Find all integers $\,a,b,c\,$ with $\,1<a<b<c\,$ such that \[ (a-1)(b-1)(c-1)  \] is a divisor of $abc-1.$

Let $M=\frac{abc-1}{(b-1)(c-1)}$ is natural, which is divisible by $a-1$. Now we will write the intervals of $M$.

1) $M>\frac{abc-ac}{(b-1)(c-1)}=\frac{ac}{c-1}>a$. So we have $M\geq a+1$.
2) $M<\frac{abc}{(b-1)(c-1)}=(a+\frac{a}{b-1})\frac{c}{c-1}\leq (a+1)(\frac{c}{c-1} =a+1+\frac{a+1}{c-1}\leq a+2$.
Which implies to $M=a+1$.
As we know $\frac{M}{a-1}=1+\frac{2}{a-1}$ is natural thus a=2 or a=3.
Two more cases:
1) If a=2 so from $M=a+1$ ew have $c=3+\frac{5}{b-3}$, from here we get b=4,c=8.
2) If a=3 analogously $c=4+\frac{11}{b-4}$ and from this we get b=5,c=15.

davron

A bit of a different approach that's lengthier:

[hide="Solution"]

First off make the substitutions $a=x+1, b=y+1, c=z+1$.  This gives us:
 \[ xyz|((x+1)(y+1)(z+1)-1)\]
\[\implies xyz|(xyz+xy+xz+yz+x+y+z)\]
\[\implies xyz|(xy+xz+yz+x+y+z)\]
\[\implies  1|(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz})\]

Now, note that since $1<a<b<c$ or the equivalent $2\le a<b<c$, we get $1\le x<y<z$.  Therefore the maximum possible value for $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}$ is when $x=1, y=2, z=3$ giving us a value of $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=2\frac{5}{6}$.  Henceforth $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}=1,2$.  Before considering these two cases separately, we use some minor bounding.  Assume $x,y,z\ge 4$.  However, this gives a maximum sum of $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{4*5}+\frac{1}{4*6}+\frac{1}{5*6}<\frac{3}{4}+\frac{3}{16}<1$ absurd.  Henceforth at least one of $x,y,z$ is less than or equal to $3$.   Since $x$ is the smallest of the three, we know $x\le 3$.

For the case that the sum equals $1$, we get:

$x=3$
\[ \implies \frac{1}{3}+\frac{1}{y}+\frac{1}{z}+\frac{1}{3y}+\frac{1}{3z}+\frac{1}{yz}=1\]
\[ \implies yz+3z+3y+z+y+3=3yz\]
\[ \implies \&\text{contradiction mod 2}\]

$x=2$
\[ \implies \frac{1}{2}+\frac{1}{y}+\frac{1}{z}+\frac{1}{2y}+\frac{1}{2z}+\frac{1}{yz}=1\]
\[ \implies 3y+3z+2=yz \]
\[\implies  (y-3)(z-3)=11\]
\[ \implies (y,z)=(14, 4), (4,14)\]
$z>y$ so the solution set here is $(x,y,z)=(2, 4, 14)$ which gives $(a,b,c)=(3, 5, 15)$

$x=1$
\[ \implies 1+\frac{1}{y}+\frac{1}{z}+\frac{1}{y}+\frac{1}{z}+\frac{1}{yz}=1 \]
\[ \implies \text{absurd} \]

--

For the sum equaling $2$ case, we get:

$x=3$
\[\implies \frac{1}{3}+\frac{1}{y}+\frac{1}{z}+\frac{1}{3y}+\frac{1}{3z}+\frac{1}{yz}=2\]
\[  \implies yz+3z+3y+z+y+3=6yz\]
\[ \implies 5yz-4z-4y=3\]
\[\implies (5y-4)(z-\frac{4}{5})=\frac{31}{5}\]
\[\implies (5y-4)(5z-4)=31\]
\[ \implies (y,z)=(7, 1), (1,7)\]
$z>y$, so $(x,y,z)=(3,1,7)$ but then we still have to order more since $y>x$ to give $(x,y,z)=(1,3,7)$ (we'll see later why we had to order when we examine $x=1$).  This gives $(a,b,c)=(2,4, 8)$.

$x=2$
\[ \implies \frac{1}{2}+\frac{1}{y}+\frac{1}{z}+\frac{1}{2y}+\frac{1}{2z}+\frac{1}{yz}=2\]
\[ \implies 3y+3z+2=3yz \]
\[\implies  \text{contradiction mod 3}\]

$ x=1$
\[ \implies 1+\frac{1}{y}+\frac{1}{z}+\frac{1}{y}+\frac{1}{z}+\frac{1}{yz}=2\]
\[ \implies 2y+2z+1=yz\]
\[ \implies (y-2)(z-2)=5 \]
\[\implies (y,z)=(7, 3), (3,7)\]

$z>y$ so we get $(x,y,z)=(1,3,7)$ as before giving the pair $(a,b,c)=(2,4,8)$.

Hence, our solutions are $(a,b,c)=\boxed{(2,4,8), (3,5, 15)}$ :).[/hide]

The most difficult process:

${(a-1)(b-1)(c-1) \mid abc-1-(a-1)(b-1)(c-1)=\sum{ab}-\sum{a}
\Rightarrow c \le \frac{1-a-b+2ab}{ab-2(a+b)+1}}$.
We may now see that
$a \ge 7 \Leftrightarrow 2ab-7(a+b)+3 \ge 14(a+b)-98-7(a+b)+3=7(a+b)-95 \ge 0
\Leftrightarrow c=\frac{1-a-b+2ab}{ab-2(a+b)+1} \le 4$.(The second line follows from the fact that $(a-7)(b-7) \ge 0$
Contradiction.So we are left to check the five cases $a=2,3,4,5,6$,which is long but not too boring.

[quote=Myth]Actually, ehsan2004 asked us "is it easy or not", he didn't asked us to solve the problem  :P  :P  :P
So I can answer to his question: IT IS EASY  :D[/quote]

o pidr kimga tawavossan abjag'ini chiqarib qoyaman sangamas bunaqa gaplani qiliw

it is very easy problem :x=a-1  y=b-1  z=c-1 :P

[quote=mortezamoradi]I think it's a new solution for this problem.
 [hide="new solution"]$1 < a < b < c$ , $abc-1=k(a-1)(b-1)(c-1)$

if $k=1 \implies abc-1=(a-1)(b-1)(c-1)\implies ab+bc+ca=a+b+c$
but we know that $ab+bc+ca>a+b+c$. so it can't happen.

if $k \not=1$, so we have $2$ conditions:

$condition1)$At least one of the $a,b,c$ is even$\implies abc-1$ is an odd number$\implies$ all of the $a,b,c$ are even and $k$ is an odd number.

$I) a\ge 4 \implies a> \frac{7+\sqrt {33}}{4} \implies (a-\frac{7}{4})^2>\frac{33}{16}\implies a^2-\frac{7}{2}a+1>0$
$\implies 3(a^2-2a+1)>a^2+a+1\implies 3(a-1)^3>a^3-1\implies 3(a-1)^3\ge a^3$
$\implies (a-1)\sqrt[3]{3}\ge a$ , by the same proof for $b,c$ we have: $(b-1)\sqrt[3]{3}\ge b , (c-1)\sqrt[3]{3}\ge c$
$\implies(a-1)\sqrt[3]{3}(b-1)\sqrt[3]{3}(c-1)\sqrt[3]{3}\ge abc \implies\boxed{3(a-1)(b-1)(c-1)>abc-1}$
but we know that :$abc-1=k(a-1)(b-1)(c-1) , k \not= 1$ ,k is an odd number $\implies\boxed{abc-1\ge 3(a-1)(b-1)(c-1)}$
so it can't happen.

$II)a=2,b\ge 6\implies b>3+\sqrt{6} \implies (b-3)^2>6 \implies 3(b-1)^2>2b^2$
$\implies (b-1)\sqrt{3}>b\sqrt{2}$ ,by the same proof for $c$ we have:$(c-1)\sqrt{3}>c\sqrt{2}$
$\implies (b-1)\sqrt{3}(c-1)\sqrt{3}>b\sqrt{2}c\sqrt{2}\implies \boxed{3(b-1)(c-1)>2bc-1}$
but we know that :$2bc-1=k(2-1)(b-1)(c-1) , k \not= 1$ ,k is an odd number $\implies\boxed{2bc-1\ge 3(b-1)(c-1)}$
so it can't happen.

$III)a=2,b=4 \implies 2*4*c-1=3k(c-1)\implies c-1|8c-1 , c-1|8(c-1)$
$\implies c-1|(8c-1)-8(c-1)=7 \implies c-1= 1 or 7\implies c=2or8 ,c>b>a=2$
$\implies \boxed {a=2,b=4,c=8}$

$condition2)$all of the $a,b,c$ are odd.

$I)a\ge 5 \implies a>\frac{5+\sqrt{21}}{2} \implies (a-\frac{5}{2})^2>\frac{21}{4} \implies a^2-5a+1>0$
$\implies 2(a-1)^2>a^2+a+1 \implies 2(a-1)^3>a^3-1 \implies 2(a-1)^3\ge a^3$
$ \implies (a-1)\sqrt[3]{2}\ge a$ , by the same proof for $b,c$ we have: $(b-1)\sqrt[3]{2}\ge b , (c-1)\sqrt[3]{2}\ge c$
$\implies(a-1)\sqrt[3]{2}(b-1)\sqrt[3]{2}(c-1)\sqrt[3]{2}\ge abc \implies\boxed{2(a-1)(b-1)(c-1)>abc-1}$
but we know that :$abc-1=k(a-1)(b-1)(c-1) , k \not= 1 \implies\boxed{abc-1\ge 2(a-1)(b-1)(c-1)}$
so it can't happen.

$II)a=3,b\ge 9\implies b>4+\sqrt{12} \implies (b-4)^2>12 \implies 4(b-1)^2>3b^2$
$\implies 2(b-1)>b\sqrt{3}$ , by the same proof for $c$ we have: $2(c-1)>c\sqrt{3}$
$\implies 2(b-1)2(c-1)>b\sqrt{3}c\sqrt{3}\implies \boxed{4(b-1)(c-1)>3bc-1}$
but we know that :$3bc-1=k(3-1)(b-1)(c-1) , k \not= 1\implies\boxed{3bc-1\ge 4(b-1)(c-1)}$
so it can't happen.

$III)a=3,b=7\implies 3*7*c-1=k(3-1)(7-1)(c-1)\implies 21c-1=12k(c-1)$
$\implies 3|21c-1 , 3|21c\implies 3|-1$ . so it can't happen.

$IIII)a=3,b=5 \implies 3*5*c-1=k(3-1)(5-1)(c-1)\implies 15c-1=8k(c-1)$
$\implies c-1|15c-1,c-1|15(c-1)\implies c-1|(15c-1)-15(c-1)=14\implies c-1=1,2,7,14$
$\implies c=2,3,8,15$ and $c>b>a=3,c=odd\not=8$
$\implies \boxed{a=3,b=5,c=15}$[/hide][/quote]

Correct

Solution from OTIS office hours:

The answers are $(2,4,8)$ and $(3,5,15)$ which are seen to work.

[b][color=red]Claim:[/color][/b]  	The numbers $(a,b,c)$ have the same parity.
[i]Proof.[/i]  	Otherwise $(a-1)(b-1)(c-1)$ is even but $abc-1$ is odd. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	We have $a < 4$.
[i]Proof.[/i]  	Since 	\[ 2 < \frac{abc}{(a-1)(b-1)(c-1)} 		< \frac{a}{a-1} \cdot \frac{a+2}{a+1} \cdot \frac{a+4}{a+3} \] 	this implies the result. $\blacksquare$

At this point let $x=a-1$, $y=b-1$, $z=c-1$; then the condition rewrites as \[ xyz \mid xy+yz+zx+x+y+z \] and we know $(x,y,z)$ are same parity.
We consider two cases. [list] 	[*]If $x=1$ then the equation reads  	\[ yz \mid 2(y+z)+1. \] 	This fails if $yz > 2(y+z)+1 \iff (y-2)(z-2) > 5$, 	so check $(y,z) \in \left\{ (3,5), (3,7) \right\}$.
[*]If $x=2$ then let $y=2r$ and $z=2s$; 	we get the equation \[ 4rs \mid 2rs + 3r + 3s + 1. \] 	[list] 		[*]If $4rs = 2rs+3r+3s+1$ then $(2r-3)(2s-3) = 11$, 		which gives $r=2$ and $s=7$, which works. 		[*]Otherwise we have $8rs > 2rs+3r+3s+1$, 		so the equation fails. 	[/list] [/list]

Using a function and bounding that makes this problem a casework I think I have done the same problem in some imo i don't remember the year

The only solutions are $(2, 4, 8)$ and $(3, 5, 15)$ which both work.

[b][color=#f00]Claim:[/color][/b] $a<4.$
[b]Proof:[/b] For $a \ge 4,$ we have $$1 < \frac{abc-1}{(a-1)(b-1)(c-1)} < \frac{a}{a-1} \cdot \frac{b}{b-1}\cdot \frac{c}{c-1} \le \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5}=2.$$ This is a contraction because there are no integers between $1$ and $2.$ $\Box$
Now, we consider $2$ cases.

