number theory
Diophantine equation
Miscellaneous Problems

a) Two positive integers are chosen. The sum is revealed to logician $A$, and the sum of squares is revealed to logician $B$. Both $A$ and $B$  are given this information and the information contained in this sentence. The conversation between $A$ and $B$ goes as follows: $B$ starts

B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` Now I can tell what they are.'

What are the two numbers?

b) When $B$ first says that he cannot tell what the two numbers are, $A$ receives a large amount of information. But when $A$ first says that he cannot tell what the two numbers are, $B$ already knows that $A$ cannot tell what the two numbers are. What good does it do $B$ to listen to $A$?

A similar problem was considered in Kvant journal in 1977 N 3 (in Russian):
http://kvant.mccme.ru/1977/03/mnogo_bitov_iz_nichego.htm
In that problem the logicians A and B are given respectively the sum and the product of two unknown positive numbers whose sum is below 100, and their dialog goes as follows:
A: I cannot determine what the numbers are.
B: I knew that you cannot determine them.
A: Then I know these numbers.
B: Then I know them as well.

This problem is intriguing!

Number the sentences in conversation:
  <<1>>  B: I can't tell what the numbers are.
  <<2>>  A: I can't tell what the numbers are.
  <<3>>  B: I can't tell what the numbers are.
  <<4>>  A: I can't tell what the numbers are.
  <<5>>  B: I can't tell what the numbers are.
  <<6>>  A: I can't tell what the numbers are.
  <<7>>  B: Now I can tell what the numbers are.

Denote the sum S and sum of squares N. S-list will be a list of all possible sums a + b, N-list a list of all possible sums of squares a^2 + b^2. Initially, the S-list contains all positive integers >1, N-list all positive integers expressible as sum of two positive integer squares.

The section <<i>> below describes the information gained after the line <<i>> in the conversation was spoken.

<<1>> N is not expressible as the sum of positive integer squares uniquely. N-list is correspondingly adjusted (numbers expressible as sum of two positive integer squares uniquely are thrown out). S-list is adjusted – sums are thrown out, that don’t allow for summands a, b, such that a^2 + b^2 is on the current N-list.

<<2>> S allows for more values of a, b (a + b = S), such that a^2 + b^2 belongs to the N-list. S-list is adjusted (sums are deleted not allowing for more values of a, b (a + b = the sum), such that a^2 + b^2 belongs to the N-list).

<<3>> N allows for more values of a, b (a^2 + b^2 = N), such that a + b is on the S-list. Those integers that don’t are deleted from the N-list.

<<4>> S allows for more values of a, b, such that a^2 + b^2 belongs to the N-list. Those integers that don’t are deleted from the S-list.

<<5>> N allows for more values of a, b, such that a + b belongs to the S-list. Those integers that don’t are deleted from the N-list.

<<6>> S allows for more values of a, b, such that a^2 + b^2 belongs to the N-list. Those integers that don’t are deleted from the S-list.

<<7>> At this moment there is only one pair of positive integers a, b, such that a + b is on the S-list and a^2 + b^2 = N. If N-list contains more values, they need to be checked to see if they allow for unique solution.

If some bounds are imposed on a,b, I guess the solution could be found by computer or manual deleting from the S- and N-list. I haven't really thought beyond this outline yet, at this moment solving the problem in full generality seems difficult... I'd love to see the intended solution.

[quote="Peter"]a) Two positive integers are chosen. The sum is revealed to logician $A$, and the sum of squares is revealed to logician $B$. Both $A$ and $B$  are given this information and the information contained in this sentence. The conversation between $A$ and $B$ goes as follows: $B$ starts

B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` I can't tell what they are.'
A: ` I can't tell what they are.'
B: ` Now I can tell what they are.'

What are the two numbers?
[/quote]

Let $A_i(s)$ be the set of candidate pairs of integers that $A$ can derive after the $i$-th claim of $B$, assuming that $A$ is given the sum $s$.
Similarly, let $B_i(t)$ be the set of candidate pairs of integers that $B$ can derive after the $i$-th claim of $A$, assuming that $B$ is given the sum of squares $t$.

Clearly, we have
\[A_0(s) = \{ [x,y]\,:\,1\leq x\leq y,\,x+y=s \}\]
\[B_0(t) = \{ [x,y]\,:\,1\leq x\leq y,\,x^2+y^2=t \}\]

The $i$-th claim [i]B: ` I can't tell what they are.'[/i] basically means that $|B_{i-1}(t)|>1$, while the $i$-th claim [i]A: ` I can't tell what they are.'[/i] means that $|A_i(s)|>1$.

Therefore, we have the following recurrence:
\[A_{i+1}(s) = \{ [x,y]\in A_i(s)\,:\,|B_i(x^2+y^2)| > 1\}\]
\[B_{i+1}(t) = \{ [x,y]\in B_i(t)\,:\,|A_{i+1}(x+y)|> 1\}\]

The whole dialog means that in the sequences
\[A_0(s)\supset A_1(s) \supset A_2(s) \supset A_3(s)\]
\[B_0(t)\supset B_1(t)\supset B_2(t)\supset B_3(t)\]
all sets contain more than one element, except for $B_3(t)$ which contains a single element.

So far, I've found only one such pair of integers $[8,9]$, which results in the following sequence of sets:
\[A_0(8+9) = \{ [1, 16], [2, 15], [3, 14], [4, 13], [5, 12], [6, 11], [7, 10], [8, 9] \}\]
\[A_1(8+9) = A_2(8+9) = A_3(8+9) = \{ [3, 14], [4, 13], [8, 9] \}\]
\[B_0(8^2+9^2) = B_1(8^2+9^2) = B_2(8^2+9^2) = \{ [1, 12], [8, 9] \}\]
\[B_3(8^2+9^2) = \{ [8, 9] \}\]

It remains unclear how to prove that there are no other suitable pairs.

Below is my attempt to prove that $[8,9]$ is the only solution to the original problem. My proof still relies on a certain unproved statement which, however, sounds plausible from heuristic perspective and can be studied independently of the original problem.

[quote="maxal"]Therefore, we have the following recurrence:
\[A_{i+1}(s) = \{ [x,y]\in A_i(s)\,:\,|B_i(x^2+y^2)| > 1\}\]
\[B_{i+1}(t) = \{ [x,y]\in B_i(t)\,:\,|A_{i+1}(x+y)|> 1\}\]
[/quote]

These recurrences can be given a different interpretation: a pair $[x,x']$ from $A_i(x+x')$ belongs to $A_{i+1}(x+x')$ iff there exists a pair $[y,y']\ne [x,x']$ such that $[y,y']\in B_i(x^2+x'^2)$ (in particular, $x^2+x'^2=y^2+y'^2$).
Similarly, a pair $[x,x']$ from $B_i(x^2+x'^2)$ belongs to $B_{i+1}(x^2+x'^2)$ iff there exists a pair $[y,y']\ne [x,x']$ such that $[y,y']\in A_{i+1}(x+x')$ (in particular, $x+x'=y+y'$).
Call such a pair $[y,y']$ a [i]certificate[/i] for the membership $[x,x']\in A_{i+1}(x+x')$ or $[x,x']\in B_{i+1}(x^2+x'^2)$, respectively.

For the only element $[b_3,b'_3]\in B_3(t)$, there exists a certificate $[a_3,a'_3]\in A_3(b_3+b_3')$, for which there exists a certificate $[b_2,b'_2]\in B_2(a_3^2+a'_3^2)$, for which there exists a certificate $[a_2,a'_2]\in A_2(b_2+b'_2)$, and so on, ending up with the certificate $[b_0,b'_0]\in B_0(a_1^2+a'_1^2)$. That is, we have a chain of certificates:
\[[b_3,b'_3]\rightarrow [a_3,a'_3] \Rightarrow [b_2,b'_2] \rightarrow [a_2,a'_2] \Rightarrow [b_1,b'_1] \rightarrow [a_1,a'_1] \Rightarrow [b_0,b'_0]\]
where $\rightarrow$ connect pairs with equal sums, $\Rightarrow$ connect pairs with equal sums of squares, and no two adjacent pairs are equal.

Since $|B_2(t)|>1$, there exists an element $[c_2,c'_2]\in B_2(t)$ such that $[c_2,c_2']\ne [b_3,b'_3]$ with its own chain of certificates:
\[[c_2,c'_2]\rightarrow [d_2,d'_2] \Rightarrow [c_1,c'_1] \rightarrow [d_1,d'_1] \Rightarrow [c_0,c'_0]\]
but since $[c_2,c'_2]\not\in B_3(t)$, it is impossible to extend this chain of certificates into
\[[c_2,c'_2]\rightarrow [d_2,d'_2] \Rightarrow [c_1,c'_1] \rightarrow [d_1,d'_1] \Rightarrow [c_0,c'_0] \rightarrow [j,j'] \Rightarrow [k,k'].\]
In other words, this means that for every representation of the number $m=c_0+c'_0$ as the sum of two positive integers $m=j+j'$, $j\leq j'$, the number $j^2+j'^2$ has an unique unordered representation as the sum of squares of two positive integers, with exacty one exception $[j,j']=[c_0,c'_0]$ (since $c_0^2+c'_0^2=d_1^2+d'_1^2$).

Here comes the unproved statement that such integers $m$ (I've exhaustively tested all $m\leq 10^5$) are limited to the set:
\[ \{ 8, 9, 10, 15 \}\]
with the exceptional pairs $[c_0,c'_0] = [1,7], [1,8], [5,5], [5,10]$, respectively. They correspond to $[d_1,d'_1] = [5,5], [4,7], [1,7], [2,11]$, respectively. So, we can unroll back the possible elements in the certificate chain for $[c_2,c'_2]\in B_2(t)$:
\[[d_1,d'_1] \in \{ [5,5], [4,7], [1,7], [2,11] \}\]
\[[c_1,c'_1] \in \{ [1, 10], [1, 12], [1, 9], [2, 6], [2, 8], [2, 9], [3, 10], [3, 5], [3, 7], [3, 8], [4, 4], [4, 6], [4, 9], [5, 6], [5, 8], [6, 7] \}\]
\[[d_2,d'_2] \in \{ [2, 9], [6, 7], [8, 9] \}\]
\[[c_2,c'_2] \in \{ [1, 10], [1, 12], [1, 16], [2, 11], [2, 15], [3, 10], [3, 14], [3, 8], [4, 13], [4, 7], [4, 9], [5, 12], [5, 6], [5, 8], [6, 11], [7, 10] \} \]
and since $[c_2,c'_2]$ and $[b_3,b'_3]$ both belong to $B_2(t)$, we have $b_3^2 + b_3'^2 = c_2^2 + c'_2^2$ and thus
\[[b_3,b'_3] \in \{ [1, 8], [5, 10], [6, 13], [8, 11], [8, 9] \}. \]
However, it is easy to check that
\[B_3(1^2+8^2) = B_3(5^2+10^2) = \emptyset\]
\[B_3(6^2+13^2) = \{ [3,14], [6,13] \}\]
\[B_3(8^2+11^2) = \{ [4,13], [8,11] \}\]
\[B_3(8^2+9^2) = \{ [8,9] \}\]
out of which only the last satisfies the requirement $|B_3(t)|=1$. This consideration unambiguously derives the pair $[8,9]$ as the solution (under the assumption of yet unproved claim mentioned above).

I thought of this way:

Cant=0 in binary and can=1 therfore the code becomes 1000000=8*8 and from the opposit side =0000001=1 there fore no,s are 0,8

a)
Let's define a [url=http://en.wikipedia.org/wiki/Binary_relation]binary relation[/url] on the natural numbers as $a\rightarrow b$ iff there exists natural numbers $u,v$ with $a =u+v$ and $b = u^2 + v^2$. We'll use the notations $a\rightarrow b$ and $b\leftarrow a$ interchangeably. Note that every un-ordered pair of natural numbers $\{u,v\}$ can be uniquely identified by another ordered pair $(a,b)$ of natural numbers satisfying $a\rightarrow b$. In other words, the relation $a\rightarrow b$ is asserting that the pair of information available to the logicians is consistent. We will now generalize this notation with a parameter. Let $a_{\overrightarrow{k}} b$ denote that $(a,b)$ is consistent given that the logicians have exchanged their "I can't tell" assertions (at least) $k$ times (starting with B). The case $k=0$ is the special case above, the state before any assertions were made, i.e. $a\rightarrow b \iff a_{\overrightarrow{0}} b$. For example $3_{\overrightarrow{0}} 5$, but $3\not _{\overrightarrow{1}} 5$ since 5 can be written as sum of squares in an unique way and B would know the answer on his first question if this pair was indeed the chosen one. On the other hand $9_{\overrightarrow{1}} 65$ is true since B wouldn't be able to distinguish between the initial cases $9\rightarrow 65$ and $11\rightarrow 65$. However $9_{\overrightarrow{2}} 65$ is false because of all the ways of writing $9$ as sum of natural numbers, only $(8,1)$ admits a relation of the form $8^2+1^2 = 7^2 + 4^2$. Hence $(9,65)$ cannot be the initial pieces of information if the logicians exchange "I can't tell" twice.

