Find all polynomial $P(x)$ satisfying:$deg P=2005$ and $P(x^2+1)=(P(x))^2-1\forall x$

[quote="nhat"]I don't think that we need $deg P=2005$ because i can prove that the problem doesn't exist the polynomial satisfies this condition for all degree of $P(x)$ :)  :)  :)[/quote]

$P(x) = \phi$  :rotfl:

Edit:  Denote a series, $x_0 = 0, x_n = x_{n-1}^2 + 1$.  From the problem condition we denote a second series $y_0 = P(0) = k, y_n = P(x_n) = P(x_{n-1})^2 - 1 = y_{n-1}^2 - 1$ for all integers $n$.  Simply pick the first $2007$ values and you have a system of $2007$ linear equations in $2006$ variables for the coefficients - no solution.  Now it remains only to prove that none of the equations are linearly dependent, which I really don't want to do  :D

Sorry! Because i forgot the case $deg P(x)=0$ :(  :(


Congratulations to Yufei - formerly quite active Canadian forum member and moderator, who achieved the very high honour of Putnam fellow (top 5 in North America) in this past December's Putnam Competition for undergraduates.

Unfortunately, MIT did not sense his genius in their presence.  They picked the wrong team again, and Princeton's team prevails.   :D

Congratulations, Yufei! :clap2:  :omighty:

Congratulations, Yufei!

Why am I not surprised?

ignorant americans!  :mad:

how could they not pick a brilliant canadian mind like yufei!

ya MIT did that last year to,..

P.S. Congrats!

Congrats Yufei!!!!!!!!!

Make Canada proud!!!!

Go Yufei!

Just a question: which university is Yufei at now?

I wanted to go there, but I've been rejected :'(

Congratulations!

Wait - 4 of the 5 Putnam fellows are from MIT and they only came third?

congradz Yufei^^
(nneonneo: i'm rejected too >.<
but put it this way: MIT did not even pick Yufei in their team, so they certainly hv problem choosing people)  :rotfl:

wow!~!~!~!~ Congrats!!!! That's so cool!!!! :clap:

Absolutely amazing as always! Congrats Yufei!!!


Let $ a,b,c$ be the sides of a triangle with $ a\plus{}b\plus{}c\equal{}2$
[b]Prove or disprove[/b] that,
\[ \sum_{cyc} \frac{a\plus{}b\minus{}c}{c^2} \le 12
\]

Not true. We can make one side arbitrarily small and the other two equal. For example, 0.9, 0.9, and 0.2.

[quote="Moonmathpi496"]Let $ a,b,c$ be the sides of a triangle with $ a \plus{} b \plus{} c \equal{} 2$
[b]Prove or disprove[/b] that,
\[ \sum_{cyc} \frac {a \plus{} b \minus{} c}{c^2} \le 12
\]
[/quote]

It seems that the inverse inequality is true (with changing the constant). If $ a,b,c$ are the sides of a triangle with $ a \plus{} b \plus{} c \equal{} 2$, then

\[ \sum_{cyc} \frac {a \plus{} b \minus{} c}{c^2} \geq \frac{9}{2}
\]

Sergey Markelov
markelov@mccme.ru


number theory
least common multiple
geometry
FTW
probability
3D geometry
sphere

After [b]both[/b] of isabellas marathons were locked. I'm going to started a new one. Let's keep this running until 90 pages like the previous marathon. Remember, no complex algebra. Don't spam! And try to post new problems.

 :D  :D
___________________________________________________________________________________________________________

PROBLEM ONE: Bob walked into a store and saw something he liked. He spent half his money, and then 10 dollars more.
He saw something else he liked, and decided to buy that. So he spent half of the remainder of his money. And then 10 dollars more, but then he had no money, :o oh noes! How much money did he have after he bought the first item?

$ \frac{1}{2}x \plus{} 10 \plus{} \frac{1}{4}x \plus{} 10 \equal{} x$

$ \frac{\minus{}1}{4}x \equal{} \minus{}20$

$ x \equal{} 80$

Bob had $ \$80$ originally, so he had $ \boxed{\$30}$ left after buying the first item.

PROBLEM TWO: What is the sum of the smallest 3 common multiples of $ 12$ and $ 15$?

[quote="isabella2296"]$ \frac {1}{2}x \plus{} 10 \plus{} \frac {1}{4}x \plus{} 10 \equal{} x$

$ \frac { \minus{} 1}{4}x \equal{} \minus{} 20$

$ x \equal{} 80$

Bob had $ \$80$ originally, so he had $ \boxed{\$30}$ left after buying the first item. [/quote]

i don't think that's correct.
think about it
if he spent half his money and 10 dollars more on the last item, then the 10 dollars must be the other half of his money. therefore, after buying his first item, he should have 20 dollars left. i'm not sure if my solution is right, but o well

[quote]PROBLEM TWO: What is the sum of the smallest 3 common multiples of $ 12$ and $ 15$? [/quote]

LCM of $ 12$ and $ 15$ is $ 60$, so the smallest 3 common multiples would be $ 60$, $ 120$, and $ 180$
$ 60 \plus{} 120 \plus{} 180 \equal{} 360$

PROBLEM THREE
Jill went to the fair on Saturday and spent half her allowance. The next day, Jill went shopping with her friends. She was going to buy a dress, but she found out she didn't have enough money. So she borrowed $ \$$$ 12$ from her friends to buy it. If the money she had left only paid for $ \frac {3} {4}$ of the cost of the dress, then how much money did she have before the fair?

i'm not sure if this is the level for MOEMS, so if it isn't, tell me and i'll post something easier (or harder if it's too easy)

If the money Jill had left was $ \frac{3}{4}$ of $ \$12$, then she had $ \$9$. $ 9 \times 2 \equal{} \boxed{18}$.

PROBLEM FOUR: Mr and Mrs. Wilson have 3 children: Jessica, Jeremy, and Jenna. The average of Jessica and Jeremy's age is $ 11$. The average of all three children's ages is $ 10$. How old is Jenna?

um..
no, not quite the answer
the problem said that she borrowed 12 dollars from her friends, and it also said that she paid for $ \frac {3} {4}$ of it, so 12 dollars must've been $ \frac {1} {4}$ of the price. if 12 dollars if 1/4 of the price, then 36 dollars is 3/4 of the price, which is what she paid for it. since that is half her money, she had 72 dollars to start out with

ANSWER TO PROB 4

8, the average age of all three children is 10, so their ages must add up to 30
the average age of the first two is 11, so their ages must add up to 22
the last child's age would be the difference between those two numbers
$ 30 \minus{} 22 \equal{} 8$

don't have another prob yet

Oh, oops, I misinterpreted your problem. I guess I'll post a problem.

PROBLEM FIVE: Find the LCD of 403152 and 6432.

SOLUTION TO PROBLEM FIVE:

Factor both:
$ 403152 \equal{} 2^4 \cdot 3 \cdot 37 \cdot 227$
$ 6432 \equal{} 2^5 \cdot 3 \cdot 67$.
So the LCD is
$ 2^5 \cdot 3 \cdot 37 \cdot 67 \cdot 227 \equal{} \boxed{54022368}$

PROBLEM SIX: Mr. Jones has a car and a truck.  The truck averages 15 mpg on the highway and 12 mpg in the city.  His car averages 32 mpg on the highway and 24 in the city.   He drove his truck 17 miles on the highway and 3 miles in the city, and the car 12 miles ont he highway and 5 in the city.  If gas is $ \$3.95$ per gallon, how much money did Mr. Jones spend on gas?  Round to the nearest cent.

I think you mean the car and the truck, not the car and the car?

Please clarify. Thanks.

I see no confusion so heres the answer:

[hide] He used 1 29/30 of a gallon, so 59/30 times 3.95 233.05/ 30= 7.63833333 endless.[/hide]

I think its wrong and my solution is horrible I know

 :(  :(  :(

THere's no confusion now, because i edited it.

Please explain how you got to 59/30 gallons.

[hide="The Solution To Problem Six"]

Truck: Highway - $ \frac{17}{15}$
           City - $ 2$
Car: Highway - $ \frac{8}{3}$
        City - $ \frac{24}{5}$

(The above chart is the number of gallons.)

We take each number in the above chart and multiply it by 3.95, then add our results together.

The answer is $ \boxed{\33.97}$.

I know I got it wrong, I just know it.

[/hide]

PROBLEM SEVEN: Find the least common factor of 403152 and 6432.

Using factorization from my other problem:
$ 403152\equal{} 2^4 \cdot 3 \cdot 37 \cdot 227$
$ 6432\equal{}2^5 \cdot 3 \cdot 67$
So the least common factor is $ 16 \cdot 3\equal{}\boxed{48}$

And to your answer to problem 6: He doesn't use 2 gallons driving his truck in the city! he uses 1/4, and for the car, i think you have a lot of your fractions flipped.

Oops. My brain's on vacation, and I'm not functioning properly today.  :(

PROBLEM EIGHT: My brother is, at this moment, one year younger than twice my age. When I was 3, his age was also a prime number. When I was 4, my brother was 2 years younger than 4 times my age. What is the number of years between our ages?

[quote="AIME15"]Using factorization from my other problem:
$ 403152 \equal{} 2^4 \cdot 3 \cdot 37 \cdot 227$
$ 6432 \equal{} 2^5 \cdot 3 \cdot 67$
So the least common factor is $ 16 \cdot 3 \equal{} \boxed{48}$[/quote]

Sorry, that's incorrect. The question asked for the [b]least[/b] common factor, which is always $ \boxed 1$.

[quote="isabella2296"]PROBLEM EIGHT: My brother is, at this moment, one year younger than twice my age. When I was 3, his age was also a prime number. When I was 4, my brother was 2 years younger than 4 times my age. What is the number of years between our ages?[/quote]

AHHHH! New solution!

When you were 4, then your brother was $ 4 \cdot 4 \minus{} 2 \equal{} 14$ years old. Then the difference between your ages is $ \boxed {10}$. Yeesh! That's all the information you need to know!!!!

[b]New problem[/b]
Given that a tank holds 8 more gallons of oil when it is $ \frac 67$ full than when it was $ \frac 47$ full, what is the capacity of the tank, in gallons?

Yup. That one was tricky.

Forgot to mention.

Please just post one problem at a time in the future - thanks.  :)

1. $ x \plus{} x \plus{} 1 \plus{} x \plus{} 2 \plus{} x \plus{} 3 \equal{} 5600$

$ 4x \plus{} 6 \equal{} 5600$

$ x \equal{} 1398.5$

So we take the smallest integer greater than this, 1399, and find the sum of

$ 1399 \plus{} 1400 \plus{} 1401 \plus{} 1402 \equal{} \boxed{5602}$

2. There are 3 ways the get an even number, so our answer is $ \boxed{\frac{1}{2}}$.

3. First we list the ways to get a sum of 7.

1,6
2,5
3,4
4,3
5,2
6,1

There are 6 ways. The number of ways to roll two die is 36, so our answer is $ \boxed{\frac{1}{6}}$.

New Problem: Find the number of license plates that exist made up of 3 letters followed by 3 digits, if each letter is distinct.

[quote="tonypr"]
New problem
Get this one and reward yourself with a cookie  :)
if $ x \plus{} \dfrac{1}{x} \equal{} 4$ what's the value of $ x^5 \plus{} \dfrac{1}{x^5}$? [/quote]

$ x \equal{} a, \dfrac 1x \equal{} b$

$ a \plus{} b \equal{} 4, ab \equal{} 1$

$ (a \plus{} b)^3 \equal{} a^3 \plus{} b^3 \plus{} 3ab(a \plus{} b)$
$ 4^3 \equal{} a^3 \plus{} b^3 \plus{} 12$
$ a^3 \plus{} b^3 \equal{} 64 \minus{} 12 \equal{} 52$

$ (a \plus{} b)^5 \equal{} a^5 \plus{} b^5 \plus{} 5ab(a^3 \plus{} b^3) \plus{} 10(ab)^2(a \plus{} b)$
$ (a \plus{} b)^5 \equal{} a^5 \plus{} b^5 \plus{} 5*52 \plus{} 40$
$ 4^5 \equal{} a^5 \plus{} b^5 \plus{} 300$
$ a^5 \plus{} b^5 \equal{} 1024 \minus{} 300 \equal{} 724$

[quote="isabella2296"]
New Problem: Find the number of license plates that exist made up of 3 letters followed by 3 digits, if each letter is distinct.[/quote]

Edit : oops, distinct

$ 26*25*24*10^3*6C3 \equal{} 312000000$

answer : 312000000

<New Problem>

Find $ x$ if $ x \equal{} \frac {1}{2} \plus{} \frac {1}{6} \plus{} \frac {1}{12} \plus{} ... \plus{} \frac {1}{90} \plus{} \frac {1}{110}.$

Sorry, tonypr's question was unsuitable for MOEMS level.

But give yourself a cookie anyway, Bachukas.  :lol:

I have one.

The product of two positive whole numbers is 2005. If neither
number is 1, what is the sum of the two numbers?

We have a systems of equations.

$ x \plus{} y \equal{} 84$

$ x \minus{} y \equal{} 98$

Adding our equations yields

$ 2x \equal{} 182$

$ x \equal{} 91$

Thus, $ y$ is equal to $ \boxed{\minus{}7}$.

New Problem: Lucy is reading a book. She reads the same amount of pages each day, and it takes her a week to finish it. On a piece of paper, she records the page number she stops at. Lucy's mom thinks that the numbers on the list show the number of pages she reads each day. Adding the numbers up, Lucy's mom believes that there are 1344 pages in the book!!! How many pages are there REALLY in the book?

EDIT:

lol, i misunderstood the prob.

x+2x+...+7x=28x=1344, x=28

28*7=196

answer : 196 pages

<New Problem>

For what value of n does $ 7! \equal{} 2^4*3^2*n$?

Sorry, Fantasy, your answer to my problem was incorrect. Anyone else want to try?

Fantasy has corrected his solution. n=35
Prove that $ n^3 \minus{} n$ divides by 6, n- positive integer

$ 1344 \div 28 \equal{} 48$ not 28, so the answer is $ \boxed{336}$.

Bachukas, your problem is too hard for MOEMS level.

