to $(s-4)(s+1) \leq 0$, where $s \leq 4$ is trivial.

[hide="Okay, I found a different solution to the first one than everybody:"]
By the power mean inequality we have: $\frac{(a+b)+(c+d)}2\geq \sqrt{\frac2{\frac1{(a+c)^{2}}+\frac1{(b+d)^{2}}}}$
or $(a+b)+(c+d)\geq 2\sqrt{\frac2{\frac1{(a+c)^{2}}+\frac1{(b+d)^{2}}}}=2\sqrt{\frac2{\frac{(a+c)^{2}+(b+d)^{2}}{(a+c)^{2}(b+d)^{2}}}}$
$=(a+c)(b+d)\sqrt{\frac8{4+2ac+2bd}}=(a+c)(b+d)\sqrt{\frac4{2+ac+bd}}\geq (a+c)(b+d)$
The first statement is just AM-Inverse Root Square Mean, the last inequality follows from AM-GM: Take $\frac{a^{2}+c^{2}}2\geq ac, \frac{b^{2}+d^{2}}2\geq bd$. Now add them and $2\geq ac+bd$. This means the last chunk of the inequality: $\sqrt{\frac4{2+ac+bd}}$ is always greater than or equal to $1$, so we are done![/hide]
Of course, $(a+b)(c+d)=ab+bc+cd+da$...
Oh by the way from the very first post, the step I didn't see was where he used AM-HM (or C.S.) at the very end.

[quote="pvthuan"]The following is even harder:

Let $a,b,c,d$ be nonegative real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=4,\] prove that \[a+b+c+d\geq\frac23(ab+bc+cd+da+bd+ca).\] [/quote]

Also, I think it must be the same as adding symmetrically the original inequality by switching around a, b, c, d...

sorry, can you explain what you mean? also what's this I just heard about your mock usamo is that soon?

The first problem, note $ab+bc+cd+da=(a+c)(b+d)$, so we are done.
The second one which is very famous, is known as Vietnam MO 1996.


One hundred points are chosen in the co-ordinate plane. Show that at most $ 2025\equal{}45^2$ rectangles with vertices among these points have sides parallel to the axes.

[hide]Take an arbitrary point $ P$. Assume that $ m$ points lie on the same absicca as $ P$ and $ n$ points lie on the same ordinate as $ P$. There are $ 99\minus{}m\minus{}n$ other points which do not lie on the same absicca and ordinate as $ P$. So the maximum number of rectangle with $ P$ as one of its vertices is $ 99\minus{}m\minus{}n$. If $ m\plus{}n\ge17$, then the maximum number of such rectangles is at most 81. If $ m\plus{}n<17$, the maximum number of rectangle with $ P$ as one of its vertices is at most $ mn<m(17\minus{}m)<8\cdot9\equal{}72<81$. So, for any point $ P$, the maximum number of rectangles with $ P$ as one of its vertices is 81. Thus the maximum number of rectangles in total is $ \frac{81\cdot100}4\equal{}2025$. Q.E.D.[/hide]


Greetings! I am hosting a mock MATHCOUNTS Masters' Round Contest on AoPS. For more information, see my thread in the MATHCOUNTS forum at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=205407.

Please stick to mathcounts stuff in teh Mathcounts forum

The reason why I posted it here as well is so that more people will see and sign-up for it. The masters' round IS targeted towards middle-schoolers, but contains middle/high school math, so it's fine if high schoolers participate in it too.


hi im applying to mathcamp and i had a couple questions...
1) how much aid is given? is it still possible to receive some aid if your income level is medium-ish?
2) i noticed that the application specifically says in one single envelope put rec letters, application, problem/solutions, etc. however, is it okay if rec letters are sent separately by the teacher/recommendation writer?
thanks, :)

I'm not sure what "medium-ish" is, but Mathcamp has very good scholarships, and you should apply for one if you will need help affording camp.

My family's income is nearly at the bottom of that range, and Mathcamp gave me a substantial scholarship. Plus, I just found out that I got the USAMTS scholarship, and that helps even more. (I'm not sure if I'm going to MC this year, as I'm applying to another program, but with this scholarship the costs for both programs are about the same so I won't be unable to go to MC just because we can't afford it. Yay! :lol: )

And, for your second question, you probably should just email MC (where to email with questions about application is on their website.)

As others have said, we definitely have substantial scholarships.  It's not based strictly on income: we also take special circumstances into account.  For reference, in past years, there was an automatic deduction for anyone whose income was under $ \$$100,000 annually.  We got rid of the automatic tuition deduction to make the scholarship process simpler, but that should be a sign that we take making camp affordable very seriously!

As to the recommendation letters: you can mail them separately if you need to, but it's easier for us to collect them if they come in one package.


AMC USA(J)MO USAMO algebra polynomial calculus integration

Proof by induction.

It is obvious that 7+1=8 has 3 nondistinct prime factors.