Note that, we can express the above reasoning in the following assertions:
\[\forall \text{ even } k, \ a_{\overrightarrow{k+1}} b\iff \exists a' \neq a, a_{\overrightarrow{k}} b _{\overleftarrow{k}} a'\]\[\forall \text{ odd } k, \ a_{\overrightarrow{k+1}} b\iff \exists b' \neq b, b_{\overleftarrow{k}} a_{\overrightarrow{k}} b' \]If we expand the above recurrence one step, we obtain the following:
\[\forall \text{ even } k, \ a_{\overrightarrow{k+2}} b\iff \exists a_1\neq a, b_1\neq b\neq b_2, {b_1}_{\overleftarrow{k}}a_{\overrightarrow{k}} b _{\overleftarrow{k}} {a_1}_{\overrightarrow{k}}b_2\]
similarly for the odd case, and so on if we keep expanding. Now there is no requirement for $b_1$ to be distinct from $b_2$ in the above expression to hold true. If $b_1 = b_2$, then the path can be extended in both direction arbitrarily, and the relation $a_{\overrightarrow{k}}b$ holds true for all k, so is $a'_{\overrightarrow{k}}b$ and $a_{\overrightarrow{k}}b'$. in other words, the logicians would never find out the original numbers after arbitrary many assertions of ignorance! We state this case slightly more formally by introducing another notation. We define a symmetric relation $a\leftrightarrow c$ to mean that there exists $b$ such that $a\rightarrow b \leftarrow c$. This symmetric relation defines an undirected graph $G$ on the natural numbers.

Let us define a cycle on this undirected graph to mean a sequence $v_1\leftrightarrow v_2\leftrightarrow \cdots v_n \leftrightarrow v_1$ such that $v_i\neq v_{i+2}$ for all $i$ modulo $n$. The condition that "the input $a\rightarrow b$ is not resolvable within any finite amount of exchange between the logicians" is equivalent to the characterization that $a\leftrightarrow c$ is part of a cycle on $G$, for some $c$ satisfying $b\leftarrow c$. Hence we will concentrate on the relation $\leftrightarrow $ from this point. It surprisingly turns out that graph $G$ is sufficiently well connected and only finitely many of its vertices are not part of a cycle. Showing that almost all vertices of it is part of some cycle implies that every edge between some two of those vertices is also part of some cycle (by stitching together the two cycles by a bridge).

Right now we use the famous sum of squares identity
\[(ac+bd)^2+(ad - bc)^2 = (ac - bd)^2 + (ad + bc)^2\]
The unrdered pairs $\{ac+bd, ad - bc\},\{ac - bd, ad + bc\}$ are distinct iff $a,b,c,d > 0, a/b > c/d > 1$ (WLOG). Hence for all such quadruples $(a,b,c,d)$ there exists an edge between the following two natural numbers
\[a(c+d) + b(c-d) \leftrightarrow a(c+d) - b(c-d)\cdots (1)\]
This also gives a complete description of the graph since the identity generates all solutions to the sum of squares diophantine equation (well known). See end of the solution to get a computer generated list-of-neighbors representation of the graph.

It's easy to observe that odd numbers have only odd neighbors and even numbers have only even neighbors, and $x\leftrightarrow y\implies 2x \leftrightarrow 2y$. So It is sufficient to show that all but finitely many odd numbers are part of a cycle, and some power of 2 is part of a cycle. It's not hard to observe that $32\leftrightarrow 34 \leftrightarrow 38\leftrightarrow 32$ is a cycle. eg. $(a,b,c,d) = (3,1,6,5)\implies 32\leftrightarrow 34 $, similarly, we have $(4,2,10,8), (5,3,4,3)$ proving the other two edges. So it remains to prove that all but finitely many odd numbers are part of some cycle.

If we plug in $c = d +1, b = 1$ in statement $(1)$, we observe that $a(2d+1)\pm 1$ is connected for all $a>1$ except the case $(a,d)=(2,1)$ (for which we have $a/b = c/d$, which is not allowed). Note that every even integer other than $6$ or powers of $2$, can be written as $a(2d+1)$ satisfying $a>1$ and $(a,d)\neq (2,1)$. Hence $2n-1\leftrightarrow 2n+1$ for all positive integers $n$ except for $n = 1, 2, 3, 4, 8, 16, \cdots $.

We also have the fact that $2^n-1\leftrightarrow 2^n+5$ proven by the quadruplet $(2, 1, 2^{n-2}+2, 2^{n-2}-1)$ for all $n\ge 5$. Similarly $(2, 1, 2^{n-2}+1, 2^{n-2}-2)\implies 2^n+1\leftrightarrow 2^n-5$ for all $n\ge 5$. So we have just proven the existence of cycles
\[2^n-1\leftrightarrow 2^n+5\leftrightarrow 2^n+3\leftrightarrow 2^n+1\leftrightarrow 2^n-5\leftrightarrow 2^n-3\leftrightarrow 2^n-1\]
It is easy to stitch together the paths from $2^n+1$ to $2^{n+1}-1$ with these cycles to show that all odd numbers $\ge 33$ are part of cycles. There's another base cycle $19\leftrightarrow 21\leftrightarrow 23\leftrightarrow 19$ which proves that all odd numbers $\ge 19$ are in fact part of cycles. We know the same holds for even numbers $\ge 38$. We can manually check that it is in fact true for even numbers $\ge 22$.

Coming back to the original problem, lets assume that the initial pair corresponds to $a\rightarrow b$. Since B cannot resolve $b$ on the first try, there exist $c$ such that $a\rightarrow b \leftarrow c$ or $a\leftrightarrow c$. If $a\leftrightarrow c$ is part of some cycle, none of them will ever be able to figure out the numbers. So one of $a,c$ has to be $\le 20$. We only have finitely many possibilities in this case. All edges consisting of one vertex $\le 20$ can be checked. In fact from the fact that $c_{\overrightarrow{6}}b$ and $c\not _{\overrightarrow{7}}b$, the edge $c\rightarrow b$ must be extendable in both direction by exactly $5$ edges, but not more. In the graph $G$, the path is contracted in length by half, and the edge $a\leftrightarrow c$ is extendable in both directions by exactly $2$ additional edges.
\[15\leftrightarrow 13\leftrightarrow 17\leftrightarrow 19\leftrightarrow 21\leftrightarrow \cdots\]
In this case $a = 17, b = 145, c = 19$. None of the other edges in $G$ satisfy this condition.
The pair $17\rightarrow 145$ corresponds to the initial values $(8,9)$.

b)
It's true that B receives no new information from the first answer of A. However, if he didn't listen to A's answer, A wouldn't be able to tell whether B's second answer would have changed if he did listen to A (In this case it wouldn't, but A doesn't know that). The act of listening to A before answering second time gives A more information on his second answer, which wouldn't be the case if he didn't listen.
[code]
#####################
## It is only required to compute by hand upto 20 and
## a few other special cases to solve the problem
#####################
8 [10]
9 [11]
10 []
11 [13]
12 []
13 [15, 17]
14 [16, 18]
15 []
16 [20]
17 [19]
18 [22, 24]
19 [21, 23, 25]
20 [22]
#####################
## All vertices (and edges) belong to cycles below this point
#####################
21 [23, 27]
22 [26, 28]
23 [25, 27, 29, 31]
24 [26, 30, 32]
25 [27, 29, 31]
26 [28, 30, 34]
27 [29, 33]
28 [32, 36, 38]
29 [31, 37, 39]
30 [34, 36]
31 [35, 37, 39, 41]
32 [34, 38, 40]
33 [35, 37, 39, 43, 45]
34 [36, 38, 44, 46]
35 [37, 41, 43, 45]
36 [42, 44, 48]
37 [39, 41, 43, 47]
38 [40, 42, 46, 50, 52]
39 [41, 45, 49, 51, 53]
40 [44, 48, 50][/code]


Let $ x$ and $ y$ be positive real numbers, such that $ x^3 \plus{} y^3 \equal{} 4x^2$. What is the largest possible value of $ x \plus{} y$?

denote $ y \equal{} xk$ for some k positive real , then the condition $ x^3 \plus{} y^3 \equal{} 4x^2$ can be rewrited as $ x \plus{} xk^3 \equal{} 4$ , therefore $ x \equal{} \frac {4}{1 \plus{} k^3}$
we need to find the maximum of
$ P \equal{} x \plus{} xk \equal{} \frac {4(1 \plus{} k)}{1 \plus{} k^3} \equal{} \frac {4}{k^2 \minus{} k \plus{} 1} \equal{} \frac {4}{(k \minus{} \frac {1}{2})^2 \plus{} \frac {3}{4}} \leq \frac {16}{3}$
the equality occurs when $ k \equal{} \frac {1}{2} ; x \equal{} \frac{32}{9} ; y \equal{} {16}{9}$

That's the official solution, but i'd like to know how we know which problems we can substitute $ y \equal{} kx$ for some real number in and which problems we can't?  :)

Look at the conditions on $ x, y$.  If $ x$ can be zero then we need a separate case to handle those values.  We also need to keep track of what the possible signs fo $ k$ are.  When $ x$ and $ y$ are positive reals, we can always write $ y \equal{} kx$ for some positive real $ k$.  As to when this is [i]useful[/i], experience is the only guide :).  (It tends to turn homogenous equations or inequalities into nonhomogenous equations or inequalities with one fewer variable, but it obviously can also be used when we don't start with something homogenous.)

Thanks, I definitely need to look at more problems of this sort for the experience then  :)


Let n be the largest integer that is the product of exactly 3 distinct prime numbers, d, e, and 10d+e where d and e are single digits.  What is the sum of the digits of n?

Enjoy!

Edit: Jesus, why do I keep thinking this is the intermediate forum. :(

I think I'll move this one over to the Intermediate forum.

[hide]if d and e are single digit primes the only choices for them are 2,3,5, or 7  letting d=7 and e=3 that makes. So n=7*3*73=1533. the sum of the digits is 12. [/hide]

And how do you know that that is the largest?


analytic geometry
graphing lines
slope
trigonometry
vector
probability

[quote="ith_power"]A small body is sliding on an inclined plane of angle α. The coefficient of kinetic friction μ is larger than tan α, thus the body set into motion in an arbitrary direction will come to a halt after a while. How does the direction of the body’s velocity immediately before stopping depend on the body’s initial direction and speed?
(Dezs˝o Varga)[/quote]
Let $ v_1 \hat i \plus{} v_2 \hat j$ be the velocity of the body with the plane of the slope being the $ x \minus{} y$ plane (see figure, top view),
The retardation due to $ \mu mg \cos \alpha$ will affect magnitudes of $ v_1$ and $ v_2$ but not their directions. $ g \sin \alpha$ will act downwards on $ v_1$ changing its direction but not its magnitude. $ g \sin \alpha$ will affect $ v_2$ but in general, its direction will be parallel to $ \hat j$.
Now, the final direction will be that of either of $ v_1$ or $ v_2$, whichever becomes zero later.
The answer will be a nasty equation but I believe it is do-able.

[quote="Rajiv"]
Now, the final direction will be that of either of $ v_1$ or $ v_2$, whichever becomes zero later.
The answer will be a nasty equation but I believe it is do-able.[/quote]
how about the case when both become zero simultaneously

[quote="ith_power"]how about the case when both become zero simultaneously[/quote]
What you can do in that case is find $ s_x$ and $ s_y$, the displacements, and the direction is $ tan^{ \minus{} 1} \left( \frac {s_y}{s_x}\right)$ with the horizontal.

[quote="ith_power"]i don't get it :(  [b]parallel[/b] what?? :huh:  :maybe:[/quote]
Sorry not parallel. I've edited the post. See it.

[quote="Rajiv"][quote="ith_power"]how about the case when both become zero simultaneously[/quote]
What you can do in that case is find $ s_x$ and $ s_y$, the displacements, and the direction is $ |tan^{ \minus{} 1} \left( \frac {s_y}{s_x}\right)$ with the horizontal.[/quote]

thats very trivial but how to do that??
[hide="my approach"]
get differential equations of v1 and v2.
now as the body tends to stop, v1/v2=$ \frac {dv_1}{dv_2}\equal{}\frac{a_1}{a_2}$
a=acceleration[/hide]

[quote="Rajiv"][quote="ith_power"]

[quote="I"]
What you can do in that case is find $ s_x$ and $ s_y$, the displacements, and the direction is $ |tan^{ \minus{} 1} \left( \frac {s_y}{s_x}\right)$ with the horizontal.[/quote]

thats very trivial but how to do that??

[/quote]
That is simple. Simce $ t_1 \equal{} t_2$, we can use $ s \equal{} ut \plus{} 0.5at^2$.

For $ v_1$, $ a \equal{} \mu mg \cos \alpha$.
For $ v_2$, $ a \equal{} g \sin \alpha \plus{} \mu mg \cos \alpha$.
Remember that the particle stops moving along $ \hat j$ once it stops as $ \tan \alpha < \mu$. Hence, if the particle is set into motion upwards, it shall not descend again.[/quote]
its not that simple.

direction of friction will be changing always. :wink:  :wink: its not upwards or down wards, its backwards to the direction of velocity.

[quote="ith_power"][quote="Rajiv"]Since we have resolved the velocity into components, we need not worry about directions. And as I said earlier, the friction will not change the direction of the velocity (it will only affect the direction in which the velocity vector points finally, an issue which has been solved earlier in this thread) but will only change magnitude of the velocity.[/quote]
but friction will act upon the horizontal component too.
you need to resolve friction force too.[/quote]
The friction force will retard both $ v_1$ and $ v_2$. And I [b]hav[/b]e resolved it. So whichever of $ v_1$ or $ v_2$ becomes zero later is the final direction of teh velocity vector, ie: $ 0$ or $ \frac{\pi}{2}$ with the horiz.
The probability of $ v_2$ reducing to $ 0$ is higher as a component of mg acts on it too.
Howevre, if both become zero simultaneously, $ v_2$ will be pointing up, and $ v_1$ rightwards. the magnitudes can be found to calculate the angle.

