Ramsey Theory
geometry
combinatorics
Extremal combinatorics
right triangle
IMO
IMO 1983

Let $ABC$ be an equilateral triangle and $\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$, and $CA$ (including $A$, $B$, and $C$).  Determine whether, for every partition of $\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle.

The answer is positive; there always a class of the partition will contain the three vertices of a right-angled triangle.

First notice that if there are at least two points of the same color on a side, then all the points in $\mathcal{E}$ which project orthogonally onto points of that color on the respective side must have the opposite color if we are to have no monochromatic right-angled triangles.

Now assume we can find a side  on which there is at most one point bearing one of the colors (blue, say; we take the two colors to be red and blue). Then, by the observation above, it's obvious that the other two sides contain three blue vertices of a right-angled triangle, and we're done (all the points of those two sides are blue, except maybe for the endpoints which they have in common with the other side and the point which projects orthogonally onto the unique blue point on the third side). We can now assume all three sides contain at least two points of each color. Take $U\in BC$ s.t. $BC=3BU$. $U$ projects orthogonally onto $V\in CA$, and $V$ projects orthogonally onto $T\in AB$. We can easily see that $T$ projects orthogonally onto $U$. Again, by the observation in the previous paragraph, the points $U,V,T$ must have different colors if we are to have no monochromatic right-angled triangle, and this is impossible (we only have two colors).

Does this work?
Let $X\cup Y=\mathcal E$ and $X\cap Y=\emptyset$. Note that one of these subsets contains the neighborhood of either $A$, $B$, or $C$. WLOG $X$ contains the neighborhood of $A$. Let $D,E$ be points in $X$ such that $ADE$ is equilateral. Then, if $F$ is the midpoint of $AE$, $\angle AFD$ is right.


search
quadratics
number theory unsolved
number theory

Find all $ m,n\in\mathbb{Z}$ such that

$ 3mn \minus{} 1|m^3 \plus{} n^3$

This is a problem from ISL, year 199*. Search on forum or Kalva.

I not found this problem in ISL. Someone known this problem? I've heard it need elliptic curves to solve.

Take any odd prime divisor p of 3mn-1. (p,3mn)=1. p divides $ 3m(m^3 \plus{} n^3) \equal{} 3m^4 \plus{} n^2 \equal{} 0(modp)$ which implies -3 is a quadratic residue mod p. So we have $ p \equal{} 1(mod3)$ by quadratic reciprocity.
So , unless $ 3mn \minus{} 1$ is even, all prime divisors are 1(mod3) implying $ 3mn \minus{} 1$ itself to be 1(mod3), contradiction.
The case when both m and n are odd remains to be considered. Im unable to move further, I would happily welcome any help


If f(x)=cosx then proof this:
[img][url=http://imageshack.us][img]http://img505.imageshack.us/img505/8954/int3cl6.gif[/img][/url][/img]
for n=1,2,3,....

Kunny is right.
If you induct on n, then the base case will be $n = 1$.
If $n = 1$, then $cos(x+\frac{\pi}{2}) =-cos(\frac{\pi}{2}+x-\pi) =-cos(x-\frac{\pi}{2}) =-cos(\frac{\pi}{2}-x) =-sinx$
and $f'(x) = \frac{d}{dx}cosx =-sinx$. Thus the base case holds.

Suppose now that $f^{n}(x) = cos(x+\frac{n\pi}{2})$.
Then by chain rule we have
$f^{n+1}=-sin(x+\frac{n\pi}{2})$.
$cos(x+\frac{(n+1)\pi}{2}) = cos ((x+\frac{\pi}{2})+\frac{n\pi}{2}) = cos(x+\frac{\pi}{2})cos(\frac{n\pi}{2})-sin(x+\frac{\pi}{2})sin(\frac{n\pi}{2}) =-[sinxcos(\frac{n\pi}{2})+cosxsin(\frac{n\pi}{2}) ]=-sin(x+\frac{n\pi}{2})$.
Thus $f^{n+1}(x) = cos(x+\frac{(n+1)\pi}{2}).$ Q.E.D


topology
search
real analysis
real analysis unsolved

Let A denote the plane minus 4 points on the X axis. Let B denote the plane minus 4 points of arbitary configuration. Are A and B homemorphic? I'm also interested in the following question, what are the possible images of a line under a homeomorphism? sorry if it has been posted before,i did search though.

The answer is yes; given an $n$-tuple of distinct points $A_{1},\dots, A_{n}$ and another $n$-tuple of distinct points $B_{1},\dots, B_{n}$, one can find a homeomorphism $F\colon \mathbb R^{2}\to \mathbb R^{2}$ such that $F(A_{k})=B_{k}$. First you find bounded connected open sets $U_{1},\dots U_{n}$ with disjoint closures such that $A_{k},B_{k}\in U_{k}$ for each $k$. Then mess with each $U_{k}$ separately.

For the second question: do you mean the image of a line under a homeomorphism of the line, or under a homeomorphism of the plane?

i dont understand. can you elaborate just a little more please?

for the second question i mean image under a homemorphism of the plane

Here is another approach. An ordered $m$-tuple of points in $\mathbb R^{n}$ can be identified with a point in $\mathbb R^{mn}$. Let $X$ be the set of all $m$-tuples $\mathcal{A}=(A_{1},\dots,A_{m})\in\mathbb R^{mn}$ such that all $A_{k}$s are distinct. Claim #1: $X$ is a connected open subset of $\mathbb R^{mn}$.

Let's say that two $m$-tuples $\mathcal{A},\mathcal{B}\in X$ are equivalent if there is a homeomorphism $f\colon\mathbb R^{n}\to\mathbb R^{n}$ such that $f(A_{k})=B_{k}$ for $k=1,\dots,m$. Claim #2: this is indeed an equivalence relation.

Claim #3: each equivalence class is an open subset of $X$. Once this is established, the result follows at once: there is only one equivalence class because $X$ is connected. To prove Claim #3, consider $\mathcal{A}\in X$ and let $d=\min_{i\ne j}\parallel A_{i}-A_{j}\parallel$. Suppose that $\mathcal{B}\in X$ is such that the distance between $\mathcal{A}$ and $\mathcal{B}$ in $\mathbb R^{mn}$ is less than $d/2$. For each $k=1,\dots,m$ the closed ball $\mathbb B_{k}=\mathbb B(A_{k},d/2)$ contains the point $B_{k}$ in its interior. Define $f$ within the ball $\mathbb B_{k}$ as follows: $f(A_{k})=B_{k}$; $f(x)=x$ for $x\in\partial \mathbb B_{k}$, and $f$ is affine on each radius of $\mathbb B_{k}$. This is a homeomorphism of $\mathbb B_{k}$ onto itself. Finally, let $f$ be the identity outside of the (disjoint) union of $\mathbb B_{k}$s.

Your second question (about images of a line) is answered by the [url=http://en.wikipedia.org/wiki/Jordan-Sch%C3%B6nflies_theorem]Jordan–Schönflies theorem[/url]

Now try the converse:

If $S,T$ are finite subsets of $\mathbb R^{n}$ such that $\mathbb R^{n}\setminus S$ is homeomorphic to $\mathbb R^{n}\setminus T$ (not necessarily through a homeomorphism extending to the whole of $\mathbb R^{n}$), then $S$ and $T$ have the same number of elements.

thanks. i'll try this new problem given by grobber and get back

[quote="mlok"]
Your second question (about images of a line) is answered by the [url=http://en.wikipedia.org/wiki/Jordan-Sch%C3%B6nflies_theorem]Jordan–Schönflies theorem[/url][/quote]

Doesn't it concern only the case of the image of a closed curve $S^{1}$ ? In case of any line, not necessarily closed, i think we have more "images" of our lines.

The difference disappears if we use the one-point compactification of $\mathbb R^{n}$ (i.e. $S^{n}$). A line is just a circle that passes through $\infty$.


If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before.

If $ x$ is an integer, prove that $ x^3\minus{}x$ is divisible by 3.

[hide="See if you know the answer."]$ x^3 \minus{} x \equal{} x(x^2 \minus{} 1)$
$ \Rightarrow x(x \minus{} 1)(x \plus{} 1)$

As it is three consecutive number so it is divisible by 3! and hence divisible by 3.[/hide]

Please hide your answers using the [code][hide][/code] command.

Alternatively, you can
[hide]brute force it by observing it in mod 3. Brute force is a legit choice here, since there are only three values to try.[/hide]

In retrospect, I should have changed the title of this thread. It's too big of a hint.

[quote="Nerd_of_the_Ages"]brute force it by observing it in mod 3.[/quote]

This is essentially what you do when you claim that three consecutive numbers must contain a multiple of $ 3$; I would hardly consider it brute forcing.  :wink:

Another problem very similar to this that is solved along the same lines is:

(Zeitz) Show that $ n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)$ is never a perfect square for all natural $ n$.

[quote="fishythefish"]If you've seen the problem before, don't answer it. It's more fun if you haven't seen it before.

If $ x$ is an integer, prove that $ x^3 \minus{} x$ is divisible by 3.[/quote]

By fermat, we have $ x^3\equiv x\pmod{3}\Longleftrightarrow 3|(x^3\minus{}x)$

[quote="jvenezuela716"]
Another problem very similar to this that is solved along the same lines is:

(Zeitz) Show that $ n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)$ is never a perfect square for all natural $ n$.[/quote]

Could someone help me out on this one?  :oops: The only way I know is to expand $ n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)\plus{}1$ and prove that it is always a square by factoring... but that isn't quite "solved along the same lines".

[quote]That is the solution in Zeitz; it is similar in that both problems require you to factor the expression into a certain form that allows you to make an argument based on the factored form.[/quote]
well, i coudnt do that way. but here is my solution
[hide]let digits be a,b,c,d  and they are consecutive integers, so either a&d or b or c are multiples of 3
similarly only a&c or b&d are multiples of 2 and one of them in each case is a multiple of 4
after analyzing combinations it will be observed one digit will always remain which is not multiple of 2,3,4
 so that digit is a multiple of 5 or >5. but another multiple of same no. in the four consecutive digits is not possible.  so, it is not a square as in squares multiples are in pairs.
[/hide]

@analytic: The four consecutive integers don't have to be digits.

Another related problem involving the product of consecutive integers: prove that the product of $ n$ consecutive positive integers is always divisible by $ n!$. Warning: I do not know of a solution similar to the ones above.

sorry i meant numbers by digits, but anyway i think my solution is incorrect or incomplete

[quote]prove that the product of  consecutive positive integers is always divisible by . Warning: I do not know of a solution similar to the ones above.[/quote]
 here is my solution
[hide]well consider the product of first n numbers it is divisible by n! or equal to it. but here we are considering the product of any n integers , so we may have to shiftour set of numbers starting from 1 to any other no.
  however, the multiples which were originally in span of first n no. will also be multiples of this product because we know that any no. n , its next multiples will be in span of next n numbers or ny multiple willbe in the span of n numbers.

hence proved[/hide]

[quote="analytic"]
let digits be a,b,c,d  and they are consecutive integers, so either a&d or b or c are multiples of 3
similarly only a&c or b&d are multiples of 2 and one of them in each case is a multiple of 4
after analyzing combinations it will be observed one digit will always remain which is not multiple of 2,3,4
 so that digit is a multiple of 5 or >5. but another multiple of same no. in the four consecutive digits is not possible.  so, it is not a square as in squares multiples are in pairs.
[/quote]

Your last argument is not valid; consider $ 121\cdot 122 \cdot 123 \cdot 124$. Only $ 121$ is not divisible by one or more of $ \{2,3,4\}$, but it is itself a perfect square of the number greater than five.

@jvenezuela716: analytic already admitted that the solution given was "incorrect or incomplete"

I know, I just wanted to clarify the error; I mean no malice intent.

leoxnlin , if you have any other soluton than mine of that n factorial question , do post
i would like to see other approach to the question


A pyramid with a square base is cut by a plane that is parallel to its base and is 2 units from the base. The surface area of the smaller pyramid that is cut from the top is half the surface area of the original pyramid. What is the altitude of the original pyramid?

[hide]oh okay well if we say that the altitude of the larger one is $ h$, we know that the ratio of similitude of the smaller one to the larger one is $ h\sqrt{\frac{1}{2}}$ = $ \frac{\sqrt{2}h}{2}$ now just solve for $ h$ in,

$ h \minus{} h\frac{1}{\sqrt{2}} \equal{} 2$

$ h \equal{} 4 \plus{} 2\sqrt{2}$

me thinks. i might have made a calc. error as i have no scratch paper with me[/hide].

so to generalize if two pyramids are similar, then are the ratios of their sides equal to the square root of the ratio of their surface areas?
if its is cognos you are right

yes, if the ratio of the surface areas is $ n$, then the ratio of any side length of one to any side length of the other is $ \sqrt{n}$. if someone wants to state that more clearly, then do it because my explanation is badly worded.