[b]Case 1:[/b] $a=2.$
We want $$\frac{2bc-1}{(b-1)(c-1)} \in \mathbb{N}, 2<b<c.$$
We have $$1 < \frac{2bc-1}{(b-1)(c-1)}<\frac{2bc}{(b-1)(c-1)} \le 2 \cdot \frac{3}{2} \cdot \frac{4}{3}=4.$$ So, we must have $\frac{2bc-1}{(b-1)(c-1)}=3,$ since its always odd. If $b \ge 6,$ we have $1< \frac{2bc-1}{(b-1)(c-1)} < \frac{2bc}{(b-1)(c-1)} \le 2 \cdot \frac{6}{5} \cdot \frac{7}{6}=\frac{14}{5} <3.$ So, $b<6 \implies b \in \{3, 4, 5\}.$

[b]Subcase 1:[/b] $b=3.$
If $b=3,$ we have $$\frac{6c-1}{2c-2}=3 \implies 6c-1=6c-6 \implies \text{ contradiction.}$$

[b]Subcase 2:[/b] $b=4.$
If $b=4,$ we have $$\frac{8c-1}{3c-3}=3 \implies 8c-1=9c-9 \implies c=8 \implies (a, b, c)=(2, 4, 8).$$

[b]Subcase 3:[/b] $b=5.$
If $b=5,$ we have $$\frac{10c-1}{4c-4}=3 \implies 8c-1=12c-12 \implies c=\frac{11}{2} \implies \text{ contradiction.}$$
Now if you follow the same steps for the other case, you will end up with the solution $(3,5,15),$ and these are the only $2$ solutions. $\blacksquare$

Note first that:
$$1>\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac a{a-1}\cdot\frac b{b-1}\cdot\frac c{c-1}\le2\cdot\frac32\cdot\frac43=4$$
since $a\ge2$, $b\ge3$, and $c\ge4$, from which it follows that either $\frac{abc-1}{(a-1)(b-1)(c-1)}=2$ or $\frac{abc-1}{(a-1)(b-1)(c-1)}=3$.

Suppose that $a>3$. Then we obtain the stronger bound:
$$\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac a{a-1}\cdot\frac b{b-1}\cdot\frac c{c-1}\le\frac43\cdot\frac54\cdot\frac65=2,$$
a contradiction. So either $a=2$ or $a=3$.

[b]Case 1:[/b] $a=2$
Then $\frac{2bc-1}{(b-1)(c-1)}$ is an integer, which implies that $b$ and $c$ are both even since the numerator is odd. So $b\ge4$ and $c\ge6$. Also, since $\frac{2bc-1}{(b-1)(c-1)}$ is odd, it must be $3$. By SFFT, $(a,b,c)=\boxed{(2,4,8)}$.

[b]Case 2:[/b] $a=3$
Then $\frac{3bc-1}{2(b-1)(c-1)}$ is an integer. Since
$$\frac{3bc-1}{2(b-1)(c-1)}<\frac32\cdot\frac b{b-1}\cdot\frac c{c-1}\le\frac32\cdot\frac43\cdot\frac54<3$$
the integer must be $2$. Finally SFFT produces the only other solution $(a,b,c)=\boxed{(3,5,15)}$.

[hide = Proof] We use the substitution $x = a - 1, y = b - 1, z = c - 1$. Then the fraction \[\frac{abc - 1}{(a - 1)(b - 1)(c - 1)} = \frac{(x + 1)(y + 1)(z + 1) - 1}{xyz} = 1 + \frac{x + y + z + xy + yz + zx}{xyz}.\] Now we have that as $\frac{x + y + z + xy + yz + zx}{xyz} \geq1$, that \[\sum_{cyc}^{3} x + \sum_{cyc}^{3} xy \leq xyz + \sum_{cyc}^{3} xy < 3xyz.\] Now we have going back the original fraction that $\frac{x + y + z + xy + yz + zx}{xyz} < 3$. Now as the problem requires integer values we have to solve $x + y + z + xy + yz + zx = xyz$. Also if $x \geq 3$, then the original fraction is less than $1$. Thus $x \leq 2$.

The first equation $(x =2)$ yields $(y - 3)(z - 3) = 11 \implies y = 4, z = 14$.

The second equation $(x = 1)$yields $(y - 2)(z - 2) = 5 \implies y = 3, z = 7$. Now undoing the substitutions yields the two triples \[(a, b, c) \in \{(3, 5, 15), (2, 4, 8)\}.\] $\blacksquare$ [/hide]

Let
\begin{align*}
x&=a-1\\
y&=b-1\\
z&=c-1\\
0<x<y<z
\end{align*}

$$xyz \vert xyz + xy + xz + yz + x + y + z$$,
So we find all $x,y,z$ with $xyz \vert xy + xz + yz + x + y + z$
Here an important step is to note the difference in degree and the contrary inequality,
$$xyz \leq xy + xz + yz + x + y + z$$
Which can be used for bounding.

Since $x,y,z$ are distinct positive integers,
$$xy + xz + yz + x + y + z \leq \dfrac{17}{6} xyz$$
With equality when $x=1,y=2,z=3$.

Thus it suffices to consider
$$xyz = xy + xz + yz + x + y + z$$
And
$$2xyz = xy + xz + yz + x + y + z$$

Consider $xyz = xy + xz + yz + x + y + z$
If $x \geq 6$, $xyz \geq 6yz > xy + xz + yz + x + y + z$, no soln.

If $x=5$, we want to solve $4yz=6y+6z+5$.
However $4yz \geq 24z > 6y+6z+5$, so no soln.

If $x=4$, we want $3yz=5y+5z+4$.
However $3yz \geq 15z > 5y+5z+4$, so no soln.

If $x=3$, we want $2yz=4y+4z+3$.
However $2yz \geq 8z \geq 4y+4z+4 > 4y+4z+3$, no soln.

If $x=2$, we want $yz=3y+3z+2$.
If $y \geq 6$, $yz \geq 6z \geq 3y+3z+3 > 3y+3z+2$, no soln.
If $y=5$, $5z=17+3z$, $z \notin \mathbb{Z}$, no soln.
If $y=4$, $4z=14+3z$, giving $z=14$ and the soln set (2,4,14).
If $y \leq 3$, $yz \leq 3z < 3z + 3y + 2$, no soln.

If $x=1$, $0=2y+2z+2$, no soln.

Now consider $2xyz=xy+xz+yz+x+y+z$.

If $x \geq 3$, $2xyz \geq 6yz > xy+xz+yz+x+y+z$, no soln.

If $x=2$, we want $3yz=3y+3z+2$
However $3yz \geq 9z > 3y+3z+2$, no soln.

If $x=1$, we want $yz=2y+2z+1$.
If $y \geq 4$, $yz \geq 4z \geq 2z+2y+2 > 2y+2z+1$, no soln.
If $y=3$, $3z=2z+7$ gives $z=7$, and the soln set $(1,3,7)$.
If $y \leq 2$, $yz \leq 2z < 2y+2z+1$, no soln.

Thus the only solns to $(x,y,z)$ is
$$(x,y,z) = (2,4,14), (1,3,7)$$,
Corresponding to
$$(a,b,c) = (3,5,15), (2,4,8)$$.
$\square$

[hide=500th post to an IMO problem!]first of all we have $1<\frac{abc-1}{(a-1)(b-1)(c-1)}< \frac{abc}{(a-1)(b-1)(c-1)}$\\

[b][color=#f00]Claim:-[/color][/b]$a< 4$\\

[color=#000]pf:-[/color] FTSOC assume $a\ge 4$\\

we have $\frac{abc}{(a-1)(b-1)(c-1)}\le 2 \implies 1<\frac{abc-1}{(a-1)(b-1)(c-1)}<2$ clearly this is not possible as $\frac{abc-1}{(a-1)(b-1)(c-1)}$ is an integer , hence we must have $1<a<4$ $\blacksquare$\\

[color=#f00][b]Claim:-[/b][/color] $(a,b,c) \in \{(2,4,8),(3,5,15)\}$ is the only set of solution\\

[color=#600]pf:-[/color] well since we have $a=2$ or $3$ as the only possible working value , we work with:\\

[b][u][/u][color=#000]case 1:-[/color][/b]$a=2$\\

then we have $z=\frac{2bc-1}{(b-1)(c-1)}\in \mathbb{Z}$\\

also since $b>2$ and $c>3$ we have $z=\frac{2bc}{(b-1)(c-1)}=3$ or $4$\\

[color=#000][b][u]subcase 1.1:-[/u][/b][/color] $z=3$\\

from this case we get $(b-3)(c-3)=5\implies b=4,c=8$ which works\\

[color=#000][b][u]subcase 1.2:-[/u][/b][/color]$z=4$\\

clearly from this case we get $(2b-4)(c-2)=3$ which really don't works\\

[b][u][color=#000]case 2:-[/color][/u][/b]$a=3$\\

we mark $p=\frac{3bc-1}{(b-1)(c-1)}$ so since $b>3 , c>4$ we get $3<p\le 5$ so $p=4$ or $5$\\

[b][u][color=#000]subcase 2.1:-[/color][/u][/b]$p=4$\\

from this case we get $(b-4)(c-4)=11\implies b=5 , c=15$ which works\\

[b][u][color=#000]subcase 2.2:-[/color][/u][/b]$p=5$\\

we really don't get any $b,c$ to be working\\

hence clubbing all the results we got from cases we get $\boxed{(a,b,c) \in \{(2,4,8),(3,5,15)\}}$ as the only solution set $\blacksquare$[/hide]

Expand $(a-1)(b-1)(c-1)$ into $abc-ab-bc-ac+a+b+c-1$
Then, $\frac{abc-1}{abc-ab-bc-ac+a+b+c-1}$ simplifies to $1+\frac{ab+bc+ac-a-b-c}{abc-ab-bc-ac+a+b+c-1}$
$\frac{ab+bc+ac-a-b-c}{abc-ab-bc-ac+a+b+c-1}$ then simplifies to $\frac{a(b-1)+b(c-1)+c(a-1)}{(a-1)(b-1)(c-1)}$
Even more simplifying gives $\frac{a}{(a-1)(c-1)}+\frac{b}{(a-1)(b-1)}+\frac{c}{(b-1)(c-1)}$, which is less than $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ which is less than $3$ so $\frac{abc-1}{(a-1)(b-1)(c-1)}$ is less than $4$ but also greater than $1$.
Therefore, $\frac{abc-1}{(a-1)(b-1)(c-1)}$ is equal to $2$ or $3$.
Solve we get that $a=2$ or $a=3$ since $a>3 \implies \frac{abc-1}{(a-1)(b-1)(c-1)}<2$ which makes $\frac{abc-1}{(a-1)(b-1)(c-1)}$ neither equal to $2$ nor equal to $3$.
Plugging in $a=2$, we get $b=4$ and $c=8$ by solving $\frac{2bc-1}{(b-1)(c-1)}=3$ and no solutions for $\frac{2bc-1}{(b-1)(c-1)}=2$
Plugging in $a=3$, we get $b=5$ and $c=15$ by solving $\frac{3bc-1}{2(b-1)(c-1)}=3$ and no solutions for $\frac{3bc-1}{2(b-1)(c-1)}=2$
Therefore, the only solutions for $(a,b,c)$ are $(2,4,8)$ and $(3,5,15)$

I resolved this one.

The answer is $\boxed{(2,4,8)}$ and $\boxed{(3,5,15)}$, which work.

Let $x = a-1, y = b-1, z = c-1$.

We have $xyz \mid xy + yz + zx + x + y + z$.

First note that $xy > x + y + 1$ for any integers $x \ge 3, y \ge 3$ because it's true for $(3,3)$ and stays true when you add $x$ or $y$ by $1$.
Claim: We have $xyz > xy + yz + zx + x + y + z $ whenever $x \ge 3$.
Proof: This is because it's true for $x = 3, y = 4, z = 5$, and adding $1$ to any variable keeps it true since it changes the LHS by $xy$ (for example) and the RHS by $x + y + 1$, and $xy > x + y + 1$ for $x, y \ge 3$.  $\square$

This means that either $x = 1$ or $x = 2$.

Case 1: $x=1$
Then $yz\mid 2(y+z) + 1$. We can do a similar trick as above for proving that $yz > 2(y+z) + 1$ for $y\ge 4$ since it's true for $(4,5)$ and when you increase either variable by $1$ it stays true. Therefore, either $y = 2$ or $y = 3$.

If $y =2$, then $2z \mid 2z + 5$, bad.

If $y = 3$, then $3z\mid 2z + 7$, so we must have $z = 7$, which works.

Hence we have $(x,y,z) = (1,3,7)$, which corresponds to $(a,b,c) = (2,4,8)$. $\square$

Case 2: $x = 2$.
Then $yz \mid 3(y+z) + 2$. We see that $yz > 3(y+z) + 2$ for $(y,z) = (6,7)$ and for $y,z$ that large, increasing either variable by $1$ results in an increase more than in the RHS (3), so it's true for any $y \ge 6$. Hence we must have $y\in \{3,4,5\}$.

If $y = 3$, then $3z\mid 3z + 11$, bad.

If $y = 4$, then $4z\mid 3z + 14$, so $z\mid 14$ and $z\equiv 2\pmod 4$, so $z = 14$.

If $y = 5$, then $5z\mid 3z + 17\implies z = 17$, which doesn't work.

Therefore, $(x,y,z) = (2,4,14)$, so $(a,b,c) = (3,5,15)$.


integration
calculus
calculus computations

Hello, i am trying to solve this nonlinear ODE

$ y'y'' \equal{} \minus{} 1$

can someone help me?

p.s ok... my solution is $ 2y'y'' \equal{}  2 \Rightarrow y' \equal{}\pm\sqrt {2(c_1 \minus{} x)}\Rightarrow y(x) \equal{} \pm \frac {2\sqrt {2}}{3}\sqrt {(c_1 \minus{} x)^3} \plus{} c_2$

I'm just starting to learn how to solve these, but maybe the following would help?

