quadratics algebra polynomial AMC AIME USA(J)MO USAMO

For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$, define $\Delta A$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$, whose $n^\text{th}$ term is $a_{n+1}-a_n$. Suppose that all of the terms of the sequence $\Delta(\Delta A)$ are $1$, and that $a_{19}=a_{92}=0$. Find $a_1$.

[hide]For starters, we see that $(a_3-a_2)-(a_2-a_1)=a_3-2a_2+a_1=1$. Similarly, $a_{n+2}-2a_{n+1}+a_n=1$, so that $a_{n+2}=1+2a_{n+1}-a_n$. Now, we let $a=a_1$ and $b=a_2$, so that we're trying to solve for $a$.

Notice that the next few terms are the sequence are $a_3=1+2b-a$, $a_4=3+3b-2a$, and etc, where the nth term $a_n$ is $a_n=\frac{(n-1)(n-2)}{2}+(n-1)b-(n-2)a$.

Because $a_{19}=a_{92}=0$, we get two equations involving $a$ and $b$. $17a-18b=153$ and $90a-91b=4095$. Solving, $a=a_1=\boxed{819}$ [/hide]

[hide]More direct way of doing it is to express $ \Delta A$ first, call the difference between $ a_1$ and $ a_2$ to be b, then the next difference is b+1, then b+2...then $ a_{19} \equal{} a_1 \plus{} 18b \plus{} (17)(17 \plus{} 1)/2$ and $ a_{92} \equal{} a_1 \plus{} 91b \plus{} (90)(90 \plus{} 1)/2$ Solving, we find that b = -54, and $ a_1 \equal{} 819$.[/hide]

[hide]Since the second row of finite differences is constant, we know that the sequence must be created by a quadratic $ f(n) \equal{} a_n$. We're given that two roots are $ 19$ and $ 92$, so we know $ f(n) \equal{} k(n\minus{}19)(n\minus{}92)$. Taking the finite differences, we find that the second row is $ 2k$, so $ k\equal{} \frac{1}{2}$, whence we get our answer of $ .5(1\minus{}19)(1\minus{}92) \equal{} 819$. [/hide]

could you solve this problem by finding the traditional closed form of the recursion? I found the characteristic polynomial of the recursion to be $ (x\minus{}1)^3\equal{}0\longrightarrow x\equal{}1$ thus the root x=1 is of multiplicity 3.
Thus we have $ a_n\equal{}c_1(1)^n\plus{}c_2n(1)^n\plus{}c_3n^2(1)^n$ plugging in what we know for $ a_{19}$ and $ a_{92}$, we get the two equations $ 0\equal{}c_1\plus{}19c_2\plus{}841c_3$ and $ 0\equal{}c_1\plus{}92c_2\plus{}92^2c_3$ we are trying to find $ a_1\equal{}c_1\plus{}c_2\plus{}c_3$. i don't know if you can solve from here. does my way work?

You have two equations and you need a third. You just need to verify that the second finite differences are equal to 1. This gives you $ 2c_3\equal{}1$.

what do you mean the 2nd finite differences are equal to 1?
thanks for helping btw!

Let $ a_1, a_2, a_3, a_4, \ldots$ be a series. The first differences of these are given by $ b_j \equal{} a_{j\plus{}1} \minus{} a_j$

So, we have $ b_1, b_2, b_3, \ldots \equal{} a_2 \minus{} a_1, \; a_3 \minus{} a_2, \; a_4 \minus{} a_3, \ldots$

The second differences are given by $ c_j \equal{} b_{j\plus{}1} \minus{} b_j$

So, we have $ c_1, c_2, \ldots \equal{} b_2 \minus{} b_1, \; b_3 \minus{} b_2, \ldots \equal{} a_3 \minus{} 2 a_2 \plus{} a_1, \; a_4 \minus{} 2a_3 \plus{} a_2, \ldots$

What he's saying is that every term of the series $ c$ is equal to $ 1$.

But how did you figure out that the constant will be (n-1)(n-2)/2?

[quote="pkerichang"][hide]
since $ a_3 \minus{} a_2 \minus{} (a_2 \minus{} a_1) \equal{} 1$, we see that $ a_3 \minus{} 2a_2 \plus{} a_1 \equal{} 1$ and so on that $ a_{n \plus{} 2} \minus{} 2a_{n \plus{} 1} \plus{} a_n \equal{} 1$.  or $ a_{n \plus{} 2} \equal{} 1 \plus{} 2a_{n \plus{} 1} \minus{} a_n$.  let $ a_1 \equal{} a$ and $ a_2 \equal{} b$ and try few terms, we find a pattern that goes like $ 1 \plus{} 2b \minus{} a, 3 \plus{} 3b \minus{} 2a, 6 \plus{} 4b \minus{} 3a, \dots$.  so $ a_n \equal{} \frac {(n \minus{} 1)(n \minus{} 2)}{2} \plus{} (n \minus{} 1)b \minus{} (n \minus{} 2)a$ since $ a_{19} \equal{} a_{92} \equal{} 0$, we see that:

$ 17a \minus{} 18b \equal{} 153$
$ 90a \minus{} 91b \equal{} 4095$

solving for $ a$ we find that $ a_1 \equal{} 819$.  QED.

 [/hide][/quote]

Note that $ 3 \equal{} 1\plus{}2, 6 \equal{} 1 \plus{} 2 \plus{} 3, 10 \equal{} 1 \plus{} 2 \plus{} 3 \plus{} 4,$ etc. So for $ n$ it's just the sum of the first $ n\minus{}2$ natural numbers. I think he should prove the pattern always holds first before writing "QED". LOL  :wink:

Wow, I totally had no memory of seeing this problem before during the practice (this is 1992 AIME #8 btw)...
but here's what I did during the practice:
$ a_{n\plus{}2}\minus{}2a_{n\plus{}1}\plus{}a_n\equal{}1\implies\minus{}a_{91}\plus{}a_{90}\equal{}1$

$ a_{90}\equal{}2a_{91}\plus{}1 ............(1)$

$ a_{91}\minus{}2a_{90}\plus{}a_{89}\equal{}a_{91}\minus{}2(2a_{91}\plus{}1)\plus{}a_{89}\equal{}1$

$ a_{89}\equal{}3a_{91}\plus{}3 ............(2)$

I continued this to get $ a_{88}\equal{}4a_{91}\plus{}6$ etc and this clearly follows the pattern the above posters noted...

so I just assume the pattern is consistent and get $ a_{19}\equal{}73a_{91}\plus{}36*73\equal{}0\implies a_{91}\equal{}\minus{}36$
and $ a_1\equal{}91a_{91}\plus{}45*91\equal{}91(45\minus{}36)\equal{}819$...

If this sort of problems appear on, say, USAMO, would you be docked points for assuming the consistency of the pattern (as what i did)?

anyways, how would you prove the pattern though (of course just looking at the first few terms wouldn't be rigorous)?

You can use the fact that $ a_{k\minus{}1}\minus{}2a_k \plus{} a_{k\plus{}1} \equal{} 1$ for all $ k$, and use induction.

Here's an easy solution:

Write out and add first $ k\minus{}1$ terms of the second finite difference sequence:
$ a_3\plus{}a_1\minus{}2*a_2\equal{}1$
$ a_4\plus{}a_2\minus{}2*a_3\equal{}1$
…
…
…
$ a_k \plus{} a_{k\minus{}2} \minus{} 2*a_{k\minus{}1} \equal{} 1$
$ a_{k\plus{}1} \plus{} a_{k\minus{}1} \minus{} 2*a_k \equal{} 1$

Adding the above $ k\minus{}1$ equations we get:

$ \boxed{(a_{k\plus{}1} \minus{} a_k) \equal{} k\minus{}1 \plus{} (a_2\minus{}a_1)} \minus{}\minus{}\minus{} (1)$

Now taking sum $ k \equal{} 1$ to $ 18$ in equation $ (1)$ we get:
$ 18*(a_1\minus{}a_2) \minus{} a_1 \equal{} 153 \minus{}\minus{}\minus{} (2)$

Now taking sum $ k \equal{} 1$ to $ 91$ in equation $ (1)$ we get:
$ 91*(a_1\minus{}a_2) \minus{} a_1 \equal{} 4095 \minus{}\minus{}\minus{} (3)$

$ 18* (3) \minus{} 91*(2)$ gives $ a_1\equal{}\boxed{819}$.  :wink:

Ehh let me elaborate on Phelpedo's solution because all of you are (well this was 5 years ago so you guys all did use) semi-complicated solutions although you could probably do this problem by using ideas learned in my 8th grade math class.

Firstly notice that the [b]second difference[/b] of $(a_1, a_2, a_3 \cdots)$ are the same so therefore the numbers must follow a Quadratic Pattern.

So $a_x = ax^2 + bx + c$

Let's first look at $y=ax^2 + bx + c$. If $x=0$ then $ax^2 + bx + c = c$. If $x=1$, then $ax^2 + bx + c = a + b + c$. If $x=2$, then $ax^2 + bx + c = 4a + 2b + c$ and finally if $x=3$, then $9a + 3b + c$

Let's lay out our terms: $c, a+b+c, 4a+2b+c,9a+3b+c$. Firstly, let's look at the first differences. That would be $(a+b+c - c, 4a+2b+c - a - b -c, 9a+3b+c - 4a - 2b -c) \rightarrow (a+b, 3a+b, 5a+b)$

Now let's look at the second differences. The second difference is $(3a + b - a - b, 5a + b - 3a - b) \rightarrow (2a, 2a)$. For now (especially since this is an AIME problem) we can assume that the second difference of a quadratic is always going to be $2a$ (easy to show which I'll do below)

We already have that the second difference is $1 \rightarrow 2a = 1 \rightarrow a = \frac{1}{2}$

Now we can proceed to bash this out but, oh wait! We already have $2$ of the roots, 19 and 92!

Therefore the equation is $a_x = \frac{1}{2}(x-19)(x-92)$

Plugging in $x=1$ gives $a_1 = \frac{1}{2}*(-18)*(-91) = (-9)*(-91) = \boxed{819}$

[hide="random stuff"]

so for $x-1, x, x+1$ for a quadratic $ax^2 + bx + c$ we want to show that their second differences are always going to be $2a$.

Simply plug the values in and subtract $ax^2 + bx + c - a(x-1)^2 - b(x-1) - c =  2ax - a + b$

Also $a(x+1)^2 + b(x+1) + c - ax^2 -bx - c = 2ax +a + b$

And now when you subtract them the second difference will always be $2ax + a + b - 2ax + a - b = 2a$ as desired.

[/hide]

Middle School Classroom Techniques FTW

[hide=Solution]Let $a_1=a,a_2=b.$ From the conditions, we have $$a_{n-1}+a_{n+1}=2a_n+1,$$ for all $n>1.$ From this, we find that
\begin{align*}
a_3&=2b+1-a \\
a_4&=3b+3-2a\\
a_5&=4b+6-3a,
\end{align*}
or, in general, $$a_n=(n-1)b+\frac{(n-2)(n-1)}{2}-(n-2)a.$$ Note: we can easily prove this by induction. Now, substituting $n=19,92,$ we find that
\begin{align*}
0=&18b+\frac{17\cdot18}{2}-17a\\
0=&91b+\frac{90\cdot91}{2}-90a\\
b=&\frac{17a-\frac{17\cdot18}{2}}{18}=\frac{90a-\frac{90\cdot91}{2}}{91}.
\end{align*}
Now, cross multiplying, we find that
\begin{align*}
91\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\
1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\
73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\
a=\frac{18\cdot91}{2}=9\cdot91=\boxed{189}.
\end{align*}[/hide]


Hi all mathlinkers, this is a little problem which cause some problems for me,

1- Proove that it exists a unique polynom $H_n \in \mathbb{Q}[X]$ such that : \[ \forall x \in \mathbb{R} , \frac{d^n}{dx^n}\left(e^{-x^2}\right) = (-1)^n H_n(x)e^{-x^2}  \]
(It is done)
2- Calculate $H_0, \dots H_3$. (It is done)

3 and 4  not done:(.
3- Proove : $H_n=2XH_{n-1}-2(n-1)H_{n-2}$
4- Proove : $H^{\"}_{n}-2XH^{'}_{n-1}+2nH_n=0$.

Thanks!

Sorry, it was very easy, I m stupid :rotfl:

Sorry


Thank you for viewing my post!!!! I am very grateful for any help that you may be able to give.

Here are 3 questions that I just am not sure how to answer. If you know the correct solutions please help me out.   Thank you :)  :oops:

Which of the below will not cause the demand for X, a normal good, to increase?
	a)	A decrease in the price of X.
	b)	A change in consumer taste in favor of product X.
	c)	An increase in consumer incomes.
	d)	An increase in the price of a substitute good.
	e)	A decrease in the price of a good that is a complement to X.

.	If the own-price elasticity of demand for gasoline is -.30 and there is a 10% decrease (-.10) in the price of gasoline, this will cause the quantity of gasoline demanded to:
	a)	Increase by .30 (30%).
	b)	Increase by .03 (3%).
	c)	Decrease by .03 (-3%).
	d)	Decrease by .30 (-30%).

A black market is:
	a)	Something that happens when producers sell goods for a greater price than the government mandated price ceiling.
	b)	A characteristic of a surplus or excess supply condition.
	c)	Legal but frowned upon by economists who feel it violates consumer sovereignty.
	d)	None of the above.

I am very interested in economics, especially using calculus to solve some problems about economics but my friends, your questions are really  good. Let's analyze and find the right answers.
It's very hard to have a perfect answer ,

 For the first question, actually, a) makes the movement along the demand curve. Other answers make the demand curve shift to the right and also allow the demand to increase.
Opps! There is no negative value in price elasticity of demand. The right answer for this question is d)
Keep on going with the last question, the answer is none of the above because a) seems to be right but not enough clue about black market.
Have a nice day! :)

hello everyone,

actually i have a very important work to do using stata 9.but unfortunately i'm new to that software.i was just wondering,if i could find anyone to help me out.

Please reply,it's very important.