Do you have another, easier one? I recommend reading the sticky to find out what MOEMS level is.

How about this problem:

How many five-figure numbers are, which can be divided by 4 and digits are: 1,2,3,4,5 (digits can't repeat)

last two digits = 12, 24, 32, 52

6*4=24

answer : 24

<New Problem>

What is the series of 5 prime numbers that form a arithmetic sequence?

P.S. oops, sry. I forgot to say that find the SMALLEST sequence.

Prime numbers:

It is not easy to write a solution of this problem and I don't have enough time. Answer: 5; 5+6d ; 5+12d; 5+18d; 5+24d; there d- positive integer indivisable by 5;


calculus
integration
calculus computations

A particle moves along the x-axis so that its velocity at any time $ t \geq 0$ is given by $ v(t) \equal{} 5te^{\minus{}t}\minus{}1$.  At $ t\equal{}0$, the particle is at position $ x\equal{}1$.  What is the total difference traveled by the particle from $ t\equal{}0$ to $ t\equal{}4$?

You may want to show us what you have done so far in order for us to get a better picture of where you stand on these problems. The set up here is the same as with your other entry [the one I replied to a few seconds ago] only you will use integration by parts method to solve the integral.

[quote="akech"]... only you will use integration by parts method to solve the integral.[/quote]

Actually, I think a calculator will be necessary to compute the integral (or at least the bounds of integration) since the particle changes directions somewhere in between $ t \equal{} 0$ and $ t \equal{} 4$.

[hide="hint:"]Total distance travelled by an object moving in one-dimension between times $ t_1$ and $ t_2$ is given by: $ \int_{t_1}^{t_2} |v(t)| \,dt$. If you have a calculator available you can approximate the integral numerically using fnInt().[/hide]


In here, you will find stunning displays of three dimensional art.

Here's a picture of my artistic take on an erupting volcano.

Here's another picture of my artistic take on an erupting volcano, fully rendered in OpenGL, with some liberal Photoshop usage.

I'll post a tutorial on how do make these later, if anyone expresses interest.

The base texture that I'm wrapping around the OpenGL spheres in the previous pictures. This gets wrapped around the globe, I then shrink the globe, texture the globe and sky, and then add various lights and particle emitters. This gets Photoshopped and filtered into the previous images you've already seen.

Note that this must be viewed at full size to get the full effect. Otherwise, it looks like a bunch of ugly dots.

I really liked putting the negative space into this one. You can imagine that the actual set fills in the space I left white, and the colored area represents items OUTSIDE of the solution set.

This one is particularly visually stunning to me, because it showcases the pink background against the green solution set.

To me, this fractal (my own formula) looks like the eye of a hurricane, so I colored it to represent the ominous overtones of a raging storm. The purple represents anger and fury, and the white represents calmness in the center.

This fractal (similar formula to the last) looks a bit like the eye of the Ancient Egyptian god Horus.

I made this one appear on a grassy OpenGL field, and to me, it looks like a double spiral staircase.

For something like this I recommend NuCalc.

LOL at that.
This IS NuCalc  :lol:
That's what I use for graph pictures ;-)


For fixed m and variable k, there are multipliers a and b such that f(m+k)=af(m+1)+bf(m), form a conjecture as to the nature of a and b in terms of k. Prove this conjecture.

btw, i THINK that a=f(k) and b=f(k-1) but i cannot prove it

f(n) is the n th Fibonacci term where f(1)=f(2)=1

I've seen this same problem on here before, followed by one about a*f(n)^2 + b*f(n + 1)^2 being a fibonacci number for all n.  Try a search for it.  (I think it may have been in the advanced section rather than in pre-olympiad, but I'm not positive.)


Suppose Farmer Joe has 8 paddocks on his farm. He moves his cows to a new paddock every day, so every 8 days the cows return to where they were originally. If he has 13 cows and 5 acres per paddock, how many cow-days does he get per year? A cow-day is the amount 1 cow eats in 1 day.

[quote="uberl33tmage"]for some reason i get [hide]37960[/hide][/quote]

first of all, there are 365 days in a year. 37960 is about 130 years..i think

[hide]i got 7[/hide]

Wouldn't the answer just be[hide]$13\times 365=4745$ because he has 13 cows and so gets 13 cow-days each day for 365 days[/hide]

What does "amount a cow eats" mean? In acres? Then samath's answer multiplied by 5 would be correct...

"Amount a cow eats" simply means "amount a cow eats".  The problem does not tell you this, and you don't need this information.  I think the only necessary number in that problem is 13 cows.


I'm a newbie and I've been analyzing this problem many times and could not figure out how i did wrong.

Heres how i thought of it...
I went by columns and split it into many cases.

Column 1 has 6c3 aka 20 cases.
Column 2 can be 1 of 4 cases.
1. All 3 shades are on the same row as those shaded in col 1.
2. 2 shades on the same row.
3. 1 shade on the same row
4. 0 shade on the same row.

Let o = shaded, x = blank and p = possible squares

For the first case, there is only 1 way to make that arrangement and 1 way to construct col 3 and 4. so 20*1*1*1 = 20.
ooxx
ooxx
ooxx
xxoo
xxoo
xxoo

For the second case, there is 3c2 * 3c1 ways to make that combination. Then for the 3 col there are 4 spaces where you do not violate the 2 per row rule. so thats * 4c3. The fourth row only has 1 combination so 20*3*3*4*1 for case 2.

ooxx
ooxx
oxpx
xopx
xxpx
xxpx

Third case there are 3c1 * 3c2 ways to fill col 2.  Row 3 has 5 choices this time and col 4 is still predetermined, netting 20*3*3*10*1 for case 3

ooxx
oxpx
oxpx
xopx
xopx
xxpx

Fourth case there is 3c3 way to fill out the second row but 6c3 ways to fill the 3rd and still 1 way to fill the fourth netting 20*1*20*1.

By adding all the cases i get 2940. I obviously over counted like crazy but i don't see how the cases are suppose to overlap. Can anyone explain this for me? Thanks!

The problem is you can't have three blanks in one row. So not all possible ways to choose the first 3 columns (by your method) will work.

Sorry for being unclear but the 4th row was constructed all with xxx because there will only be 1 way to construct it and it is based on the set up for the 3rd column.

Its either you misunderstood me or i misunderstood you so please provide an example to illustrate your point. Sorry -_-.

[quote]For the second case, there is 3c2 * 3c1 ways to make that combination. Then for the 3 col there are 4 spaces where you do not violate the 2 per row rule. so thats * 4c3. The fourth row only has 1 combination so 20*3*3*4*1 for case 2.[/quote]

What if you had something like this to begin with (the first two columns)

11
11
10
01
00
00

And then, according to your 4C3 argument, one of the possible choices for the third column would be

110
110
101
011
001
000

However, now there is [i]no[/i] way you can color the fourth row and satisfy the conditions of the problem.

(There may be similar flaws in your other cases -- double check them)

If you want to see a solution that does not involve case analysis, check the AoPSWiki for an outline of my PIE solution (Solution 3, it's kind of late right now so I'll fill it out more later, or if someone who understands it well would like to fill in the details, they are encouraged to).

Thanks randomdragon, damn, i should have seen that!!

Noodle, I also used this approach on the contest. :lol:   Check your work for cases 2 and 3.

[hide="case 2"]For case 2, there are 3c2=3 ways to shade the 2 squares on the same row and there is 3c1=3 way to then shade 1 out of the 3 squares that is on a row without any shaded squares.  Now for the 3rd column: 2 of the squares in this column have to be shaded (otherwise all rows would not have at least 2 shaded squares).  For the last shaded square in the 3rd column there are 2 possible rows (so 2 ways to choose the last square in the 3rd column).  This leaves 1 way to shade the last column (convince yourself why).  Thus, there are 20*3*3*2=360 ways for this case (20 ways to shade 1st column).[/hide]

[hide="case 3"]For case 3, there is 3c1=3 ways to choose the 1 square to shade on one of the same rows as the shaded squares in the 1st column.  Then, there are 3c2=3 ways to choose 2 other squares to shade in this column.   Now for the 3rd column: the row that does not have any shaded squares yet must have a shaded square here.  Then, there are 4c2=6 ways to choose the other 2 squares to shade in this row.  As with case 2, this leaves 1 way to shade the last column.  For this case there are 20*3*3*6=1080 ways.

I got the same number of ways for your other cases.  Adding them up gives: 20+400+1080+360=1860 ways total.[/hide]


Prove that there is an infinite number of ...uhm prime numbers :D .

It has been done many times before.

[hide] Assume that we have a finite number of primes $P_{1}, P_{2}.........P_{n}$. Consider the number $Q=P_{1}\times P_{2}\times P_{3}\times............. \times P_{n}+1$. Since $Q$ cannot be prime number because it is not in the list we conclude that $Q$ is a composite number. However, $Q$ can't be composite either for it cannot be divided by any of the primes $P_{1}, P_{2}.........P_{n}$. A contradiction, thus our original assumption of a finite number of prime numbers is false. Therefore, there exists an infinite number of primes.

$QED$.[/hide]

yes, Franz Joseph has shared with us Euclid's proof for this. Though there are many more, his is by far the most famous.

I got this one myself, but I'm sure it's quite famous
[hide]
The probability that a randomly picked number is divisible by a prime p is $\frac{1}{p}$. Therefore, the probability that a number is not divisible by any prime is $(1-\frac{1}{2})(1-\frac{1}{3})...$. Since we know that this probability is 0, there must be an infinite number of primes.[/hide]


AMC
AMC 10
AMC 12
geometry
AMC 10 B
MATHCOUNTS
Pythagorean Theorem

*walks to Virginia*

It is in November. I think you mean the 2007 AMC 8.

*walks back to New York*

*walks to New York and knocks on 1=2's door*

No I meant the AMC10 and/or AMC12  :D

*walks back to Virginia*

not from virginia, but i thought the test was easy, but hard. only one i didn't know how to do was #25, so i didn't attempt it. made careless mistakes on #2 and #21.

to me, it was a lot harder to see the concepts in some of the problems this year than previous years. i totally blanked during the pythagorean theorem problem (#18, i think?) and i felt stupid when i realized you just substitute.

lot of geometry on this test.

i thought it was actually a pretty easy year.

just blanked out on some probs

lol, yeah it was pretty easy.  i got lucky though and guessed # 23 right.  I also guessed # 25..... wrong.  all the others were doable.  :D

other than a couple of problems, it was easier than usual
but the "couple of problems" were a bit harder than the ones that were used before (minus that one problem from last year on the amc10b #25)
i also notice that no one hear is actually from virginia...
and yes, we can discuss it

[quote="1=2"]*walks to Virginia*

*clenches teeth*

shut up, shut up, we can't discuss it yet...[/quote]

its ok, the AMC forum says we can now.

"Walks to virginia, realizes he's already in virginia"
Yeah, i may not have made it.  I think i got a 118.5 on the AMC 10, compared with a 126 from last year.  So i think this one was harder.

Hello

I'm seeing some of Kevin's classmates here...

I guess #24 felt like a mathcounts countdown "what is the units digit of" kind of problem.
None of the problems required really any fancy or obscure math tricks and #25, which I missed, had a few short solutions.


A, B and C are positive integers.
Determine all possible values of  C,  for which  the equation:
$(2^{A})- (5^{B})= C$ ,  exactly two distinct solutions.

A, B and C are positive integers.
Determine all possible values of  C,  for which  the equation:
2^A– 5^B= C ,  exactly two distinct solutions.


Let n be a natural number. Prove that there exist a set with n members such that sum of members of any subsets of it isn't a perfect square.

 ;)

There is a very nice and well-known idea tht we can apply here:

Just take $a$ to be a prime number larger than $\frac{n(n+1)}2$, and take our set to be $\{a,2a,\ldots,na\}$.


I'm looking at participating at ARML, HMMT and possibly PMC but I'm not sure how worthwhile it will be for me.  I'm currently able to solve about half the problems on an AIME.  Will I be able to do well in these competitions or will it be a waste of time?

I think ARML is a more for team compare to AIME.
ARML is mostly, a team test, there is only an individual round and take only a little time.

I think AIME questions are harder than ARML.

They're different kinds of competitions. My opinion is that in general, the questions are easier than AIME, but time is much more a factor in ARML.

Personally, I like ARML because I don't get to meet other math people that often, since there isn't much math interest in this area (other than the GPML meets).

ARML questions are much more about speed than AIME questions, so ARML questions require the ability to gain "insight" into a problem very quickly. (Average amount of time to solve an ARML question = approx. 5 mins, average amount of time to solve an AIME question = approx. 12 mins.) That being said, late ARML individual questions are sometimes similar in difficulty to middle AIME questions, and late ARML Team questions are similar in difficulty to some late AIME questions. I've seen students who can solve around half of AIME questions get about 5 or 6 on the ARML individual. But both contests teach you to use the tools of elementary mathematics to solve difficult problems, so for studying purposes, the questions of both contests are [i]decent[/i] preparation for the other.

Note that ARML is moving to a 10-question individual section. I'm not sure how this will change difficulty.


The surface area of a cube is numerically equal to twice its volume.  Find the length of a diagonal of the cube.

[quote="rd5493"][quote="ProtestanT"][hide]
Let $s$ = side length
$6s^{2}=2s^{3}$
$s=3$
a diagonal on one face is $3\sqrt{2}$
a diagonal through the whole cube is a right triangle with $3\sqrt{2}$ and $3$ as legs, so the diagonal is $3\sqrt{3}$
Answer: $3\sqrt{3}$
[/hide][/quote]

actually no, remember, it said [b]TWICE[/b] its volume.
The answer is... :ninja:  :ninja: "Let's take him out before he gives out the answer"[/quote]

What are you talking about?