At least I feel so!

[quote="Rajiv"]
Howevre, if both become zero simultaneously, $ v_2$ will be pointing up, and $ v_1$ rightwards. the magnitudes can be found to calculate the angle.

[/quote]

now that's the point,the real tough part in the question, isn't it?
one more thing you wrote earlier that if it goes up, it nevrer returns. but suppose while going upwards v2 becomes 0 but v1>0. instantaneously friction changes direction to become backwards to v1. then there is net force mg acting downwards along v2 so v2 will change again.

[quote="I"]The $ \tan$ condition.[/quote]
If a particle were placed on a plane of inclination $ \alpha$, what is the minimum value of $ \alpha$ for which the particle begins to move if the co-efficient of friction is $ \mu$?

If you can answer that, you'll understand why the $ \tan$ condition is provided.

Its wonderful that I see a problem, from the Rudolf Ortvay competition here  :D  :D
If I may ask, how did you guys get to know about the competition?

[quote="Rajiv"]Friction opposes motion. If $ v_2$ becomes zero, there bill be no friction force along $ \minus{} \hat j$.
Imagine if the particle were thrown straight up. It will go to some height, stop.... and stop. Can you appreciate the condition $ \tan \alpha < \mu$?

Now if the particle were thrown horizontally, it will not move down the plane? Why? The $ \tan$ condition.[/quote]
yeah. you are right if it is thrown straight up. but my problem is here. will it stop or move on??

[quote="ith_power"]but my problem is here. will it stop or move on??[/quote]
I dont get your question. Please re-phrase your question.

@rajiv
 see you must give solution for general case, when a particle is thrown at general angle $ \theta$ to vertical upwards direction. then the particle may come back after going to the topmost point of the path.

[quote="Thaakisfox"]Its wonderful that I see a problem, from the Rudolf Ortvay competition here  :D
If I may ask, how did you guys get to know about the competition?[/quote]
just googled for some physics problems. :oops:

[quote="ith_power"]@rajiv
 see you must give solution for general case, when a particle is thrown at general angle $ \theta$ to vertical upwards direction. then the particle may come back after going to the topmost point of the path.[/quote]
I have given the solution for general, when the particle is projected with $ \sqrt {v_1^2 \plus{} v_2^2}$.
I was just trying to explain that the particle can't come down if thrown upwards.

Do you get what I mean???

[quote="I"]
Now if the particle were thrown horizontally, it will not move down the plane
[/quote]

 :roll:

[quote="Rajiv"][quote="ith_power"]@rajiv
 see you must give solution for general case, when a particle is thrown at general angle $ \theta$ to vertical upwards direction. then the particle may come back after going to the topmost point of the path.[/quote]
I have given the solution for general, when the particle is projected with $ \sqrt {v_1^2 \plus{} v_2^2}$.
I was just trying to explain that the particle can't come down is thrown upwards.

do you get what I mean???

 :roll:[/quote]
no francly i dont. the body cant come down if it is thrown[u][b] straight[/b][/u] upwards, but can come down if not thrown vertically straight. :maybe:

[quote="ith_power"]
no francly i dont. the body cant come down if it is thrown[u][b] straight[/b][/u] upwards, but can come down if not thrown vertically straight.[/quote]
Not at all unless given a velocity with a component in $ \minus{} \hat j$ direction.

Look, if the body is moving towards your right, horizontally (top view) the force on it down the plane, pulling it down is $ mg \sin \alpha$ and the force of limiting friction is $ \mu mg \cos \alpha$.
Clearly, $ \mu mg \cos \alpha > mg \sin \alpha$, borrowing from the $ \tan \alpha < \mu$ condition. This condition is the most important part of the problem, although may appear innocuous, neglecting it can make your answer go horribly wrong.

Hope you get me this time!!! :oops:

[quote="Rajiv"]
Look, if the body is moving towards your right, horizontally (top view) the force on it down the plane, pulling it down is $ mg \sin \alpha$ and the force of limiting friction is $ \mu mg \cos \alpha$.
Clearly, $ \mu mg \cos \alpha > mg \sin \alpha$, borrowing from the $ \tan \alpha < \mu$ condition. This condition is the most important part of the problem, although may appear innocuous, neglecting it can make your answer go horribly wrong.

Hope you get me this time!!! :oops:[/quote]

here is where the problem lies. the friction is [b]acting towards left [/b]when body is moving right. so there is a net force downwards.

[quote="ith_power"]here is where the problem lies. the friction is [b]acting towards left [/b]when body is moving right. so there is a net force downwards.[/quote]

What? A force downwards? How is that?
[hide="An explanation:"]If you mean the $ mg \sin \alpha$, then it is comfortably balanced by the $ \mu mg \cos \alpha$.

Look, the force due to gravity down the plane cannot exceed, nor equal $ \mu mg \cos \alpha$.

Watch how:
$ \begin{align*} & \tan \alpha < \mu \\
& \sin \alpha < \mu \cos \alpha \\
& \color{red}{\boxed{(mg) \sin \alpha < (mg) \mu \cos \alpha}}$

$ \cdots$Multiplying by $ (mg)$ on both sides.

Is it clear to you that the component of force due to gravity down the slope [b]cannot[/b] accelerate it down the plane in presence of friction.
[/hide]
Hope my diagram clarifies everything.

[quote="ith_power"]THE PICTURE IS VALID IF AND ONLY IF THE BODY IS MOVING DOWNWARDS IN THE DIRECTION OF $ mgsin\alpha$. IN ALL OTHER CASES FRICTION IS NOT IN THE STRAIGHT LINE(ALONG OR OPPOSITE) $ mgsin\alpha$. IT CANNOT BALANCE $ mgsin\alpha$ IN ANY OTHER CASE.[/quote]
Do you mean that the body can move down the plane even in cases where $ v_2$ is not directed towards $ \minus{}ve Y$ direction???

In my $ 1^{st}$ figure, the top view one, the $ \minus{}ve Y$ direction denotes the direction [b]down[/b] the slope.

[quote="Rajiv"]
Do you mean that the body can move down the plane even in cases where $ v_2$ is not directed towards $ \minus{} ve Y$ direction???

In my $ 1^{st}$ figure, the top view one, the $ \minus{} ve Y$ direction denotes the direction [b]down[/b] the slope.[/quote]
 i mean that the body    [b]"never"[/b]  moves [b]down[/b](in -ve y direction) the plane unless initial $ v_1$ is 0.

[quote="Rajiv"][quote="ith_power"] i mean that the body [b]"never"[/b] moves [b]down[/b](in -ve y direction) the plane unless initial $ v_1$ is 0.[/quote]
Ok, I get what you mean. I didnt think about that :oops: . I suppose I was confused :wallbash: . But what I said earlier, that the body cannot be accelerated downwards once $ v_2$ becomes $ 0$ hold true. Do you agree???[/quote]

sorry but i don't :wink:
 the reason is as i said earlier, v1 may not become zero at that moment .so direction of friction becomes opposite to v1 i.e. horizontal. now the only force changing v2 is mgsin$ \alpha$. and it will produce downward acceleration to change v2. then again the body starts moving downwards or strictly speaking gets a component of velocity in -ve y direction. :)

[quote="Rajiv"][quote="ith_power"]the reason is as i said earlier, v1 may not become zero at that moment .so direction of friction becomes opposite to v1 i.e. horizontal. now the only force changing v2 is mgsin$ \alpha$. and it will produce downward acceleration to change v2. then again the body starts moving downwards or strictly speaking gets a component of velocity in -ve y direction. :)[/quote]
Do you mean that the body will be accelerated down with $ mg \sin \alpha$? :huh:  No, surely not! The friction will oppose motion in any direction, be it $ \hat j$ or $ \hat i$.
At least you do agree with that? That is the basic, fundamental definition of friction.[/quote]
its neither $ \hat i$ nor $ \hat j$ but a combination of both.

[quote="ith_power"]its neither $ \hat i$ nor $ \hat j$ but a combination of both.[/quote]
PARADOX????

You must say, "Its $ \hat i$ and $ \hat j$, in fact, a combination of both."

Lets see what other members on the forum suggest. They may have a better, flawless method to go about it...

Well ...
see that just before stopping , V->0
Hence V1^2 +V2^2 ->0

thus
V1->0 and V2->0

hence v1/v2=d(v1)/d(v2) (By L hospital's rule) = (md(v1)/dt)/(md(v2)/dt)=F1/F2
where F1 and F2 are the forces on the object in the respective directions

but v1/v2=dy/dx=tana

hence writing the differential equations and solving for a
we get a as pi/2 or -(pi/2)

hence the body either moves in the upward direction or in the downward direction before stopping

uppward- when we give initial velocity along y only
downward - at any other angle ..

[quote="rohan2007"]uppward- when we give initial velocity along y only
downward - at any other angle ..
[/quote]
Thats what I said...vaguely! :D


(Vasc) Let $ n \ge 3$ be an integer and let $ a_1,a_2,\ldots,a_n$ be nonnegative real numbers with $ a^2_1\plus{}a^2_2\plus{}\cdots \plus{}a^2_n \equal{}1.$ Prove that
$ \sum^n_{i\equal{}1}a^3_i \plus{}\frac{6}{(n\minus{}2)\left(\sqrt{n}\plus{}1\right)} \sum_{1 \le i<j<k \le n} a_ia_ja_k \le 1.$

Dear friends, I wait for the solutions using classical inequalities as Cauchy Schwarz or AM-GM or something like that.

[quote="arqady"][quote="can_hang2007"](Vasc) Let $ n \ge 3$ be an integer and let $ a_1,a_2,\ldots,a_n$ be nonnegative real numbers with $ a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n \equal{} 1.$ Prove that
$ \sum^n_{i \equal{} 1}a^3_i \plus{} \frac {6}{(n \minus{} 2)\left(\sqrt {n} \plus{} 1\right)} \sum_{1 \le i < j < k \le n} a_ia_ja_k \le 1.$
[/quote]
For the proof enough to check only $ a_2 \equal{} a_3 \equal{} ... \equal{} a_n$ and a case $ a_n \equal{} 0,$ which is obvious. :wink:[/quote]
For $ n\equal{}4,$ the inequality becomes:

If $ a,b,c,d \ge 0,a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1,$ then
$ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}abc\plus{}bcd\plus{}cda\plus{}dab \le 1.$

To prove it, we can use the Cauchy Schwarz Inequality as follows:
$ (a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}abc\plus{}bcd\plus{}cda\plus{}dab)^2\equal{}[a(a^2\plus{}bc)\plus{}b(b^2\plus{}cd)\plus{}c(c^2\plus{}da)\plus{}d(d^2\plus{}ab)]^2 \le (a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2[(a^2\plus{}bc)^2\plus{}(b^2\plus{}cd)^2\plus{}(c^2\plus{}da)^2\plus{}(d^2\plus{}ab)^2]\equal{}a^4\plus{}b^4\plus{}c^4\plus{}d^4\plus{}(a^2\plus{}c^2)(b^2\plus{}d^2)\plus{}2a^2bc\plus{}2b^2cd\plus{}2c^2da\plus{}2d^2ab \le a^4\plus{}b^4\plus{}c^4\plus{}d^4\plus{}(a^2\plus{}c^2)(b^2\plus{}d^2)\plus{}a^2(b^2\plus{}c^2)\plus{}b^2(c^2\plus{}d^2)\plus{}c^2(d^2\plus{}a^2)\plus{}d^2(a^2\plus{}b^2)\equal{}(a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2\equal{}1.$

But for $ n>4,$ how can we use the Cauchy Schwarz? Dear friend, please think of it.

[quote="can_hang2007"](Vasc) Let $ n \ge 3$ be an integer and let $ a_1,a_2,\ldots,a_n$ be nonnegative real numbers with $ a^2_1 \plus{} a^2_2 \plus{} \cdots \plus{} a^2_n \equal{} 1.$ Prove that
$ \sum^n_{i \equal{} 1}a^3_i \plus{} \frac {6}{(n \minus{} 2)\left(\sqrt {n} \plus{} 1\right)} \sum_{1 \le i < j < k \le n} a_ia_ja_k \le 1.$

Dear friends, I wait for the solutions using classical inequalities as Cauchy Schwarz or AM-GM or something like that.[/quote]
This inequality is [b]Problem 6.39[/b], page 324 of new book (where it is presented with a Cauchy-Schwarz solution):

Vasile Cirtoaje, Vo Quoc Ba Can, Tran Quoc Anh, [url=http://canhang2007.wordpress.com/2009/12/08/new-book-inequalities-with-beautiful-solutions-v-cirtoaje-v-q-b-can-t-q-anh/][b][i]Inequalities with Beautiful Solutions[/i][/b][/url], GIL publishing house 2009.


trigonometry
ratio
geometric sequence
geometry unsolved
geometry

Let $s \geq 3$ be a given integer. A sequence $K_n$ of circles and a sequence $W_n$
of convex $s$-gons satisfy:
\[ K_n \supset W_n \supset K_{n+1}  \] for all $n = 1, 2, ...$
Prove that the sequence of the radii of the circles $K_n$ converges to zero.