Let $ y^{'} \equal{} \frac{dy}{dx} \equal{} w$
Then
$ y^{''} \equal{} \frac{dw}{dx}$
Dividing both sides by $ y^{'}$, we have
$ \frac{dw}{dx} \equal{} \frac{\minus{}1}{w}$ which is solvable in the following manner:

$ \int w dw \equal{} \minus{}\int 1 dx$
$ \frac{w^2}{2} \equal{} \minus{}x \plus{} C_1$
$ w \equal{} \pm \sqrt{2C_1 \minus{} 2x}$
Let $ 2C_1 \equal{} C_2$
$ w \equal{} \pm \sqrt{C_2 \minus{} 2x}$
Now that we've got this, substitute back in for $ w$
$ \frac{dy}{dx} \equal{} \pm \sqrt{C_2 \minus{} 2x}$
We'll deal with the positive root for now, and add the $ \pm$ later.
$ \int 1 dy \equal{} \int (C_2 \minus{} 2x)^{\frac{1}{2}} dx$
$ y \equal{} \pm \frac{1}{3} (C_2 \minus{} 2x)^{\frac{3}{2}} \plus{} C_3$

This is basically the same as your solution, except I've concatenated the constant C_1 into C_2, and hence the multiplier on the final term is different as a result.

I'm not seeing where you have problems, except as mlok already pointed out you need a negative.


Find the smallest positive integer, $n$, which can be expressed as the sum of distinct positive integers $a,b,c$ such that $a+b,a+c,b+c$ are perfect squares.

Ideally $A^2=a+b,B^2a+c,C^2=b+c$ would be consecutive squares (bicose our number is $(A^2+B^2+C^2)/2$

$A \geq 2$.
$2^2,3^2,4^2$ and $4^2,5^2,6^2$ can't work (sum must be even).
$3^2,4^2,5^2$ does not works
$5^2,6^2,7^2$ works !!

$n = 55$ is smallest


geometry
3D geometry
icosahedron
parallelogram
trigonometry
trig identities
Law of Cosines

Two aliens are situated on opposite vertices A and B of a planet shaped like a regular icosahedron, with edge 1. What is the length of the shortest path from A to B along the surface of the planet ?

Can you tell me please what FMO means. In will also be nice if you tell us from which year, round, grade etc. the problem is.  :)

Welcome to ML/AoPS.  :)

FMO means Flanders Mathematical Olympiad. It's from the second round of the 2000-2001 FMO. You can find a lot of problems of these competitions on the site of FMO.

Nice problem!  Here's my solution:

[hide]
The path from A to B can go through either 3 or 4 faces of the icosahedron, depending whether it goes through another vertex on its way or not.
If it does, we can unfold the three faces into three equilateral triangles on top of one another in a plane and the shortest distance between the two points is the straight line connecting them, of length $3\sqrt3/2$, or approximately 2.598.
If it doesn't, unfold the four triangles it goes through similarly and we have a parallelogram with sides 1 and 2 and A and B are opposite corners.  The shortest path is the diagonal, let's say of length c.  Then by law of cosines,
$c^2=1^2+2^2-2*1*2*cos(120)=7$
$c=\sqrt7$, approx. 2.646
Thus the shortest path is $3\sqrt3/2$.

[/hide]


Hi there, I'm new to the forums and I was hoping someone could help me with a few questions. I hope this is the right forum for this question.

1) Show that 3^90 is logically equivalent to 1 (mod 91)

2) Use Fermat's Little Theorem ( n^(p-1) is equiv. to 1(modp) where p is a prime that does not divide n ) to show that if 11 divides n+1, then 11 divides N^57 - N^3

I think I did #1 correctly. I computed [(3^16 mod91)^5 (3^8 mod91) (3^2 mod 91)]mod 91 and got 1.

For #2 I subsitiuted  p=11, n = a + 1 a =21. And used the theorem to show that there were no remainders. Is there a way to calculate this for any value? Rather than just some arbitrary value?

Thank you in advance for any help you can offer.

For 1 there's a less brute-force solution: thus 91=7.13, we just have to prove that
$3^{90}\equiv 1 (mod 7)$, which follows from Fermat's litle diretly because 6 divides 90
and that
$3^{90}\equiv 1 (mod 13)$ which follows from $3^3 \equiv 27 \equiv 1 (mod 13)$.
because both 3 and 7 divide $3^90-1$, so 91 divides it too.

If i understand the second well, we have
$n\equiv -1 (mod 11)$ and have to show
$n^{57}-n^3\equiv 0 (mod 11)$ which follows from
$-1-(-1)\equiv 0 (mod 11)$.
This seems too easy, are you sure about this problem?

To solve such problems, it's good to know congrouences.
http://mathworld.wolfram.com/Congruence.html


algorithm
induction
number theory unsolved
number theory

suppose that n is positive integer.
in each time we sum n with the sum of its even digits in decimal .
prove that after some of this algorithm the sequence will be constant.
for example:
$ 12 \rightarrow 14 \rightarrow 18 \rightarrow 26 \rightarrow 34 \rightarrow 38 \rightarrow 46 \rightarrow 56 \rightarrow 62 \rightarrow 70$

Fun problem!

[hide="informal explanation"]
Let $ f(n) \equal{} n \plus{}$ sum of even digits of $ n$ in base 10. The problem is to show that for each $ n$, the sequence $ f(n)$, $ f(f(n))$, $ f(f(f(n)))$ etc. eventually stabilizes.

First, note that the claim is true for $ n<10$:
$ 1 \to 1$, as well as of course $ 3 \to 3$, $ 5 \to 5$, $ 7 \to 7$ and $ 9 \to 9$
$ 2 \to 4 \to 8 \to 16 \to 22 \to 26 \to 34 \to 38 \to 46 \to 56 \to 62 \to 70$
$ 6 \to 12 \to 14 \to 18 \to 26 \to \ldots \to 70$ (we've already dealt with 26).

Also, notice that with these starting points, the sequence hits a fixed point smaller than 100. This is important!

Now, notice that if $ n < 100$ then $ f(n) < 110$ (this isn't hard to prove). This means that if we start with any number below 100, the sequence either hits a fixed point below 100 or it jumps to some number between 100 and 109. But starting from such numbers, the sequence behaves exactly like the sequences beginning in $ 0, \ldots 9$, since the 1 in the 100's place doesn't change anything. [b]If [/b]it was true that such a sequence could surpass 200, then we would be in trouble since that 2 in the beginning would change the behavior. But we just saw that all sequences starting under 10 never pass 100, so the sequence starting in $ \{100, \ldots, 109\}$ never passes 200, and in particular it must stop.

Therefore, we can conclude: If $ n < 100$, the sequence stops below 200.

Consider now the starting points $ n < 1000$. It's even easier than before to show that if $ n < 1000$ then $ f(n) < 1100$. Therefore, for all $ n < 1000$, the sequence either stops or it lands somewhere between 1000 and  1100. But starting with these numbers, the sequence behaves just like the sequences starting under 100, since the 1 in the 1000's place doesn't do anything. We just saw that all such sequences freeze before they reach 200. So, all sequences starting between 1000 and 1100 freeze before they reach 1200, and finally:

If $ n < 1000$, the sequence stops below $ 1200$.

The pattern should be clear by now...
[/hide]

[hide="More formal proof"]

Definition: for a natural number $ n$ let $ n \equal{} a_0\plus{}10a_1\plus{}100a_2\plus{}\ldots 10^k a_k$ be the base-10 expansion of $ n$. Let $ e(n) \equal{} \sum_{i, 2|a_i} a_i$ be the sum of the even digits of $ n$, and let $ f(n) \equal{} n\plus{}e(n)$.

Lemma: If $ n < 10^k$ for $ k > 1$ then $ f(n) < 10^k \plus{} 10^{k\minus{}1}$.

Proof: If $ n < 10^k$ then $ n$ has $ k$ digits, and the sum of the even digits is at most $ 8k$, so $ f(n) \leq n\plus{}8k$. If $ k>2$ then $ 8k < 10^{k\minus{}1}$ holds trivially. For $ k\equal{}2$ the bound is $ f(n) \leq n\plus{}16$, so it's enough to check that $ 98 \to 106$, $ 96 \to 102$, $ 94 \to 98$ all remain below 110. For smaller numbers the bound of $ n\plus{}16$ suffices to show that $ f(n) < 110$. QED

We now prove by induction on $ k$ that if $ n < 10^k$ then the sequence $ a_1 \equal{} n$, $ a_2 \equal{} f(n)$, etc. with $ a_{m\plus{}1} \equal{} f(a_m)$ satisfies $ a_m < 10^{k\plus{}1}$ for all $ m$. In particular, it must be constant from some point.

For $ k\equal{}1$ this is done "by hand" by checking the sequences starting with 2, 4, 6 and 8. Suppose now that the claim is true with $ k\minus{}1$, and we'll prove it for $ k$.

Let $ a_1 \equal{} n < 10^k$ be given. The sequence $ a_1, a_2, \ldots$ either remains below $ 10^k$ in which case it must stabilize, or at some point we have $ a_m < 10^k$ and $ a_{m\plus{}1} \equal{} f(a_m) \geq 10^k$. By the lemma, $ a_{m\plus{}1} < 10^k \plus{} 10^{k\minus{}1}$. Let $ b_1 \equal{} a_{m\plus{}1} \minus{} 10^k$, and define again $ b_2 \equal{} f(b_1)$, $ b_3 \equal{} f(b_2)$ etc.

The base-10 expansion of $ a_{m\plus{}1}$ is identical to that of $ b_1$ except that it has an additional 1 as the leftmost digit. Therefore $ e(a_{m\plus{}1}) \equal{} e(b_1)$, so $ a_{m\plus{}2} \equal{} 10^k \plus{} b_2$ and in general $ a_{m\plus{}j} \equal{} 10^k \plus{} b_j$ holds as long as $ b_j$ doesn't exceed $ 10^k$.

By the induction hypothesis, the sequence $ b_1, b_2, \ldots$ satisfies $ b_j < 10^k$ for all $ j$, so the equation $ a_{m\plus{}j} \equal{} 10^k\plus{}b_j$ remains true, and in particular the sequence $ a_m$ never exceeds $ 2*10^k < 10^{k\plus{}1}$. This is what we needed to prove - QED.
[/hide]


find all functions such that for any polynom p we have : f(p(x))=p(f(x))

If $f(a)\neq a$ for some $a$ then choose $p(x)=\frac {(f(0)+1)(x-a)} {f(a)-a}$ so,$p(f(a))=f(p(a))+1$.

you must write: $ p(x)= \frac{(f(p(a))+1)(x-a)}{f(a)-a} $

[quote="kam"]because $ p(a)=0$[/quote]

chose p(x)= ax^n , a, reel number, n naturel number.Take x=1.Check that f(1)= 1, then f(a)=a.


for $ n\equal{}1$ we have $ 3^1\equal{}1^1\plus{}2^1$ so 3 is good.
for $ n\equal{}2$ we have $ 5^2\equal{}3^2\plus{}4^2$ so 5 is good.
for $ n\equal{}3$ we have $ 6^3\equal{}3^3\plus{}4^3\plus{}5^3$ so 6 is good.
I also found good numbers for $ n\equal{}4,5,6,7,8$.

is my conjecture:

for all $ n$ there exists $ n$-th power which is a sum of some $ n$-th powers, not all of them equal 1.

true or not?

Any big number $ A$ can be represented in form $ A \equal{} \sum_i x_i^n$.
But if you interested only case, when $ x_i \equal{} i \plus{} a$ and $ A \equal{} m^k \equal{} \sum_{i \equal{} a}^b i^k$, then it may be false for big k.

[quote="knoppix"]for $ n \equal{} 1$ we have $ 3^1 \equal{} 1^1 \plus{} 2^1$ so 3 is good.
for $ n \equal{} 2$ we have $ 5^2 \equal{} 3^2 \plus{} 4^2$ so 5 is good.
for $ n \equal{} 3$ we have $ 6^3 \equal{} 3^3 \plus{} 4^3 \plus{} 5^3$ so 6 is good.
I also found good numbers for $ n \equal{} 4,5,6,7,8$.

is my conjecture:

for all $ n$ there exists $ n$-th power which is a sum of some $ n$-th powers, not all of them equal 1.

true or not?[/quote]

I Think you should redefine your question because
$ k^n\equal{}(k\minus{}1)^n\plus{}(k^n\minus{}(k\minus{}1)^n)\cdot1^n$

[quote="knoppix"][quote="Rust"]Any big number $ A$ can be represented in form $ A \equal{} \sum_i x_i^n$.[/quote]
How can it be proved?

Do you mean "any big number" as "for given $ n$ there is $ C \equal{} c(n)$ such that any $ A > C$ can be represented" ?
If yes, is this constant known or we know only that it exists?

Rafikafi, $ k^n \minus{} (k \minus{} 1)^n$ need not be a $ n$-th power  :wink:[/quote]

It doesn't have to. What I meant is actually if $ k > 2$ then $ k \minus{} 1\neq  1$
And you have on the right side once the n-power of $ k \minus{} 1$ and $ (k^{n} \minus{} (k \minus{} 1)^{n})$ times the n-power of $ 1$, so not all equal to $ 1$

oh, yes  :blush:

in my examples all numbers are different, so let's add this assumption.
what's the answer now?

There's the [url=http://en.wikipedia.org/wiki/Waring%27s_problem]Waring's problem[/url] I know of which is quite similar, but if all have to be diffrent things are getting more complicated, but its probably true.


AMC
AIME
trigonometry
USA(J)MO
USAMO
geometry
AIME I
\/closed

Ok, so is woot really worth the money for me to take sophomore year? i can consistently get like 7-9 on aime (8 this year) and i know it says that at 5 you should be ready for it but is it really wise to take it this early? or would i be better suited for junior year?

That depends on how much effort you're willing to put into it.  The value of taking it earlier is that you'll be less busy in school; I took WOOT junior year, which turned out to be the busiest year of my high school experience, so I wasn't able to devote as much time to it as I would've liked.  If you're really willing to commit to it, or if you don't plan on being significantly busier junior year than sophomore year, it might be a good idea to take it earlier.