Now consider $7^{{7}^{k+1}}+1=(7+1) \frac{7^{7}+1}{7+1}... \frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$

Therefore it suffices to show $\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$ is composite.

Let $x=7^{7^{k}}$.

The expression becomes:

$x^{6}-x^{5}+x^{4}...+1$ which factors as $(x^{3}+3x^{2}+3x+1)^{2}-7x(x^{2}+x+1)^{2}$

Since x is an odd power of 7, 7x is a perfect square. This is the desired factoring.

So THAT'S how you're supposed to factor it. Very nicely done.

wait, i think you have a +1 missing in the numerator and denominator of that fraction, right?

Wow, how did you see that crazy factoring?

I nearly beat my brains out for several hours trying to find a factoring of $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1$. Damn you, USAMO. What was I supposed to do, guess random polynomials until something worked? Or automatically know? God, my brain is so fried right now.

:omighty:

You didn't just get that out of guessing did you?  I spent an hour staring at it and failed miserably.

ug

the key was to realize that it would only factor for x of that certain form

but that still looks like a difficult factoring to find...

Yeah, I tried $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1$ a whole bunch and then realized that that is prime (43) when $x=2$... so I gave that up and didn't find a solution. Arrgghh, I thought of difference of squares and also of $(x+1)^{6}$ but never got this factorization.

Wow
Good job

I don't think anyone from TJ got that one >_<

[quote="TheAmazingOne"]I nearly beat my brains out for several hours trying to find a factoring of $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1$. - you, USAMO. What was I supposed to do, guess random polynomials until something worked? Or automatically know? God, my brain is so fried right now.[/quote]
OMFG, so did I! Darn, wasted about 3.5 hrs just on that factorization and never got it.  :(

This reminds me of a comment I made to Neal Wu the other day...

"In olympiad solutions, there are always manipulations that seem like somebody pulled it out of their a**."

lol

Hm... I said that the polynomials $(x-w)(x-w^{2})(x-w^{4})$ and $(x-w^{3})(x-w^{5})(x-w^{6})$ where $w^{7}=-1$ were integral-coefficiented... is that right?

At some point I figured out that since the roots of $x^{7}+1$ are the seven roots of -1, that meant that the roots of $\frac{x^{7}+1}{x+1}$ are the six nonreal roots of -1. So I got it to

$x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1 = \left(x^{2}+2\left(\cos \frac{\pi}{7}\right)x+1\right) \left(x^{2}+2\left(\cos \frac{3\pi}{7}\right)x+1\right) \left(x^{2}+2\left(\cos \frac{5\pi}{7}\right)x+1\right)$

And when I saw that all three factors were irrational I nearly cried. After trying all kinds of modular arithmetic tricks with prime numbers and getting nothing I finally said, "screw this," and spent a while thinking about dinosaurs that can undergo binary fission and basically not achieving anything there, either.

oh wow... so that's what's special about 7.. kept thinking I was supposed to use 7=2^3-1 somehow
my thought process was like
if this thing factors, it must be something like difference of squares or cubes
if it factors, then we would instantly be done, the solution's probably not going to be that short
nah, it probably doesn't factor, let's try using contradiction and orders

wasn't bothering to look for short solutions anymore after doing #2 and #4

given this solution, how many points will one get if they got all the steps except that last one?

In fact, $x^{6 \cdot 7^{k}}-x^{5 \cdot 7^{k}}+x^{4 \cdot 7^{k}}-x^{3 \cdot 7^{k}}+x^{2 \cdot 7^{k}}-x^{7^{k}}+1$ is an irreducible cyclotomic polynomial. I knew this, and I also knew that the result didn't generally apply for primes other than 7. For some reason that did not cause me to try particularly hard for an algebraic manipulation. Instead, I spent most of my time trying a whole bunch of number theory that was probably too advanced to appear on the USAMO.

aargh! i just realized, i failed to figure out a factorization because I COMPUTED 512 IN BASE SEVEN INCORRECTLY!!! WHY MUST I SUCCUMB TO STUPID MISTAKES ON [b]OLYMPIAD[/b] PROBLEMS?!  :wallbash_red:

at least that might be good news, as the graders seeing the partial credit i put down will probably understand that i was looking for a factorization based on a computational error, and hopefully be generous enough to put me over the ~>14 bottleneck... maybe...?