That is correct. For similar shapes or constructions or etceteras, if the ratio of linear lengths is 1:n, the ratio of areas (of faces or whatever) is 1:n^2 and the ratio of volumes is 1:n^3.


symmetry
number theory
prime factorization

Nice question!

[hide="Hint"]Pay special attention to the words, "in lowest terms."[/hide]

[quote="xirtamax"]How many rational numbers between 0 and 1, when written in lowest terms have
a numberator and denominator whose product is exactly 25![/quote]

I remember almost the same problem appeared in Regional Maths Olympiad (RMO)(India) in 94.

Expressing 25! as a product of powers of distinct primes,

we have 25! = 2^22 * 3^10 * 5^6 * 7^3 * 11^2 * 13*17*19*23

So there are 9 distinct prime factors of 25! viz. 2,3,5,7,11,13,17,19,23.

Now given the conditions of the problem,
we have to construct all fractions of the form   m/n where m<n and m is coprime to n and mn=25!

Now looking at the prime factorization of 25! ,we observe that the no. of fractions will be the sum of all distinct combinations of 9 distinct objects

  (1/2 ) * summation of C(9,r) where r ranges from 0 to 9

= (1/2) * 2^9 = 2^8

The factor 1/2 is because of the fact C(9,r) = C(9,9-r)

We need to be careful about the wording here; the answer as stated is infinitely many.

[quote="JBL"]How do you figure?  There are only finitely many pairs of factors of 25! -- how could that possibly get you infinitely many rationals?[/quote]

Anyway so what's the correct ans?? I'm getting it as 2^8 as explained in one my posts

[quote="santosh_blr"][quote="blahblahblah"]When I post at 1:15 in the morning, I guess I think that, say, 2/4 and 3/6 are different numbers  :blush:[/quote]

True that 2/4 and 3/6 are different fractions but then when [b]reduced to lowest terms[/b] they are both equal to 1/2.

And the conditions of the problem state that the fractions should be in their lowest terms.[/quote]

Yes, I know that, as is pretty clear from the writing in the post.

[quote="mathfanatic"][hide]25! includes prime factors from 2 to 23; there are 9 such primes.  For each prime, we have two choices: it can be placed in the numerator or the denominator.  (All of the powers of any prime must go together, since otherwise the fraction will not be in lowest terms.)  Thus there are 2^9 choices.  But half of these numbers have the numerator greater than the denominator, in which case the number will be > 1, so there are only 2^9/2 = 2^8 = 256 numbers.[/hide][/quote]

It's worth noting that it took me about 10 minutes to figure out that half of the numbers would indeed have the numerator greater than the denominator, oddly enough.

Exploiting symmetry is a lot better than comparing primes  :D

i find it rather odd how that both this problem and the other problem he posted "counting problem" both appear on the intermediate counting midterm challenge,

alos, that i believe the challenge came out on monday after class, but im not too sure about this last bit. Can Dpatrick clarify this?

[quote]figure out that half of the numbers would indeed have the numerator greater than the denominator, [b]oddly [/b]enough.   ....Exploiting symmetry ....[/quote]

You mean "[b]evenly[/b]". right :)?


Let $%Error. "eqalign" is a bad command.
{\& \cos a = \frac{{\sin B}}{{\sin A}},\cos b = \frac{{\sin C}}{{\sin A}},\cos (a+b) = \sin B.\sin C$
Prove that $\tan^{2}A = \tan^{2}B+\tan^{2}C$


This reminds me a bit of the 2, 2, and 5 thread...

[quote="skierkid_13"]Jim is as old as his brother was when Jim was as old as his brother was when Jim was as old as his brother is now.  Jim and his brother are both less than 100 years old.  How old are they?[/quote]

I'm pretty sure that's contradictory.

If right now, Jim is as old as his brother was, that means his brother is older. However, you then state that at some point, Jim was as old as his brother is now. That would mean Jim is older.

...

Yea......this question is confusing. You know, if you read this for a long time, you will become dizzy.

NEVER!!!I will never give up on a problem!

*I give up* :ewpu:

"Jim is as old as his brother was" doesn't that imply Jim is older than his brother?
Then it says, "Jim was as old as his brother is now" which implies his brother is older than him!?

Which I already brought up, but yeah, this is a faulty problem.

Sorry i omited something in my typing.  The problem should say: "Jim is [u][b]17[/b][/u] times as old as his brother was when Jim was as old as his brother was when Jim was as old as his brother is now. . ."

Ah. Now I can plug in 1 through 5 and see which works.

[hide="my solution"]Let Jim's current age be $J$ and his brother's current age be $B$. I found it worked best to work backwards so I started with "When Jims was as old as his brother is now." This was $J-B$ years ago. So then his brother had been $B-(J-B)=2B-J$. So Jim was that old $J-(2B-J)=2J-2B$ years ago. So his brother's age at that point was $B-(2J-2B)=3B-2J$. Jim is currently 17 times as old as that so $J=17(3B-2J)=51B-34J$. Now, moving the Js all onto one side you get $35J=51B$ so $J=51$ and $B=35$.[/hide]


probability
floor function
function
algebra

If A is the set of all natural numbers, and N is the subset of A up to a certain arbitrary integer n, what is the probability that a random number m within N is a perfect square?

The solution's below:
 [hide]                [|n|]
               _____
                  n[/hide]

Is M always an integer?

EDIT: BTW how did you do this problem?  :huh:

I got
[hide]
$\frac{\lfloor\sqrt{n}\rfloor}{n}$

Let's say the greatest square less than  or equal to n is $a^{2}$.  Then since $n>a^{2}$, $\sqrt{n}>a$.  Because $\sqrt{n}$ has to be less than $a+1$, the greatest integer less than $\sqrt{n}$ is $a$.  $a$ is also the number of squares less than n, so the answer is $\frac{\lfloor\sqrt{n}\rfloor}{n}$.
[/hide]
Which seems a little different from your answer.

EDIT again: OK, there's my solution. And I've figured out how to do the Floor function in Latex.

Can somebody post a solution instead of only answers?

I realised that the number of perfect squares within m would have to be less than m, unless m was a perfect square, so that's how I thought of using the greatest integer down function.

Wait, never mind, I get it. It's actually pretty easy.

next time, if you post an answer call it answer, not solution.  solution typically refers to how you get an answer.


limit
real analysis
real analysis theorems

Let $p_{n1}, p_{n2}, \cdots, p_{nn}$ be nonegative real numbers whose sum equal to $1$. if
\[a_{n}= p_{n1}b_{1}+p_{n2}b_{2}+\cdots+p_{nn}b_{n}\]
converges to $a$ as $n \rightarrow \infty$ whenever $b_{n}\rightarrow a$  as $n \rightarrow \infty$. Prove that $\lim_{n \rightarrow \infty}p_{nk}=0$ for $k=1,2, \cdots$.

Let $p_{n1}=1$ and $p_{nk}=0$ for $k>1$. Then $a_{n}=b_{1}$ for all $n$, hence $(a_{n})$ is a convergent sequence regardless of $(b_{n})$. Yet, $\lim_{n\to\infty}p_{n1}\ne0$.

Maybe you wanted to add that $a=b$?

Oh sorry...i've edited my post..

i meant $b_{n}$ converge to the same limit as $a_{n}$

Hint: look at what happens when only one term of the sequence $(b_{n})$ is not equal to zero.

yes i did that. i if $\lim p_{ni}=r$ not equal to zero then i consider $\lim \frac{b_{n}}{p_{ni}}$ but i don't know how it goes??

thanks

i even don't know how to prove that $\lim p_{ni}$ exist.  :blush:
of course they are bounded.

please help...

Fix an integer $k$. Let $b_{k}=1$ and let $b_{n}=0$ whenever $n\ne k$. The sequence $(b_{n})$ has limit $0$. Therefore, by our assumption $\lim_{n\to\infty}a_{n}=0$. On the other hand, $a_{n}$ is equal to ...

But that just an example??   :maybe:

pardon my stupidness..

[quote="bolpoin"]... converges to $a$ as $n \rightarrow \infty$ whenever $b_{n}\rightarrow a$...[/quote]
means that [b]for any[/b] convergent sequence $b_{n}$ the sequence $a_{n}$ has the same limit. You can use this information by choosing $(b_{n})$ to be any convergent sequence you like.

hmm... to be honest i thinks its strange..
but indeed i can't resist the power of logic.

Thanks.

does anyone have another proof?

The theorem you mentioned above is a particular case of Toeplitz theorem.  Another proof maybe by using Toeplitz theorem.

 Just an idea: How about taking $p_{nj}: =\max{i}p_{ni}$ and then try to prove that $\lim_{n \rightarrow \infty}p_{nj}=0$. So the conclusion follow from pinching theorem... but i don't know to continue that at this moment.


calculus
integration
inequalities
function
real analysis
real analysis unsolved

Integrating by parts we get that

$ \int\limits_{0}^{1}xf'(x)dx\equal{}f(1)\minus{}1\equal{}\alpha$,

and

$ \int\limits_{0}^{1}x^{2}f'(x)dx\equal{}f(1)\minus{}2\equal{}\beta$.

Thus

$ (\beta\minus{}\alpha)^{2}\equal{}1\equal{}\left(\int\limits_{0}^{1}(x\minus{}x^{2})f'(x)dx\right)^{2}\leq \int\limits_{0}^{1}(x\minus{}x^{2})^{2}dx\cdot \int\limits_{0}^{1}f'(x)^{2}dx$.

It follows that

$ \int\limits_{0}^{1}f'(x)^{2}dx\geq 30$.

[quote="bogdanl_yex"]Nice solution...But in which situation we have inequality? :D[/quote]

$ f'(x) \equiv x\minus{}x^2$.

Bogdan,

Can you tell me please where can I find the Romanian National Olympiad problems that were on the shortlist?

Thanks

I have a brochure from SSMR were are the problems...

$ \int_{0}^{1}f(x)dx \equal{} \int_{0}^{1}xf(x)dx \equal{} 1$
________________________________________________________________

I was just playing with these relations and I have found something easy to prove:
Prove that it exists $ c\in(a,b)$ such that $ f(c)\equal{}\pi/3$

[quote="Zamfirmihai"]$ \int_{0}^{1}f(x)dx \equal{} \int_{0}^{1}xf(x)dx \equal{} 1$
________________________________________________________________

I was just playing with these relations and I have found something easy to prove:
Prove that it exists $ c\in(0,1)$ such that $ f(c) \equal{} \pi/3$[/quote]

it is $ (0,1)$, not $ (a,b)$. sorry :oops:


Show the product of two real roots of x^4+x^3-1=0 is a root of x^6+x^4+x^3-x^2-1=0

Let $ y_a\equal{}x_ix_j,i<j,a\equal{}1,2,...,6,i\equal{}1,2,3,4$, were $ x_i$ are roots of $ x^4\plus{}x^3\minus{}1\equal{}0$.
Therefore $ \sigma_1\equal{}\sum_i x_i\equal{}\equal{}\minus{}1,\sigma_2\equal{}\sum_{i<j}x_ix_j\equal{}0\equal{}\Sigma_1\equal{}\sum_a y_a.$
$ \Sigma_2\equal{}\sum_{a<b}y_ay_b\equal{}\sigma_1\sigma_3\minus{}\sigma_4\equal{}1$,
$ \Sigma_3\equal{}\sum_{a<b<c}y_ay_by_c\equal{}s_2\sigma_4\plus{}\sigma_3^2\equal{}\minus{}1$,
$ \Sigma_6\equal{}\sigma_4^3\equal{}\minus{}1$,
$ \Sigma_4\equal{}\Sigma_6 \Sigma_2'(\frac{1}{y_a}\equal{}\sigma_4^3(\sigma_1'\sigma_3'\minus{}\sigma_4')\equal{}\sigma_4(\sigma_1\sigma_3\minus{}\sigma_4)\equal{}\minus{}1$,
$ \Sigma_5\equal{}\Sigma_6\Sigma_1'\equal{}\sigma_4^2\sigma_2\equal{}0$.
Therefore $ y_a$ are roots of $ y^6\plus{}y^4\plus{}y^3\minus{}y^2\minus{}1\equal{}0$


Could anyone help me complete all chemical reactions :(
aN2H4 + HgCl2 ->
b) N2H4 + HNO2 ->
c) N2H4 + K2Cr2O7 + H2SO4 ->
d) N2H4.H2SO4 + CuSO4 + NaCl ->
e) N2H4.HCl + SNCl2 + HCl -> NH4Cl +.........

Is this homework?  :wink:  :D
[hide="Hint"]
First try to identify which are oxidation and reduction reactions (assuming you know them). Then try to write the half reactions and balance them :)
[/hide]

yes it 's my homework:( Tomorrow I have to send it to my teacher :(
Could you post a full clearly solutions  sorry I am very stupid :(

No you arent! Do you know about oxidation and reduction reactions? And do try and make a start! Ive seen your problems and solutions in the inequalities forum! They are great! :coolspeak: Then why this lack of confidence?


