calculus
integration
logarithms
real analysis
real analysis unsolved

Find $ \int \frac{1}{lnx}\left(1\minus{}\frac{1}{lnx}\right)\,dx$

$ \int \frac{1}{\ln x}\left(1\minus{}\frac{1}{\ln x}\right)\ dx$

$ \equal{}\int \frac{\ln x\minus{}1}{(\ln x)^2}\ dx$

$ \equal{}\int \frac{\ln x\minus{}x\cdot\frac{1}{x}}{(\ln x)^2}\ dx$

$ \equal{}\int \left(\frac{x}{\ln x}\right)'\ dx$

$ \equal{}\frac{x}{\ln x}\plus{}Const.$

This integral is very simple but is nice. My solution is the following:
$ \int \frac{1}{lnx}\left(1\minus{}\frac{1}{lnx}\right)\,dx$ may be represented in th following way:
$ \int \frac{1}{lnx}\,dx \minus{} \int \left(\frac{1}{lnx}\right)^{2}\,dx$
Integration by parts ot the first integral gives that it is equal to:
$ \frac{x}{lnx} \plus{} C \plus{}  \int \left(\frac{1}{lnx}\right)^{2}\,dx \minus{} \int \left(\frac{1}{lnx}\right)^{2}\,dx$
So the final answer is: $ \frac{x}{lnx} \plus{} C$


Define a sequence $ (a_n), n\geq1$ by $ a_1\equal{}1,a_2\equal{}2$ and $ a_{n\plus{}2}\equal{}2a_{n\plus{}1}\minus{}a_n\plus{}2$ for $ n\geq 1$. Prove that for any $ m$, $ a_ma_{m\plus{}1}$ is also a term in the sequence.

we have $ a_n\equal{}n^2\minus{}2n\plus{}2$ so $ a_ma_{m\plus{}1}\equal{}a_{m^2\minus{}m\plus{}2}.$

[quote="hell_ever"]Define a sequence $ (a_n), n\geq1$ by $ a_1 \equal{} 1,a_2 \equal{} 2$ and $ a_{n \plus{} 2} \equal{} 2a_{n \plus{} 1} \minus{} a_n \plus{} 2$ for $ n\geq 1$. Prove that for any $ m$, $ a_ma_{m \plus{} 1}$ is also a term in the sequence.[/quote]

[hide]$ a_{n\plus{}2}\minus{}a_{n\plus{}1}\equal{}a_{n\plus{}1}\minus{}a_n\plus{}2$, so the second difference between terms is constant, so the recurrence satisfies a quadratic, and you can solve for the quadratic and get $ f(x)\equal{}x^2\minus{}2x\plus{}2$. Then you can directly show that $ f(x)f(x\plus{}1)\equal{}f(x^2\minus{}x\plus{}2)$, and since $ x^2\minus{}x\plus{}2$ is an integer, $ f(x^2\minus{}x\plus{}2)\equal{}$ exists in $ (a_n), n\geq1$. [/hide]


geometry
invariant
ratio
geometry proposed

What are the invariants of affine geometry?

The most useful invariants are:
Parallelism
Coincidence (collinearity and concurrency)
Ratio of lengths along the same direction
Ratio of areas.


Prove that for every $ k\in\mathbb{N}$ exists $ n\in\mathbb{N}$ sutch that  $ n\cdot2^k \plus{} 17$ is a perfect square.

for $ k\equal{}0$ we take $ n\equal{}8$
for $ k\equal{}1$ we take $ n\equal{}4$
we suppsoe that $ \exists n_k\in\mathbb{N}: \ n_k2^k\plus{}17\equal{}x^2$ is perfect square.
$ (x^{2^k\plus{}1})^2\equal{}(n_k2^{k}\plus{}17)^{2^k\plus{}1}\equiv 17^{2^k\plus{}1}\plus{}C_{2^{k\plus{}1}}^{1}n_k2^{k}17^{2^k}(mod \ 2^{k\plus{}1})\equiv 17(mod\ 2^{k\plus{}1})$
then $ \exists n_{k\plus{}1}\in\mathbb{N}: \ n_{k\plus{}1}2^{k\plus{}1}\plus{}17\equal{}(x^{2^k\plus{}1})^2$ is perfect square.

Generalization:  given a monic polynomial with integer coefficients $ f(x)$ such that $ f(a) \equiv 0 \bmod p^k$, then $ f(x) \equiv 0 \bmod p^{k\plus{}1}$ has a solution if $ f'(a) \not \equiv 0 \bmod p^k$.

(There's a more specific way to state this lemma, but I've forgotten it.)  Here we have $ f(x) \equal{} x^2 \minus{} 17$.

[quote="t0rajir0u"]Generalization:  given a monic polynomial with integer coefficients $ f(x)$ such that $ f(x) \equiv 0 \bmod p$ has a solution, then $ f(x) \equiv 0 \bmod p^k$ has a solution for all $ k$ if and only if $ f'(x)$ does not vanish $ \bmod p$.

(There's a more specific way to state this lemma, but I've forgotten it.)  Here we have $ f(x) \equal{} x^2 \minus{} 17$.[/quote]
what do ou mean by $ f'(x)$ does not vanish $ \bmod p$

Sorry, I have stated the lemma incorrectly.  (The statement may still be wrong...)

I have found the correct statement!

[b]Hensel's Lemma:[/b]  Let $ F(x) \in \mathbb{Z}[x]$, let $ p$ be a prime, and let $ a_{0}\in \mathbb{Z}$.  Let $ \tau$ be the highest power of $ p$ dividing $ F'(a_{0})$.  Given that there exists $ r > \tau$ such that $ r \plus{} \tau$ is the highest power of $ p$ dividing $ F(a_{0})$, there exists an integer $ a_{1}$ such that $ a_{1}\equiv a_{0}\bmod p^{r}$ and $ p^{\tau \plus{} r \plus{} 1}| F(a_{1})$.

When $ \tau \equal{} 0$, we have the result I originally attempted to state.


well i saw a proof in griffith about the diraction of magnetic field due to an infinitely large steady current carrying plate in x direction .well i could argue on the normal basis of an current element being at equal distances and component cancellingout...that the field has to be in y direction.
but then he gave a beautiful proof arguing on the line that if there were field in z direction then by biot savrt law reversing current would reverse it's direction
upto this everything was ok
then he writes but the field in z direction can't possibly depend on current direction in xy plane
why is this?

I'm not sure if I fully understand your question, but I think that this is because we can rotate the x and y axis around z axis and this can't change the field in z direction because the physical situation does not change. So in other words the field in z direction can't possibly depend on current direction in xy plane :)

[quote="Pardesi"]then he writes but the field in z direction can't possibly depend on current direction in xy plane
why is this?[/quote]

For an infinite sheet of current in the xy plane the component of the magnetic field in the z direction is always zero, no matter what is the direction of the current in the plane. To see this, think on the infinite sheet of current as an infinite set of parellel wires oriented in the direction of the current. Now, consider we are to calculate the field at a point P at a certain distance from the plane: from the wire picture of the plane, I think its easy to see that wires that are equidistant from point P create in P magnetic fields whose vertical components cancel and whose horizontal components add, and so the resultant magnetic field for all plane can only be also in the xy plane. You can repeat this reasoning for every direction in the xy plane to conclude that the z component must be always zero.

golduck can u explain a bit more i think u understood...

carcul i could argue on the grounds u did but what i wanted is a non component proof..as i have written already

Well, the magnetic field does not depend on how we choose our axis (the nature does not care about it :) ). Also the magnetic field in some distance z above the surface we are talking about, does not depend on our position in x and y (because the current carrying plate is infinite). If we now choose the axis as you said and then change the directions of x and y axis the magnetic field (so also the component in z direction) won't change. But changing the directions of x and y axis is the same as changing the direction of the current. However, as we know from Biote-Savart law, changing the direction of the current should change the magnetic field vector (and also the component in z) into opposite. That proves that the magn. field in z is 0.

golduck i am not sure i am getting that right or not let me tell u what i have understood
u choose an axis say x along current flow no say the field exists in z direction and now change the direction of x-axis and now still the field remains same becuse magnteic field is ofcourse independent of our choice of axis and then what?

Then changing the direction of x axis is the ame thing as changing the current direction. If we change current direction the direction of magnetic field changes everywhere ($ dB=\frac{\mu_{0}I dl}{4\pi r^{2}}$ changing $ I$ to $ -I$ changes $ dB$ to $ -dB$) so now we found that the field in z should change to opposite. So if we found that the field shouldn't change and also that it should change into opposite - so the field in z must be 0.

ok the biot savart part is right but how does one one get that the field direction doesn't change

hmm, you said that yourself:
[quote="pardesi"]u choose an axis say x along current flow no say the field exists in z direction and now change the direction of x-axis and now still the field remains same becuse magnteic field is ofcourse independent of our choice of axis and then what?[/quote]


AMC
AIME
geometry
circumcircle
ratio
geometric transformation
rotation

From the way tetrahedr0n's question is worded, I would assume he means $\bigtriangleup ABF$, especially since AIME asks for integer answers. But you know what they say about assuming things...(and this could be from Kalva).

It is from Kalva, and I know I solved this problem, and I remember the triangle business. But then again... I lost my copy of the 2002 AIME, unfortunately.

Let $P$ be the intersection of $AD$ and $CE$.

$CP: PE=2: 1\longrightarrow CP=18, PE=9$
$AP: PD=2: 1\longrightarrow AP=12, PD=6$

$\triangle AEF\sim \triangle CEB$ (because all angles are the same), so $\frac{\triangle AEF}{\triangle CEB}=\left(\frac{AE}{CE}\right)^2=\frac{16}{81}$

In the same way, $\frac{\triangle BEF}{\triangle CEA}=\frac{16}{81}$, so $\frac{\triangle AFB}{\triangle ABC}=\frac{16}{81}$

Now all we have to find is the area of $\triangle ABC$, which is equivalent to $6\times \triangle AEP$.