This is an inmediate consecuence of the following fact: if inside a circle you have a convex s-gon, and inside this s-gon you have another circle, then the radius of the inner circle is at most $\frac{1+\cos\frac{\pi}{s}}{2}$ times the radius of the outer circle. So the radius of $K_1, K_2, \dots$ form a sequence bounded by above for a geometric sequence of ratio less than zero.
The mentioned fact is easy to prove. Consider inside a circle of radius 1 another circle of radius $r$, $r>\frac 1 2$, then the circle concentric with the first one and radius $2r-1$ is also inside both circles. But if this radius $2r-1>\cos\frac\pi s$ is easy to prove that any chord of the outermost circle subtending an angle of at least $\frac{2\pi}n$ intersects the innermost circle, so that $2n$ segments between the outermost and the innermost circles cannot close to form an s-gon.


What is a+b?     $ \sqrt{\frac{3}{2} \cdot \frac{4}{3}\cdot \frac{5}{4}\cdot \frac{6}{5}   \cdots  \frac{a}{b}} \equal{} 3$

[quote=5849206328x]This simplifies to $ \sqrt{\frac{a}{2}}\equal{}3 \Rightarrow a\equal{}18$, so $ b\equal{}17$ and $ a\plus{}b\equal{}\boxed{35}$[/quote]

how did you get to the $ \sqrt{\frac{a}{2}}=3 \Rightarrow a=18$ in the first place

[size=20]silently laughs at 11 year bump[/size]


Don't click the hidden content until you have heard the woman
and tried what she says.

http://www.birthdaycards.com/pickup?ID=A222-C9Z3-R6R6-THR

The woman says your result is:
[hide]109,890 or 0[/hide]

There is only one number determined by that algorithm...

me@home,

you are in error, but so too is the post of trisect.  Here are two examples to show that there is not just one number (result):
[hide]For example, let the number be 12321 (which has at least 3 different digits).  Reverse it and get 12321.  Subtract them to get 0.  Reverse 0 to get 0.  Finally, add 0 to 0 to get the result of 0.  For a different example, let the number be 12312.  Reverse the digits to get 21321.  Subtract:  21321 - 12312 =
9009.  Reverse 9009 to get 9009.  Add 9009 to 9009 to get
18018 for the result.  So a result can be 0 or 18018 or
109890 (so far).  Other possible results?[/hide]

I thought that one was slightly different (you mean the $\sqrt{53}$?)

No, this isn't "slightly different."

It has two poems AND an algorithm to work through.
Also, the woman this time is determining YOUR final number, not YOU guessing hers.


Imagine a magnetic dipole on the plane of the page of with $N$ turns, a clockwise current of $i_{0}$, an internal resistance of $R$, a moment of inertia of $I$, and an area of $A$.  A uniform magnetic field of strength $B$ points eastward, supplying a torque on the magnetic dipole.

$\tau = iNBA\sin(\theta)$

This is what one would learn in basic physics.  This is correct, however how would one go about finding the angular velocity of the dipole at a particular angle or even better yet the angle with respect to time of the dipole?  The difficulty with this problem for me is that the changing magnetic flux would alter the current.

$i=i_{0}-\frac{1}{R}\frac{d\phi_{B}}{dt}= i_{0}-\frac{1}{R}BA\frac{d\cos(\theta)}{dt}=i_{0}+\frac{BA\sin(\theta)}{R}\dot\theta$

$I\alpha = I\ddot\theta = NBA\left(i_{0}+\frac{BA\sin(\theta)}{R}\dot\theta\right)\sin(\theta)$

This disgusting second order nonlinear DE is probably nonsolveable, but my question is: is this DE the correct representation for this problem? and is there another way to find this angle as a function of time considering both torque on the loop and induced current?

First of all you the flux through the coil is not $ BA\cos \theta$ it is $ BNA\cos \theta$. Next the Faraday's law tell you that the electromotive force(which is only a Lorentz force effect) over a loop is simply the the change of flux of magnetic field with negative sign:
$ EMF =-\frac{di}{dt}L-\frac{d(BNA\cos \theta)}{dt}=-\frac{di}{dt}L-BNA \sin \theta \frac{d\theta}{dt}$, where $ L$ is the self-inductance of the coil and is equal to $ \frac{\mu_{0}I N^{2}A}{l}$
Also,
$ EMF = iR$
So,
$ iR+\frac{di}{dt}L=BNA\sin \theta \frac{d\theta}{dt}$.
From the torque thing we have as you have written
$ I\frac{d^{2}\theta}{dt^{2}}= iBNA \sin \theta$


trigonometry function calculus derivative

[tex]
f(x) = \alpha_1 \cos (\beta_1 x) + \alpha_2 \sin (\beta_2 x);
[/tex]
What is the Maximum and Minimum Value of Function [tex]f(x)[/tex], in terms of [tex]\alpha_1, \alpha_2, \beta_1, [/tex] and [tex] \beta_2?[/tex]

hello, the first derivative is given by
$ f'(x)\equal{}\minus{}\alpha_1\beta_1\sin(\beta_1x)\plus{}\alpha_2\beta_2\cos(\beta_2x)$
i think we can't solve the equation $ f'(x)\equal{}0$ by an algebraic way.
Sonnhard.

The case of $ \beta_1 \beta_2 \equal{} 0$ is uninteresting. Consider two cases:

[b]Case 1:[/b] $ \beta_1 / \beta_2 \in \mathbb{Q}$

[b]Case 2:[/b] $ \beta_1 / \beta_2 \not \in \mathbb{Q}$


Determine the smallest non-zero positive whole number constituted entirely by threes and zeroes ( but not containing any leading zeroes) which is divisible by   2,078,304.

[hide="hint?"]
2078304's factorization is $2^5 \cdot 3 \cdot 21649$.

Hmmm... 21649 seems to be prime, at least for all the values I've tried (up to 71).
[/hide]

[quote]On Tue May 09, 2006 3:46 am  mathnerd314  wrote:  hint?[/quote]

2078304's factorization is 3*(2^5)*21649 and 21649 is a prime number.

Accordingly, at the outset,  determine the smallest non-zero number constituted entirely by ones and zeroes  which is divisible by 21649.

Note that any given power of 2 and 21649 are relatively prime.

The remainder of the result  will follow.


Is there a website that offers pdf files of previous National Competition questions?

you can find some places....but mathcounts expects you to order them

totally zealous had a website with a few nationals
i'm also pretty sure biffanddoc had an upload site with like 10 nationals
idk if they're working now

Oh yeah... I never noticed that before...

aznwithabrain, could you link to your websites, because I can't find Totally zealous

it was some common free website slash totallyzealous (i remember, because i had...

oh, i remember

i *think*

it might be like [url]http://www.freewebs.com/totallyzealous[/url]

...to fix some memory problems with it earlier...

so ya

jorian

um what
this is like the 20389472194th of these topics i've seen
READ THE STICKY


Find all positive integers $ n$ such that $ n^{2006}\plus{}n\plus{}1$ is prime number.

[quote="delegat"]Find all positive integers $ n$ such that $ n^{2006} \plus{} n \plus{} 1$ is prime number.[/quote]
It's quite easy. Note that $ n^{2006}\plus{}n\plus{}1 \equal{} (n^{2006}\minus{}n^{2})\plus{}(n^{2}\plus{}n\plus{}1)$   and $ n^{2}\plus{}n\plus{}1| n^{3}\minus{}1 | n^{2004}\minus{}1$
So $ n\equal{}1$is unique solution .

It is very nice and simple Congratulations and many thanks for solution.


we assign either "+" , "-" to each members of the sequence   of consecutive integers from 1 to 2005 and then add them all. What is the smallest  possible non -negative value of the sum of all signed members.

Put positive signs before $1, 2, 3, ..., 998, 999, 1001, 1002, ..., 1416, 1417, 1855$. Put negative signs before the rest. Then the sum is $1$.
Of course there are many other alternate assignings of the signs.
I think this problem's level is somewhere between intermediate and pre-olympiad.


solve the equation $a^{3}+b^{3}=22c^{3}$ in positive integers.

it's easy to show that $a,b,c\equiv 0(mod 9)$ then $a=9k_{1}, b=9k_{2}, c=9k_{3}$ where $k_{1}, k_{2},k_{3}\in \mathbb{N}$. Follow this method,(infinit decrease) so that the equation has not answer over $\mathbb{N}$

[quote="Hawk Tiger"]Sorry,I can't agree with you,ehsan2004.
If $a\equiv-1,b\equiv+1(\mod 9)$
$c\equiv 0 \mod 9$
then there is not a contraduction.
So I don't think we can prove it easily.[/quote]

oops  :oops:

I forgot it easily. Well, you can solve it only for this manner   :blush:

with infinite decrease we can show that $a,b$ have to be odd...

by the same reasoning a and b can't be multiples of 11
Also 7 divides (abc) and 9 divides c

[quote="Hawk Tiger"]I want to ask a question to suweijie,Have you solved this hard problem?
Or are you sure this problem can be solved by using elementary Number theory?[/quote]
Sorry that I haven't solve it yet.This problem is from the book called Dr.Riemann's Zeros.The author of it solved the problem in half a term's time.Surely it's a hard problem,but I think there must exist an elementary method to solve it. :lol:

[quote="suweijie"]no one? :maybe:[/quote]

I think it is really hard,please help me!

[quote="ZetaX"]If you accept "nonelementary" solutions, it's not that hard (see your book  :wink: ). And it's quite obvious that it's rather hard if you needlessly restrict to "elementary" methods.[/quote]

But i don't have this book :(
Can you tell me "nonelementary" solutions? :wink:


Determine the last three digits of the number
$ 2009^{2008^{2007}}$

Mhm, so where are the mistakes in my solution? I still believe that my method itself is valid  :wink:

Yes, but I think $ 2007\neq4k\plus{}2$...

Aww, those stupid mistakes  :oops:
Thanks for pointing it out. Some corrections:
Line 6 should be changed.
Of course $ 2007 \equal{} 4k \plus{} 3$ etc.
And now starting all over again the last three lines of transformations it should give a proper result  :wink:

P.S. What was that Canadian problem you based that one upon?

[quote="mazur89"]$ 2007\equiv7(mod20)$ so $ 2008^{2007}\equiv8^7 \equal{} 2^{21}\equiv2(mod25)$.
So $ 2008^{2007}\equiv352(mod400)$ because $ 16|2008^{2007}$.
So $ 2009^{2008^{2007}}\equiv9^{352} \equal{} 3^{704}\equiv3^{304}(mod1000)$
$ 2009^{2008^{2007}}\equiv3^{304}\equiv3^4 \equal{} 81(mod125)$ and $ 2009^{2008^{2007}}\equiv3^{304}\equiv1(mod8)$ so the last three digits are $ 081$.[/quote]

Can you please explain me how did you calculate all these? :(

Do you have general pattern for finding last three digits of x^y?


How do you think when you solve these kind of problems.

[quote="7h3.D3m0n.117"]Normally, you identify a specific pattern (multiply by 2, add by 3, etc.) and then you apply that pattern onto the sequence.[/quote]
Yes. It depends whether the sequence is geometric or arithmetic.

or recursive...

it could also represent a polynomial of 2nd degree or higher

Step 1 is always to identify what kind of sequence it is, arithmetic, geometric, generated by a polynomial, etc.  then, u can use that to find the next number in the sequence

When the last number in the sequence, you can find out the amount of numbers in the sequence.

[b]Arithmetic:[/b] $ a\plus{}d(n\minus{}1)\equal{}\text{last number}$ Or you can use this for any number.

[b]Geometric: [/b]$ a\cdot d^{(n\minus{}1)}\equal{}\text{last number}$, also you can use this for any number.

Oh, $ a$ is the first number in the sequence. $ d$ is the rate.


Ok, so first off, even though I didn't get into the schools I really wanted to go(Cornell, Columbia, UPenn), I am still lucky to be able to make this decision...

I am majoring in biology and my intent is to go to medical school...

These are the schools that I am deciding between

UC San Diego
UCLA
UC Berkeley
Boston University
NYU (waitlisted)
[I don't understand how I was waitlisted at NYU biology but accepted to UCSD biology]

I've lived in California since I was 4, and I wanted to go to college in the east coast. New York really impresses me and I love the city life. The thing is...the three UCs I listed have amazing biology programs. UCSD, I've read, has one of the best in the nation. However, I've heard that social life at UCSD is pretty dull....
I'm looking to have fun at college, but that's along with rigorious studying...

Another factor is...When getting into med school, a high GPA is worth its weight in gold. Should I go to a school a little less prestigious in biology in hopes of acheiving a higher GPA....or should I go to the best possible school and bear with the toughest competition.

Here are my academic listings, just in case you were wondering...
Indian Male born in Philadelphia

1490 SAT (800 M, 690 V)
800 SAT IIC
720 SAT Chem
660 SAT Writing

3.83 unweighted GPA
4.2 weighted
3 in AP English 11 and in AP US History
4 in AP Chem

over 150 hours volunteering at local hospital
over 200 hours teaching at local martial arts school
7 years in martial arts
active in some clubs....no huge leadership role in any...

I'm nothing spectacular, so I'm lucky to get into the schools I did...I don't know how I will go against the competition...Though no matter where I go, I will apply myself 100%

I'm just trying to decide
If anyone has some insight or any comments at all, I'd really like to read them

I lived for 4 years in New York/New Jersey.  I also went to college in New Jersey.

We moved to a town just outside San Diego 6 years ago and won't be moving back any time soon.