[i](~> means "slightly greater than". of course, the explanation defeats the purpose of the symbol, but whatever.)[/i]

I agree, it's a beast factorization, but I got it. Actually, my method was much more roundabout, but got at the same factorization.
I enjoyed this problem a lot, and the induction on it, but like someone said earlier, it was mostly algebra, not number theory.
For people trying to factor x^6-x^5+x^4-x^3+x^2-x+1, by the rational root theorem, the only possible rational roots are 1 and -1, niether of which work. From this, I saw that there had to be another way, and found it.
Overall, as someone who has never gone to a math summer program, learned any official proof-writing skills, or studied at all for the USAMO (just AIME), I'm happy. I got 6 or 7 on each of 1 and 5, and probably at least 1 on 4 and 6, and maybe even 2. But I'm a senior, so there's no chance at MOP.
Btw, (I know I should look at the thread to this first, but it's late and I'm tired), 4 seemed easy to get the correct answer on, but very difficult to rigorously prove without just explaining with logic (which I tried to do).

ok so instead of complaining about it, how about lets figure out how we could have come up with it...some people did...

first, it is natural to say that x=7^7^n

furthermore, we know that we must factor (x^7+1)/(x+1)

i suppose there are mulitple ways to show something is composite...you really should know that a major way is some contrived factorization

i was lazy and did not write out the expansion once...that was obviously stupid on my part

the whole 7 thing kind of makes sense to me...but i ono it is still pretty like wtf

omg, i almost got it!
i just didn't know how to factor it out crazy like that :(

[quote="scorpius119"][quote="Phelpedo"]How does the base 7 representation help you with this factorization?[/quote]

Basically, i tried searching for a difference of squares that would factorize $\frac{7^{7^{n}}+1}{7^{7^{n-1}}+1}$ (by far the most common way of proving something not prime, including olympiad problems, is finding an explicit factorization). After pulling out the prime factors of $\frac{7^{7}+1}{8}$ (which were 113 and 911), i found the nontrivial difference of squares
\[\frac{7^{7}+1}{8}=512^{2}-399^{2}\]
(These numbers come from $512=\frac{911+113}{2}$ and $399=\frac{911-113}{2}$.) Now observe that $512=1331_{7}$ and $399=1110_{7}$, which are nice in base 7 and this suggests a more general factorization; the larger square probably being of the form $(x^{3}+3x^{2}+3x+1)^{2}=(x+1)^{6}$. So we consider the difference
\[(x+1)^{6}-(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1) \]
and it turns out to factor nicely and we get another square for $x=7^{7^{n-1}}$. (For the quartic $x^{4}+2x^{3}+3x^{2}+2x+1$, you can use a substitution like $u=x+\frac{1}{x}$ to pull out the quadratic factors.)

(Unfortunately for me, i incorrectly got $512=1311_{7}$ and tried to deal with
\[(x^{3}+3x^{2}+x+1)^{2}-(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1) \]
but this turned out to give a much stranger quartic, which was nonsense.)[/quote]

The $1331_{7}$ works out nicely, but I don't see how $1110_{7}$ led to you $(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1)$ or to $-7x(x^{2}+x+1)^{2}$, which is the right answer.

The natural way is to generalise it ;)
See http://www.mathlinks.ro/Forum/viewtopic.php?t=146020 , b), $p=7$.

[quote="MellowMelon"][quote="Phelpedo"]
The $1331_{7}$ works out nicely, but I don't see how $1110_{7}$ led to you $(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1)$ or to $-7x(x^{2}+x+1)^{2}$, which is the right answer.[/quote]
Remember we're trying to find an answer in the form of $(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1) = a^{2}-b^{2}$. At this point we have $a$: it's $(x+1)^{3}$. So he was solving for $b$ by considering $b^{2}= a^{2}-(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1)$. After expanding and simplifying all the algebra, you end up with $7x^{5}+14x^{4}+21x^{3}+14x^{2}+7x$, which ought to be pretty easy to factor to the desired $7x(x^{2}+x+1)^{2}$. Dunno if it's totally necessary to even think about $1110_{7}$,[/quote]

Ah, right, I forgot what the original problem was :wink:  Actually, you can experiment and find that the statement is not true for other primes, so the $1110_{7}$ might lead you to $(x^{2}+x+1)*x$ or $(x^{2}+x+1)*7$.

Just an interesting note: For anyone who has Engel, look up Number Theory Problem 139 on page 136  :wallbash_red:

The problem is to show that
\[\frac{5^{125}-1}{5^{25}-1}\]
is composite.

[quote="carpo"][quote="Altheman"]furthermore, we know that we must factor (x^7+1)/(x+1)[/quote]
If I knew that, the problem would have been much easier. . . . I spent a lot of time trying actual number theory.[/quote]

Win.

[quote="krustyteklown"]Just for a bit of further discussion, any theories on how this problem would've been scored partially? I really don't see them taking off more than 1 point for rigor GIVEN that you got the factorization. Can't see more than 1 or 2 points being given on a failed attempt too. Any 3-5 scores on this problem at all?[/quote]
I suspect that, if you gave the factorization, you would lose one point if you did not show that the smaller term in the difference of squares is greater than $1$. I suppose you could lose additional points if your factorization was slightly incorrect (e.g. a reversed sign) or if your proof was exceptionally poorly written. I expect that only one point or less was awarded for setting up the induction without the factorization.


algebra
polynomial
function
complex analysis
complex analysis unsolved

(a) Show that there exists a sequence of polynomials $ \left(p_n\right)$ such that $ p_n(z) \to 0$ if $ \Im z > 0$, $ p_n(z) \to \minus{}1$ if $ \Im z < 0$.