$AE=AP=12$ and $EP=9$, so using Hero's formula,
$\triangle AEP=\sqrt{\frac{33}{2}\left(\frac{33}{2}-12\right)\left(\frac{33}{2}-12\right)\left(\frac{33}{2}-9\right)}$
$=\sqrt{\left(\frac{33}{2}\right)\left(\frac{9}{2}\right)\left(\frac{9}{2}\right)\left(\frac{15}{2}\right)}$
$=\frac{27\sqrt{55}}{4}$

Therefore, $\triangle ABF=\left(\frac{16}{81}\right)(6)\left(\frac{27\sqrt{55}}{4}\right)=8\sqrt{55}.$ So our answer is $8+55=63.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=222[/img]
In the diagram above, $I$ is the centroid of the triangle. Since $CE$ is a median, $BE=AE=12$. Since $CE=27$ and medians intersect each other in a ratio of $2: 1$, we have $CI=18$, $IE=9$, $AI=12$, $ID=6$. Thus, triangle $AEI$ has sides $12, 12, 9$ for an area of $\sqrt{\left(\frac{33}{2}\right)\left(\frac{9}{2}\right)\left(\frac{9}{2}\right)\left(\frac{15}{2}\right)}=\frac{27\sqrt{55}}{4}$.
By Power of a Point, $(FE)(CE)=(AE)(CE)$, so $FE=\frac{(12)(12)}{27}=\frac{16}{3}$.
Then $\bigtriangleup AFE$ has area $\frac{[AIE]\left(\frac{16}{3}\right)}{9}=4\sqrt{55}$ since the two triangles $AFE$ and $AIE$ share the same height and both have their bases on some part on line segment $FI$.
Finally, $AFB=\frac{[AFE]*24}{12}=2*4\sqrt{55}=8\sqrt{55}$ since the two triangles $AFB$ and $AFE$ share the same height and have their bases on some part of line segment $AB$.
Thus, our desired answer is $\boxed{8\sqrt{55}}$.

EDIT: I guess I was a little late. :(

Consider the equilateral triangle in the complex plane.
That is, $(a,11)$ represents the point $a+11i$ and $(b,37)$ represents the point $b+37i$.
Then $b+37i$ is a $60^\circ$ counterclockwise rotation of $a+11i$ about the origin, or $(a+11i)\text{cis}(60^\circ)=b+37i$
\[\left(a+11i\right)\left(\frac{\sqrt{3}i}{2}+\frac{1}{2}\right)=b+37i\]
\[\frac{a}{2}+\frac{\sqrt{3}ai}{2}+\frac{11i}{2}-\frac{11\sqrt{3}}{2}=b+37i\]Setting the real and imaginary parts equal:\[a-11\sqrt{3}=2b\]\[\sqrt{3}a+11=74\]From the second equation, $a=\frac{63}{\sqrt{3}}=21\sqrt{3}$. Then we obtain $21\sqrt{3}-11\sqrt{3}=2b$ or $b=5\sqrt{3}$ after plugging back into the first equation.
Finally, $ab=\left(21\sqrt{3}\right)\left(5\sqrt{3}\right)=\boxed{315}$.

That's correct :first:

Next problem, please.

I hope these problems are above #10, because I am stumped on both.

Couldn't you just do it by setting the distances equal?

[hide]
$a^2+121=(b-a)^2+676=b^2+1369$
[/hide]

[quote="frt"]2. Find $ab$ if $(0,0)$, $(a,11)$, $(b,37)$ is an equilateral triangle.

(Source: 1994 AIME)[/quote]

Using the distance formulas: $\sqrt{(0-a)^2+(-11)^2}=\sqrt{(a-b)^2+(11-37)^2}\longrightarrow-2ab+b^2=-555$

Then, $\sqrt{b^2+37^2}=\sqrt{(a-b)^2+(11-37)^2} \longrightarrow b^2+693=a^2-2ab$, then substituting with the first equation we get $b^2-a^2=-1248$, I don't know if I'm on the right track with any of this though :(

[b]Allan Z: [/b]
That's how I did it and I got $704$ also, so I think it's correct.

[b]miyomiyo:[/b]
... that post just sorta came outa nowhere lol ...

[quote]
[b]miyomiyo:[/b]
... that post just sorta came outa nowhere lol ...
[/quote]
Yeah, I know.  Everybody's worrying about the AIME....  I mean, it's just for fun, right?[/b]

[quote="miyomiyo"]
Yeah, I know.  Everybody's worrying about the AIME....  I mean, it's just for fun, right?[/quote]
no our lives depend on it
if we do bad we'll be like "ZOMG I SUCK" and then we'll head directly to the train station

[quote="gotztahbeazn"]ehh lets get this marathon back up haha

AIME 2003 #13

A bug moves around a wire triangle. At each vertex it has $1/2$ chance of moving towards each of the other two vertices. What is the probability that after crawling along $10$ edges it reaches its starting point?[/quote]

[hide="Alternately"] Let the original point be $0$, and the other two points be $1$ and $2$.  Then we can model the movement of the bug as adding either $1$ or $2$ to a counter, and taking the value of the counter $\bmod 3$.

The generating function is

$F(x) = (x+x^{2})^{10}= x^{10}(1+x)^{10}$

And we want to find the sum of every third coefficient.  This is

$\frac{ F(1)+F(\omega)+F(\omega^{2}) }{3}= \frac{1024+1+1}{3}= 342$

So, of course, the answer is $\boxed{ \frac{342}{1024}= \frac{171}{512}}$. [/hide]
This provides a good rigorous explanation of [b]gotztahbeanz[/b]'s idea.  I believe it is also isomorphic to the matrix approach.

[b]AIME 2002, #14[/b]

A set $\mathcal{S}$ of distinct positive integers has the following property: for every integer $x$ in $\mathcal{S},$ the arithmetic mean of the set of values obtained by deleting $x$ from $\mathcal{S}$ is an integer.  Given that 1 belongs to $\mathcal{S}$ and that 2002 is the largest element of $\mathcal{S},$ what is the greatest number of elements that $\mathcal{S}$ can have?

[quote="cincodemayo5590"][quote="t0rajir0u"]
[b]AIME 2002, #14[/b]

A set $\mathcal{S}$ of distinct positive integers has the following property: for every integer $x$ in $\mathcal{S},$ the arithmetic mean of the set of values obtained by deleting $x$ from $\mathcal{S}$ is an integer.  Given that 1 belongs to $\mathcal{S}$ and that 2002 is the largest element of $\mathcal{S},$ what is the greatest number of elements that $\mathcal{S}$ can have?[/quote]
[hide="Hmm..."]
Let $n$ be the maximum possible number of integers in $\mathcal{S}$, and let $x_{1},x_{2},\ldots x_{n}$ be those integers. Then, looking at the arithmetic mean of those integers, we see that $\sum_{i=1}^{n}x_{k}\equiv x_{i}\bmod n-1$ for any $x_{i}$ in $\mathcal{S}$, which means that each $x_{i}$ must be equivalent to one another in $\bmod (n-1)$. Now, this means that we can only pick one integer out of every period of $n-1$, but since we're picking $n$ integers to be in $\mathcal{S}$, we will pick from a total of $n^{2}-n$ integers, each equivalent both to $2002$ and $1 \bmod n-1$.
But note that also this means that $2002 \equiv 1$, and this occurs only with mods of the factors of $2001=667 \cdot 3=3 \cdot 23 \cdot 29$. These factors are equal to $n-1$, so the largest $n$ from this list is $30$. Seeing that $30$, when plugged into our expression $n^{2}-n$, does not exceed $2002$, this is the maximum number of integers in $\mathcal{S}$. $\boxed{030}$. [/hide]

[b]1998, #15[/b]
Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i,j)$ and $(j,i)$ do not both appear for any $i$ and $j$. Let $D_{40}$ be the set of all dominos whose coordinates are no larger than $40$. Find the length of the longest proper sequence of dominos that can be formed using the dominos of $D_{40}$.[/quote]

thats a cool problem...i think we need to bust out some K_40...too bad we cannot have an eulerian cycle on K_40..., but we do on K_39...and for those other 40 edges, its not so bad to see that half of them work...so 761

What's bust out some $K_{40}$?  :oops:

[hide]

For each domino, you need to have a pair to transition into it and a pair to transition out of it $\implies$ you can transition to any other number, and since you need to transition in and out, we need an even number.

Largest even under 40 is 38, so make each number transition to 38 other numbers. $\implies$ Length is $(40)(38)/2=760$

And then you add $1$ to complete it $\implies \boxed{761}$
[/hide]

Altheman is making explicit graph theory that you used implicitly.  I believe $K_{40}$ is the [url=http://mathworld.wolfram.com/ConnectedGraph.html]connected graph[/url] with $40$ vertices, which only has vertices of odd degree and therefore does not have an [url=http://mathworld.wolfram.com/EulerianCircuit.html]Eulerian circuit[/url].  But $K_{39}$ does.