In a state system of the airlines is such that each city is connected by air with no more than 3 and from any city can go to any other city with no more than one transfer.What is the greatest possible number of the cities in this country?

[quote="enndb0x"]$ 4$

[b]ee"[/b][/quote]

What..?

And also, by one transfer, do you mean one route only or that you can make one stop? Like, going from City A to City C, which aren't connected, can you stop in City B, which is connected to both A and C?

I understood the problem  that planes can not stop ,and they can not go to the cities which are not connected , so the bigest number should be $ 4$ :wink:

We can go to cities that is not conected only with 1 stop.NO more.


[b]If $ \theta$ is acute and $ \sin{\frac{\theta}{2}}=\sqrt{\dfrac{x-1}{2x}}$, then find $ \tan\theta$ in terms of $ x$[/b]

[hide="outline for books solution"]using the identity, $ \sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$, they solved for $ \cos \theta$ and then plugged that into $ \tan^2x+1=\sec^2x$[/hide]

[hide="outline for my plan of action"]find $ \cos \theta$ like they did and then plug that back into the original identity to find $ \sin\frac{\theta}{2}$. then use the identity $ \tan\theta=\frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}$. would that work? i got some crazy answer when i did it.

[/hide]

Well, you'd have to find $ \cos \frac{\theta}{2}$, and since that involves taking a square root you need to carefully consider what quadrant you're working in.

or you could find $ cos\theta$, which is [hide]$ \frac{1}{x}$[/hide]  and then draw a triangle with the side opposite to $ \theta$ as $ x$ and the adjacent $ \sqrt{1\minus{}x^2}$.  therefore $ tan\theta$ is [hide]$ \frac{x}{\sqrt{1\minus{}x^2}}$[/hide]

[u]Huge hint[/u]
[hide]
Put x = sec$ \alpha$[/hide]


number theory
greatest common divisor
geometry
superior algebra
superior algebra unsolved

Hi, here I have  a philosophical question more than problem itself  :huh:... consider the following problem:

[b]P[/b]: Let R be a commutative ring with unity. If $ a_1f_1 \plus{} \ldots \plus{} a_nf_n \equal{} 1$ in $ R$ then for any  $ N > 0$ then  $ b_1f_1^N \plus{} \ldots \plus{} b_nf_n^N \equal{} 1$.

this result if very useful in algebraic geometry, and very easy to probe in that context since $ a_1f_1 \plus{} \ldots \plus{} a_nf_n \equal{} 1$ means that for $ X \equal{} Spec(R)$,
\[ X \equal{} D(f_1)\cup\cdots \cup D(f_n)
\]
($ Spec(R)$ the set of prime ideals of $ R$ and $ D(f)$ the primes in $ R$ which doesn't contains $ f$)
and since $ D(f) \equal{} D(f^N)$  then we are done.

Now the last step involves zorn's lemma:
\[ X \equal{} D(f_1^N)\cup\cdots \cup D(f_n^N)
\]
if $ I \equal{} f_1^NR \plus{} \ldots \plus{} f_n^NR$ is a proper ideal then exists a prime $ \mathfrak{p}\in X$ such that $ I\subseteq \mathfrak{p}$ which is a contradiction.

[b]Now my question:[/b]
The problem [b]P[/b] can be reduced to the case $ N \equal{} 2$, and it looks quite elementary... so it's really needed zorn's lemma? I proved by hand for $ n \equal{} 2,3$, but for  the general case I got lost. it would be weird if there is an $ n$ such that the problem can be solved by hand for $ n$ but its needed zorn's lemma for $ n \plus{} 1$.. what do you think about it?

Zorn's lemma and the axiom of choice are never needed for finite constructions; they're only relevant when we have infinitely many objects.

But the construction he used is not using finitely many objects.


geometry
circumcircle
geometric transformation
homothety
geometry unsolved

Let O be the center of circumcircle c of acuted-triangle ABC. Another circle c* with center O* tangents internally circle c at point A and side BC at point D. Also circle c* intersect  the lines AB and AC respectively at points E and F, and lines OO* and EO* respectively at points K and A,  G and E. Lines BO and KG intersect at H. Prove DF^2=AF x GH.

im very interested to see the solution with homothety maybe that will help me to learn how to solve problems using it.

See here:
http://www.mathlinks.ro/viewtopic.php?p=1705283#1705283

Best regards,
sunken rock

Let $O$ be the center of circumcircle $c$ of acuted-triangle $ABC$. Another circle $c'$ with center $O'$ is tangent internally to circle $c$ at point $A$ and side$ BC$ at point $D$. Also circle $c'$ intersect  the lines $AB$ and $AC$ respectively at points $E$ and $F$, and lines $OO'$ and $EO'$ respectively at points $K$ and $A$,  $G$ and $E$. Lines $BO$ and $KG$ intersect at$ H$. Prove $DF^2=AF \cdot GH$.


You chould look at some of the other pages on that website too, particularily "The Usefullness of Mathematicians"

Which of course reminds me of:
An engineer, a physicist and a mathematician were traveling by train in Scotland. Glancing from a train window, they observed a black sheep in the middle of a field. "How interesting," observed the engineer, "all scottish sheep are black!" To which the physicist responded, "No, no! Some Scottish sheep are black!" The mathematician, looking desperate, replied: "But what are you saying?. We can just state that in Scotland there exists at least one field, containing at least one sheep, at least one side of which is black."

(copy pasted from http://ochoa.mat.ucm.es/~arrondo/jokes.html)

[quote="mirebeka"]Actually, that last one reminded me of this one:

A Mathematician, a Biologist and a Physicist are sitting in a street
cafe watching people going in and coming out of the house on the
other side of the street.

First they see two people going into the house. Time passes.
After a while they notice three persons coming out of the house.
The Physicist: "The measurement wasn't accurate.".
The Biologists conclusion: "They have reproduced".
The Mathematician: "If now exactly 1 person enters the house
then it will be empty again."[/quote]

I got a good chuckle out of that one.

This one is brilliant:

Method of natural functions:

The lion, having spent his life under the Sahara sun, will surely have a tan.

Induce him to lie on his back; he can then, by virtue of his reciprocal tan, be cot.

That was a good one!
But in mirebeka's joke, there is a logical error a seasoned mathematician is not expected to make:
He assumes the house empty to begin with.

Actually, I think that they all know or assume that the house was empty to start out with, because all three would be making the same mistake.

/bump 14 year bump last post was before I was born :o

[quote=RedFireTruck]how did you find this[/quote]

Search function


Do you think there are different styles of learning math?  Or do all smart math students pretty much learn math in the same way.

What are the different styles?  Strengths and weaknesses?

I know of this particular style of maths in India that's called Vedic Maths.....it's basically quite a different way of approaching anything from the basic stuff like addition or subtraction to stuff like integration.{I mean, in vedic maths, you dont learn that integration is the opposite of differenciation.....integration is done from first principles...(the method is quite similar to the method that some russian books on integration from first principles refer to.}
But then I guess most of us learn maths in the same way......Here in India, most people consider Vedic maths as an addition to normal math so that you do some stuff a bit faster than by doing it the conventional way............learning math in a completely different way is not really something I have seen....

I think that each math student have his/her method... someone want simply to "destroy" a problem, others just take math as a religion, other try to see the problem as a problem in the real life, etc...

[quote]I think that each math student have his/her method... someone want simply to "destroy" a problem, others just take math as a religion[/quote]

But then, in the end, don't almost all of us approach the problem in the same way ? At most there will be about two to three ways in which each one of us look at things....

I mean, if you think for a while, you'll realise that each one of us might quite have a different attitude towards the problem , but then isn't the [b]Mathematical approach[/b] to the problem rather similar across the globe ?

I think there are many styles of learning math and many more of applying it to problems.  I took AP Statistics last year an all of our tests consisted of old AP problems.  Throughout the whole year I did 2-3 of them the way that the solutions explained them.  On many occasions I had to prove my solutions to my teacher because they were so different, but still correct.  It might just be me, but from my experience there are definately many styles of learning and applying math.


AMC
USA(J)MO
USAMO
complex analysis
complex analysis unsolved

Let $C_{1}, C_{2}, C_{3}$ be disjoint circles in the plane.  Prove that there is a fourth circle (or possibly straight line) which is perpendicular to the given three.

(I realize that this could double for a USAMO-type problem, but I am thinking about this problem from a conformal mapping perspective.  With that said though, I welcome solutions from any perspective).

It seems that the following argument gives a proof:
take the complement of two circles, say $ C_{1}$ and $ C_{2}$, and map it onto annulus by the appropriate Moebius map (always possible). Then $ C_{3}$ becomes a circle inside the annulus (not encompassing the origin). It seems that the radius of the outer circle passing through the center of the image of $ C_{3}$ provides the required line.  Take the inverse Moebius map to get back to the original plane and the image of the radius will be the circle or straight line having the required property (since angles are preserved under confomal maps and circles are preserved under Moebius maps).


function
real analysis
real analysis solved

Prove or disprove the following statements:
(a) There exists a monotone function $f : [0, 1] \rightarrow [0, 1]$ such that for each $y \in [0, 1]$ the equation $f(x) = y$ has uncountably many solutions $x$.
(b) There exists a continuously differentiable function $f : [0, 1] \rightarrow [0, 1]$ such that for each $y \in [0, 1]$ the equation $f(x) = y$ has uncountably many solutions $x$.

(a) If $f^{-1}(y)$ has at least two points, then it's a non-degenerate compact interval. This means that it contains a non-empty open interval, and these intervals are disjoint for different $y$'s. This means that given a monotonic function $f$ we can find at most countably many $y$ s.t. $|f^{-1}(y)|\ge 2$, so this disproves the assertion.

(b) Again, we can disprove the assertion. Moreover, we can show that the set consisting of those $y$ for which $f^{-1}(y)$ is uncountable has null measure.

By Sard's Lemma, the set $S=\{f(x)\ |\ f'(x)=0\}$ has null measure. Now take any $y\in f([0,1])\setminus S$ (assume $f$ is not constant, to exclude the trivial case when $f([0,1])$ has only one element and there are no such $y$), and assume that $f^{-1}(y)$ is uncountable. Since $f'$ is continuous, for each $x$ s.t. $f(x)=y$ we can find $\varepsilon=\varepsilon(x)>0$ s.t. $|f'|>0$ on $I_x=(x-\varepsilon,x+\varepsilon)$. This means that $f$ is injective on $I_x$. Now choose such an interval $I_x$ for each $x\in f^{-1}(y)$, and for each $I_x=(x-\varepsilon,x+\varepsilon)$ let $J_x=\left(x-\frac\varepsilon 2,x+\frac\varepsilon 2\right)$. The intervals $J_x$ are disjoint for different $x\in f^{-1}(y)$, but we also know that there are uncountably many of them (one for each $x\in f^{-1}(y)$). This is provides the desired contradiction.


geometry
geometric transformation
reflection

prove that:

1) if P and Q are two conjugate points to a circle ((with the center O)(O,P,Q are colinear)),and M be a variable point

on the circle,the triangles OMP and OMQ are simillar to each other.

2)if R and S are another two conjugate points(O,P,Q,R,S are colinear) to the circle:

angle PMR=angle QMS   or    angle PMR+angle QMS=180

still nobody :!: . guys its not too hard. :o

[color=blue][b](1)[/b][/color] P, Q are conjugate WRT circle (O) means P, Q are symmetric with respect to the circle (O) or P is the reflection (inversion) of Q in the circle (O). By the basic property of inversion, $OP \cdot OQ = r^2$, where r is the radius of (O). Hence, $\dfrac{OP}{r} = \dfrac{r}{OQ}$. Since $M \in (O)$, $OM = r$ and $\dfrac{OP}{OM} = \dfrac{OM}{OQ}$. In addition, the triangles $\triangle OPM, \triangle OMQ$ have the angle $\angle O$ common. Therefore, the triangles $\triangle OPM \sim \triangle OMQ$ are similar by SAS.

[color=blue][b](2)[/b][/color] The triangle pairs $\triangle OPM \sim \triangle OMQ$,  $\triangle ORM \sim \triangle OMS$ are similar (see (1)). Therefore,

$\angle OMP = \angle OQM,\ \ \angle OMR = \angle OSM$

For example, assume that the points P, R lie inside of the circle (O) and the points O, P, R follow on the line OP in this order. Then the points Q, S lie outside of the circle (O) and the points O, S, Q follow on the line OP in this order. Hence,

$\angle PMR = \angle OMR - \angle OMP = \angle OSM - \angle OQM = \angle QMS$

The last equality holds because the sum of 2 internal angles $\angle Q + \angle M$ of the triangle $\triangle QMS$ is equal to the external angle at the remaining vertex S. If the points P, R' lie inside of the circle (O) but on the on the opposite sides of the center O, then $\angle PMR' = 180^\circ - \angle QMS'$ is deduced similarly:

$\angle PMR' = \angle OMR' + \angle OMP = \angle OS'M + \angle OQM = 180^\circ - \angle QMS'$

The last equality holds because the sum of 3 internal angles $\angle Q + \angle M + \angle S'$ of the triangle $\triangle QMS'$ is equal to $180^\circ$.