[quote="ProtestanT"][hide]
Let $s$ = side length
$6s^{2}=2s^{3}$
$s=3$
a diagonal on one face is $3\sqrt{2}$
a diagonal through the whole cube is a right triangle with $3\sqrt{2}$ and $3$ as legs, so the diagonal is $3\sqrt{3}$
Answer: $3\sqrt{3}$
[/hide][/quote]
Yeah, now its correct

i guess you read my solution when i had the mistake, and quoted it when i edited the mistake.
but either way, the mistake(s) are gone now.
:D

[hide] if $s$ is the side length of the cube, then the volume of the cube is $s^{3}$, and the surface area of the cube is $6s^{2}$ so you do $2s^{3}$  = $6s^{2}$ so we get $s$ = $3$ so the space diagonal of the cube is $3\sqrt{3}$[/hide]


inequalities
trigonometry
inequalities unsolved

If $a,b,c>0$ and $A,B,C$ are angles in rad prove that

$\frac{a}{b}cosA+\frac{b}{c}cosB+\frac{c}{a}cosC\geq \frac{3}{4}$  .

what abt a,b,c
and \[ A,B,C \]

do they form part of triangle or not :?:

else LHS can be negative

http://www.mathlinks.ro/Forum/viewtopic.php?t=72297&postorder=asc&start=40
arpit goel you may see here there is a connection between a,b,c and A,B,C.
silouan do you really think this is easier to prove then the initial form of the inequality?i don't think so...

Why should the two inequalities be equivalent?

Also, if a, b, c are the sidelengths of a triangle and A, B, C are its angles, then

$\frac{a}{b}\cos A+\frac{b}{c}\cos B+\frac{c}{a}\cos C\geq\frac34$

is wrong for a = 2, b = 12, c = 12.

Silouan, did you mean some different inequality?

You may have meant

$\frac{a}{b}\cos B+\frac{b}{c}\cos C+\frac{c}{a}\cos A\geq\frac32$.

But this is easy: It is known that $c=a\cos B+b\cos A$ (in fact, the sine law yields $a : \sin A = b : \sin B = c : \sin C$, so this is equivalent to $\sin C=\sin A\cos B+\sin B\cos A$, but this follows from $\sin C=\sin\left(A+B\right)$, since C = 180° - (A + B) by the sum of angles in a triangle). Thus, $a\cos B=c-b\cos A$, so that

$\frac{a}{b}\cos B=\frac{a\cos B}{b}=\frac{c-b\cos A}{b}=\frac{c}{b}-\cos A$.

Thus, using the $\sum$ notation for cyclic sums, $\sum\frac{a}{b}\cos B=\sum\left(\frac{c}{b}-\cos A\right)=\sum\frac{c}{b}-\sum\cos A$. But AM-GM yields $\sum\frac{c}{b}\geq 3\sqrt[3]{\frac{c}{b}\cdot\frac{a}{c}\cdot\frac{b}{a}}=3$, and we know that $\sum\cos A\leq\frac32$. Thus, $\sum\frac{a}{b}\cos B=\sum\frac{c}{b}-\sum\cos A\geq 3-\frac32=\frac32$, qed.

  dg

i did not say that the inequality posted here in this form is equivalent with the other one but i am sure(look at the title)that this is what siluoan meant(of course with some additional conditions for a,b,c and A,B,C if what  puuhikki wrote is right-i did not check)

[quote="darij grinberg"]Why should the two inequalities be equivalent?[/quote]

Because
\begin{eqnarray*}\\ \frac{a}{\sqrt{b^2 + 15ca}}+\frac{b}{\sqrt{c^2 + 15ab}}+\frac{c}{\sqrt{a^2 +15bc}} &=& \frac{a}{\sqrt{b^2}\sqrt{ 1+ \frac{15ac}{b^2}}}+\frac{b}{\sqrt{c^2}\sqrt{1 + \frac{15ab}{c^2}}}+ \frac{c}{\sqrt{a^2}\sqrt{1 +\frac{15bc}{a^2}}}\\ &=&\frac{a}{b\sqrt{ 1+ \frac{15ac}{b^2}}}+\frac{b}{c\sqrt{1 + \frac{15ab}{c^2}}}+ \frac{c}{a\sqrt{1 +\frac{15bc}{a^2}}}\\ \end{eqnarray*}

Now tan is continuos and goes from 0 to infinity in the range $[0,\pi/2]$ as well as tan^2. By intermediate value theorem, for all $a,b,c>0$ there exists $\alpha,\beta,\gamma \in [0,\pi/2]$ such that $\text{tan}^2\alpha=\frac{15ac}{b^2},\text{tan}^2\beta=\frac{15ab}{c^2},\text{tan}^2\gamma=\frac{15bc}{a^2}$. Therefore
\begin{eqnarray*} \frac{a}{b\sqrt{ 1+ \frac{15ac}{b^2}}}+\frac{b}{c\sqrt{1 + \frac{15ab}{c^2}}}+ \frac{c}{a\sqrt{1 +\frac{15bc}{a^2}}}&=&\frac{a}{b\sqrt{1+\text{tan}^2\alpha}}+\frac{b}{c\sqrt{1+\text{tan}^2\beta}}+\frac{c}{a\sqrt{1+\text{tan}^2\gamma}}.\\ \end{eqnarray*}
Now $\frac{1}{\sqrt{1+\text{tan}^2x}}=\text{cos }x$ for all $x\in ]0,\pi/2[$.

Puuhiki, you have a very nice idea. But unfortunately, these angles in Pardoosung's inequality should have some restrain conditions I think.  :(


How did everyone do? I think I got a 138, screwed up #25 and guessed wrong on #21. I'll start posting the problems tonight.

It felt pretty tough, as I had very little time to check my work and I usually had 20+ minutes to do so on practice 10s. (I took the 12A).

I think I got a 144... didn't count the time when michael is running and the truck passes him for number 25...  That's ok though.

[quote="RisingStar"]Can someone post answers?[/quote]

Yes, can somebody please post the answers? My school somehow didn't has a solutions packet for the amc 10b and I am very nervous to see what I got. Please help. Thanks

21 and 25 are, respectively, C and B. There is a topic with all the answers.

[quote="the future"]well it depends what the answers were for 21 and 25, if they were C and B then I got a 144...

I think I did all the other ones right

number 24 was the hardest problem in my opinion...

but this test had 4 geometry questions!?!?  thats my favorite and best subject so that was a bummer...

and the 10A had like 8 geo questions[/quote]
Which failed me. The 12B, IMO, had a bunch of geometry.

[quote="the future"]well it depends what the answers were for 21 and 25, if they were C and B then I got a 144...

I think I did all the other ones right

number 24 was the hardest problem in my opinion...

but this test had 4 geometry questions!?!?  thats my favorite and best subject so that was a bummer...

and the 10A had like 8 geo questions[/quote]

24 wasn't that bad. I had a protractor and a ruler, so I just drew it to scale and measured the angle :D.

Also there were so many As and Bs in a row (like 10 straight As or Bs)

which is totally differnet frmo the norm.

ALso only 1 D...I took the 12A and that test had 11Ds..

Even without a protractor, 24 isn't that bad, see the solutions in the topic.

how are the answer choices determeined?

like for number 19, wouldnt you say that the distractors are not very good becuase once you figure out the 48pi part (120 degrees) the rest is trivial...

In my opinion

This AMC 10B was much easier than the AMC 10A
I mean I'm not very good at math
and I still only missed one and left one blank

Good thing I didn't have time to check my answers though
then I think I would have wasted time checking 9 to 16 because it was all A's and B's

did anyone find number 17 to be surprisingly easy?

compared to like 10-16?

[quote="mathgeniuse^ln(x)"]I got 144 again

Third year in a row.

Sigh.[/quote]

Didn't you get 146.5 two years ago?

don't worry neil...at least u stayed at 144

my amc 10 score actually went down by 12 pts...from 144

wtf


topology
analytic geometry
abstract algebra

Use the Mayer - Vietoris to compute the homology groups of the space $ X$ obtained by gluing two solid tori along their boundary as following : Let $ D^2$ be the unit disk and $ S^1$ be the unit circle, $ A\equal{}S^1\times D^2$, and $ B\equal{}S^1\times D^2$, then $ X$ is the quotient space obtained by identifying $ (z,w)\in A$ with $ (zw^3, w)\in B$ for all $ (z,w)\in S^1\times S^1$

I am getting stuck on how to find $ H_{1}(X)$, since I am not sure how the crazy map goes.

Can some one explain for me? thanks

Does the "crazy map" mean the boundary map?

Let $ T : \equal{} S^1 \times S^1$. Then $ X$ is formed as the pushout of $ f : T \to A$ and $ g : T \to B$ given by $ f(z,w) \equal{} (z,w)$ and $ g(z,w) \equal{} (zw^3,w)$.

That being said, the reason you might be having trouble is that Mayer-Vietoris holds for open inclusions of subspaces, and the way you have phrased things, while $ X$ is covered by $ A \cup B$ it isn't true that $ A$ and $ B$ are open in $ X$. However, we can replace $ X$ with the "homotopy pushout" $ Y : \equal{} A \cup (T \times I) \cup B$, where $ (T \times 0)$ is glued to $ A$ along $ f$ and $ (T \times 1)$ is glued to $ B$ along $ g$. Since the inclusions $ T \to A$ and $ T \to B$ are "nice enough" (cofibrations), we have that $ X$ is homotopy equivalent to $ Y$. So it suffices to compute the homology of $ Y$ instead -- the nice thing being that $ Y$ is covered by open sets $ U$ and $ V$ whose intersection deformation retracts onto $ T \times (1/2)$.  Now, we can at least make things precise.

Then we have a long exact sequence

$ H_1(T) \to H_1(U) \oplus H_1(V) \to H_1(Y) \to H_0(T) \to H_0(U) \oplus H_0(V) \to H_0(Y) \to 0$.

Since the map $ H_0(T) \to H_0(U) \oplus H_0(V)$ is injective ($ T$ is path connected) we have that $ H_1(Y) \to H_0(T)$ is a zero map; therefore the sequence is

$ H_1(T) \to H_1(U) \oplus H_1(V) \to H_1(Y) \to 0.$

Therefore $ H_1(Y)$ is determined by the image of $ H_1(T)$. The map $ H_1(T) \to H_1(U) \oplus H_1(V)$ is $ f_* \times g_*$, so we are left to compute these maps. It's pretty clear that $ f_*$ just kills the second coordinate of $ H_1(T)$ -- computing $ g_*$ is more interesting, so let's do that explicitly. (What I mean by $ f_*$ should follow analogously if you follow similar steps).

Let $ h : B \to S^1$ be given by $ h(z,w) \equal{} z$. Then $ h$ is a homotopy equivalence as it factors $ B \equal{} S^1 \times D^2 \to S^1 \times * \to S^1$, since $ D^2 \to *$ is a homotopy equivalence and $ S^1 \times * \to S^1$ is a homeomorphism. Then $ (h \circ g) : S^1 \times S^1 \to S^1$ is given by $ (h \circ g)(z,w) \equal{} zw^3$.

As $ H_1(T) \equal{} \mathbb{Z} \oplus \mathbb{Z}$, elements can be represented in the form $ (a,b)$ where $ a$ and $ b$ are integers. Then $ (h \circ g)_*(a,b) \equal{} a \plus{} 3b$.

So, we have that the map $ f_* \times g_*$ sends $ (a,b)$ to $ (a,a\plus{}3b)$, and $ H_1(Y)$ is the cokernel of this homomorphism, which is $ \mathbb{Z} / 3$. So $ H_1(X) \equal{} \mathbb{Z} / 3$.


search
combinatorics unsolved
combinatorics

A white plane is partitioned onto cells (in a usual way). A finite number of cells are coloured black. Each black cell has an even (0, 2 or 4) adjacent (by the side) white cells. Prove that one may colour each white cell in green or red such that every black cell will have equal number of red and green adjacent cells.

[quote="Baronov"]Search for Russian olympiad (final round) 2005 10-th grade problem 8.[/quote]

Well, actually that thread is without solution, and I don't want to revive it, so I post it here.  :)

Reviving old threads is better than posting new ones!

  darij

Furthermore, you've posted this question in 2 distinct places on the forum, making this doubly redundant.

Solution outline:

Consider all graphs where white squares bordering black squares are vertices, and for each pair of adjacent white and black squares, the white square shares an edge with exactly one other white square bordering that black square.

Now if this graph is bipartite, we're done. If it's not, then consider such a graph containing the fewest odd cycles, where no two non-consecutive elements in a cycle are allowed to share an edge. Then among these, consider the one with the largest odd cycle. Argue that this odd cycle can be made even larger, reaching a contradiction. That last argument is kind of annoying but I think it works.

[quote="probability1.01"] Argue that this odd cycle can be made even larger, reaching a contradiction. [/quote]

Please show me how..  :blush:

I coldun't find out that part too :oops: , can you explain more please?!
thanks

[quote="Summerburn"]Well, or is it obvious? I don't see...[/quote]

It's not, but it's kind of hard to explain without a diagram. I'll try to explain a little bit. So if there is an odd cycle, note that there must be at least two consecutive white squares in the cycle such that they are on opposite sides of a black square. Actually, this condition must happen an odd number of times.

Now if the black square we have is part of a "ring" of black squares, we must enter and leave the ring in the cycle, so there is another black square in the ring which is "jumped over" as we go through the cycle. (Another way to think about it... the white squares have positive and negative regions). Therefore, these black squares form a pair, so there must be at least one black square with consecutive white squares in the cycle on opposite sides which is not a part of such a pair.

Such a black square, then, must be all alone, bordering no other black squares. Now if we have WBW, where the W's are white squares in our cycle, consider the white squares above and below the B. I'm pretty sure it's true that if we swap the edges, so that instead of

[code]
 W
W+W
 W
[/code]

we have

[code]
  W
 /
W   W
   /
  W
[/code]

then we either make a larger odd cycle or reduce the number of odd cycles.


[quote="Slizzel"]In this case we have:

$\frac{1}{a^2-bc+1}\leq1\Leftrightarrow a^2-bc+1\geq1$[/quote]

I can't understand.