[quote=d343seven]In a five-team tournament, each team plays one game with every other team.  Each team has a $50\%$ chance of winning any game it plays.  (There are no ties.)  Let $m/n$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime positive integers.  Find $m+n$.[/quote]

[b]My solution. [/b] First we note that the five team play a total of $4+(4-1)+(4-2)+(4-3)+(4-4)=10$ games. Because each games have two possible results then we have, $\underbrace{2 \times 2 \times \cdots \times 2}_{10 \;times}=2^{10}$ possible results for the tournament. there are five ways to choose an undefeated team. Say team $A$  win all four of its games. Then each of the remaining six games has two possible results for a total of $2^6$ possible results. Because at most one team one team can be undefeated, there are $5 \times 2^6$ tournaments that produce an undefeated team. A similar argument shows that $5 \times 2^6$ of the $2^{10}$ possible tournaments produce a winless team. However, these probabilities are not mutually exclusive. It is possible to have exactly one undefeated team and exactly one winless team in the same tournament. There are $5 \times 4=20$ such two-team permutations. Say team $A$ is undefeated and team $B$ is winless. there are $7$  [hide]not $8$ because $A$ and $B$ play against each other! [/hide] games in which either team $A$ or team $B$ or both teams play. The results of these seven games are decided. Each of the remaining three games left has two possible results for a total of $2^3$ tournaments. In other word, $5 \times 2^5$ of the $2^{10}$ tournaments have both an undefeated team and a winless team. Thus, according to [b]The Inclusion-Exclusion principle[/b], there are \[ 2^{10}-2 \times 5 \times 2^6+5 \times 2^5=17 \times 2^5.\]
tournament results in which there is neither an undefeated team nor a winless team. Therefore, the required probability is $\frac{17 \times 2^5}{2^{10}}=\frac{17}{32}.$ Thus we have, $\boxed{m+n=17+32=49.}$

[quote=tetrahedr0n]In triangle $ ABC$ the medians $ \overline{AD}$ and $ \overline{CE}$ have lengths 18 and 27, respectively, and $ AB \equal{} 24$.  Extend $ \overline{CE}$ to intersect the circumcircle of $ ABC$ at $ F$.  The area of triangle $ AFB$ is $ m\sqrt {n}$, where $ m$ and $ n$ are positive integers and $ n$ is not divisible by the square of any prime.  Find $ m \plus{} n$.[/quote]

It's well known that the centroid $G$ (usually denoted by $G$), divides the median into $2:1$ ratio. Thus $AG=12,GD=6$ and $CG=18, GE=9$. From power of a point on $E$, we have $AE\times EB=CE\times EF \implies EF=\frac{16}{3}$. Note that $AE=AG$, hence $AGE$ is isosceles. Since $E$ is the midpoint of $AB$, $\triangle AFE$ and $FEB$ have the same area. Thus $[AFB]=2\times [AFE]=2 \times  EF \times EA\times \sin \angle AEF$. But $\sin \angle  AEF = \sin AEG = \frac{\sqrt{24^2-9^2}}{2}$. Plugging the values gives $8\sqrt{55}$. So the answer is $8+55=63$

[quote=d343seven]In a five-team tournament, each team plays one game with every other team.  Each team has a $50\%$ chance of winning any game it plays.  (There are no ties.)  Let $m/n$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime positive integers.  Find $m+n$.[/quote]

aka 1996 aime p6
[hide=solution]we apply pie by counting the complement. we know that the probability that there is a winless team is 5(1/2)^4=5/16 because there are 5 teams and an equal probability of a win and a loss. by symmetry the probability for an undefeated team is 5/16 as well. the probability that there is one undefeated team and one winless team is 5(1/2)^4*4(1/2)^3=20/128=5/32. by pie the probability that there is at least 1 undefeated team or 1 winless team is (5/16+5/16-5/32)=15/32 => ans=17/32=49.[/hide]

An insanely good barycentric problem. Here is a full, step by step solution outlining my usual process for a barybash (as well as some more specific details for those that don't know much about bary).

[quote=2002 AIME I, Problem 13]In triangle $ ABC$ the medians $ \overline{AD}$ and $ \overline{CE}$ have lengths 18 and 27, respectively, and $ AB \equal{} 24$.  Extend $ \overline{CE}$ to intersect the circumcircle of $ ABC$ at $ F$.  The area of triangle $ AFB$ is $ m\sqrt {n}$, where $ m$ and $ n$ are positive integers and $ n$ is not divisible by the square of any prime.  Find $ m \plus{} n$.[/quote]

First, we consider the given conditions. We can get $2$ equations in terms of $a,b$ by using the distance formula, and since we already know $c$, we can figure out the sidelengths. This allows us to find $F$ by intersection, and we can finish with area formula + Herons.

We now begin the bash. Apply barycentric coordinates on $\triangle ABC$. We know that $D=\left(0, \frac{1}{2}, \frac{1}{2}\right), E=\left(\frac{1}{2}, \frac{1}{2}, 0\right)$. We can now get the displacement vectors $\overrightarrow{AD} = \left(1, -\frac{1}{2}, -\frac{1}{2}\right)$ and $\overrightarrow{CE}=\left(-\frac{1}{2}, -\frac{1}{2}, 1\right)$. Now, applying the distance formula and simplifying gives us the two equations
\begin{align*}
2b^2+2c^2-a^2&=1296 \\
2a^2+2b^2-c^2&=2916. \\
\end{align*}

Substituting $c=24$ and solving with algebra now gives $a=6\sqrt{31}, b=3\sqrt{70}$. Now we can find $F$. Note that $CE$ can be parameterized as $(1:1:t)$, so plugging into the circumcircle equation and solving for $t$ gives $t=\frac{-c^2}{a^2+b^2}$ so $F=(a^2+b^2:a^2+b^2:-c^2)$. Plugging in for $a,b$ gives us $F=(1746:1746:-576)$. Thus, by the area formula, we have $$\frac{[AFB]}{[ABC]}=
\left|\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
\frac{97}{162} & \frac{97}{162} & -\frac{16}{81}
\end{matrix}\right|=\frac{16}{81}.$$ By Heron's Formula, we have $[ABC]=\frac{81\sqrt{55}}{2}$ which immediately gives $[AFB]=8\sqrt{55}$ from our ratio, extracting $\boxed{63}$.

Solution to the first problem: After length chasing using PoP, we can then note that the area of AEF is $\frac{16}{81}$ of the area of CEB and the area of BEF is $\frac{16}{81}$ of the area of CEA, so thus the area of $\triangle ABF$ is $\frac{16}{81}$ of the area of $\triangle ABC$. Thus, we need $\triangle ABC$'s area, which is not immediately apparent. However the area of one of the little triangles created by two medians is equal to one-sixth of the area of the triangle, so the area of $ABC$ is six times the area of $AEP$ which comes out to $\frac{81\sqrt{55}}{2}$. Multiplying by $\frac{16}{81}$ gives $8\sqrt55 \implies \boxed{63}$.

As centroid divides the median in the ratio of 2:1
Then the values of CG=18 and GE=9
Now let the <AEC=X
AND <ECA=Y
THE TRIANGLES ECA AND EBF ARE SIMILAR TO EACH OTHER HENCE BY SIMILARITY OR THALES THEOREM 12/27=EF/12 HENCE EF=16/3
NOW APPLYING COS(X)=81+144-144/2(9)12
COSINE RULE IN TRIANGLE. EGA
GIVING COS(X)=3/8 AND SIN(X)=√55/8
AREA OF TRIANGLE BFE =(12)(16/3)1/2√55/8 WHICH SIMPLIFIES TO
4√55
WE KNOW A MEDIAN DIVIDES THE AREA INTO TWO EQUAL PARTS HENCE THE AREA OF TRAINGLE BFA=2(4√55)=8√55
AS A+B=8+55=63


Columbia
geometry
geometric transformation

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You can always request a new forum / subforum if you think that a certain category of people can use it. Ask any of us (the admins) in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21296]this topic[/url].

We welcome your suggestions about this, so feel free to express them in this topic.

With regards,
AoPS - MathLinks Admins

[quote="Ravi B"][quote="Valentin Vornicu"]or just go straight forward to usergroups menu above, and select the Maryland usergroup and click join. [/quote]
Hmm, when I click on the Usersgroups button, I don't see all the local groups.  The only groups I see (besides the ones I am already in: Local Moderators and NYC) are Main Administrators, Administrators, Moderators, and VIP.  Perhaps the feature you mentioned hasn't been implemented yet?[/quote]Good point, I had forgotten to do that too :P ... it's done at this point. Joining usergroups is only required for North American (US & probably in the future Canada) local communities. For all the other (national mostly) communities, you should at least be able to speak / write / understand some of the language that it is used there.

[quote="siuhochung"]i don't know if it is good to have hong kong separated from china for the forum stuff.... indeed hong kong and china send different teams in imo, the level of students are different and the way to select students to imo are different... so.. to me, it is better to have two forums separated.[/quote]those forums are for people that find it [b]very hard[/b] to express in English. You can use Chinese there (as oposed to the main forums). There will be different subforums created, but first we'd like to see some activity before creating more forums.

What about the South American Region?  :D
Not forget!!

Here, almost countries speak spanish like: Argentina, Colombia, Peru, etc.
and portugues in Brazil!!

If there weren't enough people in the Forum...
we can participate into the regional community... at the first time!! :)

Thanks!!

OK I will also make a Portughese community if you guys think that you would use it :)

I just found the spanish community and...
 it's ok.

 i was talking about my brazilian partners, however
 they entered in this community wihtout problems!! :)

thanks for all!!!

What there should be is a World page, with a translation required for the four major languages - English, Spanish, French, and German. This should be easy enough given modern free translation sites like

translate.google.com :idea:


Can this differential equation be solved symbolically?

$ y' \equal{} y^2 \minus{} xy$

What are the best software programs for symbolically solving ordinary Diff. Eqs?

Just for the record, here's what Mathematica gave as the answer:

$ y[x]\to \frac{2 e^{\minus{}\frac{x^2}{2}}}{2 C[1]\minus{}\sqrt{2 \pi } \text{Erf}\left[\frac{x}{\sqrt{2}}\right]}$

hello, Maple 12 gives the following answer
$ y \left( x \right) \equal{}\minus{}2\,{\frac {{{\rm e}^{\minus{}1/2\,{x}^{2}}}}{\sqrt {\pi
}\sqrt {2}{{\rm erf}\left(1/2\,\sqrt {2}x\right)}\minus{}2\,{\it \_C1}}}$
Sonnhard.