(b) Show that there exists a sequence of polynomials $ \left(p_n\right)$ such that $ p_n(z) \to 0$ if $ z \neq 0$ and $ p_n(0) \equal{} 1$ for all $ n$.

Apply the complex version of the Stone-Weierstrass Theorem to any function satisfying the given properties on successively larger compact subsets of $ \mathbb{C}$

Could you make that more explicit, please? This is quite new for me...


geometry
calculus
ratio
Pythagorean Theorem

1) Given two circles $\omega_1$ and $\omega_2$ with equal radius $r$ such that both are externally tangent to each other and a straight line, and circles $\omega_3  ,  \omega_4 \ldots \omega_n$ are drawn such that all are tangent to the line, to the circle drawn before it, and to $\omega_1$. If these circles are drawn to infinity, is it possible to find the total area ecompassed by $\omega_1 + \omega_2 \ldots \omega_\infty$ in terms of $r$? If so, what is it?

2) In essence the same problem (i think): If $\omega_1$ and $\omega_2$ have different radii $r_1 > r_2$, and the circles are drawn so that they always are tangent to the line, the circle before, and $\omega_1$, what is the total area in terms of $r_1$ and $r_2$?

:D

I don't know if this helps, but there was some relationship between the radii. If I remember correctly, it's $\frac{1}{\sqrt{r_3}}=\frac{1}{\sqrt{r_1}}+\frac{1}{\sqrt{r_2}}$.

[quote="demji"]i don't see any non-calculus way to do it.  Basically i see two possible ways to draw it.  Either $\omega_{3}$ tangent to the first two or just tangent to the first and the other ones between the last one and one and the line.  It would be helpful if you could identify which drawing you want as they give different answers.  Thanks[/quote]

i don't see how calculus factors in...but i agree that there are 2 possibilities for the next circle...consider the 3 circles, 1 between the 2 larger ones...consider if you started with one small and one large circle and you made the other large circle...note that this does not work when $r_1=r_i$, but that happens a finite number of times...

this is pretty simple...let the radii of the circles be a,b, and the smaller one is c...draw some perpendicular lines, use the pythagorean theorem then we get a system, and we find the result that frt found (verified)...this is for constructing progressively smaller circles

to find the radius of the large one, just interchange r_1 and r_3 (see above for why)

oh, what i meant is that $\omega_3$ is tangent to $\omega_{1,2}$ and the line... so that all circles drawn afterwards are in the area between the first two circles and the line. does that make sense?

What i think is that in case..wen the first circle is tangent to both the circles and to the line also..den the radius of that is r/4..if further circles are drawn den .dere radii will be in G.P. with the common ratio 1/4..I dont knw hw to prove this..but its a conjecture ..

[url=http://en.wikipedia.org/wiki/Descartes'_theorem]This may be useful[/url]


geometry
rectangle
perimeter
algebra
system of equations

For some reason i keep doing poorly on these questions and my final is coming up on thursday  :|

The length of a rectangle is 24 more feet than 3 times its width. The perimeter is 320 feet. Set up equations and show your work to get the length and width. [6 pts]

[hide="Solution"]Let length be represented by $l$, and let width be represented by $w$.
From the problem, we have the system of equations:
$l = 3w+24$
$2l+2w = 320$
Let's find the width first.
By substitution, we insert equation one everywhere we see an $l$ in equation two:
$2(3w+24)+2w = 320$
Distribute, combine like terms
$8w+48 = 320$
Subtract 48, divide by 8
$w = 34$
Now we go and put that value back into one of the original equations.
$l = 3(34)+24$
$l = 126$
Now we have a length and a width, which is what the problem wanted.
Do a quick check for yourself, and see that the answer is indeed
$\boxed{L=126; W=34}$
[/hide]

Oh, and Q.E.D. stands for Quod Erat Demonstrandum, which translates to "That which was to be demonstrated."
It's usually appended to the end of a proof, not a simple algebraic solution such as this.


prove that (x_n)_n is convergent , where x_n verifies the condition:
(1+1/n)^(n+x_n)=1+1/1!+1/2!+1/3!+...+1/n!.

Moubi... can you pls give a solution with Landau's tehnic because i tryed to do this with Landau and the result was different?? thanks :D:D:D

g(x) = o(f(x)) means that g(x)/f(x) - > 0 as x - > +oo.
Be careful to not confuse with g(x) = O(f(x)) which means that g(x)/f(x) is bounded when x - > +oo.

In our case, g(x) = o(1) means that g(x) - > 0.

Pierre.


I'm just wondering, what made you personally choose math as a major?