I'm taking WOOT this year, and I'd say that WOOT really gets its worth from the personal effort you put into it. WOOT offers you plenty of resources: practice Olympiads and AIMEs, problem sets, and handouts, but it's up to you personally to spend the time and make use of it. Like during winter break, several of us would go on every night and work through ISL problems -- that was definitely a great learning experience.

So if you're going to be lazy, WOOT's not well worth your money, but with 8's on the AIME, I think you have the ability to learn quite a bit from the class.

I was a junior last year but I could only get 2s and 3s on practice AIMEs (4 on the actual 2008 I). I took WOOT this year, even though I didn't meet the suggested AIME scores, because it was going to be my senior year and I wouldn't have another opportunity to take it (and because senior year is significantly less busier in terms of school work than all of the other high school years in my opinion). Through practice AIMEs (both WOOT and previous ones), I consistently got 9s in practice and 10 on the actual 2009 I. Personally, I think that your decision to take WOOT or not shouldn't really be based on whether you know enough math or not: you can learn the math over the summer if you really put in the effort. Just consider whether you will have the time to devote to the class to get your money's worth.

also, how much of intermediate trig + complex numbers is covered in aops vol. 2? i'm trying to decide if i should get the book once it comes out. if it turns out most of it is covered in vol. 2, then i'm considering taking woot this year.

to get ready for woot, what should i review? i'm 3/4 done with aops 2, should i do interm counting or nt?

[quote="sunehra"]I'm assuming by "book" you are referring to the Precalculus book that will be coming out. AoPS Vol. 2 only has 1 chapter on complex numbers and another chapter on just trig. What it covers is probably good enough for the AIME, but I guess the precalculus book will go more in-depth as did the other introduction/intermediate AoPS texts.[/quote]
Actually, the Trig/Complex numbers book. Mr. Rusczyk says he should be done around the end of summer.

@simo: i am also very tempted to take woot this year, but i'm sort of afraid i wont be ready for it. according to jvenezuala, it's more about the effort that you put into woot than preparation before it. among your two choices, they both seem to have much of the material already in aops vol. 2, just with more problems. otherwise, i would go with interm. algebra.

Oh, and could somebody tell me the format of a chapter in aops vol. 2? for example, i know the format in the introduction/intermediate series is:

Lesson, ~10 Exercises after each lesson, ~20-30 Review problems at the end of the chapter along with a bunch of challenge problems.

[quote="sunehra"]I'm assuming by "book" you are referring to the Precalculus book that will be coming out. AoPS Vol. 2 only has 1 chapter on complex numbers and another chapter on just trig. What it covers is probably good enough for the AIME, but I guess the precalculus book will go more in-depth as did the other introduction/intermediate AoPS texts.[/quote]

That's correct -- the Interm level books are all much more in-depth than AoPS2.  Someone who finishes AoPS Vol 2 will still have quite a bit to learn from the Intermediate books.  For example, I'm just finishing a chapter on complex number geometry, which isn't really addressed much at all in AoPS2.

simo: Either of those classes would be a fit pre-WOOT (or alongside it).  The three groups of WOOT are only for the practice olympiads -- each practice olympiad has three levels, and you can switch which level you do from test to test.  (You are given all three tests and solution sets, so you will have access to all the testing material.)

qwerty: Vol2 is not as in-depth as the Interm books.  The individual chapters are structured as a survey of a subject, with examples and exercises scattered through the text, followed by a list of chapter-end problems, many of which are drawn from contests.  This is somewhat similar to the newer books, with the big exception that in the newer books, we give questions *before* the lessons, to give you a chance to develop the lesson on your own, rather than having us simply tell you the math.  That way, you discover the math, and it becomes yours, not ours.

I very much agree with the assertions that what you get out of WOOT will mainly depend on what you put into it.

i was wondering: what is the structure of a woot class? do you do problems or do you leave the problems for the practice olympiads and just do concepts? does it start out with problems like the intro and interm books? also: i'm scared about the fact that i can't solve a usamo problem yet and will be lost.

If you've taken any other AOPS classes, it's basically the same with a more heavy emphasis on doing problems and less on concepts. Concepts are almost entirely covered in handouts; the class shows how they may be applied.

The problems are not just on the practice olympiads. There are problems done in class, posted on the message board (tons), in the practice olympiads, in the handouts, and occasionally done in the chatroom that's open 24/7. For a good reason: to improve [i]you do problems[/i].

If you can qualify for USAMO but can't solve anything at that level, I believe you are actually among WOOT's target audience.

As a side- my parents are really reluctant about paying for woot, so i'm weighing my options here. which one do people think will be the best for my situation.
1. take the aime class (i can get like a 9 right now on aime, so i'm wondering if the class will still help me).
2. finish aops 2, go through acops, and do a TON of problems.
3. take interm number theory, go through the interm counting book, and also do a TON of problems.
I still might do woot, but im just trying to see if some other cheaper options might still help.

Oh shoot. I just realized that I'm going to be a sophomore next year too, and WOOT is something I've always wanted to do, I just didn't realize it was already going to be sophomore year!

Hmm, right, the reasons here to take WOOT sophomore year are obvious. I estimate that I'll have about half as much homework then as I will in junior year, and I fear senior year might be a little late for anything in particular. I'll have to ask my parents as well, and I noticed that its a lot cheaper to register before July 15th.

 :lol: Lol, I guess I'll just have to ask them to register me for WOOT rather than buying me a cell phone  :P

[quote="simo14"]
1. take the aime class (i can get like a 9 right now on aime, so i'm wondering if the class will still help me).[/quote]

Though I've never taken the AIME class, I don't see how you could benefit from it if you are getting 9s now. A 120/9 will almost definitely get you into USAMO, at which point you should be focusing on Olympiad problems.

To the contrary, the AIME Problem Series Classes focus a lot more on the later problems (11-15) than the earlier and medium problems, and so still would improve chances of making USAMO (stupid mistakes are killer!).

As for the three options:

I'd pick the second option, as it should hopefully give you a broader range of knowledge, rather than focusing on specific topics.  The AIME Problem Series would also give you this, but you didn't add "and do a TON of problems" to the end of it :P

so i'm just wondering about the usefullness of acops as it applies to usamo and aime problems. because i've seen the table of contents and almost all of it is covered in aops 2.. like what's the structure of the chapters and about what level of difficulty/what type are the problems? Also, what would be the advantages and disadvantages of going through acops instead of just doing late aime and usamo problems?
Also, what kind of work load does woot have (yes, my parents are finally considering it)? Like how many message board problems do you get per week/day etc.
Thanks again

ACoPS goes a bit beyond AoPS2 in general problem solving strategies, and has some more challenging problems that include strategies that are common to the USAMO.

As for WOOT, it's largely what you make of it---it can be a lot of work if you want to take the time to do it.  Each of the 16 classes comes with around a dozen message board problems.  There are also 3 "Problem Of The Day"s each week during most of the course to challenge the students and inspire discussion.  There are extra problems in the handouts throughout the course, and there are 8 practice tests of various types (mostly practice olympiads).

[quote="simo14"]ok, so to get me to ~12 aime next year (4 pt increase) and able to solve 1 usamo problem, how much would the weekly time commitment be?

Also, I heard that there were 3 groups of woot. could someone explain each of them please?

@qwertythecucumber: the structure is concept, ~5 excercises, concept, ~5 excercises,concept, ~5 excercises.... and then ~20-25 tough problems arranged from easy aime to very easy olympiad/ hard aime.[/quote]

There are three groups of WOOT? Also I don't remember having so many problems last year.

1. The three groups are probably referring to the three sections of the practice olympiads. That's a misleading distinction to make though, since you can switch which section you do as you please and nowhere else do they separate people like that.

2. simo was responding to a question about AOPS2, not WOOT.


I'm sure I missed something, because this has nothing to do with primes (at least that I can see), but here goes:
$f(x)=4x$ or $f(x)=\sin(4x)$ or $f(x)=4\sin(x)$.
$g(x)=2x$ or $g(x)=\sin(2x)$ or $g(x)=2\sin(x)$.
$f(0)=0$, $f'(0)=4$, and $\lim_{x\to 0} \frac{g(x)}{f(x)} = \frac{1}{2}$.

Jimmy: there's a way to say what you wanted to say. Go look at my "big-O and litle-o" topic.

We know that (as $x\to0$), $f(x)=4x+o(x)$ and $g(x)=2x+o(x).$ We also then know that $f(0)=0$ and $f'(0)=4.$ If we insist that $g(0)=0$ so that $g$ is continuous at 0, then $g'(0)=2.$

As for the quotient,

$\frac{g(x)}{f(x)}=\frac{2x+o(x)}{4x+o(x)}= \frac{1+o(1)}{2+o(1)}\to\frac12.$

Thanks Kent, the "big-O, little-o" makes this problem much clearer, but I still can't see what this has to do with primes.

[quote="spartansludge"]f is diferentiable at zero and

lim        f(x)/x=4      lim         g(x)/x=2
x ->  0                    x->  0

find f(0), f'(0), lim        g(x)/f(x)
                     x -> 0[/quote]

Okay.  It's clear that $f(0)=0$, because otherwise $f(x)/x$ would go to $\infty$ as $x\to 0$

Next $f'(0)=\lim_{x\to 0} [f(x)-f(0)]/x=f(x)/x=4$

Finally $\lim_{x\to 0} g(x)/f(x)=[g(x)/x]/[f(x)/x]=4/2=2$

So we're done.

What Jimmy is writing is a bit nonsensical; it suffices to consider those examples on a multiple choice test but isn't very enlightening otherwise, and it certainly doesn't offer any sort of proof.


inequalities
limit
calculus
calculus computations

Suppose that the sequence $ {a_n}$ satisfies the conditions:
$ a_n > 0$, $ a_n(1 \minus{} a_{n \plus{} 1}) > \frac {1}{4}$ for $ n\in\mathbb N$.
Prove that $ a_n > \frac {1}{2}$ for $ n\in\mathbb N$.

Suppose that for every positive integer $ n$ we have $ a_n > 0$ and $ a_n(1 - a_{n + 1}) > \frac {1}{4}$.

[b]1.[/b]This implies that $ 1 - a_{n + 1} > 0$. Thus we have the stimation $ 0 < a_n < 1$ for every positive integer $ n$.

[b]2.[/b] Using the AM-GM inequality to the positive numbers $ A: = a_n$ and $ B: = 1 - a_{n + 1}$, we get
\[{ {\color{blue}{\left (\frac {a_n + 1 - a_{n + 1}}{2}\right )^2}}\ge a_n(1 - a_{n + 1})  {\color{blue}>\frac {1}{4}}}\]
which leads to the fact that $ \color{blue}{a_n > a_{n + 1}}$ for every positive integer $ n$.

[b]3.[/b] From this we deduce that the sequence $ \{ a_n\}_{n = 1}^{\infty}$ is a decreasing sequence of positive numbers. Hence, there exists the limit of this sequence, say $ \lim_{n\to\infty}a_n = a\in [0,1)$.

[b]4.[/b] The limit can be computed exactly from the second assumption. We obtain
\[ a - a^2 - \frac {1}{4}\ge 0.\]
Now we have $ a = \frac {1}{2}$.

[b]5.[/b] The proof finishes the fact that
\[ a_1 > a_2 > \cdots > a_n > \cdots > a = \frac {1}{2}.\]

Can you prove that $ a_n>\frac{1}{2}$ without finding the limit of $ a_n$?

Well, let $ a\equal{}\inf_n a_n$. Then we still have $ a(1\minus{}a)\ge\frac 14$  (because $ a_n$ can be made arbitrarily close to $ a$ and $ a_{n\plus{}1}\ge a$), whence $ a\equal{}\frac 12$. Since the inequality $ a_n(1\minus{}a_{n\plus{}1})>\frac 14$ is strict and $ 1\minus{}a_{n\plus{}1}\le\frac 12$, we must have $ a_n>\frac 12$.


1. Many numbers can be made by adding two or more consecutive terms of the arithmetic sequence 2, 5, 8, 11,... . Two such examples are $ 7\equal{}2\plus{}5$ and $ 24\equal{}5\plus{}8\plus{}11$. What is the smallest number that can be made in at least two different ways by adding consecutive terms of this sequence?

2. What point on the line $ y\equal{}2x$ is closest to the point (0, 5)?

3. A square is inscribed in a right triangle with legs of 8 units and 15 units. If two of the vertices of the square lie on the hypotenuse and the other two vertices of the square lie on the legs of the triangle, what is the length of a side of the square? Express your answer as a common fraction.

[quote="happyme"]1. Many numbers can be made by adding two or more consecutive terms of the arithmetic sequence 2, 5, 8, 11,... . Two such examples are $ 7 \equal{} 2 \plus{} 5$ and $ 24 \equal{} 5 \plus{} 8 \plus{} 11$. What is the smallest number that can be made in at least two different ways by adding consecutive terms of this sequence?

2. What point on the line $ y \equal{} 2x$ is closest to the point (0, 5)?

3. A square is inscribed in a right triangle with legs of 8 units and 15 units. If two of the vertices of the square lie on the hypotenuse and the other two vertices of the square lie on the legs of the triangle, what is the length of a side of the square? Express your answer as a common fraction.[/quote]

[hide]
2.
(2.5,5)[hide][/hide][/hide]

[quote="Ihatepie"][quote="happyme"]1. Many numbers can be made by adding two or more consecutive terms of the arithmetic sequence 2, 5, 8, 11,... . Two such examples are $ 7 \equal{} 2 \plus{} 5$ and $ 24 \equal{} 5 \plus{} 8 \plus{} 11$. What is the smallest number that can be made in at least two different ways by adding consecutive terms of this sequence?

