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The sum of the base-$ 10$ logarithms of the divisors of $ 10^n$ is $ 792$. What is $ n$?

$ \textbf{(A)}\ 11\qquad
\textbf{(B)}\ 12\qquad
\textbf{(C)}\ 13\qquad
\textbf{(D)}\ 14\qquad
\textbf{(E)}\ 15$

[hide]
$ 10^n\equal{}2^n5^n$.  Thus the number of divisors of $ 10^n$ is $ (n\plus{}1)^2$.  If $ n$ is odd, then these form $ \frac{(n\plus{}1)^2}{2}$ pairs that multiply to $ 10^n$.

The sum of the logarithms of the divisors is the same as the logarithm of the product of the divisors, so we have, if $ n$ is odd,

$ 10^{n(n\plus{}1)^2/2}\equal{}10^{792}$

It is easy to verify that $ n\equal{}11$ satisfies this, so we don't need to check the case where $ n$ is even.  Answer is $ A$.
[/hide]

No need to divide into even and odd cases. The sum of the logarithms is

$ \sum_{j\equal{}0}^n\sum_{k\equal{}0}^n\log(2^j5^k)\equal{}\sum_{j\equal{}0}^n
\sum_{k\equal{}0}^nj\log 2\plus{}k\log 5$

$ \equal{}\sum_{j\equal{}0}^n\sum_{k\equal{}0}^nj\log 2\plus{}\sum_{k\equal{}0}^n\sum_{j\equal{}0}^nk\log 2$

$ \equal{}((n\plus{}1)\log 2)\sum_{j\equal{}0}^nj\plus{}((n\plus{}1)\log 5)\sum_{k\equal{}0}^nk$

$ \equal{}\frac{n(n\plus{}1)^2}2\left(\log 2\plus{}\log 10\right)\equal{}\frac{n(n\plus{}1)^2}2.$

So $ n(n\plus{}1)^2\equal{}1584.$

That's something a little bigger than $ n^3$ being a little bigger than $ 10^3.$ $ n\equal{}11$ should be the first or second case you try.

The product of all the divisors of $ k\equal{}p_1^{e_1}\ldots p_n^{e_n}$ is simply $ k^{n/2}$. (Use Gaussian pairing.)

[quote="tenniskidperson3"]What about 16? $ 1\times2\times4\times8\times16 = 1024$ and $ 16 = 2^4$ but $ 1024\ne16^{\frac {1}{2}}$.[/quote]

The solutions above involve the fact that we're dealing with a power of 10.  So $ 2^{4}$ doesn't necessarily have to fall into the pattern.

These types of problems with product or sum of the divisors are pretty famous...

im surprised this was number 23 on the 12.

[quote="tenniskidperson3"]What about 16? $ 1\times2\times4\times8\times16 = 1024$ and $ 16 = 2^4$ but $ 1024\ne16^{\frac {1}{2}}$.[/quote]

Oh wait, sorry, I mis-posted. It should be $ k^{\text{number of divisors of k}/2}$.

[hide="Solution"]The sum of the base-10 logarithms is equal to the base-10 logarithm of the product of all divisors.

We know there are an equal number of 2's and 5's, so let's just count the total number of 2's.
For any particular exponent of 5 (0 through n), the exponent of 2 may range from 0 through n, which sum to $ \frac{n(n\plus{}1)}{2}$

Multiply by the $ n\plus{}1$ values for the exponent of $ 5$ to get $ \frac{n(n\plus{}1)^2}{2} \equal{} 792 \iff n^3 \plus{} 2n^2 \plus{} n \equal{} 1584$

Since $ 12^3 > 1584$, the only legitimate answer choice is $ \boxed{\textbf{(A)} \; 11 }$[/hide]

As soon as you notice that $ \frac{n(n\plus{}1)^2}{2} \equal{} 792$, and $ 11|792$, we must have $ n\equal{}11$.

[quote="t0rajir0u"]Still wrong.  The correct expression is $ k^{\lfloor d(k)/2 \rfloor}$ (consider what happens when $ k$ is a square).[/quote]
I think you want to omit the "floor" operation.  Because if we have a perfect square, its square root is an unpaired factor.  The square root contributes exactly "half" as much as any pair of factors.

:oops:

It seems I was thinking of something else.

How would we find the base-$ 10$ logarithms of the divisors of $ 10^{n}$?  Thanks in advance!

You can't find the exact base ten logarithms of any integers besides the powers of ten. In this problem, you needed to cleverly combine the divisors like 2 and 25 since you don't have their logs. The easiest way to do this is to take each pair of factors that multiply to $ 10^n$ and sum their logs beforehand. You know you'll get $ n$, or that the average is $ n/2$ for all factors of $ 10^n$ (including perhaps $ 10^{n/2}$ if $ n$ is even, which is otherwise uncombined). The rest of the problem is counting how many factors there are, which isn't hard (you can try small cases and come up with a pattern or use common number theory to make the prediction).

[quote="314.math"][quote="E^(pi*i)=-1"]If $ n$ is odd, then these form $ \frac {(n \plus{} 1)^2}{2}$ pairs that multiply to $ 10^n$.[/quote]

How do we know this?  Thanks![/quote]

The number of divisors of $ a_1^{e_1}a_2^{e_2}...a_n{e_n}$ is $ (e_1 \plus{} 1)(e_2 \plus{} 1)...(e_n\plus{}1)$. $ 10^n\equal{}2^n 5^n$, so it has (n+1)(n+1) divisors. We can pair $ 2^a 5^b$ and $ 2^{n\minus{}a}5^{n\minus{}b}$. Since each pair contains 2 elements, and there are $ (n\plus{}1)^2$ elements in total, there are $ \frac{(n\plus{}1)^2}{2}$ pairs.

[quote="pythag011"][quote="314.math"][quote="E^(pi*i)=-1"]If $ n$ is odd, then these form $ \frac {(n \plus{} 1)^2}{2}$ pairs that multiply to $ 10^n$.[/quote]

How do we know this?  Thanks![/quote]

The number of divisors of $ a_1^{e_1}a_2^{e_2}...a_n{e_n}$ is $ (e_1 \plus{} 1)(e_2 \plus{} 1)...(e_n \plus{} 1)$. $ 10^n \equal{} 2^n 5^n$, so it has (n+1)(n+1) divisors. We can pair $ 2^a 5^b$ and $ 2^{n \minus{} a}5^{n \minus{} b}$. Since each pair contains 2 elements, and there are $ (n \plus{} 1)^2$ elements in total, there are $ \frac {(n \plus{} 1)^2}{2}$ pairs.[/quote]

What does this notation mean: $ a_1^{e_1}a_2^{e_2}...a_n{e_n}$?  Thanks!

[quote="314.math"][/quote][quote="pythag011"][quote="314.math"][quote="E^(pi*i)=-1"]If $ n$ is odd, then these form $ \frac {(n + 1)^2}{2}$ pairs that multiply to $ 10^n$.[/quote]

How do we know this?  Thanks![/quote]

The number of divisors of $ a_1^{e_1}a_2^{e_2}...a_n{e_n}$ is $ (e_1 + 1)(e_2 + 1)...(e_n + 1)$. $ 10^n = 2^n 5^n$, so it has (n+1)(n+1) divisors. We can pair $ 2^a 5^b$ and $ 2^{n - a}5^{n - b}$. Since each pair contains 2 elements, and there are $ (n + 1)^2$ elements in total, there are $ \frac {(n + 1)^2}{2}$ pairs.[/quote][quote="314.math"]

What does this notation mean: $ a_1^{e_1}a_2^{e_2}...a_n{e_n}$?  Thanks![/quote]

It is the prime factorization. For example $ 108=2^2*3^3$

[quote="pythag011"][/quote][quote="314.math"][/quote][quote="pythag011"][quote="314.math"][quote="E^(pi*i)=-1"]If $ n$ is odd, then these form $ \frac {(n + 1)^2}{2}$ pairs that multiply to $ 10^n$.[/quote]

How do we know this?  Thanks![/quote]

The number of divisors of $ a_1^{e_1}a_2^{e_2}...a_n{e_n}$ is $ (e_1 + 1)(e_2 + 1)...(e_n + 1)$. $ 10^n = 2^n 5^n$, so it has (n+1)(n+1) divisors. We can pair $ 2^a 5^b$ and $ 2^{n - a}5^{n - b}$. Since each pair contains 2 elements, and there are $ (n + 1)^2$ elements in total, there are $ \frac {(n + 1)^2}{2}$ pairs.[/quote][quote="314.math"]

What does this notation mean: $ a_1^{e_1}a_2^{e_2}...a_n{e_n}$?  Thanks![/quote][quote="pythag011"]

It is the prime factorization. For example $ 108 = 2^2*3^3$[/quote]

So how would you write $ 108 = 2^2*3^3$ in $ a_1^{e_1}a_2^{e_2}...a_n{e_n}$ form?  Thanks!

Okay, I'm not going to pyramid, but I'm responding to the above confusion.

First of all, there is a typo. It should read $ a_1^{e_1}a_2^{e_2}\cdots a_n^{e_n}$. But I prefer the notation $ p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ because it's more obvious that $ p$ refers to primes and we already have $ n$ in the problem.

Here are some examples:
$ 108 \equal{} 2^2\cdot3^3$ (so $ p_1 \equal{} 2$, $ e_1 \equal{} 2$, $ p_3 \equal{} 3$, $ e_2 \equal{} 3$ and $ k \equal{} 2$)
$ 2008 \equal{} 2^3\cdot251$ (so $ p_1 \equal{} 2$, $ e_1 \equal{} 3$, $ p_2 \equal{} 251$, $ e_2 \equal{} 1$ and $ k \equal{} 2$)
$ 10^n \equal{} 2^n\cdot5^n$ (so $ p_1 \equal{} 2$, $ e_1 \equal{} n$, $ p_2 \equal{} 5$, $ e_2 \equal{} n$ and $ k \equal{} 2$)

I hope that clears up any confusion. The formula for the number of divisors of any number given its prime factorization is one of the basic facts to know in number theory for the AMC, so you should figure out how to derive it if you don't know already. But as I wrote earlier, you can just use small cases to establish the pattern:

1 has 1 divisor; 10 has 4 divisors (1, 2, 5, and 10); 100 has 9 divisors (1, 2, 4, 5, 10, 20, 25, 50 and 100); 1000 has 16 divisors (1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500 and 1000). What does 1-4-9-16 indicate? Squares, in particular $ (n\plus{}1)^2$.

[hide="Solution with recursion"]
Define $ a_n$ as the sum of the base-10 logarithms of the divisors of $ 10^n$.

$ \sum_{d|10^n}\log_{10}d\equal{}\log_{10}\prod_{d|10^n}d$ by properties of logarithms.

We see that $ a_1\equal{}2$, as the divisors of $ 10$ are $ 1,2,5,10$.

Now, $ a_n$ must include all the numbers found in $ a_{n\minus{}1}$ because any divisor of $ 10^{n\minus{}1}$ will divide $ 10^n$.

In addition to those, we get new divisors of the form $ 2^n\cdot 5^0$, $ 2^n\cdot 5^1$, and so on, up to $ 2^n\cdot 5^n$. The product of these contributes $ \log_{10}(2^{n\cdot n}5^{1\plus{}2\plus{}\cdots\plus{}n})$.

By symmetry, we also get $ \log_{10}(5^{n\cdot n}2^{1\plus{}2\plus{}\cdots\plus{}n})$.

Together, this contributes $ n^2\plus{}\frac {n(n\plus{}1)}2$ new divisors.

Therefore, $ a_n\equal{}a_{n\minus{}1}\plus{}n^2\plus{}\frac{n(n\plus{}1)}2$.

$ \equal{}n^2\plus{}(n\minus{}1)^2\plus{}\frac{n(n\plus{}1)}{2}\plus{}\frac{(n\minus{}1)n}{2}$, and so on.

So, $ a_n\equal{}\sum k^2\plus{}\frac 12 \sum k^2\plus{}k\equal{}\frac{n(n\plus{}1)^2}{2}$

$ n(n\plus{}1)^2\equal{}792\cdot 2\equal{}2^43^211$

$ 2^43^2\equal{}(2^23)^2\equal{}12^2$

$ n(n\plus{}1)^2\equal{}11\cdot 12^2$, so $ n$ is clearly $ \boxed{\textbf{(A)}\ 11}$.
[/hide]

[quote="TZF"]
Multiply by the $ n\plus{}1$ values for the exponent of $ 5$ to get $ \frac{n(n\plus{}1)^2}{2} \equal{} 792 \iff n^3 \plus{} 2n^2 \plus{} n \equal{} 1584$

Since $ 12^3 > 1584$, the only legitimate answer choice is $ \boxed{\textbf{(A)} \; 11 }$[/hide][/quote]

I don't understand why you multiply by n+1,can someone explain?

You will notice that the sum of the logs of each pair of divisors becomes $n$.

Also, $10^n=2^n \cdot 5^n$
Therefore, the total number of divisors is $(n+1)(n+1)=(n+1)^2$, if $n$ is odd.

The number of pairs of divisors is $\frac {(n+1)^2}{2}$
So, the sum of the logs of all the divisors is $\frac {n(n+1)^2}{2}$

$\frac {n(n+1)^2}{2}=792$

So, $\boxed {n=11}$


If my school offered it, I would definitely try 12B... (having taken 10A)
A little variety would be nice.

12B - only reason i took 10A is to shoot for the perfect score; unfortunately i failed, but oh well. im confident that i'll make at least 1 mistake no matter which test i take so theres no point in taking the 10B.

12B (My school only offers B), because I want to make the olympiad. =)

Yeah, I think the AMC 12 is in fact much easier to qualify with than the AMC 10.  Then again, it has been a while since I've done Mathcounts/AMC 10 level problems.

They have different questions, but they are supposed to be similar in difficulty.  The B theoretically only exists for the people who can't take the A on the day of the test for various reasons, but a lot of people take both the A and the B to increase their chances of qualifying for the AIME.  Can't blame them--I would do the same if I could :D.

12 tests a bit more in depth information while the 10 is similar to MC level problems. I took 12A and am taking 12B because that is the only thing my school offers.

[quote="captcha000"]My teacher wants us to take AMC-10 on both...by us i mean all freshmen and sophomoes...[/quote]

I'd ask her to make an exception if you'd like to do 12 on one of them.

I made an acceptable score on 12A, so I'm not taking a B test.