I came into college thinking I would be Math and then turned to Computer Science.  Now, going into my senior year, it looks like I should be easily getting a double major in both.  So, I can provide my insight into why I chose to [i]not[/i] major in Math initially and why I'm actually ending up doubling in it.

So why did I choose Computer Science over Math?  It's because I go to MIT.  OK, not entirely, but for a huge part, yes.  Every other person I knew was doing something really exciting and amazing using their Computer Science skills.  Yes, math can provide a similar excitement and amazement but what really converted me was that with compsci I could do stuff that was easy to explain to people.  It's easier for me to spread this excitement and to apply the skills I've learned to areas where people will actually benefit directly from it.

Now, why am I doubling in math?  It's a little bit of a coincidence.  The first semester of my compsci major, I could only take one compsci class, so I took three math classes along with it.  Then last semester, two of my compsci classes were joint in the math department.  And so this year, I'll be taking three more math classes to finish up my degree.  I'm definitely taking abstract algebra I and II, so I just need one more.

But that probably doesn't tell you [i]why[/i] I like math.  It's a beautiful subject.  I'm interested in unraveling the inner workings of our world.  Math is also a little idealistic and philosophic, and that's something I find interesting.  Though, it has also made a huge impact on other branches of science and life (e.g. geometry, calculus, probability, etc.).


invariant
combinatorics proposed
combinatorics

The first quadrant is divided into unti squares. We are allowed to perform the following move: if the square $(x,y)$ has a token, while squares $(x,y+1),(x+1,y)$ are empty, then we take the token on $(x,y)$ and put a token on each of the other two squares.

Initially, we have tokens on $(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)$. Can we clear these six squares by a sequence of moves?

How about if we start with a single token at $(1,1)$? This is also part of the problem, but I hadn't managed to solve it when I posted the first message. I think I have now.

Is the bottom left square $(0,0)$ or $(1,1)$ ?

It's $(1,1)$, but if you look at the problem carefully, you'll see there's no difference, because (assume the bottom left was $(0,0)$) initially we have no tokens on $(x,0)$ or $(0,y)$, and by the defined moves we can't put any tokens there, so... :).

a) Assign to the case $(x,y)$ the value $ \frac 1 {2^{x+y-2}}$.
Then, an operation does not change the total sum of the token which are the quadrant.
But :
- At the beginning, the sum is $1+ 2\times \frac 1 2 + 3 \times \frac 1 4 = \frac {11} 4$.
- After a finite number of moves, if the six initial squares are empty, the sum is at most $\frac 5 4 $, since the sum of the values of all the squares for the infinite board is $\sum_{k=1}^{+\infty} \frac k {2^{k-1}} = 4$.

b) What is the question? To clear the six cells too?

Pierre.

Yes, the question for b) is the same: can the six cells be cleared?

I did the exact same thing, you know :). Isn't that strange?

I already saw other such problems with the same type of invariant.
It looks like a one from Conway.

Pierre.

We can use the same idea for b) : we can only include one square on each of the first row and column, so we lose at least 1/8 on each. This brings the maximum of the invariant down to 1, which is what we started with on the first square.


geometry
angle bisector
geometry unsolved

Let ABC be a triangle with <B>90 and  <A<60.Let P be a point on the side  AB such that <CPB=60.Let D be the point on CP which also lies on the interior angle bisector of  <A.If <CBD=30,prove that CP is a trisector  of  angle  ACB.

[quote="ssupermath"]Let ABC be a triangle with <B>90 and  <A<60.Let P be a point on the side  AB such that <CPB=60.[/quote]

So $P=A$ :?:

[color=darkblue]In the your enunciation what is the role of the point $D$ ?? I wanton. Maybe so ![/color]

[quote="ssupermath"][color=darkred]Let $ABC$ be a triangle with $B>90^{\circ}$ and  $A<60^{\circ}$. Let $P$ be a point on the side $[AB]$ so that $m(\widehat{CPB})=60^{\circ}$. Let $D$ be the point on $[CP]$ so that $m(\widehat{ CBD})=30^{\circ}$. Prove that $[AD$ is a trisector of the $\widehat{BAC}$.[/quote][/color]

I think it is clear.D points is also on the bisector of <BAC.

Please clarify your problem, because of there is not meaning in the notations, specially after your figure.

We can have the configuration of the triangle $%Error. "bigtriangle" is a bad command.
PAC,$ such that $\angle PAC < 60^{o}$ and $\angle APC = 120^{o}$ and the point $B,$ movable in the extension of $AP,$ such that $\angle ABC > 90^{o}.$

So, the constant segment line $CP,$ doesn't to be the trisector of a non constant angle.

As Virgil Nicula said, I don't understand also, the role of the point $D.$

Kostas Vittas.