The fraction $\frac{1}{a^2-bc+1}$ to less (or equal) then 1 it's denominator should be $\geq1$  :)

[quote="Slizzel"]In this case we have:

$\frac{1}{a^2-bc+1}\leq1\Leftrightarrow a^2-bc+1\geq1$
$\frac{1}{b^2-ac+1}\leq1\Leftrightarrow b^2-ac+1\geq1$
$\frac{1}{c^2-ab+1}\leq1\Leftrightarrow c^2-ab+1\geq1$

Summing the 3 relations $\frac{1}{a^2-bc+1}+\frac{1}{b^2-ac+1}+\frac{1}{c^2-ab+1}\leq3\Leftrightarrow a^2+b^2+c^2\geq\frac{1}{3}$. If $a=b=c$ it's ok :), else... i don't know  :D[/quote]
Sorry but this is totally wrong :
It's true that  if we have $\frac{1}{a^{2}-bc+1}\leq 1$ and $\frac{1}{b^{2}-ca+1}\leq 1$ and $\frac{1}{c^{2}-ab+1}\leq 1$ this means that $\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3$
But if we had $\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3$
It doesn't necesserly mean that $\frac{1}{a^{2}-bc+1}\leq 1$ and $\frac{1}{b^{2}-ca+1}\leq 1$ and $\frac{1}{c^{2}-ab+1}\leq 1$

our ineq is equivalent to:

$\sum\frac{a(a+b+c)}{3a(a+b+c)+2}\geq\frac{1}{3}$

what can we  do ????????/ :(
 :D  :(

let $x=a(a+b+c)$

why not?????????? :D

Homogenizing, the ineq is equivalent to
\[
\sum_\mathrm{cyc} \frac{1}{a(a+b+c)+2(ab+bc+ca)} \leq \frac{1}{ab+bc+ca}
\]
Multiply both sides by $2(ab+bc+ca)$,
\[
\sum_\mathrm{cyc} \frac{2(ab+bc+ca)}{a(a+b+c)+2(ab+bc+ca)} \leq 2
\]
Subtracting from 3, the above ineq is equivalent to
\[
\sum_\mathrm{cyc} \frac{a(a+b+c)}{a(a+b+c)+2(ab+bc+ca)} \geq 1
\]
Now Cauchy
\[
LHS = (a+b+c)\sum_\mathrm{cyc} \frac{a^2}{a^2(a+b+c)+2a(ab+bc+ca)} \geq \frac{(a+b+c)^3}{\sum_\mathrm{cyc} [a^2(a+b+c)+2a(ab+bc+ca)] } = 1
\]

hei, very nice!! :D

and this is my proof: :P

our ineq is equyvalent to:

$\sum\frac{a}{3a(a+b+c)+2}\geq\frac{\frac{1}{3}}{a+b+c}$

by cauchy:

$lhs\geq\frac{(a+b+c)^2}{3(a+b+c)(a^2+b^2+c^2)+2(a+b+c)}=\frac{\frac{1}{3}}{a+b+c}$

which is true:

$\frac{a+b+c}{3(a^2+b^2+c^2)+2}=\frac{\frac{1}{3}}{a+b+c}$

multifly we get $ab+bc+ca=1/3$ ;)

it's cost me alot of time :D  :D

[color=blue]
From the moment I saw this inequality, the next thing that came into my mind was this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=85768]inequality proposed by Titu Andreescu in G.M in 2001[/url].

    Let $a, b, c>0$ such that $ab+bc+ca=\frac{1}{3}.$ Prove that:
\[\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ca+1}+\frac{c}{c^{2}-ab+1}\geq\frac{1}{a+b+c}. \]
[/color]

 [i] Solution.[/i] Let us apply the well-known inequality Cauchy-Schwarz. We have:
\[(a(a^{2}-bc+1)+b(b^{2}-ca+1)+c(c^{2}-ab+1))(\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ca+1}+\frac{c}{c^{2}-ab+1})\geq (a+b+c)^{2}, \]
so we get that
\[\frac{a}{a^{2}-bc+1}+\frac{b}{b^{2}-ca+1}+\frac{c}{c^{2}-ab+1}\geq\frac{(a+b+c)^{2}}{a^{3}+b^{3}+c^{3}-3abc+a+b+c}. \]
Now we only need to prove that $\frac{(a+b+c)^{2}}{a^{3}+b^{3}+c^{3}-3abc+a+b+c}\geq\frac{1}{a+b+c}.$ But this inequality is in fact equality because using the well-known identity $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$, we deduce that $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}-ab-bc-ca,$ so we are done. Now let's deduce this chinese problem from this one:
     We have:
\[\sum\frac{a}{a^{2}-bc+1}\geq\frac{1}{a+b+c}\]
But this is wquivalent with:
\[\sum\frac{a^{2}+ab+ac}{a^{2}-bc+1}\geq 1, \]
or
\[\sum\frac{(a^{2}-bc+1)+(ab+bc+ca-)}{a^{2}-bc+1}\geq 1 \]
or
\[3+\sum\frac{ab+bc+ca-1}{a^{2}-bc+1}\geq 1 \]
or
\[\sum\frac{1}{a^{2}-bc+1}\leq 3. \]

when i see this kind of inequalities i know that almost everytime  there a "nice" solution but i do not have the time to find it :P so i find only stupid solutions. :oops:  i will only say the main steps since i do not have the time to post a full solution right now.denote $x=ab,y=bc,z=ca \Rightarrow a^2=xz/y.b^2=xy/z,c^2=yz/x$.then the conditions rewrites as $x+y+z=1/3$ and our ineq rewrites as $3\ge y/(xz-y^2+y) +.......)$.now we put $x=a/3(a+b+c),y=.....$(note that a,b,c are different from the initial ones-you may also put $a_1,a_2,a_3$ instead of $a,b,c...$)
finally our ineq is equivalent with $1\ge b(a+b+c)/(ac-b^2+3b(a+b+c))+.........$ and that is also equivalent with $(ac-b^2)/(ac+3ab+3bc+b^2)+.....\ge0$ and that you easily kill be clearing everything out and then some muerhead

As a note, I'd like to point out that this inequality can be "dumbassed".

The desired inequality is equivalent to $ \sum_{cyc}\frac{ab\plus{}ac\plus{}bc}{a^2\plus{}3ab\plus{}3ac\plus{}2bc} \leq 1$, which is equivalent to $ (ab\plus{}ac\plus{}bc)\sum_{cyc}(a^2\plus{}3ab\plus{}3ac\plus{}2bc)(b^2\plus{}3bc\plus{}3ab\plus{}2ac) \leq (a^2\plus{}3ab\plus{}3ac\plus{}2bc)(b^2\plus{}3bc\plus{}3ab\plus{}2ac)(c^2\plus{}3ac\plus{}3bc\plus{}2ab)$, or, expanding

$ 5\sum_{sym}a^4b^2 \plus{} 11\sum_{sym}a^3b^3 \plus{} 5\sum_{sym}a^4bc \plus{} 76\sum_{sym}a^3b^2c \plus{} 147a^2b^2c^2 \leq 6\sum_{sym}a^4b^2 \plus{} 10\sum_{sym}a^3b^3 \plus{} \frac{13}{2}\sum_{sym}a^4bc \plus{} 75\sum_{sym}a^3b^2c \plus{} 144a^2b^2c^2$.

Cancelling terms, we get the equivalent

$ \sum_{sym}a^3b^3 \plus{} \sum_{sym}a^3b^2c \plus{} 3a^2b^2c^2 \leq \sum_{sym}a^4b^2 \plus{} \frac{3}{2}\sum_{sym}a^4bc$, which is evident by weighted AM-GM.

I think the inequality can also be done with the SOS method.

Let $ X \equal{} a \plus{} b \plus{} c$ and $ Y \equal{} ab \plus{} ac \plus{} bc$.

The inequality can be rewritten as (through some manipulation)

$ \sum (a \minus{} b)^2(2Y^2 \plus{} \frac {X^2}{2}(ac \plus{} bc \minus{} ab))$

or

$ \sum (a \minus{} b)^2S_c$ for the $ S_a, S_b, S_c$ chosen as above.

WLOG, $ a\geq b \geq c$, so it is clear that $ S_a$, $ S_b, S_b \plus{} S_c \geq 0$, and that $ (a \minus{} c) \geq (a \minus{} b)^2$ so we are done.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=474878&hilit=Macedonia+inequality]Macedonia National Olympiad 2009[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=474875&p=3632920#p3632920]Macedonia National Olympiad 2009 - Problem 4[/url]
The following inequality is also true.
Let $a$, $b$, $c$ be nonnegative reals such that $\sqrt{ab}+\sqrt{bc}+\sqrt{ca} = \frac{1}{3}$. Prove that
\[\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3 .\]

[quote=billzhao]Let $a$, $b$, $c$ be nonnegative reals such that $ab+bc+ca = \frac{1}{3}$. Prove that
\[\frac{1}{a^{2}-bc+1}+\frac{1}{b^{2}-ca+1}+\frac{1}{c^{2}-ab+1}\leq 3 \][/quote]

So here's my long solution and probably brainless

the inequality can be rewritten as
$$\sum_{a,b,c}\frac{a^2-bc}{a^2-bc+1} \ge 0$$
$$\iff \sum_{a,b,c} (a^2-bc)(b^2ca+1)(c^2-ab+1) \ge 0$$
$$\iff \sum_{a,b,c} (a^2-bc)(b^2c^2-b^3a+b^2-c^3a+a^2bc-ca+c^2 -ab) \ge 0$$
$$\iff \sum_{a,b,c} (-a^3b^3-a^3c^3-b^3c^3 + a^2b^2+a^2c^2 +abc(a^2+b^2+c^3)-a^3b-a^3c-b^3c-bc^3 +a^2 -bc +abc(b+c)) \ge 0$$

Setting $a+b+c = u$, $v=ab+bc+ca=\frac{1}{3}$, $w=abc$
It can be easily seen $a^3b^3+b^3c^3+c^3a^3 = v(v^2-3uw)+3w^2$
and $\sum_{a,b,c}(a^3b+a^3c+b^3c+bc^3) = 2(u^2v-2v^2-uw)$
Replacing this and other values, the inequality reduces to
$$\iff -3(v(v^2-3uw)+3w^2)+2(v^2-2uw)+3w(u^3-3uv+3w)  -  2(u^2v-2v^2-uw) + u^2 - 2v -v +2uw \ge 0$$
$$ \iff -3v^3+6v^2+3u^3w-2u^2v+u^2-3v \ge 0$$

Replacing $v= \frac{1}{3}$

$$\iff 3u^2+27u^3w \ge 4$$

Now there are 2 cases.

[b]Case 1 :[/b] $a,b,c$ form sides of a triangle
Then using Schur's inequality,
$a^3+b^3+c^3 + 3abc \ge ab(a+b) + bc(b+c) + ca(c+a)$
$\implies u^3+9w \ge uv$
$\iff 27w \ge 4u-3u^3$

$3u^3 + 27u^3w - 4 \ge 3u^2 + u^3(4u-3u^3) - 4 = (3u^2 - 4)(1-u^4)$
Since $a,b,c$ form sides of a triangle,
$3u^2 - 4 \le 0$ and obviously $1-u^4 \le 0$

$$\implies 3u^3 + 27u^3w \ge 4$$
QED

[b]Case 2 :[/b] $a,b,c$ don't form the sides of a triangle.

Normalizing, we are left to prove
$$\iff u^2v^2 + u^3w \ge 4v^3$$
Since $a,b,c$ are  not the sides of a triangle,
Without loss of generality,
$$a \ge b+c \ge \frac{u}{2}$$
$$u^3a \ge \frac{u^4}{2} \ge 4.5 v^2 \ge 4v^2$$
$$\implies u^3w \ge v^24bc \ge v^2(4bc - (b+c-a)^2)$$
$$\implies u^2v^2 + u^3w \ge 4v^3$$

QED

[quote=Kunihiko_Chikaya]The equality holds when $a=b=c=\frac{1}{\sqrt{3}}$.[/quote]

Then, ab + bc + ca = 1, not $\frac{1}{3}.$

First, we homogenize to get \[\frac{1}{a^2 - bc + 1} = \frac{3(ab+bc+ca)}{a^2 + 3ab +3ac + 2ab} = \frac{3}{2} - \frac{\frac{3}{2}(a)(a+b+c)}{a(a+b+c)+2(ab+bc+ca)}.\] Substituting this expression into the inequality means that it suffices to prove \[\iff \sum_{cyc}\frac{a}{a(a+b+c)+2(ab+bc+ca)} \geq \frac{1}{a+b+c}.\] By Cauchy-Schwarz, we have
\begin{align*}
\sum_{cyc}\frac{a}{a(a+b+c)+2(ab+bc+ca)} &\overset{\text{Cauchy}}{\geq} \frac{(a+b+c)^2}{\sum_{cyc} a^2(a+b+c) + 2a(ab+bc+ca)} \\
&\geq \frac{(a+b+c)^2}{(a^2+b^2+c^2)(a+b+c) + 2(a+b+c)(ab+bc+ca)} \\
&\geq \frac{(a+b+c)^2}{(a+b+c)(a^2+b^2+c^2 + 2ab+2bc+2ca)} \\
&\geq \frac{1}{a+b+c}.
\end{align*}


inequalities
calculus
integration
inequalities proposed

let $ a,b,c$ is positive prove that:
$ 5(\frac {1} {2a \plus{} b} \plus{} \frac {1} {2b \plus{} c} \plus{} \frac {1}{2c \plus{} a}) \plus{} \frac {1}{a \plus{} b \plus{} c} \le \frac {4}{9}( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}) \plus{} 4(\frac {1}{2b \plus{} a} \plus{} \frac {1}{2a \plus{} c} \plus{} \frac {1}{2c \plus{} b})$

It's obvious after using integral method.
For you, tuantam1lan.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \frac{1}{a\plus{}b}\plus{}\frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{9}{a\plus{}b\plus{}c}\geq6\left(\frac{1}{2a\plus{}b\plus{}c}\plus{}\frac{1}{a\plus{}2b\plus{}c}\plus{}\frac{1}{a\plus{}b\plus{}2c}\right)\]

[quote="arqady"]It's obvious after using integral method.
For you, tuantam1lan.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \frac {1}{a + b} + \frac {1}{a + c} + \frac {1}{b + c} + \frac {9}{a + b + c}\geq6\left(\frac {1}{2a + b + c} + \frac {1}{a + 2b + c} + \frac {1}{a + b + 2c}\right)
\]
[/quote]
my proof:
assume $ p=a+b+c=1$.inequality equivalent:
$ \sum_{cyc}{\frac{1}{a+b}+\frac{3}{a+b+c}-\frac{6}{a+b+2c}) \ge 0}$
<=> $ \sum_{cyc}{\frac{1}{3-c}+1-\frac{6}{c+3}) \ge 0}$
<=>$ \sum_{cyc}{\frac{7c-6-c^2}{9-c^2}} \ge 0$<=>$ \sum_{cyc}{(7c-6-c^2)(9-b^2)(9-c^2) \ge 0}$(1)
let $ q=ab+bc+ca$;$ r=abc$.we have:
(1)<=>$ 243-63\sum{a(b^2+c^2)}+21abc+12\sum{a^2b^2}+27\sum{a^2}-3a^2b^2c^2 \ge 0$
<=>$ -3r^2+138r-3(4q-9)(9-q)-27(4q-9) \ge 0$
apply shur inequality third and founth degree we have:
$ r \ge \frac{4q-9}{3},r \ge \frac{(4q-9)(9-q)}{18}$
hence $ LHS \ge -3r^2+138r-54r-81r=3r(1-r) \ge 0$
equality hold when $ a=b=c=1$or $ a=b$,$ c=0$

[quote="arqady"]It's obvious after using integral method.
For you, tuantam1lan.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \frac {1}{a \plus{} b} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {9}{a \plus{} b \plus{} c}\geq6\left(\frac {1}{2a \plus{} b \plus{} c} \plus{} \frac {1}{a \plus{} 2b \plus{} c} \plus{} \frac {1}{a \plus{} b \plus{} 2c}\right)
\]
[/quote]
If we denote:$ x\equal{}a\plus{}b;y\equal{}b\plus{}c;z\equal{}c\plus{}a$ then we need prove:$ \sum\frac1{x}\plus{}\frac{18}{x\plus{}y\plus{}z}\geq 6\sum\frac1{x\plus{}y}$
Ineq can prove by integarls.
If we need prove by method p,q,r then we will calculation  simple.