I'm surprised to hear that social life at UCSD is not so great - the weather is phenomenal.  I would take social life complaints with a grain of salt - I think if you want to hang out and have a good time, you'll find a home at any decent-sized school.

I don't know if anyone cares, but...

I just came back from a southern California visit, where I toured/walked around myself both the UCLA and UCSD campus

I've pretty much elminated UCLA from my choices, simple because I like UCSD much much better. The six college system seems like it will benefit my trying to get involved and their biology program is a plus. I was not feeling the LA campus as much as SD....
So visits are good....

I'm going to Berkeley next weekend, and hopefully make a decision a little after...

If i were you, I would weight a little towards UC berkeley and Boston U. rather than SD or LA. UC Berkeley, although its quality has gradually diminished in recent years, still has high reputation and is still one of the nation's top university. Boston U. is great as well.


Perfect Square
number theory
prime numbers
IMO Longlist

Let $n+1, n \geq 1$ positive integers be formed by taking the product of $n$ given prime numbers (a prime number can appear several times or also not appear at all in a product formed in this way.) Prove that among these $n+1$ one can find some numbers whose product is a perfect square.

Take the $n+1$ vectors of length $n$ with elements $0,1$, which we regard as vectors with entries in $\mathbb F_2$, the field with $2$ elements, formed as follows: the $i$'th vector has $1$ on the $j$'th coordinate if the prime indexed $j$ appears with an odd exponent in it, and $0$ if it appears with an even exponent. Since $\mathbb F_2^n$ has dimension $n$ and we have $n+1$ vectors, they are linearly independent, meaning that we can form a non-trivial linear combination with null sum. The product of the numbers corresponding to the vectors in this linear combination have a perfect square product, as can easily be seen.

typo: linearly independent -> linearly dependent


Let f be a function $f(x)=a_0x^3+a_1x^2+a_2x+a_3,\ a_0\ne 0,\ x\in R$ and a line $d$ for which $ d\cap {G_f} =(P_k |k\in \overline {1,3}).$ For any $k\in\overline{1,3}$ the tangent at $G_f$ in the point $P_k\in {G_f}$ meets once more again the graph $G_f$ in the point $Q_k.$ Prove that the point $Q_k,\ k\in \overline{1,3}$ are collinear.

I have never thought it was you who invented the fact that $P_1+P_2+P_3=N\Longrightarrow \left(N-2P_1\right)+\left(N-2P_2\right)+\left(N-2P_3\right)=N$... ;)

  darij

Here is my solution.

Since we can set the coefficient of $x^2$ as $0$ by parallel displacement toward in the $x$ direction,

we are to consider $f(x)$ as $f(x)=a_0x^3+b_1x+b_2\ (a_0\neq 0)$.

Supposed that the cubic function $y=f(x)$ has the three intersection points

$P_k(\alpha_k,f(\alpha_k))\ (k=1,2,3)$

with the given line $d:y=m(x)$, we have $f(x)-m(x)=a_0(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$,

in comparing the coefficients of $x^3$ and $x^2$ for both sides, we get

$\alpha_1+\alpha_2+\alpha_3=0,\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1=\frac{b_1}{a_0}\cdots [1]$.

Let $Q_k(\beta_k,f(\beta_k))$ and $y=l_k$ be the line passing through two points $P_k$ and $Q_k$, we have

$f(x)-l_k(x)=a_0(x-\alpha_k)^2(x-\beta_k)$, thus in similar way above we obtain $2\alpha_k+\beta_k=0\Longleftrightarrow \beta_k=-2\alpha_k\cdots[2]$.

Now for three points $Q_k(\beta_k,f(\beta_k))$,

we have $\frac{f(\beta_2)-f(\beta_1)}{\beta_2-\beta_1}=a_0(\beta_2^3-\beta_1^3)+b_1(\beta_2-\beta_1)$

$=a_0(\beta_2^2+\beta_2\beta_1+\beta_1^2)+b_1=a_0\{(\beta\1+\beta_2)^2-\beta_1\beta_2\}+b_1$

$=a_0[\{-2(\alpha_1+\alpha_2)\}^2-(-2\alpha_1)(-2\alpha_2)]+b_1=4a_0(\alpha_3^2-\alpha_1\alpha_2)+b_1\ (\ \because [1],[2]\ )$

where from $[1]$, we have $\alpha_1\alpha_2=\frac{b_1}{a_0}-\alpha_3(\alpha_1+\alpha_2)=\frac{b_1}{a_0}+\alpha_3^2$.

Plugg this into the above equation we obtain $\frac{f(\beta_2)-f(\beta_1)}{\beta_2-\beta_1}=-3b_1$.

In the same way we obtain $\frac{f(\beta_3)-f(\beta_1)}{\beta_3-\beta_1}=-3b_1$,which completes the proof.

kunny

[b]Lemma. [/b]The points $X_k(x_k,f(x_k)),\ k\in \overline{1,3}$ are collinear$\Longleftrightarrow x_1+x_2+x_3=-\frac{a_1}{a_0}.$

[b]Solution of the problem.[/b] The points $P_k,\ k\in \overline {1,3}$ are collinear $\Longleftrightarrow p_1+p_2+p_3=-\frac{a_1}{a_0}. $ But the points $P_k,P_k,Q_k $are collinear, i.e. $2p_k+q_k=-\frac{a_1}{a_0},\ k\in \overline {1,3}.$ Thus, $2(p_1+p_2+p_3)+(q_1+q_2+q_3)=-\frac{3a_1}{a_0}$, i.e$- \frac{2a_1}{a_0}+(q_1+q_2+q_3)=-\frac{3a_1}{a_0}$, i.e. $q_1+q_2+q_3=-\frac{a_1}{a_0}$, i.e. the points $Q_1,Q_2,Q_3$ are collinear.


We know the following to be true:

    \indent 1.  $ Z$ and $ K$ are integers with $ 500 < Z < 1000$ and $ K > 1$;

    \indent 2.  $ Z$ = $ K \times K^2$.
\\What is the value of $ K$ for which $ Z$ is a perfect square?

$ Z\equal{}K^3$, and $ Z$ is a perfect square, so it must be a perfect sixth power. The only perfect sixth power between 500 and 1000 is 729. From this, we get $ K\equal{}9$.


rotation
integration
calculus
calculus computations

The region in the first quadrant bounded by the circle $x^2+y^2=a^2$, and the chord x+y=a is rotated through 360 degrees about this chord.
Find the volume of the solid formed

Draw the line $x+y=t$ where $t\in\ [-\sqrt{2}a,\sqrt{2}a]$
The distance from the origine to the line $x+y=t$ is $\frac{t}{\sqrt{2}}$
The value of the corde intersecting the circle $x^2+y^2=a$ and the line $x+y=t$ is $2b$ where $a^2=\frac{t^2}{2}+b^2$
Thus $S_{b}=\frac{\pi}{2}(2a^2-t^2)$
So $V=\int S_{b}dt=\frac{\pi}{2}\int_{-\sqrt{2}a}^{\sqrt{2}a}(2a^2-t^2)dt=\pi\frac{4\sqrt{2}}{3}a^3$
Hope did not make mistake :?


Does anyone know where I can find answers and/or solutions to any of the old USAPhO screening tests? The old tests are located at http://www.compadre.org/psrc/evals/olympiad.cfm

Thanks,
sran

The answers are included in the PDF files.  Contact AAPT for the ones with no answers.


Find the minimal value of
\[
A = (s+5-3|\cos t|)^2 + (s-2|\sin t|)^2
\]
where $s$, $t$ are arbitrary real numbers.

Try to find a solution as beautiful as possible.

The minimum is 2; It's square of distance from quarter of ellipse (that is in the quarter-plane x>=0,y>=0) and line y=x-5; Among the lines that intersect the quarter-ellipse and are parallel to y=x-5, the line y=x-3 is closest to y=x-5, and the distance between them is  :sqrt: 2

Hmm... Im a bit rusty, but does anyone see what is wrong with this?

Rearrange the entire thing so that it is quadratic in R:
2R^2 + R(10-6cos-4sin) + 5cos^2 - 30cos +29.
We can glean two things from this: discrimant >= 0 and the min value has x at  -b/2a (where ax^2+bx+c = 0 is standard for this quick formula).

Substituting the -b/2a value into the quadratic, I obtained the actual min to be (cos^2 - 4sin^2 - 30cos + 33 - 12cossin + 20sin)/2, which is just
-(discriminant)/2, which would be 0. Where is the error in this?


Diophantine equation
number theory unsolved
number theory

$ x,y$ integer number . solve: $ 41x^{2}\minus{}23y^{2}\equal{}1$

Note that $ 8^2\equal{}41\plus{}23$
So we get $ (8x\plus{}1)(8x\minus{}1)\equal{}23(x^2\plus{}y^2)$
$ x$ is $ \plus{}3$ or $ \minus{}3$ $ ($mod $ 23)$
Putting $ x\equal{}23k\plus{}3$ we get $ 943k^2\plus{}246k\plus{}16\equal{}y^2$ and
Putting $ x\equal{}23k\minus{}3$ we get $ 943k^2\minus{}246k\plus{}16\equal{}y^2$ and
In both cases only solution is $ k\equal{}0$
So $ x\equal{}\plus{}3$ or $ \minus{}3$ and $ y\equal{}\plus{}4$ or $ \minus{}4$

Why these two equations only have solution k=0?

Is this strange because I think if a Pell equation has one solution, it should have infinite solutions :maybe:

LEMMA:Let $ p$  be a prime number with $ p\equal{}3mod4$ and $ p/x^2\plus{}y^2$ then $ p/x$ and $ p/y$
We have $ (x,y)\equal{}1$ and $ (8x\plus{}1)(8x\minus{}1)\equal{}23(x^2\plus{}y^2)$ but because $ 8x\minus{}1\equal{}3mod4$ there is prime $ p\equal{}3mod4$ and $ p/8x\minus{}1\equal{}>p/23(x^2\plus{}y^2)\equal{}>p\equal{}23$(because $ (x,y)\equal{}1$) so $ |x|\equal{}3$...


algebra
polynomial
quadratics
calculus
integration
modular arithmetic
summer program

For how many prime numbers(p) we can find an integer(m) satisfying the following equalities?
$m^{3}+3m-2\equiv0(mod p)$
$m^{2}+4m+5\equiv0(mod p)$

Hint[hide]Use the second equation first. Realize that when factoring a trinomial with x^2 as the first term and 4x as the second, that 5+np, where n is any integer, must have paired factors that add up to 4. That will limit down the number of possible primes. Then simply sub into the first equation.[/hide]

[quote="mna851"][hide]Write the second equation as $\ m(m+4) \equiv-5 \mod p$.  This has a solution if $\ m \equiv-5 \mod p$.  Then plugging into the first equation $\-108 \equiv 0 \mod p$ so $\ p$ must be equal to $\ 2$ or $\ 3$.[/hide][/quote]
But $-5\cdot(-5+4)= 5$.  :maybe:

[hide="A useful theorem"] [b]Lagrange's Theorem:[/b]  An integer polynomial of degree $n$ has at most $n$ roots $\bmod p$ for $p$ a prime. [/hide]
[hide="Another useful idea"] Completing the square in the second equation may yield useful results. ;) [/hide]

Ok so using torajirou's second suggestion (I'm not sure how to apply the first one) we get $\ (m+2)^{2}\equiv-1 \mod p$ so $\ (m+2)^{2}=kp+1$ and $\ k= \frac{(m+1)(m+3)}{p}$.  Now we can use case work.

[b]Case I:[/b] $\ m+1 = pa$ then $\ (pa-1)^{3}+3(pa-1)-2 \equiv-6 \equiv 0 \mod p$ so $\ p$ can be $\ 2$ or $\ 3$.

[b]Case II:[/b] $\ m+3 = pa$ then $\ (pa-3)^{3}+3(pa-3)-2 \equiv-38 \equiv 0 \mod p$ so $\ p$ can be $\ 2$ or $\ 19$.

So $\ p$ can be equal to $\ 2$, $\ 3$ or $\ 19$.

[quote="mna851"]$\ k= \frac{(m+1)(m+3)}{p}$.  [/quote]

You didn't need my suggestion to arrive at that.  It's also not true.

It should've been pretty easy to see that $p = 3, 19$ could not have worked.  [hide="because of my second suggestion..."] $\exists m : (m+2)^{2}\equiv-1 \bmod p \Leftrightarrow p \equiv 1, 2 \bmod 4$ (well-known result of quadratic reciprocity).  [/hide]

In my mathematics olympiad training weekend the following very similar was said to be Lagrange's theorem:

Let $p$ be prime and $f$ a polynomial of degree $n$ with integer coefficients. Then either the congruence $f(x)\equiv 0 \pmod p$ has at most $n$ noncongruent solutions mod $p$ or the coefficients of $f$ are $\equiv 0\pmod p$. t0rajir0u's form of the Lagrange's theorem is false: $px$ has infinitely many roots modulo $p$.

[quote="ZetaX"][quote="t0rajir0u"][hide="A useful theorem"] [b]Lagrange's Theorem:[/b]  An integer polynomial of degree $n$ has at most $n$ roots $\bmod p$ for $p$ a prime. [/hide][/quote]
Lagrange¿ Do you have a reference for that, since I just know it as nameless theorem for general integral domains...  :maybe:[/quote]

The reference is for example W.Sierpinski and his book "Number Theory".