Look [url=http://en.wikipedia.org/wiki/Riccati_equation]here[/url] for your first question.


trigonometry ARML LaTeX geometry incenter quadratics Vieta

You can use a triangle $36^\circ-36^\circ-108^\circ.$ Golden section.
The answer is [hide]$\cos 36^\circ =\frac{1+\sqrt{5}}{4}.$[/hide]

Can you explain a little more how you know this? I'm lost  :D

O.K. Set regular pentagon $ABCDE,$ let $P,Q$ be the intersection points of the diagonals $BE$ and $AC,\ AD$ respectively. Remark the similarity of two triangles $PAB$ and $ABE.$

I e-mailed this to my arml coach, sorry about the latex, i don't feel like reworking it  :P

--Solution 1--
(all in degrees)
Denote triangle ABC such that <A=36, <B=<C=72, D,E,F, are on BC, CA, AB, such that AD, DE, CF are angle bisectors, respectively, also the incenter is I
<EBC=72/2=36=<EBA, <BEC=180-36-72=72
<FCE=72/2=36, <EIC=180-36-72=72
also AE=BE since  <BAC=<EBA=36
by AA similarity, triangle ABC ~ BEC ~ CIE
AB/BC=BE/EC=CI/IE
WLOG BD=1, and AB=x, then BC=2=BE=AE, EC=AB-2=x-2=CI=BI
x/2=2/(x-2)=(x-2)/IE
x^2-2x-4=0
x= 1+\sqrt{5}

cos <IBD=cos 36=1/[x-2]=1/[ \sqrt{5}-1]= [\sqrt{5}+1]/4

cos ABD=cos 72= 1/x=1/[1+\sqrt{5}]=[ \sqrt{5}-1 ]/4

--Solution 2--
consider f(x)=x^5-1
the roots are e^{2i\pi*k/5}, for k=-2,-1,0,1,2
easily, we factor out k=0, x=1, to get
g(x)=x^4+x^3+x^2+x+1
notice how the roots that correspond to k=-2,2 are reciprocals, consider a quadratic with those roots, it clearly is x^2+ax+1 using vieta's coefficient sums, also the roots with k=-1,1 are also reciprocals, so let their quadratic be x^2+bx+1, thus

x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1)

=x^4+(a+b)x^3+(2+ab)x^2+(a+b)x+1

is an identity in x, so the coefficients match up, so
a+b=1, and 2+ab=1 or ab=-1,
a quadratic with roots a,b is
x^2-x-1=0 by vieta, solving a,b=(1+-\sqrt{5})/2
the roots of x^2+bx+1 sum to cis72+cis-72=2cos72
note cos72 is in the first quadrant, and b is the opposite of the sum, the sum of the roots is negative, making b negative
b=[ 1-\sqrt{5} ]/2= -2\cos 72
cos 72=[ \sqrt{5}-1 ]/4
similarly, -a/2=cos 144=[ -1-\sqrt{5} ]/4

[quote="pieterminate"]Find cos 36 in terms of $\frac{a+\sqrt{b}}{c}$ where a, b, and c are natural numbers.
[hide="hint"]
use a 36-72-72 triangle[/hide][/quote]

[hide]

Construct a $36-72-72$ triangle. Label the triangle ABC with B & C being at the legs. Bisect $C$ onto side $AB$ at point $D$. now $AD = CD = CB = 1$ then we have $AB = \frac{\sin{36}}{\sin{72}} + 1 = \frac{\sin{72}}{\sin{36}}$ thsi simplifies into $4\cos^2{36} = 1 + 2\cos{36}$ which solves into $\displaystyle \frac{1 + \sqrt{5}}{4}$ [/hide]

[quote="scorpius119"][hide="yet another geometric solution"]
Let $\phi$ be the diagonal length of a regular pentagon $ABCDE$ with side length 1. By Ptolemy's Theorem in quadrilateral $ABCD$, $(1)(1)+(1)(\phi)=(\phi)(\phi)\Rightarrow \phi^2-\phi-1=0\Rightarrow \phi=\frac{1+\sqrt{5}}{2}$ (since $\phi$ is positive).

Drawing a perpendicular from $B$ to diagonal $AC$, $\cos 36^\circ =\frac{\phi}{2}=\boxed{\frac{1+\sqrt{5}}{4}}$.
[/hide]

[hide="and a pure trig solution"]
Let $P=4\cos 36^\circ \cos 72^\circ$. Then $P\sin 36^\circ=\sin 108^\circ$ after using double angle formulas. Since $\sin 108^\circ=\sin 72^\circ$, $4P=1$.

But also if $x=\cos 36^\circ$ then $P=4x(2x^2-1)$. So $8x^3-4x-1=0$. Letting $y=2x$, $0=y^3-2y-1=(y-1)(y^2-y-1)$.

Since $y\neq 1$, for $1=2\cos 60^\circ$, $y^2-y-1=0\Rightarrow y=\frac{1+\sqrt{5}}{2}\Rightarrow x=\boxed{\frac{1+\sqrt{5}}{4}}$.
[/hide][/quote]

i love the geometric solution!!!

[quote="4everwise"]Too difficult for Getting Started.

From the hint you provided, I'm pretty sure that you got this from the last problem in AoPS Volume 2 Chapter 2.

I remember this problem. ;)

AoPS 2 generally goes in the Intermediate Forum.[/quote]

To me, I don't have any AoPS books. ;)

An outline of another trigonometric solution which may be instructive

Using the identity

                            COS(5X) = 16.(COSX)^5 - 20.(COSX)^3 + 5.(COSX)

and noting that 5.18 =90 we see that COS18 satisfies the equation 16.COS18)^4 - 20.(COS18)^2 +5 =0
NOw solve this equation for (COS18)^2  and calculate 2.(COS18)^2 -1 which is COS36 using the identitity COS2X =2.(COSX)^2 -1

err...non geometric solution that i took a stab at a while ago

$x^5-1=(x-1)(x^4+x^3+x^2+x+1)$, since $x=1$ is obviously not the solution we need, we let $x^4+x^3+x^2+x+1$ factor into $(x^2+ax+1)(x^2+bx+1)$, if we expand this, we note that $a+b=1$ and $ab=-1$, to find the values of $a$ and $b$, we can use:

$x^2+x-1$, which yields the values for $a$ and $b$ as $\frac{-1 \pm \sqrt{5}}{2}$.

Plugging either of these values into the first equation, we extract the real parts of the roots, which is $\frac{-1+\sqrt{5}}{4}$, which we deduce as $\cos{72}$, since it can't be a negative.

Thus, $\cos{36}=\sqrt{\frac{3+\sqrt{5}}{8}}$

We can also use $\cos{(2x)}=-\cos{(3x)}$ where $x=36^\circ$.
If you remember the triple angle formula for cosine, the rest is pretty simple. (Even if you don't, you can use $\cos (a+b)=\cos a\cos b-\sin a\sin b$ to get the formula.)

[quote="4everwise"]Too difficult for Getting Started.

From the hint you provided, I'm pretty sure that you got this from the last problem in AoPS Volume 2 Chapter 2.

I remember this problem. ;)

AoPS 2 generally goes in the Intermediate Forum.[/quote]

What?  This was in my trig class a year ago, that isn't too hard! (even though only 2 people ;)  solved it :) )

And I was thinking a almost pure geometric solution, construct a line that makes another 72-72-36 triangle.
but correct, the answer is: [hide]$\frac{1+\sqrt{5}}{4}$[/hide]

How did you get to your second step: 16x^5-20x^3+5x+1=0 ?  Elaborate?  :roll:

$\cos 5\theta=16\cos^{5}\theta -20\cos^{3}\theta +5\cos\theta$

I think $\theta$ or another variable should have been used in the above post because there is already a value for $x$.

Since $\cos 5x=-1$:

$16\theta^{5}-20\theta^{3}+5\theta+1=0$

You can find the formula for $\cos (nx)$ and $\sin (nx)$ (for any positive integer $n$) by using De Moivre's Theorem ;)

http://www.upal.ca/adeel/problems/category/trig.html

See "Trigonometry" for the expansions of $\cos (nx)$ and $\sin (nx)$.


geometry
incenter
circumcircle
IMO Shortlist
ratio

Let $ABC$ be a triangle, and $I$ its incenter. Consider a circle which lies inside the circumcircle of triangle $ABC$ and touches it, and which also touches the sides $CA$ and $BC$ of triangle $ABC$ at the points $D$ and $E$, respectively. Show that the point $I$ is the midpoint of the segment $DE$.

It has certainly been posted before. You can use Casey's theorem and the transversal theorem. Do you have a different solution, Pascual?

[hide="Solution by Sailor"]
Solution by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=320]Sailor[/url]:

That's a nice property already used(and proved) many times on the forum.

One easy approach is the following:
Let $k_c$ be tangent to $k$ at $X$.
The homotety of center $X$ that sends $k_c$ into $k$ will send the point $E$ into the midpoint $A'$ of the arc $\widehat{BC}$ (which doesn't contain point $C$) of the circumcircle of $\triangle{ABC}$. Analogously, it will send the point $D$ into the midpoint $B'$ of the arc $\widehat{AC}$.

Now, from the [i]Pascal's Theorem [/i]applied to the hexagon $AA'XB'BC$ it follows that the points $D,I,E$ are collinear.
Consequently, $I$ is the midpoint of $[DE]$.
[/hide]

well i basically used caseys to find the ratio between incircle and the other tangent circle, but a friend of mine did it without caseys with a bash of trigonometry...I was wondering if there is a solution by inversion????