With Thanks


Following people will get 3 weeks from today to submit their proof with drawing (required for Problem 2).

chess64
The Original Pi Guy
Shogia
dangvh
BornForMath

Maximum point is 2.

1. For $a,b,c,i,j,j \geq 0$ and $p \geq q$ with $r \geq s$, prove that

$2(1+i)^j + 4a^3+4b^3+4c^3+2(pr+qs) \geq 2+2a^2b+2a^2c+2ab^2+2b^2c+2ac^2+2bc^2 + 2(ps+qr) + 2ij$

2. $\triangle ABC$ is an acute angled, meaning $\angle A, \angle B, \angle C < 90^\circ$. $A'$ is a point on semicircle diameter BC (so BC is the diameter of semicircle drawn from that segment) and it is opposite of $\angle A$. Same thing is happened to $B'$ and $C'$. Prove whether it is possible to construct the semicircles such that $AB' = AC', BC' = BA', CA' = CB'$ and if it is possible, give detailed proof with sketched drawing (it can be made from computer). On the drawing, you must give prove on the three equality that is suggested above.

Solution MUST be LATEXED. You have 3 weeks. That's enough time to learn and make your solution in LaTeX.

I have the same question.  Proofs are definetely not my strong point though.  Should be very interesting.

There are three options.

1. You may send to me by Private Message (below the message you'll see "Attach an document")

2. You may mail me the image to g89unit@gmail.com

That address has name "Silver Falcon." My personal mail is not used for my contests (for my friends and more personal stuff).

3. You may IM me and direct connect and send me the image by sending "File."

My sn is jwec05.

More professional way is:

Put the image in PDF document and send the whole document by attaching it in Private Message. If you made the LaTeX document with image on it, any way is actually fine.

on number 1, do a,b,c,i,j have to be integers?

As far I know, the restriction for them to be integers is not necessary.

[quote="Xantos C. Guin"][quote="Silverfalcon"]
1. For $a,b,c,i,j,j \geq 0$ and $p \geq q$ with $r \geq s$, prove that

$2(1+i)^j + 4a^3+4b^3+4c^3+2(pr+qs) \geq 2+2a^2b+2a^2c+2ab^2+2b^2c+2ac^2+2bc^2 + 2(pr+qs) + 2ij$[/quote]

does $i=\sqrt{-1}$?
or is $i$ a variable?
I'm just wondering this for fun.[/quote]

No, whenever you see i, it equals -1^1/2 which is an imaginary number.

but here i is a variable, because i >= 0

[quote="chess64"]Well, good luck everyone! I'm not doing it.... I have no idea how to do any of these problems. Oh well, at least I made it to round 3 with the highest score :)[/quote]

Same here, I hate proofs...so good luck  :coolspeak:

In number 1, are all the variables (except p,q,r,s) greater than or equal to 0? Or just j?

I was thinking the same thing.  I dont think that the inequality works if  we set each of a, b, and c equal to -1; j equal to 1/2; i equal to -1; while p=q and r=s.

Also on number 2 are we proving this for ALL triangles and can the semi circle be inside the triangle or must it be outside.

I would still like to know for the second one whether we are proving this for all triangles as well as whether or not the semicircle is inside or outside the circle.

[quote="The Original Pi Guy"]all the variables in that segment are greater then 0. That's what that notation means.[/quote]

Right.

For the second problem, my solution has semicircles outside the triangle. So, please send the solution with the semicircles outside the triangle.

Since many questions were asked for #2, here are some must-do things in solution:

1. Construct a drawing
2. Explain how you constructed your drawing
3. Using the things you did in Step 2, prove the equality for all triangles.

How many people advance to the finals? 2?

I'm hoping it's two so it's easier for me to determine the winner.

COP 4th Round 3 has now officially ended (to my clock since that's how I started three weeks ago).

There were only two people who sent the answers. So they will be ones who are going to be competing.  :D

They are:

The Original Pi Guy
BornForMath

Congratulation!!!


trigonometry
calculus
function
logarithms
algebra
polynomial
analytic geometry

I think this would be a good place for math jokes. My precalc teacher told me this last year:

$\displaystyle\frac{\sin x}{n}=6$, because if we cancel the $n$'s, we have...
$six=6$.

Any more?

Here's one that I first heard from one of my math teachers

Calculus joke!

So, a bunch of functions decide to have a party.  Everyone is there - $\sin x, \cos x, \ln x$ - and everyone's having a lot of fun except $e^x$, who is alone in a corner and having no fun at all.  $\sin x$ goes up to him and asks, "Why don't you integrate yourself into the party?"

[hide="And he responds..."]
"It wouldn't matter."
[/hide]

Edit:  And here's a great little joke.  Consider a $26^{th}$ degree polynomial $P(x) = (x-a)(x-b)(x-c)...(x-z)$.  What is $P(x)$?

[hide="And the answer is..."]
$P(x) = 0$.  Notice that $P(x) = (x-a)(x-b)(x-c)...(x-x)(x-y)(x-z)$ and $x-x = 0$.  :)
[/hide]

Haha, I've heard the $e^x$ one before. This isn't really a joke, but it's kind of funny for representational purposes
$\sqrt{-1}$  $r=1-3\sin\theta$ math.
[hide="Explanation"]$\sqrt{-1}=i$ $r=1-3\sin\theta$ is the graph of a heart in polar coordinates soo....
I heart math.[/hide]
Edit: here's another one I know...
$(12 + 144 + 20 + 3 * 4 ^ .5) / 7 + 5 * 11 == 9 ^ 2 + 0$
[hide="Translated"]A dozen, a gross, and a score, plus three times the square root of four, divided by 7, plus 5 times 11, is nine squared, and not a bit more.[/hide]

Another: It is a mathematician who doesn't know the difference between a donut and a coffee cup.

One my brother made up...
What did the 1 say to the 10?
"Hey 1, who's the fat guy?"

Proof that all positive integers are interesting:

Proceed by contradiction.
Assume that there is a subset of the integers $A$ that is all of the uninteresting numbers. By the well-ordering principle, $A$ has a least element. Thats interesting! Contrdiction.

An engineer, a physicist, and a mathematician are on a trip in Scotland, and see a black sheep on standingn on a hill.
Engineer: Wow! All sheep are black in Scotland!
Physicist: No, no, no. All this proves is that there is at least one hill in Scotland, that has at least one sheep standing on it that is black.
Mathematician: On this side at least.

I've got one:

Heisenberg is flying down the highway in his car when a cop pulls up behind him and turns on his lights. Heisenberg exits and pulls into a gas station and waits for the cop to walk up next to the car. The cop asks him, "Do you know how fast you were going?" Heisenberg replies," No, but I know exactly where I am."

Statistician jokes:

A statistician is boarding an airplane when a bomb is discovered in his bag. When officials question him, he says, "Studies show that the probability of a bomb being aboard an airplane is $\frac{1}{1000}$. However, the probability of two bombs being aboard an airplane is $\frac{1}{1000000}$. So I am much safer.

Three statisticians were duck hunting when they saw a duck. The first shot the duck and missed right in front of it. The second shot the duck and missed right behind it. The third shouted, "Got it!"

A statistician is someone who can have his head in ice-cold water and his feet in the oven and say, "On the average, I am very comfortable."

BTW, shouldn't this be in G&FF?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4554]How do math teachers drink alcohol[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53836]A Mathematician's Love Letter[/url]

There are 10 kinds of people, those who understand the binary numbers and those who don't
 :?:  :o

[hide="Explanation"]The number $10$ in the binary system is the decimal number $2$
 ;) [/hide]

[quote="t0rajir0u"]Calculus joke!

So, a bunch of functions decide to have a party.  Everyone is there - $\sin x, \cos x, \ln x$ - and everyone's having a lot of fun....
[/quote]

I know it in a different way...

While the functions have the party, suddenly $lnx$ comes into the room,  horror-stricken and shouts :
"Leave fast, the derivatives found us and they are coming here to differentiate us!"  :o

Panic, squeals, and the functions they try to leave as faster they can.
Only one function, remained cool and continued drinking.

- Well, didn't you hear what I said, says the Logarithm, leave fast, the derivatives are coming!

 - Oh, come on! I don't worry about it. I am $e^x$    :lol:  :D

Seen on a T-shirt:

$\sqrt{\frac{-1}{64}}$

[hide="Explanation:"]
i over 8
[/hide]

[quote]Oh, come on! I don't worry about it. I am $e^x$[/quote]
A version of this joke (my favorite):
.... So panick strikes, functions are runnig helter skelter..  $e^x$ asks  a constant .. "why  every one is running?"
Constant says " You see that's the one mean differential operator ... nothing is left if he gets through me" .. and he dashes off..
"He won't bother [b]me[/b]!"  $e^x$ muses  to no one in particular.

Few seconds later: (When the operator is within the striking distance)
"Hello,  I am $e^x$     "   He says in a cool voice .
"Hello, I'm [hide]$\frac{\partial } {\partial y}$   [/hide]
:)

Great programmers don't die; they just GOSUB without return.
2+2=5 for extremely large values of 2.

From the 2005 MathPath yearbook:
2 is NOT!! equal to 3... not even for extremely large values of 2.
Therefore, 2+2 is NOT!! equal to 5... not even for extremely large values of 2.

Top reasons not to do math homework?
I got stuck proving that the World Series converged.
It was Isaac Newton's birthday.
I could only get arbitrarily close to my textbook - I couldn't actually ever reach it.

Any others?

I locked my homework in the trunk, but a four-dimensional dog got in and ate it.

I could have sworn I put my homework in a Klein bottle, but this morning I couldn't find it.

[quote="JesusFreak197"]I could have sworn I put my homework in a Klein bottle, but this morning I couldn't find it.[/quote]Have you checked inside my avatar yet?

I got a donut and a cup of coffee and then I got stuck figuring out which one to dunk.

Expand:
(3x+2y)^8

Answer:
( 3 x + 2 y ) ^ 8
(  3  x  +  2  y  )  ^  8
(   3   x   +   2   y   )   ^   8

its kinda lame but whatever

It ruined it because there was nothing in the expression that indicated that you would copy 3 of them, spaced vertically apart.

Best if it stayed at

( 3 x + 2 y ) ^ 8

[quote="planetmagicdust"]pick up line:

If you're x^2, I wish I was 1/3x^3 so I could be the area under your curves.[/quote]

Welcome to the forum.

In the future, please refrain from reviving old topics (this one is from over a year ago). Thanks.


Hello everyone,

I'm interested in doing some practise problems so that come time for the USAPhO in January, I'll be well prepared.  I think that it goes without saying, that the best way to understand Physics is to just simply work problems.

So, I'm wondering if anyone here has any idea what are some good problem books?  Ideally, I'm looking for a book that's progressive; that'll start easy so that I can work and improve my Physics problem solving skills up to the harder problems.  For example, I wouldn't want to start cold feet on Olympiad problems, but I wouldn't want a book that only deals with easy problems (e.g. F = ma, F = 6, m = 2, what's a?...ok a bit of an exaggeration there) throughout either.

Any ideas?

Also a side question:  How much knowledge of Physics is needed exactly to do Physics competitions?  Is such knowledge easily attainable by doing problems, or would reading a supplementary text be necessary?

Thanks!

Try searching this forum and seeing different topics. This question has been asked many times before. Any ways, this  is a more specific question than the others. SO I'll recommend "200 Puzzling Physics Problems " by Gnadig and Published by Cambridge University Press. See amazon.

The review on that book indicated that the level of the problems are generally at the level of IPhO or higher...and that concerns me.

Then again...the review says the book is suitable for freshman/sophomore honours physics courses...hmmm....

Have you used the problem book yourself?  How was it?

Any other thoughts?


Putnam
function
logarithms
calculus
derivative
LaTeX
real analysis

Find all differentiable functions $f: (0,\infty)\rightarrow(0,\infty)$ for which there is a positive real number $a$ such that

$f'\left(\frac{a}{x}\right)=\frac{x}{f(x)}$

for all $x>0$.

[hide="here is my solution"]
$f'(x)=\frac{a}{xf\left(\frac{a}{x}\right)}$

$g(x)=f(x)f\left(\frac{a}{x}\right)$

$g'(x)=f'(x)f\left(\frac{a}{x}\right)+f(x)\left(\frac{-a}{x^{2}}\right)f'\left(\frac{a}{x}\right)$

so $g'(x)=0$ and $g(x)=k$ for some constant.

$f(x)f(a/x)=k$

$f(a/x)=\frac{k}{f(x)}$

$f'(a/x)=\frac{x}{f(x)}=\frac{-kf'(x)}{f(x)^{2}}$

$xf(x)=-kf'(x)\longrightarrow\frac{-f'(x)}{f(x)}=\frac{x}{k}$

integrating both sides yields

$-\ln(f(x))=\ln\left(\frac{1}{f(x)}\right)=\frac{x^{2}}{2k}+C$ so we get

$f(x)=\frac{A}{e^{Bx^{2}}}$ where $B=\frac{1}{2k}$

i am wondering if i made a mistake or if this solution is correct.
[/hide]

Please do not open new threads.

this problem has already been discussed inf ull:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=383299#p383299

[quote="me@home"][quote="maokid7"]
2nd-3rd lines:

$g(x)=f(x)f\left(\frac{a}{x}\right)$

$g'(x)=f'(x)f\left(\frac{a}{x}\right)+f(x)f'\left(\frac{a}{x}\right)$
[/quote]

Don't you have to also use the chain rule, i.e. $\frac{d}{dx}\left( f\left(\frac{a}{x}\right) \right) = \frac{-a}{x^{2}}\cdot f'\left(\frac{a}{x}\right)$ ?
I think you forgot that when you took the derivative of $g$...[/quote]

thanks for pointing that out. i guess i got lazy when typing my work in latex, but i did do that and still ended with the same answer>


calculus
derivative
function
symmetry
rotation
analytic geometry
graphing lines

A large square of side length s contains two smaller squares of side lengths 6 and 9.  The two smaller squares overlap in at most a line segment or point.  What is the least possible value of s?  Justify your answer.