EDIT: Sorry dear ssupermath and all. I was understanding wrong the problem. Now I see that it is OK.


ratio
geometry
rectangle
3D geometry
percent

9. seven points are equally distributed on a circle. how many non-congruent quadrilaterals can be drawn with vertices chosen from among the seven points?

11.find the product of all numbers of the form (1-1/n)^-1 for 2<=n<=15

27.three players play 4 rounds of a game and end with a tie score of 80. a different player wins each of the final three rounds. when a player wins, s/he doubles his or her points, and the 2 losers each must subtract the amount the winner gains from their scores. what was the highest score at the end of round 1?

10. the ratio of the length of a rectangle to it's width is the same as that of the diagonal to the length. if the width is 2, how many units are in the length of the diagonal?

27. the dimensions of a rectangular block are 30 cm, 16 cm, and 12 cm. it is made from cubes which are each 1 cm^3 in colume. if the outside of the block is painted red, what percent of the block i smade up of cubes with exactly two sides painted red? express your answer to the nearest tenth.

4. there are positive integers k, n, and m such that 19/20<1/k+1/n+1/m<1. what is the smallest possible value of k+n+m?

this is the test I'm doing next, so I'll post the answers when I'm done with it (probably today)


geometry
geometric transformation
rotation
calculus
3D geometry
analytic geometry
graphing lines

1) [hide] Cone with height 3 and radius 4. 16pi*3/3=16pi.
[/hide]
2) [hide] Cone with height 4 and radius 3. 9pi*4/3=12pi.
[/hide]
3) [hide] A cone with height 4 radius 3, base to base with a cone with height 2 radius 3. 12pi+4pi*3/3=16pi.
[/hide]
4) [hide] A cylinder with height 6 radius 3, with two cones with height 4 radius 3 and height 2 radius 3 subtracted out. 36pi*3-16pi=92pi.
[/hide]
5) [hide] The line that goes through (4,0) and (1,4) has slope -4/3, so the equation of the line is y=-4/3(x-4), or y=-4/3(x)+16/3. When x is 0, y is 16/3, so that's the height of the cone. So the cone is 256pi/9. Then you have to subtract the little cone at the top with radius 1 and height 16/3-12/3=4/3, which has a volume of 4pi/9. That's 28pi total. Then you have to subtract the cylinder with radius 1 and height 4, which is 4pi, so the answer is 24pi.
[/hide]

[quote="Treething"]Integrals and rotation ax-mmph grmbl grr =X[/quote]
What?

I think it means... grrrrr.... I have no clue. At least that's what it seems to me.

[quote="Gyan"][hide][quote]Please put your solutions in spoiler. If you use calculus, you're being boring[/quote]
Pappus , I presume, is not boring  ;) [/hide][/quote]
What's pappus? :huh: And will someone please tell me if my solutions are right?

Your answers look correct in terms of method.  Your calculations don't make sense to the calculator-less/scratch paper-less reader (me).

[quote="JBL"]Compute the volumes of the following figures:

1) The figure formed when the triangle with vertices (0,0), (3, 0) and (0, 4) is rotated around the x axis.
2) The figure formed when the triangle with vertices (0, 0), (3, 0) and (0, 4) is rotated around the y axis.
3) The figure formed when the triangle with vertices (0, -2), (3, 0) and (0, 4) is rotated around the y axis.
4) The figure formed when the triangle with vertices (3, -2), (0, 0) and (3, 4) is rotated around the y axis.
5) The figure formed when the triangle with vertices (1, 0), (4, 0) and (1, 4) is rotated around the y axis.

Please put your solutions in spoiler.  If you use calculus, you're being boring :)[/quote]

1> 16pi

2> 12pi

3>18pi

4>36pi

5>24pi

The solutions involve simple symmetry considerations of conic solids .. ;)

Hope the above ans. are correct ..haven't recheck my calculations

Bother, I knew my answer for 4 was suspiciously large. I seem to have inverted height and radius. In fact, I also inverted height and radius for the second cone in 3. Santosh_blr's answers are right.

[quote="LynnelleYe"]
What's pappus? :huh: [/quote]

Pappus  way to find the area of  rotational  volueme is to [hide]multiply the  2-d area with the length of the orbit of the centroid .[/hide]   For example  let us take Q.2 .

[hide]The area of  trainagle  is
[hide]$(1/2)\times 3 \times 4  =  6$[/hide]

The centroid is  at [hide]
$(\frac {0+3+0} {3}$  , $ \frac {0+0+4} {3})  $ or  $  { (1,4/3) }$[/hide]

And it will travel  (when rotated along y axis):
[hide]$2 \pi  (1) =  2 \pi $[/hide]

So volume is:
[hide]$6 ( 2 \pi ) = 12 \pi $[/hide]

Some may find it neat, but using Calculus is, of course, much easier (and, if you ask me, less boring :)
[hide](BTW if you can find the area  of the triangle , all you have to find is the average value of x-cordinates, multiply this product   with $2\pi$ )
[/hide]
[/hide]

Just to add to the note above,  one [b]very[/b] popular question is to  ask: "find the centroid of a  semi-circular plate"

The calculus is not hard, but  one easy way to find this is ....