I have a different proof using the following Lemma:
[b]Lemma[/b]: For a circle $S_1$ which is internally tangent to $S_2$ at X, let $A,B\in S_2$ s.t. AB is tangent to $S_1$ at a point $C$. Then CX bisects the arc AB.

From this lemma, it is enough to prove the following:
Let D,E be the points on CA and CB respectively s.t. $DE\perp CI,I\in DE$. Let also lines AI and BI meet the circumcircle at another point A',B' respectively. Then it suffices to show $\angle IA'E+\angle IB'D=\angle C$. This is easy by considering the similarity between triangles IBA' and IAB'.

Solution by inversion...

Let $o, i, j$ be respectively Circumcircle, Incircle, ThirdCircle.

In inversion in $i$ lines $AB, BC, CA,$ and $o$
will give four congruent circles, call them $c', a', b', o'$.

$o'$ passes through points of intersection of pairs $(a',b'), (b',c'), (c',a')$ distinct from $I$
(call them $K, L, M$ respectively).

Angle bisector of $\angle ACB$ is perpendicular bisector of $DE$.
$J$ (center of $j$) lies on this pernpedicular bisector.

So it is enough to prove that $J$ lies on $DE$ or in other words $DI || IE$.

Image of $j$ in this inversion is a circle $j'$ to which $a', b', o'$ are tangent internally.
So $J'$ (center of $j'$) is point $K$.
Let $(a'), (b')$ be centers of $a'$ and $b'$ and $X$ point of intersection of angle bisector of $\angle ACB$ and $(a')(b')$.
Now because $D'(a') = (a')K$ we get $D'I || (a')X$
($D'$ and $E'$ are images of $D$ and $E$)
Similary $E'I || (b')X$
Thus $E'I || D'I$
And $EI || DI$, just as we wanted.

This problem has already been posted for at least 3 times.
As far as I remember, there are solutions making use of Pascal, Ceva, etc.

Well-known lemma:
If circle $K$ is tangent internally to the circle $L$ at point $F$ and the tangent to the smaller circle at $G$ cuts the bigger circle in $X, Y$, then the line $FG$ bisects the $\angle XFY$.

So, let $F$ be the tangency point of the circle to the circumcentre of triangle $ABC$, and let $A', B'$ be the middle points of arcs $BC$ and $AC$ respectively (the arcs which don't contain $A$/$B$). Then, from the lemma the lines $FD$ and $FE$ cuts the circumcircle of $ABC$ in $B'$ and $A'$ respectively. Note also that $I$ is intersection of the lines $AA'$ and $BB'$, so from the Pascal Theorem for the hexagon $A, F, B, A', C, B'$ the points $D, E, I$ are collinear.
Now, it's also obvious that $I$ is the midpoint of $DE$ since $CI$ is the angle bisector in the isosceles triangle $CDE$.

[quote="grobber"]It has certainly been posted before. You can use Casey's theorem and the transversal theorem. Do you have a different solution, Pascual?

[hide="Solution by Sailor"]
Solution by [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=320]Sailor[/url]:

That's a nice property already used(and proved) many times on the forum.

One easy approach is the following:
Let $ k_c$ be tangent to $ k$ at $ X$.
The homotety of center $ X$ that sends $ k_c$ into $ k$ will send the point $ E$ into the midpoint $ A'$ of the arc $ \widehat{BC}$ (which doesn't contain point $ C$) of the circumcircle of $ \triangle{ABC}$. Analogously, it will send the point $ D$ into the midpoint $ B'$ of the arc $ \widehat{AC}$.

Now, from the [i]Pascal's Theorem [/i]applied to the hexagon $ AA'XB'BC$ it follows that the points $ D,I,E$ are collinear.
Consequently, $ I$ is the midpoint of $ [DE]$.
[/hide][/quote]

Sorry to revive the thread, but I have some questions

1) How do you know that the homothecies actually hit the midpoints of the chords?
2) How does Pascal's imply that D,I,E are collinear? Doesn't pascal's just say that if you extend the sides the intersections are concurrent? o.O

Dear Mathlinkers,
see also for this nice of Deprez,
http://perso.orange.fr/jl.ayme   vol. 4 A new mixtilinear incircle adventure p. 15
Sincerely
Jean-Louis

I hope my solution is correct.... it is too easy so I feel queaky about this. :oops:
Please rectify my errors. :(
[hide="My Solution"]
Let the circle drawn be denoted by $ (P)$. It is easily proven that $ I$ lies on $ DE$.
[b]Construction[/b]
Take points $ X$ and $ Y$ respectively on $ DC$ and $ EC$ so that they are the midpoints of these respective segments. Now we join : $ IX; IY; XY$.
[b]Proof: [/b]
Observe that since $ CD$ and  $ CE$ are tangents to the circle $ (P)$ hence $ CD\equal{}CE$ also since $ X$ and $ Y$ are the midpoints of these tangents; from triangle $ ECD$ we have  $ XY \parallel DE$
Now we consider $ \triangle IEY$ and $ \triangle IXY$. These have;
\[ \measuredangle EIY\equal{}\measuredangle IYX; \measuredangle IYE\equal{}\measuredangle YIX\]
Also side $ IY$ is common so from AAS rule of congruency these two triangles are congruent.
So $ IE\equal{}IY$ and we already have $ IE\parallel IY$. Therefore $ IEYX$ is a $ \parallel^{\text{gm}}$

$ IE\equal{}XY$ and $ XY \equal{} \frac 12 DE$ so that $ IE\equal{} ID$
And, sice $ I$ lies on $ ED$ we are done. :)
[/hide]

It is not strange the topic be posted many times, is a classical result, one of the oldest references is an english problem book by Rev. Radford 1925. It has been also studied by Leo Bankoff in Crux 1983, pp.2-7 (A mixtilinear adventure).

Dear F. and Mathlinkers,
the reference I gave in my article is
Deprez E., Mathesis X (1890) 67-68.
It is right ?
Sincerely
Jean-Louis

Dear J-L and Mathlinkers,
yes, the reference is right. Mathesis 1890 p. 67-68 gives the solution by Deprez to a more complicated problem proposed by De Longchamps.

[quote="Potla"]I hope my solution is correct.... it is too easy so I feel queaky about this. :oops:
Please rectify my errors. :(

Let the circle drawn be denoted by $ (P)$. It is easily proven that $ I$ lies on $ DE$.
[/quote]

Could you please explain how it is proven that $ I$ lies on $ DE$ ?

Thanks.

Just for convenience, here's the official wording of the problem:
[quote="1993 G1"]Consider the triangle $ABC$, its circumcircle $k$ with center $O$ and radius $R$, and its incircle with center $I$ and radius $r$. Another circle $k_c$ is tangent to the sides $CA,CB$ at $D,E$, respectively, and it is internally tangent to $k$. Show that the incenter $I$ is the midpoint of $DE$.[/quote]
Let circle $k_c$ have center $P$ and radius $t$ and let it touch circle $k$ at point $X$. Let the midpoint of $DE$ be $M$ and the midpoint of arc $AB$ be $N$. Clearly line $CMPN$ is the angle bisector of $\angle{C}$ and $O,P,X$ are collinear. From right triangle $\triangle{CDP}$, and the Law of Sines on $\triangle{CAN}$, we get (set $\angle{A}=2\alpha$, etc.)
\[PC=\frac{t}{\sin\gamma},\qquad CN=2R\cos(\alpha-\beta).\]From the power of point $P$ with respect to circle $k$, we see that
\begin{align*}
t(2R-t)=R^2-(R-t)^2&=R^2-PO^2\\
&=PC\cdot PN=PC(CN-PC)=\frac{t}{\sin\gamma}\left(2R\cos(\alpha-\beta)-\frac{t}{\sin\gamma}\right),
\end{align*}so
\[2R\sin^2\gamma-t\sin^2\gamma=2R\sin\gamma\cos(\alpha-\beta)-t\]and
\[t\cos^2\gamma=t(1-\sin^2\gamma)=2R\sin\gamma(\cos(\alpha-\beta)-\sin\gamma)=4R\sin\alpha\sin\beta\sin\gamma=r.\]It's simple to verify that
\[CM=CI=\frac{r}{\sin\gamma},\]so we're done.

Consider the first problem of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=155800&start=0]this.[/url] Two proofs have been provided. One by inversion and the other by using the lemma. If we consider the inversion proof, we can see that since $I$ maps to $I_A$, $M$ maps to $M'$, $N$ maps to $N'$ and we also know that $I_AM'=I_AN'$ which means $\frac{IM\cdot r^2}{IN\cdot AM}=\frac{IN\cdot r^2}{AI\cdot AN}\Longrightarrow IM=IN$ as required.

[b][u]Generalisation:[/u][/b]
Let $ABC$ be a triangle and consider the circle $\Omega$ which is tangent to the [b]rays[/b] $AB, AC$  in the points $M, N$ respectively and which is externally tangent in the point $T$ to the circumcircle of the triangle $ABC$. Prove that the excenter of $\Delta ABC$ opposite to $A, I_A$ is the midpoint of $MN$.
Proof is given [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=155800&start=0]here[/url] as a generalisation.

This is a very common and famous problem.The most common solutions are using homothety,but I shall present a solution using Casey.

Let the smaller circle be tangent to the circumcircle at $X$.Then by Casey's theorem to the circles $A,B,BXE,C$ we have
$AB*CE+AC*BD=AD*BC \Rightarrow \boxed{AD=AE=\frac{2bc}{a+b+c}}$.Thus if the midpoint of $DE$ is $I'$ we have $AI'=\frac{2bc}{a+b+c}\cos\frac{A}{2}$.It is also well known that $AI=\frac{2bc}{a+b+c}\cos\frac{A}{2}$ so $I \equiv I'$ as desired.