[hide]

15.  Since the conditions of the problem dictate that the smaller squares can only abut as if they had solid edges, and tilting the squares only increases the value of s (since the diagonals are longer than the sides), the minimum value occurs when the squares are flush against each other, creating an answer of 15.

[/hide]

Your argument is incomplete.  What if we say tilt one square so that the midpoint of one its sides is at a vertex of the other square?  How do you prove that such an arrangement is no better?

If we take this approach, tilting the square of side length 6 to a 45 degree angle, and arranging it as you said, the new height/width will be as follows:

The hypoteneuse of a 45-45-90 triangle is always s:rt2:, so the distance from midpoint to adjacent midpoint of the smaller square is 3:rt2:.  This means that a square with sides 3:rt2: shares a vertex and the same orientation as the larger square.  Since the diagonal of the smaller square is 6:rt2: (using the same concept as above), and the square within it has a height of 3:rt2:, it means that the remaining length of the vertical diagonal is 3:rt2:.  However, only half of that lenght lies above the height 'covered' by the larger square, so height of the smaller square above the larger square is 9:rt2:/2.  Combined with the height of the larger square, that makes the height of the encompassing square at least 15.36 .

But, if you tilt the smaller square, it will approach its minimum 'height' as the square approaches the same orientation of the larger square.

Okay, good, you have ruled out the specific arrangement that I suggested.  But what about other arrangements?  Your last paragraph is more of a conjecture than a proof.  How do you know that some angle in between 0 and 45 degrees won't lead to a length smaller than 15?

Okay then...

Taking the smaller square, a line drawn from the midpoint of a side of the square to an opposite vertex (in this case creating a right triangle with sides 3-6-:sqrt:(45)) will create an angle of approximately 63.435 degrees at the midpoint.  The height of the smaller square above the larger one can then be defined as  :sqrt:(45) sin(116.565 - x), with x being the angle between the 'base' of the smaller square and the top of the larger one.
Taking the derivative of the height formula gives us  :sqrt:(45) cos(116.565 - x).  The derivative equals zero when x is 26.565, however this is a maximum of the function, not the minimum (if you doubt this, check it yourself).  So, the minimum of the function must lie at one of its endpoints, in this case, 0 or 45.  Substituting a 0 into the original height function yields a value of approximately 6, and substituting a 45 into that functions yields a value of approximately 6.364.  This indicates that the minimum height occurs when the angle between the larger square and the smaller square is 0.

Okay, nice work.  One last thing.  Your argument is for the case when the midpoint of one side of one square is a vertex of the other square.  What if the intersection point is not a midpoint?  Could that case be any better?

(Note to self: next time just make a complete proof to begin with)

Because of the convenient symmetry of a square (all sides have an equal length, so the whole can be rotated and the same things will still apply), what can be applied to the top of the large square can also be applied to the side that shares the vertex that connects to the smaller square.

Shifting the square so that more of the base lies above the vertex than below redefines the equation from above ( :sqrt:(45) sin(116.565 - x)) to be, for  :sqrt:(45) < z :le: 6:sqrt:(2), z sin(180 - arcsin(6/z) - x).

z sin(180 - arcsin(6/z) - x), given the conditions of z set above and treating x as a constant (because I intend to prove that the function will be greater at any value of x between 0 and 90), attains a maximum at z = 6:sqrt:(2) because:

y = z sin(180 - arcsin(6/z) - x)
dy/dz = sin(180 - arcsin(6/z) - x) + z cos(180 - arcsin(6/z) - x) (1/:sqrt:(1-(6/z):^2:)) (-6z^(-2)) = 0
sin(180 - arcsin(6/z) - x) = cos(180 - arcsin(6/z) - x) (1/:sqrt:(1-(6/z):^2:)) (6/z)

Unfortunately, this equation never holds true with the given values of z.
Taking the original equation and substituting for the endpoints of z, one gets 6:sqrt:(2) sin(135 - x) and :sqrt:(45) sin(116.565 - x).

6:sqrt:(2) sin(135 - x) > :sqrt:(45) sin(116.565 - x)
sin(135 - x) > sin(116.565 - x)
135 - x > 116.565 - x
135 > 116.565  -  true statement, but useless
sin(x) = sin(180 - x)
sin(135 - x) > sin(63.435 + x)
135 - x > 63.435 + x
71.565 > 2x
35.7825 > x

In the range 0 < x < 35.7825, 6:sqrt:(2) sin(135 - x) has a range of 6 to 8.3757, and 6:sqrt:(2) sin(135 - x) has a range of 6 to 6.5562.  Because the secant line of 6:sqrt:(2) sin(135 - x) has a greater slope after the intersection at x = 0, and the nature of the sine curve is the same (both resultant angles within the sine functions are between 90 and 135), 6:sqrt:(2) sin(135 - x) > 6:sqrt:(2) sin(135 - x).

Impressive work.  Congratulations!  The proof is a little longer than I would have guessed, but it looks good to me.

Here's the history of the problem.  One of the folks at the tutoring company I work for put the problem on a draft of one of our practice SAT I tests.  I tried to solve the question, and realized after 4 minutes that I didn't see a simple solution, although I guessed 15.  Anyway, if a problem takes me more than 1 or 2 minutes, that means it is too hard for an SAT, so we threw out the problem.  (It turns out the test writer had an incomplete argument for why the answer should be 15.)  Since a math problem is a terrible thing to waste, I figured I would post the problem here.

I wouldn't consider this exactly a "hard" problem, unfortunately what would normally be construed as common sense sometimes doesn't translate very well into a proof.

Once again, we have a *little* room to wiggle, based on definition.  If this problem were to have come from a book full of "trick" problems, my answer would not be 15!  As written, I actually agree the answer is 15, but the following illustration and minor wording changes would lead to a trick question.

What is a "square"?  Is it the lines?  Is it the area?  Does it depend on context?  What is meant by "overlap"?  Does it mean that the lines overlap?  Does it mean the areas overlap?

Hehe.  By "squares" I meant including the interior, but you're right that it sometimes means just the boundary.  The same goes for "circle"; some authors use "circle" for the boundary and "disk" for the whole thing.  Same for "sphere"; many authors use "sphere" for the boundary and "ball" for the whole thing.


[quote="warut_suk"]Walking to a river, a boy has a 5-litre and a 7-litre container. How can he measure 1 litre?[/quote]

  Fill the 5-litre cup and pour it into the 7-litre cup.  Once again fill the 5-litre cup and pour it into the 7-litre cup, but this time 3-litres of water will remain in the 5-litre cup.  Empty the 7-litre cup.  Pour the 3-litres in the 5-litre cup into the 7-litre cup.  Fill the 5-litre cup and pour whatever you can into the 7-litre cup.  Since the 7-litre cup only has 4 litres left of water, [b] 1 litre of water will remain in the 5-litre cup [/b]


parameterization
linear algebra
matrix
inequalities
linear algebra open

Given a (n-k)-by-(n-k) matrix X11 is nonsingular, how do you go about to parameterize all n-by-n nonsingular matrices X of the form
[ X11  X12 ]
[ X21  X22 ].

More generally, for any given X11, how do you go about to parameterize all n-by-n nonsingular matrices X of the form
[ X11  X12 ]
[ X21  X22 ].

I have a feeling that the Gram-Schmidt process has something to with this, but cannot quite put it in the picture.

The problem in the background is, how to parameterize all n-by-n nonsingular matrices X such that rank(X-A)=k, where A is given.

All the questions above come from my way of attacking a more fundamental problem in the background: how do you parameterize the pair (H,G) such that the quotient of two determinants
det([A, H ; G, A+H-G ])   / det( -A-H+G )
is positive. The Matrices A,H,G are all n-by-n, and A is given.

A little help please … anyone? Thank you.
K. Yam
 :?

Arr, I see the first problem can be solved by a variation of LDU decomposition.


conics
ellipse
linear algebra
matrix
inequalities unsolved
inequalities

Here is the problem for which V.I. Arnold hasn't a short proof.

Consider an ellipse in $\mathbb R^n$ with semi-axes $a_1\leq a_2\leq ...\leq a_n$ and its cut by hyperplane, which contains the ellipse's center. It is also ellipse, suppose its semi-axes are $b_1\leq b_2\leq ...\leq b_{n-1}$. Prove that \[a_1\leq b_1\leq a_2\leq b_2\leq...\leq b_{n-1}\leq a_n.\]

It is nice, isn't it?

I guess, it should be a well known result.

A very quick answer: it seems to be very similar to the known fact that eigenvalues of $n\times n$ orthogonal matrix and its main $(n-1)\times (n-1)$ minor interlace. See references and generalizations in Semen Malamud's paper in http://ru.arxiv.org/abs/math.CV/0304158
It has a short proof for sure

For a (relatively) short solution click [hide="here"] One can use the fact that $a_k$ is just the minimum over all planes $L$ of dimension $k$ of the distance from the origin to the farthest from the origin point of the section of the ellipsoid by $L$. To show that $b_{k-1}\leqslant a_k$ just notice that there exists a plane $L$ of dimension $k$ such that for it that largest distance is $a_k$ while its intersection with our hyperplane is some $k-1$-dimensional plane contained in our hyperplane, so that largest distance is at least $b_{k-1}$. To prove that $b_k\geqslant a_k$ just choose the $k$-dimensional plane contained in our hyperplane for which $b_k$ is the distance to the farthest point in the section of the $n-1$-dimensional ellipsoid. Then, since it is also a $k$-dimensional plane in the $n$-dimensional ellipsoid, this distance is at least $a_k$. [/hide]
The problem is very nice and, indeed, related to the fact about matrices mentioned by Fedor. It is "well known", of course, though rarely formulated in this purely geometric way. I wonder what Arnold's solution is.


geometry
trapezoid
MATHCOUNTS
rectangle
cyclic quadrilateral

an isosceles trapezoid has diagnols of 40 and bases of 30 and 50. what is the area of the trapezoid.

I think it asks for it as a decimal but i have it in simplest radical form.
is the answer $ 400\sqrt{7}$?

The answers and solutions are currently on the MathCounts website.

[quote="mountain.dew"]Let ABCD be an isosceles trapezoid with AD = BC. Let the altitude from B intersect DC at E and the altitude from A intersect DC at F. Also, draw BD, a diagonal. Since ABEF is a rectangle and $ \triangle ADF$ and $ \triangle BEC$ are congruent, DF = 10, FE = 30, and EC = 10. However, in right $ \triangle BED$, BD (the hypotenuse) = DE = 40. This is a contradiction.

I think you made a mistake in wording the problem or something.[/quote]

what contradiction are you talking about.  theres no contradiction?

From the given problem, there is one unique solution.

ALthough the given problem isnt the same as the one used on mathcounts state round (thus different answer), there is still only one unique solution to the problem thats given.

i stated the problem wrong its:

An isosceles trapezoid has legs of length 30 cm each, two
diagonals of length 40 cm each and the longer base is 50 cm.
What is the trapezoid’s area?

i said the shorter base was 30 not the legs

[hide="hint"] Okay, when I did this problem at state, first I drew the perpendicular lines down from A and B to some points E and F.The area of triangle ADC is $ \frac{30*40}{2}=600$ It is also $ \frac{50*height}{2}$, so the height is 24.After that, you just need to find AB, and then it's simple.[/hide]

[url=http://www.analyzemath.com/Geometry/trapezoid_problems.html]This[/url] should be helpful. Scroll down until you find the problem similar to this one.

Go to [url=http://www.mathcounts.org/webarticles/articlefiles/510-07_S_Solutions.pdf]this[/url], and find the exact problem incase you feel lazy again. :wink:

[quote="ordinary boy"][quote="mountain.dew"]What's wrong with my thinking then?[/quote]

well, first, you said BD(hypotenuse)=40.  Um no, thats an invalid assumption.[/quote]
that's what it said in the problem.. (diagonals of length 40)

Solution to actual problem:
[hide]Suppose we have isosceles trapezoid ABCD. Draw BD, a diagonal of length 40. So, triangle BDC is a right triangle (Pythagorean triples). Its area is $ \frac{30*40}{2}= 600$. Drawing the perpendicular from B to DC and calling the intersection E, it is also a height of triangle BDC. So, $ 600 = \frac{50*BE}{2}$. We find BE, the height of the trapezoid, to be 24.
In right triangle BEC, $ EC = 18$ by Pythagorean Theorem. Drawing the perpendicular from A to DC and calling the intersection F, $ DF = EC = 18$. So, $ FE = AB = 14$. So, the area of ABCD is $ \frac{(14+50)24}{2}= 768$.[/hide]

Don't know if this is allowed or whatever since its the mathcounts forum, but if you know it, you can just use ptolmey's and then brahmagupta (sp?) since you know its cyclic.

yeah.. but that's probably not mathcounts level

The actual problem (2007 State Target #6):

[quote="Mathcounts"]An isosceles trapezoid has legs of 30 cm each, two diagonals of 40 cm each and the longer base is 50 cm. What is the trapezoid's area?[/quote]

Solution found [url=http://www.mathcounts.org/webarticles/articlefiles/510-07_S_Solutions.pdf]here[/url].

[quote="Math Geek"]Wait, so using Ptolemy's Theorem, you would get that the product of the diagonals minus the product of the bases equals 100, so the equal legs of the trapezoid would be 10 each.  Then we find that the semiperimeter is 50, and using Bramagupta's Theorem (Man, that's a cool name), we find that the area of the cyclic quadrilateral is sqrt(40)(40)(20)(0) when we find the square root of the product of the differences between the semiperimiter and the side lengths.  Hmmm....  I think I made a mistake.  That's strange.  How the heck can a quadrilateral have area of 0? :huh:[/quote]

THe original question posted is different then the actual mathcounts question.

this is the real question

[b]An isosceles trapezoid has legs of 30 cm each, two diagonals of 40 cm each and the longer base is 50 cm. What is the trapezoid's area?[/b]
Note the legs are 30, not one of the bases


geometry
circumcircle
rectangle
trapezoid
geometry unsolved

[img]http://i60.photobucket.com/albums/h1/Lorax42/HW3-2.jpg[/img]

We have triangle ABC with squares constructed on the outside of each side.  Connect the corners of the squares and find the midpoints of these new lines.  When we creat a line from MB' to B, this line is perpendicualr to line AC, ect.