Hi...I've been trying to solve this problem for 30minutes but without a result.

f and g can be derived.
You can create a new function [tex]h(x) = (f(x))^2 + (g(x))^2[/tex]
You know that:

[tex]f(0) = 2[/tex] and [tex]g(0) = 1[/tex]
[tex]f'(x) = g(x)[/tex] and [tex]g'(x) = -f(x)[/tex]

Decide what h'(x) is and use the result to show that h(x) = 5 for every x.

h(x) = 5 for every x = h(x) => h(x) a constant (it's easy to se that)....That means h'(x) = 0......But I could see this without having to do anything with g(x) and f(x)....So the question is...Is there any other way to solve it?

um you kinda need to use the given condition f'(x) = g(x), g'(x) = -f(x)

just differentiate both sides with respect to x, pluggin to get h'(x) = 0

then pluggin 0 into the original equation to get h(0) = 5.


Am vazut ca mai exista tipuri de exercitiu "Unde este gresala?"

Fie sirul $ (a_{n})_{n\geq 1}$, deifnit prin $ a_{1}\equal{}1$ si $ a_{n\plus{}1}\equal{}\sqrt{a_{1}\plus{}a_{2}\plus{}...a_{n}}$.
Asta e toata ipoteza:D, iar acum o prelucrare:
Se observa ca $ a_{n\plus{}1}^{2}\minus{}a_{n}^{2}\equal{}a_{n}$. Dand lui $ n$ valorile $ 1,2,...,n,n\plus{}1$ in aceasta relatie de recurenta si adunand vom obtine $ a_{n\plus{}1}^{2}\minus{}a_{1}^{2}\equal{}a_{1}\plus{}a_{2}\plus{}...a_{n}$ care este in contradicite cu ipoteza.

Unde este gresala?:)

sirul (an)n>=1 este definit prin an+1=(a1+a2+...+an)^(1/2) oricare ar fi n>=2.
Asa ca si relatia (an+1)^2-(an)^2=an este adevarata doar pentru n>=2.
Pentru n=1 am avea (a2)^2-(a1)^2=1-1=0 care nu este =1=a1.

Av vrut sa va pun un document World minuscul ca attachment zipuit minusccul(6951B) si n-am reusit, fiindca depasisem totalul de attachmente de 10MB!
vi-l trimit ca un attachment intr-un mesaj personal.


x and y are nonnegative integers. Prove that the equation $14x^{2}+15y^{2}=7^{1990}$ has no solutions.
This is from AoPs vol.2

[hide]
assume on the contrary that there is a solution, and let it be $(x_{1}, y_{1})$.  Then since $14x_{1}^{2}$ and $7^{1990}$ are both divisible by 7, $15y_{1}^{2}$ must also be divisible by 7.  So $y_{1}= 7y_{2}$.  Substituting and dividing both sides by 7 we obtain $2x_{1}^{2}+105y_{2}^{2}= 7^{1889}$.  Similarly, $x_{1}$ must be divisible by $7$ so $x_{1}= 7x_{2}$, and again we get $14x_{2}^{2}+15y_{2}^{2}= 7^{1888}$.  Continuing we see we'll arrive at $2x_{n}^{2}+105y_{n}^{2}= 7$, which has no solutions, contradiction.[/hide]

thanks a lot pkerichang, very nice solution. :)

i think this is easy with modulo 3
this is equal to 2x^2 = 1 modulo 3 which has no solution

well...that's right...but I think the USAMTS problem making people intends it to be a infinite descendent problem tho (but as long as it's right who cares :D)


If a standard six-sided die is rolled twice, what is the probability that the result of the second roll is greater than or equal to the result of the first roll? Express your answer as a common fraction.

[hide=" click to reveal hidden content"]

$ 36$ possible outcomes: $ 6$ where the two numbers are equal, $ 15$ where the first is less than the second and $ 15$ where the first is greater than the second.

Therefore $ \frac{15}{36}\equal{} \frac{5}{12}$.

Hope this is helpful.

[/hide]


The integer numbers from $1$ to $2002$ are written in a blackboard in increasing order $1,2,\ldots, 2001,2002$. After that, somebody erases the numbers in the $ (3k+1)-th$ places i.e. $(1,4,7,\dots)$. After that, the same person erases the numbers in the $(3k+1)-th$ positions of the new list (in this case, $2,5,9,\ldots$). This process is repeated until one number remains. What is this number?

[code]

int a[2003],i,j;
for(i=1;i<=2002;i++)a[i]=i;
while(n>3){n=0;
                 for(i=1;i<=2002;i+=3)a[i]=0;
                 //erase numbers
                 for(i=1;i<=2002;i++)
                          if(a[i]!=0&&a[i-1]==0)
                          {n++;
                           j=i;
                           while(!a[j--]);a[j]=a[i];a[i]=0
                          }
                //moves numbers together after erase
                //and counts numbers
                }
//now one should just
//show a to see the last 2 numbers
[/code]

[hide="hint"]Try to prove that in the nº prossece then we have a period of $ 3^n$ númber[/hide]

It is actually 2,6,10. Working backwards we can reach the desired number. :)
[hide]A number with position
a) $3m$ goes to position $2m$ after erasing
b) $3m+2$ goes to position $2m+1$

Talking about the last number
it can get to position $1$ only if it comes from position $2$
it can get to position $2$ only if it comes from position $3$
it can get to position $3$ only if it comes from position $5$
it can get to position $5$ only if it comes from position $8$
it can get to position $8$ only if it comes from position $12$
it can get to position $12$ only if it comes from position $18$
it can get to position $18$ only if it comes from position $27$
it can get to position $27$ only if it comes from position $41$
it can get to position $41$ only if it comes from position $62$
it can get to position $62$ only if it comes from position $93$
it can get to position $128$ only if it comes from position $192$
it can get to position $192$ only if it comes from position $288$
it can get to position $288$ only if it comes from position $432$
it can get to position $432$ only if it comes from position $648$
it can get to position $648$ only if it comes from position $972$
it can get to position $972$ only if it comes from position $1458$

Since $1.5\times 1458\ge 2003$, $1458$ is the number we were looking for. [/hide]


calculus
integration
geometry
rectangle
3D geometry
pyramid
calculus computations

To derive the formula for the definite integral for the area of a surface of revolution, why doesn't this work?

$\delta{A}= 2\pi (y+\delta{y})\delta{x}$
$A = 2\pi y\ dx$

In other words, when finding the formula for the volume of revolution we just take the sum of the volume of many cylinders, why can't we just take the sum of the surface areas of the same cylinders?

Jimmy

[quote]When we want the area under a curve, we take the sum of the areas of small rectangles. Do you see why taking the sum of just the heights of these small rectangles would not yield the area? This should answer your question.[/quote]
Using integration, I thought that's exactly what we do do to find the area under a curve. We certainly don't calculate arc lengths.

jmerry

[quote]That's not an accurate approximation of the area unless the surface looks like a cylinder locally. Look at what happens when you try this for a cone; the slant makes a large difference.[/quote]

I understand what you're saying, but why is it we can ignore this slant when calculating the volume of a cone formed by rotating a triangle about an axis? We assume that, in the limit, each individual strip approximates a cylinder and this works out fine. But, when is comes to the surface area, we have to account for the arc lengths for each individual strip.

Surface area in three dimensions is sensitive to small irregularities in a way that volume isn't. The same is true of arclength and area in two dimensions.

For example, consider this building: a pyramid containing a basketball arena, and surfaced with corrugated metal. How much volume is inside that pyramid? We can estimate that accurately by just measuring the width and height and calculating the volume of the equivalent flat-sided pyramid, since the part of the volume near the surface is comparatively tiny. How much metal is that on the surface? To answer that, we need to know more about the corrugation- the area is considerably larger than that of a flat-sided pyramid of the same size.

[quote="jmerry"]For example, consider this building: a [color=blue][blue][/color] pyramid containing a basketball arena, and surfaced with corrugated metal. [/quote]
Hey! I know that building. (And so do you.)  :lol:

jmerry:
[quote]Surface area in three dimensions is sensitive to small irregularities in a way that volume isn't. The same is true of arclength and area in two dimensions. [/quote]

I see. I have never seen or heard this explained anywhere else before. Most textbooks fail to mention this. Surprising, as it seems to me a pretty important distinction.
You say area in two dimensions is sensitive to small irregularities in the curve, but we ignore these when summing rectangles, surely?

What jmerry said about two dimensions is that arclength is sensitive to small-scale fluctuations and area isn't.

Kent

Yeah, I see now. Thanks for putting me straight.


geometry
rhombus
ratio
3D geometry
sphere

Is the goat outside of the field or inside of the field?  Either way, you'll be finding the area of the annulus.

no sorry if i was not very clear. but the other end of the goat's string is tied to a point on the cirle.

[quote="maokid7"]ok i have been trying to solve this problem but am not able to.  i was wondering if anyone could please help me.

1) a goat is tied to the fence of a circular field.  the field (cirlce) has a radius of 40 feet. if the the goat's string is n feet, and he is able to graze exactly half the area of the field, then what is the value of n.[/quote]

Just find half the area of the field, which is $\frac{40^2}{2}\pi$ and then make a circle inside the bigger circle and say that the radius$^2\pi$ (the goat's string) is equal to that area $800\pi$ and solve for that.

EDIT: Sorry, misunderstood the question, the goat is tied to the fence, or on the circle.

ok it appears that i did a horrible job of describing the problem. So i will just type it over in a more general forn.  :oops:  :P
ok here it is:
there is a given circle with a radius of 40. this cirlce passes through the center of another circle, which has a radius of n. if the area of the shared reginon is equal to half the area of the circle with radius 40, then what is the value of n.

i hope this has made thing more clear.

the radius of the other circle, or the length of the goat, is $40+\sqrt{3}$ Atleast that's what I got, here's how I did it, the two circles intersect each other but you have to find the radius of the other circle. So when they intersect, it goes through half of the circle, and then you can make a rhombus with the radius ($40$) of the first circle and see that it's composed of  $30,60,90$ because the chord which is the diameter of the other circle bisects the radius of the other circle so you have $40$ as the hypotenuse and $20$ of the leg you need, and another $20$ for the other triangle adjacent to it, so so far you have the length as $20+20$ but you also have to find the litle part that went outside the half way mark of the first circle, to find that, you can extend the triangles to $45,45,90$ and since one of the legs as already found out is $20\sqrt{3}$, you can see that the length is $20$ and $20\sqrt{3}$ so $20+20\sqrt{3}$

[quote="maokid7"]gosh i suck at this it would be so much easier if i knew how to add a picture to this (please tell me so that i may, it will save both of us time) but in the problem  n (the other radius) is less than 40.[/quote]

Perhaps I did it wrong,  :? but how can it be less to $40$ because the other circle has to go through the half way mark of the first circle to get half the area, so it has to be greater than $40$, maybe I'm just really  confused :?  :(

[quote="maokid7"]o sorry i didnt mean 40 i meant 80  :P  sorry
but could you please show me how you got this answer  :)[/quote]

I don't really have pics but I'll try:

Take the two circles and intersect them, take the radius $40$ of the first circle and extend them outwards on opposite sides and then close the gap with two lines of equal length on the top, thus it forms a rhombus, and since the chord which is the diameter of the second circle is bisected by the radius of the first circle when the radius is put in a $90$ degree angle, you can see that the rhombus is formed by $4$ right triangles that are $30,60,90$ then you find that the legs that bisect the diameter of the second circle each are $20 \sqrt{3}$ and if you extend the right triangles and make them $45,45,90$ which goes all the way till it meets the edge of the second circle, you will see that it is $20\sqrt{3}$ plus the $20$, so here's my mistake so far: the radius of the other circle is [u]$20\sqrt{3}+20$[/u].

I might be wrong, anybody else wanna go at it? :P

thanks for the help. but im sorry that i was not able to follow your work. but thanks for trying to help.

The problem is not so easy as it seems. In fact for any reasonable ([tex]0\le n\le 2[/tex]) string length/field radius ratio [tex]n=\frac{r}{R}[/tex] the clean/whole area ratio [tex]N[/tex] is given by the formula:

[tex]N=\frac{1}{\pi}(2-n^2)\cdot \arcsin\frac{n}{2}+\frac{n^2}{2}-\frac{n}{2\pi}\sqrt{4-n^2}[/tex]

Now you have to set [tex]N=\frac{1}{2},\ R=40[/tex] and solve for [tex]r[/tex]. I don't seem to see any other way except doing it by some numerical method.

Surprisingly, the following 3D problem formulated in the same way seems to be much easier.