[hide]Assuming it is distance $h$  from the center then:

$2 \pi h (\pi r^2 /2) =  (4 /3) (\pi r^3) $ and  solve for h.[/hide]

BTW here is  Pappus theorem  [url=http://mathworld.wolfram.com/PappussCentroidTheorem.html]( I googled it)[/url]

Yes, that is the "other" Pappus theorem. The more popular Pappus theorem is [url=http://mathworld.wolfram.com/PappussHexagonTheorem.html]here[/url] (at least in my opinion.)


algorithm
pigeonhole principle
number theory
relatively prime
combinatorics proposed
combinatorics

Starting from a positive integer $n$, we can replace the current number with a multiple of the current number or by deleting one or more zeroes from the decimal representation of the current number. Prove that for all values of $n$, it is possible to obtain a single-digit number by applying the above algorithm a finite number of times.

There is a nice solution to this...

Since no one seems to be posting a solution, I'll post the one I have.

It is a classical problem that every integer has a multiple of the form $111\cdots\>100\cdot\>0$ (this can be proved using the pigeonhole principle). Removing the zeroes, and multiplying the rest by 19, we obtain a number of the form $2111\cdots\>109$ or 209. Removing the 0, and multiplying by 9 gives 261 or a number of the form $190\cdot\>071$. Removing the zeroes, we see that we only have to treat the two cases 261 and 1971. An easy (but tedious!) calculation proves the result.

well, i had a partial solution in the case $(3,n)=1$,which i did intend to post sometime but....

basically the ideas involved were rather interesting.note firstly that the number $n$ we start with can be made relatively prime to 10 by dividing out all the powers of 2 and 5(that is possible since it is equivalent to multiplying the number by 5 or 2 as the case may be and then dividing by 10(throwing away the last zero).

now if the number is a prime $p$ that satisfies $\large(\frac{10}{p}\large)=-1$ i.e. 10 is no a square modulo $p$ then $10^{(p-1)/2}\equiv -1(\mod p)$ so that the algorithm would reduce $p$ to $11$.
for 11 the following chain does the job;
$11\rightarrow 506\rightarrow 56\rightarrow 28\rightarrow 14\rightarrow 7$.

now the general idea is this: if we can get a multiple of $n$ which "includes" a prime of that sort(i.e. you can get such a prime after throwing away a few zeroes ) then you are through.

now a prime whose last three digits are $011$ will certainly do the job(since then such a prime will be $3(\mod 8),1(\mod 5)$ and so $\large(\frac{2}{p}\large)=-1,\large(\frac{5}{p}\large)=1$.

to do this suppose there exists a $t$ such that the sum of digits of $t$ is divisible by $n$,its last three digits are 011 and $(t,n)=1$. then consider numbers of the form $10^tnk+t$; for some large enough $k$ such a number will be prime(Dirichlet's theorem). in this case we are through again. to see this, note that since $(n,10)=1$ there exists some $k$ such that $10^k\equiv 1(\mod n)$.since the sum of digits of $t$ is divisible by $n$, we can get a multiple of $n$ to be $\tilde{t}$ which is $t$ along with some extra zeroes(this is because of the above observation-$10^k\equiv 1(\mod n)$).
then $10^{\tilde{t}}nk+\tilde{t}$ is a multiple of $n$ and the algorithm reduces it to $10^tnk+t$ which is a prime of the required sort(choosing $k$ sufficiently large).
 thus finally all we need is a $t$ of the aforementioned type.this is easy since as $n>10$(else there is nothing to prove) there exists a multiple of $n$ of the form $a=11\cdots1$ and so there is also a multiple of $n$ which has only ones and with the number of ones also being a multiple of $n$(for instance $a\cdot (100\cdots01)$ with an appropriate number of zeroes).now consider $t=111\cdots 2011$.this will keep the sum of digits a multiple of $n$ and if $k|t,a$ then $k|t-a=900$ but by assumption that is not possible unless $k=1$ and that proves the last part.


Show that $-1 \leq x \leq 1 \thickspace \rightarrow \thickspace \cos\left[\arcsin(x)\right] = \sin\left[\arccos(x)\right]$

Well, if x is positive, then everything is in quadrant I, and we can see that $\arcsin{x}$ is the complement of $\arccos{x}$.  We also know that $\cos{\frac{\pi}{2}-\theta}=\sin{\theta}$.  Therefore, it holds there.  The same thing applies with $x\le0$.  The angles are in quadrants II and IV, but the sin and cosine resulting are again the same.

Both sides = sqrt(1-x^2) using cos arcsin x = sqrt(1-(sin arcsin x)^2) = sqrt(1-x^2) and similarly for sin arccos x.