2. What point on the line $ y \equal{} 2x$ is closest to the point (0, 5)?

3. A square is inscribed in a right triangle with legs of 8 units and 15 units. If two of the vertices of the square lie on the hypotenuse and the other two vertices of the square lie on the legs of the triangle, what is the length of a side of the square? Express your answer as a common fraction.[/quote]

[hide]
2.
(2.5,5)[/quote]

To finish that hide add a [/hide] where you want it to stop hiding the text.

How did you get that point?

I just used common sense to figure it out.

I could've used lots of math, but in competitions, you have to go fast.

Is it right though?

Frankly, I have no clue what the right answers are to most of the problems I post, we just get a popular vote and everyone posts their logic/solutions and we scrutinize each other. I used logic to reach at my conclusion (same one as yours), but I was wondering if there was a sure-fire way.

[hide=" # 2"]we want the point closest to (0,5), so we find the equation of a line that's perpendicular to y=2x and goes through the point (0,5).  That line is y=-x/2+5.  To find the point, we solve the two equations.  By solving, x= 5/2, y= 5

[b](2.5, 5)[/b]

[/hide]

[hide="#3 is a classic"]Let it be ABC with AC=15, BC=8, and AB=17. Let the square touch AC at D and let CD=y.

then $ \frac{y}{s}\equal{}\frac{15}{17}$
and $ \frac{15\minus{}y}{s}\equal{}\frac{17}{8}$
adding, you get $ \frac{15}{s}\equal{}\frac{409}{136}$
Solving, $ s\equal{}\frac{2040}{409}$

umm that doesn't look right...but im pretty sure i have the right strategy[/hide]

[quote="unimpossible"][hide="#3 is a classic"]Let it be ABC with AC=15, BC=8, and AB=17. Let the square touch AC at D and let CD=y.

then $ \frac {y}{s} \equal{} \frac {15}{17}$
and $ \frac {15 \minus{} y}{s} \equal{} \frac {17}{8}$
adding, you get $ \frac {15}{s} \equal{} \frac {409}{136}$
Solving, $ s \equal{} \frac {2040}{409}$

umm that doesn't look right...but im pretty sure i have the right strategy[/hide][/quote]

Hmm...

Your first equation looks fine, but in your second one, I don't get how $ s$ can be similar to the bottom leg of the triangle.

[quote="happyme"][quote="the future"][quote="unimpossible"]

Solving, $ s \equal{} \frac {2040}{409}$

umm that doesn't look right...but im pretty sure i have the right strategy[/hide][/quote]

Thats what I got.

happyme try finding some angles interms of another one.[/quote]

And thats supposed to mean...  :huh:[/quote]

try find angles in terms of another, exactly what it means...like angle chasing?

[quote="happyme"]Ohhh, I thought interms was some kind of vocabulary (like centroid).

 :ninja:[/quote]

well anyways you can see the x and 90-x form similars


Discuss NCAA Football here.

What did you guys think of the USC-Notre Dame game?  I watched basically the whole thing and got the impression that USC got outplayed pretty badly.  Notre Dame definitely should have won.

So, my rankings are as follows:
1. Virginia Tech
2. Texas
3. USC
4. Notre Dame
5. UCLA
6. Miami
7. Texas Tech
8. Georgia
9. Alabama
10. Penn State

UVA beat Florida State.  That is the extent of my thoughts right now.  Why yes, I am from Charlottesville, why do you ask?

[quote="yif man12"]Texas Tech?!?! Have you passed the hash-pipe around a couple times too many? They haven't beaten anybody![/quote]

hash-pipe wooo

wow, and UNC beat Miami. How about that.

I agree more or less with the schools in the top 10, but I have to say USC would be my number 1 choice, then Texas.

And what do you mean Texas Tech hasn't beaten anybody?? They're undefeated! haha. I think the AP poll even has them somewhere in the bottom half of the top 10.  I think it's kinda early to tell now... but I am sure Texas and USC will remain top 5.  :D

And oh man. that notre dame/USC game was a thriller O_O.

[quote="yif man12"]The only half-decent team they've beaten is Nebraska. I'd rate a good 1 loss team like Penn State or Oregon, or even Notre Dame ahead of them. If Texas Tech had Notre Dame's schedule, they'd be 0-6.[/quote]

yeah idk, i heard notre dame and texas tech's schedule were easier, so thats why they're like number 9 and 10 even though they have good records.  who is your number one pick?

I love USC, but they are really in a first half slump right now.  What a game against Notre Dame!! Texas would finish USC i think :mad:

1.Texas
2.USC
3.Notre Dame
4.Virginia Tech
5.Georgia
6.Alabama
7.UCLA
8.Miami(FL)
9.LSU
10.Penn State

Notre Dame may be 4-2, but they have had the toughest games of any team in NCAAF.  I rank them 3.  What is Texas Tech doing in the top 10 in all the polls?????Just because they are 6-0??? They have not beaten any teams in top 25 i think.  I still root for USC!!!!!!!!

i like USC. but you have to admit the new coach for notre dame has done a whole lot for the team, almost beat No.1

texas tech trouncing USC?!?! :huh:  :rotfl:

no way..

I guess we'll see how good Texas Tech really is against Texas.  I predict it will be a pretty close game with Texas coming out on top. (unfortunately)

I don't think it will even be that close.

1) USC
2) Va Tech
3) ND
4) Texas

they're very close, but in the end, i think the best team this year is USC.


Given $101$ distinct non-negative integers less than $5050$ show that one can choose four $a, b, c, d$ such that $a + b - c - d$ is a multiple of $5050$

This seems to be very easy.
We have those integers in an increasing order: $0\leqslant a_1<a_2<...<a_{101}<5050$. Let us look at the differences between consequent terms of the sequence $a_n$. If we suppose all those differences to be pairwise different, we have:
$a_{101}=a_1+(a_2-a_1)+(a_3-a_2)+...+(a_{101}-a_{100})\geqslant 0+1+2+3+...+100=5050$ - a contradiction.
So there have to be such $i,j$ that: $a_i-a_{i+1}=a_k-a_{k+1}$, whence $a_i+a_{k+1}-a_{i+1}-a_k=0$. And that's it.

The problem condition should say DISTINCT $a,b,c,d$ as per the version here http://imomath.com/othercomp/Pol/PolMO399.pdf
Therefore tram's solution does not consider the fact that perhaps $a_3-a_2\equiv a_2-a_1\pmod {5050}$, which does not work because it is not $4$ distinct numbers.

Here is a solution which I hope works, the main idea is to consider differences anyways.

For all ordered pairs $(a_i,a_j)$ where $i\ne j$ and $1\le i,j \le 101$ consider the difference $a_i-a_j\pmod {5050}$. Suppose two differences are equal, that is $a_i-a_j\equiv a_k-a_l\pmod {5050}$. If $i,j,k,l$ are distinct then we have $a_i+a_l\equiv a_k+a_j\pmod {5050}$ as desired. Therefore some two must be equal. Suppose $i=k$, then we have $a_j\equiv a_l\pmod {5050}$ which is impossible as $0\le a_j,a_l<5050$. The same applies if $l=j$. Therefore we have either $i=l$ or $j=k$. If both of these are true we have $a_i-a_j\equiv a_j-a_i \pmod {5050}$, meaning that $a_i-a_j\equiv 2525\pmod {5050}$. There are at most $50$ such pairs $(a_i,a_j)$. If $i=l$, then $2a_i\equiv a_j+a_k\pmod {5050}$. Hence for every $a_i$ we have a pair $(a_j,a_k)$ such that $a_j+a_k\equiv 2a_i\pmod {5050}$, this means that the differences $a_j-a_i\equiv a_i-a_k \pmod {5050}$, $a_k-a_i\equiv a_i-a_j\pmod {5050}$ are equal. Therefore there are $50$ pairs of equal differences of the first kind which are double counted and then $202$ pairs of equal differences of the second kind which are double counted. Note there are $10100$ differences in total, hence the total number of distinct differences is $10100-50-202>5050$, contradiction.

Edit: Ok apparently there's another way to do this problem using sums as well, in this thread: http://artofproblemsolving.com/community/q1h290894p1571457


$ x,y,z,t\in \mathbb{R}$   :?:
If so there are infinitely many answers.

There are infinitely answers :
$ x \equal{} 1$ , $ y \equal{} 1 \plus{} 6k^{3}$ , $ z \equal{} 1 \minus{} 6k^{3}$ , $ t \equal{} 6k^{2}$  :!:


In Colombia, they taught us a useful Lemma, the problem is that the teachers do not know the name of the Lemma. So they told us we have to prove it every time  use it. The prove I know uses induction, and it is very boring to write it. So i wanted to know if there's a name for these Lemma.

so here it is:

Let $p>2$ be prime.
if $p^{\alpha} || a-b$,  $p \nmid a$ and $p^{\beta} || k$
$\rightarrow$ $p^{\alpha+\beta}||a^k - b^k$

Iff $p=2$
then it is
$2^{\alpha}||\frac{a^2 - b^2}{2}$,  $2\nmid a$ and $2^{\beta}|| k$
$\rightarrow$ $p^{\alpha + \beta}||a^k - b^k$

I don't know if it's the rigth place to post these.

The place is totally right  :)

Some time ago, someone called it after Hensel in http://www.mathlinks.ro/Forum/viewtopic.php?t=27757 and deleted it then (the statement itself was repeated in posts 10 and 11).
Also Alex. Belov called it after him in IMO preparation last year, so possibly it is really called so.

Thanks ZetaX, I hope I find the name of this Lemma so that I don't have to prove it every time I use it. I've seen much IMO problems wich are much easier to prove if you know the Lemma.


Taken from Nassau County Math League, this year, round 3.

A ladder is resting against a wall (top and base both touching the wall, base on the floor). If the base is moved 4 feet from the wall, it moves the top of the ladder down the wall by one-fourth the length of the ladder. How long (in exact number of feet) is the ladder?

is the answer suppose to be a whole number? cause i'm not quite getting one.

The problem doesn't appear to specify integer solutions.

Some information must be missing because I don't think this question as it is has a unique answer.

It does have a unique solution.. (I get [hide]:sqrt:(16/7)[/hide]). The bit that I misread at first was that initially the base of the ladder is also touching the wall.

ah! I misread that too. I agree with your answer then.

frodo wrote:I got[hide] 16 :sqrt: 7/7[/hide]

You sure you don't mean [hide]4 :sqrt: 7 / 7? Thats the same thing as what I got, just arranged differently..[/hide]

The answer is  [hide]16 :sqrt: 7  /7. [/hide]

I also misread the question.

By the way, if the base had not been touching the wall initally, would it have been possible to solve the problem?

Oh whoops, forgot to multiply my x by 4..

um is this the same thing? i got [hide]16 :sqrt:(1/7)[/hide]

Yeah. Just multiply top and bottom by :sqrt:7 and you'll get the same thing.

oh yeah i forgot to rationalize the denominator


Let $2n-1$ points be lie in a circle.Color $n$ points in black.Show that:One can find $2$ black points such that there exist one in two arcs which determined by them in the circle contains different $n$ points exactly!

It's easy :lol:  :lol:
You can see IMO1990,problem 2,harder a lot:


Show that $n^{\pi(2n)-\pi(n)}<4^{n}$ for all positive integer $n$.

\[n^{\pi(2n)-\pi(n)}< \prod_{n<p\leq 2n\atop p\ \mbox{\footnotesize is prime}}p \leq{2n\choose n}< (1+1)^{2n}= 4^{n}.\]

[quote="Peter"]Show that $ n^{\pi(2n) \minus{} \pi(n)} < 4^{n}$ for all positive integer $ n$.[/quote]

Sorry, what is $ \pi(n)$ ?  :blush:.


vector
linear algebra
matrix
counting
distinguishability
linear algebra unsolved

Show that 'the dimension of' the vector space of ploynomials of degree $ \le d$ in $ n$ variables is equal to the binomial coefficient
$ \begin{pmatrix} n \plus{} d \\
d \end{pmatrix} \equal{} \frac {(n \plus{} d)!}{d! n!}$.

[quote="Bradley"]Show that the vector space ... is equal to the binomial coefficient[/quote]  That would be quite something, indeed!  More seriously, though, this is just a simple counting problem: this vector space has a very natural basis (the basis of monomials) and so all you have to do is to count the number of monomials of degree at most $ d$ in $ n$ variables.  This has something to do with nonnegative integer solutions to the equations $ x_1 \plus{} \ldots \plus{} x_n \equal{} k$ for $ k \equal{} 0, 1, \ldots, d$.

Oh! I'm sorry.
I corrected my question.

Number of monomials of degree $ \le d$ in $ n$ variables $ \equal{} \begin{pmatrix} n\plus{}d\\d \end{pmatrix} ??$  Why$ ??$

First, note that we can add another variable and make it homogeneous; the number of monomials of degree $ \le d$ in $ n$ variables is the same as the number of monomials of degree equal to $ d$ in $ n\plus{}1$ variables.

This is one of several standard counting problems- in this case, putting $ d$ indistinguishable objects in $ n\plus{}1$ distinguishable slots. We can get the binomial coefficient by encoding these arrangements.

An example of the encoding: let $ n\equal{}3,d\equal{}6$. The sequence aababbaaa represents $ x_1^2x_2^1x_3^0x_4^3$, or $ x_1^2x_2$ when we go back down to three variables. The length of the blocks of "a"s determine the exponent of each $ x_i$, as separated by the "b"s. We could arrange the six copies of "a" and three of "b" in any order, for $ \binom{6\plus{}3}{3}$ total ways.


A standard deck of 52 cards has 13 ranks (Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King) and 4 suits ($ \spadesuit$, $ \heartsuit$, $ \diamondsuit$, and $ \clubsuit$), such that there is exactly one card for any given rank and suit.  Two of the suits ($ \spadesuit$ and $ \clubsuit$) are black and the other two suits ($ \heartsuit$ and $ \diamondsuit$) are red.  The deck is randomly arranged. What is the probability that the top card is a 5?