Given two spheres [tex]s_1,\, s_2[/tex] the latter having its center on the surface of the first one, the radii respectively [tex]r_1,\, r_2[/tex]. Let the volume of the intersection of [tex]s_1[/tex] and [tex]s_2[/tex] be [tex]V_c[/tex] and the volume of [tex]s_1[/tex] be [tex]V_1[/tex].

Find [tex]N=\frac{V_c}{V_1}[/tex] in terms of [tex]n=\frac{r_2}{r_1}[/tex].


Prove that every integer >= 12 is the sum of two composite numbers.

I have proved this by considering various case. Is there any simple way?

Let's take only the primes (it's obvious that a composite number $n=ab>4$ we can see it as $ar+as$, with $b=r+s$): but every prime $>11$ can be seen as $9+2t$, and both $9$ and $2t$, if $t>1$ are composite

Simply observe that $ 3 $ divides one of $ n - 4 $, $ n - 6 $,  $ n - 8 $

A partial solution of Goldbachs conjecture says that every even integer greater $2$ is the sum of a prime and a number that is prime or product of two primes.
So I want to ask the question (since the theorem given above is very difficult to prove):
Are there similar statements that can be proved 'elementary'¿


Prove that

If $\binom{m}{n}=\binom{m}{p}$ then $n=p$ or $m=n+p$.

Can we solve completely the equation

$\binom{a}{b}=\binom{c}{d}$ ?

No, not really. Just write out pascels triangle to see why.

proving that if $\binom{m}{n}=\binom{m}{p}$ then $n=p$ or $m=n+p$ is simple.
but i don't quite understand what you mean by solving for:
$\binom{a}{b}=\binom{c}{d}$

I've tried this, but I think it's too complicated for such a simple thing! (I've never found a proof of this fact in a book, tough it's oftenly used).
Suppose $n>p$, so $n=p+h$, $h\in\mathbb{N}^{*}$. Then
\[1=\frac{\binom{m}{n}}{\binom{m}{p}}=\frac{(m-p)\cdot\cdot\cdot(m-p-h+1)}{(p+h)\cdot\cdot\cdot(p+1)}\]
Numerator and Denominator are the product of $h$ consecutive natural numbers, then, if $m-p><p+h$, it would be Numerator $><$ Denominator. Hence $m-p=p+h$, that is, $m=2p+h=n+p$.

Well you could use logic to prove that.  The number of ways to choose 2 items from a group of 5 is the same as the number of ways to not choose 3, and thus the same as the number of ways to choose 3.  You also can show, by the definition of mCn, that the only possibilities are if n=p or m=n+p

Ok..
The question WASN'T about the part

$n=p$ or $m=n+p$ THEN $\binom{m}{n}=\binom{m}{p}$

I was asking about the other part, that is,

if $\binom{m}{n}=\binom{m}{p}$ THEN $n=p$ or $m=n+p$

I think I solved it now anyways but it seems hard

By comparing two terms next to each other, you would find out that it's strictly increasing up to the midway, then strictly decreasing after passing the midway.

Yeah, i was trying to figure out a proof for that...
I'm sure you could show that with the n!/(n-x)!x! thing though...


$ a,b,c\geq 0$ . Prove that :
$ a\sqrt{b^2\plus{}4c^2}\plus{}b\sqrt{c^2\plus{}4a^2}\plus{}c\sqrt{a^2\plus{}4b^2}\leq \frac{3}{4}(a\plus{}b\plus{}c)^2$

[quote="can_hang2007"][quote="conan_naruto236"]$ a,b,c\geq 0$ . Prove that :
$ a\sqrt {b^2 \plus{} 4c^2} \plus{} b\sqrt {c^2 \plus{} 4a^2} \plus{} c\sqrt {a^2 \plus{} 4b^2}\leq \frac {3}{4}(a \plus{} b \plus{} c)^2$[/quote]
It's my problem, see my hint here :) http://www.mathlinks.ro/viewtopic.php?t=194955[/quote]
Thank you very much .
But can you write your solution's lemma 2 ?
[b]Lemma 2 :[/b]
          $ \sum_{cyc} \frac {2b \plus{} c}{4b^2 \plus{} 3bc \plus{} c^2} \geq \frac {27}{8(a \plus{} b \plus{} c)}$


geometry
3D geometry
pyramid
vector
calculus
integration

While sitting around waiting for interior decorators to finish up in the scroll repository, a bunch of bored and mischievous workers decided that they'd build a catapult. When they finally finished, it was able to shoot a projectile straight up into the air. So. in a move which they later regretted, they decided to launch one of the pyramid blocks, which had a density of 2400 kilograms per cubic meter.

When the block crashed back into the catapult, smashing it to pieces, it was determined to have a velocity of 20 meters per second. Assuming that gravity in Neopia is exactly 9.90 meters per second squared, the block hit the catapult at the same elevation as it was launched from the catapult, and there was, miraculously, no air resistance acting on the block, how high, in centimeters, did the block fly above the catapult? Please round to the nearest centimeter and please, do not try this at home.

Help me with this please

[quote="236factorial"]Hmmmm.... lenny conundrum....

[hide]$V_f^2=V_i^2+2ad$

$V_i$, the initial velocity from the air falling, is 0.

$V_f$, the final velocity, was 20 m/s

$a=9.9$.

We want to find $d$.

$400=2(9.90)(d)$.

$d=2020 cm$. [/hide][/quote]

good job...i just wanted to say that $a=-9.9$ and the velocity is $-20$

it doesnt make a difference but oh well

[quote="jli"]good job...i just wanted to say that $a=-9.9$ and the velocity is $-20$

it doesnt make a difference but oh well[/quote]

Oh yeah... too long since I've solved vector problems....

but negative divide by negative equals positive, so I got lucky  :D

Well, distance is always positive, so...

And how did you learn that formula? Is it used for other things besides Physics? I only learned it (along with $V_F=V_o+at$ and $x=V_ot+\dfrac{1}{2} at^2$) in my AP Physics class.

Really? I'm not sure what level my physics textbook is on, but there sure are these four equations called something like the kinematic equations, which I learned from an internet printout and my textbook.

Are you just taking standard Physics, then, or are you actually in AP Physics? They would definitely teach the formulas even if it's not the AP class, since they're very integral to Physics; I just didn't realize you were taking it.

BTW, thinking about it, this probably should have been in the Physics forum, since Physics is not exactly a part of Getting Started. :P

I don't take physics, I just read stuff. I don't know when AP Physics is, but I'm sure it's not 8th grade  :(

Actually, it was at my middle school... but it's the only middle school in the country that offers it.


Compute: $ (\sqrt{5} \plus{} \sqrt{7})(\sqrt{5} \minus{} \sqrt{7})$.

hello,expanding the product after the binomial formula we get $ \sqrt{5}^2\minus{}\sqrt{7}^2\equal{}5\minus{}7\equal{}\minus{}2$
Sonnhard.

It is easy to watch the following way of writing more.
$ \left(\sqrt{5}\right)^2 \plus{} \left(\sqrt{7}\right)^2$

[hide="A little more explanatory"]
Recall the identity \[ a^2\minus{}b^2\equal{}(a\minus{}b)(a\plus{}b).\] In this case, $ a\equal{}\sqrt{5}$ and $ b\equal{}\sqrt{7}.$ Using our identity in reverse, we have \[ \left(\sqrt{5}\plus{}\sqrt{7}\right)\left(\sqrt{5}\minus{}\sqrt{7}\right)\equal{}\left(\sqrt{5}\right)^2\minus{}\left(\sqrt{7}\right)^2\equal{}5\minus{}7\equal{}\boxed{\minus{}2}.\][/hide]

[quote="Kouichi Nakagawa"]It is easy to watch the following way of writing more.
$ \left(\sqrt {5}\right)^2 \plus{} \left(\sqrt {7}\right)^2$[/quote]
Incorrect. Use $ (a\plus{}b)(a\minus{}b)$

Sorry.
I wrote it incorrectly simply.
It is a parenthesis to be important.
Furthermore, my post assumed that I said for Sonnhard.

The following expression transformation

$ (x\plus{}a)(x\plus{}b)\equal{}x^2\plus{}(a\plus{}b)\plus{}ab$
If $ b\equal{}\minus{}a$, $ (x\plus{}a)(x\plus{}b)\equal{}x^2\plus{}(a\minus{}a)\plus{}a\times(\minus{}a)\equal{}x^2\minus{}a^2$
So, $ (a\plus{}b)(a\minus{}b)\equal{}a^2\minus{}b^2$

We can lead a middle school students in common-sense terms more.


$\int\frac{1}{1-tan{\theta}}d\theta$

I tried conjugate, rewriting into cosine and sine, partial fractions...can't seem to get anywhere.. please help!

Thanks!

Switch it to sins and cosines.
You get cos (x) / (cos x - sin x) dx

Multiply by conjugate, cos x + sin x, on top and bottom
You get [(cos x)^2 + sin x cos x] /  [(cos x)^2 - (sin x)^2]

Make [(cos x)^2 - (sin x)^2] ---> cos 2x

Now split it up into two integrals

[(cos x)^2] / [cos 2x] dx +
[sin x cos x] / [cos 2x] dx

for the first one, change [(cos x)^2] to:
[1 + (cos 2x)] / 2 and you end up with
[1 + (cos 2x)] / [2 cos 2x] dx
Then split this up again into [sec 2x] / 2 dx + 1/2 dx
Integrate (Remember that integral of sec u is ln |sec u + tan u|) to get first half of the answer, which is
(1/4 ln |sec 2x + tan 2x| + 1/2 x) I think.

for the second one, change [sin x cos x] to --> 1/2 sin 2x to get
[sin 2x] / [2 cos 2x] dx. Then do a u substitution with
u = cos 2x. Integrate for the second half of the answer, which is
(-1/4 ln |cos 2x|) I think.

Final Answer: 1/4 ln |sec 2x + tan 2x| + 1/2 x - 1/4 ln |cos 2x| + C

Hope I didn't make any mistakes >_<, but I think this should work. You can combine ln terms if you want as well as replace back the x's with theta's.

Maybe there's a faster way though... I donno. :?:

Thanks for the help!  Unfortunately, I got the solution a few minutes after I posted this message.. I did it the same way you did.

And why isn't this the right forum?  Isn't this forum for calculus problems?

Thanks again!

No, that'd be the [url=http://www.artofproblemsolving.com/Forum/index.php?f=66]calculus forum[/url] ;-)


probability geometry integration calculus

Alice and Bob both go to a party which starts at 5:00. Each of them arrives at a random time between 5:00 and 6:00. What is the probability that the number of minutes Alice is late for the party plus the number of minutes Bob is late for the party is less than 45? Express your answer as a common fraction.

Hello
Probability of the number of minutes Alice is late for the party plus the number of minutes Bob is late for the party is less than 45
=$ \frac{Area of triangle}{Area of square}$
=$ \frac{9}{32}$
Thank u.

This is a continuous rather than discrete problem. So the solution is not $ \sum_{0\le i<45}\frac{1}{60}\cdot\frac{45\minus{}i}{60}\equal{}\frac{23}{80}$ but $ \int_0^{45}\frac{1}{60}\cdot\frac{45\minus{}x}{60}dx\equal{}\frac{9}{32}$.

What do these symbols mean? $"\sum \int"$

$\sum$: See [url=http://www.artofproblemsolving.com/Wiki/index.php/Sigma_notation]here[/url].

$\int$: Something you learn in calculus

You don't need to use fancy notations

If we let 1 unit be one minute, the area of the shaded region is $\frac{45^2}{2}$ square units and the whole area is 3600 square units. Therefore, the probability that a randomly chosen point will land in the shaded region is $\frac{45^2}{2\cdot 3600}=\boxed{\frac{9}{32}}$.

But wouldn't this method of finding the answer end up counting that the total number of late minutes is equivalent to 45?
Like, Alice arrives on 5:45 and Bob on 5:00, which would be equal to 45 minutes late, which is not "less than 45" as the problem states.

[quote=Mogmog8][hide=@above]Since we are counting area, the boundaries are negligible (they contribute no area).[/hide][/quote]

Ohhhhhh!
Thank you!


Given $0<x_1<1$ and $x_{n+1}=x_n+\frac{x_n^2}{n^2}$
1/ Prove that exist $k>0$ such that : $x_n<k$  $\forall n\ge 1$

First [tex]x_n<nx_1 \forall n\ge 2[/tex]
hence [tex]x_m<m-1[/tex] for some [tex]m>1[/tex]
Now we have
[tex] x_{n+1}>x_n>0,n=1,2,...[/tex]
hence [tex]\frac{1}{x_n}-\frac{1}{x_{n+1}}=\frac{x_n}{x_{n+1}n^2}<\frac{1}{n^2}<\frac{1}{n-1}-\frac{1}{n},n=2,3,...[/tex]
therefore [tex]\frac{1}{x_m}-\frac{1}{m-1}<\frac{1}{x_n} \forall n>m[/tex]
 ...Q.E.D


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

[quote="manuel"]Prove every torsion free module over a PID is projective.[/quote]

If K is the field of fractions of a PID R and R≠K, then the R-module K is torsion-free and not projective.