I know examples, but I want to know the [b][u]why[/u][/b] of how someone can get an extraneous solution with an absolute value equation.

For example: $ |2z \minus{} 3| \equal{} 4z \minus{} 1$

Yields: $ z \equal{} \minus{} 1$ and $ z \equal{} \frac {2}{3}$, however when checked, only $ \frac {2}{3}$ fits...

Again, [b][u]why[/u][/b] does this happen?

It is because of the absolute value symbol. It works just like a squared siymbol.
(|4z-3|)^2=(4z-3)^2
we see that in this process.
x^2=y means that +/-sqrty are the roots.
The squaring destroys the absolute value symbol so we forget to check if it fulfills the original equation.

[quote="cmpgm88"][quote="varunrocks"]\(|4z-3|)^2=(4z-3)^2
we see that in this process.
x^2=y means that +/-sqrty are the roots.
The squaring destroys the absolute value symbol so we forget to check if it fulfills the original equation.[/quote]

I'm not quite sure I understand.... Is there a graph of some sort you could show me, or another problem (or why the posted problem yields that solution)?[/quote]You probably have seen $ |x|\equal{}\begin{cases}x&\text{if }x\ge0\\\minus{}x&\text{if }x<0\end{cases}$.

Using this definition of absolute value, then $ |2z\minus{}3|\equal{}\begin{cases}2z\minus{}3&\text{if }2z\minus{}3\ge0\\\minus{}(2z\minus{}3)&\text{if }2z\minus{}3<0\end{cases}$.

If $ 2z\minus{}3\ge0\Rightarrow z\ge\frac32$, then $ 2z\minus{}3\equal{}4z\minus{}1$ or $ z\equal{}\minus{}1$. Is this $ z$ in the corresponding interval?

If $ 2z\minus{}3<0\Rightarrow z<\frac32$, $ \minus{}(2z\minus{}3)\equal{}4z\minus{}1$ or $ z\equal{}\frac23$. Is this $ z$ in the corresponding interval?

The one that failed is the extraneous root.

well you are doing it wrong

l2z-3l=4z-1

well you take 2z-3 to the other side

0=4z-1+2z-3

you might be wondering why I did not change it to -2z+3
well because of the absolute value sign you do not change it but instead you take off the absolute value sign.

0=6z-4
4=6z
z=4/6
z=2/3

and also if you plug in -1 it doesnt work
let us try it

l-1(2)-3l=4(-1)-1
l-2-3l=-4-1
l-5l=-5
you cannot forget about the absolute value sign
5=-5
which is not true thus
Z=2/3

So, if I understand correctly, there is only one solution to this kind of equation?

Not necesseraly. The reason there is sometimes an extraneous solution is when you have

$ |ax \plus{} b| \equal{} cx \plus{} d$

and

$ cx \plus{} d <0$

for that extraneous solution.

How would you know then, if $ cx\plus{}d$ is less than zero?

By plugging in the number you get as a solution.

Make an inequality:

$ cx \plus{} d < 0$

$ x < \minus{} \frac{d}{c}$

But you would have to find that value of $ x$ first so I guess might as well just find them and plug them back in.

[quote="calculatorwiz"]well you are doing it wrong

l2z-3l=4z-1

well you take 2z-3 to the other side

0=4z-1+2z-3

you might be wondering why I did not change it to -2z+3
well because of the absolute value sign you do not change it but instead you take off the absolute value sign.

0=6z-4
4=6z
z=4/6
z=2/3

and also if you plug in -1 it doesnt work
let us try it

l-1(2)-3l=4(-1)-1
l-2-3l=-4-1
l-5l=-5
you cannot forget about the absolute value sign
5=-5
which is not true thus
Z=-5[/quote]

I see that this can get one solution that will work, but since this only gives one answer, how can I know that the other one is wrong without working it as seen in the first post?


Prove that:There exist infinitely many $ n$ such that for every prime divisor $ p$ of $ n^2\plus{}3$ we can find $ k,k^2<n$ and $ p|k^2\plus{}3$

[quote="hxy09"]Prove that:There exist infinitely many $ n$ such that for every prime divisor $ p$ of $ n^2 \plus{} 3$ we can find $ k,k^2 < n$ and $ p|k^2 \plus{} 3$[/quote]
Let X be the set all such numbers . We will show thatI if $ n\in X$ then $ n^2 \minus{} n \plus{} 3 \in X$ too . Indeed , we have :
\[ (n^2 \minus{} n \plus{} 3)^2 \plus{} 3 \equal{} (n^2 \plus{} 3)((n \minus{} 1)^2 \plus{} 3)\]
Take any prime divisor p of $ (n^2 \minus{} n \plus{} 3)^2\plus{}3$ then p is a divisor of $ n^2 \plus{} 3$ or $ (n \minus{} 1)^2 \plus{} 3$ .
If $ p|n^2 \plus{} 3$ , then we can choose k such that $ p|k^2 \plus{} 3$ and $ k^2 < n$ (because $ n\in X$ ) . But we also have $ n < n^2 \minus{} n \plus{} 3$ so $ k^2 < n^2 \minus{} n \plus{} 3$ .
If $ p|(n \minus{} 1)^2 \plus{} 3$ then we can easily choose $ k^2 \equal{} (n \minus{} 1)^2 < n^2 \minus{} n \plus{} 3$ .
Now our purpose is to find a number that satisfy condition ,but it is easy to see that $ n \equal{} 5$ belongs to X . Therefore we can construct an infinite set that all elements belong to X .
\[ x_1 \equal{} 5; x_{n \plus{} 1} \equal{} x_n^2 \minus{} x_n \plus{} 3\]


[hide="Solution"]
$ 4x^2\minus{}81\equal{}(2x)^2\minus{}9^2\equal{}(2x\plus{}9)(2x\minus{}9)$
[/hide]


ARML probability geometry calculus integration parallelogram LaTeX

1. Suppose that f(x) = ax + b where a and b are real numbers. Given that f(f(f(x))) = 8x + 21, what is the value of a + b?

2. If :theta: is an angle between 0o and 90o for which cos (:theta:)*cos (2:theta:) = 1/4, what is the value of :theta: in degrees?

3. A box contains 4 fair coins and 6 biased coins. Whenever a fair coin is flipped, it comes up heads with probability 0.5. Whenever a biased coin is flipped, it comes up heads with probability 0.8. A coin is randomly chosen from the box and then flipped. If the result is heads, what is the probability that it was the biased coin?

4. Let ABCD be a convex quadrilateral, and let O is the intersection of the diagonals AC and BD. The areas of  :tri: AOB and :tri: COD are respectively 9 and 4. What is the minimum possible area of the quadrilateral ABCD?

5. From a class of 12 students, 3 are chosen to form a math contest team. The team is required to include at least one boy and one girl. If exactly 160 different teams can be formed from the 12 students, what is the absolute value of the difference between the number of boys and the number of girls in the class?

6. A point (x,y) is called integral if both x and y are integers. How many integral points are inside the parallelogram whose four sides are on the lines x = 100, x = 300, y = x/3 + 0.1, and y = x/3 + 0.6?

7. If sin x + sin y = 1/:rt5: and cos x + cos y = 2/:rt5: what is the value of cos (x-y)?

8. Let a be the integer 333...333 (2003 digits) and let b be the integer 666...666 (2003 digits). What is the 2004th digit (counting from the right) that appears in the product ab?

7:

[hide]

square both equations to get

sin :^2: x + 2 sin x sin y + sin :^2: y = 1/5

and

cos :^2: x + 2 cos x cos y + cos :^2: y = 4/5

adding the two equations yields

1 + 2(sin x sin y + cos x cos y) + 1 = 1

2 + 2 ( cos (x-y) ) = 1

cos (x-y) =-1/2

[/hide]

gauss202 wrote:2. If :theta: is an angle between 0o and 90o for which cos (:theta:)*cos (2:theta:) = 1/4, what is the value of :theta: in degrees?

[hide]

cos( :theta: ) * cos(2 :theta: ) = 1/4

cos( :theta: ) * (2cos :^2: ( :theta: ) -1) = 1/4

2cos :^3: ( :theta: ) - cos( :theta: ) = 1/4

Let cos( :theta: ) = x.

2x :^3: - x = 1/4

8x :^3: - 4x - 1 = 0

(2x+1)(4x^2-2x-1) = 0.

x = -1/2  or  ( :rt5: + 1)/4  or  (1 -  :rt5: )/4.

Since 0 <  :theta: < 90, 0 < x < 1. Thus we reject -1/2 and (1 -  :rt5: )/4, leaving x = ( :rt5: +1)/4. Thus  :theta: = 36 :^o:

[/hide]

[quote="Rep123max"]
Is there away to do it with no checking?[/quote]

Yes. I'll give the answers after some more people have worked on them. But use spoiler for your answers for now.

Rep123max wrote:Ok, I put it in spoiler, but the Latex still shows up though.

Spoiler:
[hide]3. A box contains 4 fair coins and 6 biased coins. Whenever a fair coin is flipped, it comes up heads with probability 0.5. Whenever a biased coin is flipped, it comes up heads with probability 0.8. A coin is randomly chosen from the box and then flipped. If the result is heads, what is the probability that it was the biased coin?

I am not sure on this problem. What I did was find the total probability of getting a head which was [tex]\frac{4}{10}\times .5+\frac{6}{10}\times .8=\frac{17}{25}[/tex]. Then I took the probability of getting a biased coin and getting a heads, and divided that by the total probability of getting a heads which is [tex]\frac{12}{25}\div \frac{17}{25}=\frac{12}{17}[/tex] so I think the answer is [tex]\frac{12}{17}[/tex].[/hide]

This is correct.  P(A | B) = P(A and B)/P(B)

#5:

[hide]

Let x be the number of boys (or girls, it doesn't matter).  First, we must choose at least one boy and one girl: x(12-x).  Then, choose one of the 10 left and divide by 2 for overcounting.  Therefore, 5x(12-x) choices.

5x(12-x) = 160

The rest is pretty simple.[/hide]

I used spoiler as much as possible (spoiler does not hide LaTeX)

gauss202 wrote:1. Suppose that f(x) = ax + b where a and b are real numbers. Given that f(f(f(x))) = 8x + 21, what is the value of a + b?

[hide]f(x) = ax + b
f(f(x)) = a(ax+b) + b = a^2*x + ab + b
f(f(f(x))) = a(a^2 x + ab + b) + b = a^3 x + a^2 b + ab + b

Since f(f(f(x))) = 8x + 21,
a^3 = 8 and a^2*b + ab + b = 21
So a = 2.  Plugging this into the second equation,
(2)^2 b + (2)b + b = 21
7b = 21
b = 3

a + b = 2 + 3 = 5[/hide]

Quote:2. If :theta: is an angle between 0o and 90o for which cos (:theta:)*cos (2:theta:) = 1/4, what is the value of :theta: in degrees?

[hide][tex]\cos \theta \cos (2 \theta) = \frac{1}{4}[/tex]

By the double-angle formula for cosine:
[tex]\cos \theta (2 \cos^2{\theta} - 1) = \frac{1}{4}[/tex]
[tex]2 \cos^3 \theta - \cos \theta = \frac{1}{4}[/tex]
[tex]2 \cos^3 \theta - \cos \theta - \frac{1}{4} = 0[/tex]

Let [tex]y = \cos \theta[/tex].  Then the equation becomes 2y^3 - y - 1/4 = 0.  Multiplying by 4, we get 8y^3 - 4y - 1 = 0.  By the rational root theorem, any rational roots must be in the set {1, -1, 1/2, -1/2, 1/4, -1/4, 1/8, -1/8}

1 is no good.
-1 is no good.
1/2 is no good.
-1/2 is good.

Dividing 8y^3 - 4y - 1 by (y+1/2), we are left with 8y^2 - 4y - 2 = 0.  Applying the quadratic formula, we find that [tex]y = (4 \pm \sqrt{16 + 64})/16[/tex] [tex]= (4 +- 4*sqrt(5))/16 = (1 \pm \sqrt{5})/4[/tex].

So [tex]\cos \theta = - \frac{1}{2}[/tex] or [tex]\frac{1 + \sqrt{5}}{4}[/tex] or [tex]\frac{1 - \sqrt{5}}{4}[/tex].

But since [tex]0^o \leq \theta \leq 90^o[/tex], [tex]0 \leq \cos \theta \leq 1[/tex].  So the only possible solution is [tex]\cos \theta = [tex]\frac{1 + \sqrt{5}}{4}[/tex].  I can't imagine how you would find [tex]\theta[/tex] without a calculator, though.[/hide]

Quote:3. A box contains 4 fair coins and 6 biased coins. Whenever a fair coin is flipped, it comes up heads with probability 0.5. Whenever a biased coin is flipped, it comes up heads with probability 0.8. A coin is randomly chosen from the box and then flipped. If the result is heads, what is the probability that it was the biased coin?

[hide]0.4 of the time, a fair coin will be flipped, in which case the probability of heads is 0.5.  So the total frequency of [heads from fair coin] is 0.4*0.5 = 0.2.

0.6 of the time, a biased coin will be flipped, in which case the probability of heads is 0.8.  So the total frequency of [heads from biased coin] is 0.6 * 0.8 = 0.48.

In all, [frequency of heads] is 0.2 + 0.48 = 0.68.  0.48/0.68 = 12/17 of the time, this comes from the biased coin, and 0.2/0.68 = 5/17 of the time, it comes from the fair coin.  So the probability that it was the biased coin is 12/17.  [/hide]

Quote:4. Let ABCD be a convex quadrilateral, and let O is the intersection of the diagonals AC and BD. The areas of  :tri: AOB and :tri: COD are respectively 9 and 4. What is the minimum possible area of the quadrilateral ABCD?

[will return to later]

Quote:5. From a class of 12 students, 3 are chosen to form a math contest team. The team is required to include at least one boy and one girl. If exactly 160 different teams can be formed from the 12 students, what is the absolute value of the difference between the number of boys and the number of girls in the class?

[hide]Suppose there are b boys, so that there are 12-b girls.

Case 1: The team has two boys and one girl.

There are b ways to pick the first boy, b-1 ways to pick the second boy, and 12-b ways to pick the girl.  In all, there are b(b-1)(12-b) ways.  But we counted each possibility twice, since we counted both boy A, boy B and boy B, boy A.  So there are b(b-1)(12-b)/2 ways.

Case 2: The team has two girls and one boy.