There are $ 4$ fives, and $ 52$ cards. $ \frac{4}{52}\equal{}\frac{1}{13}$

[hide=Solution]Or, there are four equal sets of $13$ cards with one $5$ in each of them[/hide]

bruh what r u even trying to say, also try not to bump 14 yr old posts...

[hide=sol]There are $4$ fives and $52$ total cards, so the answer is $\frac{4}{52}=\frac{1}{13}$[/hide]


a + 2(x + 1) = s what is x + 1 in terms of s and a

solution(s):

x + 1 = s - a / 2

and

x + 1 = s / a + 2

I know the first equation ir right but I'm not 100% sure about the second.

are you sure the second one is correct??

the second solution is wrong, because in order for $x+1=\frac{s}{2+a}$, then $s$ must be equal to $(a+2)x+a+2$, which is not true.

er...this would be easier to read in latex, but if you dont know it, thats ok...im taking that he meant that  $s/a$ was one term, but that still isnt true because if you divided everything by $a$, that gives $\displaystyle1 + \frac {2(x+1)} a\ = \frac sa\ $ so then you subtract 1 and multiply everyhthing by $\displaystyle\frac a2\ $ giving $\displaystyle\frac s2\ -\frac a2\ $ which is your first answer


calculus
integration
function
real analysis
limit
derivative
real analysis unsolved

Hi,

Suppose that f is riemann integrable on[a,b] whenever b>a and a is fixed. Let int [size=75]from a to inf.[/size] of f(x)dx=lim [size=75]b->inf [/size]f(x)dx.
How can I use integration by parts here? I am confused because the text says to show that integration by parts can be applied to this function?

Thanx for any help.[/i][/u]

Let me guess: the text is Principles of Mathematicial Analysis by Rudin, and the problem is chapter 6 exercise # 9? If so, the apply the same proof used in Theorem 6.22:

Suppose that for every $b\geq a,$ $F$ and $G$ are differentiable functions on $\left[ a,b\right],$ $F^\prime = f\in \mathrm{R}$ and $G^\prime = g\in \mathrm{R}$ on $\left[ a,b\right],$ and that both $\lim_{b\to\infty}\int_{a}^{b}f(x)G(x)dx$ and $\lim_{b\to\infty}F(b)G(b)$ exist and are finite. Now apply integration by parts to the integral: that is apply the fundamental theorem of calculus to the derivative of the product $F(x)G(x)$ as follows: since
\[\int_{a}^{b}\left[f(x)G(x)+F(x)g(x)\right] dx = \int_{a}^{b}\frac{d}{dx}\left[F(x)G(x)\right] dx \]

\[= \left[F(x)G(x)\right]_{x=a}^{b}=F(b)G(b)-F(a)G(a) \]
we have that
\[\int_{a}^{b}F(x)g(x)dx = F(b)G(b)-F(a)G(a)-\int_{a}^{b}f(x)G(x)dx \]
which is the usual integration by parts formula, now let $b\to\infty$ to get
\[\lim_{b\to\infty}\int_{a}^{b}F(x)g(x)dx = \lim_{b\to\infty}\left(F(b)G(b)-F(a)G(a)-\int_{a}^{b}f(x)G(x)dx\right) \]

\[=\lim_{b\to\infty}F(b)G(b)-F(a)G(a)-\lim_{b\to\infty}\int_{a}^{b}f(x)G(x)dx \]
where the last equality follows by the existence and finiteness of the limits involved, and the existence of the third limit is implied by the existence/finiteness of the of two.

I just wanted to say that your guess was right and thank you much for your answer

Romy


Find all homogeneous linear recursive sequences such that there is a $ T$  such that $ a_n\equal{}a_{n\plus{}T}$ for each $ n$.

With no restrictions on the coefficients, the characteristic polynomial has to have only roots of unity as its roots (each with multiplicity $ 1$).  With rational or integer coefficients, the characteristic polynomial has to be a product of distinct cyclotomic polynomials.  This is a straightforward corollary of well-known results on such sequences.


Have you heard of freerice? [url]http://www.freerice.com[/url]

Oh yes I have.

My friend Rahul showed it to me about a year ago, and then proceeded to tell it to everyone who he knew. When he told Mr. Christman, the woodshop teacher, Mr. Christman liked it a lot, and then introduced it to all his students, who told their friends. So now half the school knows about it one way or another.

yeah. evilhamster told me (and i mean told me in real life)

I guess he/she/it has something against you.  :rotfl:

well this has been posted thrice before. that may be why.


A regular hexagon of side 7 is surrounded on each side by regular hexagons of the same size adjacent to its sides.  What is the perimeter of the resulting figure?

[hide]I get 126. If you take and make a little hexagon piece (this will truely only work if your shape is a perfect hexagon) and surround it with 6 others, each hexagon you've surrounded the original one with will have 3 sides as part of the perimeter. You have 18 sides as part of the perimeter, with each having a length of 7 = 126. [/hide]

[hide]
126. There are 6 hexagons around the original. 3 sides show on each. 18 sides in total. 18*7=126.
[/hide]


Let $ P(x)$ is polynomial deg $ n$ with real coefficial. Prove that: Equation $ P(x)\equal{}2^x$ has at most $ n\plus{}1$ real root.

We have:Let $ Q(x) \equal{} P(x) \minus{} 2^x$,then $ Q^{(n)} (x) \equal{} C \minus{} 2^x.(ln2)^n$(C const) has at most one root,then $ Q(x)$has at most n+1 root.


calculus
integration
real analysis
real analysis unsolved

let $ t \in \mathbb{R}$ and consider $ g(t) \equal{} \int\limits_{\mathbb{R}} e^{ \minus{} \frac {x^2}{2}} \cos(tx) ~ d \lambda(x)$. why do we have $ g(t) > 0$? (I need this to show $ g(t) \equal{} \sqrt {2 \pi} e^{ \minus{} \frac {t^2}{2}}$, which would then follow from the ode $ dg/dt \equal{} \minus{} t g$.)
I have already managed to prove $ \int\limits_{\minus{}\pi/2}^{\infty} e^{ \minus{} \frac {x^2}{2}} \cos(tx) ~ d \lambda(x) > 0.$

Can you evaluate
\[ \int_{ \minus{} \infty }^{ \plus{} \infty } {e^{ \minus{} \frac{{x^2 }}
{2}} e^{iz} d\lambda \left( x \right)} .
\]

yes this is $ e^{i z} \sqrt {2 \pi}$ ... how does this help? we have

$ g(t) \equal{} \int\limits_{\mathbb{R}} e^{ \minus{} x^2/2} e^{ \minus{} i t x} d \lambda(x)$.

[quote="-oo-"]yes this is $ e^{i z} \sqrt {2 \pi}$ ... how does this help? we have

$ g(t) \equal{} \int\limits_{\mathbb{R}} e^{ \minus{} x^2/2} e^{ \minus{} i t x} d \lambda(t)$.[/quote]

Sorry, Can you evaluate
\[ \int_{ \minus{} \infty }^{ \plus{} \infty } {e^{ \minus{} \frac {{x^2 }} {2}} e^{itz} d\lambda \left( x \right)} .
\]

no otherwise I would not post this problem ...

Use the ODE that you already have: since $ dg/dt\equal{}\minus{}tg$, it follows that $ \frac{d}{dt}\left(e^{t^2/2}g\right)\equal{}0$, so $ g$ has constant sign.


floor function
logarithms
inequalities
Divisibility Theory

Show that for all prime numbers $p$, \[Q(p)=\prod^{p-1}_{k=1}k^{2k-p-1}\] is an integer.

[quote="nicetry007"]For $ p \equal{} 2$, it is trivially true. Let $ p \;\geq\; 3$.

$ Q(p) \equal{} \prod_{i \equal{} 1}^{p \minus{} 1}k^{2k \minus{} p \minus{} 1} \equal{} \left( \frac {\prod_{i \equal{} 1}^{p \minus{} 1}k^{k}}{\left( (p \minus{} 1)! \right)^{(p \plus{} 1)/2}}\right)^{2}$

It is enough to show that the term inside the parenthesis is an integer.

Let $ q$ be a prime  less than $ p$.  The exponent of $ q$ in the denominator is given by

$ \frac {(p \plus{} 1)}{2}\left (\lfloor\frac {p \minus{} 1}{q}\rfloor \plus{} \lfloor\frac {p \minus{} 1}{q^{2}}\rfloor \plus{} \cdots \plus{} \lfloor\frac {p \minus{} 1}{q^{t}}\rfloor \right)$

In the numerator, the terms that are multiples of $ q$ are as follows:

$ q^{q}, (2q)^{2q}, \cdots, \left( \lfloor\frac {p \minus{} 1}{q}\rfloor q \right)^{\lfloor\frac {p \minus{} 1}{q}\rfloor q}$

Factoring out a $ q$ from all these terms gives us

$ q^{(q \plus{} 2q \plus{} \cdots \plus{} \lfloor\frac {p \minus{} 1}{q}\rfloor q )} \equal{} q^{\frac {q}{2}\lfloor\frac {(p \minus{} 1)}{q}\rfloor \left( \lfloor\frac {p \minus{} 1}{q}\rfloor \plus{} 1 \right)}$

In all, there are $ \lfloor\frac {p \minus{} 1}{q}\rfloor$ multiples of $ q$ in the numerator. Of these  $ \lfloor\frac {p \minus{} 1}{q^{2}}\rfloor$ are multiples of $ q^{2}$.

Factoring out a $ q$ from these terms gives us

$ q^{(q^{2} \plus{} 2q^{2} \plus{} \cdots \plus{} \lfloor\frac {p \minus{} 1}{q^{2}}\rfloor q^{2})} \equal{} q^{\frac {q^{2}}{2}\lfloor\frac {p \minus{} 1}{q^{2}}\rfloor \left( \lfloor\frac {p \minus{} 1}{q^{2}}\rfloor \plus{} 1 \right)}$.

Continuing in this fashion, we can compute the exponent of $ q$ in the  numerator as follows:

$ \frac {q}{2}\lfloor\frac {p \minus{} 1}{q}\rfloor \left( \lfloor\frac {p \minus{} 1}{q}\rfloor \plus{} 1 \right) \plus{} \frac {q^{2}}{2}\lfloor\frac {p \minus{} 1}{q^{2}}\rfloor \left( \lfloor\frac {p \minus{} 1}{q^{2}}\rfloor \plus{} 1 \right) \plus{} \cdots \plus{} \frac {q^{t}}{2}\lfloor\frac {p \minus{} 1}{q^{t}}\rfloor \left( \lfloor\frac {p \minus{} 1}{q^{t}}\rfloor \plus{} 1 \right)$.

Comparing the exponent of $ q$ in the numerator and the denominator , it suffices to show

$ \frac {(p \plus{} 1)}{2}\left(\lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \right) \leq \frac {q^{i}}{2}\lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \left( \lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \plus{} 1 \right)$.

It is enough to show

$ (p \plus{} 1) \leq q^{i}\left( \lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \plus{} 1 \right) \;\text{\;(i.e.)\;}\; \frac {(p \plus{} 1)}{q^{i}}\leq \lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \plus{} 1$.

Suppose $ q^{i}$ divides $ p \plus{} 1$. Then $ \lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \equal{} \frac {(p \plus{} 1)}{q^{i}} \minus{} 1 (i.e.) \frac {(p \plus{} 1)}{q^{i}} \equal{} \lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \plus{} 1$.

Suppose $ q^{i}$ does not divide $ p \plus{} 1$. Then $ \lfloor\frac {(p \plus{} 1)}{q^{i}}\rfloor \equal{} \lfloor \frac {p \minus{} 1}{q^{i}}\rfloor$   (i.e.) $ \frac {(p \plus{} 1)}{q^{i}} \equal{} \lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \plus{} f$ where $ f$ is a fraction less than $ 1$.
Hence, $ \frac {(p \plus{} 1)}{q^{i}}\leq \lfloor\frac {p \minus{} 1}{q^{i}}\rfloor \plus{} 1$.

Therefore, $ Q(p)$ is an integer. In fact, a perfect square.[/quote]

[i]Nice problem .