[quote="math10"][quote="arqady"]It's obvious after using integral method.
For you, tuantam1lan.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \frac {1}{a \plus{} b} \plus{} \frac {1}{a \plus{} c} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {9}{a \plus{} b \plus{} c}\geq6\left(\frac {1}{2a \plus{} b \plus{} c} \plus{} \frac {1}{a \plus{} 2b \plus{} c} \plus{} \frac {1}{a \plus{} b \plus{} 2c}\right)
\]
[/quote]
If we denote:$ x \equal{} a \plus{} b;y \equal{} b \plus{} c;z \equal{} c \plus{} a$ then we need prove:$ \sum\frac1{x} \plus{} \frac {18}{x \plus{} y \plus{} z}\geq 6\sum\frac1{x \plus{} y}$
Ineq can prove by integarls.
If we need prove by method p,q,r then we will calculation  simple.[/quote]
thanks you very much

[quote="math10"]
If we denote:$ x \equal{} a \plus{} b;y \equal{} b \plus{} c;z \equal{} c \plus{} a$ then we need prove:$ \sum\frac1{x} \plus{} \frac {18}{x \plus{} y \plus{} z}\geq 6\sum\frac1{x \plus{} y}$
Ineq can prove by integarls.
[/quote]
After using an integral method we need to prove that $ \sum_{cyc}(a^6\plus{}2a^2b^2c^2\minus{}3a^3b^3)\geq0,$ which is wrong. :wink:

[quote="tuantam1lan"][quote="arqady"][quote="math10"]
If we denote:$ x \equal{} a \plus{} b;y \equal{} b \plus{} c;z \equal{} c \plus{} a$ then we need prove:$ \sum\frac1{x} \plus{} \frac {18}{x \plus{} y \plus{} z}\geq 6\sum\frac1{x \plus{} y}$
Ineq can prove by integarls.
[/quote]
After using an integral method we need to prove that $ \sum_{cyc}(a^6 \plus{} 2a^2b^2c^2 \minus{} 3a^3b^3)\geq0,$ which is wrong. :wink:[/quote]
i think it true when x,y,z is sides of a triangle[/quote]
Of cause!  :D


number theory
Diophantine equation
number theory open

Does this Diophantine equation have integer solutions $ x,y,z$ $ xyz \neq 0$ that $ x^4 \plus{} 2y^4 \equal{} z^2$
I think this equation has infinitely many solutions(it is easy to prove if there exists one solution,it will ''create"  infinitely many solutions)
So can somebody help me to find one?  :)

$ (x^4\plus{}y^4)^2\equal{}x^4z^2\plus{}y^8$.
Pythagorean triples.

[quote="Taoli"]$ (x^4 \plus{} y^4)^2 \equal{} x^4z^2 \plus{} y^8$.
Pythagorean triples.[/quote]
Right,but can you continue and present a full proof?Thank you. :)

So Pc can help you to find one.

To solve it completely is a perfect practice.

I got this solution (I don't think it's right  :oops: )
We have $ (x^2z)^2\plus{}(y^4)^2\equal{}(x^4\plus{}y^4)^2$ and $ (x^2z,y^2)\equal{}1$
Case 1: $ y$ is an odd
Then there exists 2 intergers m,n and $ m>n>0, (m,n)\equal{}1, m\minus{}n$ is an odd
and $ y^4\equal{}m^2\minus{}n^2\minus{}>(y^2)^2\plus{}n^2\equal{}m^2, x^2z\equal{}2mn, x^4\plus{}y^4\equal{}m^2\plus{}n^2$
so there exists 2 intergers $ k,l$, $ k>l>0, (k,l)\equal{}1, k\minus{}l$ is an odd
-> $ y^2\equal{}k^2\minus{}l^2, m\equal{}k^2\plus{}l^2, n\equal{}2kl$
Moreover, we have $ x^4\equal{}2n^2\equal{}8k^2l^2$ is a perfect square (Wrong !!!)
Case 2: $ y$ is an even
We have $ y^4\equal{}2mn, x^2z\equal{}m^2\minus{}n^2, x^4\plus{}y^4\equal{}m^2\plus{}n^2$
->$ x^4\equal{}(m\minus{}n)^2$ -> $ x^2\equal{}m\minus{}n$ ->$ 0<z\equal{}\frac{m\minus{}n}{m\plus{}n}<1$ (Wrong !!!)
So there is no root of this equation
 :(

Sorry,$ z$should be $ m\plus{}n$
But we can continue mathboy2710's proof and use Fermat's method to get our contradiction :)


Use the formula $180(n-2)$ and set it up as:

$180(n-2)-\frac{180(n-2)}{n}=2190$.

$n=15.3$, so $n=16$. But shouldn't it come out even :? Did I do something wrong?

He never said that the polygon was regular... ;)

Any number bigger than 15 will work, as a 15 sided polygon will have 2340 total degrees. Since the angle taken out wasn't specific, we can take out any angle big enough to satisfy the other condition.

He said it was a convex polygon...so none of the angles can be $\geq 180^\circ$.  That means there is some limit.  I haven't calculated out the problem yet, though, so I don't know the answer.

It'd have to be 15 then.
16 would have 2520 total degrees, and the subtracted angle would have to be 330, which isn't possible.

[quote="4everwise"]If the sum of all the angles except one of a convex polygon is $2190^\circ$, then how many sides does the polygon have?[/quote]

$180(n-2)-2190 < 180$
$180(n-2) < 2370$
$n = \lfloor \frac{237}{18} \rfloor+2$

Hmm, i don't have a calc, but I think $237/18 \approx 13$, so $15$?

The answer must be 15.

[hide]a 15 sided figure has interior angles totaling 2340, so the other angle must be 150 degrees. This is possible. A 16 sided figure has interior angles meausring a total of 2520, meaning that the other angle must be 330, but that is impossible since it's a convex polygon. A 14 sided figure has interior angles totaling 2160, not enough for 2190. [/hide]


induction
linear algebra
matrix
algebra
function
domain
linear algebra unsolved

Prove that $rank[\lambda I-A$  |  $B]=n$ for any $\lambda\in{\mathbb C}$ [b]iff[/b] $rankR=n$, where $R=[B$ | $AB$ | $A^2B$ | ... | $A^{n-1}B]$, $A\in{\mathbb R}^{n \times n}$,$B\in{\mathbb R}^{n \times m}$,$n$,$m\in{\mathbb N-\{0\}}$.

 :D

Realy, nobody is trying to solve this task?  :huh:  :D       IT IS FREE !!!  :P  :D

....and who tell what's the name of this theorem will get extra 10 point bonus ... :P  :P  :P  :D

[quote="Andrei Berechet"]Really, only a control engineer cares for the Popov - Belevitch - Hautus criterion. Check your Systems Theory courses and you'll find your answer... or laboratory practice notes ;)[/quote]

hey , i know about this. i just put it here for the others to see if they care about this theorem  :P  :D

Hey guys,you are discussing this theorem and no result.I'm wondering if any of you can post a solution.I tried to solve it,but failed. :(

[quote="Andrei Berechet"]Ovidiu, you just have to open Prof. Oara's TS II lecture notes in slide 21 of the second chapter and post the proof of the criterion.

However, it is a pretty simple proof once you have some knowledge of systems theory results, and also not very interesting if you're not into linear systems, that's why I haven't posted it myself.[/quote]

yes, i know  :D ... check http://riccati.pub.ro/   :P  for lectures about control systems  :D

Jeane,post it here on the Forum. :w00tb:

ok, here is the half of the sollution:

$a)$ Let $R=[B | AB |...|A^{n-1}B]=n$. We have to prove that $rank[\lambda I_n - A$   $B]=n$, for any $\lambda \in {\mathbb C}$.
Well, let asseume that $rank[\lambda I_n - A$   $B]<n => \exists q \neq 0$ such that $q^T(\lambda I_n - A)=0$ and $q^TB=0$,
so we get $\lambda q^T=q^T A$ and by induction we get $q^T A^k = {\lambda}^k q^T$, for any integer $k\geq 1$ .
Multiplying by $B$ in right part, and using $q^T B=0$ we get $q^T A^k B= {\lambda}^k q^T B =0$ , so we get finnaly
$rank[B|AB|...|A^{n-1}B]<n$ which is a contradiction.

(qed)

P.S.: For the $b)$ part i mean the part were we have to prove $rank[\lambda I_n - A$   $B]=n => rank[B|AB|...|A^{n-1}B]=n$,
i can solve but to understand it you have to read some books about control systems  :D  :lol:  :D  :roll:.

OBSERVATIONS::::
In the expression $rank[\lambda I_n - A$   $B]=n$, $A$ is not multiplying with $B$. The matrix $M=[\lambda I_n - A$   $B]$
represents the concatenations of the matrices $\lambda I_n - A$  AND $B$.

[quote="cezar lupu"]Jeane,I read the first part of your solution and it is clearly that I have to study some Control engenieering.Maybe you'll help me in recommeding some books. ;)[/quote]

ok, cezare..., start and learn from oara and stefan courses on http://riccati.pub.ro/ then try to use google to
find some speciality books in control systems  :P  :D

[quote="Andrei Berechet"]@jianuovidiu: Yup, actually we've studied some Systems Theory together once, when Paul was away (with me as teaching assistant) and I was in the class with Prof. Stefan when you took the exam. If I rememeber correctly, you were the guy who went to that TS contest in Serbia  :? Btw, what was your grade in TSII?

@cezar lupu: Systems theory is mainly applied mathematics, and it's relevant to you only in the extent in which you're intrested in this particular domain (control systems, financial and all sorts of phaenomena modelling,...) For an excellent introduction to the subject, I recommend the URL Ovidiu quoted (read the lecture notes for Systems Theory I and II and for Optimization Techniques).[/quote]

nice to meet u here :D

p.s.: my mark was 10.00 at TS2  :D  :D  :D  :P  :roll:


Two different numbers are selected at random from the set $\{2,3,5,7,11,13,17,19,23,29\}$. What is the probability that their sum is 24?

A) 1/30
B) 2/45
C) 1/15
D) 1/10
E) 2/15

As usual, leave this problem to beginners if this is too easy for you. :)

MMPC: Michigan Mathematics Prize Competition

[hide="Hint"]
The number of subsets you can make from a set of $n$ terms is $2^n$
[/hide]

Your hint is not useful for this problem. We're only looking at two-element subsets.

[hide]The pairs (5,19), (7,17), and (11,13) are good, and there are 10 choose 2 or 45 ways total. the probability is 1/15. [/hide]

[hide]There are three groups whose sum is $ 24 $. They are (5,19); (11,13); (7,17)

The combination of chosing 2 numbers out of 10 numbers are : 10C2.

So the probability to get the sum of 24 is : $ \frac{3}{10C2}=\frac{1}{15} $

The answer is $ C $[/hide]

The pairs (5,19), (7,17), and (11,13) can have the sum of 24 while there are 10*9/2 = 45 choices. so the answer is 3/45 = 1/15. [color=red][b]C[/b][/color]


how many minutes do we get to do 1 individual round?
are calculators allowed?
thanks.

The answers to all of your questions are [url=http://web.mit.edu/hmmt/www/rules.shtml]here[/url].


Given positive integers a and c and integer b,prove that there exists a positive integer x such that $ a^{x}\plus{}x\equiv b(mod c)$ :?:

See here http://www.mathlinks.ro/Forum/viewtopic.php?p=354533#354533 .


calculus
integration
inequalities
geometry
trigonometry
calculus computations

Find the volume of given solid bounded by z = x, y = x, x + y =2, and z = 0
I dont know how to find limit of x, y. Help me, please.

These are four planes, which determine a tetrahedron.

Only two of those planes have $ z$ in the equation, so let's look at $ z$. We have $ z\ge 0$ and $ z\le x$ (all intersections of these planes are in the closed first octant), so that's our $ z$ range. We get one more piece of information from that: for there to be anything, we must have $ x>0$.
Now, the base is the triangle bounded by the three lines $ x\equal{}0,y\equal{}x,x\plus{}y\equal{}2$ in the $ xy$-plane. Can you figure that out?

The range is
$ 0\le z\le x,\,\,but\,\,0\le x\le y\,\,or\,\,0\le x\le 2\minus{}y,\,\,or\,\,0\le x\le 2$ ? and why do we have that?

All three of those inequalities are true. None describes the region completely on its own.