How would I begin to prove this?

See the following problems:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=22917[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=118200[/url]

EDIT: Sorry, those are different problems. See the problem [url]http://www.mathlinks.ro/viewtopic.php?t=172109[/url] and use orthocenter H of the $ \triangle ABC$ instead of the circumcenter O.

Let H be the orthocenter of the $ \triangle ABC,$ let $ HC \perp AB$ meet $ A_CB_C$ at M and let parallel to HC through $ A_C$ meet $ CB_C$ at P.

$ \frac {CB_C}{CA_C} \equal{} \frac {BC}{AC} \equal{} \frac {a}{b}$

$ \angle A_CPC \equal{} \angle MCB_C \equal{} \angle B,\ \angle PCA_C \equal{} \angle C\ \Longrightarrow$
$ \triangle A_CPC \sim \triangle ABC\ \Longrightarrow \frac {CP}{CA_C} \equal{} \frac {a}{b}$

$ CB_C \equal{} CP,$ C is the midpoint of $ B_CP,$ $ HC \equiv CM \parallel PA_C$ is a midline of $ \triangle A_CB_CP,$ M is midpoint of $ A_CB_C,$ $ M \equiv M_C.$

Nice to see the [b][size=100]yetti’s[/size][/b] come-back.

We denote as $ B',$ $ C',$ the intersection points of the line segment $ BC,$ from the circumcircles of the squares ( or similar rectangles ), $ ACB_{c}B_{a}$ and $ ABC_{b}C_{a},$ respectively.

So, we have $ B_{a}B'\perp BC,$ $ C_{a}C'\perp BC$ and $ AB' \equal{} AC'$ $ ,(1)$ $ ($ from $ \angle AB'C' \equal{} \angle AB_{c}C \equal{} \angle \omega \equal{} \angle AC_{b}B \equal{} \angle AC'B'$ where $ \angle \omega \equal{} 45^{o}$ in the case of the proposed problem with erected squares $ ).$

From $ (1),$ we conclude that the [b][size=100]A-[/size][/b]altitude of $ \bigtriangleup ABC,$ as the midperpendicular of the segment $ B'C',$ passes through the midpoint of the segment $ B_{a}C_{a}$ $ ($ from the trapezium $ B_{a}B'C'C_{a}$ $ )$ and the proof is completed.

Kostas vittas.


geometry
rectangle
combinatorial geometry
combinatorics unsolved
combinatorics

There are 5 points in a rectangle (including its boundary) with area 1, no three of them are in the same line. Find the minimum number of triangles with the area not more than $\frac 1{4}$, vertex of which are three of the five points.

minimum is 2? (If I didn't miss any case.)

Divide the rectangle into 2 equal parts using a line parallel to one of the sides. One can show that if 3 points lie inside one of the 2 equal parts, then the area of the triangle with these 3 points as vertice as area <=1/4. Do this for the 2 orientations, then either 3 points lie in a quater rectangle, or there are 2 triangles with area <=1/4, in which case we are done. If 3 points lie in a quater rectangle, let them be A,B,C, and let another point be D. One can show that the area of the convex hull of A,B,C,D is at most 1/2. Then since there exists 2 triangles out of ABD,ACD,BCD that add up to at most this convex hull, so another triangle has area <=1/4. Thus, 2 is minimum.
Finally, note that 2 is possible. (see diagram)

your solution is not correct.

this diagram not contains 2 trianles with area not more than 1/4.

This problem is not easy.

One Idea is Complex geometry. (crazy idea)  :blush:

You are wrong, Lopes!

The diagram contains exactly two small triangles.


Exits on a highway are number consecutively from 1-50. The distance from exit 41 to exit 50 is 100 km. If each exit is atleast 6km from the next exit, what is the longest possible distance in kilometer, between any 2 consecutive exits from exit 41-50.

 :)

[hide]There are nine exits. If 8 of them are 6 apart, that's a total of 48, leaving 52km left.[/hide]

[quote="The QuattoMaster 6000"]Not 9 exits, as you counted, there are ten.  However, there are 9 distances between consecutive exits.  Here they are:
1. 41 to 50
2. 41 to 42
3.43 to 44
4.44 to 45
5.45 to 46
6. 46 to 47
7. 47 to 48
8. 48 to 49
9. 49 to 50[/quote]
41 to 50 isn't one, 42 to 43 is.

Oh, yes.  Sorry, about that!  I can't edit it now, though.

[quote="davidli"][quote="jhredsox"][hide]
nine exits, 8 are 6 apart, total of 48, 52km left (the answer)

[/hide]-jorian[/quote]

Ug, couldn't you have at least tried to make yours sound different?[/quote]

i tried to sum mine up..........thought no one would notice (didn't want to write the EXACT same thing)
:P
-jorian

Well then how about just don't post?

Geez, it's really annoying when someone posts a problem and 3-5 people post the EXACT same solution right after another.

I realize this is spam, but hopefully some people will get the message and not make posts that are on the borderline between spam and non-spam just to increase their post count.

Mods: You can delete this post if you feel it's necessary.


21. Two cards are dealt from a deck of four red cards labeled A, B, C, D and four green cards labeled A, B, C, and D. A winning pair is two of the same color or two of the same letter. What is the probability of drawing a winning pair?

[hide]EDIT: Is it just 1/7+3/7?[/hide]

Say that the first card is a red A (it doesn't matter). The second card must be a green A, a red B, a red C, or a red D. Thus, the total probability is $ \boxed{\frac47}$.


vector
integration
calculus
calculus computations

Question is:

Compute the flux of the vector field, \vec{F} , through the surface, S.
\vec{F} = 7\vec{r} and S is the part of the surface $ z \equal{} x^2 \plus{} y^2$ above the disk $ x^2 \plus{} y^2 \leq 4$ oriented downward.

My answer so far is

$ \int\limits_R (7x\vec{i} \plus{} 7y\vec{j} \plus{} (x^2 \plus{} y^2)\vec{k}) \cdot (2x\vec{i} \plus{} 2y\vec{j} \minus{} k) dA$
$ \int\limits_R (14x^2 \plus{} 14y^2 \minus{} (x^2 \plus{} y^2)) dA$
$ \int\limits_R (13x^2 \plus{} 13y^2) dA$
$ 13 \int\limits_R (x^2 \plus{}y^2) dA$
$ 13 \int^{2\pi}_0\int^4_0 r^3 dA$
$ 13 \int^{2\pi}_0 64 dA$

is this correct?


How do you prove that a function is surjective and/or injective?

For example, I was looking at 1992 IMO #2:

$ f(x^2\plus{}f(y)) \equal{} y \plus{} f(x)^2$ for all x,y, for a function $ f:$ R-->R.

The solution started off with "It is easy to see that f is both injective and surjective."  I don't see how it is "easy to see" yet.

Thanks

[quote="kyyuanmathcount"]How do you prove that a function is surjective and/or injective?

For example, I was looking at 1992 IMO #2:

$ f(x^2 \plus{} f(y)) \equal{} y \plus{} f(x)^2$ for all x,y, for a function $ f:$ R-->R.

The solution started off with "It is easy to see that f is both injective and surjective."  I don't see how it is "easy to see" yet.

Thanks[/quote]

You prove something is injective by definition: If f(a)=f(b), then a=b.   :)

So here we have $ f(x^2 \plus{} f(a)) \equal{} a \plus{} f^2(x)$ and $ f(x^2 \plus{} f(b)) \equal{} b \plus{} f^2(x)$. If we assume $ f(x^2 \plus{} f(a)) \equal{} f(x^2 \plus{} f(b))$, then $ a \equal{} b$, so it's injective

It's definitely not easy to see it's surjective (for me at least, as I haven't found that yet...).

For injectivity, assume, for the purpose of showing a contradiction, that there exist different $ a$ and $ b$ for which $ f(a)\equal{}f(b)$. Then consider $ f(x^2\plus{}f(a))$ and $ f(x^2\plus{}f(b))$ for a fixed $ x$.
For surjectivity, in this case,  you can construct $ c$, for any real number $ k$, such that $ f(c)\equal{}k$. Fix $ x$ and do something with $ k\equal{}y\plus{}f(x)^2\equal{}f(x^2\plus{}f(y))$. (Solve for $ y$ in terms of $ x$ and $ k$. Then $ k\equal{}f(blah)$, where $ blah$ is something involving $ k$ and a fixed $ x$.)

There is [i]never[/i] a good reason to quote the entire post preceding yours.

Edit: Bictor was faster.

For any $ c \in \bf R$, take $ y \equal{} c \minus{} f(x)^2$ to get that $ f(blah) \equal{} c$ for some $ blah$, so $ f$ is surjective.

I'd make one small alteration to Anaxerzia's argument: assume $ f(a) \equal{} f(b)$.  Then $ f(x^2 \plus{} f(a)) \equal{} f(x^2 \plus{} f(b))$ and so $ a \equal{} b$.  If you start by assuming that $ f(x^2 \plus{} f(a)) \equal{} f(x^2 \plus{} f(b))$ then you can only deduce injectivity for the set of values of the form $ x^2 \plus{} f(y)$.  Now, it happens that because $ f$ is surjective this includes all real values, but it's not [i]a priori[/i] clear that this is the case.


How many positive integers less than or equal to 100 are not square numbers, Fibonacci numbers, or triangular numbers?

Solution: You find the number of perfect squares, fibonacci numbers, and triangle numbers of in the list, take away repeats, and subtract it from the number of positive integers there are less than or equal to 100

[quote="Art of Owna"]What are triangular numbers?[/quote]

[quote="Wolfram MathWorld"]The triangular number $T_{n}$ is a figurate number that can be represented in the form of a triangular grid of points where the first row contains a single element and each subsequent row contains one more element than the previous one. This is illustrated above for $T_{1}=1$, $T_{2}=3$, .... The triangular numbers are therefore 1, 1+2, 1+2+3, 1+2+3+4, ..., so the first few triangle numbers are 1, 3, 6, 10, 15, 21, ... [/quote]

Other information can be found at:

[url]http://mathworld.wolfram.com/TriangularNumber.html[/url]

i brute forced it an got

[hide]74[/hide]

i doubt im right though.  probably messed up while counting.  1, another 1, another 1, ...

Apparently my pointless problem causes troubles.  No correct answers yet.  Cognos you missed some overlaps.

Frost

[quote="cognos599"][hide]the squares are

$1$ $4$ $9$ $16$ $25$ $36$ $49$ $64$ $81$ $100$

the fibonacci that are not squares

$2$ $3$ $5$ $8$ $13$ $21$ $34$ $55$ $89$

the triangular number that are not squares or numbers in the fibonacci sequence

$6$ $10$ $15$ $28$ $36$ $45$ $55$ $66$ $78$ $91$

if you add all of them up you get $10$ + $9$ + $10$ = $29$

then you do $100$ - $29$ = $\fbox{71}$[/hide]

there is probably an easier way but this is more accurate.[hide][/hide][/quote]

cognos:
[hide]
You counted 55 twice: as a fibonacci number and a triangular number.

Then the answer should be $100-28=\fbox{72}$.

EDIT: Thanks for the tip, leetspeak.  The answer should actually be $\fbox{73}$, I hope.
Btw I looked again, and I found out why it's 73(I hope).  Cognos counted 36 twice too.  I seem to have missed that while reviewing.
[/hide]

That's wrong again.

See this is what pointless brute force does to one's mind
IT SUCKS OUT BRAINS

[hide="what I did"]This just helps organization-wise, make a 10 by 10 grid of the numbers from 1-100 and cross out numbers.[/hide][/hide]

[quote="13375P34K43V312"]YEs, that's' the correct answer.[/quote]

Do you think there will be a problem like this or will all these problems not be so burte-force like.

[quote="frost13"]THey certainly wont all be like that, but I remember specifically a problem about how many numbers less than 100 could(or couldn't) be the sum of 2 squares.  Totally brute force, really depends on how well you organize yourself!![/quote]

I just did that problem today, and messed up 4 times because of my handwriting  :ninja:

[quote="frost13"]THey certainly wont all be like that, but I remember specifically a problem about how many numbers less than 100 could(or couldn't) be the sum of 2 squares.  Totally brute force, really depends on how well you organize yourself!![/quote]

Are you sure it wasn't difference of two squares? Because with difference, there's actually a nice solution.

Yeah it was sum.  It was a state team which I think was discussed on this forum before states.  It was another that required nice clean work:-)


Find all sequences of positive integers $\{a_n\}_{n=1}^{\infty}$, for which $a_4=4$ and
\[\frac{1}{a_1a_2a_3}+\frac{1}{a_2a_3a_4}+\cdots+\frac{1}{a_na_{n+1}a_{n+2}}=\frac{(n+3)a_n}{4a_{n+1}a_{n+2}}\]
for all natural $n \geq 2$.