   AMM problem always takes my interest . :lol:

  [u] [b]Solution :[/b][/u]

  Fristly , we will recall an well-known identity :  $ \prod_{k \equal{} 1}^{n \minus{} 1}k! \ \equal{} \ \prod_{k \equal{} 1}^{n \minus{} 1} k^{n \minus{} k} \ \ (*) \ \ (n \ge 2)$

  Indeed , $ \prod_{k \equal{} 1}^{n \minus{} 1}k! \ \equal{} \ (1 ) . (1.2)....(1.2.....(n \minus{} 1))$

we can see that  in this product , the numbers $ 1$ apprears $ n \minus{} 1$ times , the numbers $ 2$ appears $ n \minus{} 2$ times ,....,

the numbers $ n \minus{} 1$ appears only one time . So , we can deduce that :

  $ \prod_{k \equal{} 1}^{n \minus{} 1}k! \ \equal{} \prod_{k \equal{} 1}^{n \minus{} 1}(n \minus{} k)! \ \equal{} \ 1^{n \minus{} 1} . 2^{n \minus{} 2} \cdots (n \minus{} 1)^1 \ \equal{} \ \prod_{k \equal{} 1}^{n\minus{}1} k^{n \minus{} k} \ \ (**)$ , Done

  Now return to our problem :

  We can rewrite $ \mathbb{Q} (p)$ as follows :

  $ \mathbb{Q} (p) \equal{} \prod_{k \equal{} 1}^{p \minus{} 1} k^{2k \minus{} p \minus{} 1} \equal{} \frac {\prod_{k \equal{} 1}^{p \minus{} 1} k^{k \minus{} 1} }{ \prod_{k \equal{} 1}^{p \minus{} 1} k^{p \minus{} k} }$

  $ \equal{} \frac {\prod_{k \equal{} 1}^{p \minus{} 1} k^{k \minus{} 1} \cdot \prod_{k \equal{} 1}^{p \minus{} 1} k^{p \minus{} k} }{ \left(\prod_{k \equal{} 1}^{p \minus{} 1} k^{p \minus{} k} \right)^2} \ \equal{} \ \frac {\prod_{k \equal{} 1}^{p \minus{} 1} k^{p \minus{} 1} }{\prod_{k \equal{} 1}^{p \minus{} 1}(p \minus{} k)! \cdot \prod_{k \equal{} 1}^{p \minus{} 1}k! }$ ( Using $ (**)$ )

  $ \equal{} \frac { ((p \minus{} 1)!)^{p \minus{} 1}}{\prod_{k \minus{} 1}^{p \minus{} 1} k! (p \minus{} k)! } \ \equal{} \ \frac {1}{p^{p \minus{} 1}} \cdot \frac { (p!)^{p \minus{} 1}}{\prod_{k \minus{} 1}^{p \minus{} 1} k! (p \minus{} k)! } \equal{} \frac {1}{p^{p \minus{} 1}} \cdot \prod_{k \equal{} 1}^{p \minus{} 1} \frac {p!}{ k! \cdot (p \minus{} k)!}$

   So , we can have the beautiful identity : $ \mathbb{Q} (p) \equal{} \prod_{k \equal{} 1}^{p \minus{} 1} \frac {\binom{p}{k} }{p}$

   We have known that : if $ p$ is a prime , then :  $ p | \binom{p}{k} \ \ \forall \ \ k \equal{} \overline{ 1 ; p \minus{} 1}$

$ \rightarrow \frac {\binom{p}{k} }{p} \in \mathbb{N} \ \ \forall \ \ k \equal{} \overline{ 1 ; p \minus{} 1} \rightarrow \mathbb{Q} (p) \in \mathbb{N}$ , QED[/i]

Let $q$ a prime smaller than $p$, $n=\lfloor{\log_q p}\rfloor$ and $j$ be an integer, $1\leq j\leq n$.

We start from this obvious fact: $\left\lfloor{\frac{p}{q^j}}\right\rfloor+1>\frac p{q^j}$. This gives that $q^j\left (\left\lfloor\frac{p}{q^j}\right\rfloor+1\right )$ is strictly greater that $p$, which is to say, greater or equal than $p+1$.

Multiplying this inequality by $\left\lfloor{\frac{p}{q^j}}\right\rfloor$ gives:

$q^j\left (\left\lfloor{\frac{p}{q^j}}\right\rfloor^2+\left\lfloor\frac{p}{q^j}\right\rfloor \right ) \geq (p+1)\left\lfloor\frac{p}{q^j}\right\rfloor$, or

$2q^j\sum\limits_{i=1}^{\left\lfloor\frac{p}{q^j}\right\rfloor} i \geq (p+1)\left\lfloor\frac{p}{q^j}\right\rfloor$ (*)

and this is true for every positive integer $j$.

We'll define $\mu_q(m)$ for every positive integer $m$ as the greatest integer $r$ such that $q^r$ divides $m$. We shall now compute:

$\mu_q\left (\prod\limits_{k=1}^{p-1} k^{2k} \right )=\sum\limits_{k=1}^{p-1} 2k\mu_q(k)=\sum\limits_{j=1}^n \sum\limits_{i=1}^{\left\lfloor\frac{p}{q^j}\right\rfloor} 2ijq^j$. (**)

Note that we can extend the inner sum to $\left\lfloor\frac{p}{q^j}\right\rfloor$ (and not only to $\left\lfloor\frac{p-1}{q^j}\right\rfloor$ ) because $q^j$ cannot divide $p$.

To finish, we compute
$\mu_q\left (\prod\limits_{k=1}^{p-1} k^{p+1} \right )=\sum\limits_{k=1}^{p-1} (p+1)\mu_q(k)=\sum\limits_{j=1}^n \sum\limits_{i=1}^{\left\lfloor\frac{p}{q^j}\right\rfloor} (p+1)j=\sum\limits_{j=1}^n j(p+1)\left\lfloor\frac{p}{q^j}\right\rfloor$. (***)

(*), (**) and (***) finish the proof.

The sums in $j$ in (*) and (**) may need some further details: the terms of a sum like $\sum\limits_{k=1}^{p-1} f(k)\mu_q(k)$ are 0 when $k$ is not multiple of $q$, $f(k)$ when $k$ is multiple of $q$ but not of $q^2$, $2f(k)$ when $k$ is multiple of $q^2$ but not of $q^3$ and $jf(k)$ when $k$ if multiple of $q^j$ but not of $q^{j+1}$. So, we make $i$ span the numbers that gives less than $p$ when multiplied by $q^j$ and $k=iq^j$.


How does the Alcumus score work? It's really weird...

The has been asked numerous times; just look at one of the previous threads (or hover your mouse over that question mark next to Alcumus Score on your report).

When you miss a question, does the program let you redo it sometime later so that you can get a correct? And does this affect your Alcumus score by raising it, or does it still count it as wrong since you did it wrong the first time?

You will be able to redo questions when you have exhausted all questions that are currently active.

If you get these questions correct the second time around your Alcumus score will be affected, generally in the positive direction.


How many people will be invited to take the USAMO this year?

i would guess same as last year, around 500.

About 500.  We are currently finalizing the text on the 2008 USAMO Teachers' Manual.  When it is posted you can read the complete text.  In the meantime, read page 4 of last year's at
http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-USAMO-TM/TM-USAMO,2007.pdf
This year's will be unchanged except for date and data changes.

Steve Dunbar
MAA Director, American Mathematics Competitions


All these are identical (unless there are typos)...

x^2 + ((2-y)/3)x + 3 = 0

x^2 + (2/3-y/3)x + 3 = 0

x^2 + (\frac{2}{3}-\frac{y}{3}) x + 3 = 0

x^2 + \frac{(2-y)}{3}x + 3 = 0

x^2 + \frac{(2-y)}{3} * x + 3 = 0

3 + x^2 + \frac{(2-y)}{3}x + 5 = 2

What I need is something that could take any two of these equations and tell me if they are equal or not. I could possibly do some LaTeX parsing beforehand to make it look like a regular equation if need be.

Note that in the above examples there are differences not only in polynomial order, but in the way the fractions are written. Also, the above is just one set of equivalent equations. I would need this to work with any expression or equation.

I would think something like Sage or Mathematica's api stuff could do something like this. However, I don'tt know how to implement it. I played with Sage some, but I could never get it to compare properly. I just kept getting syntax errors.

Anyone have any thoughts on how to solve this issue?

Just evaluate at 2 or three random points. :) (unless you want to say that $ x^2 \plus{} y^4 \equal{} \minus{} 1$ is equivalent to $ x^4 \plus{} y^2 \equal{} \minus{} 1$; here I don't know what to do).

The question is not really well-posed: it will help if you define inambiguously which expressions/equations you consider equivalent and which you don't.

There are an infinite number of expressions and equations, therefore it would be impossible to list all those which should be considered equivalent and which should not be.

The case is simply this:

An expression or equation is stored in a database.

A user manually enters an expression or equation.

I want to be able to do a comparison to determine if they are equivalent.

Here is a very basic case, even more basic than the example posted in the first problem.

Let's say an answer is stored in a database as (a - b) / c = 4.

A user can enter (for example):
   a / c - b / c = 4

   (-b + a) / c = 4

   (a - b) * (1 / c) = 4

   (a / c - b / c) - 4 = 0

All these are equivalent.

However, if the user entered a - b / c = 4 (no parentheses), it would not be equivalent.

I feel like I just restated what I said in my first post. Sorry, I don't know how to explain it any better.

[quote="fedja"](unless you want to say that $ x^2 \plus{} y^4 \equal{} \minus{} 1$ is equivalent to $ x^4 \plus{} y^2 \equal{} \minus{} 1$; here I don't know what to do). [/quote]
These would definitely not be equivalent.

OK, let me offer you a multiple choice then:

First, let us reduce all the equations to the form $ F(x,y)\equal{}0$ (this is more or less routine parsing).
Which of the following will be considered equivalent to it:
a) $ 10 F(x,y)\equal{}0$
b) $ 20 F(x,y)^5\equal{}0$
c) $ 30 F(x,y)^4\plus{}F(x,y)^2\equal{}0$
d) $ \frac{40 F(x,y)}{F(x,y)^2\plus{}10}\equal{}0$
e) $ \frac{50 F(x,y)}{F(x,y)\minus{}1}\equal{}0$
g) $ F(x,y)[F(x,y)\plus{}1]\equal{}0$

Also, do you have a fixed expression you want to compare the user input to or you need to compare 2 user inputs?

[quote="fedja"]OK, let me offer you a multiple choice then:

First, let us reduce all the equations to the form $ F(x,y) \equal{} 0$ (this is more or less routine parsing).
Which of the following will be considered equivalent to it:
a) $ 10 F(x,y) \equal{} 0$
b) $ 20 F(x,y)^5 \equal{} 0$
c) $ 30 F(x,y)^4 \plus{} F(x,y)^2 \equal{} 0$
d) $ \frac {40 F(x,y)}{F(x,y)^2 \plus{} 10} \equal{} 0$
e) $ \frac {50 F(x,y)}{F(x,y) \minus{} 1} \equal{} 0$
g) $ F(x,y)[F(x,y) \plus{} 1] \equal{} 0$

Also, do you have a fixed expression you want to compare the user input to or you need to compare 2 user inputs?[/quote]

I would need to think about which of the expressions above would be equivalent. For simplicities' sake, let's assume all would be considered equivalent since F(x,y) = 0.

For your second question, it would be a fixed expression, stored in a database, which it would be compared to. However, being able to compare two users' entries may be beneficial at some point in the future.

As I said, verification of expressions is easy: plug a few ugly points in and verify that the results are the same within 100 times the machine error or something like that.

Equations are tougher. It seems like you want them to be equivalent if they have the same set of solutions in $ \mathbb C^n$. Checking one inclusion is easy when the answers are compared to something fixed: again, plug in a few ugly points that are solutions of the correct equation into the one the user submitted and see if you get the right thing. It is the other inclusion (i.e., verification that the submited equation has no extra solutions) that can give a big headache. If you want to be 100% sure, you'll have to somehow solve the user equation and this can make your verification a very slow process. You can do something reasonable with purely algebraic expressions, if I remember it right, but as soon as you throw in exponents and logarithms, it becomes transcendentally hard. The best thing that comes into my head is fixing all variables but one and comparing the number of roots in a circle on the complex plane of some moderate size using the argument principle. This is still fairly fast and should be quite reliable.

Symbolic computation attempts will, probably, lead nowhere (unless you restrict yourself to pure algebra and use some rather advanced methods but they are too time consuming). Indeed, I suspect that you will have trouble even with the equivalence of $ xy\minus{}1\equal{}0$, $ x\equal{}\frac 1y$, and $ y\equal{}\frac 1x$ if you follow that way.

hello, your first equation is equivalent to
$ x^2\plus{}\frac{2\minus{}y}{3}x\plus{}3\equal{}0$
$ x^2\plus{}\left(\frac{2}{3}\minus{}\frac{y}{3}\right)x\plus{}3\equal{}0$
$ x^2\plus{}\frac{2}{3}x\minus{}\frac{xy}{3}\plus{}3\equal{}0$.
Sonnhard.


May anh cho em hoi lam cach nao de doc duoc sach o mathlinks?

Tội nghiệp không, sao không ai trả lời em vậy.  :D Anh nghĩ người ta quảng cáo sách để bán chứ không cho đọc chùa.
Em muốn đọc thì order (mua) thử xem, nhưng theo anh thấy chỉ toàn là sách dở hơi không thôi.

hê hê,hiển nhiên rồi,Việt Nam xem chùa quen rồi,ra thế giới rộng lớn dễ gì ăn không của họ,gì cũng tiền cả,sao em không thử làm một cái Master Card rồi thực hiện giống như mua bán trên mạng eBay ấy,thông báo trước,công ti và web nước ngoài rất không thích thực hiện mua bán qua mạng với Việt Nam,nên mua qua người wen ở nước ngoài thì tốt hơn. ;)

Làm gì có sách trên này mà đọc.Mà nếu em đang học cấp 3 thì không cần phải đọc những sách trên này đâu.Cứ đọc tốt mấy quyển "lưu truyền nội bộ" là tốt rồi.Còn nếu cấn tài liệu toán thì liên hệ với anh cũng được.Cứ gửi vào hòm thư nguyenkims@gmail.com


Let a,b,c be positive real number such that
$9+3abc=4(ab+bc+ca)$
Prove that $a+b+c\ge 3$

[quote="spdf"]...In fact we can assume [tex] a+b+c=3[/tex]...[/quote]
why???

[quote="zxingtan"]expand $(3a-4)(3b-4)(3c-4)$,and let $t=a+b+c$[/quote]
Your solution is new,please complete it :)

[deleted]
[I've made a stupid mistake :blush: ]

Sorry but if this proof by contradiction, can you please detail it better.

[quote="spdf"]Not so hard.Assume that [tex] a+b+c\leq\ 3[/tex],we will prove that [tex]4(ab+bc+ca)-3abc\leq\ 9[/tex].In fact we can assume [tex] a+b+c=3[/tex].Put [tex] a= 3x;b=3y;c=3z[/tex].Your inequality becomes:if [tex] x+y+z=1[/tex] then [tex]4(xy+yz+zx)-9xyz\leq\  1[/tex](well-known inequality)[/quote]

Thank you very much. :)

Another solution:
Take [tex] a=x+1;b=y+1;c=z+1 [/tex],then we must prove that:
[tex] x+y+z \geq 0 [/tex] when [tex] 5(x+y+z)+xy+xz+yz=3xyz [/tex].
We consider three case:

Case 1: [tex] xyz \geq 0 \Rightarrow \frac{(x+y+z)^2}{3}+5(x+y+z) [/tex]
            [tex] \geq 5(x+y+z)+xy+xz+yz=3xyz \geq 0 \Rightarrow x+y+z \geq 0 [/tex]

Case 2: [tex] x \geq 0; y \geq 0; z \leq 0 [/tex].
Assume that  [tex] x+y+z \leq 0 \Rightarrow -x \geq y+z [/tex].
[tex] 5(x+y+z)=yz(3x-1)-x(y+z) \geq -4yz+(y+z)^2=(y-z)^2 \geq 0 [/tex].