Draw the triangle, then pick one variable to be on the outside with static bounds and one on the inside with bounds depending on the outer variable. One choice is significantly more convenient than the other.

But most importantly- [i]draw the triangle[/i].

When I draw the triangle, I see there are two regions:
$ {{D}_{1}}\equal{}\left\{ \left( x,y \right)|0\le x\le 1,x\le y\le 2\minus{}x \right\},{{D}_{2}}\equal{}\left\{ \left( x,y \right)|y\le x\le 2\minus{}y,0\le y\le 1 \right\}$

No, there aren't two regions. One of those is not actually enclosed by the given lines.

And the other problem: How to find range of $ \varphi$ when calculating the area enclosed by $ r\equal{}n.\cos \varphi$ and $ r\equal{}\cos \left( n\varphi  \right)$

[quote="chuong01"]And the other problem: How to find range of $ \varphi$ when calculating the area enclosed by $ r \equal{} n.\cos \varphi$ and $ r \equal{} \cos \left( n\varphi \right)$[/quote]
Please help me. Thanks


You can give 4-8 numbers to make 24. Only 1 and 2 digit numbers, decimals allowed. Negatives also allowed. Exponents and roots NOT allowed.

Starting: $ 1/2,3,3,4,6,36$

um...

theres other 24 game going on already.

The stuff is slightly different this time.
There are 4-8 numbers, not all are positive integers

[quote="vallon22"]guys, plz play on the other 24 game thread in this forum
we do not need 100 games going on if they are similar
thanks[/quote]

You are right, but it's too easy.


algebra
polynomial
geometry
geometric transformation
reflection
number theory proposed
number theory

Everyone knows the lemma :
[quote]
If $ p$ is a prime number and $ q$ is prime such that divides $ \frac {x^{p} \minus{} 1}{x \minus{} 1}$, then $ q\equiv 1 (\mod p)$ or $ p \equal{} q$.
[/quote]

Prove the general form:
If $ n$ is a nature number and $ q$ is prime such that divides $ \frac {\Phi_{n}(x)}{x \minus{} 1}$, then $ q\equiv 1 \mod n$ or $ q|n$. :)

[quote="BG Yoda"]Prove
If $ n$ is a nature number and $ q$ is prime such that divides $ \frac {\Phi_{n}(x)}{x \minus{} 1}$, then $ q\equiv 1 \mod n$ or $ q|n$. :)[/quote]
Yimin Ge, [url=http://reflections.awesomemath.org/2008_2/cyclotomic.pdf][i]Elementary Properties of Cyclotomic Polynomials[/i], Mathematical Reflections 2008/2[/url].

Or just http://www.mathlinks.ro/viewtopic.php?t=126566  :wink:


let triangle ABC satisfy $sinA+sinB+sinC\leq 1$
prove that min {A+B,B+C,C+A}$<30^{o}$

well wlog we can say C is the obtuse angle so we just gotta prove that

$A+B\leq30$

let's do this by contradiction or w/e it's called

if $A+B>30$ then $C<150$

and then $\sinh$ will be greater than $0.5$

and the greatest possible value of $\sinh+\sinh$ will also be greater than $0.5$

so that doesn't work.

and just to make sure, if $C \geq 150$ then $\sinh \leq 0.5$
and $\sinh+\sinh \leq 0.5$ also

it all works out

and i'm not coherent right now so sorry

[hide]If you think geometrically it  become really obvious..
[hide="hint"]Think of a  triangle inside a circle whose diameter is 1.[/hide][/hide]

ugh gyan's a genius .-.

i think your solution works a lot better than mine

[quote="Treething"]well wlog we can say C is the obtuse angle so we just gotta prove that

$A+B\leq30$

let's do this by contradiction or w/e it's called

if $A+B>30$ then $C<150$

and then sinC will be greater than 0.5

and the greatest possible value of sinA+sinB will also be greater than 0.5

so that doesn't work.

and just to make sure, if C $\geq$ 150 then sinC $\leq 0.5$
and sinA+sinB $\leq$ 0.5 also

it all works out

and i'm not coherent right now so sorry[/quote]
it's a joke ,it's all cọnjecture
[quote]if A+B>30 then C<150

and then sinC will be greater than 0.5[/quote]
are you sure?

[quote="C'est lavie"]i hope on other hand[/quote]
Not sure what the comment meant exactly .. in any case, more explanation  of the hint given above:
[hide]Draw a  circle passing through $ABC$,(let  $O$ is the center of the circle)  if this radius is $R$ then:
$\frac{a}{\sin A}= \frac{b}{\sin B}=\frac{c}{\sin C}= 2R$ tells us for that triangle  $(1/2R)(a+b+c) \le 1 \implies (a+b+c) \le 2R$
WOLG assume $a$ is the largest side,  and remembering  ($b+c>a$
we have $R \ge a+b+c > 2a \implies a<R$
Since the chord $BC$ is less than half the diameter (Or  $\angle BOC <60^{o}\implies \angle A > 150^{o}\implies B+C < 30^{o}$

(Of course, it takes much longer to explain than visualize)

Of course, the direct way is  quite simple too..)

[hide]We start with:
$\sin A = \sin (\pi-(B+C)) = \sin (B+C)$  and

$\sin (B+C) = \sin B \cos C+\sin C \cos B < \sin B+Sin C$
(Both sin and cos values are less than or equal to  1,   cosine is always less than 1 for a proper triangle )
so
 $1 \ge \sin A+\sin B+\sin C > \sin A+\sin A = 2 \sin A$
$\implies \sin (A < (1/2)$

(This means $A$ is either greater than $150^{o}$ or less than $30^{o}$)
(since all three angles can not be less than $30^{o}$, one of them (say  $A$ - assuming it to be the largest) has to be greater than $A>150^{o}$
This $\implies B+C < 30^{o}$

[/hide][/hide]


Let $G$ be the centroid of a triangle $ABC$, and $M$ be the midpoint of $BC$.  Let $X$ be on $AB$ and $Y$ on $AC$ such that the points $X$, $Y$, and $G$ are collinear and $XY$ and $BC$ are parallel.  Suppose that $XC$ and $GB$ intersect at $Q$ and $YB$ and $GC$ intersect at $P$.  Show that triangle $MPQ$ is similar to triangle $ABC$.

Sketch: It is enough to prove $PQ||BC$, $MP||AB$, $MQ||AC$. By Ceva $AM$, $BY$,$CX$ concur - denote this point by $N$. From this we get by reversed Ceva (for point $N$ in triangle $GBC$) and Thales $PQ||BC$. Call intersetion of $CG$ and $BG$ with sides $C'$ and $B'$ respectively. By Menelaos in CC'X and line GB we have $3QX=CQ$ and by Thales $3GP=PC$. Now:

$\frac{C'P}{PC}=\frac{C'G+GP}{PC}=\frac{1}{3}+\frac{CG}{2PC}=\frac{1}{3}+\frac{GP+PC}{2PC}=1$ Hence $MP||AB$ and similarly $MQ||AC$. QED

Also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=43338 .

  darij

Since $XY||BC$, then from Ceva, it follows that $AM, BY$ and $CX$ are concurrent. Let $\alpha=\angle{BCP}$ and $\beta=\angle{ACP}$. Then from the Sine Theorem, $\frac{BP}{\sin\alpha}=\frac{BC}{\sin(\angle{BPC})}$ and $\frac{PY}{\sin\beta}=\frac{YC}{\sin(\angle{YPC})}=\frac{\frac{1}{3}AC}{\sin(\angle{BPC})}$. So $\frac{BP}{PY}=\frac{BC\sin\alpha}{\frac{1}{3}\sin\beta}$. Denote $C_{1}$ by the midpoint of $AB$. Using similar arguments, $\frac{BC_{1}}{C_{1}A}=1=\frac{BC\sin\alpha}{AC\sin\beta}$. So $\frac{BP}{PY}=3$. From Ceva for $\triangle{BGC}$ it results that $QP||BC||XY$. So $\frac{BQ}{QG}=3$. Let $B_{1}$ be the midpoint of $AC$. It results that $Q$ is the midpoint of $BB_{1}$. So $QM||AC$. Using similar arguments $PM||AB$. So $\triangle{ABC}$ and $\triangle{MPQ}$ have parallel sides. So they're similar.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1991Problem1

Vo Duc Dien

suppose $CG$ intersect $AB$ at $R$ and $BG$ intersect $AC$ at $S$ :
lemma 1:$M$and$Q$and$R$ are on same line.
proof of lemma 1:
on triangle $GBC$ we must proof:
$\frac{RG}{RC} . \frac{CM}{MB} . \frac{BQ}{QG} =1 $
so we must proof $\frac{BQ}{QG} = 3$ we know that :
$\frac{BQ}{QG} = \frac{CB}{GC} . \frac{\sin QCB}{\sin QCG} \rightarrow  \frac{BQ}{QG}= \frac{CB}{GC} . \frac{GC}{GX} = 3$
end of proof of lemma1.
proof of problem:
so $QMP=\angle A$ by menlaues on triangle $RMC$ and point $B,Q,G$ and also $SMB$ and point $C,P,G$ so we know that:
$\frac{MQ}{AC} = \frac{MP}{AB}$
end of proof of problem.
oh excuse me but im so lazy and this solution is so short!

Let $\Delta MM_BM_C$ be the medial triangle WRT $\Delta ABC$ and then apply Ceva on $\Delta ABC$, $\Delta GBC$, $\Delta BM_CC$ and $\Delta BM_BC$ to obtain $\Delta MPQ$ $\sim$ $\Delta ABC$

@below I didn't expect a guy like you to bash such a beautiful geo :noo:

This shows the power of bary :P
As usual take $A,B,C=(1,0,0),(0,1,0),(0,0,1)$.Then obtain $M=(0,\dfrac{1}{2},\dfrac{1}{2}),X=(\dfrac{1}{3},\dfrac{2}{3},0),Y=(\dfrac{1}{3},0,\dfrac{2}{3}),G=(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}),P=(\dfrac{1}{4},\dfrac{1}{4},\dfrac{1}{2}),Q=(\dfrac{1}{4},\dfrac{1}{2},\dfrac{1}{4})$.We are ready for computation.Since $|PQ|^2=-a^2yz-b^2xz-c^2xy$ where $PQ$ is the displacement vector, so its easy to see $|PQ|^2=\dfrac{a^2}{16},|MQ|^2=\dfrac{b^2}{16}$.Thus $MPQ\sim ABC$ with similarity factor $\dfrac{1}{4}$.This center of similarity is $AM\cap BP \cap CQ$.

[hide=Proof]
Let $M$ be the midpoint of $AB$and $N$ the midpoint of $AC$. Let $A''M$ meet $BG$ at $X$. Then $X$ must be the midpoint of $A''M$ (an expansion by a factor $2$ center $B$ takes $A''M$ to $CA$ and $X$ to $N$). Also $\frac{BX}{BN} = \frac{1}{2}$ and $\frac{BG}{BN} = \frac{2}{3}$, so $XG = \frac{BX}{3}$. Let the ray $CX$ meet $AB$ at $Z$. Then $ZX$ = $\frac{CX}{3}$. (There must be a neat geometric argument for this, but if we take vectors origin $B$, then $BX = \frac{BN}{2} = \frac{BA}{4}$ + $\frac{BC}{4}$, so $BZ = \frac{BA}{3}$ and so $XZ = \frac{1}{3}$ ($\frac{BA}{4} - \frac{3BC}{4}) = \frac{CX}{3}$.) So now triangles $\triangle BXC$ and $\triangle ZXG$ are similar, so $ZG$ is parallel to $BC$, $Z$ is $B'$, and $X$ is $C''$. But $A''X$ is parallel to $AC$ and $\frac{1}{4}$ its length, so $A''C''$ is parallel to $AC$ and $\frac{1}{4}$ its length. Similarly $A''B''$ is parallel to $AB$ and $\frac{1}{4}$ its length. Hence $A''B''C''$ is similar to $\triangle ABC$.[/hide]

Surprised this solution hasn't been posted yet!  Maybe it's fundamentally similar to some of the others.

First remark by similarity ratios that $\tfrac{PC}{PG} = \tfrac{BQ}{QG} = 2$, so $PQ\parallel BC$.

Let $D$ and $E$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively.  Triangles $EGY$ and $EBC$ are homothetic, as are triangles $PGY$ and $PCB$; looking at both centers of homothety shows that $E$, $P$, and $M$ are collinear.  Likewise, $D$, $Q$, and $M$ are collinear.