[i]Peter Boyvalenkov[/i]

[quote="imortal"]sorry, I don't see contradiction with $a_4=4$.

well you still may have
$a_1=16$
$a_2=1$
$a_3=3$
$a_4=4$
$a_5=10$
$a_6=6$
and the first three sums will be ok.
but then a_7 will not be an integer, which is the contradiction[/quote]
In this way we should try all possible versions for contradictions and come to the conclusion $a_n=n$

yep, but the only two versions are $a_1=1$, $a_2=2$ or
$a_1=16$, $a_2=1$ so it is not a problem :)


[quote="ductrung"]For any positive real numbers $a,b$ and $c$,
\[\frac{a^{2}+bc}{a(b+c)}+\frac{b^{2}+ca}{b(c+a)}+\frac{c^{2}+ab}{c(a+b)}\le 1+\frac{1}{3}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right) \]
[/quote]

It is equivalent to
$(a-b)^{2}(b-c)^{2}(c-a)^{2}+a^{3}b^{3}+a^{4}bc-a^{3}b^{2}c-a^{2}b^{3}c+ab^{4}c-a^{3}bc^{2}-ab^{3}c^{2}+a^{3}c^{3}-a^{2}bc^{3}-ab^{2}c^{3}+b^{3}c^{3}+abc^{4}\geq 0$

which is true .
(The other side could be also proved by expandind more easily sing only Muirhead)

For LHS Let $P(a,b,c)= 2a^{2}-b^{2}-c^{2}$. WLOG $a\geq b\geq c$ we have that $P(a,b,c) \geq P(b,a,c) \geq P(c,a,b)$ and
$\frac{1}{2a(b+c)}\geq \frac{1}{2b(a+c)}\geq \frac{1}{2c(a+b)}$. then by chebisev.
\[\sum_{cyc}\frac{a^{2}+bc}{a(b+c)}\geq \frac{1}{2}\sum_{cyc}\frac{b+c}{a}\Leftrightarrow \sum_{cyc}\frac{P(a,b,c)}{2a(b+c)}\geq 0 \]
If $A=\sum_{cyc}\frac{b+c}{a}$ then it is easy to show that $A \geq 6$ (by Cauchy). so then $\frac{A}{2}\geq \sqrt{\frac{3A}{2}}$.
so $\sum_{cyc}\frac{a^{2}+bc}{a(b+c)}\geq \sqrt{\frac{3A}{2}}$

ductrung, can you give some hint to prove the RHS, thanks.

[quote="Beginner"]For LHS Let $P(a,b,c)= 2a^{2}-b^{2}-c^{2}$. WLOG $a\geq b\geq c$ we have that $P(a,b,c) \geq P(b,a,c) \geq P(c,a,b)$ and
$\frac{1}{2a(b+c)}\geq \frac{1}{2b(a+c)}\geq \frac{1}{2c(a+b)}$.[/quote]
Hi Beginner,
If $a\ge b\ge c$, then
\[\frac{1}{2a(b+c)}\leq \frac{1}{2b(a+c)}\leq \frac{1}{2c(a+b)}\]
What more, if it were true that
\[\sum\frac{a^{2}+bc}{a(b+c)}\ge\frac{1}{2}\sum\frac{b+c}{a}\]
then it would contradict the RHS !

For the RHS I use
\[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\le \frac{1}{2}+\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}\]

[quote="ductrung"]For any positive real numbers $a,b$ and $c$,
\[\sqrt{\frac{3}{2}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)}\le \frac{a^{2}+bc}{a(b+c)}+\frac{b^{2}+ca}{b(c+a)}+\frac{c^{2}+ab}{c(a+b)}\le 1+\frac{1}{3}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right) \]
[/quote]
For the left, using the known $(x+y+z)^{2}\ge 3(xy+yz+zx)$, it suffices to prove that
\[2\sum \frac{a^{2}+bc}{a(b+c)}\cdot \frac{b^{2}+ca}{b(c+a)}\ge \sum\frac{b+c}{a}\]
But it is equivalent to
\[(a-b)^{2}(b-c)^{2}(c-a)^{2}\ge 0 \]
which is trivial. :)

[quote=ductrung]For any positive real numbers $a,b$ and $c$,
\[\sqrt{\frac{3}{2}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)}\le \frac{a^{2}+bc}{a(b+c)}+\frac{b^{2}+ca}{b(c+a)}+\frac{c^{2}+ab}{c(a+b)}\le 1+\frac{1}{3}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right) \][/quote]

[img]http://s9.sinaimg.cn/middle/006ptkjAzy76sy2cbQ448&690[/img]

[quote=ductrung]For any positive real numbers $a,b$ and $c$,
\[ \frac{a^{2}+bc}{a(b+c)}+\frac{b^{2}+ca}{b(c+a)}+\frac{c^{2}+ab}{c(a+b)}\le 1+\frac{1}{3}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right) \][/quote]

The following inequality is stronger and easier :
For any positive real numbers $a,b$ and $c$,
\[ \frac{a^{2}+bc}{a(b+c)}+\frac{b^{2}+ca}{b(c+a)}+\frac{c^{2}+ab}{c(a+b)}\le \frac{1}{4}\left(6+\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right) \]


$ f: \mathbb N \times \mathbb Z \rightarrow \mathbb Z$  satisfy the given conditions

$ a)$ $ f(0,0)\equal{}1$ , $ f(0,1)\equal{}1$ ,

$ b)$ $ \forall k \notin \left\{0,1\right\}$  $ f(0,k)\equal{}0$ and

$ c)$ $ \forall n \geq 1$ and $ k$ , $ f(n,k)\equal{}f(n\minus{}1,k)\plus{}f(n\minus{}1,k\minus{}2n)$

find the sum $ \displaystyle\sum_{k\equal{}0}^{\binom{2009}{2}}f(2008,k)$

$ f(n,k) \equal{} \#\{X|2\sum_i x_i \equal{} k \ or k \minus{} 1\}$, were $ X$ is subset $ \{1,2,...,n\}$.
Therefore $ f(n,n(n \plus{} 1) \plus{} 1 \minus{} k) \equal{} f(n,k)$. Let $ F(n,k) \equal{} \sum_{i \equal{} 0}^k f(n,i)$, then $ F(n,n(n \plus{} 1) \plus{} 1 \minus{} k) \plus{} F(n,k) \equal{} 2^{n \plus{} 1}$.
It give $ F(n,\frac {n(n \plus{} 1)}{2}) \equal{} 2^n.$

[quote="sinankaral53"]$ f: \mathbb N \times \mathbb Z \rightarrow \mathbb Z$  satisfy the given conditions

$ a)$ $ f(0,0) \equal{} 1$ , $ f(0,1) \equal{} 1$ ,

$ b)$ $ \forall k \notin \left\{0,1\right\}$  $ f(0,k) \equal{} 0$ and

$ c)$ $ \forall n \geq 1$ and $ k$ , $ f(n,k) \equal{} f(n \minus{} 1,k) \plus{} f(n \minus{} 1,k \minus{} 2n)$

find the sum $ \displaystyle\sum_{k \equal{} 0}^{\binom{2009}{2}}f(2008,k)$[/quote]
Turkish solution

[hide=Solution]
We create a [b]bijection[/b]. Let $ A_m = \{ 1, 2\times 1, ..., 2 \times m \} $ and $ S(X) $ denote the sum of the elements of set $ X$. We claim that $ \boxed{ f(n,m) \text{ counts the number of ways to pick } Q \in A_n \text{ such that } S(Q) = m} $. Call such a set $ Q $ [i]good[/i].

We induct to show this. Base case $ n = 0$ is easy to verify against conditions a) and b).

For $ n = \ell +1 $ notice that $ 2( \ell +1) $ is either in $ Q $ or not, so that helps us split into either $ f(\ell, m) $ in the first and $f(\ell, m - 2(\ell+1))$ for the latter. Combining gives the recursion in c).

So we want to keep the sum of our elements in the bound $ [ 0, \binom{2009}{2} ] $. Remark that $S(A_{2008}) = 2 \binom{2009}{2} + 1 $. As such, if $ Q$ is a [i]good[/i] set, then $ A_{2008} \setminus Q $ is not. So we have a bijection between sets that are [i]good[/i] and those that are not. Thus, our evaluated sum is just $ \boxed{ \frac{2^{2009}}{2} = 2^{2008}} $.
[/hide]


geometry
vector
calculus
derivative
Asymptote
analytic geometry
graphing lines

This problem is easy but nobody solved it.
point $A$ moves in a line with speed $v$ and $B$ moves also with speed $v'$ that at every time the direction of move of $B$ goes from $A$.We know $v \geq v'$.If we know the point of beginning of path of $A$, then $B$ must be where at first that $B$ can catch $A$.

I, for one, simply can't understand the text. For example, what does "at every time the direction of move of $B$ goes from $A$" mean?

Could you write this again in other (more intelligible :)) words/phrases?

Well excuse me for bad explaining. Well I mean  that at each time if person $A$ stops and person $A$ continue its path with its direction at this time it catch $A$.
I hope you have understood what I want to say.If you haven't ask me I explain more.

Well as is shown on figure direction of the speed vector of $B$ at each time is aline passing through $A$.Speed of $A$ is $v$ and speed of $B$ is $v'$,and $v \geq v'$.
Now we know on what line $A$ moves and where is its start point.
Find all points that if $B$ start its move at this point it catches $A$.

Yes, I understood it the second time you explained it. I just don't think it's easy :). Thanks for taking the time to make it clear for me.

No solution!!! Well people think easily this problem is very simple.

First of all, let's find the trajectory of $B$
Well, it's not strictly necessary, but it's nice :
let $(x_A(t), y_A(t))$ be the position of $A$ at time $t$ and let, wlog, $x_A(t)=0\ \forall t$ and $y_A(t)=v_At$; similarly let $(x_B(t),y_B(t))$ be the positione of $B$ at time $t$.
Let me now use $x$ instead of $x_B(t)$ and $y$ instead of $y_B(t)$.

Thanks to the fact that $B$ has constant speed, we can write
\[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=v_B^2\]
or
\[\left(\frac{dx}{dt}\right)^2\left[1+\left(\frac{dx}{dy}\right)^2\right]\qquad (1)\]

Then, we may state that
\[v_At-y=\frac{dy}{dx}x\]
because $B$ "follows" $A$.
Taking the derivative in t, we get
\[v_A-\frac{dy}{dt}=\frac{d^2y}{dx^2}\frac{dx}{dt}x + \frac{dy}{dx}\frac{dx}{dt}\qquad (2)\]

Now, getting $\frac{dx}{dt}$ from (2) and substituting in (1) we get
\[1+\left(\frac{dy}{dx}\right)^2=\frac{v_B^2}{v_A^2}x^2\left(\frac{d^2y}{dx^2}\right)^2\]

Separating the variables we get (let $k=v_B/v_A$):
\[\frac{dy}{dx}=\frac{1}{2}\left(Cx^{-1/k}-\frac{1}{C}x^{1/k}\right)\]

By knowing that, at $x=x_B(0)$, $\frac{dy}{dx}=y_B(0)/x_B(0)$ we can find C.

Integrating again, we get, if $k\neq 1$:
\[y=\frac{1}{2}\left[\frac{Ck}{1-k}x^{1-1/k}+\frac{k}{C(1+k)}x^{1+1/k}\right] + D\]
where we can know $D$ by imposing the value of $y_B(0)$.

Know, for evry choice of $C$ and $D$, if $0<k<1$, this curve has an asymptote for $x=0$, unless, obviously, $x_B(0)=0$

If $k=1$, the equation is more pleasant:
\[y=\frac{1}{2}\left(\frac{x^2}{C}-C\log(x)\right) + D\]
which, again, has an asymptote in $x=0$, unless, obviously, $x_B(0)=0$.

The problem is more difficult if $A$ moves on a curve, instead of a straight line...I don't thinks it's solvable in the most general case...even if $A$ moves along a circle, the calculations grow horribly.

Well I see your solution is very huge.I'll send a shorter solution soon

[quote="EvaristeG"]
First of all, let's find the trajectory of $B$
Well, it's not strictly necessary, but it's nice :
[/quote]
This means that mine is not the shortest way...I only thought that it'd be interesting to find this trajectory...
A shortest solution can be:
let $A$ be in the point $(0,0)$ and move on the straight line $x=0$;

1) if $B$ is, at the beginning, in a point of the halfplane $x\leq 0$ he won't catch
$A$ unless he too starts from $(0,0)$; this can be easily seen from the fact that $B$ moves on a curve (not necessarily a straight line) and that $v_B\leq v_A$;

2) if $B$ is in the halfplane $x>0$, we have to find out if it's possible to get to a moment when $A$ and $B$ have the same $y$ and different $x$s; if it is possible, $B$ won't catch $A$ (as we are in the previous case).

Let now suppose that, in hypotesis of case 2), $B$ reaches $A$, then the tangent to the trajectory of $B$ will be the line $x=0$ and $y_B\geq y_A$; but if at a given time $y_B\geq y_A$, the tangent to the trajectory of $B$ will be a line with negative slope; since we suppose that $y_B(0)>y_A(0)$, in the moments befor $B$ reaches the line $x=0$  we will have $(y_B-y_A)/(x_B-x_A)\leq 0$, but this would imply that at two given times $t_1>t_0$ we would have $y_A(t_1)<y_A(t_0)$, but this is impossible.
(If you draw it and think is easier....it's been terribly hard to write it in a semi-intelligible way)

Only an intuitive sketch.

Lets suppose that the prey start from P the hunter from H and it catch the prey at K.

The line PK is the trajectory of  the prey  and the arc (HHK) is the trajectory of the hunter.

As Vp > Vh, the angle in K between the two trajectory must be greater or equal to 90 , otherwise is easy to see that the hunter cant catch the prey.

As Vp > Vh, |PK| > |(HHK)| >|HK| then it holds that <KHP > <HPK (*).

But in some intermediate instant the positions of hunter and prey H and P are s.t. the (*) can be respected, then we have a contradiction.

Then the only good positions of the hunter are those on the line (and in front) of the  prey.

Suppose at time $t$, $A$ and $B$ are at points $f(t)=(x_t,y_t), g(t)=(x'_t,y'_t)$.
Suppose at each time $y_t=0$ and $x_t$ is increasing now We say that $B$ must be on the line $y=0$ and $x'_0 \geq 0$.Well with this conditions ,it's obvious that $B$ can catch $A$.
Now suppose the infinity point $\infty_0=(0,+\infty)$.It is obvious that at time $t$ if the angle $<BA \infty_0$ is bigger than $\frac \pi{2}$ then $B$ can't reach $A$ because $y_t > y'_t$ it is easy to prove this. Now Suppose $h(t)=<BA \infty_0$ at the time $t$.
We prove that for every time $t$ that $0<h(t)< \frac \pi{2}$ then there is a time $t'$ that $h(t') \geq 2h(t)$.At time $t$ suppose perpendicular bisector of $AB$ intersect the line $y=0$ at $A'$.Easily $B$ can't catch $A$ till it reaches to point $A'$.And we can prove easily that when $A$ reach at point $A'$ in time $t '$ then $B$ reach to point $B'$ in the triangle $ABA'$ .Now $h(t')=<B'A' \infty_0 \geq <BA' \infty_0 =2<BAA'=2<BA \infty_0=2h(t)$.
Well now problem is obvious and eveything is solved.