[b][color=red](1.)
$x \in \left[ { - 1,1} \right] \Rightarrow \arcsin x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] \Rightarrow \cos \left( {\arcsin x} \right) \in \left[ {0,1} \right] \Rightarrow $
$\mathop  \Rightarrow \limits_{\mathop {\cos \left( {\arcsin x} \right) \ge 0}\limits^ \uparrow  } \cos \left( {\arcsin x} \right) =  + \sqrt {1 - \sin ^2 \left( {\arcsin x} \right)}  \Rightarrow $
$ \Rightarrow \cos \left( {\arcsin x} \right) =  + \sqrt {1 - x^2 } $

(2.)
$x \in \left[ { - 1,1} \right] \Rightarrow \arccos x \in \left[ {0,\pi } \right] \Rightarrow \sin \left( {\arccos x} \right) \in \left[ {0,1} \right] \Rightarrow $
$\mathop  \Rightarrow \limits_{\mathop {\sin \left( {\arccos x} \right) \ge 0}\limits^ \uparrow  } \sin \left( {\arccos x} \right) =  + \sqrt {1 - \cos ^2 \left( {\arccos x} \right)}  \Rightarrow $
$ \Rightarrow \sin \left( {\arccos x} \right) =  + \sqrt {1 - x^2 } $

From (1.) and (2.) we have:
$\cos \left( {\arcsin x} \right) = \sin \left( {\arccos x} \right) = \sqrt {1 - x^2 } ,\forall x \in \left[ { - 1,1} \right]$
[/color][/b]


find all pairs (p;q) of positive integers such that p+q and pq+1are both powers of 2

Hint
[hide]$(p,q) = \left(2^n - 1,2^n + 1\right)$ is a solution for all $n$
Can you prove that these are the only ones? :)[/hide]

Oops, in my "hint" I forgot the trivial case where one of $p$ and $q$ is equal to 1  :oops:


A birdbath is designed to overflow so that it will be
self-cleaning.  Water flows in at the rate of 20 milliliters per minute and
drains at the rate of 18 milliliters per minute.  One of these graphs shows
the volume of water in the birdbath during the filling time and
continuing into the overflow time.  Which one is it?

[asy]/* AMC8 2002 #6 Problem */
size(4inch,1inch);
for ( int x = 0; x <= 4; ++x )
{
    draw((x*1.5,1)--(x*1.5,0)--(x*1.5+1,0));
    label(rotate(90)*scale(0.7)*"Volume", (x*1.5-.2,.5));
}
label("$A$", (.5,0), S);
label("$B$", (2,0), S);
label("$C$", (3.5,0), S);
label("$D$", (5,0), S);
label("$E$", (6.5,0), S);
draw((0,0)--(.5,.5)--(1,.5),linewidth(1));
draw((1.5,0.6)--(1.8,0.6)--(2.5,0), linewidth(1));
draw((3,0)--(4,.75), linewidth(1));
draw((4.5,.8)--(5.5,.8), linewidth(1));
draw((6,0)--(6.5,.5)--(7,0), linewidth(1));[/asy]

Provide the correct letter (A, B, C, D, or E) answer.

$ A$ as the water will first rise and then stay the same as the water input is greater than the output but the size of the birdbath cannot hold more.


[b]Jill is given the job of cutting a pie for herself and five friends. After she cuts the pie she will be the third to choose a slice. The smallest slice she can cut is 10% of the total pie. If each person selects the largest possible slice on his or her turn, what is the largest percent Jill can cut for herself? Give your answer to the nearest percent.[/b]

[quote="Ihatepie"][hide]

If she gives 4th and 5th 10 %

we are left with 80% for 3 people

80/3= 26 2/3, which to the nearest percent in 27[/hide][/quote]

That's how I did it, but if you think about it, you can NOT have 27%...I think..

26 is possible, but not 27. Just saying.. I know that 27 is right, I think..

well it says round to the nearest percent, so that is what u do

Yeah, MATHCOUNTS doesn't really seem to care much about precision or significant figures. I mean, precision or significant figures isn't really that important in the real world.  :roll:

wow, good job degreen, i think he's rite.  There are 6 ppl.  So, the last 3 get 10% each, leaving 70%.  That gives 23 1/3% or 23% for Jill.

[quote="dgreenb801"]It says Jill AND five friends. Thus there are six people. So the last three would get 10% each, while the remaining 70% would be divided by three to make 23%, right?[/quote]
Good point. Thanks for pointing that out.


factorial
number theory unsolved
number theory

Let abcd be four digits number.Is there any solution for
abcd=a!+b!+c!+d! .Why?

Let's consider a solution of the problem, if any. It is easy to see that no digit is $8$ or $9$, that at most one is $7$ and that two of them are at least $6$.
Thus there only remains few cases which can be checked by hand to verify that, in fact, there is no solution.

Pierre.