There are 12-b ways to pick the first girl, 11-b ways to pick the second girl, and b ways to pick the boy.  In all, there are (12-b)(11-b)b ways.  Again, we overcounted by a factor of 2, so there are (12-b)(11-b)b/2 ways.

In all, there are b(b-1)(12-b)/2 + (12-b)(11-b)b/2 = b(12-b)[(b-1)+(11-b)]/2 = b(12-b)[10]/2 = 5b(12-b) ways to form the team.  So 5b(12-b) = 160.

b(12-b) = 32
12b - b^2 = 32
b^2 - 12b + 32 = 0
(b-4)(b-8) = 0
So b = 4 or b = 8, in which case 12-b = 8 or 12-b = 4.  In either case, the absolute value of the difference is 4.[/hide]

Quote:6. A point (x,y) is called integral if both x and y are integers. How many integral points are inside the parallelogram whose four sides are on the lines x = 100, x = 300, y = x/3 + 0.1, and y = x/3 + 0.6?

[later]

Quote:7. If sin x + sin y = 1/:rt5: and cos x + cos y = 2/:rt5: what is the value of cos (x-y)?

[later]

Quote:8. Let a be the integer 333...333 (2003 digits) and let b be the integer 666...666 (2003 digits). What is the 2004th digit (counting from the right) that appears in the product ab?

First, let's look at a*6 (that is, 333....333 * 6).

Code:
   3333..333
x          6
-------------
1999...99998

Each time we multiply 3*6, we get 18.  For every 3*6 after the first one, there is a 1 carried over from the previous 3*6, so the result is 19.  The 9 is placed, and the 1 is carried over to the next digit.  In the end, we get 19999.....9998.  But how many 9s are there, exactly?  To figure this out, we do the following:

Code:         18   (2003 of these)
        18            |
       18             |
      18              V
     18
............
   18
+ 18
-----------------
199..999998

There are 2003 18s, since there are 2003 3s and each one was multiplied by a 6.  Every '8' after the first one becomes a '9' after a 1 is carried over from the previous 18.  So there should be 2002 9s.  The number a*6 is then:
19999...9998 (a 1, followed by 2002 9's, followed by an 8).

Since b = 66666....6 (2003 6s) = 6*10^2002 + 6*10^2001 + ... + 6*10 + 6, we can think of the product a*b as a*6*10^2002 + a*6*10^2001 + ... + a*6*10 + a*6.  But we just found what a*6 is!  So the expression a*b becomes:

Code:             199999..98 (2002 9s)
            199999..98  (")
           199999..98   (")
          199999..98    (")
     ...........
+ 199999...98            (")
-----------------------

The 1st digit (ones place) of the resulting number is just 8.
The 2nd digit is 8+9 = 17 --> 7, carry the 1
The 3rd digit is 8+9+9+1 = 27 --> 7, carry the 2
The 4th digit is 8+9+9+9+2 = 37 --> 7, carry the 3
....
The nth digit is 8+9+9+...+9 (n-1 9's) + (n-2) = 8+9(n-1)+(n-2) = 8+9+9(n-2)+(n-2) = 17+10(n-2) = 7+10(n-1).  So the end result is a 7, carry n-1.

This works up to the 2003rd digit, as seen below:

Code:         |
         V
        199999...98 (2002 9s)
       199999...980 (")
      199999...9800 (")
     199999...98000 (")
    ...........
+ 1999..980000...00 (2002 9s, 2002 0s)
-------------------------------------
?????????7777777...777

At this 2003rd digit, the end result is a 7, and 2003-1 = 2002 is carried over to the next digit.

At the 2004th digit, we have 2002 9's, plus a 1, plus 2002 from the previous carry.  2002*9+2002+1 = 2002*10+1.  The 20020 part is carried over, and the 1 remains.  Hence, the answer is 1.

[Edited to fix some silly arithmetic.  2002-1 is not 2002, and in the previous method, I forgot that there was one fewer 8 than normal.  ]

mathfanatic, [hide]number 2 can be done another way, where you don't have to find the arccos of that number. And number 8, I don't know where you went wrong, but it's not right.[/hide]

Quote:4. Let ABCD be a convex quadrilateral, and let O is the intersection of the diagonals AC and BD. The areas of  :tri: AOB and :tri: COD are respectively 9 and 4. What is the minimum possible area of the quadrilateral ABCD?

spoiler:[hide]

Let [AOD]=x and [BOC]=y, then x/4=9/y so xy=36.  Then [ABCD]=x+y+13 >= 2 :sqrt: (xy) + 13 = 25 by AM-GM.

Answer: minimum possible area of the quadrilateral ABCD is 25.

[/hide]

mathfanatic wrote:Is the answer [hide]1[/hide] any better?

Yes. That's right. But can you think of a faster way you could have done it?

Quote:2. If :theta: is an angle between 0o and 90o for which cos (:theta:)*cos (2:theta:) = 1/4, what is the value of :theta: in degrees?

It's been found that cos(:theta:)=( :rt5: +1)/4 above.

spoiler:

[hide]It's easy to verify that  :rt2: /2 < (:rt5: +1)/4 <  :rt3: /2, so 30o < :theta: < 45o, since 0o < :theta: < 90o. Hence 60o < 2:theta: < 90o and 90o < 3:theta: < 135o.

cos(3:theta:) + cos(:theta:)=2cos(2:theta:)cos(:theta:)=1/2, so cos(3:theta:)=(1- :rt5: )/4. cos(2:theta:)cos(:theta:)=1/4 implies that cos(2:theta:)=( :rt5: -1)/4. Therefore cos(2:theta:) = -cos(3:theta:)=cos(180o - 3 :theta: ). Hence 2 :theta:=180o - 3 :theta:,  :theta: =36o.[/hide]

That's correct fuzzylogic, but there's still a much faster way to get to that answer. (Hint: [hide]Multiply both sides of the original equation by 4sin (:theta:)[/hide])

ooh, gauss, your hint is a much faster way! Thanks.

Quote:6. A point (x,y) is called integral if both x and y are integers. How many integral points are inside the parallelogram whose four sides are on the lines x = 100, x = 300, y = x/3 + 0.1, and y = x/3 + 0.6?
[hide]We seek integral values of x, 100<x<300, such that an integer y lies between x/3+.1 and x/3+.6.  Since x is an integer, the fractional part of x/3 is either zero, .3333, or .6666.  Clearly, only the last value results in a possible integral value of y.  Therefore, x :equiv: 2 mod 3.  There are 67 values of x between 100 and 300 that are  :equiv: 2 mod 3.[/hide]

Quote:8. Let a be the integer 333...333 (2003 digits) and let b be the integer 666...666 (2003 digits). What is the 2004th digit (counting from the right) that appears in the product ab?

[hide]Write 111...111 (2003 1's) as (10^2003-1)/9.  Then the desired product is this number squared times 18, or (10^2003-1) :^2: *2/9.  Expand to get (10^4006-2*10^2003+1)*2/9.  The number in parentheses is 9999.....980000...001 (the number of 9's is irrelevant, and there are 2002 zeros).  Dividing this by 9, we get 111...11088...89 (again, an irrelevant number of ones, and 2002 eights).  The zero in this last number is in the place we are concerned with, and it clearly changes to a 1 when the number is multiplied by 2.[/hide]


function
induction
algebra
functional equation
algebra proposed

[quote="Math pro"]find$ f: Q \plus{} \minus{} > Q \plus{}$ such that
i.$ f(x \plus{} 1) \equal{} f(x) \plus{} 1$
ii.$ f(x^3) \equal{} f(x)^3$[/quote]
Are you sure,because only your i) we will find $ f(x) \equal{} kx$ (cauchy equation ) and with ii) we find $ k \equal{} 1$
Post your sourse

[quote="Math pro"]you are wrong,think again all names :lol:[/quote]
I think you should check my notice before post any comment([hide="to mathpro "]because if not,many people will think we are spam OK [/hide],Cauchy equation,do you know it,Ley $ f(1)\equal{}k$ so $ f(x)\equal{}kx$ for all $ x\in \mathbb Q$
I think i dont need to explain more because you are good at solving function equation

check care fully,all names! :wink:
from $ f(x\plus{}1)\equal{}f(x)\plus{}1$
It 's quite hard to ->$ f(x)\equal{}kx$

[quote="Math pro"]check care fully,all names! :wink:
from $ f(x \plus{} 1) \equal{} f(x) \plus{} 1$
It 's quite hard to ->$ f(x) \equal{} kx$[/quote]
FOR you,mathpro
You can $ f(x) \equal{} k(x) for x\in \mathbb N$ Ok
Then find $ f(\frac {1}{n}) \equal{} \frac {1}{n}$OK
Then $ f(x) \equal{} kx for x\in \mathbb Q$ OK
Dont post more,meet me for any explain

[quote="Allnames"]You can $ f(x) \equal{} k(x) for x\in \mathbb N$ Ok[/quote]
What you can actually find is that $ f(n) \equal{} n \plus{} f(0)$ for $ n \in \mathbb{N}$.  The Cauchy functional equation is a different statement.

[quote="Allnames"]Then find $ f(\frac {1}{n}) \equal{} \frac {1}{n}$OK[/quote]
That's not an immediate consequence of either i) or ii).

At first,I think everyone should solve this problem,it is easier but have the same way:
Find $ f: Q\minus{}>Q$ such that:
i.$ f(x\plus{}1)\equal{}f(x)\plus{}1$
ii.$ f(x^2)\equal{}f(x)^2$

[quote="Math pro"]find$ f: Q + - > Q +$ such that
i.$ f(x + 1) = f(x) + 1$
ii.$ f(x^3) = f(x)^3$[/quote]
It is not hard problem .
From condition we have $ f(1)=1$ .
Induction we have : $ f(x+n)=f(x)+n,\forall n\in \mathbb{N}$
Therefore $ f(n)=n$
Let $ x=\frac{p}{q},\gcd(p,q)=1$
Chose n such that $ q^3|n$ then we have :
\[ f((x+n)^3)=f(x^3)+f(3n^2x)+f(3nx^2)+f(n^3)=(f(x)+n)^3
\]
Therefore :
\[ 3n^2x+3nx^2=3n^2f(x)+2f(x)^2n
\Leftrightarrow 3n^2(f(x)-x)+3n(f(x)-x)=0
\]
This equality hold for infinite n ,so $ f(x)=x,\forall x\in \mathbb{Q}^{+}$

As [b]TTsphn[/b] mentioned it is easy to establish that $ f(n)\equal{}n$ for all $ n\in\mathbb{N}$.
So our goal is to show that the same answer holds for all rational numbers,what is indeed not so hard...
For this we only need to consider what follows from this obviously true equation:
\[ f\left(\left(\frac{p}{q}\plus{}q^2\right)^3\right)\equal{}\left(f\left(\frac{p}{q}\right)\plus{}q^2\right)^3
\]

[quote="TTsphn"][quote="Math pro"]find$ f: Q + - > Q +$ such that
i.$ f(x + 1) = f(x) + 1$
ii.$ f(x^3) = f(x)^3$[/quote]
It is not hard problem .
From condition we have $ f(1) = 1$ .
Induction we have : $ f(x + n) = f(x) + n,\forall n\in \mathbb{N}$
Therefore $ f(n) = n$
Let $ x = \frac {p}{q},\gcd(p,q) = 1$
Chose n such that $ q^3|n$ then we have :
\[ f((x + n)^3) = f(x^3) + f(3n^2x) + f(3nx^2) + f(n^3) = (f(x) + n)^3
\]
Therefore :
\[ 3n^2x + 3nx^2 = 3n^2f(x) + 2f(x)^2n \Leftrightarrow 3n^2(f(x) - x) + 3n(f(x) - x) = 0
\]
This equality hold for infinite n ,so $ f(x) = x,\forall x\in \mathbb{Q}^{ + }$[/quote]
exactly,tanpham! :D  here is my solution,too.
the solution for the second problem
esay to see that $ f(1)=1$
Induction we have:$ f(x+n)=f(x)+n  n\in \mathbb{N}$
so $ f(n)=n n\in \mathbb{N}$
Let $ x=\frac{p}{q}$ with $ (p,q)=1$ choose $ qln$
Then by the same way we have done :wink:


function
factorial
counting
distinguishability
algebra
linear equation

Deteremine the number of $ n$ digit numbers with all digits odd, so that $ 1$ and $ 3$ each occur a nonzero, even number of times.

first of all, we require $ n \ge 4$ so that both $ 1$ and $ 3$ can occur a non-zero even number of times.

Our choices of digits are $ 1$, $ 3$, $ 5$, $ 7$, and $ 9$; so, without the restriction that $ 1$ and $ 3$ occur an even number of times, we have $ 5^n$ numbers (from the 5 choices for each digit).

But we have to deal with the restriction.  $ 3^n$ of our numbers use only $ 5$'s, $ 7$'s, and $ 9$'s; $ 4^n$ use no $ 1$'s, and $ 4^n$ use no $ 3$'s.  By the principle of inclusion-exclusion, we have $ 5^n \minus{} 2\cdot 4^n \plus{} 3^n$ numbers containing at least one $ 1$ and $ 3$.

getting stuck here - trying a new approach.

Let $ a_n$ be the number of $ n$ digit numbers with only odd digits and a positive even number of $ 1$'s and $ 3$'s.

For our initial condition, we have $ a_4 \equal{} \binom{4}{2} \equal{} 6$ because we have four digits to choose from, two of which must be $ 1$'s and the other two of which must be $ 3$'s.

Then, for $ a_n$, we can take any number counted toward $ a_{n \minus{} 1}$ and add in a $ 5$, $ 7$, or $ 9$ in any position, or we can take any of the numbers counted towards $ a_{n \minus{} 2}$ and add in a pair of $ 1$'s or $ 3$'s.  The former gives $ 3na_{n \minus{} 1}$ possibilities, as we choose 1 of 3 digits we can add in, and we can place this digit in n different positions.  The latter gives $ 2 \binom{n}{2} a_{n \minus{} 2}$, as we choose to either insert a pair of $ 1$'s or $ 3$'s, and have $ \binom{n}{2} \equal{} \frac {n(n \minus{} 1)}{2}$ positions to place them in.  Thus our recursive formula is $ a_n \equal{} 3na_{n \minus{} 1} \plus{} n(n \minus{} 1)a_{n \minus{} 2}$.

We now must add in $ a_3 \equal{} 0$ as an additional initial condition.