For the Sawayama's Lemma in $\triangle ABC $ and cevian " $ BC $ ". Then the chord $ ED $ passes through the incenter $ I $ of $ \triangle ABC \Rightarrow I$ is midpoint of $ DE $

[img]http://i.imgur.com/AHtUOSf.png?1[/img]

Let the point $M$ to be the midpoint of $DE$. Suppose that the circle touches the circumcircle of $ABC$ in $P$. Prove that $MDCP$ and $MEBP$ are concyclic. Then we are done.

Does this approach work?

[hide=My Approach]Suppose $\overline{CI} \cap \overline{DE} = I'$. We will prove $I=I'$ by showing $I'$ satisfies the properties of an angle bisector and we will show $\overline{I'D}=\overline{I'E}$. Note that since $\overline{CD}$ and $\overline{CE}$ are tangents from point $C$ to the circle, it follows that $\overline{CD}=\overline{CE}$. By Angle Bisector Theorem, $\frac{\overline{CD}}{\overline{CE}}=\frac{\overline{I'D}}{\overline{I'E}}\implies \overline{I'D}=\overline{I'E}$. So $I'$ is the midpoint of segment $DE$. Now all that is left to show is that the angle bisector of either $\angle A$ or $\angle B$ intersects $\overline{DE}$ at $I'$ and that will be enough to claim $I'=I$. WLOG let's show the angle bisector of $\angle B$ intersects $DE$ at $I'$. Since $\overline{CI}$ is both a median and angle bisector of triangle $CDE$ and we know $CD=CE$, we can conclude that $CI' \perp DE$. This means that line $CI'$ passes through the center of the circle. Now let $\angle I'BC = \alpha$ and show $\angle B = 2\alpha$. I'm trying to angle chase a bunch here but am getting no where :(.[/hide]

Can someone give me a hint on how to proceed?

Thank you,
book_learner

Above  can't work because you are  not using that the circle touch the circumcircle of $\triangle ABC$.

[hide=Hint]Let $R$ be the  touch point.Let $DR$ intersect $\odot ABC$ at $F$.Prove that $F$ is midpoint of arc $AC$[/hide]

Hey can someone check my solution? The other solutions used complex stuff like Casey's theorem, some homothety (haven't learnt this), and some other inversion stuff. But I think my solution only uses elementary theorems which makes me really wonder if I even have it right :)

[b][color=#00f]Claim:[/color][/b] $ID=IE$

[b][color=#f00]Proof:[/color][/b] Because $I$ is on the angle bisector of $\angle{ACB}$ we know that $ID=IE$.

[b][color=#00f]Claim:[/color][/b] $\triangle{DWI} \cong \triangle{EWI}$

[b][color=#f00]Proof:[/color][/b] We have $\triangle{DIC} \cong \triangle{EIC}$ because of SSS. So we know that
$\angle{WID}=\angle{WIE}=90^\circ$. Thus we have $\triangle{DWI} \cong \triangle{EWI}$ by SAS (or by SSS with Pythagorean Theorem).

[b][color=#00f]Claim:[/color][/b] $I, D, E$ are all collinear

[b][color=#f00]Proof:[/color][/b] Now, labeling $\angle{DWI}=\angle{EWI}=\alpha$ and $WI=x, ID=IE=y$ we have \[ \frac{ID}
{\sin(\alpha)}=\sqrt{x^2+y^2} \implies ID=\sin(\alpha)\sqrt{x^2+y^2} \] by Law of Sines. Then doing Law of Sines on $\triangle{WDE}$ we have \[ \frac{DE}{\sin(2\alpha)}=\frac{\sqrt{x^2+y^2}}{\cos(\alpha)} \implies DE=\frac{\sqrt{x^2+y^2}}{\cos(\alpha)} \cdot \sin(2\alpha) \]
Then because $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ we have \[ DE=2\sin(\alpha)\sqrt{x^2+y^2}=2ID=ID+IE \] which implies $I, D, E$ are all collinear.

Because we have already proved that $ID=IE$ we have just proved that $I$ is the midpoint of $\overline{DE}$.

(Here’s an FYI to anyone who has solved this problem.) [url=https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2020-02/mr_2_2020_mixtilinear.pdf]This handout[/url] utilizes a more natural approach to solve this question and prove other properties of the mixtlinear incircle.

lol i didn't even know what a mixtilinear thing was when i "solved" it. also bump, can somebody check my sol to this problem? its @2above

Cute problem.

By symmetry we can assume that the sides are instead $AB$ and $AC$. Under $\sqrt{bc}$ inversion the $A$-mixtilinear incircle goes to the $A$-excircle. Then $D$ and $E$ go to the points of contact of the $A$-excircle with the sides $AB$ and $AC$ (call these $D'$ and $E'$). Finally, it's well known that $(A,M;I,I_A)=-1$ where $M$ is the midpoint of the arc $BC$ not passing through $A$ which implies that $I$ goes to $I_A$ (inversion induces harmonic bundles). Then $AD'I_AE'$ is cyclic so $I$ lies on $DE$. We are done. $\blacksquare$


inequalities
function
inequalities unsolved
algebra

[quote="silouan"]Write it as $48(a^3+b^3+c^3+d^3)\geq 8(a^2+b^2+c^2+d^2)(a+b+c+d)+(a+b+c+d)^3$
Then use Muirhead and Holder.[/quote]
 :o I think only Muirhead is already OK,isn't it?
(Because I did it by Muirhead only,just expand all terms...)

With using both Muirhead and  Holder you have not so much expand.That's all

[quote="Beat"]What is Muirhead?[/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15896]Muirhead's theorem[/url] is a very famous generalization of AM-GM inequality.

[quote="tkhtn"][quote="Soarer"]$6(\sum a^3) \ge \frac{3}{2}(\sum a^2)$ (Chebyshev)
[/quote]
Why?? I don't see Trebúsep here.  :oops:[/quote]

By using Chebychev we have $(a+b+c+d)(a^2+b^2+c^2+d^2)\leq 4(a^3+b^3+c^3+d^3)$

.But  $a+b+c+d=1$  etc

Oh my, :D sory :blush: .
I don't remember: $a+b+c+d=1$

i've got a different solution  :D

from cauchy-schwarz we have that
\[ \sum a^3=(\sum a)(\sum a^3 )\geq (\sum a^2)^2 \]

then we just need to show that

\[ 6x ^2-x-\frac{1}{8}\geq 0 \]

where $x=\sum a^2$.

but

\[ 6x^2-x-\frac{1}{8}=\frac{1}{8}(12x+1)(4x-1) \]

and from cauchy-schwarz again we have that $4x\geq (a+b+c+d)^2=1$ :lol:

hy

i wonder what were the results of the selection test?

how many people solved 3 or more problems?

seems to be 1 full mark, several (should be ~5) 21 or above.

Hallo Vasile!

 How are you ? Our book is ready , but the last correction are still on the road.When will be ready your new book in Inequalities?

 Have a nice day - babis

[quote="Vasc"]More interesting is the following:
Let $a,b,c,d$ be non-negative real numbers such that $a+b+c+d=4$. Prove that
$5(a^2+b^2+c^2+d^2) \ge a^3+b^3+c^3+d^3+16$.[/quote]

Write the inequality of the form $\sum (3-a)(a-1)^2 \ge 0$

If $a \leq 3$, done.

Else, write the inequality as $(a+c+d)(b-1)^2 + (a+b+d)(c-1)^2 + (a+b+c)(d-1)^2 \ge (b-1)^2+(c-1)^2+(d-1)^2+(a-3)(a-1)^2$
By Cauchy-Schwarz, $(3a-9)((b-1)^2+(c-1)^2+(d-1)^2) \ge (a-3)(a-1)^2$.
However, we also have $a - (3a-9) = 9-2a$, and $(9-2a)(b-1)^2 \ge (b-1)^2$ because $a \leq 4$, same for $c,d$.

QED

[quote="Vasc"]More interesting is the following:
Let $ a,b,c,d$ be non-negative real numbers such that $ a+b+c+d=4$. Prove that
$ 5(a^{2}+b^{2}+c^{2}+d^{2}) \ge a^{3}+b^{3}+c^{3}+d^{3}+16$.[/quote]

Also, the following is more interesting than the original one

If $ a,b,c,d$ are positive and $ a+b+c+d=1$ then
\[ 12(a^{3}+b^{3}+c^{3}+d^{3})+1\ge 7(a^{2}+b^{2}+c^{2}+d^{2}). \]

I wait for a nice solution to it.

Yes,you are right. My inequality has equality when $ a=b=c=d$ or $ a=b=c,d=0$ and VASC inequality has equality when $ a=b=c=d$ or $ a=4,b=c=d=0.$

By Cauchy-Shwarz Inequality, (a^3+b^3+c^3+d^3)(a+b+c+d)=(a^3+b^3+c^3+d^3)>=(a^2+b^2+c^2+d^2)^2.
So, 12(a^3+b^3+c^3+d^3)-7(a^2+b^2+c^2+d^2)+1>=12(a^2+b^2+c^2+d^2)^2-7(a^2+b^2+c^2+
+d^2)+1=(3(a^2+b^2+c^2+d^2)-1)(4(a^2+b^2+c^2+d^2)-1)=M.
Left to prove, that m>=0. As, Average Square of a,b,c and d is more or equal to Average Arithmetical of a,b,c and d, a^2+b^2+c^2+d^2>=(a+b+c+d)^2/4=1/4. So, 4(a^2+b^2+c^2+d^2)-1>=0. So, we have to prove, that a^2+b^2+c^2+d^2>=1/3, but we can choose a,b,c and d so, that 1/4<a^2+b^2+c^2+d^2>1/3. For those numbers, the following inequality doesn't take place.