Case 3: [tex] x \leq 0; y \leq 0; z \leq 0 [/tex].Observe that [tex] -x,-y,-z \in [0,1] [/tex] then:
[tex] 0=5(x+y+z)+xy+yz+xz-3xyz \leq 5(x+y+z) +2(xy+xz+yz) \leq 5(x+y+z)+\frac{2(x+y+z)^2}{3} \Rightarrow x+y+z \geq 0 [/tex].
So we have done. :)

[quote="minhkhoa"][quote="zxingtan"]oh,certainly
for $(t-4)^3>=(3a-4)(3b-4)(3c-4)$[/quote]
Why?? ;)[/quote]

$A_{3}>=G_{3}$

there are some mistake in AnhCuong's solution
[quote="Anh Cuong"]Case 2: [tex] x \geq 0; y \geq 0; z \leq 0 [/tex].
[tex] 5(x+y+z)=yz(3x-1)-x(y+z) \geq -4yz+(y+z)^2 [/tex].
[/quote]
$yz\le 0 $ and $y+z$is not always positive

[quote="zxingtan"]
$A_{3}>=G_{3}$[/quote]
but to get that do we have to have
$3a-4\ge 0$
$3b-4\ge 0$
$3c-4\ge 0$


Find the corresponding regression line, quadratic model, and exponential model for the following data:

(0,98)

(4,50)

(8,32)

(12,26)

(16,24)

(20,23)

Also, I would appreciate it if someone answered my problem here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=286452

Thanks in advance.

Most scientific calculators can do this if you stick in the data.  TI-83 and above definitely has it.

I got a linear regression equation of $ y\equal{}\minus{}3.279x\plus{}74.952$, a quadratic regression of $ y\equal{}.334x^{2}\minus{}9.953x\plus{}92.75$, and an exponential regression of $ y\equal{}72.157(.933)^{x}$.

Sorry if this isn't where you wanted me to put it. :)


Hi all!!!! I was wondering if anyone knows what where the medal cutoffs in IMO 1995, held in North York... Thanks in advance!!!!!!!!.

This doesn't really belong into this forum, anyway here are the cutoffs obtained by comparing two statistics:

gold $\geq 37$ points
silver $\geq 29$ points
bronze $\geq 19$ points

  darij

Gold: 36 points... that was what I wanted... thanks!


Source: AHSME 1973 #8

If 1 pint of paint is needed to paint a statue 6 ft. high, then the number of pints it will take to paint (to the same thickness) 540 statues similar to the original but only 1 ft. high, is
$\text{(A)} \ 90 \qquad \text{(B)} \ 72 \qquad \text{(C)} \ 45 \qquad \text{(D)} \ 30 \qquad \text{(E)} \ 15$

[hide="Correct answer"]If a statue is $6$ times shorter, then its surface area is $6^2=36$ times smaller. Hence we need $540/36=15$ pints of paint.[/hide]

[hide]For every n times smaller an object is than its original, its area is $n^2$ times smaller. Thus each of the statues need only $\frac{1}{36}$ pints of paint.

Since there are 540 statues, we need $540(\frac{1}{36})=15$. [/hide]


Let $ A , B$ be two Lebesgue measurable sets of finite measure. For each $ x\in\mathbb{R}$ we define :
 $ f(x) \equal{} m(A\cap(B \plus{} x))$. Where $ B \plus{} x \equal{} \{b \plus{} x: b\in B\}$
Prove that :
 $ \int_{\mathbb{R}}fdx \equal{} m(A)m(B)$ where m is Lebesgue measure

thanks in advance

[u]Some hint[/u]. Express $ m(A\cap (B\plus{}x))$ as an integral, and then substitute it into $ \int_{\mathbb{R}}f$.  Then, what can you say about the order of the integrals?  What can you say about the measure of a translated set?


If points A',B',C' belong to the incircle too, then this is just Steinbard's Theorem!

I think it is Steinbar[b]t[/b] :huh:
btw, I remember Darij wrote an article about it and posted it on his website.

[quote="Mathx"]the incircle of triangle ABC touches BC,AC,AB at  D,E,F.
A',B',C' are points such that $DA',EB',FC'$ are concurrent Prove that $AA',BB',CC$' are concurrent too.[/quote]

Indeed, if the points A', B', C' lie on the incircle of triangle ABC, then it follows from the Steinbart theorem; see Theorem 2 in the note "Variations of the Steinbart Theorem" on [url=http://de.geocities.com/darij_grinberg]my website[/url] (thanks, Sailor and Mecrazywong, for the publicity!). If the points A', B', C' lie on the lines EF, FD, DE, respectively, then it follows from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3536]the Cevian Nests theorem[/url]. If the points A', B', C' are just any arbitrary points, then your result is not true.

  Darij

I Couldn't go to Darij Site.
Can anyone explain what is Steinbart theorim.


geometry
3D geometry
tetrahedron
circumcircle
geometry unsolved

In the following quadrilateral, let $\alpha=\beta=\gamma=\theta=45.$ Prove that the other angles equal to 45, too.

[quote="shobber"][quote="rem"]is this a top view of a tetrahedron or the intersection of diagonals of a quadrilateral? :?[/quote]
It is just a point inside a quadrilateral. Not a tetrahedron nor diagonals.[/quote]

Yes it is.

Sorry I am really stupid. :wallbash_red:  :oops:

I don't know what do you mean by prove others are 45 too.any way I think problem is:
[color=blue]In quadrilateral $ABCD$ we have $\angle FAD=\angle FDC=\angle FCB=\angle FBA=45^{\circ}$ prove that ABCD is a square.[/color]

we have $\angle FAD=\angle FDC=\angle FCB=\angle FBA=45^{\circ}$ draw a perpendiculer line to AF at A that intersect circumcircle of $\triangle ABF,\triangle AFD$ at $I , H$

we have $\angle AIF=ABF=FAD=HFD=DAH$ hence $FI || AD$ note that $FI$ pass trough circumcenter of $AFB$ like this we have $FI || BC$ therefore $AD || BC$ and $AB||DC$ hence $ABCD$ is parallelogram.
the rest is easy.


[quote="Phelpedo"]Is there an interesting form for all primitive solutions to
\[x^{2}+y^{2}+z^{2}=w^{2}\]
[/quote]
$x=u^{2}-v^{2}-t^{2},\quad y=2uv,\quad z=2ut,\quad w=u^{2}+v^{2}+t^{2}$???


Suppose July of year $ N$ has five Mondays. Which of the following must occur five times in August of year $ N$? (Note: Both months have $ 31$ days.)

$ \textbf{(A)}\ \text{Monday} \qquad
\textbf{(B)}\ \text{Tuesday} \qquad
\textbf{(C)}\ \text{Wednesday} \qquad
\textbf{(D)}\ \text{Thursday} \qquad
\textbf{(E)}\ \text{Friday}$

The three days repeating five times in either of these months are the first or last 3 days. So the last day of July must be Monday, Tuesday, or Wednesday, making the first 3 days of August range from Tues-Thurs, Wed-Fri, or Thurs-Sat. The only one present in all of these is Thursday $ \boxed D$.


This inequality is wrong by substituting x=y=1 and u=v=2, L.H.S.=4/3<3/2=R.H.S.
It's wrong even if you change the sign, by substituting x=y=u=1 and v=2.

is this really wrong? looking in none source [url=https://imomath.com/othercomp/Pol/PolMO293.pdf]imomath[/url]  the wording is the same

but in [url=https://www.mimuw.edu.pl/~chel/Olimp/2etang93.html]a polish one[/url] it is different

I believe the wording in the next comment is the correct wording, as it written in the second source above (the polish one), unless a solution is found I have reasons to doubt

[quote=parmenides51]If $ x,y,u,v$ are positive real numbers, prove the inequality :
\[ \frac {xy \plus{} xv \plus{} uy \plus{} uv}{x \plus{} y \plus{} u \plus{} v} \geq \frac {xy}{x \plus{} y} \plus{} \frac {uv}{u \plus{} v}
\][/quote]

Inequality needs to prove equivalent to : $$xy+uv+vx+uy \ge \frac{xy(x+y+u+v)}{x+y}+\frac{uv(x+y+u+v)}{u+v}$$
$$\iff vx+uy \ge \frac{xy(u+v)}{x+y}+\frac{uv(x+y)}{u+v}$$
$$\iff \frac{vx+uy}{(x+y)(u+v)}\ge \frac{xy}{(x+y)^2}+\frac{uv}{(u+v)^2}$$
$$\iff \frac{x}{x+y}\left(1-\frac{u}{u+v}\right)+\frac{u}{u+v}\left(1-\frac{x}{x+y}\right)\ge \frac{x}{x+y}\left(1-\frac{x}{x+y}\right)+\frac{u}{u+v}\left(1-\frac{u}{u+v}\right)$$
$$\iff \left(\frac{x}{x+y}-\frac{u}{u+v}\right)^2\ge 0 $$

[quote=PWojtaszewski]If $ x,y,u,v$ are positiv real numbers, prove the inequality :
\[ \frac {xu \plus{} xv \plus{} yu \plus{} yv}{x \plus{} y \plus{} u \plus{} v} \geq \frac {xy}{x \plus{} y} \plus{} \frac {uv}{u \plus{} v}
\][/quote]
https://artofproblemsolving.com/community/c6h1868053p12660399


Are you serious? 68/75 is high, especially for a middle schooler (unless you're a n00b like you, jk). That's an average of freaking 23.

[quote="SplashD"][quote="bpms"]Why is the AIME qualifying score through USAMTS so high? Doesn't almost everyone who qualifies for the AIME through USAMTS do so through the AMC as well?[/quote]

68/75 isn't really [i]that[/i] high.

AMC is a contest that your school pays for while USAMTS is free, so that might be one reason.

Also, some people's school might not participate in AMC, so this gives them another way of qualifying for the AIME[/quote]
Like my school... But why is that so high? I would think that almost all of those people can do the AMC's...

Well if you're tenth grade or below you only need like 8...

[quote="bpms"][quote="joml88"]Qualifying through USAMTS is harder than through AMC and it puts you at a 100 AMC score.  If you're getting 68+ on USAMTS, you probably have USAMO aspirations and should therefore take the AMC unless you plan on getting 12+ on the AIME.[/quote]
I am in 8th grade, my school doesnt do the AMCs and I have like 0 chance of qualifying through USAMTS because I got below a ten on the first round (although I am hopefully going to triple that score in this round).[/quote]

It [i]is[/i] still possible to take the AMC if your school doesn't offer it.  Just talk to your math teacher about getting them to offer it.

Or you could find another school nearby that offers it (e-mail the AMC director for a list), including colleges/universities on [url=http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml]this list[/url]

Or if there aren't any near by, you could give [url=http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-AMC,CU+Bro.pdf]this flyer[/url]  to a nearby college/university to get them to offer it.

Or you could just pay for it yourself and have a teacher or someone proctor you (I'm not exactly sure about the rules for proctoring though).  This is actually what I did  last year.  I was in a club, and I just had the advisor who was an english teacher that taught in a seperate program from the rest of the school proctor me.

But yea, just because your school doesn't offer the AMC right now doesn't mean that you aren't able to take it.

[quote="h_s_potter2002"][quote="bpms"][quote="joml88"]Qualifying through USAMTS is harder than through AMC and it puts you at a 100 AMC score.  If you're getting 68+ on USAMTS, you probably have USAMO aspirations and should therefore take the AMC unless you plan on getting 12+ on the AIME.[/quote]
I am in 8th grade, my school doesnt do the AMCs and I have like 0 chance of qualifying through USAMTS because I got below a ten on the first round (although I am hopefully going to triple that score in this round).[/quote]

It [i]is[/i] still possible to take the AMC if your school doesn't offer it.  Just talk to your math teacher about getting them to offer it.

Or you could find another school nearby that offers it (e-mail the AMC director for a list), including colleges/universities on [url=http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml.]this list[/url]

Or if there aren't any near by, you could give [url=http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-AMC,CU+Bro.pdf]this flyer[/url]  to a nearby college/university to get them to offer it.

Or you could just pay for it yourself and have a teacher or someone proctor you (I'm not exactly sure about the rules for proctoring though).  This is actually what I did  last year.  I was in a club, and I just had the advisor who was an english teacher that taught in a seperate program from the rest of the school proctor me.

But yea, just because your school doesn't offer the AMC right now doesn't mean that you aren't able to take it.[/quote]
http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml is a broken link

Your link doesn't work, but mine does...

The problem is that h_s_potter's link has a period on the end.


[color=blue]¿Alguien se fijó en la [i]Memberlist[/i]? ¿Qué pudo suceder para que alguien, supuestamente, haya enviado tantos mensajes? Según mi parecer, ha de ser la acción de un [i]hacker[/i]. ¿Qué opinan al respecto?[/color]

[color=blue]Es realmente extraño. Por lo visto, a determinadas horas aparecen dos personas que enviaron supuestamente 2^24-1 y 2^24-4 mensajes supuestamente (por supuesto, en la lista, no están escritos de esta forma). Cuando quise averiguar qué mensajes enviaron, decía el foro que no estaba habilitada esta opción.

¿Quién tiene una idea de lo que pudo haber sucedido?[/color]