Therefore $\triangle ABC\sim\triangle MED\sim\triangle MPQ$.


trigonometry
geometry
inradius
incenter
ratio
area of a triangle
Heron\u0027s formula

In a triangle ABC, prove that
$tan\frac{A}{2} = \frac{r_a}{s}$
$tan\frac{A}{2}tan\frac{B}{2} + tan\frac{B}{2}tan\frac{C}{2} + tan\frac{C}{2}tan\frac{A}{2} = 1$
$tan\frac{A}{2}tan\frac{B}{2}tan\frac{C}{2} = \frac{r}{s}$
where $r_a$ denotes the exradius of the circle opposite to A, $r$ the inradius and $s = \frac{a+b+c}{2}$

[color=blue] [b](1)[/b] [/color] Let $R_a, S_a, T_a$ be the tangency points of the excircle $(E_a)$ against the vertex $A$ with the lines $a = BC, b = CA, c = AB$. Since the segment lengths of tangents form any vertex to the two corresponding tangency points are equal,

$AT_a = AB + BT_a = c + BR_a$
$AS_a = AC + CS_a = b + CR_a$
$AT_a + AS_a = c + b + BR_A + CR_A = c + b + a = 2s$
$AT_a = AS_a = s$

Since $AE_a$ is a bisector of the angle $\measuredangle A$ and $E_aT_a \perp AT_a$,

$\tan{\frac{\widehat A}{2}} = \frac{E_aT_a}{AT_a} = \frac{r_a}{s}$

[color=blue] [b](2)[/b] [/color] Since $\frac{\measuredangle C}{2} = \frac \pi 2 -\left(\frac{\measuredangle A}{2} + \frac{\measuredangle B}{2}\right)$,

$\tan{\frac{\measuredangle C}{2}} = \frac{1}{\tan{\left(\frac{\measuredangle A}{2} + \frac{\measuredangle B}{2}\right)}} = \frac{1 - \tan{\frac{\widehat A}{2}} \tan{\frac{\widehat B}{2}}}{\tan{\frac{\widehat A}{2}} + \tan{\frac{\widehat B}{2}}}$

$\tan{\frac{\widehat A}{2}} \tan{\frac{\widehat B}{2}} + \tan{\frac{\widehat B}{2}} \tan{\frac{\widehat C}{2}} + \tan{\frac{\widehat C}{2}} \tan{\frac{\widehat A}{2}} =$

$= \frac{\tan{\frac{\widehat A}{2}} \tan{\frac{\widehat B}{2}} \left( \tan{\frac{\widehat A}{2}} + \tan{\frac{\widehat B}{2}} \right) + \tan{\frac{\widehat B}{2}} \left( 1 - \tan{\frac{\widehat A}{2}} \tan{\frac{\widehat B}{2}} \right) + \left( 1 - \tan{\frac{\widehat A}{2}} \tan{\frac{\widehat B}{2}} \right) \tan{\frac{\widehat A}{2}}}{\tan{\frac{\widehat A}{2}} + \tan{\frac{\widehat B}{2}}} =$

$= \frac{\tan{\frac{\widehat B}{2}} + \tan{\frac{\widehat A}{2}}}{\tan{\frac{\widehat A}{2}} + \tan{\frac{\widehat B}{2}}} = 1$

[color=blue] [b](3)[/b] [/color] Let $R, S, T$ be the tangency points of the incircle $(I)$ with the sides $a = BC, b = CA, c = AB$. The triangle area $\triangle$ is equal to

$\triangle = \frac 1 2 IR \cdot BC + \frac 1 2 IS \cdot CA + \frac  1 2 IT \cdot AB = \frac r 2 (a + b + c) = rs$

The distance from a triangle vertex to the tangency point with the incircle is equal to the difference between the triangle semiperimeter and the opposite side:

$2s = a + b + c = BR + CR + CS + AS + AT + BT = 2(BR + CS + AT)$

$AS = AT = s - (CS + BR) = s - (CR + BR) = s - a$

and similarly, $BT = BR = s - b$, $CR = CS = s - c$. Hence,

$\tan{\frac{\widehat A}{2}} = \frac{r}{s - a}$,  $\tan{\frac{\widehat B}{2}} = \frac{r}{s - b}$,  $\tan{\frac{\widehat C}{2}} = \frac{r}{s - c}$

$\tan{\frac{\widehat A}{2}} \tan{\frac{\widehat B}{2}} \tan{\frac{\widehat C}{2}} = \frac{r^3}{(s - a)(s - b)(s - c)}$

Using Heron's formula for the triangle area $\triangle = \sqrt{s(s - a)(s - b)(s - c)}$,

$\tan{\frac{\widehat A}{2}} \tan{\frac{\widehat B}{2}} \tan{\frac{\widehat C}{2}} = \frac{r^3 s}{\triangle^2} = \frac{r^3s}{r^2s^2} = \frac r s$

[quote="Myth"](2) also has a geometry proof, doesn't it? ;)[/quote]
Let $a = BC, b = CA, c = AB$ be the triangle sides. The incenter $I$ is the center of mass of the material points with masses $a, b, c$ at the vertices $A, B, C$. The distances from the vertices to the incenter are

$AI = \sqrt{bc \tan{\frac \beta 2} \tan{\frac \gamma 2}}$

$BI = \sqrt{ca \tan{\frac \gamma 2} \tan{\frac \alpha 2}}$

$CI = \sqrt{ab \tan{\frac \alpha 2} \tan{\frac \beta 2}}$

and the moment of inertia of the material points at $A, B, C$ with respect to the incenter $I$ (the center of mass) is

$J_I = a \cdot AI^2 + b \cdot BI^2 + c \cdot CI^2  = abc \left(  \tan{\frac \beta 2} \tan{\frac \gamma 2} + \tan{\frac \gamma 2} \tan{\frac \alpha 2} + \tan{\frac \alpha 2} \tan{\frac \beta 2} \right)$

Let $K$ be the intersection of the bisector of the angle $\alpha = \measuredangle CAB$ with the side $a = BC$. Length of the angle bisector $AK$ is given by

$AK^2 = bc \left( 1 - \frac{a^2}{(b + c)^2} \right)$

The angle bisector divides the side $a = BC$ in the ratio of the adjacent sides $\frac{CK}{BK} = \frac{b}{c}$ and the incenter $I$ divides the angle bisector $AK$ in the ratio $\frac {AI}{IK} = \frac{b + c}{a}$. Hence,

$BK = \frac{c}{b + c} \cdot CB = \frac{ac}{b + c}$

$CK = \frac{b}{b + c} \cdot CB = \frac{ab}{b + c}$

$IK^2 = \frac{a^2}{(a + b + c)^2} \cdot AK^2 = \frac{abc \cdot a}{(a + b + c)^2}  \left( 1 - \frac{a^2}{(b + c)^2} \right)$

The moment of inertia with respect to the point $K$ is

$J_K = a \cdot AK^2 + b \cdot BK^2 + c \cdot CK^2 =$

$= abc \left( 1 - \frac{a^2}{(b + c)^2} \right) + abc\ \frac{ac}{(b + c)^2} + abc\ \frac{ab}{(b + c)^2} =$

$= abc \left( 1 - \frac{a^2}{(b + c)^2} \right) + abc\ \frac{a}{b + c}$

On the other hand, this moment of inertia is equal to

$J_K = J_I + (a + b + c) \cdot IK^2 = J_I + abc\ \frac{a}{a + b + c} \left( 1 - \frac{a^2}{(b + c)^2} \right)$

Consequently,

$J_I = J_K - (a + b + c) \cdot IK^2 =$

$= abc \left( 1 - \frac{a^2}{(b + c)^2} \right) + abc\ \frac{a}{b + c} - abc\ \frac{a}{a + b + c} \cdot \left( 1 - \frac{a^2}{(b + c)^2} \right) =$

$= abc \left( 1 - \frac{a^2}{(b + c)^2} \right) \frac{b + c}{a + b + c} + abc\ \frac{a}{b + c} = $

$= abc\ \frac{b + c - a}{b + c} + abc\ \frac{a}{b + c} = abc$

As a result,

$J_I = abc \left(  \tan{\frac \beta 2} \tan{\frac \gamma 2} + \tan{\frac \gamma 2} \tan{\frac \alpha 2} + \tan{\frac \alpha 2} \tan{\frac \beta 2} \right) = abc$

which implies that

$\tan{\frac \beta 2} \tan{\frac \gamma 2} + \tan{\frac \gamma 2} \tan{\frac \alpha 2} + \tan{\frac \alpha 2} \tan{\frac \beta 2} = 1$

Is this a geometric solution ?!?

WoW! You killed me!  :lol:  :lol:  :lol:

Anyway: $\tan A/2\tan B/2=\frac{4r^2}{(b+c-a)(a+c-b)}$ - sum up and apply two formulas for the area  of the triangle: Heron's formulae and $S=pr$.

That's all.

[quote="Myth"]WoW! You killed me!  :lol:  :lol:  :lol: [/quote]

Oh no!  Myth is dead!

Well, I suppose that in the grand scheme of things, that makes me significantly smarter.  I've moved up a notch.


We have infinity chessboard with positiv real numbers such that each number is equal $\frac{1}{4}$ of sum four neighbouring numbers. Prove that all numbers on this chessboard are equal. This task is easy when we have to prove this for positive integers :) (because exist smallest element on this chessboard).

Hm.. Then, I don't think this is an NT problem..  :maybe:


Let $ p(x)\equal{}x^{n} \plus{}x^{n\minus{}1}\plus{}...\plus{}x ,n>1$ and $ x$ is Real number.Prove that the remainder when $ p(x)$ is divided by $ p_{1}(x)\equal{}x^{2}\minus{}2x$ is polynom of the form $ kx$ when $ k$ is natural number.

 I hope you understand the conditions of the problem  :blush:

[hide="Solution"]$ p(x) \equal{} (x^2 \minus{} 2x)q(x) \plus{} r(x)$.
$ \deg r < \deg (x^2 \minus{} 2x) \equal{} 2\implies r(x) \equal{} ax \plus{} b$.

$ x^n \plus{} x^{n \minus{} 1} \plus{} \cdots \plus{} x(1 \minus{} a) \minus{} b$ is divisible by $ x^2 \minus{} 2x$, but $ x|x^2 \minus{} 2x$, so $ x|x^n \plus{} \cdots \minus{} b$, which means $ b$ must be divisible by $ x$ as well. So let $ b \equal{} cx$.

Then $ r(x) \equal{} ax \plus{} cx \equal{} (a \plus{} c)x \equal{} kx$, Q.E.D.[/hide]

Edit: Oh, didn't see that $ k$ had to be natural...  :oops:

If u assumed $ b$ is constant, you can't switch it to being a monomial. Besides, you need to prove that $ a\plus{}c$ is natural.

[hide]$ P(x) \equal{} x(x \minus{} 2)Q(x) \plus{} ax \plus{} b$

Putting $ x \equal{} 0$ we get $ P(0) \equal{} b$. But $ P(0) \equal{} 0$, hence $ b \equal{} 0$ and $ R(x) \equal{} ax$. Now put $ x\equal{}2$ to get $ P(2)\equal{}2a\implies a\equal{}{P(2)\over 2}\equal{}2^{n\minus{}1}\plus{}2^{n\minus{}2}\plus{}\dots\plus{}2\plus{}1\equal{}{2^n\minus{}1\over 2\minus{}1}\equal{}2^n\minus{}1$. Therefore $ R(x)\equal{}(2^n\minus{}1)x$. QED[/hide]


Does anyone have the USPhO exam from 2009?  I was wondering where did you guys get all these problems.

They were on the webassign website as per directions from [url]http://www.aapt.org/physicsteam/2010/registration.cfm[/url], but they are no longer there because the exam window has started. If anyone has them in PDF form, maybe they could post it here.


Hey,

Does anyone have an XBOX Live game that comes with one of those free 2-month trials?  If so, I would really appreciate it if you could PM me the number on it.  I can send you something in return, just specify what you want.

i have some

though u'd have to wait til friday evening as they're at my dad's house (i go there for x-mas)

-jorian

i don't really need it so i can probably give it for free :D

-jorian

Anyone else?  I'm getting really bored without Halo!  lol I should probably practice some math but...

[quote="jhredsox"]I can get them tonight

I might not have any, I might have thought I had more than I really do, I'll check

-jorian[/quote]

OK, thanks in advance!

oh, wait, right now im at my dad's house

i can't get out of bed (very sorry) so i'll look early tomorrow

-jorian


Let $ A,B,C$ be the digits of the prime number $ AAABBBC$ such that $ A\minus{}B\equal{}4$ .

Find $ A,B,C$.

Thanks .

(I wasnt sure if it fits here or not, so I apologize if it doesnt).

There are fairly few viable options, so exhaustion seems a straightforward method.  There are in fact two solutions.

It seems that this is not a good question.
By taking $ mod 2,3,5$ we know $ C\equal{}1$, but then , if as JoeBlow say, there are two solutions, we will need to divide the solutions by quite a number of primes to ensure that they are really solutions to the question.


linear algebra
matrix
algorithm
combinatorics unsolved
combinatorics

Hi,

I guess it should not be in the algorithm section.. But anyways here it is. The name above was chose by me and i hv every reason to believe that i did not choose it right.

Anyways, here is the problem. It got me interested and frustrated (for i could not solve it). Hope to get some help frm u guys.

[b]The problem:

The statement is dazzingly simple. It asks u to find the number of ways in which u can put 0's in a m*n matrix such that every row and every column of the matrix contains an odd number of zeros.
[/b]

[b]Below i tell u my thoughts. U can skip this if u want[/b] --- But spare it a look if u can. I focused on the special case of square matrices (n*n) matrices...Lets say for the moment that n is odd and is greater than or equal to 5.

Clearly we have factorial(n) ways in which we keep exactly one zero in every row and every column. Then we have the other way in which we fill the entire matrix..Now we come to ways in which we can have some row/column with 3 zeroes ...and thats where i am stuck...

I tried another approach wherein I started with an already filled n*n matrix...Then I planned to drop some entries to see if I can guide myself to a better path...I could not.
I think I have tried both the ways - filling an initially empty matrix and dropping values off an initially filled matrix - but I have not made any headway...

[b]Please Try solving this and let me know ur thoughts..[/b]

Here's a (simpler) problem you may want to look at first:

[b]How many ways can you fill an $ n \times n$ matrix with 0's such that every row contains an odd number of zeroes?[/b]

I cannot even solve this.... :noo: ....Now u guys solve them both...

Anyways, i will tell u what stumped me..Earlier I thought i just need to look at the number of ways u can fill a single row with odd number of zeroes for a particualr n, lte sya n = 5....Then u can hv 1/3/5 number of zeroes in a single row which means u hv 3 choice for a single row...so i thought 3^5 is the answer for this case...but i was wrong...

U can hv 3 zeroes or 1 zero in a single row in a variety of ways...and that negates my solution..

So i am again stuck :noo: Please solve this one..

$ 2^{(m\minus{}1)(n\minus{}1)}$

We use 1 instead of 0 in this proof.

Give the upper left (m-1)(n-1) subgrid an arbitrary configuration.

All squares except lower right are trivially determined. The sum mod 2 of the last row and last column at this point are the same as the sum of all 1's in the (m-1)(n-1) subgrid mod 2, so therefore the last row at this point=last column at this point mod 2, so the lower right corner square is then determined uniquely as well.