[quote="sprmnt21"]

But in some intermediate instant the positions of hunter and prey H and P are s.t. the (*) can be respected, then we have a contradiction.
[/quote]

here is the reason of that:

as the tangent h'(t) to the arc (HH'K) varies with continuity, there is a point H' s.t. H'P' is parallel to HK[this result, if I'm not wrong is due to Lagrange]. If <H'KH is less than <HKoo it results that |PK| < |KH'| < arc(H'K) and we are done. If, instead, <H'KH >  <HKoo we have <H'KH > 2<HKoo.

Then we can apply the same argument to the new position [P', H', K]  and son on. But, as the angle in K between PK and the arc(HH'K) is greater than 90 and due to the fact that 2^n <HKoo > 90 for some n,
it is sure that we will have <H_nKH less than <HKoo.

Let $ABC$ be an arbitrary triangle.
$B$ moves towards $C$ with constant speed size $\alpha$ at each time.
$C$ moves towards $A$ with constant speed size $\beta$ at each time.
$A$ moves towards $B$ with constant speed size $\gamma$ at each time.
i) Determine $\alpha,\beta,\gamma$ dependent on triangle $ABC$ so that triangle $ABC$ be similar to itself at each time.
ii) How long does it take for them to meet? (dependent on $a,b,c$)
iii) In this case, what is the point $P$ at which $A,B,C$ meet each other?
iv) How many times do they go around $P$?

Doesn't any one answer my question? ;)
It's a good problem.


geometry
perimeter
rectangle
number theory
relatively prime
perpendicular bisector

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD$. The crease is $EF$, where $E$ is on $AB$ and $F$is on $CD$. The dimensions $AE=8$, $BE=17$, and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C;
draw(F--D--A--E);
draw(E--B--C--F, linetype("4 4"));
filldraw(E--F--Cp--Bp--cycle, white, black);
pair point=( 12.5, 35/3 );
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
label("$B^\prime$", Bp, dir(point--Bp));
label("$C^\prime$", Cp, dir(point--Cp));[/asy]

alan, I don't understand your argument after you say $B'D=x$.

[b]EDIT:[/b] Nevermind.

[quote="JesusFreak197"]alan, I don't understand your argument after you say $B'D=x$.[/quote]

perhaps a picture would help?  the method is basically expressing two equal lengths in different ways and equating them to solve for $x$, and then plugging back in.

"Dictionary of Lengths"
AB=8, EB=17, B'E=17, AB'=15. FC=3, FD=22 (since AB=25).
Now let B'D=X.
$X^2+22^2=(15+x)^2+3^2$
Simplify so x=25/3. Add all lengths to get 290/3 which is 293.

similar triangles (all similar to AEB' with ratio 8:15:17). also, we can do some trig by extending EF, BC and B'C' and using the half-tangent formula

[b]Walk-Through:[/b]
[list]
[*] Let $K = B'C' \cap CD$.
[*] Pythag on $AB'E$ gives $AB' = 15$.
[*] Now, since $AB'E \sim C'KF$, we have $$FK = \frac{C'F}{AE} \cdot B'E = \frac{51}{8}$$ so $$DK = (8 + 17) - \left( 3 + \frac{51}{8} \right) = \frac{125}{8}.$$
[*] Moreover, because $AB'E \sim DKB'$ also holds, $$B'D = \frac{DK}{AB'} \cdot AE = \frac{25}{3}.$$
[*] Thus, $$P = 2 \left( \left(15 + \frac{25}{3} \right) + (8 + 17) \right) = \frac{290}{3}.$$
[*] Extract $m + n = \boxed{293.}$
[/list]


Ellen baked $ 2$ dozen cupcakes of which half contained chocolate, two-thirds contained raisins, one-fourth contained chocolate chips, and one-sixth contained nuts.  What is the largest possible number of cupcakes that had none of these ingredients?

We have $ 24 \div 2 \equal{} 12$ that contain chocolate, $ 24 \times \frac{2}{3} \equal{} 16$ that contain raisins, $ 24 \div 4 \equal{} 6$ that contain chocolate chips, and $ 24 \div 6 \equal{} 4$ that have nuts. This makes a total of $ 12 \plus{} 16 \plus{} 6 \plus{} 4 \equal{} 38$ cupcakes, but we only have 24 cupcakes in total! That means $ 38 \minus{} 24 \equal{} \boxed{14}$ cupcakes don't have any of these ingredients.

I don't think that's right izzy. That counts how many cupcakes that have more than one topping.

[hide="what I think"]
We can assume that all $ 16$ of the cupcakes with raisins have every other topping. That means there are $ 24\minus{}16\equal{}\boxed{8}$ cupcakes with no toppings (maximum).[/hide]

[quote="RoFlLoLcOpT"]I don't think that's right izzy. That counts how many cupcakes that have more than one topping.

[hide="what I think"]
We can assume that all $ 16$ of the cupcakes with raisins have every other topping. That means there are $ 24 \minus{} 16 \equal{} \boxed{8}$ cupcakes with no toppings (maximum).[/hide][/quote]

This is correct:
4 nut+chocolate chip+raisins+chocolate
2 chocolate chip+raisings+chocolate
6 chocolate+raisins
4 just raisin

Make a total of 16 cupcakes. 8 remain.

We see that the largest amount of cupcakes with some type of the ingredients is 2/3, that means all the other toppings can be included in this one. Thus, 1/3 x24= 8 cupcakes.


Consider an infinite sequence of positive real numbers $a_{1}, a_{2}, \ldots$ and a positive real $\alpha$ such that

(1) $a_{i}= \sqrt[i]{\alpha}$ for $i = 1, 2, \ldots, n$;

(2) $a_{k+n}= (a_{k}a_{k+1}\cdots a_{k+n-1})^{\frac{1}{n}}$ for $k = 1, 2, \ldots$.

Does this sequence converge? If so, what is its limit?

Speaking of which, could you upload the other BATH tests?  I wanted to post a few of the questions I wrote, but I've forgotten them...  :oops:

[hide="Anyway, solution"] The sequence is clearly bounded above and below, although I am not sure if it's monotonic.  Anyway, it almost definitely converges.  Yeah.  So:

Let $b_{i}= \ln a_{i}$.  Let $\beta = \ln \alpha$.  Then

$b_{i}= \frac{\beta}{i}, i = 1, 2, ... n$
$b_{k+n}= \frac{b_{k}+...+b_{k+n-1}}{n}, k = 1, 2, ...$

For $n$ consecutive positive integers we have the identities

$nb_{k+n}= b_{k}+...+b_{k+n-1}$
$nb_{k+n+1}= b_{k+1}+...+b_{k+n}$
...
$nb_{k+2n-1}= b_{k+n-1}+... b_{k+2n-2}$

Adding and collecting like terms:

$b_{k}+2b_{k+1}+3b_{k+2}+... nb_{k+n-1}= b_{k+n}= 2b_{k+n+1}+3b_{k+n+2}+...+nb_{k+2n-1}$

In particular, for $k = mn+1$ we have the identity

$b_{mn+1}+2b_{mn+2}+...+nb_{(m+1)n}= b_{(m+1)n+1}+2b_{(m+1)n+2}+...+nb_{(m+2)n}$

In other words, let $s_{m}= b_{mn+1}+...+nb_{(m+1)n}$, where

$s_{1}= \beta+\frac{2 \beta}{2}+...+\frac{n \beta}{n}= n \beta$

Then

$s_{m+1}= s_{m}\Leftrightarrow s_{m}= s_{1}\forall m \in \mathbb{N}$

If $\lim_{k \to \infty}b_{k}= b$, then

$\lim_{m \to \infty}s_{m}= n \beta = \frac{n(n+1)}{2}b \implies$
$b = \frac{2 \beta}{n+1}$

Hence

$\lim_{k \to \infty}a_{k}= \boxed{ \alpha^{ \frac{2}{n+1}}}$ [/hide]


calculus
integration
logarithms
trigonometry
calculus computations

Evaluate $ \int_{0}^{\pi} \ln (1\plus{} \cos x)\ dx$.

hello, the value of this integral is $ \minus{}\pi\ln(2)$.
Sonnhard.

Correct, Sonnhard, but is there a nice way of proving it?

[quote="AndrewTom"]Evaluate $ I\equal{}\int_{0}^{\pi} \ln (1 \plus{} \cos x)\ dx$.[/quote]

Set : $ x\equal{} \pi \minus{}t$

* $ I\equal{}\int_{0}^{\pi} ln(1\plus{}cos(\pi\minus{}t))dt\equal{} \int_{0}^{\pi}ln(1\minus{}cosx)dx$

* $ 2I\equal{}\int_{0}^{\pi} ( ln(1\plus{}cosx)\plus{}ln(1\minus{}cosx))dx\equal{}\int_{0}^{\pi} ln(1\minus{}cos^{2}x)dx$

$ \ \equal{}\int_{0}^{\pi}ln(sin^{2}x)dx\equal{}2\int_{0}^{\pi}ln(sinx)dx\equal{}\minus{}2\pi ln(2)$

$ \Rightarrow I\equal{} \minus{}\pi ln(2)$

Done :)

Thanks materazi. Very nice. I have the following:

Let $ I(\alpha) \equal{} \int_{0}^{\pi} \ln(1 \plus{} \alpha \cos x)\ dx$.

Then $ \frac{dI(\alpha)}{d\alpha} \equal{} \int_{0}^{\pi} \frac{\cos x}{1\plus{}\alpha \cos x}\ dx$

$ \equal{} \frac{\pi}{\alpha} \minus{} \frac{\pi}{\alpha\sqrt{1\minus{}\alpha^{2}}}$.

Now $ I(0) \equal{} 0$, so $ I \equal{} I(1) \equal{} \int_{0}^{1} \frac{dI(\alpha)}{d\alpha}\ d\alpha$

$ \equal{}\int_{0}^{1}\frac{\pi}{\alpha} \minus{} \frac{\pi}{\alpha\sqrt{1\minus{}\alpha^{2}}}\ d\alpha$

$ \equal{}\pi[\ln \alpha \plus{} \ln \frac{1\plus{}\sqrt{1\minus{}\alpha^{2}}}{\alpha}]_{0}^{1}$

$ \equal{}\minus{}\pi\ln 2$


Εστω $ a_n$ κυρτη ακολουθια δηλαδη $ 2a_n \leq a_{n\plus{}1}\plus{}a_{n\minus{}1}$

Να αποδειχθει οτι:

$ lim(a_{n\plus{}1}\minus{}a_n)\equal{}0$

[quote="AlexandrosG"]Εστω $ a_n$ κυρτη ακολουθια δηλαδη $ 2a_n \leq a_{n \plus{} 1} \plus{} a_{n \minus{} 1}$

Να αποδειχθει οτι:

$ lim(a_{n \plus{} 1} \minus{} a_n) \equal{} 0$[/quote]

Γενικά γνωστό, αλλά χρειάζεσαι διαφορετική διατύπωση (π.χ. $ a_n$ φθίνουσα), αλλιώς η $ n^2$ είναι κυρτή και χτυπάει...

Cheerio,

Durandal 1707

Sorry για την παραλειψη.:blush:
Η ακολουθια ειναι επιπλεον φραγμενη.
Πως μπορουμε τωρα να δειξουμε οτι οι διαφορες διαδοχικων ορων της τεινουν στο 0?

[quote="AlexandrosG"]Η ακολουθια ειναι επιπλεον φραγμενη.[/quote]

Ok, τότε αρκεί να δείξεις ότι η ακολουθία είναι φθίνουσα ;)

Αν υπάρχει $ a_{n}\geq a_{n\minus{}1}$ και θέσεις $ \delta \equal{} a_{n\plus{}1}\minus{}a_n$, τότε $ a_{n\plus{}m}\geq a_n \plus{} m\delta$, άτοπο αφου η ακολουθία είναι φραγμένη. Άρα, η $ a_n$ είναι φθίνουσα και φραγμένη $ \implies$ συγκλίνει...

Πιο εύκολα (μιας και δε ζήτησες να δειχθεί ότι η $ a_n$ συγκλίνει), αν θέσεις $ b_n \equal{} a_n \minus{} a_{n\minus{}1}$, η $ b_n$ θα είναι αύξουσα και άνω φραγμένη από το $ 0$ (αφού η $ a_n$ είναι φθίνουσα). Αν $ b_n < \minus{}\delta<0$, καταλήγεις πάλι σε άτοπο επειδή η $ a_n$ είναι φραγμένη και τελείωσες.

Hope this helps,

Durandal 1707


algebra
polynomial
function
linear algebra
matrix
induction
linear algebra unsolved

Let $ x$, $ y$, $ z$ be three real numbers satisfying that
\[ \{\begin{array}{r r r} x + y + z & = & 0 \\
x^2 + y^2 + z^2 & = & 2 \\
x^3 + y^3 + z^3 & = & 0\end{array}
\]
Prove that for all integer $ n$, we have $ x^{2n + 1} + y^{2n + 1} + z^{2n + 1} = 0$.

$ x^2\plus{}y^2\plus{}z^2\equal{}0 \Rightarrow x\equal{}y\equal{}z\equal{}0$.

I believe that $ x,y,z \in \mathbb{C}$
  Let $ x,y,z$  are the eigenvalues of  $ A \in M_3{(\mathbb{C})}$
and put $ \delta_1 \equal{} x \plus{} y \plus{} z,\delta_2 \equal{} xy \plus{} yz \plus{} xz, \delta_3 \equal{} xyz$ from Suppose we have $ \delta_1 \equal{} \delta_2 \equal{} \delta_3 \equal{} 0$
Characteristic equation of  $ A$ is : $ A^3 \minus{} \delta_1A^2 \plus{} \delta_2A \minus{} \delta_3I_3 \equal{} 0$  then $ A^3 \equal{} 0$ then $ A^{2n \plus{} 1} \equal{} 0,\forall n \in \mathbb{N^{*}}$ implies
$ Tr(A^{2n \plus{} 1}) \equal{} x^{2n \plus{} 1} \plus{} y^{2n \plus{} 1} \plus{} z^{2n \plus{} 1} \equal{} 0$
Generall $ x^k \plus{} y^k \plus{} z^k \equal{} 0, \forall k \in \mathbb{N},k \geq 3$

So many ways to finish this one...