[quote="manuel"]Prove every torsion free module over a PID is projective.[/quote]You might want to add the assumption that the module be finitely generated to make the claim become true? ;)


geometry
3D geometry
modular arithmetic
calculus
integration
Gauss
number theory

Does anyone have a proof of Fermat's polygonal number theorem..that all positive integers are expressible as the sum of at most n n-gonal numbers, first proved by Cauchy.

moreover, does anyone know how to to prove that x^3=y^2+2 has only one solution in integers and that is (3,2). This is an Fermat equation elliptic mayhaps ...and i cant find solution online..

Vinh :D  :D

I think I solved this one at PROMYS.

Use cases in y^2= x^3-2 and each time yields a non-integer expression except 2 times where it yields a solution for x and a +/- solution for y. Check your answer, since the solution is 3,5 and 3,-5 i think.

y^3 = x^2+2

first note, that by looking mod 4, we see that x must be odd.

then y^3 = (x+ :sqrt: -2)(x- :sqrt: -2), and lucky for us Z[ :sqrt: -2] is a unique factorization domain.

lemma 1: if x is odd then (x+ :sqrt: -2) is rel. prime to (x- :sqrt: -2)

proof: suppose some p prime in Z[ :sqrt: -2] divides both.  Then p also divides their difference or 2 :sqrt: -2.  But then N(p)|N(2 :sqrt: -2) = 8.  At the same time, N(p)|x^2+2 which is odd, so N(p) = 1, a contradiction since N(p) = 1 => p is a unit.

Now that we know that (x+ :sqrt: -2) and (x- :sqrt: -2) are relatively prime, each must be a cube in order for their product to be a cube.  So

(x+ :sqrt: -2) = (a+b :sqrt: -2)^3 = (a^3-6ab^2)+(3a^2b-2b^3) :sqrt: -2.
for a,b in Z.
equating real and imaginary parts, we see that 3a^2b-2b^3=1 but this would force b to be  :pm: 1 since b divides both sides.  Trying both we get that (a,b) must be one of the following (1,1) (-1,1).  then x must be either -5 or 5

plugging in, the only solutions are (5,3) and (-5,3).

He wrote $x^3=y^2+2$, but then he said that the only solution is $3,2$, and that's what made me think that he meant $x^3=y^2+1$. If I remember correctly, that had a solution similar to yours, but in $\mathbb Z[i]$, of course.

I think a proof for Fermat polygonal number theorem can be found from Nathanson, M. B. "A Short Proof of Cauchy's Polygonal Number Theorem." Proc. Amer. Math. Soc. 9, 22-24, 1987.

Indeed. I found the proof from that paper. Unfortunately I don't know if there are some copyright problems so I don't know if I'm allowed to write the proof here.

Well, I've read Nathonson and must admit that the prove of the whole theorem is a bit longer (and most time nasty computation...).
When you not copy it word by word but just the ideas etc., not the exact wording, there should be no problem with copyright.

OK, here comes a sketch I found from the Nathanson's article. If $m\geq 1$ the polygonal number of order $m+2$ is a number of the form
\[ p_m(k)=\frac{m}{2}(k^2-k)+k  \]
where $k=0,1,\cdots$. Now there are so called Pepin's and Dickson's tables which shows it is enough to prove only the case $n\geq 120m$. The proof is based on the following Cauchy's lemma:

For odd positive integers $a$ and $b$ such that $b^2<4a$ and $3a<b^2+2b+4$ we can find nonnegative integers $s,t,u$ and $v$ such that
\[ a=s^2+t^2+u^2+v^2  \]
and
\[ b=s+t+u+v.  \]

Using this result we can prove the following two theorems

Thm 1. Let $m\geq 3$ and $n\geq 120m$. Then $n$ can be expressed as the sum of $m+1$ polygonal numbers of order $m+2$ and at most four of the numbers greater than 1.

Thm 2. For $m\geq 3$ and $m$ odd, every sufficient large integer is the sum of four polygonal numbers of order $m+2$ and that one of the number is $0$ or $1$.

Hope this helps.

[quote="vinh"]Does anyone know how to to prove that x^3=y^2+2 has only one solution in integers and that is (3,2).[/quote]

A nice generalization for this: Let $d>1$, square-free $d\equiv 2 \pmod{4}.$ Assume that the class number of $\mathbb{Q}(\sqrt{-d})$ is not divisible by $3$. Then $y^2=x^3-d$ has an integral solution iff $d$ is of the form $3t^2\pm 1,t\in \mathbb{N}$. The solutions are then $(t^2+d,\pm t(t^2-3d))$.

[quote="srulikbd"]well I could only find http://links.jstor.org/sici?sici=0002-9939%28198701%2999%3A1%3C22%3AASPOCP%3E2.0.CO%3B2-Y&size=LARGE&origin=JSTOR-enlargePage
where can I find gauss proof for the 3 case?[/quote]
Nathanson: Additive Number Theory, The Classical Bases, starting from the page 27.


trigonometry
geometry
trig identities
Law of Sines

This might be a bit easy:
Prove that the largest angle of a triangle is the angle opposite the longest side.

here's how to prove that the angle opposite the longest side is larger than any other angle.

since AC is the longest side, we can draw a segment BD such that CD=CB and D is on segment AC
therefore :ang: CDB= :ang: DBC
since :ang: CDB> :ang: BAC
hence :ang: DBC> :ang: BAC
therefore :ang: ABC> :ang: BAC

the problem can be solved in this way.
This solution is one of my all-time favorite solutions. It's neat and clear, easy to understand, but really involves some advanced topics.

Law of Sines gives us the answer pretty easily. Let $A$ be the largest angle. Suppose $B<A$ and $\sin{B}>\sin{A}$. Clearly $\frac{\pi}{2} < A < \pi$. Then $\pi-A < B < \frac{\pi}{2}$. But then $A+B > A+(\pi-A) = \pi$ which is a contradiction.

So if $B<A$, $\sin{B} < \sin{A}$. By Law of Sines, $a/\sin{A} = b/\sin{B}$ so $a>b$ and $a$ is the longest side.

[quote="paganinio"][quote="paladin8"]Law of Sines gives us the answer pretty easily. .[/quote]
1,The Law of Sines itself is based on the 'largest angle of a triangle is the angle opposite the longest side.' theorem, isn't it??
2,Even if it isn't, I am sure 99% people will think that my solution is EASIER.[/quote]

1) Do you know a proof of the law of sines? (e.g, no matter which altitude we drop, the areas are all the same)

2) It's not a competition; having multiple solutions is always good.


I want to collect some of the hardest ever MC problems here, to use as a practice resource.
I want them from all subjects.

Rules:
1. By difficulty, i don't mean calculations. I mean difficulty of finding the way to solve it.
2. If using real MC problems, change the numbers around.
3. If using original problems, keep them MC level.
4. Post solutions hidden on the question post.
5. Yes, I said solutions, not answers. Show ALL steps.
6. Obviously, no 2009 questions. (at least until they allow it in the forum.)
7. If using real MC problems, mention from what round and year.
8. Please use LaTex and asymptote. NO GEOGEBRA!

If this actually comes together, It could be an amazing resource for MC.

Thank you.
Gracias.
Danke.
(insert french thank you)
(insert japanese thank you)

[quote="AIME15"]*cough* MATHCOUNTS MARATHON *cough*[/quote]

\me agrees.  Some of the problems posted are pretty hard...(for me anyway)

I just want a quick...
you know what. Your right i am stupid for posting this.


conics
hyperbola
geometry
geometry unsolved

The condition can be expressed as $ PB \minus{} PC \equal{} b \minus{} c, PA \minus{} PC \equal{} a \minus{} c,$ which implies that $ P$ is one of the intersections of the three hyperbolas $ \mathcal{H}_a,\mathcal{H}_b,\mathcal{H}_c$ with foci in two vertices and passing through the third one. Precisely, it is the intersection of the three branches that do not pass through $ A,B,C.$ In order to construct such point, let $ X,Y,Z$ be the tangency points of the incircle $ (I)$ with $ BC,CA,AB$ and let $ (A),(B),(C)$ be the circles centered at $ A,B,C$ with radii $ AZ,BX,CY,$ respectively. Indeed, centers $ S'$ of the circles internally tangent to $ (B),(C)$ satisfy $ S'B \minus{} S'C \equal{} CX \minus{} BX \equal{} b \minus{} c$ $ \Longrightarrow$ $ S' \in \mathcal{H}_a.$ Hence the wanted point $ P$ is identical to the center of the second Soddy's circle $ ({S_0}'),$ which is internally tangent to $ (A),(B),(C).$

could you tell me please what's wrong whith this Soddy construction?

 :(


If $a,b,c$  are positive real number prove

$\frac{a^{2}+2bc}{b^{2}+c^{2}}+\frac{b^{2}+2ac}{c^{2}+a^{2}}+\frac{c^{2}+2ab}{a^{2}+b^{2}}> 3$.

This was already posted and solved by Muirhead. I think a solution without Muirhead could be a fantastic challenge for all maniacs of inequalities.

I just saw this in the other thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=1066 . Note that the inequality can be sharpened to

$\frac{a^2+2bc}{b^2+c^2}+\frac{b^2+2ca}{c^2+a^2}+\frac{c^2+2ab}{a^2+b^2}>3$

for positive reals a, b, c.

Here is a pretty nice solution. Using the $\sum$ notation for cyclic sums, the inequality in question rewrites as

$\sum\frac{a^2+2bc}{b^2+c^2}>3$.

This is equivalent to $\sum\frac{a^2+2bc}{b^2+c^2}-3>0$. But

$\sum\frac{a^2+2bc}{b^2+c^2}-3=\sum\left(\frac{a^2+2bc}{b^2+c^2}-1\right)=\sum\left(\frac{a^2}{b^2+c^2}-\frac{\left(b-c\right)^2}{b^2+c^2}\right)$
$=\sum\frac{a^2}{b^2+c^2}-\sum\frac{\left(b-c\right)^2}{b^2+c^2}$.

Thus, we have to show that $\sum\frac{a^2}{b^2+c^2}-\sum\frac{\left(b-c\right)^2}{b^2+c^2}>0$, i. e. that $\sum\frac{a^2}{b^2+c^2}>\sum\frac{\left(b-c\right)^2}{b^2+c^2}$.

Now, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in the Engel form[/url], we have

$\sum\frac{a^2}{b^2+c^2}=\sum\frac{\left(a^2\right)^2}{a^2b^2+c^2a^2}\geq\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2b^2+c^2a^2\right)+\left(b^2c^2+a^2b^2\right)+\left(c^2a^2+b^2c^2\right)}$
$=\frac{\left(a^2+b^2+c^2\right)^2}{2\left(b^2c^2+c^2a^2+a^2b^2\right)}$.

Hence, it remains to prove that $\frac{\left(a^2+b^2+c^2\right)^2}{2\left(b^2c^2+c^2a^2+a^2b^2\right)}>\sum\frac{\left(b-c\right)^2}{b^2+c^2}$, i. e. that

$\left(a^2+b^2+c^2\right)^2>2\left(b^2c^2+c^2a^2+a^2b^2\right)\sum\frac{\left(b-c\right)^2}{b^2+c^2}$.

Now,

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

The HM-GM inequality, applied to the numbers $b^2$ and $c^2$, yields $\frac{2b^2c^2}{b^2+c^2}\leq\sqrt{b^2c^2}=bc$; thus,

$2\left(b^2c^2+c^2a^2+a^2b^2\right)\sum\frac{\left(b-c\right)^2}{b^2+c^2}=\sum\left(\frac{2b^2c^2}{b^2+c^2}+2a^2\right)\left(b-c\right)^2\leq\sum\left(bc+2a^2\right)\left(b-c\right)^2$.

Hence, instead of proving $\left(a^2+b^2+c^2\right)^2>2\left(b^2c^2+c^2a^2+a^2b^2\right)\sum\frac{\left(b-c\right)^2}{b^2+c^2}$, it will be enough to show the stronger inequality $\left(a^2+b^2+c^2\right)^2>\sum\left(bc+2a^2\right)\left(b-c\right)^2$.

With a bit of calculation, this is straightforward; here is a longer way to show it without great algebra:

$\sum\left(bc+2a^2\right)\left(b-c\right)^2=\sum\left(a\left(a+b+c\right)-\left(c-a\right)\left(a-b\right)\right)\left(b-c\right)^2$
$=\sum a\left(a+b+c\right)\left(b-c\right)^2-\sum\left(c-a\right)\left(a-b\right)\left(b-c\right)^2$
$=\left(a+b+c\right)\sum a\left(b-c\right)^2-\left(b-c\right)\left(c-a\right)\left(a-b\right)\underbrace{\sum\left(b-c\right)}_{=0}$
$=\left(a+b+c\right)\sum a\left(b-c\right)^2=\left(a+b+c\right)\sum a\left(\left(b-c\right)\left(b-a\right)+\left(c-a\right)\left(c-b\right)\right)$
$=\left(a+b+c\right)\left(\sum a\left(b-c\right)\left(b-a\right)+\sum a\left(c-a\right)\left(c-b\right)\right)$
$=\left(a+b+c\right)\left(\sum c\left(a-b\right)\left(a-c\right)+\sum b\left(a-b\right)\left(a-c\right)\right)$
$=\left(a+b+c\right)\sum\left(b+c\right)\left(a-b\right)\left(a-c\right)$.