[quote="puuhikki"] t0rajir0u's form of the Lagrange's theorem is false: $px$ has infinitely many roots modulo $p$.[/quote]

Ah, sorry.  I think the original statement I heard at PROMYS was given for monic polynomials (which works out fine in prime moduli).


How many four-digit positive integers are multiples of 3?

the rule in the solution works for seven, just ignore the remainder that happens.
P.S. sorry for the bump/revive

[quote=DawnTanner]WAIT was this post in 2009????????????????????????[/quote]

How did you get this from 2009? Probably alot of scrolling ig.

nah, you just have to get this problem on alcumus and there's a button that says "discuss" and you can click it and go to this thread

[quote=bbojy][quote=DawnTanner][quote=meiaiya]I don't think so bc that would take way too much scrolling, which I doubt anyone would do. Maybe they found their profile and checked their feeds (idk if that works actually)[/quote]

no, I just found it in the world feed no scrolling[/quote]

thats cuz i revived it (sorry)[/quote]

no it’s fine if it’s an alcumus problem :)

[quote=NSAoPS][quote=ProbaBillity][hide="My other solution"]
First, find the number of 4 digit integers. This is 9000. Now, because I know that 9999 is divisible by 3, I can see the pattern (in mod 3) $ 1,2,3,1,2,3,1,2,3,...$ This pattern occurs 9000 times, so there are 3000 1's, 3000 2's, and $ \boxed{\text{3000 3's}}$.
[/hide][/quote]

I was just around born then lol[/quote]

when gamebot posted, I wasn't even born lol  ;)

[hide][hide][hide]We can make a list:
$1002$, $1005$, $1008$. . .$9996$, $9999$
Then we divide the whole list by $3$, leaving
$334$, $335$, $336$. . .$3332$, $3333$.
We subtract $333$ from each number, which gives us [u]3000 numbers[/u].[/hide][/hide][/hide]

[hide][hide][hide]We can use (basic(basic(basic(basic)))) complementary counting for this problem.
We know that there are $\frac{9999}{3}=3333$ numbers divisible by $3$ from $1-9999$, and we know that there are $\frac{999}{3}=333$ numbers divisible by $3$ from $1-999$, and so the answer is $3333-333=\boxed{3000}$[/hide][/hide][/hide]

[hide][hide][hide]BTY : 2009 was 5 years before my birth[/hide][/hide][/hide]


complex analysis
complex analysis theorems

If a series $\sum_{n=0}^{\infty}c_{n}z^{n}$ has radius of convergence $R\in(0,\infty)$, what can be said about the subset of $|z|=R$ on which it is convergent?

It is easy to produce examples where this set is

- empty
- the whole circle
- all but $n$ points on the circle, for any $n\in\mathbb{N}.$

There have to be other possibilities.

It's exactly the same as the corresponding problem for trigonometric series. What is known about the possible sets of (pointwise) convergence of a Fourier series?


Ring Theory
inequalities
inequalities theorems

Can some one show me what it is?

It's used here by [u]ehku[/u](the third post)
http://www.mathlinks.ro/viewtopic.php?t=173331

and also here in the second post
http://www.mathlinks.ro/viewtopic.php?t=147090

Thanks a lot  :)

hmm I don't know, what you are talking about. (it is something like P inequality of degree 3 maybe?) Hungkhtn would help I guess.
for fun I wanted to quote two lines from Darij

[quote]I have an beautiful proof using WTF. Of course, it is trivial using OMG, but this is advanced mathematics and thus the proof stinks. Don't ask me what this all means, it's top secret and will be published on my gravestone. [/quote]
lol  :)


An interesting inequality that I have found and solved a while ago. I forgot where got it. Good luck.

Show that for all positive reals $ a, b, c,$

$ (\frac{a\plus{}2b}{a\plus{}2c})^3\plus{}(\frac{b\plus{}2c}{b\plus{}2a})^3\plus{}(\frac{c\plus{}2a}{c\plus{}2b})^3\geq 3$

[quote="geniusbliss"]
$ \sum_{cyclic}\frac { (a \plus{} 2b)^3 } {(a \plus{} 2c)^3} \ge \frac {(\sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c})^3}{9}$
therefore to prove,
$ (\sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c})^3 \ge 27$,
or to prove,
$ \sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c} \ge 3$
subtract $ 1$ from each term,
we get,
$ 2(\sum_{cyclic} \frac {b \minus{} c}{a \plus{} 2c}) \ge 0$
divide the whole by 2
we get,
$ \sum_{cyclic} \frac {b \minus{} c}{a \plus{} 2c} \ge 0$
now again add one to each term to get,
$ (a \plus{} b \plus{} c)(\sum_{cyclic} \frac {1}{a \plus{} 2c}) \ge 3$
$ Q.E.D$[/quote]

well instead of subtracting 1 from each term, you can just add 1 to each term and apply AM-HM from there, which will give you the desired result.


number theory
relatively prime
prime numbers
number theory proposed

prove that product of 4 consecutive numbers is divisible by 24.

exhibit 3 integers which are relatively prime but not relatively prime in pairs

expessing in modulo remainders.first (mod4)
combinations are 0,1,2,3 or rearrange them but there will be a multiple of 4 and a separate multiple of 2
so, divisible by 8.
now, (mod3)
combinations are 0,1,2,1 or something else but there is always a multiple of 3. so , multiple of 3

combining we get multiple of 8*3=24

hi analytic......

sorry couldnt understand ur proff.... kindly expand... is there any other proof apart form congruences

i am explaining for mod 3, similarly try it for mod4

three consecutive numbers are of the form 3k, 3k+1, 3k+2 i.e multiple of 3 , and leaving remainder 1,2.
and of course they can arranged as per required like puttinf second term first but that wont have any effect on our solution.

we see that one of these is always a multiple of 3

similarly do, it by mod 4. and combine both

just understand that in four numbers 2 are ven and of those evens is a multiple of 4 . so, if u multiply u  get 4*2=8 i. multiple of 8
i m not that good at expressing myself.

for the second problem, basic form of the numbers are:
pq  ,  qr,   pr
            where p, q, r are primes.
it can moulded into other possibilities but this is one method.

now just put any three prime numbers for p,q,r and u get the answer. so, there exist infinite possibilities as primes are also infinite.


Zach Abel
Zarathustra Brady
Ryan Ko
Yi Sun
Arnav Tripathy
Alex Zhai

Congrats!!!

[quote="quantum leap"]Has it already been determined?  Zach abel wasn't a winner...Usually aren't only six of the usamo winners chosen to be part of the team?[/quote]

Well, Arnav [i]took[/i] the TST and is at MOP.  And since he made this post I'd say it's pretty much determined.

You don't necessarily have to be a USAMO winner to take the TST.

Congratulations, good luck at IMO 2006.

I have a question.

May I know how you were able to figure this out?

[quote="Dark_Shadow"][quote="quantum leap"]Has it already been determined?  Zach abel wasn't a winner...Usually aren't only six of the usamo winners chosen to be part of the team?[/quote]Blame RSI and the Canadians.[/quote]I knew the Canadians were behind this :D Congrats to all on the team. See ya all in Slovenia.

:)  good job.

EDIT: three aopsers i see

Cool, cool, I guessed three of em right.

Congrats on making it

Yep, Brian Lawrence, and also Adam Hesterberg and Sherry Gong (I think?) made it to RSI, so that's why they aren't on the list.

Probably some others too, now that I think about it.

I don't think Sherry Gong is going to RSI unless she has a different name.

[quote="maokid7"]i have a question.
has anyone ever made the US IMO team as a freshman? :o[/quote]

a couple of people have.

Gabriel Caroll and Anders Kaseorg are the most recent examples.

Reid Barton!!!
4-time gold medalist, right?

Do the numbers of the contestants on the us imo team have any significance, and what do the observers do?

[quote="maokid7"]has anyone ever made the US IMO team as a freshman?[/quote]
The 1998 team had three freshmen on it- Reid Barton, Gabriel Carroll, and Paul Valiant.

Oh, man I thought that it was sweet when both Reid Barton and Gabriel Carroll were only 2 of 4 people in the entire world to get a perfect score on the IMO.

And of course Brian Lawrence got a perfect score on the IMO last year as well which was pretty beastly.

Then, don't forget that in 1994 all six members of the team got a perfect score (42/42).

Man the memories.

[quote="MysticTerminator"]btw, what is this "observer A, B, C" thing? Just curious...[/quote]Observers are there to observe (duh) -- they're credentialed for the competitions (so that they can attend the ceremonies and such) but have no official status, for instance, with respect to the juries.

Note, for example, Vietnam has a lot of observers, as they will presumably be taking notes in advance of the 2007 IMO.

The A,B,C seems simply to indicate the accomodation status: the A is roughly on par with the team leaders, the B with the team deputy leaders, and the C with the students.

See [url]http://imo2006.dmfa.si/regulations.html[/url]

There is no 'official' team competition and yet they have listed the highest scoring teams on the [url=http://imo2006.dmfa.si/about.html]official page[/url].  I find that amusing :D

[quote="Phelpedo"][quote="quantum leap"]What are the contestant numbers (i.e. #1, #2, ...) for then?  Is #1 the team captain?[/quote]

It would be alphabetical order, except for Ryan Ko.[/quote]

Wait, does that mean Ryan Ko is the captain or something?  I can't quite understand this.

Phelpedo is just saying that it looks like alphabetical order.  It's probably just an error.  When I first viewed the page a few days ago they had some names backwards (Tripathy Arnav, for example).  Furthermore, they didn't have a picture for Ryan Ko up at that time.  So it is very likely that they meant to put it in alphabetical order but when they edited in Ryan's picture, they forgot or something.

why don't they just an "official" team competition?


Hello Support,

There are two problems that started to happen to me very recently:

(1) I use to continue looking the topics I have post in or some of those that I haven't post but I am watching. The topic replay notification is a GREAT tool for that. Nevertheless, as from yesterday around 9:00 pm (I am in GMT-5, non-day saving light time yet!) I started to receive the notification basically in blank. There is a "file" that I cannot download, but the direct link to the topic is not there, then, I have to use the search the topic. It is not that bad, but it was nicer (for me) to have the direct link there and not to be searching each topic every time.

EDIT: I include an image of what I receive in my inbox.

(2) The last three days, the forum access during the afternoon hours, say around noon to 5:00 p.m.  has been VERY slow, out of common. The access used to be very fast (and it is for me in the evening and night hours), but now, I even receive pretty often a time out when I try to go to a topic or post something. I thougt it could be matter of the OS or the browser, but it is the same if I use Fedora, Red Hat 9, KDE or Windows XP, and it is the same if I use Explorer, Netscape, Mozilla (old versions) or Mozilla Firefox. It is not the provider Either, because any other page works fine!

Thank you. Best regards,

[quote="djimenez"]By the way, the reply notifications that I have received after 11:40 today don't have the problem (1) anymore![/quote]Very peculiar. Thanks for reporting that in, anyone else having this problem? (can you please give me throught PM the headers of the blank emails to check if they were actually sent with our SMTP server?)


$ F$ is fibonacci.
Prove

$ C_n^0\plus{}C_{n\minus{}1}^1\plus{}C_{n\minus{}2}^2\plus{}...\equal{}F_{n\plus{}1}$

Looks like here $ C^k_n \equal{} {n \choose k}$. (OP: it's best to write it the second way to avoid confusion, since other posters use $ C^n_k \equal{} {n \choose k}$)

This is easy to prove combinatorially by using the "$ F_n$ counts ways to tile a rectangle with squares and dominoes" and then conditioning on how many dominoes/squares are used.

Scroll down to "Excursion" [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=217361]here[/url].


Hi everyone,

Can someone give me a hint on Explicit Form.

Find the explicit form of this particular solution:

y^-1/2 = (2+sin x)^-1  + 3/8

Thanks for your help[/quote]

hello, write you equation in the form
$ \frac{1}{\sqrt{y}}\equal{}\frac{1}{2\plus{}\sin(x)}\plus{}\frac{3}{8}$.
Building the common denominator we get
$ \frac{1}{\sqrt{y}}\equal{}\frac{8\plus{}3\left(2\plus{}\sin(x)\right)}{8\left(2\plus{}\sin(x)\right)}$
Building the reciprocal of both sides we have
$ \sqrt{y}\equal{}\frac{16\plus{}8\sin(x)}{14\plus{}3\sin(x)}$
and from here finally the desired result
$ y\equal{}\left(\frac{16\plus{}8\sin(x)}{14\plus{}3\sin(x)}\right)^2$
Sonnhard.


calculus
integration
calculus computations

Compute $ \int_0^1 x^{2n\plus{}1}e^{\minus{}x^2}dx\ (n\equal{}1,\ 2,\ \cdots)$ , then use this result, prove that $ \sum_{n\equal{}0}^{\infty} \frac{1}{n!}\equal{}e$.