Brute force:
$ 0 \equal{} x^3 \plus{} y^3 \plus{} z^3 \equal{} (x \plus{} y \plus{} z)(...) \minus{} 3xyz \equal{} \minus{} 3xyz \implies xyz \equal{} 0$
$ 0 \equal{} x^2 \plus{} y^2 \plus{} z^2 \equal{} (x \plus{} y \plus{} z)^2 \minus{} 2\sum xy \implies \sum xy \equal{} 0$
Consider $ p(t) \equal{} (t \minus{} x)(t \minus{} y)(t \minus{} z) \equal{} t^3$. $ x,y,z$ are its roots so $ x \equal{} y \equal{} z \equal{} 0$.

Sequential Way:
$ a_n \equal{} x^n \plus{} y^n \plus{} z^n$ obeys a linear recursion relation and is uniquely determined by its first 3 values, which are 0, so it is 0 for all $ n$.

Symmetric Polynomials (related to first solution) -
$ x \plus{} y \plus{} z$, $ x^2 \plus{} y^2 \plus{} z^2$, $ x^3 \plus{} y^3 \plus{} z^3$ are the first 3 symmetric polynomials of 3 variables, and the first 3 elementary polynomials ($ \sum x, \sum xy, xyz$) can be expressed with addition and multiplication of those 3 polynomials (Newton's Identities). Because the symmetric polynomials are 0 at $ (x,y,z)$, so are the elementary polynomials, and continue like in the first solution.

Generating Functions way (related to 2nd solution) -
The generating function of the sequence is $ \frac{1}{1\minus{}tx}\plus{}\frac{1}{1\minus{}ty}\plus{}\frac{1}{1\minus{}tz}\equal{}\frac{3\minus{}2t \sum x \plus{} t^2 \sum x^2}{1\minus{}t \sum x \plus{} t^2 \sum xy \minus{}t^3 xyz}\equal{}\frac{3}{1}\equal{}3$, and immediately $ x^n\plus{}y^n\plus{}z^n\equal{}0$ for $ n \ge 1$.

We can probably use the Vandermonde Matrix too. All in all, we always get $ x \equal{} y \equal{} z \equal{} 0$.

My apology to all. The condition $ x^2\plus{}y^2\plus{}z^2\equal{}0$ must be $ x^2\plus{}y^2\plus{}z^2\equal{}2$, sorry for my stupid mistake  :blush:

LOL... still, the previous ideas are applicable.

$ 0 \equal{} x^3 \plus{} y^3 \plus{} z^3 \equal{} (x \plus{} y \plus{} z)(...) \minus{} 3xyz \equal{} \minus{} 3xyz \implies xyz \equal{} 0$
WLOG $ x \equal{} 0$, and then $ (y,z) \equal{} ( \minus{} 1,1)$ up to a permutation.
Then $ x^{n} \plus{} y^{n} \plus{} z^{n} \equal{} 1 \plus{} ( \minus{} 1)^n$

Alternatively, $ a_n \equal{} x^n \plus{} y^n \plus{} z^n \implies a_{n \plus{} 3} \equal{} \lambda_1 a_{n \plus{} 2} \plus{} \lambda_2 a_{n \plus{} 1} \plus{} \lambda_3 a_{n}$ when $ \lambda_{1,2,3}$ are the coefficients of $ (t \minus{} x)(t \minus{} y)(t \minus{} z)$ up to a sign. $ \lambda_{1,3} \equal{} 0 \implies a_{n \plus{} 3} \equal{} \lambda_2 a_{n \plus{} 1}$. By induction, $ a_{2n \plus{} 1} \equal{} \cdots \equal{} a_{3} \equal{} 0$.

Alternately, letting $ p_i \equal{} x^i \plus{} y^i \plus{} z^i$, we have that $ p_1, p_2, p_3$ form an algebraic basis for the ring of symmetric functions in three variables. Any degree $ 2n\plus{}1$ term must involve either $ p_1$ or $ p_3$ (as $ p_2^k$ is always of even degree) and is therefore equal to zero under this substitution.

$ \delta_2 \equal{} xy \plus{} yz \plus{} xz \equal{} \minus{} 1,\delta_3 \equal{} 0$
Characteristic equation of $ A$ is $ A^3 \minus{} A\equal{}0$,from easy $ A^{2n \plus{} 1} \equal{} A,A^{2n} \equal{} A^2,\forall n \in \mathbb{N^{*}}$
then $ Tr(A^{2n \plus{} 1}) \equal{} x^{2n \plus{} 1} \plus{} y^{2n \plus{} 1} \plus{} z^{2n \plus{} 1} \equal{} x \plus{} y \plus{} z \equal{} 0 (n \geq 1)$,$ Tr(A^{2n}) \equal{} x^{2n} \plus{} y^{2n} \plus{} z^{2n} \equal{} x^2 \plus{} y^2 \plus{} z^2 \equal{} 2 (n \geq 2)$

We have $ x\plus{}y\plus{}z\equal{}0$, $ xy\plus{}yz\plus{}zx\equal{}\minus{}1$ and $ xyz\equal{}0$.
Let $ S_n\equal{}x^n\plus{}y^n\plus{}z^n$. We have \[ S_{n\plus{}3}\minus{}S_{n\plus{}1}\equal{}\sum x^{n\plus{}3}\plus{}\left(\sum x^{n\plus{}1}\right)\left(\sum yz\right)\\ \equal{}\left(\sum x^{n\plus{}2}\right)\left(\sum x\right)\plus{}xyz\left(\sum x^n\right)\equal{}0\]
It implies $ S_{n\plus{}3}\equal{}S_{n\plus{}1}$, so $ S_{2n\plus{}1}\equal{}S_1\equal{}0$.


Let $ f(x)$ and $ g(x)$ be two non zero polynomials such that $ f(x^2\plus{}x\plus{}1)\equal{}f(x)g(x)$. Prove that $ \deg f\equal{}\text{even}$.

[quote="Moonmathpi496"]Let $ f(x)$ and $ g(x)$ be two non zero polynomials such that $ f(x^2 \plus{} x \plus{} 1) \equal{} f(x)g(x)$. Prove that $ \deg f \equal{} \text{even}$.[/quote]
(I suppose that $ f,g$ are finite. Else we can't really say whether $ \deg f \equiv 0 \bmod 2$)
Assume that $ f(x)$ has at least one real root and let $ r$ be the biggest real root of $ f(x)$. Then $ f(r^2 \plus{} r \plus{} 1) \equal{} f(r)g(r) \equal{} 0$, hence $ r^2 \plus{} r \plus{} 1$ is also a real root. But $ r^2 \plus{} 1 \ge 1 > 0 \iff r^2 \plus{} r \plus{} 1 > r$. Contradiction. Hence $ f$ doesn't have any real roots and hence $ \deg f \equiv 0 \bmod 2$...

Edit: Sorry. I assumed that $ f(x),g(x)$ had real coefficients :blush:

Edit 2: I'm not that sorry. $ f(x) \equal{} x \minus{} i$. then $ \frac {f(x^2 \plus{} x \plus{} 1)}{f(x)} \equal{} \frac {x^2 \plus{} x \plus{} 1 \minus{} i}{x \minus{} i} \equal{} \frac {(x \minus{} i)(x \plus{} 1 \plus{} i)}{x \minus{} i} \equal{} x \plus{} 1 \plus{} i$. So $ f(x) \equal{} x \minus{} i, g(x) \equal{} x \plus{} 1 \plus{} i$ works but $ \deg f \equal{} 1$. So there is something wrong with the problem. (I guess that $ f(x),g(x)$ have real coefficients :P )

I also think that the polynomials are have real co-efficients. Thanks for your solution. :)


quadratics
modular arithmetic
superior algebra
superior algebra unsolved

[color=green]Let r be a primitive root of the prime p with p congruent to 1 (mod 4).  Show that -r is also a primitive root.[/color]

Thanks for your help!!!   :blush:

[b]Lemma:[/b]  $r$ a primitive root $\Leftrightarrow r^{ \frac{p-1}{2}}\equiv-1 \bmod p$

(Proof is left as an exercise.)

Since $p \equiv 1 \bmod 4$, we know that $\frac{p-1}{2}$ is even; hence,

$(-r)^{ \frac{p-1}{2}}\equiv (-1)^{ \frac{p-1}{2}}r^{ \frac{p-1}{2}}\equiv-1 \bmod p$

[quote="t0rajir0u"][b]Lemma:[/b]  $r$ a primitive root $\Leftrightarrow r^{ \frac{p-1}{2}}\equiv-1 \bmod p$[/quote]
You confuse quadratic residue with primitive root.
Counterexamples: $p=7, r=-1$ , $p=13 , r=8$.

it works for every prime $p$. if $p=2$, it's trivial since $r=-r$, so let's assume $p$ to be odd. we work in the field $\mathbb{F}_{p}$. since $r^{p-1}= 1$, we have $(-r)^{2(p-1)}=1$, hence $(-r)^{p-1}= \pm 1$. but $-1=(-r)^{p-1}$ would imply $-1==(-1)^{p-1}r^{p-1}= 1$ and hence $p=2$. hence $(-r)^{p}=1$. finally, $-r=1$ would imply that $r$ has order $2 \neq p$, a contradiction.

Use that $-1\equiv r^{2k}\pmod p$, k is natural.


inequalities
linear algebra
matrix
geometry
3D geometry
sphere

[hide]Cauchy-Schwarz Inequality implies
\[ \left( \frac{1}{2x} \plus{} \frac{1}{3y} \plus{} \frac{1}{6z} \right) \left( \frac{x}{2} \plus{} \frac{y}{3} \plus{} \frac{z}{6} \right) \ge \left( \frac12 \plus{} \frac13 \plus{} \frac16 \right)^2 \equal{} 1
\]
and in this case we have an equality. Therefore
\[ \frac{x/2}{1/(2x)} \equal{} \frac{y/3}{1/(3y)} \equal{} \frac{z/6}{1/(6z)} \implies x^2 \equal{} y^2 \equal{} z^2
\]
and since $ x,y,z >0$, $ x\equal{}y\equal{}z$ and we have that
\[ x \plus{} x^2 \plus{} x^3 \equal{} 14 \implies x \equal{} \boxed{2}.
\][/hide]

[hide="Answer"]
By Cauchy-Schwarz, $ \left(\frac x2 \plus{} \frac y3 \plus{} \frac z6\right)\left(\frac 1{2x} \plus{} \frac 1{3y} \plus{} \frac 1{6z}\right) \ge \left(\frac 12 \plus{} \frac 13 \plus{} \frac 16\right)^2 \equal{} 1$ with equality when $ \left\{\frac x2, \frac y3, \frac z6\right\} \equal{} c\left\{\frac 1{2x}, \frac 1{3y}, \frac 1{6z}\right\}$, or equivalently $ \frac {1}{c} \equal{} x^2,\ y^2,\ z^2$ and so $ x \equal{} y \equal{} z \equal{} m$.

From the second condition we have $ m^3 \plus{} m^2 \plus{} m \minus{} 14 \equal{} 0 \Longleftrightarrow \left(m \minus{} 2\right)((m \plus{} 3/2)^2 \plus{} 19/4) \equal{} 0$, and so $ \boxed{x \equal{} y \equal{} z \equal{} 2}$.

[/hide]

Edit:  :roll:

[quote="JRav"]You can't hope to obtain an actual number because you have three unknowns but only two equations.[/quote]

As the two posters before me have shown, this is clearly not the case.  Let's explore why.

Some geometric intuition:  the notion that $ n$ (linearly independent) equations determine $ n$ variables comes from linear algebra, where it means that the corresponding $ n \times n$ matrix is invertible.  Alternately, the dimension of the solution space decreases by $ 1$ with every (linearly independent) equation we add.  The behavior of systems of linear equations is straightforward because lines all behave the same way.

However, when the equations involved are nonlinear, more complex behavior can happen.  In particular, a single nonlinear equation in three variables describes a surface, which can be degenerate.  For example, the "surface"

$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 0$

is the point $ (0, 0, 0)$ (a sphere of radius zero).  Here, the restriction of the second equation to the octant with $ x, y, z > 0$ is the line $ x \equal{} t, y \equal{} t, z \equal{} t$, rather than a surface.  The first equation describes a cubic surface, and the intersection of a surface and a line not contained in the surface is a point (or some points).

Some algebraic intuition:  the fact that the second equation describes a line and not a surface means that it is actually two equations in one (a line is the intersection of two planes).  In general, a single equation that describes the intersection of two ([b]real[/b]) varieties $ f(x, y, z, ...) \equal{} 0, g(x, y, z, ...) \equal{} 0$ is

$ f(x, y, z, ...)^2 \plus{} g(x, y, z, ...)^2 \equal{} 0$.


Circles of radius 3 and 6 are externally tangent to each other and are internally tangent to a circle of radius 9. The circle of radius 9 has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.

[quote="lotrgreengrapes7926"][quote="bigboy82892"][hide="Solution."]Draw a line from the center of the radius 3 circle perpendicular to the radius of the radius 6 circle that connects to the tangent. you have a right triangle with two known sides and a rectangle. Find the unknown side of the right triangle and you get $\boxed{72}$.!!![/hide][/quote] I think you only found the length of the tangent, not the chord.[/quote]
Oops! Sorry! Then I don't know the answer. :blush:

Im sry to bring this back, but since Im studying for the AIME, Im trying to do all the problems I can.  For this, the answer is not posted yet and I get 224, I just want to know if this is correct.