Thus, in order to prove that $\left(a^2+b^2+c^2\right)^2>\sum\left(bc+2a^2\right)\left(b-c\right)^2$, we will show the equivalent inequality $\left(a^2+b^2+c^2\right)^2>\left(a+b+c\right)\sum\left(b+c\right)\left(a-b\right)\left(a-c\right)$. In fact, we will even show the stronger inequality

$\left(a^2+b^2+c^2\right)^2>\sum a^2\left(a-b\right)\left(a-c\right)+\left(a+b+c\right)\sum\left(b+c\right)\left(a-b\right)\left(a-c\right)$,

which is indeed stronger since $\sum a^2\left(a-b\right)\left(a-c\right)\geq 0$ by the Schur inequality.

Now, this stronger inequality can be established as follows:

$\sum a^2\left(a-b\right)\left(a-c\right)+\left(a+b+c\right)\sum\left(b+c\right)\left(a-b\right)\left(a-c\right)$
$=\sum\left(a^2+\left(a+b+c\right)\left(b+c\right)\right)\left(a-b\right)\left(a-c\right)$
$=\sum\left(\left(a^2+b^2+c^2\right)+\left(bc+ca+ab\right)+bc\right)\left(a-b\right)\left(a-c\right)$
$=\left(\left(a^2+b^2+c^2\right)+\left(bc+ca+ab\right)\right)\underbrace{\sum\left(a-b\right)\left(a-c\right)}_{=\left(a^2+b^2+c^2\right)-\left(bc+ca+ab\right)}+\sum bc\underbrace{\left(a-b\right)\left(a-c\right)}_{=a^2+bc-ca-ab<2a^2+bc}$
$<\left(\left(a^2+b^2+c^2\right)+\left(bc+ca+ab\right)\right)\left(\left(a^2+b^2+c^2\right)-\left(bc+ca+ab\right)\right)+\sum bc\left(2a^2+bc\right)$
$=\left(\left(a^2+b^2+c^2\right)^2-\left(bc+ca+ab\right)^2\right)+\left(bc+ca+ab\right)^2=\left(a^2+b^2+c^2\right)^2$,

and the inequality is proven.

  Darij

[quote="Lovasz"]Nive proof.
[quote="Anh Cuong"]...
First let $f(a,b,c)=\frac{a^2+2bc}{b^2+c^2}+\frac{b^2+2ac}{a^2+c^2}+\frac{c^2+2ab}{a^2+b^2}$
We will prove that:
$f(a,b,c)\geq \frac{2bc}{b^2+c^2}+\frac{b}{c}+\frac{c}{b}$...[/quote]
Why can you guest this, Anh Cuong?  :)[/quote]
it is pretty natural if you use mixing variables and put a = 0


Let $(I_{n})$ be a countable family of intervals. Show that we can choose from them a countable subfamily of intervals such that every point in it's covering will be covered at most by $3$ intervals of a subfamily.

Not sure I understand: what if those intervals all have in common the point $0$ ? Then no matter what countable subsequence of intervals you extract, $0$ will belong to all the intervals of this subsequence ?

Maybe in that case our subcovering will only have 3 elements? But I don't see how it works -- since there are countable, we can't pick the one that "extends farthest left" and the one that "extends farthest right." Perhaps some condition is missing? If it's a finite covering, the number is 2.


ARML
geometry
calculus
integration
trigonometry
3D geometry
Pythagorean Theorem

1.
In right triangle $ABC$, $\angle C = 90^\circ$, $c=3$, and $a+b=\sqrt{17}$.  Compute the area of the triangle.

2.
If $(a+b):(b+c):(c+a) = 6:7:8$, and $a+b+c=14$, compute the value of $c$.

3.
Let $z$ be a root of $z^5-1=0$, with $z \not= 1$.  Compute the value of

\[z^{15} + z^{16} + z^{17} + \cdots + z^{50}. \]

4.
A regular polygon of $2n$ sides is inscribed in a circle of radius 5.  The sum of the squares of the distances from any point on the circle to the vertices of the polygon is 600.  Compute the value of the integer $n$.

5.
The circles, each of radius 5, lie in perpendicular planes.  They intersect in two points, $X$ and $Y$, thus determining a common chord $\overline{XY}$.  If $XY = 8$, compute the distance between the centers of the circles.

6.
How many positive integers less than 54 are equal to the product of their proper divisors?

7.
Find all positive integers $n$, less than 17, for which $n! + (n+1)! + (n+2)!$ is an integral multiple of 49.

8.
In triangle $ABC$, $a \ge b \ge c$.  If $\displaystyle \frac{a^3+b^3+c^3}{\sin^3{A} + \sin^3{B} + \sin^3{C}} = 7$, compute the maximum possible value of $a$.

[hide="6"]

Clearly the number must be equal to the product of two primes or the cube of a prime.  These numbers less than 54 are 6,10,14,22,26,34,38,46,15,21,33,39,51,35 for the first and 8,27 for the second.  Thus 16 such numbers.

[/hide]

[hide="5"]

If a chord in a circle of radius 5 has length 8, it is 3 away from the center (Use a 3-4-5 right triangle).  So we have a right triangle with vertices the two centers and the midpoint of $XY$.  The two legs have length 3.  So the distance between the two centers is $3\sqrt{2}$.

[/hide]

[hide="2"]

We have $\frac{a+b}{b+c}=\frac{6}{7}\Rightarrow 7a+b=6c$.

 Likewise, we have $\frac{b+c}{a+c}=\frac{7}{8}\Rightarrow 7a-8b=c$.

Subtract these two equations to get $9b=5c$, or $b=\frac{5}{9}c$.

Similarly, from $\frac{a+b}{a+c}=\frac{6}{8}$, we get $2a+8b=6c$.

Add this to the second equation to get $9a=7c$, or $a=\frac{7}{9}c$.

So $\frac{5}{9}c+\frac{7}{9}c+c=14\Rightarrow \frac{7}{3}c=14\Rightarrow c=6$.

   [/hide]

[hide="1"]

By the Pythagorean Theorem, $a^2+b^2=9$.

We also see that $a^2+2ab+b^2=17$.

Subtracting equations, we get $2ab=8$.

The area of the triangle is $\frac{ab}{2}=\frac{8}{4}=2$.

[/hide]


[quote="bpms"][quote="Go Around the Tree"]It was believed by ancient greeks that the golden rectangle has the most pleasing form to the eye of all rectangles. In a golden rectangle, if w is its width and l is its length, then the ratio of its width to its length is L to w+L. Find the length of a golden rectangle with a width of 3 inches.[/quote]
I'm surprised no one has answered this...
[hide]Golden should be a dead give away but I'll do this rigorously.
$\frac{3}{l}=\frac{l}{3+l}$
$l^{2}-3l-9=0$
$l=\frac{3+\sqrt{45}}{2}$
$l=1.5 \cdot \frac{1+\sqrt{5}}{2}$
[/hide]
Hm...[/quote]

You messed up at the end.

[hide]$\frac{3+\sqrt{45}}{2}\ne 1.5 \cdot \frac{1+\sqrt{5}}{2}$.

$\frac{3+\sqrt{45}}{2}= \frac{3+3\sqrt{5}}{2}= 3 \cdot \frac{1+\sqrt{5}}{2}= 1.5 \cdot (1+\sqrt{5})$[/hide]


Hi, could anyone help me with this please? Thanks a lot!

1.) Two hallways, one 8 feet wide and the other 6 feet wide, meet at right angles. Determine the length of the longest ladder that can be carried horizontally from one hallway into the other.

Ah, a classic. Just to give the other form:

A (reasonably tall) building with a vertical wall stands on level ground. There's an 8 foot high fence 6 feet from the wall. What is the length of the shortest ladder that you can lean against the wall with its feet on the ground outside the fence?

These two problems have the same answer.

The problem is interesting as well in somewhat greater generality. Ask the same problem, but with the widths of the hallways $a$ and $b.$ (Or the fence has height $a$ and distance $b$ from the wall.) Let $g(a,b)$ be the solution. Before we calculate $g(a,b),$ let's predict several of its properties:

1. Symmetry: $g(a,b)=g(b,a).$ (Either direction of transit in the hallway is the same problem)

2. Boundary values: $g(0,w)=g(w,0)=w.$

3. Homogeneity: for any $t>0,\,g(ta,tb)=tg(a,b).$ We'll be able to verify this from looking at our formula by thinking about units of measurement: if $a$ and $b$ are both measured in feet, then $g(a,b)$ will also have the units of feet.

4. Special midline (45 degree) case: $g(a,a)=(2\sqrt{2})a.$


if we know  7/(x+y) - 5/(x-y)= 11
and 4/(x+y) - 3/(x-y)=7  would x be 1/24 and y be -24/7?

pls and thanks =]

No, you're not right. I plugged the numbers back into the equations and they definitely don't work.

[hide="Hint"]
Try setting $\frac{1}{x+y}= a$ and $\frac{1}{x-y}= b$. Solve for $a$ and $b$, and you'll then have another system in terms of $x$ and $y$.[/hide]


What is the greatest prime factor of the sum of the arithmetic
sequence $ 1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 80$ ?

The sum of the first $ n$ numbers is $ \frac{n(n\plus{}1)}{2}$(see if you can prove it!), so $ 1 \plus{} 2 \plus{} 3 \plus{} \dots \plus{} 80 \equal{} \frac{80(81)}{2} \equal{} 3240$. The prime factorization of this number is $ 2^3 \cdot 3^4 \cdot 5$, so the greatest prime factor is $ \boxed{5}$.

[quote="AIME15"]Actually, instead of multiplying it out, we can see that $ \frac{80(81)}{2}\equal{}40 \times 81$.  Obviously, $ 3$ is the only prime factor of $ 81$, and $ 40\equal{}8 \cdot 5$, so the answer is $ \boxed{5}$.[/quote]
Just in case you need clarification, 40 = 5*8, which = 5*2^3 so the total prime factorization would be 2^3*3^4*5  :cool:
And...I think AIME15's idea is the quickest way to solve this problem


[quote="Arvind_sn"]Billy Bob plays basketball with his buddies a lot. 49% of the time, Billy Bob shoots a 3-pointer. 24% of the time, he shoots a 2-pointer. The rest of the time, he misses. What is the average value of a Billy Bob shot? Explain.[/quote]
[hide]
$(.49)(3)+(.24)(2)=\boxed{1.95}$[/hide]

[hide]It's just expected value. 0.49(3)+0.24(2)=1.95[/hide]

[quote="kheg_k"]I don t think u guys r right

[hide]$\frac{0.49 \times 3+0.24\times2}{2}=0.975$  [/hide][/quote]
Actually, when you calculate expected value, you just multiply each score by the probability of that score, and then you add them all up to get the total (no dividing after that).
For example, the average value of something with a $50\%$ chance of $0$ and a $50\%$ chance of $1$ would be $.50\times 0+.50\times 1=\frac{1}{2}$ and not $\frac{1}{4}$.

[quote="kheg_k"]I don t think u guys r right

[hide]$\frac{0.49 \times 3+0.24\times2}{2}=0.975$  [/hide][/quote]

but why would you divide by 2?

I don't have a calculator or piece of paper on me at the moment, but the answer should just be .49 *3 + .24*2 :pilot:

[quote="kheg_k"]but since there are two shots, wouldn t u divide by two?

Remember, he misses the shot the rest of the time

Aravind_SN asked for average points per shot.[/quote]
No, you don't divide by $2$. The answer is
[hide]$0.49\times 3+0.24\times 2$ since you multiply each score by its probability and add them all up. For example, if you have a half chance of scoring $1$ and a half chance of scoring $0$ your expected value would be $0.5\times 1+0.5\times 0=0.5$ and not $0.25$.
Another way to look at it is that out of $100$ tries, he scores $3$ points $49$ times, $2$ points $24$ times, and $0$ the rest of the time, so his expected value is
$\frac{49\times 3+24\times 2}{100}=1.95$.[/hide]

Just try substituting values if he shoots 100 shots. You'll see

[quote="kheg_k"][quote="nebula42"][quote="kheg_k"]but since there are two shots, wouldn t u divide by two?