$ I_{n}\equal{}\int_{0}^{1}x^{2n\plus{}1}e^{\minus{}x^{2}}dx\equal{}\allowbreak \minus{}\frac{1}{2e}\plus{}n\int_{0}^{1}x^{2n\minus{}1}e^{\minus{}x^{2}}dx\equal{}\allowbreak \minus{}\frac{1}{2e}\plus{}nI_{n\minus{}1}$

$ \frac{I_{n}}{n!}\minus{}\frac{I_{n\minus{}1}}{(n\minus{}1)!}\equal{}\allowbreak \minus{}\frac{1}{2n!e}$

$ \frac{I_{n\minus{}1}}{(n\minus{}1)!}\minus{}\frac{I_{n\minus{}2}}{(n\minus{}2)!}\equal{}\allowbreak \minus{}\frac{1}{2(n\minus{}1)!e}$

$ ......$

$ \frac{I_{1}}{(1)!}\minus{}\frac{I_{0}}{0!}\equal{}\allowbreak \minus{}\frac{1}{2\left( 1!\right) e}$

Hence $ \frac{I_{n}}{n!}\minus{}\frac{1}{2}\plus{}\frac{1}{2}e^{\minus{}1}\equal{}\minus{}\frac{1}{2}e^{\minus{}1}\left( \frac{1}{n!}\plus{}\frac{1}{(n\minus{}1)!}\plus{}.....\plus{}\frac{1}{1!}\right)$

Let $ n\rightarrow \infty ,$

$ \vspace{1pt}\minus{}\frac{1}{2}\plus{}\frac{1}{2}e^{\minus{}1}\equal{}\minus{}\left( \frac{1}{2}e^{\minus{}1}\right) \left( \sum_{k\equal{}0}^{\infty }\frac{1}{k!}\minus{}1\right) \Rightarrow \sum_{k\equal{}0}^{\infty }\frac{1}{k!}\equal{}e$

I have nothing to add more than this. :lol:


can you chage your username? I'm just wondering because in the edit your profile section it has a space with your original username.

I don't think so.  A while back, I tried doing that and nothing happened.

We generally discourage people switching their usernames around, but if you really want to, send me an email.


Given square $ABCD$ with side length $a$ and another square $PQRS$ with side length $b$ where $\sqrt{2}a\geq b$ . If $PQRS$ is centered at $A$ and is free to rotate around it . Prove that no matter how you rotate it , the area of the $PQRS$ which is inside $ABCD$ is always constant .

 :)

[hide]The area of the part of $PQRS$ that is inside $ABCD$ (call it $X$) is actually always $\frac{1}{4}$ of $PQRS$.

Draw the diagonals of $PQRS$. $X$ is now divided into two parts by one of the diagonals. It is easy to see that you can "move" one of the two parts so that the entire shape exactly fits into one of the  small right triangles created by the two diagonals. The area of the small right triangle is $\frac{1}{4}$ of $PQRS$, and so is $X$.  (This is basically because of the symmetry of the shape. Note that the "central angle" ($\angle BAD$) is always $90^\circ$ no matter how you rotate the shape.)[/hide]


Does anyone ever feel really down sometimes for no reason at all?  or just really really tired or have shoulder pains?  What do you do about it?  At first I thought I wasn't sleeping enough but now I am sleeping 8 hours a day and I still feel really tired.

My guess is that you are either 1) having mood swings or pmsing (though i doubt it would be that...) or 2) you are really stressed out.  Being stressed out can make you feel really tired.  My reccomendation is to exercise a lot and do yoga or something, it'll relax the muscles and make you feel less tense.

If you're having shoulder pains without any activity, then it's probably stress-related, or maybe a swollen lymph node? If it is with activity, it's possible you have tendonitis. I don't know what would cause a swollen lymph node; tendonitis is annoyingly easy to get and hard to get rid of. (The best solution is to rest the shoulder for a while and take anti-inflammatory medication whenever you're going to exert it / when there's pain.) These are just possibilities though.

I don't really see why you're seeking medical help on a math forum, especially with such a vague description of your ailment... :p

It looks like you're really depressed or stressed out. Relax, take a week or two off, have fun.

try stretching the shoulder muscles, it helped me.


limit
integration
calculus
inequalities
real analysis
function
logarithms

I hate to post this, but it is driving me insane. Suppose the following four conditions

i. $ \mu (X)<\infty$,
ii. $ \{f_n\}$ are uniformly integrable
iii. $ f_n\to f$ a.e.
iv. $ |f|<\infty$ a.e.

The theorem says that we can then prove $ \lim \int_X |f_n\minus{}f|d\mu\equal{}0$. The first step is to prove that $ f\in L^1(\mu)$. I feel like this is probably ridiculously easy, but I've been coming to it on and off for about a week now. I keep trying to chop up $ X$ into manageable pieces, but I seem to keep getting chunks or absolute value signs in the wrong place.

My main goal was to show that $ f_n$ is Cauchy in $ L^1(\mu)$ and then invoke completeness to get $ f\in L^1$. Not sure if this is the best way to go, though.

My most successful (in that it provided interesting things to manipulate) attempt was to let $ \varepsilon>0$, then by uniform integrability we can choose a $ \delta>0$ such that $ \big|\int_E f_nd\mu\big|<\varepsilon$ whenever $ \mu(E)<\delta$. Then use Egorov to say that there exists a set $ A\subset X$ such that $ f_n\to f$ uniformly on $ A$ and $ \mu(X\setminus A)<\delta$.

This provided a nice $ X\equal{}A\cup (X\setminus A)$ where we have uniformly converging on $ A$ and the integrals of $ f_n$ on $ X\setminus A$ are less than $ \varepsilon$. No matter how I fiddled, I couldn't seem to control it properly, though.

Ignore everything I said if you know of a better way.   :huh:

Also, I think I can get the rest if I get $ f\in L^1$.

Since $ |f|<\plus{}\infty$ a.e., $ f$ is uniformly bounded outside a set of measure $ <\delta$. Integrate $ |f_n|$ over that exceptional set and use Fatou.

Ah. I knew that condition had to play a role somehow. Let $ N\in\mathbb{R}$ and $ A\subset X$ such that $ |f(x)|<N$ on $ A$ and $ \mu(X\setminus A)<\delta$ (where $ \varepsilon$ and $ \delta$ are as in the first post).

Then

\begin{eqnarray*}
\int_X fd\mu & \leq & \big|\int_X fd\mu \big| \\
  & = & \big|\int_A f +\int_{X\setminus A} f \big| \\
  & \leq & \big|\int_A f\big| +\big|\int_{X\setminus A} f\big| \\
  & < & N\mu(A) + \big|\int_{X\setminus A}\liminf f_n \big| \\
  & \leq & N\mu(A) + \liminf \big|\int_{X\setminus A} f_n\big| \\
  & < & N\mu(A) +\liminf \varepsilon \\
  & < & \infty
\end{eqnarray*}

Thus since $ f$ is integrable we have that $ |f|$ is integrable and hence $ f\in L^1$. Where starting at the Triangle Inequality step we have in order TI, bounded on $ A$ property (and equality for liminf), Fatou, uniformly integrable, then everything is finite.

Anything questionable there?

The idea is right but you'd better work with $ |f_n|\to |f|$ from the very beginning since Fatou is valid for non-negative functions only, so the way you apply it is formally wrong. Of course, you should also note that uniform integrability of $ f_n$ implies that of $ |f_n|$ but that is rather simple: just take a set $ E$ over which you want to integrate and consider the parts of $ E$ where the function is positive and negative separately.

Ah, that was dumb. That is exactly what was troubling me. For some reason I didn't think to try to jot down whether or not uniform integrability of $ \{f_n\}$ implies UI of $ \{|f_n|\}$. This was where the awkwardness of absolute values I was talking about in the first post came from.

Thanks. I got it now, but also, you can show that this theorem implies Dominated Convergence, but is there an example where you would know the uniform integrability of $ \{f_n\}$ but not have a function that dominates it?

Certainly. Just take $ f_n(x)\equal{}\begin{cases}\frac{1}{n\log n},&|x|<n\\0,& |x|>n\end{cases}$


geometry
3D geometry
number theory proposed
number theory

Prove that $\forall n\ge 10$ there exists a perfect cube between $n$ and $3n$.

Consider the condition that there is no perfect cube between $n$ and $3n$
Then, $k^{3}< n < 3n < (k+1)^{3}\Rightarrow (k+1)^{3}> 3n > 3k^{3}\Rightarrow 2k^{3}-3k^{2}-k-1 < 0$ for a positive integer $k$.
But, $2k^{3}-3k^{2}-k < 0$ for $k = 1,2$ and $2k^{3}-3k^{2}-k-1 > 0$ for all $k \geq 3$ in which the condition can't happen... The conclusion follows..  :)


function
integration
logarithms
limit
calculus
calculus computations

1. A population is modeled by a function P that satisfies the logistic differential equation dP/dt=P/5 (1 - P/12).
a) If P(0)=3, what is lim t-->infinity P(t)? If P(0)=20, what is lim t-->infinity P(t)?
I don't get it, like why they give me the value of function P, for example, P(0)=3, P(0)=20. Does it help to find the limits? I don't think so.
2. An object moving along a curve in the xy-plane is at position (x(t), y(t)) at time t, where dx/dt= tan(e^-t) and dy/dt=sec(e^-t) for t>=0. At time t=1, the object is at position (2,-3).
a) Is there a time t>=0 at which the object is on the y-axis? Explain why or why not. My teacher did like this: Because dx/dt velocity is positive, then it's not on y-axis. But how the velocity relates to this problem? When we talk about position, are we concerning about distance?
Thanks a lot for help.

[hide="1)"]
I believe the differential equation is $ \frac {dP}{dt} \equal{} \frac {P}{5}\left( 1 \minus{} \frac {P}{12}\right)$

and so, the solution will be given by $ \int_a^P \frac {60dP}{P(12 \minus{} P)} \equal{} \int_0^t dt$, where $ a$ is $ P(0)$. Notice here, that the value $ P(0)$ (the initial condition) is necessary to get a well defined function $ P(t)$.

now, observe that $ \frac {60}{P(12 \minus{} P)} \equal{} \frac {5}{P} \plus{} \frac {5}{12 \minus{} P}$

Hence, $ 5\left( \ln \frac {P}{a} \minus{} \ln \frac {12 \minus{} P}{12 \minus{} a}\right) \equal{} t$

this gives $ P(t) \equal{} \frac {12ae^{\frac {t}{5}}}{12 \minus{} a\left(1 \minus{} e^{\frac {t}{5}}\right) }$

So $ \lim_{t\rightarrow\infty}P \equal{} 12$ which is independant of the value of $ a$.
[/hide]

[hide="2)"]
Suppose the object does reach the $ Y$ axis at some time $ t$. Then the average $ x$-velocity till that time is
$ \left\langle v_x\right\rangle \equal{} \frac {x(t) \minus{} x(0)}{t} \equal{} \frac { \minus{} x(0)}{t}$
So if after solving the differential equation for $ x(t)$ we get that $ x(0)$ is positive, we'll have negative average velocity although actual velocity is always positive, which is impossible. Thus, the only way it can cross the $ Y$ axis is if $ x(0)\leq 0$
[/hide]

Hi, Thank you very much. I got the second one. But for the first one, do you mean that P(0)=3 and P(0)=20 are useless in this problem? But what's the point of giving us? so that we know the range, like from what number to what number? Well, I still think it's useless because finally i get: P+12-P=e^(t/5), so it's 12=e^(t/5) no matter what the value of t is, like infinity, the answer is still 12?

Yes, in this particular problem, they are useless. But you can see that it is just out of luck. If we'd got $ P(t) \equal{} \frac {12a^2e^{\frac {t}{5}}}{12 \minus{} a\left(1 \minus{} e^{\frac {t}{5}}\right) }$ or something like that, we would have needed the value of $ a$. In any case, we do need $ a$ to get the actual expression for $ P(t)$

But I don't get the fact why you did like: ln P/a - ln 12-P/12-a.   Are we supposed to just put the value a into P? But why you use P divided by a?  I didn't, so I got the final answer as I told you before, which all P was eliminated.

$ a$ is 3 in the first case and 20 in the second case.
Because I integrated $ \int_a^P \frac {60dP}{P(12 \minus{} P)} \equal{} \int_0^t dt$ from constants $ 0$ and $ P(0)$ to 'variables $ P(t)$ and $ t$, I got an expression that gave $ P(t)$ in terms of $ t$ (that line onwards, $ P$ always standes for $ P(t)$)

And sorry I skipped some steps,
$ \int_a^P \frac {60dP}{P(12 \minus{} P)}$

$ \equal{} \int_a^P \left( \frac {5}{P} \plus{} \frac {5}{12 \minus{} P}\right) dP$

$ \equal{}5\left( \ln P\minus{}\ln a\minus{}\ln (12\minus{}P)\plus{}\ln (12\minus{}a)\right)$

$ \equal{}5\left( \ln \frac {P}{a} \minus{} \ln \frac {12 \minus{} P}{12 \minus{} a}\right)$


1. If a,b,c are real number st a+b+c =0, abc=2 and c>0 give all possible values for c

I made a polynomial (x-a)(x-b)(x-c) but it doesnt get the right answer?

2. Any pair of 2 people can be classified as friends ornot friends. In any group of 4 students from a class of 8 students, one can always find 2 people who are friends. If the professor wanted to invite a group of students to dinner, with the restriction that any pair of students from the group he invites must be friends, what is the largest size group that he is guaranteed to be able to invite?

a+b+c = 0
abc = 2
c > 0 (c is positive)

so either: a and b are positive -or- a and b are negative
they cant all be positive, otherwise a+b+c > 0.
None can be 0, otherwise abc = 0.
So a < 0, b < 0, c > 0.

That should start it off...