[quote="hungkhtn"]What do you mean? lasha? You mean my inequality is wrong? I feel something not true in your solution?![/quote]
 :rotfl:

[quote=leepakhin]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$. Prove that\[ 6(a^3+b^3+c^3+d^3)\ge(a^2+b^2+c^2+d^2)+\frac{1}{8}  \][/quote]
[b] Hong Kong   2006[/b]
[url=https://artofproblemsolving.com/community/c6h149168p842892]France 2007[/url]
[url=https://artofproblemsolving.com/community/c6h1531976p9218816] Turkey  2017[/url]
[url=https://artofproblemsolving.com/community/c6h514408p21256390]h[/url]

[quote=leepakhin]Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$. Prove that\[ 6(a^3+b^3+c^3+d^3)\ge(a^2+b^2+c^2+d^2)+\frac{1}{8}  \][/quote]

Let $f(x)=6x^3-x^2-\frac1{32}$. The inequality is equivalent to $f(a)+f(b)+f(c)+f(d)\ge0$. Now by tangent-line trick, we have:
$$f(x)-\frac58x+\frac5{32}=(4x-1)^2(3x+1)\ge0\Rightarrow f(x)\ge\frac58x-\frac5{32}$$
and summing this over $a,b,c,d$ yields:
$$f(a)+f(b)+f(c)+f(d)\ge\frac58(a+b+c+d)-\frac58=0$$
as desired.


Triangle $ AXY$ is similar to triangle $ ZBC$. If $ AX \equal{} 6$ cm, $ ZB \equal{} 18$ cm and $ ZC \equal{} 63$ cm, what is the length of segment $ AY$, in centimeters?

we can see that line segment ZB is similar to line segment AX
since line segment AX is 6 cm, and line segment ZB is 18 cm, then the ratio of the sides of triangle AXY to triangle ZBC is 1:3

therefore, since we know that line segment ZC is 63, line segment AY is one-third of that, or 21

OR

after finding the ratio, we could set up a proportion

$ \frac{1} {3} \equal{} \frac {x} {63}$ where x is length of line segment AY

we cross-multiply
$ 3x\equal{}63$
$ x\equal{}21$


Cards numbered with numbers $ 1$ to $ 1000$ are to be placed on the cells of a $ 1\times 1994$ rectangular board one by one, according to the following rule: If the cell next to the cell containing the card $ n$ is free, then the card $ n \plus{} 1$ must be put on it. Prove that the number of possible arrangements is not more than half a million.

I'm not sure I understand the question; suppose I put card $ 1$ in the 2nd slot, and then put card $ 2$ in the 1st slot. Then the cells next to $ 2$ are not free, so I can put $ 3$ anywhere? But that would imply there are at least $ 2^{500}$ configurations, depending on the ends we pick.


geometry
circumcircle
parallelogram
geometry proposed

[b](14th Iberoamerican 1999)[/b]

O is the circumcenter of the acute-angled triangle ABC. The altitudes are AD, BE and CF. The line EF cuts the circumcircle at P and Q. Show that OA is perpendicular to PQ. If M is the midpoint of BC, show that $AP^2 = 2 \cdot AD \cdot OM$.

$AO\perp EF$ holds because $E,F$ are the projections of $H$ on $AC,AB$ and $O$ is the isogonal conjugate of $H$. As for the second part, $2AD\cdot OM=AD\cdot AH$. On the other hand, since $AO\perp PQ$, it means that the inversion of pole $A$ and power $AP^2$ is the one that turns the circumcircle of $ABC$ into the line $PQ$, and thus $AP^2=AE\cdot AC=AD\cdot AH$, where the last equality holds because the quadrilateral $HDCE$ is cyclic.

As grobber pointed out $ PQ \perp AO$, hence $ AP \equal{} AQ$; consequently $ \angle ABP \equal{} \angle APQ$, that is, $ AP$ is tangent to the circle $ \left( BFP \right)$, consequently $ AP^2 \equal{} AF \cdot AB$ ( 1 ), but from BDHF cyclic we get $ AF \cdot AB \equal{} AH \cdot AD$ ( 2 ). It is well known that $ AH \equal{} 2 \cdot OM \left( 3 \right)$ ( if AA' is a diameter of the circle O, then HBA'C is a parallelogram and M is the midpoint of HA' too) from (1) and (2), with (3) resulting the desired relation.

Best regards,
sunken rock


Find all n such that 2^n-1 is divisble by 3 and this quotient divides m^2+1 for some even m.

Bomb

Have a look here :
http://www.kalva.demon.co.uk/short/soln/sh98n5.html

Pierre.


geometry
3D geometry
tetrahedron
sphere
conics
ellipse
parallelogram

Thanks, dear Orlando. Posted on forums:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297600]Problem 1[/url]. Find the maximal positive number $r$ with the following property: If all altitudes of a tetrahedron are $\geq 1$, then a sphere of radius $r$ fits into the tetrahedron.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297604]Problem 2[/url]. Let $OX, OY$ and $OZ$ be three rays in the space, and $G$ a point "[i]between these rays[/i]" (i. e. in the interior of the part of the space bordered by the angles $Y OZ, ZOX$ and $XOY$). Consider a plane passing through $G$ and meeting the rays $OX, OY$ and $OZ$ in the points $A, B, C$, respectively. There are infinitely many such planes; construct the one which minimizes the volume of the tetrahedron $OABC$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297607]Problem 3[/url]. Is the number
\[\sqrt[3]{\sqrt 5 + 2} + \sqrt[3]{\sqrt 5 - 2}\]
rational or irrational?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297609]Problem 4[/url]. A circle of radius 1 is placed in a corner of a room (i.e., it touches the horizontal floor and two vertical walls perpendicular to each other). Find the locus of the center of the band for all of its possible positions.

[b]Note.[/b] For the solution of this problem, it is useful to know the following Monge theorem: The locus of all points $P$, such that the two tangents from $P$ to the ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are perpendicular to each other, is a circle − a so-called Monge circle − with equation $x^2 + y^2 = a^2 + b^2$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297628]Problem 5[/url]. Given a ball $K$. Find the locus of the vertices $A$ of all parallelograms $ABCD$ such that$ AC \leq BD$, and the diagonal $BD$ lies completely inside the ball $K$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297637]Problem 6[/url]. Let $P_i (x_i, y_i)$ (with $i = 1, 2, 3, 4, 5$) be $5$ points with integer coordinates. Show that among all triangles with vertices at these points, at least three ones have integer areas.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297647]Problem 7[/url]. Given a tetrahedron $ABCD$. Let $x = AB \cdot CD, y = AC \cdot BD$ and $z = AD\cdot BC$. Prove that there exists a triangle with the side lengths $x, y$ and $z$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297655]Problem 8[/url]. Let $a$ be a non-zero real number. For each integer $n$, we define $S_n = a^n + a^{-n}$. Prove that if for some integer $k$, the sums $S_k$ and $S_{k+1}$ are integers, then the sums $S_n$ are integers for all integers $n$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2297661]Problem 9[/url]. Prove that $2^{147} - 1$ is divisible by $343$.

And added to the resources page, see [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=18&year=1973]here[/url].


In my profile, my signature wont let me have a internal link, but i think we are aloud. Any ideas??

You weren't using the proper syntax for the [url ] tag.


Hi, I am self studying Niven's introduction to the theory of numbers. I have tried some exercises but i am not sure if my proofs are correct. Please kidnly verify them for me. Thanks very much.

29.
Consider the set S of integers 1,2,...,n. Let $2^k$ be the integer in S that is the highest power of 2. Prove that $2^k$ is not a divisor of any other integer in S. Hence, prove that $\sum_{j=1}^n \frac{1}{j}$ is not an integer if $n>1$.

[hide="my solution"]
If there is a multiple of $2^k$ in ${1,2,3,...,n}$, it is at least $2(2^k)=2^{k+1}$ which contradicts with "$2^k$ is the integer in S that is the highest power of 2".

Consider $\frac{\frac{n!}{1}+...+\frac{n!}{n}}{n!}$.
Let $2^m||n!$, and $2^k$ be the integer in {1,2,...,n} that is the highest power of 2. Then obviously $2^{m-k}||\frac{n!}{2^k}$.
Since there is no multiple of $2^k$ in {1,2...,n}, we have $2^{m-k+1}|\frac{n!}{i}$, for $i \in \{1,2,...,n\}|\{2^k\}$.
Divide both numerator and denominator by $2^{m-k}$.
Now it is clear that the numerator is odd and the denominator is even.
Thus $\sum \frac{1}{j}$ is not an integer.
[/hide]

30.
Prove that in any block of consecutive positive integers there is a unique integer divisible by a higher poewr of 2 than any of the others. Then use this, or any other method, to prove that there is no integer among the $2^{n+1}$ numbers
\[ \pm \frac{1}{k} \pm \frac{1}{k+1} \pm \frac{1}{k+2} \pm ... \pm \frac{1}{k+n} \]
where all possible combinations of plus and minus signs are allowed, and where n and k are any positive integers.

[hide="my solution"]
Assume there are two integers divisible by the highest power of 2. Let them be $a(2^k)$ and $(a+1)2^k$. Since one of $a$ or $a+1$ is even, we have one of these two numbers divisible by $2^{k+1}$, which contradicts with $2^k$ be the highest power of 2.
Let $S=k(k+1)...(k+n)$, then consider
\[ \frac{\pm \frac{S}{k} \pm ... \pm \frac{S}{k+n}}{S} \]
Let $k+x$ be the unique integer which is divisible by the highest power of among $k$, $k+1$,...,$k+n$. Let $2^m||k+x$ and let $2^y ||S$, then we have $2^{y-m}||\frac{S}{k+x}$.
Same as 29, we have $2^{y-m+1}|\frac{S}{i}$, for $i \in \{k,k+1,...,k+n\}|\{k+x\}$. Divide both numerator and denominator by $2^{y-m}$. Then the numerator is odd and the denominator is even. Therefore $\sum \pm \frac{1}{k+j}$ is not an integer.
[/hide]

31.
Consider the set T of integers $1$,$3$,...,$2n-1$. Let $3^r$ be the integer in T that is the highest power of #. Prove that $3^r$ is not a divisor of any other integer in T. Hence, prove that $\sum_{j=1}^n \frac{1}{2j-1}$ is not an integer if $n > 1$.