Am i the only person who is having a problem viewing results/answers/solutions? (the buttons are inactive and no matter how much i click them, theydon't work (tried in several browsers as well.) i have a mac btw

it didn't work for me either. and i don't have a mac...

it seems to be down...
if it helps, you can do the saab mathcounts problems one-by-one
[url=http://www.saab.org/mathdrills/mathc_table.cgi]here[/url].


inequalities
LaTeX
inequalities unsolved

The positive reals $ x,y$ and $ z$ are satisfying the relation $ x \plus{} y \plus{} z\geq 1$. Prove that:
$ \frac {x\sqrt {x}}{y \plus{} z} \plus{} \frac {y\sqrt {y}}{z \plus{} x} \plus{} \frac {z\sqrt {z}}{x \plus{} y}\geq \frac {\sqrt {3}}{2}$

[i]Proposer[/i]:[b] Baltag Valeriu[/b]

[quote="Inequalities Master"]It is easy to prove by AM-GM that:
$ \frac {x}{y + z}\geq \frac {3}{2}\cdot \frac {x^{\frac {3}{2}}}{x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}}}$, etc,
So it suffices to show that:
${ \sum{\frac {x^2}{x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}}}}\geq \frac {1}{\sqrt {3}} \iff 3(x^2 + y^2 + z^2)^2\geq (x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}}})^2$
It suffices to show that:

${ 3(x^2 + y^2 + z^2)^2(x + y + z)\geq (x + y + z)(x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}}})^2$

By CS we get that:

${ (x^2 + y^2 + z^2)(x + y + z)\geq (x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}}})^2$ and

$ 3(x^2 + y^2 + z^2)\geq (x + y + z)^2\geq (x + y + z)$, so we are done.[/quote]
Nice,but it is hard (how can you find $ \frac {x}{y + z}\geq \frac {3}{2}\cdot \frac {x^{\frac {3}{2}}}{x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}}}$
My proof is only used AMGM and CS
After used CS we need to prove $ 2(\sum x)^2\ge \sum \sqrt 3 \sqrt x (z + y)$
but $ \sqrt x \le \frac{3x + 1}{2\sqrt 3}$
then the last is very cleary (with$ \sum x\ge 1$

[b]Allnames wrote[/b]:
[quote]Nice,but it is hard (how can you find  $ \frac {x}{y + z}\geq \frac {3}{2}\cdot \frac {x^{\frac {3}{2}}}{x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}}}$ [/quote]

$ \iff 3x^{\frac {1}{2}}(y + z)\leq2({x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}})}$
By AM-GM we know that ${ x^{\frac {3}{2}} + y^{\frac {3}{2}} + y^\frac {3}{2}}\geq 3x^{\frac {1}{2}}y$ and
${ x^{\frac {3}{2}} + z^{\frac {3}{2}} + z^\frac {3}{2}}\geq 3x^{\frac {1}{2}}z$

That's it. This is a well-known lemma used commonly used in proving Nesbit's inequality.

Edit: Allnames,  I don't understand your proof, especially the line with $ \sqrt {x}\leq 3x + 12\sqrt {3}$.(when do you have equality?)
Could you detail a little bit your proof?

[quote="Inequalities Master"]Allnames wrote:
[quote]Nice,but it is hard (how can you find  $ \frac {x}{y + z}\geq \frac {3}{2}\cdot \frac {x^{\frac {3}{2}}}{x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}}}$ [/quote]

$ \iff 3x^{\frac {1}{2}}(y + z)\leq2({x^{\frac {3}{2}} + y^{\frac {3}{2}} + z^{\frac {3}{2}})}$
By AM-GM we know that ${ x^{\frac {3}{2}} + y^{\frac {3}{2}} + y^\frac {3}{2}}\geq 3x^{\frac {1}{2}}y$ and
${ x^{\frac {3}{2}} + z^{\frac {3}{2}} + z^\frac {3}{2}}\geq 3x^{\frac {1}{2}}z$

That's it. This is a well-known lemma used commonly used in proving Nesbit's inequality.

Edit: Allnames,  I don't understand your proof, especially the line with $ \sqrt {x}\leq 3x + 12\sqrt {3}$.(when do you have equality?)
Could you detail a little bit your proof?[/quote]
Sorry,i have just edited(this is my $ LaTex$ mistakes,sorry) .thas line is quite clear (only AMGM with$ \sqrt x$ and $ \sqrt {\frac {1}{3}}$
After,we use $ (\sum a)^2\ge \sum ab$ and $ (\sum a)^2\ge \sum a$

Let $ x\plus{}y\plus{}z\equal{}S$.
Thus, we can rewrite the LHS as:
$ \frac{x\sqrt{x}}{S\minus{}x}\plus{}\frac{y\sqrt{y}}{S\minus{}y}\plus{}\frac{z\sqrt{z}}{S\minus{}z}$
Each $ \frac{x\sqrt{x}}{S\minus{}x}$ is convex.
Thus, by Jensen's:
$ \frac{x\sqrt{x}}{S\minus{}x}\plus{}\frac{y\sqrt{y}}{S\minus{}y}\plus{}\frac{z\sqrt{z}}{S\minus{}z}\ge 3(\frac{S/3\sqrt{S/3}}{2S/3})\equal{}\frac{3\sqrt{S}}{2\sqrt{3}}\ge \frac{\sqrt{3}}{2}$

you can prove also that $ \dfrac{x\sqrt x}{y\plus{}z}\geq \dfrac{x^2\sqrt{3(x\plus{}y\plus{}z)}}{2(x^2\plus{}y^2\plus{}z^2)}$  :D

The positive reals $ x,y,z$ and $ w$ are satisfying the relation $ x \plus{} y \plus{} z\plus{}w\geq 1$. Prove that:
$$ \frac {x\sqrt {x}}{y \plus{} z} \plus{} \frac {y\sqrt {y}}{z \plus{} w} \plus{} \frac {z\sqrt {z}}{w \plus{} x}\plus{} \frac {w\sqrt {w}}{x \plus{} y}\geq  1$$

Let $S=x+y+z\ge1$, and let $f(x)=\frac{x\sqrt x}{S-x}$. Then, $f''(x)=\frac{3(S-x)^2+4(3S-x)}{4\sqrt x(S-x)^4}>0$ since $3S=3x+3y+3z>x$. By Jensen:
$$\sum_{\text{cyc}}\frac{x\sqrt x}{y+z}=\sum_{\text{cyc}}\frac{x\sqrt x}{S-x}\ge3\frac{\frac S3\sqrt{\frac S3}}{S-\frac S3}=\frac{S\sqrt3}{2}\ge\frac{\sqrt3}2.~\square$$

sqing#8 can be solved the same way.


How would you use Cosine? I know how to use sin and stuff, but I'm not really seeing how to apply cosine.
Thanks,
Eli

Better to have this in the High School Basics forum...

Cosine works the same way as $ \sin$ except it's   $ \frac {\text{adjacent}}{\text{hypotenuse}}$.  Same for $ \cos^{ \minus{} 1}$.

Ex.

What is the value of CAB if the triangle is right?

We need to use $ \cos^{ \minus{} 1}$.  The side adjacent to angle CAB is $ 3$ and the hypotenuse is $ \sqrt {10}$ using the pythag. theorem.  Then we have $ \cos^{ \minus{} 1}\frac {3}{\sqrt {10}} \approx 18.$

[asy]draw((0,0)--(3,0)--(3,1)--cycle);
label("$A$", (0,0), W);
label("$B$", (3,0), SE);
label("$C$", (3,1), N);
label("$3$", (0,0)--(3,0));
label("$1$", (3,0)--(3,1));[/asy]

Note $ \tan^{\minus{}1}\frac13$ is a lot easier.


I'm sure this problem has surfaced before but again, can't find it.

$ \triangle XYZ$ is equilateral and $ P$ is a point in its interior. Prove that $ a \plus{} b \plus{} c$ is constant.
($ a,b,$ and $ c$ are the heights of $ \triangle PZY,\triangle PYX,\triangle PXZ$)

Do not use areas on this one, please.

EDIT: Didn't catch the exterior point part. Not sure if the problem statement is solvable.

Since when does "Euclidean geometry" not carry a notion of area?

I worded it inadequately. Can you please solve the problem?

You can rotate the triangles that contain a, b, and c as altitudes and that have bases lying on the sides of XYZ, getting that a + b + c = altitude of XYZ.

You mean rotating for example triangle $ PZY$ on top of side $ XY$ for example, and the same with other triangles? If that's the case, I don't get how it helps.
Forgive me, but could you elaborate more what you mean?

Can anybody explain this? I'd like to understand this problem.

I'd like to ask why we aren't allowed to use areas for this problem?
Areas are a very useful tool when solving geometric problems and simplify many things, such as this problem.

Using areas would directly prove your statement.

That is exactly the difficulty of this problem. Using area is trivial, but we are studying rigorous geometry, and this problem can be solved with extremely basic geometry.
There are other similar problems that I have solved using a few concepts, but I can't apply them to this one, which is why I posted this problem here, hoping someone could find this more subtle solution.

Draw segments parallel to the sides through P. Three equilateral triangles are formed, with altitudes of length a, b, and c. Furthermore, there are three larger, overlapping equilateral triangles. (Each shares an angle with the original triangle.)

Show that the height of any larger triangle is the sum of the heights of the two triangles contained in it. Then show that the height of the original triangle is the sum of the height of any larger triangle with the height of the small triangle not contained in it.

Usually, if something isn't above a certain difficulty level (like this one) you should post it in the intermediate section.

This is around the fourth suggestion I receive to post problems somewhere else. If I don't know how to solve it, how in the world should I know how hard it is? Furthermore, I don't see how it should be bothering anybody because it is in a section one step above the "correct" section.

Thank you Bictor717 for your solution and Differ for the drawing, I appreciate it.

With geometry, you can usually tell how hard a problem is by the complexity of the problem statement.

Maybe you can, but I sure can't. I will continue to post them as I do now.

Well, if everyone's telling you to post in the Intermediate forum, why not just make a habit of doing so? :wink:

I beg to [i]Differ[/i] ( :rotfl: )  , you are only the second person to tell me to post a problem in the intermediate section, and I have had around five problems removed from the intermediate and placed on pre-olympiad. I'll stick with the pre-olympiad when in doubt .

Apparently you are the only one having nightmares and unable to sleep at night about this thread placed here.  :roll:

Don't fret over problem placement; I know personally that it's not easy to pick a place to post with four levels of forums to choose from. Just make your best guess, and if it's not there the mods will move it with no hard feelings. (speaking of which, this one being moved now)


A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? (Give your answer correct to two decimal places.)

can u chow the full steps in doing this?

Not such a routine problem. The same as the "pipe around the corner" problem. There are some topics that a search of this site using words like that will turn up.

there are 2 sites
http://answers.yahoo.com/question/index?qid=20081216082305AAx5lT9
and
http://answers.yahoo.com/question/index?qid=20071114185302AAnuW0Y

but i have no idea whats going on, i cant understand how their describing the diagram, and how they get those calculations, which is why i posted here

Here is the diagram. The red line is the building wall, the green line is the fence, the black line is the ground, and the orange line is the ladder.
[asy]D((0,20)--(0,0),red+linewidth(1));
D((0,0)--(15,0),black+linewidth(1));
MC(90,"8\mbox{\,ft}",8,D((4,0)--(4,8),green+linewidth(1)),W);
MP("4\mbox{\,ft}",(2,0),8,magenta);
D((0,12)--(12,0),orange+linewidth(1));
pair O=D("O",(0,0)), W=D("W",(4,0)), L=D("L",(12,0)),
H=D("H",(4,8),NE), B=D("B",(0,12),NE);
MA("\alpha",H,L,W,1.5,9);[/asy]
Now, can you compute the ladder length $ |BL|\equal{}|BH|\plus{}|HL|$ in terms of the angle $ \alpha$?


algebra
polynomial
group theory
abstract algebra
quadratics
inequalities
superior algebra

Let $f$ be an irreducible polynomial of degree $n$, $n \geq 3$ over $\mathbb{Q}$. Let $L$ be the splitting field of $f$ and $[L : \mathbb{Q}]=n!$. Prove that if $\alpha$ is any root of $f$, then  $\mathbb{Q}(\alpha^{4}) = \mathbb{Q}(\alpha)$

Can someone finish from here? It is merely a finite calculation now, or is there even a more elegant solution?

[quote="sissa"]If $K : = \mathbb{Q}(\alpha^{4})$ is a proper subfield of $M : = \mathbb{Q}(\alpha)$, it has index $2$ or $4$. Because $Gal(L/M) = S_{n-1}\leq Gal(L/K) \leq Gal(L/\mathbb{Q}) = S_{n}$, it follows that $Gal(L/K) = A_{n}$ because there are no other subgroups of $S_{n}$ with index $< n$. Hence $n = 4, 8$ and $\alpha^{4}$ is a solution of a quadratic equation over $\mathbb{Q}$.[/quote]

$S_{n-1}$ is a maximal subgroup in $S_{n}$ for all $n\geq 2$. Hence these inequalities imply $Gal(L|M) = Gal(L|K)$ or $Gal(L|K)=Gal(L|\mathbb{Q})$.
The latter would imply $\mathbb{Q}(\alpha^{4})=K=\mathbb{Q}\Rightarrow \alpha^{4}\in\mathbb{Q}$ this is obviously wrong, because f(x) would be $x^{4}-(\alpha^{4})$ and the splitting field of such a polynomial is a divisor of $4\cdot\phi(4)=8< 24=4!$.
So we get $Gal(L|M) = Gal(L|K) \Rightarrow  K=M \Rightarrow  \mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{4})$.


conics
parabola
symmetry
analytic geometry

How do I find the equation of a parabola with this much information:

y-intercepts at (0, -1) and (0, 5) and x-intercept at (-5, 0)

Thanks!

If you plug those points into the equation
x = ay^2 + by + c, or x=a(x-k)^2 + j   ---(I forgot the traditional constants...)
then you will get a systems of equations, which is one way to solve it.

Another way it that you know that the line of symmetry is the midpoint of the two y-intercepts. Therefore, the y-coordinate of the vertex must be 3. And that would be -b/2a for the first equation.

Even easier would be the form $ x \equal{} a(y\minus{}r)(y\minus{}s)$.  What do the points we're given immediately tell us $ r$ and $ s$ are?  How can we use the last point to determine $ a$?

[quote="Hedgietto"]What do I put in for a, b, and c?[/quote]

You are trying to find a, b, and c, because those are the constants. The points you have, such as (0, -1), you would plug into x and y.

$ a$ is the coefficent of $ x^2$, and $ b$ is the coefficent of $ x$. $ c$ is the only constant.