So our generating function must satisfy $ f(x) \equal{} 3f'(x) \plus{} f''(x)$.  This will be ugly...

So I am going to take a guess that there is some solution of the form $ f(x) \equal{} e^{bx}$.  Then we have $ f'(x) \equal{} be^{bx}$ and $ f''(x) \equal{} b^2e^{bx}$.  So our equation becomes (after dividing out $ e^{bx}$) $ 1 \equal{} 3b \plus{} b^2 \implies b^2 \plus{} 3b \minus{} 1 \equal{} 0$, so $ b \equal{} \frac { \minus{} 3 \pm \sqrt {3^2 \minus{} 4*1* \minus{} 1}}{2} \equal{} \frac { \minus{} 3 \pm \sqrt {13}}{2}$

Now the $ f(x)$ we are after will be of the form $ f(x) \equal{} ae^{\frac { \minus{} 3 \minus{} \sqrt {13}}{2} x} \plus{} be^{\frac { \minus{} 3 \plus{} \sqrt {13}}{2}x}$

Now by Taylor's theorem,
$ f(x) \equal{} \sum_{i \equal{} 0}^{\infty} f^{(i)}(0)/i! x^i$
so $ a_n \equal{} f^{(n)}(0)/n!$.

$ f^{(n)}(0) \equal{} a(\frac { \minus{} 3 \minus{} \sqrt {13}}{2})^n \plus{} b(\frac { \minus{} 3 \plus{} \sqrt {13}}{2})^n$
so we have $ a_n \equal{} 1/n! (a(\frac { \minus{} 3 \minus{} \sqrt {13}}{2})^n \plus{} b(\frac { \minus{} 3 \plus{} \sqrt {13}}{2})^n)$.

solving for $ a$ and $ b$, we find $ a \equal{} 1152/(\sqrt {13}(3 \plus{} \sqrt {13})^3)$ and $ b \equal{} 1152/(\sqrt {13} ( \minus{} 3 \plus{} \sqrt {13})^3)$

But this doesn't work!!  the formula gives fractions and negative numbers!

ok I think my mistake was not making sure $ a_0$, $ a_1$, $ a_2$, and $ a_3$ are 0.  So our format for $ f$ needs to take this into account; $ f(x) \equal{} ae^{cx}\plus{} be^{dx}\plus{}p(x)$ where $ p$ is a cubic.

Why not consider teh following generating function.

Our goal is to find the coefficient of x^n of

$ (\frac{x^{2}}{2!}\plus{}\frac{x^{4}}{4!}\plus{}...)^{2}(1\plus{}\frac{x}{1!}\plus{}\frac{x^{2}}{2!}\plus{}...)^{3}$.

Notice that from taylor series this simplifies to $ (\frac{e^{x}\plus{}e^{\minus{}x}}{2}\minus{}1)^{2}(e^{x})^3$, which gives us $ \frac{e^{5x}\minus{}4e^{4x}\plus{}6e^{3x}\minus{}4e^{2}\plus{}e^{x}}{4}$

We need to find the coefficient of x^{n}, which is just $ \frac{5^{n}\minus{}4*4^{n}\plus{}6*3^{n}\minus{}4*2^{n}\plus{}1}{4*n!}$

However, notice we need to multiply this by $ n!$ as there are n objects, so the answer is $ \frac{5^{n}\minus{}4*4^{n}\plus{}6*3^{n}\minus{}4*2^{n}\plus{}1}{4}$

are you sure that is right?  It doesn't agree with my recursive formula...

Also, I think that, if you are correct, there must be some combinatoric way to do it, as I tried at first.

Also, where do you get the factorials from in your first generating function?  I am unsure about your use of factorials in the problem... please explain.

He's using the exponential generating function. This is a case where the standard generating function will not work. Why?

Suppose we set it up as $ (x^2 \plus{} x^4 \plus{} x^6 \plus{} \ldots) ^2 (1 \plus{} x \plus{} x^2 \plus{} \ldots )^3$, and took the coefficient of $ x^N$. This would tell us the number of solutions to $ 2o \plus{} 2t \plus{} f \plus{} s \plus{} n \equal{} N$; that is, the number of ways we can divide up the $ N$ digits among an even number of $ 1$'s and $ 3$'s, and the rest among $ 5,7,9$. However, it would not take into account the number of ways in which they can be ordered!

The exponential generating function (each term has a $ /n!$ in it) solves this. What he does is, in the end, he multiplies by $ N!$ because there are $ N!$ ways of arranging the $ N$ digits. But, if there are, say, $ 2t$ threes, then the $ /(2t)!$ to get rid of the ways of ordering these is [i]built into[/i] the term $ \frac {x^{2t}}{(2t)!}$ in that part of the generating function.

I hope that made some degree of sense. Try it out for a few very simple problems to see how it works. You should get a feel of when you will need the ordinary generating function (number of solutions to a linear equation) and when you will need the exponential generating function (also orders them).

Ok I think I get it now... so the factorials basically include distinguishable premutations in the problem?

My other question: what is wrong with my recursive formula, $ a_n \equal{} 3na_{n\minus{}1} \plus{} n(n\minus{}1)a_{n\minus{}2}$?

Nevermind, I think I figured it out.  my recursive formula does not take into account that not necessarily all the places in which we can place a digit are distinguishable.  For instance, if we have the number $ 13513$, placing a $ 5$ in between the $ 35$ or the $ 51$ produces the same number - and yet this number is counted twice.


Yes you can.

since $n^{3}$ and $n^{4}$ is a dominant factor.

the result is zero. because the denominator incress much more fast then the numerator

${\lim_{n\to \infty }\, \frac{\sqrt[3]{n^{4}+1}}{2 \sqrt{n^{3}+1}+1}=\lim_{n\to \infty }\, \frac{\sqrt[3]{n^{4}}}{2 \sqrt{n^{3}}+1}}$

${n=u^{2/3}}$

${\lim_{u\to \infty }\, \frac{u^{8/9}}{2 u+1}=\lim_{u\to \infty }\, \frac{u^{8/9}}{2 u}=\lim_{u\to \infty }\, \frac{1}{2 \sqrt[9]{u}}=0}$


Some board games come with special dice. One game in particular uses an 8-sided die with the numbers 0 through 7 printed on the sides. What is the probability of rolling an even number with this die?

[hide]
the numbers are
0, 1, 2, 3, 4, 5, 6, 7

2, 4, 6 are even numbers so the probability is 3/8

if zero is considered even, then it probability is 1/2
[/hide]

[hide]3/8 is correct. Where I got the problem from they don't consider 0 to be an even number. Do you think it should be counted as one :?: [/hide]

[quote="236factorial"]0 is an even number  ;)[/quote]
That's what i think too. So let's just consider the answer to be 1/2 instead of 3/8.


rotation
geometry
geometric transformation
reflection

An equilateral triangle is placed in its starting position on
a table with its vertices labeled as shown. On each move,
it can either be reflected over
a vertical axis through its top
vertex (switching the location
of the bottom two vertices), or
it can be rotated clockwise
120 degrees about its center. How
many different sequences of five
moves will return the triangle to
its original, starting position?

[asy]size(202);
currentpen = fontsize(10pt);
picture p,q,r;
pair top = sqrt(3)/3*dir(90), left = sqrt(3)/3*dir(210), right = sqrt(3)/3*dir(330);
draw(p,top--right--left--cycle); draw(q,top--right--left--cycle); draw(r,top--right--left--cycle);
label(p,"$C$",top,top); label(p,"$B$",right,right); label(p,"$A$",left,left);
label(q,"$C$",top,top); label(q,"$B$",left,left); label(q,"$A$",right,right);
label(r,"$C$",right,right); label(r,"$B$",left,left); label(r,"$A$",top,top);
label(p,"TOP",(0,1.3)); label(q,"TOP",(0,1.3)); label(r,"TOP",(0,1.3));
add(p); add(shift((1.5,0)+3.25*dir(30))*q); add(shift((1.5,0)+3.25*dir(-30))*r);
label("Starting position",(0,1.8));
draw(Label(rotate(30)*"reflection",Relative(.5),dir(120)),(1.5,0)--(1.5,0)+2*dir(30),EndArrow(4,40));
draw(Label(rotate(-30)*"rotation",Relative(.5)),(1.5,0)--(1.5,0)+2*dir(-30),EndArrow(4,40));[/asy]

For this to happen, two of the moves have to be reflections, and the other three have to be rotations. This gives us $ \binom52\equal{}10$ ways to get it back to its starting position in five moves.

Note we cannot have the sequence rsr, since this is the same as just s; we would then reduce to 2 s and 1 r.

However, it is clear that the 5 arrangments of rrrss not containing this subsequence are valid, since they contain either rrr or ss.

(That was part of the solution)

I don't get that. Like why wouldn't rsr be valid? The question says "sequences", and rsr IS a sequence.

Help!


limit
trigonometry
real analysis
real analysis unsolved

By considering $\zeta=\pi k$, $k\to\infty$.

How does that help?
is $k$ an integer?
please explain.

[quote="color"]How do you show that
\[\lim_{\zeta\to\infty}\frac{\zeta^{2}\cos{\zeta}}{\zeta^{2}+1}\neq 0 \]
[/quote]

Because for even multiples $\pi k$ where $k$ is even you get $\cos \zeta = 1$ and for odd multiples you get $\cos \zeta =-1$ thus subsequence (absusing terminology) give limit as 1 and another subsequence gives limits as -1. Which means that limit just doesnt exist.


For how many values of $ a$ is it true that the line $ y \equal{} x \plus{} a$ passes through the vertex of the parabola $ y \equal{} x^2 \plus{} a^2$?

$ \textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ \text{infinitely many}$

[hide="Answer"]The vertex of the parabola is at the point $(0,a^2)$, and $y=x+a$ passes through the y-axis at the point $(0,a)$, so we must have $a=a^2$ with $a$ positive, so the answers are $a=0,1\Rightarrow \boxed{C}$.[/hide]


The sequence $a_0, a_1, a_2, ..$ is defined by $a_0 = a, a_{n+1} = a_n + L(a_n)$, where $L(m)$ is the last digit of $m$ (eg $L(14) = 4$). Suppose that the sequence is strictly increasing. Show that infinitely many terms must be divisible by $d = 3$. For what other d is this true?

Say $a_0$ ends with $5$ or $0$. not good. in the other cases, $a_1$ will be in $\{ 2,4,6,8 \}$. The sequence $a_n$ (mod 10) will have the period $(2,4,8,6)$.

Fe take $x_1=t=$ the first number $\equiv 2 \left( \bmod ~ 10 \right)$ in $\{ a_n \}_{n \geq 0}$ , $x_{n+1} = x_n+L(x_n)$.

We will have $a_{4k+1}=t+20k$ etc.

For all $(d,20)=1$, $20k$ runs through all residues $\bmod d$, so $x_n$ will be infinitely many times divisible by $d$.

For the other $d$'s, i think we can construct counter-examples (forgive me if I'm wrong).

I cannot understand your solution, but i wanted so bad to understand this problem. Can somebody help :(  . Thank you.

i'm going to try to explain the first lines of my solution:

we see that $a_1$ is always even (it is odd+odd or even+even). if $a_1=0$ then the sequence is not increasing, so $a_1 \neq 0 \ \ \Longleftrightarrow \ \ a_0 \neq 0$ and $a_0 \neq 5$.

If $L(a_n)=2$ then $L(a_{n+1})=4$.

If $L(a_n)=4$ then $L(a_{n+1})=8$.

If $L(a_n)=6$ then $L(a_{n+1})=2$.

If $L(a_n)=8$ then $L(a_{n+1})=6$.

So $(a_n)_{n \geq 1}$ is periodic $\bmod 10$. The formulas $a_{4k+1}=t+20k$, $a_{4k+2}=\ldots$ etc. are obvious then.

Sorry for the bad explaining. Hopefully, what I wrote is correct.


trigonometry
geometry
trig identities
Law of Sines
geometry proposed

Let $M$ be a middlepoint of the side $[BC]$. Denote $x=m(\widehat {BAM})$, $y=m(\widehat {CAM})$.
Prove that there is the following relation : $\boxed {\ \frac{\sin^2x}{\sin^2y}=\frac{\sin (B-x)}{\sin (C-y)}\ }\ .$
See http://www.mathlinks.ro/Forum/viewtopic.php?t=16779 .

Applying the Law of Sines in the triangles $AMB,AMC$ we get $\frac{\sin x}{\sin y} = \frac{\sin B}{\sin C} = \frac{AC}{AB}$

Denote $k$ the circle with diameter $BC$.

$AB \cap k = \{B,R\}$
$AC \cap k = \{C,Q\}$

$AB \cdot AR = AC \cdot AQ \Rightarrow \frac{AC}{AB} = \frac{AR}{AQ}$

The triangle $MRB$ is isosceles $(\widehat{MRB}=\widehat{MBR}=B)$

$\blacksquare \ {}$ If $A$ is external point of $k$, then $A$ is also external point of the segment $BR,$ so we have
$\widehat{MRB} = x+\widehat{AMR} \Rightarrow$
$\widehat{AMR} = \widehat{MRB}-x \Rightarrow$
$\widehat{AMR} = B-x$

$\blacksquare \ {}$ If $A$ is in the interior of $k$, then
$\widehat{BAM} = \widehat{MRB}+\widehat{AMR} \Rightarrow$
$\widehat{AMR} = \widehat{BAM}-\widehat{MRB} \Rightarrow$
$\widehat{AMR} = x-B$

similarly we have $\widehat{AMQ} = C-y$ in the first case and $\widehat{AMQ} = y-C$ in the second case

at any case we have $\frac{\sin (B-x)}{\sin (C-y)} = \frac{\sin (\widehat{AMR})}{\sin (\widehat{AMQ})}$

also, at any case we have $\sin (\widehat{MAR})=\sin x,$ since $\widehat{MAR}=x$ or $\widehat{MAR}=180^o-x$
and $\sin (\widehat{MAQ})=\sin y$

now the problem is over, because if we apply the Sines Law in the triangles $AMR,AMQ$ we get

$\triangle AMR: \ \frac{\sin (\widehat{AMR})}{AR} = \frac{\sin x}{MR}$

$\triangle AMQ: \ \frac{\sin (\widehat{AMQ})}{AQ} = \frac{\sin y}{MQ}$

$\Rightarrow \frac{\sin (\widehat{AMR})}{\sin (\widehat{AMQ})} = \frac{AR}{AQ} \cdot \frac{\sin x}{\sin y} = \frac{sin^2x}{sin^2y}$

Notice that, if $A$ is [u]on[/u] the circle $k$, then the triangles $MAB,MAC$ are isosceles $(B=x, \ C=y),$ so the initial equation has $\sin(C-y)=0$ in the denominator.