Remember, he misses the shot the rest of the time

Aravind_SN asked for average points per shot.[/quote]
No, you don't divide by $2$. The answer is
[hide]$0.49\times 3+0.24\times 2$ since you multiply each score by its probability and add them all up. For example, if you have a half chance of scoring $1$ and a half chance of scoring $0$ your expected value would be $0.5\times 1+0.5\times 0=0.5$ and not $0.25$.
[color=red]Another way to look at it is that out of [/color]$100$ [color=red]tries, he scores [/color]$3$ [color=red]points[/color] $49$ [color=red]times,[/color] $2$ [color=red]points[/color] $24$ [color=red]times, and [/color]$0$ [color=red]the rest of the time, so his expected value is[/color]$\frac{49\times 3+24\times 2}{100}=1.95$.[/hide][/quote]

But if he [b]does[/b] try hundred times, then he'll try a three-pointer 50 times and a two-pointer 50 times.[/quote]

Why?
All we need to know is how often he scores, and from where.
In the 100 shots, 49 of them will score 3 points, and 24 will score 2, 27 will score 0. So in 100 points he'll score 195 points, and dividing 195 by 100, we get 1.95

[u][i][b]edit is at bottom[/b][/i][/u]

OH! I just got something ( it s not about 100 shots

Aravind_SN said:

Billy Bob plays basketball with his buddies a lot. [color=red]49% of the time, Billy Bob shoots a 3-pointer. 24% of the time, he shoots a 2-pointer.[/color] The rest of the time, he misses. What is the average value of a Billy Bob shot? Explain.

i thought that when Aravind_SN said '49% of the time, Billy Bob shoots a 3-pointer. 24% of the time, he shoots a 2-pointer.', he meant that he makes a three pointer 49% of the time and misses 59% of the time. i also tought he makes 24% of 2-pointers and misses 76% of the time.

It s obviously very hard to make [b]every[/b] two and three pointer [b]and[/b] to miss every free throw. That s why i think i am right.

EDIT: oops i misunderstood question.  i got it now.

I'll admit the wording isn't that great but I don't Where did you get 59% he misses a three-pointer?

[quote="kheg_k"][color=red]Aravind_SN said:

Billy Bob plays basketball with his buddies a lot. [color=orange]49% of the time, Billy Bob shoots a 3-pointer.[/color] 24% of the time, he shoots a 2-pointer. The rest of the time, he misses. What is the average value of a Billy Bob shot? Explain. [/color]

When he said the thing in orange, i took it as, "when shooting the three pointer, he makes the three pointer 49% of the time.(therefore missing it 51% of the time) and when shooting two pointers, he misses it 76% of the time[/quote]
But what  does that have to do with the problem (and your answer)?  :huh:

it has nothing to do with the actual answer, but my answer is based on that because i assumed that he took as many 3's as 2's and there were other flaws.

Never mind. it was a mental mistake


How many 12 letter strings are there that contain at least 3 vowels (a vowel is one of {a,e,i,o,u})?

How many bit strings of length 14 contain 7 consecutive 1’s or 7 consecutive 0’s?

How many bit strings are there of length at least 7 and at most 15?

How many bit strings contain exactly 7 zeros and 15 ones if every zero must be immediately followed by two ones?

i havent figured out how to insert images in this teaxt box so ive added one attachment.

For information on how to make the math symbols on the forum, visit [[LaTeX]] or http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690


when i cube that on a calculator i get: 2.9240698861992501093837578111111+e38
at the end there are alot of 1's so i think your answer is correct :lol: [/hide]

[quote]Did you really spend your whole time doing that by hand[/quote]
Okay “pencil and paper” (and use of dog’s paws) is a little exaggeration. :)  I used  the following[hide="Rexx program"]/*Rexx pgm - cube ends in 1985 1s */
numeric digits(6000) /* calculation is correct for 6000 digits */
[b]u = 1                            /* start with 1 */
do k=1 to 1984
v=((u**3)%(10**k))//10
u =(((11-v)*7)//10)||u  /* get the next digit */
end
say  u[/b][/hide]
using the pattern which was mentioned in the previous ..
[hide="post"].[quote].But pattern is easy to see .. 1^3 ends in 1 1s, 71^3 ends in 2 1s
471^3 ends in 3 1s ... just try to think what for what x the number X471^3 will end in 4 1s...
[/quote]
[/hide]
[hide="For example "]  471^3 =  10448[b][size=150]7[/size][/b]111
Next digit  would be $((11-7)\cdot 7 \bmod 10)  $ or $8$so you have $8471$ as the next number which will give 4 $1s$
when you take a cube of it.[/hide]

Hope that helps.

Added later: BTW  the same problem is   in pre-olympiad section too..  and  is solved there too. - See the post below!

There's now a proof of such a number's existance in the Pre-Olympiad forum.
Oh and Mathematica is great for such large calculations.

Gyan, what's this pattern you're talking about with the number of 1's?

[quote="PenguinIntegral"][quote="anirudh"][quote="PenguinIntegral"]Can the cube of a natural number end in 1985 1's?[/quote]

yes,
but i don't really understand why[/quote]
Then how do you know it does?

And I posted the problem in the pre-olympiad fourm because I wasn't getting any answers :roll:[/quote]

I know it, because the person who wrote the answer must know that it is true because if the answer was no, then it would take a lot of time for other people to solve it.

[quote="anirudh"][/quote][quote="PenguinIntegral"][quote="anirudh"][quote="PenguinIntegral"]Can the cube of a natural number end in 1985 1's?[/quote]

yes,
but i don't really understand why[/quote]
Then how do you know it does?

And I posted the problem in the pre-olympiad fourm because I wasn't getting any answers :roll:[/quote][quote="anirudh"]

I know it, because the person who wrote the answer must know that it is true because if the answer was no, then it would take a lot of time for other people to solve it.[/quote]
You could prove it false if it was false, in a relativaly short time.


[quote="k00lperson"]The point (a,b) on a Cartesian plane is reflected over the y-axis to the point at (j,k). If a +j = 0, what is the value of b? :ddr:[/quote]

Because its reflected over the y-axis, a+j would have to equal 0.  This gives us no clue for b, so unless im wrong (probably), the answer should be the set of all real numbers

umm ive had this problem before...i think ur missing one equation..

this would have to be the case, usually the answers to graphing problems aren't R

Um. Refer to this:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1285780998&t=237026

You are missing an equation.

I typed the whole problem up???

I do not understand how to do this.
The answer key says that b is 0?????
so it is not the set of natural numbers

As you wrote it, the answer is $ \mathcal{R}$. However, with the additional restriction that $ b\plus{}k\equal{}0$, the answer is $ 0$.

Note: not sure how to make the "R" the proper way. Sorry! :blush:


[hide="w00t"]
Since the secretary is and only child and Gerald has a brother than Gerald can not be the secratary.  Rose can not be the director because she makes more than the director.  So we now that Gerald is not the secratary and rose is not the director.  Does the director make more than the secratary?  If so then Rose is the chairman.  Gerald is the director and Anita is the secratary.  I think i solved it but i made an assumption that may be wrong.[/hide]

there is too little info. this problem cannot be solved!

[quote=given]Gerald,Anita,and Rose[/quote]

are they supposed to be related somehow and why are there more than 3 people if there are only 3 positions?

[quote="colts18"]There are certain positions at a buisness. Director,Chairman,and Secretary.
THe people are:Gerald,Anita,and Rose. The secratary is an only child. Rose makes more than the director and Rose's fiance is Gerald's brother.Who is what?? :stretcher:[/quote]

  From what is given we can deduce, Rose is not the director and Gerald is not the secretary.  However, this is insufficient to solve the problem.  Unless one would assume that out of the two positions chairman, and secretary, only the chairman would make more than the director; then, Rose would be the chairman Gerald the director, and Anita the secretary.

[quote="math92"][quote="236factorial"]math92, you can't assume the director makes more than the secretary (although that's usually the case. )[/quote]
well i was just assuming that becuase i thought there was an answer but i couldn't deduce it, so i made a guess. :)[/quote]

 inductive reasoning!

You can't solve anything unless you make an assumption.


The numbers 1447, 1005, and 1231 have something in common: each is a four-digit number beginning with 1 that has exactly two identical digits.  How many such numbers are there?

[hide="Solution"]Suppose the two identical digits are both one. Since the thousands digits must be one, the other one can be in only one of three digits, \[ 11xy,\qquad 1x1y,\qquad1xy1.  \] Because the number must have exactly two identical digits, $x\neq y$, $x\neq1$, and $y\neq1$. Hence, there are $3\cdot9\cdot8=216$ numbers of this form.

Suppose the two identical digits are not one. Therefore, consider the following possibilities, \[ 1xxy,\qquad1xyx,\qquad1yxx.  \] Again, $x\neq y$, $x\neq 1$, and $y\neq 1$. There are $3\cdot9\cdot8=216$ numbers of this form too.

Thus, the desired answer is $216+216=\boxed{432}$.[/hide]

[hide="Solution"] We start with the case where there are two $1$'s and the other two digits are do not repeat.  Letting $x$ denote the two mystery digits, there are $9*8=72$ mystery digits that can fill in the spaces.

However, there are $3$ ways to arrange the $1$'s, so we get $216$ possibilities here.

Next, we count the number of numbers where $1$ doesn't repeat.  Similarly, we once again get $216$ numbers.

In total, our answer is $432$.[/hide]

[hide="Answer"]We have the following possibilities:

$11ab$
$1a1b$
$1ab1$
$1aab$
$1aba$
$1baa$

Essentially, we must have three distinct digits, and we can either have the $1$ doubled or one of the other two doubled. For each of these six possibilities, we have $9\cdot 8=72$ choices for $(a,b)$ (every digit but $1$, but $a$, $b$ distinct), so our answer is $6\cdot 72=\boxed{432}$.[/hide]

[quote="4everwise"]The numbers $ 1447$, $ 1005$, and $ 1231$ have something in common. Each is a four-digit number beginning with $ 1$ that has [b]exactly[/b] two identical digits. How many such numbers are there?[/quote]

well i understood the "exactly" as in exaclty two identical digits, meaning not three identical ones or four identical ones...only two identical ones.

I never understood the exactly as something that points out exactly only one pair of identical ones..

[quote="vikas"][quote="4everwise"]The numbers $ 1447$, $ 1005$, and $ 1231$ have something in common. Each is a four-digit number beginning with $ 1$ that has exactly two identical digits. How many such numbers are there?[/quote]
I know this is an old topic but I have a question about it.

When taht question is phrased something like that does it mean some number like 1122 woulndt work becauses doesnt that have two identical digits (although looking at solutions no one considered that a number that would work)?
However it seems to me, it does have two identical digits just two pairs of them and the question never restricted that...?[/quote]
Since it has 2 identical digits they are only asking for one pair of identical digits, otherwise they would've said 2 pairs of identical digits.

In the future, please do not revive threads that are 3 years old.

[color=red][size=80]Since this thread is linked in the contest page, it's ok.[/size][/color]

[quote="CRICKET229"]In the future, please do not revive threads that are 3 years old.[/quote]
Where else should I answer this question?

[hide=Solution]
Consider a sequence of four digits instead of a four digit number. Only looking at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is $\frac{1}{10}$. This means we can find all possible sequences with one digit repeated twice, and then divide by $10$.

If we let the three distinct digits of the sequence be $a, b,$ and $c,$ with $a$ repeated twice, we can make a table with all possible sequences:

\begin{tabular}{ccc}
aabc & abac & abca \\
baac & baca & \\
bcaa && \\
\end{tabular}

There are $6$ possible sequences.

Next, we can see how many ways we can pick $a, b,$ and $c$. This is $10(9)(8) = 720$ because there are $10$ digits, and we need to choose $3$ with regard to order. This means there are $720(9) = 4320$ sequences of length $4$ with one digit repeated. We divide by $10$ to get $\boxed{432}$ as our answer. [/hide]

[hide=Solution]We take cases. The possibilities are 1 1 _ _, 1 _ 1 _, 1 _ _ 1, 1 a a _, 1 a _ a, 1 _ a a. The first case can be done in $9 \cdot 8 = 72$ ways, the second in $72$ ways, the third in $72$ ways, the fourth in $9 \cdot 1 \cdot 8 = 72$ ways, the fifth in $72$ ways, and the sixth in $72$ ways. Therefore, the answer is $72 \cdot 6 = \boxed{432}$.[/hide]

better solution: If the repeated digit is $1$, there are $\binom{3}{1} \cdot 9 \cdot 8=216$ ways. If the repeated digit is not $1$, there are $\binom{3}{2} \cdot 9 \cdot 8 = 216$ ways again. Summing gives $\boxed{432}$.

If we have the repeated digit is $1$, we have $3\cdot 9 \cdot 8=216$ ways. If it is not $1$ we still have $216$ ways, adding gives us $\boxed{432}.$

[hide=10]This is just a simple application of casework. If the repeated digit is $1$, then there are $\binom{9}{2}$ to choose the other two unique digits and $6$ ways to arrange the last three digits in the four digit number. If the repeated digit is not $1$, there are $9$ possibilities for it and $8$ for the other digit. Then, there are $3$ ways to arrange these digits. So there are $216+216=432$ such numbers.

Key idea: Casework based on repeated digit[/hide]