I've only had a very brief look, but here is a solution. Let the centres of the circles with r=3, 6 be A, B respectively. Therefore |AB|=9. By a bit of scrutiny, one realises that AB is a diameter of the big circle, and so the centre of the big circle lies 6 from A and 3 from B, and therefore 3 from the point of tangency. This is getting a bit confusing because there is no diagram, but by Pythagoras, we know that the square of HALF of the chord = 9^2 - 3^2 = 72. Therefore the square of the whole chord = 288.

Um, well, the answer on the website says its 224

Let the chord have endpoints $X$ and $Y$ (let $Y$ be nearest the circle with radius $6$).
The big circle is centered at $O$, the circle of radius $6$ at $A$ and the circle of radius $3$ at $B$.
Draw perpendiculars from points $O$ and $A$ to $O'$ and $A'$ on chord $XY$.
Draw the perpendicular from $B$ to $B'$ on $AA'$.
$OO'$ intersects $BB'$ at $O''$.

[hide]$BO''B \sim BB'A$
So,$OO'' = 2$ and $OO' = 5$. Now Pythagorean theorem tells us $(O'Y)^{2}= (OY)^{2}-(OO')^{2}\Rightarrow (O'Y)^{2}= 56$, and $(XY)^{2}= (2(O'Y))^{2}= 4(56) = 224$. [/hide]

[quote="white_horse_king88"]Let the chord have endpoints $ X$ and $ Y$ (let $ Y$ be nearest the circle with radius $ 6$).
The big circle is centered at $ O$, the circle of radius $ 6$ at $ A$ and the circle of radius $ 3$ at $ B$.
Draw perpendiculars from points $ O$ and $ A$ to $ O'$ and $ A'$ on chord $ XY$.
Draw the perpendicular from $ B$ to $ B'$ on $ AA'$.
$ OO'$ intersects $ BB'$ at $ O''$.

[hide]$ BO''B \sim BB'A$
So,$ OO'' \equal{} 2$ and $ OO' \equal{} 5$. Now Pythagorean theorem tells us $ (O'Y)^{2} \equal{} (OY)^{2} \minus{} (OO')^{2}\Rightarrow (O'Y)^{2} \equal{} 56$, and $ (XY)^{2} \equal{} (2(O'Y))^{2} \equal{} 4(56) \equal{} 224$. [/hide][/quote]

Sorry but i couldn't figure out what did you mean by $ BO''B \sim BB'A$


I guess this is the right place to put this (?)

Linguistics seems to be getting more popular, especially due to the recent development of the ILO and North American Computational Linguistics Olympiad. And there is definitely a high correlation between math and linguistics; in fact, there are many AoPSers who seem to be interested in it. So I think it would be appropriate to add a linguistics forum to arts/sciences.

Oooh oooh ooh! Did you do NAMCLO too?

Also would vocabulary-study fit into that forum? If it does and it's reasonably popular I'd advertise it among my vocab contacts :).

No, I didn't learn about it until afterwards... Where can I find the problems? I don't like the weird trend on the NAMCLO (who cares how to make the handwriting recognition program better...). The ILO looks really good, though.

I would think "vocabulary study" would fit this. I mean, I don't know exactly what that entails.

Anyway, let's hear what an admin has to say (by the way, I would certainly moderate if necessary).

Could we also get some engineering forums (like electrical engineering for example)?  I'd mod the EE one.

After (1 minute of) careful consideration, I think I'll add the two forums (Linguistics & Engineering in the Arts & Sciences area), but if they're not used we might delete them in the future.

I think that if these 2 were added, especially with the discussion in engineering as low as someone mentioned in the thread for astronomy, it might be fair to let there be an astronomy forum too?

I second that (again), especially if the mods can delete it if there's not as much activity as I had hoped in the other thread.

EDIT: Oops, I realize that what I said wasn't very clear. What I meant was that I had hoped there would be lots of activity on an astronomy board, but that I had expressed my hopes in another thread. :blush: Sorry. I was typing quickly.

:lol: Three votes.
The other thread is quiet.

I see we still have no replies from a moderator as to whether or not an astronomy board will happen...

In this thread, there were only three supporters for two new boards, and both of these boards were created. Now we have at least six people from both threads (AstroPhys, Klebian, 1=2, Kaoticity, funcia, and myself) in support of an astronomy board.

We have four times the support per board as the people in favor of engineering and linguistics, yet we've barely had a sentence from a mod on the subject. I think all of us would really appreciate a "yes" or a "no" now that we have shown we have supporters, as was asked by Valentin Vornicu in the other thread.

By mod I think you mean admin. But anyway, you've shown some interest so we might as well see what will happen.


$ \text{lg} \ x \equiv \log_{10} x \vee \log_{2} x$?

EDIT $ \text{\LaTeX}$ typo

[quote="AstroPhys"]$ lg x \equiv log_{10} x \vee \log_{2} x$?[/quote]

What does the symbol $ \vee$ means?

Answer is x=1000, but I have no idea how should I get this.

$ \vee$ means [b]or[/b]. In your problem, does $ \lg x$ mean $ \log_{10}x$ or $ \log_2x$? What do you want us to do with the expression you posted? (I'm guessing you want to solve) Please state more clearly.

If you mean $ \lg x \equal{} \log_{10}x$ then...
[hide="Solution"](I use $ \log$ instead of $ \lg$.)

Taking $ \log$ on both sides yields $ (\log x)(\log x) \equal{} \log 1000 \plus{} 2\log x$ (by the identity $ \log x^a \equal{} a\log x$.)

Or equivalently $ (\log x)^2 \minus{} 2\log x \minus{} 3 \equal{} 0$.

Which factors as $ (\log x \plus{} 1)(\log x \minus{} 3) \equal{} 0$. Implying $ \log x \equal{} \minus{} 1$ or $ 3$.

Now $ \log x\equal{}\minus{}1\implies x\equal{}10^{\minus{}1}\equal{}\frac{1}{10}$ and $ \log x \equal{} 3\implies x \equal{} 10^3 \equal{} 1000$. So the solutions are $ x\equal{}\frac{1}{10}, 1000$.
[/hide]

Yes,  $ lg x \equiv log_{10} x$, sorry. :) Thank you for solution

[quote="nayel"]

Taking $ \log$ on both sides yields $ (\log x)(\log x) \equal{} \log 1000 \plus{} 2\log x$ (by the identity $ \log x^a \equal{} a\log x$.)

[/hide][/quote]
I don't understand why

 $ x^{\lg x}\equal{}(\log x)(\log x)$?

[quote="Lampetija"]

I don't understand why

 $ x^{\lg x} \equal{} (\log x)(\log x)$?[/quote]

$ log(a^b) \equal{} blog(a)$

So $ log(x^{log(x)}) \equal{} log(x)log(x)$


AMC
AIME
blogs
geometry
3D geometry
AIME I

CHECK YOUR ANSWERS
CHECK THEM AGAIN
AND FINALLY WHEN U CANT CHECK THEM ANYMORE
CHECK FOR 5 MORE MINUTES

you have to find what errors you make, and check for them, like i check computation and overcounting very carefully

like

24+60=30

or 332/1996=332/998

etc

Be conscientious

Be scrupulous

Be mercurial

YAY all words from my aops vocab blog that i learned

all meaning

BE CAREFUL OF DETAILS!!  :lol:

it will probably help to check the problems a little while after you did them instead of just immediately after.

stay organized, write neatly, label things if it helps. use seperate sheets of paper for each one; that way you won't confuse one problem's work for another problem.

Keep different problems' work separate, check computation, and don't make assumptions (or be sure your assumptions are absolutely right).

It certainly helps to keep your work organized.  That way, if you do something like write down a term wrong from one line to the next, it's clearer to see.

Edit:  Oh, and I like to redo problems with several different methods, that's a great way of checking them, especially on counting problems.

I think that most errors do not occur in reading the problem wrong or multiplying or dividing numbers. It is when you add or subtract that most of errors come.

I did a practice AIME today and I brute forced two problems which required about 4 pages of addition (yes, my problem solving skill deteriorated :rotfl: ). I've actually made errors couple of times but I still managed to catch them right away because I focused myself to either add or subtract, never both.

Like if you have 54+27, 67-25, 71+32, 11-31. Do the adding ones first and then move on to the subtracting ones. Brain works better this way.

Also, this is only for subtraction (you can use for addition but at least I don't). If you do subtraction, add the two numbers (one you subtracted and the result) to see if you get the original number or not. This is the best way to check your work.

Anyway, humans are not perfect and mistakes do happen. It's probably easier to find mistakes than reduce them.  :yup:

[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php[/url]

 :whistling:

Be neat and organized and try not to work too fast, you might skip a step, and besides you have three hours

Recently I've been able to minimize calculation errors by checking everything mod 9, and for subtraction, I always check the calculation with addition.

Step one: make sure you read ALL of the problems correctly
Step two: check the logical process of solving the problem; is each step logical
Step three: check calculations

Disclaimer: doesn't work for me  :( I still make 1-5 errors per test on the aime :(

[quote="bpms"]Step 1: Don't be careless.
If you can get this down, you'll be fine.[/quote]

o rly?

i think you just should not make careless errors

thats how i plan to do it

Make sure you're answering the correct question.  If they ask for the radius, don't give the diameter.  If they ask for $a-b$, don't give $a+b$, etc.

I check my solution as I'm doing it, and then just after that as well.  Then, after I've finished the first 10 problems (or whichever of them I can, if one of them happened to trip me up), I come back and check every single one very carefully before going on.  Finally, if I have time, I check through all of my solutions at the end.

It helps to actually solve the problems carefully rather than relying on having to come back and check to correct them...saves time.

Basic calculation skills are a basic part of AIME performance. If you "understand how to do the problems" but still get them wrong, then you really don't understand how to do the problems. It's even more of a slight if you're making false assumptions or balancing your equations incorrectly. Not making mistakes is part of being good at math, and the AMC isn't going to give you extra points for being close to the answer.

make sure your answer makes sense like if you are cubing some large number, make sure the last digit of your answer agrees with the last digit of the cube of the last digit of the number you are cubing.

Here are, I would say, most of the most important methods:

Edit: Oh, and I like to redo problems with several different methods, that's a great way of checking them, especially on counting problems.

Read Pietermate's most, that's definitely a good checklist.

In response to Chess64, hey I use mods too!

Definitely make sure you read the problem and found what it asked for - in fact, read it twice

[color=red][size=200][u]stay organized[/u][/size][/color]

If you stay very organized, you're much likely to copy wrong or make many kinds of errors. For example, on a practice AIME, I started with $4x+3y$ and ended up with $3x+4y$, causing me to get the problem wrong. Be very organized and write neatly and use lots of paper  :)

Yeah, on the subject of saving paper, since the AIME is only 15 problems, I would recommend never having work for two different problems on the same sheet of paper.  And clearly label each sheet with what problem it is so you can come back to it later if need be.

Are you allowed to prelabel your scratch paper for the Aime, or does it have to be completely blank?

Whenever you get an answer, make sure that it is in simplest form. So far, on both practice AIMEs that I've done, I have messed up on this. Just do this whenever you get an answer, and whenever you check it. That should work for you.

A lot of people, after doing their calculation, go back with the attitude that, oh I probably have it right, but I'm checking just in case. If you do that, you will make the same mistake again.

Make sure you check [b]aggressively[/b]. If it helps, after you do a bit of checking, put the problem aside and do some different problems. Go back later when you've forgotten the specific numbers, and then [b]re-check aggressively[/b].

[quote="humDdum"]Does anyone want to offer suggestions how to get rid of dumb errors? Apparantally whenever I go back to check my answers for practice AIMEs, I realize I know how to do the problems, but still get the answers wrong... Hm... I don't even know if this makes sense anymore... Well, post suggestions please. :D[/quote]

Heyy jeean  :D

I learned this from AMP: be ultracrazyorganized. As in, work vertically, keep problem sheets separate, show out even arthithmatic (b/c that's where I make all my errors), etc.

Outside of that, maybe slowing down works... sometimes I work so quickly that I do something like 2+3=6...and on the AIMEs, there's plenty of time so yeah...

but how's running going?


If $ \alpha,\beta,\gamma,\lambda$ are the roots of the equation $ x^{4}\minus{}x^{3}\plus{}1\equal{}0$ then the value of
$ \frac{1}{\alpha\plus{}\beta\plus{}\gamma}\plus{}\frac{1}{\lambda\plus{}\beta\plus{}\gamma}\plus{}\frac{1}{\alpha\plus{}\lambda\plus{}\gamma}\plus{}\frac{1}{\alpha\plus{}\beta\plus{}\lambda}$ is

a)odd prime
b)even prime
c)composite number
d)none of these

[hide]I think it's $ d$

$ \frac{1}{\alpha\plus{}\beta\plus{}\gamma}\plus{}\frac{1}{\lambda\plus{}\beta\plus{}\gamma}\plus{}\frac{1}{\alpha\plus{}\lambda\plus{}\gamma}\plus{}\frac{1}{\alpha\plus{}\beta\plus{}\lambda} \equal{} \frac{1}{1\minus{}\alpha}\plus{}\frac{1}{1\minus{}\beta}\plus{}\frac{1}{1\minus{}\gamma}\plus{}\frac{1}{1\minus{}\lambda}$

Substitute $ \frac{1}{1\minus{}x}\equal{}y$

$ (\frac{y\minus{}1}{y})^{4}\minus{}(\frac{y\minus{}1}{y})^{3}\plus{}1\equal{}0$
$ \iff (y\minus{}1)^4 \minus{} y(y\minus{}1)^3\plus{}y^4$

The coefficient of $ y^4$ is $ 1$
The coefficient of $ y^3$ is $ \minus{}1$
So it's neither prime nor composite.[/hide]


If I click on icon above in ML mode called "LateX Help", it open a new window and write 404 Not found.

Maybe it should leads to

http://www.mathlinks.ro/Forum/index.php?f=123
or
http://www.mathlinks.ro/Forum/viewtopic.php?t=123690

nice day

conceive

If you are using the ML domain, it won't work. It only works on the aops domain name.

The old MathLinks skin is deprecated and will slowly be replaced with the new one (can be observed on the main url http://www.mathlinks.ro ).