We can rewrite the initial equation as ${\sin }^2 x \cdot \sin (C-y) = {\sin}^2 y \cdot \sin(B-x)$ to avoid this problem, and this equation holds for any position of the point $A$

[hide="Solution 2"]

We will see that it's enough to prove that $\boxed{ \cot x- \cot B= \cot y- \cot C}$

Law of  Sines in $\triangle ABM: \frac{ \sin x}{MB} = \frac{ \sin B}{AM}$

Law of  Sines in $\triangle ACM: \frac{ \sin y}{MC} = \frac{ \sin C}{AM}$

From these two equations we get $\frac{ \sin x}{ \sin y} = \frac{ \sin B}{ \sin C}$

$\frac{ \sin (B-x)}{ \sin (C-y)} = \frac{ \sin ^2x}{ \sin ^2y} \Leftrightarrow$

$\frac{ \sin (B-x)}{ \sin (C-y)} = \frac{ \sin B \sin x}{ \sin C \sin y} \Leftrightarrow$

$\frac{ \sin B \cos x-\cos B \sin x}{ \sin C \cos y-\cos C \sin y} = \frac{ \sin B \sin x}{ \sin C \sin y} \Leftrightarrow$

$\frac{ \sin B \cos x-\cos B \sin x}{ \sin B \sin x} = \frac{ \sin C \cos y-\cos C \sin y}{ \sin C \sin y} \Leftrightarrow$

$\boxed{ \cot x- \cot B= \cot y- \cot C} \ \ (1)$

Let $AP,BQ,CR$ be the altitudes of $\triangle ABC$
$D$ the midpoint of $BR$
$E$ the midpoint of $CQ$
(then the triangles $MPB,MQC$ are isosceles and $MD,ME$ are their altitudes respectively)

$(1) \Leftrightarrow \frac{AD}{MD} - \frac{BP}{AP} = \frac{AE}{ME} - \frac{CP}{AP} \Leftrightarrow$

$\boxed{\frac{AP}{MD} \cdot AD - BP = \frac{AP}{ME} \cdot AE - CP} \ \ (2)$

$2(ABM) = AP \cdot MB = MD \cdot AB\Rightarrow \boxed{\frac{AP}{MD} = \frac{AB}{MB}}$

similarly, $\boxed{\frac{AP}{ME} = \frac{AC}{MC}}$

$(2) \Leftrightarrow \frac{AB}{MB} \cdot AD - BP = \frac{AC}{MC} \cdot AE - CP \Leftrightarrow$

$AB \cdot AD - MB \cdot BP = AC \cdot AE - MC \cdot CP$

if $w$ is the circle with diameter $AM$, then $MB \cdot BP$ is the power of $B$ w.r.t. $w$.
But also this power can be expressed as $BD \cdot BA$
similarly, $MC \cdot CP = CE \cdot CA$

so we have

$AB \cdot AD - BD \cdot BA = AC \cdot AE - CE \cdot CA \Leftrightarrow$

$AB(AD - BD) = AC(AE - CE) \ \ (3)$

Let $k$ be the circle with diameter $BC$.
Then from the power of the point $A$ w.r.t. $k$ (negative or positive) we get $AB \cdot AR = AC \cdot AQ$.
If $A$ is external point of $k$, then
$ \begin{array}{l} AD-BD = AD-RD = AR \\ AE-CE = AE-QE = AQ \end{array} \}\Rightarrow (3)$  holds

If $A$ is in the interior of $k$, then
$ \begin{array}{l} BD-AD = RD-AD = AR\\ CE-AE = QE-AE = AQ \end{array} \}\Rightarrow (3)$  holds again

Accurately, the power of $A$ w.r.t. $k$ is always equal to $AB(AD-BD)=AC(AE-CE)$,
and this includes all the possible positions of $A$ (in or out of $k$, negative or positive power)[/hide]

[quote="Virgil Nicula"] [color=darkred]Let $ M$ be a point of the side $ [BC]$ . Denote $ x \equal{} m(\widehat {BAM})$ , $ y \equal{} m(\widehat {CAM})$ .

Prove that $ \boxed {\ \frac {\sin^2x}{\sin^2y} \equal{} \frac {\sin (B \minus{} x)}{\sin (C \minus{} y)}\ \Longleftrightarrow\ MB \equal{} MC\ \Longleftrightarrow\ \sin x\sin C \equal{} \sin y\sin B\ }\ .$[/color] [/quote]

[color=darkblue][b][u]Proof I (ugly).[/u][/b] Prove easily that $ \boxed {MA \equal{} MB}\ \Longleftrightarrow\ c\cdot \sin x \equal{} b\cdot\sin y\ \Longleftrightarrow\ \boxed {\ \sin x\cdot\sin C \equal{} \sin y\cdot\sin B\ }$ .

$ \boxed {\begin{array}{c} \\
\ \boxed {\sin^2x\sin (C \minus{} y) \equal{} \sin^2y\sin (B \minus{} x)} \\
 \\
(1 \minus{} \cos 2x)\sin (C \minus{} y) \equal{} (1 \minus{} \cos 2y)\sin (B \minus{} x) \\
 \\
\sin (B \minus{} x) \minus{} \sin (C \minus{} y) \equal{} \cos 2y\sin (B \minus{} x) \minus{} \cos 2x\sin (C \minus{} y) \\
 \\
4\sin\frac {B \minus{} C \minus{} x \plus{} y}{2}\sin A \equal{} [\sin (B \minus{} x \plus{} 2y) \minus{} \sin (x \plus{} 2y \minus{} B)] \minus{} [\sin (C \minus{} y \plus{} 2x) \minus{} \sin (2x \plus{} y \minus{} C)] \\
 \\
4\sin\frac {B \minus{} C \minus{} x \plus{} y}{2}\sin A \equal{} [\sin (B \minus{} x \plus{} 2y) \minus{} \sin (C \minus{} y \plus{} 2x)] \plus{} [\sin (x \plus{} A \minus{} C) \minus{} \sin (y \plus{} A \minus{} B)] \\
 \\
2\sin A\sin \frac {B \minus{} C \minus{} x \plus{} y}{2} \equal{} \sin \frac {B \minus{} C \minus{} 3x \plus{} 3y}{2}\cos\frac {B \plus{} C \plus{} x \plus{} y}{2} \plus{} \sin\frac {B \minus{} C \plus{} x \minus{} y}{2}\cos\frac {3A \minus{} B \minus{} C}{2} \\
 \\
2\sin A\sin\frac {B \minus{} C \minus{} x \plus{} y}{2} \equal{} \sin\frac {B \minus{} C \plus{} x \minus{} y}{2}\sin 2A \\
 \\
\sin\frac {B \minus{} C \minus{} x \plus{} y}{2} \equal{} \cos A\sin\frac {B \minus{} C \plus{} x \minus{} y}{2} \\
 \\
2\sin\frac {B \minus{} C \minus{} x \plus{} y}{2} \equal{} \sin \frac {B \minus{} C \plus{} 2A \plus{} x \minus{} y}{2} \plus{} \sin\frac {B \minus{} C \minus{} 2A \plus{} x \minus{} y}{2} \\
 \\
\sin\frac {B \minus{} C \minus{} x \plus{} y}{2} \minus{} \sin\frac {B \minus{} C \plus{} 2A \plus{} x \minus{} y}{2} \equal{} \sin\frac {B \minus{} C \minus{} 2A \plus{} x \minus{} y}{2} \minus{} \sin\frac {B \minus{} C \minus{} x \plus{} y}{2} \\
 \\
\sin\frac {B \minus{} C \minus{} x \plus{} y \minus{} B \plus{} C \minus{} 2A \minus{} x \plus{} y}{4}\cos\frac {B \minus{} C \minus{} x \plus{} y \plus{} B \minus{} C \plus{} 2A \plus{} x \minus{} y}{4} \equal{} \sin\frac {B \minus{} C \minus{} 2A \plus{} x \minus{} y \minus{} B \plus{} C \plus{} x \minus{} y}{4}\cos\frac {B \minus{} C \minus{} 2A \plus{} x \minus{} y \plus{} B \minus{} C \minus{} x \plus{} y}{4} \\
 \\
\sin\frac {A \plus{} x \minus{} y}{2}\sin C \equal{} \sin\frac {A \plus{} y \minus{} x}{2}\sin B \\
 \\
\boxed {\sin x\sin C \equal{} \sin y\sin B} \\
\ \end{array}}$

[b][u] Proof II (nice).[/u][/b] Denote $ N\in AC$ , $ P\in AB$ so that $ MB \equal{} MC \equal{} MN \equal{} MP$ . Using the power of the point $ A$

 w.r.t. the circle $ C\left(M,\frac a2\right)$ obtain $ c\cdot AP \equal{} b\cdot AN$ . Apply the [b]Sinus' theorem[/b] in the triangles $ APM$ and $ ANM$ :

$ \left\{\begin{array}{c} \frac {PM}{\sin x} \equal{} \frac {AP}{\sin (B \minus{} x)} \\
 \\
\frac {NM}{\sin y} \equal{} \frac {AN}{\sin (C \minus{} y)}\end{array}\right\|$ $ \implies$ $ \frac {\sin y}{\sin x} \equal{} \frac {AP}{AN}\cdot\frac {\sin (C \minus{} y)}{\sin (B \minus{} x)}$ . But $ MA \equal{} MB\Longleftrightarrow$ $ c\cdot\sin x \equal{} b\cdot\sin y$ $ \Longleftrightarrow$ $ \frac {AP}{AN} \equal{} \frac bc \equal{} \frac {\sin x}{\sin y}$ .

In conclusion, $ \frac {\sin y}{\sin x} \equal{} \frac {\sin x}{\sin y}\cdot\frac {\sin (C \minus{} y)}{\sin (B \minus{} x)}$ , i.e. $ \boxed {\ \frac {\sin^2x}{\sin^2y} \equal{} \frac {\sin (B \minus{} x)}{\sin (C \minus{} y)}}$ .[/color]

 [size=134][color=darkred][b]I dedicate these proofs to my dear Armpist ![/b] [/color][/size]


AMC
AIME
search
geometry
3D geometry
USA(J)MO
USAMO

The title sounds pretty weird, but...

I knew all the math I needed for this AIME, but my mind just went completely blank on the actual test.  I couldn't even solve #1 or #8 which I solved in less than 2 minutes when I got home.  I also mis-read #3.  I followed the advice of not doing too much math before the AIME, etc. but I still only got a 7.  Any tips for next year?

Alot of the reason people say not to do math before the test is because you're stuck in the thinking mode of whatever type of math you were doing before.

However, you said you solved the ones you went blank on immediately afterwards, so it would seem like you just got overwhelmed by stress. There's nothing you can really do for that; you could try taking several AIMEs in the same setting (silent room, 3 hours, no breaks) to get used to the setting and not be that stressed.

Also, don't spend the last couple of days...or hours....try to learn how to do a specific type of problem that has shown up before on the previous year or two. Instead, focus on your basic skills, relax, and most of all try to write in a clear, coherent manner on the test so that you can easily double check your work and make errors.

I like to make sure that all my work for different problems is on different sheets of paper to help in checking.

[quote="Elemennop"]There's nothing you can really do for that; you could try taking several AIMEs in the same setting (silent room, 3 hours, no breaks) to get used to the setting and not be that stressed.[/quote]

My proctor let us walk around, go to the bathroom, sharpen pencils, drink, and eat snacks. We just couldn't talk to each other.

K same thing here, I on average solve around 11 questions on practice AIMEs, but here I only solved 9, and I realized how to 3 of the problems(#9, 12, 15) that I didn't do after the test, and realized how easy they were. I also wasted a lot of time on #6(I didn't see the binomial coefficients, I eventually factored LHS, but I still didn't see the binomial coefficients until after the test :( ).  On #9 it was so sad, I knew exactly what to do when I was doing the test! But I choose not to do it for some reason! #12's easy but I didn't read the question! #15 I started to attempt in the last 15 minutes. I got pretty far but decided to quit on doing it!

I'm not sure what happened, maybe I'm just retarded on Tuesday, but it seems to be more complicated than that.

Everybody else: skips hard ones, does well on first ten.

Me: does first few, loses hope, skips middle ones, gets on a roll at the end, leads to believe can solve 14 easily, wastes half an hour.

NEVER lose hope.

A good night's sleep, and no math during the weekend before more than doubled my expected scores..
And try not to get into a frenzy mode the few hours before, even if the AIME tests haven't arrived by mail yet. It can be very concentration-destructive. (I almost missed the AIME entirely... my test arrived at 1 pm ON TEST DAY. Stupid DHL!!)

I do not agree with the "no math before the test" rule in the least.  I find that doing some easier problems to warm up or reading solutions to problems boosts my confidence before the test and lets me know that my brain did not suddenly lose all its knowledge somehow within a day or two.  On the other hand, if I don't do any math for a while beforehand, and the test begins and I don't immediately know how to do number 1, I tend to panic, thinking that I'm having a bad day.  Then I'll go into a frantic search for any problem that I could do easily, which ends up wasting time.
So different things work for different people.  I'd suggest trying different things on less important tests to see what works for you.

agree.....like this AIME....it's not i don't know how to do the problem..i always make dumb mistake in the calculationor last few step...like i forget to eliminate some cube # in problem 3 and forget to check the both case in # 10......

[quote="origamimasterjared"][quote="hmmm"]I'd suggest trying different things on less important tests to see what works for you.[/quote]

Like your finals...[/quote]

Obviously. :P

It's different for everyone. In my case, I'm convinced that the adrenaline of the real thing makes me faster and clearer. The confidence that came from knowing I would be my best for the real test made preparation easy for me- just get a good night's sleep and let the natural nervous reactions happen.
That adrenaline is impossible to simulate in practice- only contest experience can tell you how you will respond.

For mathematical preparation, I would recommend hard problems from any source- harder than the real test is good. You want to be familiar with the style of question you will see, but it doesn't take too many problems to get there. It is possible to over-practice, especially with a narrow focus on the test ahead.