[hide]$-(a+b)=c$
$ab=\frac{2}{c}$
$x^2+cx+\frac{2}{c}=0$
Discriminant[/hide]
This is more than enough hints


$2^{\displaystyle \sum^n_{i=1}i^3}$
    Everything here except '2' is the superscript on the top right corner of '2', but it is too large to look like a superscript!
    I tried deleting \displaystyle command in latex code and the size of the superscript could be brought down to normal. However, without this command, the summation symbol looks awful!
    Can any one find a solution out of this dilemma?

\[
2^{\sum_{i=1}^n i^3}
\]
\[
2^{\sum\limits_{i=1}^n i^3}
\]
What's wrong with either of those?  What exactly are you trying to do?

One other solution is: \[2^{^{\sum\limits_{i=1}^n{i^3}}}\]

That's really clever!  I would [i]never[/i] have thought to try that!

Wow! Coolest thing since I learned about \textstyle :) :D

$\exp\left(\ln 2\sum_{i=1}^\infty i^3\right)$ is nice too.


function
algebra
polynomial
induction
number theory proposed
number theory

Find all pairs of function $ f: \mathbb{N} \rightarrow \mathbb{N}$ and polynomial with integer coefficients $ p$ such that:
(i) $ p(mn) \equal{} p(m)p(n)$ for all positive integers $ m,n > 1$ with $ \gcd(m,n) \equal{} 1$, and
(ii) $ \sum_{d|n}f(d) \equal{} p(n)$ for all positive integers $ n$.

[quote="Raja Oktovin"]Find all pairs of function $ f: \mathbb{N} \rightarrow \mathbb{N}$ and polynomial with integer coefficients $ p$ such that:
(i). $ p(mn) \equal{} p(m)p(n)$ for all positive integers $ m,n > 1$ with $ \gcd(m,n) \equal{} 1$, and
(ii). $ \sum_{d|n}f(d) \equal{} p(n)$ for all positive integers $ n$.[/quote]

$ P(2n)\equal{}P(2)P(n)$ $ \forall$ odd $ n>1$ and so $ P(2x)\equal{}aP(x)$ $ \forall x$ (else non zero polynomial $ P(2x)\minus{}P(2)P(x)$ would have infinitely many roots).

As a consequence, we get $ P(x)\equal{}0$ or $ P(x)\equal{}x^k$ for some $ k$ but, since $ f(d)>0$ ($ 0\notin\mathbb N$), we get only $ P(x)\equal{}x^k$ for some $ k$

Using then the second formula for $ n\equal{}1$, we get $ f(1)\equal{}1$
Using then the second formula for $ n\equal{}p$ prime, we get $ f(1)\plus{}f(p)\equal{}p^k$ and so $ f(p)\equal{}p^k\minus{}1$
Using then the second formula for $ n\equal{}p^2$ with $ p$ prime, we get $ f(1)\plus{}f(p)\plus{}f(p^2)\equal{}p^{2k}$ and so $ f(p^2)\equal{}p^{2k}\minus{}p^k$

An immediate induction gives $ f(p^n)\equal{}p^{nk}\minus{}p^{(n\minus{}1)k}$ for any prime $ p$ and positive integer $ n$.

Using then the second formula for $ n\equal{}pq$ with $ p,q$ prime, we get $ f(1)\plus{}f(p)\plus{}f(q)\plus{}f(pq)\equal{}(pq)^k$ and so $ f(pq)\equal{}p^kq^k\minus{}p^k\minus{}q^k\plus{}1$ and so $ f(pq)\equal{}(p^k\minus{}1)(q^k\minus{}1)$

And so, with many successive inductions, it's rather easy to show that $ f(\prod p_i^{n_i})\equal{}\prod(p_i^{n_ik}\minus{}p_i^{(n_i\minus{}1)k})$

Hence the result :

$ P(x)\equal{}x^k$ and $ f(\prod p_i^{n_i})\equal{}\prod(p_i^{n_ik}\minus{}p_i^{(n_i\minus{}1)k})$


inequalities
search
inequalities theorems

Hello,

I have read around the forum of Chinese "Dumbass" Notation, used when "dumbassing" an inequality - however, a Google search yields nothing of the notation.  Can somebody please explain this notation?

Thanks,
Howard

This question has been answered here.
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=355856[/url]


Use your awesome strategy skills!

$ \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1 \cdot 2 \cdot 3 \cdot 1 \cdot 2 \cdot 3 \cdot 4}$

Cancel out $ 24$.

$ \frac{5 \cdot 6 \cdot 7 \cdot 8}{6}$.

Cancel out the $ 6$'s.

$ 5 \times 7 \times 8$

$ \boxed{280}$


This is really basic proof on geometry but beginners should try it. Use two-columns method on this:

Given: M is the midpoint of line AB.

Prove: State AM = MB

Isn't that the definition of a midpoint? I don't see anything to prove...

that's what i thought...

unless you mean

[hide]AM=1/2 AB (def of midpoint)

MB=1/2 AB (def. of midpoint)

AM=AB (transitive property of equality)[/hide]

but i think it is just the definition of a midpoint

how can a line have a midpoint? lines go on to inifinity, only segments have midpoints

When I first looked at this I thought what paladin thought at first.  Then I thought that maybe he meant what yifman said.  So I looked up the definition of a midpoint at mathworld.com and it says something like a midpoint is the point on a line segment that divides that segment into two equal segments.  So I don't really know what he meant to ask.  Maybe it should ask to prove that 2AM=AB and 2MB=AB.  I don't know.


function
real analysis
real analysis unsolved

Find all strictly monotonic functions $f: \mathbb{R}^+\rightarrow\mathbb{R}^+$ for which $f\left(\frac{x^2}{f(x)}\right)=x$ for all $x$.

Basically, it's been solved before [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=1984&t=49286]here[/url].

Since $f$ is strictly monotonic and surjective on $\mathbb R_+$, it must be continuous. Now $\tilde f=\exp^{-1}\circ f\circ\exp: \mathbb R\to\mathbb R$ satisfies the hypotheses in that topic, so we can conclude that $\tilde f(x)=\alpha x,\ \forall x\in\mathbb R$ for some $\alpha\in\mathbb R$, i.e. $f(x)=ax,\ \forall x>0$ for some $a>0$ ($a=\exp(\alpha)$, actually).


Find the sum of all integers between 50 and 350 that end in a 1.

source: ahsme


[quote="basil mohammed"]Let $ f(x)$ be polynomial of degree $ n$ with $ n$ different real roots $ a_1,a_2,......a_n$ in the increasing order. Let $ t_n \equal{} a_{n \plus{} 1} \minus{} a_n$ and $ b_n$ be the maximum of $ |f(x)|$ in the interval $ (a_{n},a_{n \plus{} 1})$
Is the following statement true?
If $ t_p > t_q$ then $ b_p > b_q$
If true, can the given conditions be weakened and the conjecture still hold?[/quote]

1) feeling
It's rather easy to feel that this is wrong. It would mean that $ t_p\equal{}t_q$ $ \implies$ $ b_p\equal{}b_q$. Else, wlog say $ b_p>b_q$, then just reduce $ a_{p\plus{}1}$ wih an $ \epsilon$ value and then $ t_p<t_q$ and $ b_p>b_q$

2) building a counter example :
Let $ f(x)\equal{}(x\plus{}2)x(x\minus{}2)(x\minus{}4)$ :
$ \{a_i\}\equal{}\{\minus{}2,0,2,4\}$
$ \{t_i\}\equal{}\{2,2,2\}$
$ \{b_i\}\equal{}\{16,9,16\}$
We have $ t_1\equal{}t_2$ and $ b_1>b_2$ and we just have to reduce a little bit $ t_1$ to get our counter example :

Let $ g(x)\equal{}(x\plus{}2)(x\plus{}\frac{1}{100})(x\minus{}2)(x\minus{}4)$
$ \{a_i\}\equal{}\{\minus{}2,\minus{}\frac{1}{100},2,4\}$
$ \{t_i\}\equal{}\{2\minus{}\frac{1}{100},2\plus{}\frac{1}{100},2\}$

$ \{b_i\}\equal{}\{15.8706947,9.0902023,16.0494456\}$

And $ t1<t2$ while $ b1>b2$


symmetry
algebra
polynomial
product of roots
algebra unsolved

Let $ a,b,c$ be the roots of the equation $ x^3\minus{}x^2\minus{}x\minus{}1\equal{}0$.

$ (a)$ Show that $ a,b,c$ are distinct.

$ (b)$ Show that: $ \frac{a^{1982}\minus{}b^{1982}}{a\minus{}b}\plus{}\frac{b^{1982}\minus{}c^{1982}}{b\minus{}c}\plus{}\frac{c^{1982}\minus{}a^{1982}}{c\minus{}a}$ is an integer.

[quote="moldovan"]Let $ a,b,c$ be the roots of the equation $ x^3 \minus{} x^2 \minus{} x \minus{} 1 \equal{} 0$.

$ (a)$ Show that $ a,b,c$ are distinct.

$ (b)$ Show that: $ \frac {a^{1982} \minus{} b^{1982}}{a \minus{} b} \plus{} \frac {b^{1982} \minus{} c^{1982}}{b \minus{} c} \plus{} \frac {c^{1982} \minus{} a^{1982}}{c \minus{} a}$ is an integer.[/quote]

[b](a)[/b] , Assume that $ b\equal{}a$, then product of roots $ \equal{}1 \Rightarrow c\equal{}\frac{1}{a^2}$ and clearly $ a$ must be real.
So $ 2a\plus{}\frac{1}{a^2}\equal{}1.........(1) , a^2\plus{}\frac{2}{a}\equal{}\minus{}1.........(2)$.
From $ (2)$ , $ a^3\plus{}a\plus{}2\equal{}0 \Leftrightarrow (a\plus{}1)(a^2\minus{}a\plus{}2) \Leftrightarrow a\equal{}\minus{}1$, which is not a solution to $ (1)$, contradiction, done.

[b](b)[/b] , substitute $ a^3\equal{}a^2\plus{}a\plus{}1$ repeatedly we get $ a^3\equal{}k_1a^2\plus{}k_2a\plus{}k_3$, where $ k_1,k_2,k_3$ are integers.
By symmetry, $ b^3\equal{}k_1b^2\plus{}k_2b\plus{}k_3, c^3\equal{}k_1c^2\plus{}k_2c\plus{}k_3$

Now it becomes to prove $ [k_1(a\plus{}b)\plus{}k_2]\plus{}[k_1(b\plus{}c)\plus{}k_2]\plus{}[k_1(c\plus{}a)\plus{}k_2]\equal{}2k_1(a\plus{}b\plus{}c)\plus{}3k_2\equal{}2k_1\plus{}3k_2$ is integer, which is obviously true.


Given: Parallelogram ABCD
Prove: Diagonals bisect.

All i got so far is that AB is congruent and parallel to DC, and
AD is congruent and parallel to BC by definition of Parallelogram...

[hide="Proof"]Given: Parallelogram $ ABCD$.
Prove: $ AC$ bisects $ BD$ and vice versa.

Let $ E$ be the intersection of the diagonals and $ h$ be the distance between the lines $ AB$ and $ CD$. By the definition of a parallelogram, $ AB \equal{} CD$. The heights and bases of $ \triangle ABD$ and $ \triangle ACD$ are equal, so $ [ABD] \equal{} [ACD]$. Subtracting $ [ADE]$ from both areas yields $ [ABE] \equal{} [CDE]$. Again using $ AB \equal{} CD$, we can establish that the heights of those two triangles are also equal. Combined with the fact that the triangles are similar because of their angles, this makes them congruent to each other. $ \triangle ABE\cong\triangle CDE\Longrightarrow AE \equal{} CE$. Repeat for the other diagonal.[/hide]

how do you know that the height of the triangles are the same just by AB=DC?

$ AB \equal{} CD$ shows that the bases of the triangles are the same. The height of $ ABD$ is the length of the perpendicular from $ B$ to $ CD$ and the height of $ ACD$ is the length of the perpendicular from $ C$ to $ AB$. Because $ AB$ and $ CD$ are parallel, these two lengths are the same.

ohhhh. i get it now.

can you help me with this one.
given: triangle ABC is isosceles with base BC
BY and CX are medians
BY and CX intersect at Z
provve: AYZX is a kite.

[hide="Outline"]You know $ AB\equal{}AC$, and many things can be determined about the smaller triangles created by the medians. Use basic properties and then $ \mathrm{SAS}$ to prove the congruence of $ \triangle BXZ$ and $ \triangle CYZ$. The fact that $ AXZY$ is a kite then follows.[/hide]


Can the numbers from 1 to 81 be written in a 9 × 9 board, so that the
sum of numbers in each 3 × 3 square is the same?

Really, Nobody can solve this problem !!? :huh:

I'm glad to know :) I was occasionally looking for proofs that this cannot be done, and now I know why failed... Is there a reasonable way of finding such an arrangement, or is it sheer brute force?

Yes, there is a reasonable way of finding such an arrangement.


[quote="allrighty"]1) Suppose u1, u2, u3, ...., un are orthonormal bases for R^n. Construct the matrix A that transforms each vj into uj to give Av1 = u1, ... , Avn = un

2)Find the singular value decomposition of matrix A, if A has orthonormal columns v1, ... , vn or lengths a1, ... , an.[/quote]

1. with given vj?
A[v1 v2..] = [u1 u2..] or AV=U, so A=U/V

2. A=SVD, S has orthonormal columns, D has orthonormal strings.
In those simple cases:
if A=S, then V=I, D=I
if A=D (orthonormal strings), so S=I, V=I