For the AIME in particular, cleaning up the little mistakes is very important- only a careful attention to detail and a deep understanding of the problems can get you to the very top scores and make USAMO qualification safe. As a 10th grader, my 130 AHSME/7 AIME were enough for the 200 (12th grade) or 190 (10th grade) minimal qualifying index standards, but I got lucky on the AIME with two offsetting mistakes. Without the remarkable luck of getting the right answer for the wrong reason, I would have been on the very edge. I made HM on the USAMO, which got me to MOP- and MOP experience boosted me to 145/14 the next year and 138/13 my last time. A large part of that improvement was cleaning up all of the little mistakes.

Did you make it to USAMO freshman year too?


Determine all pairs $ (a,b)$ in positive integers such that $ a^b \plus{} 1 \equal{} 777\cdots 7$ (All digits are $ 7$).

Edit:sorry,add the condition $ b\ge 2$  (5:37am)

This is equivalent to $ a^b \equal{} 77\dots 6$. We can easily check that $ (6, 1)$ and $ (6, 5)$ are solutions, and also the only solutions.

First, we check the first four terms.
6 - Yes (6, 1)
76 - No
776 - No
7776 - Yes (6, 5)

Now, we can prove that any sequence of four or more 7's followed by a 6 cannot be in the form $ a^b$. First, notice that since the last five digits of any such number is always 77776, it must be divisible by 16, or $ 2^4$, but it will NOT be divisible by 32.  Now, for this number to be in the form $ a^b$, it must be in the form $ 2^4a^4$ where $ a$ is an odd number. So dividing by $ 2^4$ we get a number in the form $ 486111\dots$ where there are $ n \minus{} 3$ 1's if there are $ n$ 7's in the original number. So we require $ 486111\dots$ to be in the form $ a^4$.

First, we can quickly check that $ 4861$ does not work because it is prime.

Now, let's examine this sequence mod 4.

$ 48611 \equiv 3 \pmod 4$
Any subsequent number in this form will end in "11", which guarantees a remainder of 3 upon division by 4.

However, 3 is not a quadratic residue mod 4. Therefore, it cannot be a quartic residue either. Therefore, there are no other solutions.

Nice solution!XD,Yongyi781.

Anyway,the key to solve this is to notice that $ 77\cdots 76$ (where $ 7$ shows up more than 3times) is divisible by $ 2^4$ but not by $ 2^5$.

My solution is to note $ a$ is even and $ b$ is odd since $ n^2\equiv 0,1,2,4 \pmod 7$ but $ a^b \equal{} 77\cdots 76 \equiv 6 \pmod 7$.
Thus we can see $ a^b$ cannot have just four factor:$ 2$,
therefore we have $ a^b\le 7776$.

[quote="Yongyi781"] We can easily check that $ (6, 1)$ and $ (6, 5)$ are solutions[/quote]
HOW can you 'easily' find those two solutions?? I think (6,5) isn't an obvious solution. I guess there's a way to find it.

Well 6 is trivial, agree?

For 7776, notice that it is divisible by 32. So divide by 32, we get 243, which happens to be $ 3^5$. And since $ 32 \equal{} 2^5$, we have that $ 7776 \equal{} 2^53^5 \equal{} 6^5$.


What did you get? could a 60 ever make AIME? what was last years qualification score?

Last year's qualifying score was a 68 over 3 rounds. I think it may be a little lower this year because of the round 2 mess. 60 is unlikely

Hmm I struggle alot with this contest for some reason.  I guess proofs/explaing your solution are totally different than just purely problem solving.

Even if I had gotten 25 on round 2 (they lost my solutions), I would still be pretty far away from the AIME qualifying score.

I had no idea that it would be that high.  I wonder if anyone actually qualfies through this but not the AMCs?  I would asusme that with this high of a qualifying score that anybody that gets that will also have passed AMCs.

i have a 60+ but failed the ten. hopefully ill do better next week  :)

AIME index for USAMTS qualification has always been 100+10*AIME. :)

[quote="the future"]Oh and about the round2 mess.

(Sorry if I am being annoying since I kind of asked this already when it first came out)

But how are they going to calculate the AIME index for the people who did not resubmit their round 2 papers.
I am one of those people, but I have no problem since I already qualified with a higher index than the 100 given on the 12 (and I probably wount make it anyways through the USAMTS method).

However, there might just be 1 or 2 people that got like 23 round 1/24 round 3 and they did not qualify for the AIME through the AMC.  For those people, how are will their AIME qualifying index be calculated?  On the website, it says it will be recalculated after round 4, but wouldnt it be unfair to some of those people?

As for me, like I said, I really do not mind which direction they go.  I am just curious because this kind of indirectly affects me.

And are there any other AoPSERs like me that did not make a copy of their round 2 solutions??  Or am I just talking about myself.[/quote]
[quote="DPatrick"]Round 3 scores and feedback are now available on the USAMTS website.

We will make an announcement about AIME qualification during the week of Feb 23-27.  We will develop a procedure so that students who did not resubmit Round 2 by mail will still be eligible to qualify for the AIME.  [b]Please do not inquire about this prior to our announcement[/b].[/quote]

Oh sorry.

I did not notice that he posted one on this website (I was just rereading the one on the USAMTS website)


function
trigonometry
geometry
3D geometry
sphere
analytic geometry
topology

We call the set $A\in \mathbb R^n$ CN if and only if for every continuous $f:A\to A$ there exists some $x\in A$ such that $f(x)=x$.
a) Example: We know that $A = \{ x\in\mathbb R^n | |x|\leq 1 \}$ is CN.
b) The circle is not CN.

Which one of these sets are CN?

1) $A=\{x\in\mathbb R^3| |x|=1\}$

2) The cross $\{(x,y)\in\mathbb R^2|xy=0,\ |x|+|y|\leq1\}$

3) Graph of the function $f:[0,1]\to \mathbb R$ defined by
\[f(x)=\sin\frac 1x\ \mbox{if}\ x\neq0,\ f(0)=0\]

I have moved this from the Algebra subforum to Topology. I hope it is right here.

1) Not CN. Simply take the map which sends every point on the sphere to its antipode.

2) CN. Let $C$ be this cross, and let $f:C\to C$ be a continuous map. Assume (for the sake of contradiction) that $f$ has no fixed point. Let $\pi_x:\mathbb R^2\to\mathbb R$ be the map sending every point to its $y$-coordinate. Consider the function $\mathbb R\to \mathbb R,\ t\mapsto\pi_x\left(f\left(t,0\right)\right)-t$. This function is clearly continuous. Its value at $t=-1$ is nonnegative (because $\pi_x\left(f\left(-1,0\right)\right)$ must be $\geq -1$). Its value at $t=1$ is nonpositive (for similar reasons). Thus, by the intermediate value theorem, it must take the value $0$ at $t=t_0$ for some $t_0\in\left[-1,1\right]$. If this $t_0$ is not zero, then $\left(t_0,0\right)$ must then be a fixed point of $f$ (because the only way we can have $\pi_x\left(f\left(t_0,0\right)\right)-t_0=0$ with $t_0\neq 0$ is when $f\left(t_0,0\right)=\left(t_0,0\right)$), contradicting our assumption that $f$ have no fixed points. Thus, $t_0$ must be zero. Then, $\pi_x\left(f\left(0,0\right)\right)-0=0$. In other words, $\pi_x\left(f\left(0,0\right)\right)=0$. So the $x$-coordinate of $f\left(0,0\right)$ must be $0$. Similarly, the $y$-coordinate of $f\left(0,0\right)$ must also be $0$. Thus, $f\left(0,0\right)=\left(0,0\right)$, again contradicting the assumption that $f$ have no fixed point. This contradiction proves our assumption wrong, and thus the cross $C$ is CN.

3) CN. Let $G$ denote the graph of $f$. Let $h:G\to G$ be a continuous map. Assume (for the sake of contradiction) that $h$ has no fixed point. Let $\pi_x:\mathbb R^2\to\mathbb R$ denote the map which takes every point to its $x$-coordinate. Clearly, $\pi_x\mid _G$ is injective and continuous.

Let me rename the function $f$ as $p$, and use the notation $f$ for the function $\left[0,1\right] \to G,\ x\mapsto \left(x,p\left(x\right)\right)$. (Sorry for this renaming, but I am too tired to fix what otherwise would be a lot of mistakes...)

For every $t\in\left[0,1\right]$, we notice that $f\left(t\right)$ is the unique point on $G$ whose $x$-coordinate is $t$. Thus, the map $f:\left[0,1\right]\to G$ is bijective (being the inverse of $\pi_x\mid _G$) and continuous on $\left(0,1\right]$ (but not on $\left[0,1\right]$).

Clearly, $\pi_x\left(h\left(f\left(1\right)\right)\right)\leq 1=\pi_x\left(f\left(1\right)\right)$. Hence, if we can find some $t>0$ satisfying $\pi_x\left(h\left(f\left(t\right)\right)\right)\geq \pi_x\left(f\left(t\right)\right)$, then by the intermediate value theorem we can find some $t_0>0$ satisfying $\pi_x\left(h\left(f\left(t_0\right)\right)\right)= \pi_x\left(f\left(t_0\right)\right)$, which yields that $f\left(t_0\right)$ is a fixed point under $h$ (since $\pi_x\mid _G$ is injective), contradicting the assumption that $h$ have no fixed point. Thus, we must have $\pi_x\left(h\left(f\left(t\right)\right)\right) < \pi_x\left(f\left(t\right)\right)$ for every $t>0$.

Since $h$ has no fixed point, we have $h\left(f\left(0\right)\right)\neq f\left(0\right)$. Thus, $h\left(f\left(0\right)\right) = f\left(u\right)$ for some $u\in\left(0,1\right]$. Consider this $u$. Then, $u>0$, so that $\pi_x\left(u\right)>0$. Now let $V$ be an open neighbourhood of $f\left(u\right)$ in $G$ such that $\pi_x\left(V\right)$ is bounded away from $0$ (such a $V$ obviously exists since $\pi_x\left(u\right)>0$). Let $U=h^{-1}\left(V\right)$. Since $h$ is continuous, it follows that $U$ is open, and clearly $f\left(0\right)\in U$ (since $h\left(f\left(0\right)\right)=f\left(u\right)\in V$).

Since $\pi_x\left(V\right)$ is bounded away from $0$, there exists some $\epsilon>0$ such that $B\left(0,\epsilon\right) < \pi_x\left(V\right)$ (this means that every element of $B\left(0,\epsilon\right)$ is smaller than every element of $\pi_x\left(V\right)$). Consider this $\epsilon$. Let $U_1=U\cap \pi_x^{-1}\left(B\left(0,\epsilon\right)\right)$. Then, $U_1$ is an open subset of $U$ still containing $f\left(0\right)$ but satisfying $\pi_x\left(U_1\right) < \pi_x\left(V\right)$. Since the topology on $G$ was defined as the restriction of the Euclidean topology on $\mathbb R^2$, it is easy to see that there exists some $q\in \left(0,1\right]$ such that $f\left(q\right)\in U_1$ (in fact, we can choose $q$ as $\frac{1}{\pi N}$ for a sufficiently high $N\in\mathbb N$). Consider this $q$. Then, $f\left(q\right)\in U_1\subseteq U=h^{-1}\left(V\right)$, so that $h\left(f\left(q\right)\right)\in V$. Thus, $\pi_x\left(h\left(f\left(q\right)\right)\right) \in \pi_x\left(V\right)$. Combined with $\pi_x\left(f\left(q\right)\right) \in \pi_x\left(U_1\right)$ (since $f\left(q\right)\in U_1$), this yields $\pi_x\left(f\left(q\right)\right) < \pi_x\left(h\left(f\left(q\right)\right)\right)$ (since $\pi_x\left(U_1\right) < \pi_x\left(V\right)$). But this contradicts the fact that $\pi_x\left(h\left(f\left(t\right)\right)\right) < \pi_x\left(f\left(t\right)\right)$ for every $t>0$. This shows that our assumption was wrong, and $h$ must have a fixed point, qed.

For similar results, see
R. H. Bing, [i]The Elusive Fixed Point Property[/i], The American Mathematical Monthly, Vol. 76, No. 2 (Feb., 1969), pp. 119-132.


Given the series $ \sum a_n.$ Let $ x_n \equal{} \frac {a_n \plus{} |a_n|}{2}$ and $ y_n \equal{} \frac {a_n \minus{} |a_n|}{2}.$ Prove or disprove that there exists a convergent series such that $ \sum x_n$ converges and $ \sum y_n$ diverges.

Let $ a_n \equal{} \minus{}1$ for every $ n$; then $ x_n \equal{} 0$ and $ y_n \equal{} \minus{}1$.

$ \sum \minus{} 1$ isn't convergent ;)

Prove that if $ \sum p_n$ converges and $ \sum q_n$ diverges, then $ \sum (p_n \plus{} q_n)$ diverges.  Your result follows from the contrapositive.

[quote="Null"]Prove or disprove that there exists a [b]convergent[/b] series such that [/quote]  Not clearly written, but certainly referring to $ (a_n)$.


Find all natural numbers n for which:

(2n + 1)|((n!)^2 - 1)

Happy New Year!

We will prove that $ 2n \plus{} 1$ is a prime.
If $ 2n \plus{} 1 \equal{} pq$ where $ p,q > 1$
So $ p\leq \frac {2n \plus{} 1}{2}$ so $ p\leq n$
So $ p|n!$ (tradition ).
It mean that $ 2n \plus{} 1$ is a prime.
The condition become :
$ p|(\frac {p \minus{} 1}{2})!^2 \minus{} 1$
From Wilson's theory  easy to check $ p\equiv 3 (\mod 4 )$


A hydrocarbon $ X$ has the formula $ C_{6}H{}_{1}{}_{4}$. $ X$ is allowed to react with limited chlorine gas in the presence of light. The mixture contained two products with different boiling points, as well as excess amount of the starting material. What is $ X$?

[hide="Answer"]The molecular formula of X tells us that it must be an open chain alkane. In radical reactions, chlorine is very reactive and so is unselective as to the type of hydrogen it reacts with: primary, secondary, and tertiary positions all react in the same way with chlorine. Also, in the experiment with X a limited supply of chlorine was used in order to prevent the formation of polichlorinated products. So, as in the chlorination of X only two monochlorinated products were obtained, that means X can only have two types of hydrogens. The only alkane with six carbons and only two kinds of hydrogens is 2,3-dimethylbutane, which we identify as compound X. [/hide]

What do you mean, two types of hydrogen? Wouldn't $ \text{3\minus{}methylpentane}$ also give two products?

[quote="BanishedTraitor"]What do you mean, two types of hydrogen?[/quote]

Chemically different types of hydrogen. You can also see how many types there are by asking yourself "How many signals would this compound show in its 1H-NMR spectrum?".
3-Methylpentane has 4 types of hydrogen (marked A, B, C, D bellow), and so it would yield four different products:

$ CH_{3A}\minus{}CH_{2B}\minus{}CH_{C}(\minus{}CH_{3D})\minus{}CH_{2B}\minus{}CH_{3A}$.

What about $ \text{2,3\minus{}dimethylbutane}$? Aren't there three types of hydrogens?

No, there are only two, because all four methyl groups are chemically (and magnetically) equivalent:

$ (CH_{3A})_{2}\minus{}CH_{B}\minus{}CH_{B}\minus{}(CH_{3A})_{2}$.

Another way of see this is two substitute every hydrogen in a molecule by a general group X, and look how many different compounds you obtain. For example, if in 2,3-dimethylbutane you substitute any hydrogen of the four methyl groups by X, you will get the same compound: 1-X-2,3-dimethylbutane. So, all twelve hydrogens belonging to those four methyl groups are equivalent. If you substitute any of the B hydrogens by X you also get the same compound: 2-X-2,3-dimethylbutane. So, the two hydrogens B are also equivalent.

:omighty: Thanks, I knew I was missing something simple.


Let $ a\sqrt {1 \minus{} bc} \equal{} p$, then $ a^2(1 \minus{} bc) \equal{} p^2$.
We wish to show
$ p \plus{} q \plus{} r \ge 2\sqrt {2}/3$
which is equivalent to
$ p^2 \plus{} q^2 \plus{} r^2 \ge 8/27$ (Is this step correct?)
But $ p^2 \plus{} q^2 \plus{} r^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \minus{} abc$
So $ min(p^2 \plus{} q^2 \plus{} r^2) \equal{} min(a^2 \plus{} b^2 \plus{} c^2)\minus{} max(abc) \equal{} 1/3 \minus{} 1/27 \equal{} 8/27$.

$ (b \plus{} c)^2\geq 4bc\Longleftrightarrow (1 \minus{} a)^2\geq 4bc$, we have $ a\sqrt {1 \minus{} bc}\geq a\sqrt {1 \minus{} \frac {1}{4}(1 \minus{} a)^2}$ for $ 0 < a < 1$.

Let $ f(a) \equal{}a\sqrt {1 \minus{} \frac {1}{4}(1 \minus{} a)^2}\ (0 < a < 1)$, $ f'(a) \equal{} \frac {3 \plus{} 3a \minus{} 2a^2}{\sqrt {4 \minus{} (a \minus{} 1)^2}} > 0$.

$ \therefore f(a)\geq \frac {17\sqrt {2}}{24}a \minus{} \frac {\sqrt {2}}{72}$.

$ \therefore a\sqrt {1 \minus{} bc} \plus{} b\sqrt {1 \minus{} ca} \plus{} c\sqrt {1 \minus{} ab}\geq \frac {17\sqrt {2}}{24}(a \plus{} b \plus{} c) \minus{} 3\cdot \frac {\sqrt {2}}{72} \equal{} \frac {2\sqrt {2}}{3}$.

[quote="kunny"]

$ \therefore f(a)\geq \frac {17\sqrt {2}}{24}a \minus{} \frac {\sqrt {2}}{72}$.

[/quote]

 Why?

a+b+c=1
or, a+b=1-c
or, a^2+2ab+b^2=1+c-2c => 4ab
or, c(1-ab)^1/2 => 3abc+2c^2-c^3
Thus, <sum>c(1-ab)^1/2 => 9abc+2(a^2+b^2+c^2)-(a^3+b^3+c^3)...... Hereafter I am still thinking. I will post the complete solution as soon as it comes to my mind.

[quote="Sunkern_sunflora"]a+b+c=1
or, a+b=1-c
or, $ a^2\plus{}2ab\plus{}b^2\equal{}1\plus{}c\minus{}2c \geq 4ab$
or, $ c\sqrt{(1\minus{}ab)}\geq 3abc\plus{}2c^2\minus{}c^3$
Thus, $ \sum {c\sqrt{(1\minus{}ab)}} \geq 9abc\plus{}2(a^2\plus{}b^2\plus{}c^2)\minus{}(a^3\plus{}b^3\plus{}c^3)$...... Hereafter I am still thinking. I will post the complete solution as soon as it comes to my mind.[/quote]

[quote="Marius Mainea"][quote="dgreenb801"].
$ p \plus{} q \plus{} r \ge 2\sqrt {2}/3$
which is equivalent to
$ p^2 \plus{} q^2 \plus{} r^2 \ge 8/27$ (Is this step correct?)
[/quote]

I don't think so.[/quote]

$ p^2 \plus{} q^2 \plus{} r^2 \ge \frac {1}{3}(p \plus{} q \plus{} r)^2$ is definitely true. It's not equivalent to the previous step though... so if you are writing the solution backwards, starting from the problem, I don't think this will work.

[quote="Marius Mainea"]Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$ Prove that

 $ a\sqrt {1 \minus{} bc} \plus{} b\sqrt {1 \minus{} ca} \plus{} c\sqrt {1 \minus{} ab}\geq\frac {2\sqrt {2}}{3}$[/quote]
By Holder we obtain:
$ \left(\sum_{cyc}a\sqrt {1 \minus{} bc}\right)^2\sum_{cyc}\frac{a}{1\minus{}bc}\geq(a\plus{}b\plus{}c)^3\equal{}1.$
Thus, it remains to prove that $ \sum_{cyc}\frac{a}{1\minus{}bc}\leq\frac{9}{8},$ which is true.

[quote]arqady wrote:
Thus, it remains to prove that \sum_{cyc}\frac{a}{1-bc}\leq\frac{9}{8}, which is true.[/quote]

Why??? I mean, can you prove it :wink: ........cause it seems impossible for me to visualize.

Sunkern_sunflora, it's very easy! After homogenization we obtain:
$ \sum_{cyc}\frac{a}{1\minus{}bc}\leq\frac{9}{8}\Leftrightarrow(a\plus{}b\plus{}c)^6\minus{}9(ab\plus{}ac\plus{}bc)(a\plus{}b\plus{}c)^4\plus{}$
$ \plus{}9abc(a\plus{}b\plus{}c)^3\minus{}9a^2b^2c^2\plus{}8(a\plus{}b\plus{}c)^3\sum_{sym}a^2b\minus{}8abc\sum_{cyc}a^2\sum_{cyc}a\geq0,$
which is equivalent to the following obvious inequality:
$ \sum_{sym}\left(\frac{a^6}{2}\plus{}5a^5b\plus{}11a^4b\minus{}a^4bc\plus{}7a^3b^3\minus{}15a^3b^2c\minus{}\frac{15a^2b^2c^2}{2}\right)\geq0.$

We obtain the important inequality:
Let $ a, b, c$ be positive real numbers satisfying the condition $ a\plus{}b\plus{}c\equal{}1$. We have the following result ( really beautiful):
$ (1\minus{}ab)(1\minus{}bc)(1\minus{}ca) \geq (\frac{8}{9})^3$
It can be proved by mixing variable.

[quote="Sunkern_sunflora"]Now I see.........it also completes the proof. Thanks arqady. Sometimes easy problems or solutions seems quite difficult to me...dunno why! Sorry arqady, ...........for disturbing you.  :oops:[/quote]

actually, I think his solution was hardly obvious, except to maybe himself :wink:

arqady's level in solving inequalities is very high... I can't even see how those sums are equivalent :(

this is why I am not very good...

But the solution of Kunny is rather easy to understand . Kunny, I don't how it works in these type of inequality. Can you give us some types of exercises using this method?

Is there any nicer way to prove this inequality?

$ \sum_{cyc}\frac{a}{1\minus{}bc}\leq\frac{9}{8}$

[quote="matex (L)(L)(L)"]Is there any nicer way to prove this inequality?

$ \sum_{cyc}\frac {a}{1 \minus{} bc}\leq\frac {9}{8}$[/quote]
Using Chebychev's ineq, we only need to prove: $ \sum\ \frac{1}{1\minus{}bc} \leq\ \frac{27}{8}$
It's very weak. After expand, it's weaker than Schur ineq  :wink:

hmmm...

Let L=1-bc, M=1-ac and N = 1-ab

By Jensen we have that $ a\sqrt(L) \plus{} b\sqrt(M) \plus{} c\sqrt(N) \ge \sqrt(aL\plus{}bM\plus{}cN)$

It follows therefore that

$ LHS \ge \sqrt(a(1\minus{}bc) \plus{} b(1\minus{}ac) \plus{} c(1\minus{}ab) \rightarrow LHS \ge \sqrt(1\minus{}3abc)$

But by AM-GM we know that $ 1/27 \ge abc$ thus at the maximal value of abc the first inequality still has to hold. Hence

$ LHS \ge \sqrt(1\minus{}1/9) \rightarrow LHS \ge 2\sqrt(2)/3$

$ Q.E.D$

Here is my solution:after use CauChy-Schwarz and chebyshev inequality,we must prove that:
$ \sum \ \frac{1}{1\minus{}bc} \le \frac{27}{8}$
After expand,it is equivalent to:$ 3\minus{}11\sum \ ab \plus{}19abc \minus{}27a^2b^2c^2 \ge 0$
But$ abc \le \frac{1}{27}$,so $ 27 a^2b^2c^2 \le abc$
By Schur inequality,we have: $ 9abc \ge 4 \sum \ ab\minus{}1$,
finally,we must prove that: $ 1 \ge 3(ab\plus{}bc\plus{}ca)$,which is obvious true :)


kid at Inglemoor, classmates with William Johnson, who participated in the USAMO in his 9th grade.

I go to Eckstein right now, but I hope I'll get into Garfield next year.

Hey Everybody,

     I'm Steven, and I'm a junior at Tahoma High School in Maple Valley, WA. I'm new to this board, and I think the only Tahoman here. So yeah, that's about it for now.

Hm.... I lived in Washington for around 5 years, anyone know the town of Chewelah?  I guess since there's no Oregon forum, I'll post here.  I go to Sprague High School in Salem, Oregon.

I'm austin b (in case u didn't catch that already) and I'm a junior Pullman HS in Pullman.

Yes, to whoever asked, I've heard of Chewelah, WA (I'm assuming u mean the place north of Spokane.) We had a baseball tournament there a few years ago

I go to Roosevelt.  :mad:
Garfield won't let me in!!! :(

[quote="schnook"]Freshmen at Bellevue High!!!!![/quote]

Are you in math club?  I don't recognize you from your posting style.

Freshman, go to northshore junior high (7-9th grade) going up to inglemoor high, taking math there

WAIT WHAT I GO THERE TOO?! WELL SORTOF STILL IN 8TH GRADE BUT...

i know but i just freaked cause thats where im going

[quote="Lalagato"][quote="573863"]Wait what i go there too?! well sortof still in 8th grade but...[/quote]
That post was 1 1/2 years ago. :)[/quote]
That was my friend... he lives down the street from me. :)


Ring Theory
superior algebra
superior algebra unsolved

Let $A$ be a UFD and $a\in A$ non-zero. Show that the nilradical of $A/aA$ is the intersection of a finite number of prime ideals.

If $P_{1},\ldots,P_{n}$ are these ideals, show that for each primce ideal $P$ of $A/aA$ there exists $i$ such that $P_{i}\subset P$


The tail of a 1-mile long train exits a tunnel exactly 3 minutes after the front of the train entered the tunnel.  If the train is moving 60 miles per hour, how long is the tunnel?

in 3 minutes, the train goes 3 miles. 1 of those miles, however, was used for the [b]whole[/b] train exiting the tunnel. so the tunnel is 2 miles long

oh good point.  sorry i thought 2 miles was used up to travel its own length but that was wrong.  Yours makes sense.  Thanks.


Find all the set of four (x,y,z,t) positive integers such that $1 + 5^x = 2^y + 2^z \cdot  5^t$.

Well, if no one else is saying anything, I'll begin.  Take the equation mod 4.  Then LHS = 2, while RHS is the sum of two powers of 2.  Thus, one of those two powers must be $2^1$, so either y = 1 or z = 1.  If y = 1,  take the equation mod 5 and 1 = 2, a contradiction.  Thus z = 1.  So our equation simplifies to
$1 + 5^x = 2^y + 2\cdot5^t$
Looking mod 5 again, we see that y must be a multiple of 4, so y = 4n and
$1 + 5^x = 16^n + 2\cdot5^t$
$5^x - 2\cdot5^t= 16^n - 1$
Clearly x > t, so factor both sides as
$5^t(5^{x-t} - 2) = (4^n -1)(4^n + 1)$
So, RHS factors as product of two numbers with difference 2.  Meanwhile, LHS factors as product of power of 5 and a number 2 less than some other power of 5.  This implies the two powers of 5 are equal, I think :).

Sure [u]z=1 [/u]and [u]y must be a multiple of four [/u]but [u]the two powers of five can't be equal[/u]


Let $ a,b,c\in \mathbb{R}$ such that $ ab \plus{} bc \plus{} ca\le 0$ . Show that:
\[ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \plus{} abc(a \plus{} b \plus{} c)\ge ab(a^2 \plus{} b^2) \plus{} bc(b^2 \plus{} c^2) \plus{} ca(c^2 \plus{} a^2)\]
[b]Remark.[/b]If we could apply Muirhead then the sign of the inequality would be reversed.

[hide]
$ (ab \plus{} bc \plus{} ca)^2 \ge 0$

$ \therefore \sum a^2b^2 \plus{} abc(a \plus{} b \plus{} c) \ge \minus{} abc(a \plus{} b \plus{} c) \equal{} \sum a^2( \minus{} bc) \ge \sum a^2(ab \plus{} ac)$

Done [/hide]


Here are problems. I was going to do it on Friday since I don't know if I'll have time so I decided to post it early and let this test be taken by honesty (every AoPS uers is good so I can trust this :D ). Anyway, please send me answers via PM by Friday 7:00 EST.

PLEASE DO NOT START DISCUSSION TILL 7:00 that day. Thank you!

Thanks for taking the time to make it  :D

My bad on one question.

#23, the sign should be < 0.

[quote]wait, we can't start until 7:00 and we have to submit them by 7:00?[/quote]

Silverfalcon meant that you shouldn't discuss the problems before 7:00 Friday since there might be people working still working on the problems.

Please note that today is the last day you can take this test. It's 3:20 EST now so you have around 4 hours to do this. I only received 4 answers so far and I like to receive more. Thanks!  :D

oh crap. i have to go at 6:00 and I won't be back until after 8:00... darnit. o well i'll start.

[quote="solafidefarms"]When will we know our scores and placement?[/quote]

This is 6:51 now and I just made the discussion thread.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=456416#456416

Talk over there. Thanks.


I don't know if this belongs here, but I couldn't find a better forum.

If you have a rubber band with a certain width and length, is there a formula for how much force(N) is exerted by the rubber band when it is stretch out a certain amount?

Like you have one of those spring scales with the S-hook at one end. You hook the rubber band on and then you stretch the rubber band to whatever length.

If you've got the tools, this sounds like a good case for collecting some empirical data just to see. As default assumtions to start from, I'd assume that the force is proportional to the length minus the rest length (Hooke's "law"). If we're trying different rubber bands, I'd also assume that the force is proportional to the resting cross-sectional area, and also that it is inveresly proportional to the resting length.

But all of that could be pretty badly wrong. Rubber bands might not be particularly good models for Hooke's law springs. The data to show that, one way or another, might not be too hard to collect.

The amount of force you apply given that you don't permanently deform the rubberband is expressed by: $F = E\cdot\frac{A\Delta \ell}{\ell}$

$E$ is Young's modulus (a property of the material, i.e. rubber, you're using).

$A$ is the cross section.

$\Delta \ell$, the distance you stetch the rubberband.

and

$\ell$ is the original unstretched length.


calculus
integration
logarithms
limit
function
calculus computations

These are from my Calculus book, I am not sure how to go about solving them or I just don't understand/can't follow the answer that the book gives...

1.

Find the values of $p$ for which the integral converges and evaluate the integral for those values of $p$.

$\int_0^1 x^p \ln {x} \, dx$

All I really know is that if $p = -1$, the integral diverges. Other than that, I do not know how to go about doing this problem...

2.
The book has a problem in two parts...

(a) Show that $\int_\infty^{-\infty} x \, dx$ is divergent.

(b) Show that $\lim_{t \rightarrow \infty} \int_t^{-t} x \, dx = 0$. This shows that we can't define  $\int_\infty^{-\infty} x \, dx = \lim_{t \rightarrow \infty} \int_t^{-t} x \, dx$

I don't understand why part a is divergent. In my mind, the graph of x cancels out no matter how far it extends, and since it extends to infinity on both sides, they should cancel out to zero. I am not understanding the concept of integrals, I think...or I am missing the idea with the limits, because I do not see the difference....

1) Integrate by parts putting $du=\ln xdx$

2) The definition of
\[ \int^{\infty}_{-\infty} f(x)dx  \]
is
\[ \lim_{b\to\infty}\lim_{a\to-\infty}\int^b_a f(x)dx  \]
This is a 'better' definition than the first because the theory of integration works better when we can impose some boundedness conditions on the functions that we're working with.  In the first example, it blows up way to fast at either end of the real line, and some of the nice theorems on integration that you will see later don't work for such functions.

Thank you for your response. I am stil not sure what to do after I integrate with parts on the first one, how do I find what values, p, bound the integral?

Also, I unfortunately didn't understand what you were saying about the second one. I understand what you did, but I do not see how that makes a difference. This, "the theory of integration works better when we can impose some boundedness conditions on the functions that we're working with. In the first example, it blows up way to fast at either end of the real line", I do not comprehend....

Theorem: $\int_{-\infty}^\infty f(x)\,dx=\int_0^\infty f(x)\,dx+\int_{-\infty}^\infty f(x)\,dx$. If either of these improper integrals diverges, $\int_{-\infty}^\infty f(x)\,dx$ diverges.

This theorem is very easy to prove using the definition blahblahblah gave.

It should be obvious that $\int_0^\infty x\,dx$ diverges.

Another reason it should be this way: try the substitution $t=x-1$. This shouldn't change the integral...


geometry
circumcircle
combinatorics unsolved
combinatorics

I can't solve this one  :( please give me idea  :?
Given a convex n-gon with no 4 vertices lying on a circle, show that the number of circles through three adjacent vertices of the n-gon such that all the other vertices lie inside the circle exceeds by two the number of circles through three vertices, no two of which are adjacent, such that all other vertices lie inside the circle.

Yes, it si hard. So I can give you just hints.
If you denote pints 1..n in order. Note [i,j]={i,i+1..j] or {i,i-1..j}.
The key is to show that if abc is a circumscribed triangle and [a,b] is non-empty, then [a,b] contains exactly one d with abd circumscribed( or something like that). The show that circumscribed and boundary triangles triangulate the polygon.
By counting pairs of segmnets, angles you must reach the conclusion.
I don't remember the sol well, you u must think urself.
I found it in Titu Andreescu's 1998-1999 problems from all-around the world.
If you can. buy it. It's a very good book

Let us call triangle [i]selected[/i] if its circumcircle contains $n$-gon.

[b]Fact 1.[/b] No two selected triangles have common inner point.
[b]Fact 2.[/b] If $ABC$ is a selected triangle and $AB$ is a diagonal of $n$-gon, then thre is another selected triangle $ABD$.
[b]Fact 3.[/b] From Fact 1 and Fact 2 we conclude that all selected triangles form a triangulation of $n$-gon.

Let $X$ be number of selected triangles, which include two adjenct sides of $n$-gon, $Y$ be number of selected triangles, which don't include sides of $n$-gon, $Z$ be number of all remaining selected triangles.

Using Fact 3 and definition of $X,Y,Z$ we immediately obtain following facts:
[b]Fact 4.[/b] $X+2Z=n$.
[b]Fact 5.[/b] $X+2Z+3Y=2(n-3)$.
[b]Fact 6.[/b] $X+Y+Z=n-2$.

So, actually, we need to prove Fact 1 and Fact 2, which are trivial :)


Fred has 10 identical oranges. In how many ways can he distribute these oranges among his 5 friends?

This is a direct application of balls and urns, so the answer is $ \dbinom{10\plus{}5\minus{}1}{5\minus{}1}\equal{}\boxed{1001}$.

Please use the search function (in this case, the AoPSWiki) before reviving threads like this.

I think would be what you want: http://www.artofproblemsolving.com/Wiki/index.php/Balls_and_urns


inequalities
geometry
circumcircle
inequalities theorems

I knew very elegant proof of Euler's Inequality.
It is the following proof.

(i) If $ x$, $ y$, $ z$ are positive, prove that
\[ (x\plus{}y)(y\plus{}z)(z\plus{}x) \ge 8xyz\]
(ii) If $ b\plus{}c>a$, $ c\plus{}a>b$, $ a\plus{}b>c$, use (i) to prove that
\[ (b\plus{}c\minus{}a)(c\plus{}a\minus{}b)(a\plus{}b\minus{}c) \le abc\]
(iii) For a triangle, circumradius is $ R$, radius of incircle is $ r$, use (ii) to prove that
\[ 2r \le R\]

Don't you think so?

[quote="Kouichi Nakagawa"]I knew very elegant proof of Euler's Inequality.
It is the following proof.

(i) If $ x$, $ y$, $ z$ are positive, prove that
\[ (x \plus{} y)(y \plus{} z)(z \plus{} x) \ge 8xyz
\]
(ii) If $ b \plus{} c > a$, $ c \plus{} a > b$, $ a \plus{} b > c$, use (i) to prove that
\[ (b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c) \le abc
\]
(iii) For a triangle, circumradius is $ R$, radius of incircle is $ r$, use (ii) to prove that
\[ 2r \le R
\]
Don't you think so?[/quote]
There are many proof for it,use algebra,geometry,At all it is very well know
Sometime,when i prove a prblem,then i can use it to prove the ineq,


vector
calculus
geometry
calculus computations

Test question, need help breaking it down, from reading some posts, it looks like ther are some crazy smart ppl here.

A nine meter verticle antenna is erected on a planar hill whose shape is the graph of x + 2y + 2z = 11
with the base of the antenna at the point on the hill where x = 1 and y = 4. A wire is to be stretched from
the top of the antenna to a control room on the ground (ie on the hill),
and it is desired to make the wire as short as possible.

Where on the hill should the control room be located? give its x, y, z coordinates.

So you're asking the projection of the antenna-top on the hill? ;)
There's nothing you need to be "smart" for that, it is a sample high school question that any student can solve if he has studied the formulas.

[ I suppose with 9 meter you mean 9 units right? since there's no "meter" elsewhere ;) ]
x=1,y=4 => 2z=11-1-8 => z=1, add 9 metres : z(antennatop) = 10
now the perpendicular vector to the plane is parallel to k*[1,2,2], if we construct a line perpendicular to the hill and through the antennatop, the place where the line intersects the hill is the requested point.

So we solve : [1+k,4+2k,10+2k] on the hill: (1+k)+2(4+2k)+2(10+2k)=11 => k=-2
so the Point is [-1,0,6] if I counted wel :)

PS: this is not college level ;)

[quote]PS: this is not college level[/quote]
It's not "really" calculus, of course. However, the standard U.S. syllabus for a three semester calculus course, using a standard textbook like Stewart, has a section either at the end of the second semester or the beginning of the third semester on vector algebra and geometry, introducing such topics as the dot and cross products of vectors and equations of lines and planes. You might say that this all should have been covered far earlier, in high school mathematics, but the truth in the U.S. is that it isn't. This question clearly comes from that portion of that course.

Thank you Peter for your simple and elegant solution.  Id also like to thank you Kent for your explanation, it is exactly true in Canada as well.

I remember commentig in a class called Multivariate Calculus that after a few weeks we hadn't done any calculus. :D


geometry
rectangle
search
3D geometry
pyramid
ARML
analytic geometry

Let $ABCD$ be a rectangle and let $P$ be a point inside the rectangle where $PA=8, PB=4, PD=7$, then what is the length of $PC$.

^ The "British flag theorem", I think it's called.

Thank you...I never knew how to solve these kind of problems but now I know...British flag theorem..hmm..never heard of it before. But how did you know that $PA^2+PC^2=PB^2+PD^2$? Is it because $PA$ and $PC$ are going in the same "direction" and the same with $PB$ and $PD$?

[quote="Jan"]About the British Flag theorem... Is there a more official name for this theorem? I can't find anything on Google. (I'm searching for a proof)[/quote]

So did I because I was trying to understand it :o , I've never heard of this before so there must be another name :|

don't search proofs, try to do it yourself first. the time you spent searching the proof you probably proved the theorem without searching (in other words, you wasted your time, sorry but that's what i felt). here is a extremely short list of hints that you need to prove this theorem

1) draw lines from $P$ perpendicular to the sides of rectangle
2) Pythagoras theorem

so guys, you should post your proof here in spoiler mode and show off your proof :lol:

I've never heard of the British flag theorem either :P Anyways,  here is how I would do the problem (you can prove the so called British flag theorem using the exact same method that I am using) (unfortunately I can't draw a picture right now so you will just have to try and draw one from my descriptions).

[hide]Start with rectangle $ABCD$ and $P$ inside.  Now draw the altitudes from $P$ in each of $\triangle APB$, $\triangle BPC$, $\triangle CPD$, and $\triangle APD$ and label the intersections of the altitudes with the sides of the rectangle as $E,F,G$ and $H$, respectively.  We label some sides $AE=HP=DG=a$, $EG=PF=CG=b$, $AH=EP=BF=c$, and $DH=GP=CF=d$.  We know from the Pythagorean theorem that

\begin{eqnarray}
a^2+d^2 &=& 7^2\\
b^2+d^2 &=& x^2\\
a^2+c^2 &=& 8^2\\
b^2+c^2 &=& 4^2
\end{eqnarray}

Thus $x^2=b^2+d^2=(1)+(4)-(3)=49+16-64=1\Rightarrow x=1.$  $\textbf{QED}$[/hide]

i think one of the old AHSME exams (1960s or 1970s) did the same thing with this geometry question, exact same method but with different numbers

Palytoxin, thanks for the hint. Actually, whit the hint of the perpendiculars, it's an easy proof to find. Here's what I got:
Look at the picture. Because the lines we drew are all perpendicular to a side of the rectangle, we can applie Pythagoras a few times to get this:
$v^{2}=z^{2}_{1}+a^{2}_{1}$
$w^{2}=z^{2}_{1}+a^{2}_{2}$
$x^{2}=z^{2}_{2}+a^{2}_{2}$
$y^{2}=z^{2}_{2}+a^{2}_{1}$

So when we make the sum of $v^{2}$ and $x{2}$, we can see it's equal to the sum of $w^{2}$and $y^{2}$.
$z^{2}_{1}+a^{2}_{1}+z^{2}_{2}+a^{2}_{2}=z^{2}_{1}+a^{2}_{2}+z^{2}_{2}+a^{2}_{1}$

[quote="joml88"]Yeah, anyone who is trying to follow my proof, the set up should be just like the one in Jan's picture.

I've seen this type of problem on last year's AHSIMC and an old Univ. of South Carolina exam.  I wouldn't doubt it was on an AHSME either.[/quote]

Yeah Joe, I got it from the 2000 USC exam.

yes, it's the british flag theorem and it holds for ANY point P in space. so if you have a pyramid with a rectangular bases, the equation still holds.  this was the key to solving a team problem on arml 2004. i posted the problem some time ago, so you can search for british flag and i'm sure you'll find it.  the most thorough proof of the british flag theorem uses coordinates, and it accounts for 3 dimensional space as well. see if any of you can prove this.

So do we have a official name? I wonder how i can translate it into Chinese...


geometry
3D geometry
complex analysis
complex analysis unsolved

The eqn z^n-1 has n roots which are called nth roots of unity.These n roots are 1,α,α^2,..........α^n-1
Q1 If ω is non real complex cube root of unity then find value of
$ \frac{\prod_{p\equal{}1}^{4}{(\omega \plus{}\alpha ^{p})}}{\prod_{q\equal{}1}^{7}{(\omega ^{2}\minus{}\alpha ^{q})}}$

Q2 If $ n \varepsilon  N$find value of $ \prod_{r\equal{}1}^{n\minus{}1}{sin(\frac{r\pi }{n})}$

Q3 find value of $ \prod_{k\equal{}1}^{n\minus{}1}{\frac{1}{1\minus{}\alpha _{k}}}$