[hide="my solution"]
If there is a multiple of $3^r$ in {1,3,...,2n-1}, it is at least $3(3^r)=3^{r+1}$, which contradicts with "$3^r$ is the integer in T taht is the highest power of 3".
Consider
\[ \frac{\frac{(2j-1)!!}{1}+...+\frac{(2j-1)!!}{2j-1}}{(2j-1)!!} \]
Let $3^k || (2j-1)!!$, and $3^r$ be the integer in {1,3,...,2j-1} with the highest power of 3, then $3^{k-r}||\frac{(2j-1)!!}{3^r}$.
Since there is no multiple of $3^r$ in {1,3,...,2n-1}, we have $3^{k-r+1}||\frac{(2j-1)!!}{i}$ for $i \in \{1,3,...,2n-1\} | \{3^r\}$.
Divide both numerator and denominator by $3^{k-r}$, then the numerator is not divisible by 3, while the denominator is. So $\sum \frac{1}{2j-1}$ is not an integer.
[/hide]

32.
Prove that
\[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}=\frac{1}{1001}+...+\frac{1}{2000} \]

[hide="my solution"]
\[
\begin{gathered}
  1 - \frac{1}
{2} + \frac{1}
{3} - \frac{1}
{4} + ... + \frac{1}
{{1999}} - \frac{1}
{{2000}} \hfill \\
   = 1 + \frac{1}
{2} + ... + \frac{1}
{{2000}} - 2\left( {\frac{1}
{2} + \frac{1}
{4} + ... + \frac{1}
{{2000}}} \right) \hfill \\
   = \frac{1}
{{1001}} + ... + \frac{1}
{{2000}} \hfill \\
\end{gathered}
\]
[/hide]

33.
Say that a positive integer n is a sum of consecutive integers if there exist positive integers m and k such that $n=m+(m+1)+...+(m+k)$. Prove that n is so expressible if and only if it is not a power of 2.

[hide="my solution"]
$n=m+...+(m+k)=\frac{(2m+k)(k+1)}{2}$.
If k is odd, then 2m+k is odd, so n can't be a power of 2.
If k is even, then k+1 is odd, so n can't be a power of 2.
If n is not a power of 2, let $n=ab$ where a is odd and b is even.
Take $k=a-1$, $m=\frac{2b+1-a}{2}$. It is easy to verify that they satisfy the condition.
So when n is not a power of 2, it is expressible as a sum of consecutive integers.
[/hide]

34.
Prove that an odd integer $n>1$ is a prime if and only if it is not expressible as a sum of three or more consecutive positive integers.

[hide="my solution"]
Assertion: If n is expressible as sum of 3 or more consecutive positive integers, then it is not a prime.
$n=m+...+(m+k)=\frac{(2m+k)(k+1)}{2}$.
When $k \ge 2$, $2m+k$, $k+1 >2$, so n is not a prime.
Thus fi n is a prime, it must not be a sum of 3 or more consecutive integers.
[/hide]

37.
If $a$ and $b>2$ are any positive integers, prove that $2^a+1$ is not divisible by $2^b-1$.

[hide="my solution"]
If $a \le b$, it is easy to check that that there is no solution.
If $a>b$, let $a=bq+r$ with $0 \le r \le b$.
$2^b-1|2^a+1$ implies $2^b-1|(2^a+1+2^b-1)=2^a+2^b=2^b(2^{a-b}+1)$.
Since ($2^{b-1}$, $2^b$)=1, we have $2^b-1|2^{a-b}+1$
Repeat this process we eventually arrive at $2^b-1|2^r+1$.
This means $2^b-1 \le 2^r+1$.
$2^{r+1} \le 2^b \le 2^r+2$.
which is impossible unless r=0 or 1.
It is easy to check that r is impossible to be 0 or 1.
[/hide]

38.
Let positive integers g and l be given with $g|l$. Prove that the number of pairs of positive integers x,y satisfying (x,y)=g and [x,y]=l is $2^k$, where k is the number of distinct prime factors of $\frac{l}{g}$.

[hide="my solution"]
We have $(\frac{x}{g}, \frac{y}{g})=1$ and $[\frac{x}{g}, \frac{y}{g}]=\frac{l}{g}$
Since $(\frac{x}{g},\frac{y}{g})=1$, $[\frac{x}{g}, \frac{y}{g}]=(\frac{x}{g})(\frac{y}{g})=\frac{l}{g}$
Now, let k be the number of prime factors of $\frac{l}{g}$. It is obvious that x/g and y/g must have distinct prime factors. So for each prime factor x/g "chose", y/g cannot choose them anymore. This gives the number of solutions equal
$C_0^k+...+C_k^k=2^k$.
[/hide]

40.
Prove that $2+\sqrt{-6}$ and $2-\sqrt{-6}$ are primes in the class C of numbers $a+b\sqrt{-6}$.

[hide="my solution"]
Let $(a+b\sqrt{-6})(c+d\sqrt{-6})=2+\sqrt{-6}$
Then $N(a+b\sqrt{-6})N(c+d\sqrt{-6})=N(2+\sqrt{-6})=10$
$N(a+b\sqrt{-6})=2$ or 5 which is not of the form $a^2+6b^2$. So such a,b can't be found and $2+\sqrt{-6}$ is a prime. Same for its conjugate.
[/hide]

44.
Prove that there are no positive integers a,b,n > 1 such that $(a^n-b^n)|(a^n+b^n)$.

[hide="my solution"]
WLOG assume (a,b)=1.
If $\frac{a^n+b^n}{a^n-b^n}$ is an integer, then $\frac{2b^n}{a^n-b^n}$ is also an integer.
As ($b^n$, $a^n-b^n$)=($b^n$,$a^n$)=1, so $a^n-b^n|2$.
When $n \ge 2$, $a^n-b^n > 2$, so there is no solution.
[/hide]

46. Prove that there are infinitely many primes by considering the Fermat numbers.

[hide="my solution"]
Lemma: Any two Fermat numbers are coprime.
Proof: assume m>n, then it is easy to show that $2^{2^n}+1|2^{2^m}-1$, So $(2^{2^m}+1,2^{2^n}+1) \ge (2^{2^m}+1, 2^{2^m}-1) =1$. So they are coprime.

Now, all fermat numbers have different prime factors. As there are infinitely many Fermat numbers, there are infinitely many primes.
[/hide]

49. If g is a divisor of each of ab, cd, and ac+bd, prove that it is also a divisor of ac and bd, where a,b,c,d are integers.
[hide="my solution"]
$g|ac+bd$ implies $cg|ac^2+bcd$. Since $g|cd$, we have $cg|ac^2$ which implies $g|ac$.
Same for bd.
[/hide]

for 49, it looks like you are using cg | bcd, which i dont think is true.

Thanks for pointing out my mistake. Then is this okay?

$g^2|(ac+bd)^2=ac^2+bd^2+2abcd$
Since $g^2|abcd$, $g^2|(ac-bd)^2$ that implies $g|ac-bd$
Combine this with $g|ac+bd$ we have the desired result.


[quote="maggot"]Let $P\in Z[x]$, $k\in N$. Show that there exists $n$ such that $k$ divides $P(1)+P(2)+...+P(n)$[/quote]

$n=k$ works.

Let $P(x)=\sum{a_{i}x^{i}}$
$\sum_{j=1}^{k}{P(j)}=\sum_{j=1}^{k}{\sum{a_{i}j^{i}}}=\sum{\left(a_{i}\sum_{j=1}^{k}{j^{i}\right)}\equiv 0\,\,mod\,k}$

I think $n=k$ doesn't work. For example $P(x)=x^{2}, k=3$.
Anyway $n=k^{2}$ will work just fine. Using $P(x+k)=P(x) (mod k)$, we get $P(i)+P(i+k)+P(i+2k)+...+P(i+k(k-1))=0 (mod k)$ for any $i\in Z$. Just sum up this equations for $i=1,...,k$ and we get what we want.

Posted before, and Number Theory by the way.

[quote="maggot"]I don't think $k$ always divide $\frac{k(k+1)}{2}$  :wink:[/quote]

Oh yes I forgot :wink:


Let $ ABC$ be a triangle with integer side lengths, and let its incircle touch $ BC$ at $ D$ and $ AC$ at $ E$. If $ |AD^2\minus{}BE^2| \le 2$, show that $ AC\equal{}BC$.

    Let $CB = a$, $CA = b$ and $AB = c$, and WLOG assume $b \geq a$.
    By the cosine rule,
    \begin{align*}
        AD^2 &= CD^2 + b^2 - 2b\cdot CD\cdot \cos C\\
        BC^2 &= CE^2 + a^2 - 2a\cdot CD\cdot \cos C
    \end{align*}
    Thus as $CD = CE$,
    $$
AD^2 - BC^2 = b^2 - a^2 + 2a\cdot CD\cdot \cos C - 2b\cdot CD\cdot \cos C = (b - a)(b + a -2\cdot CD \cdot \cos C).
    $$
    Using $CD = \frac{1}{2}(a + b - c)$,
    $$
    AD^2 - BC^2 = (b - a)(b + a - (a + b - c) \cos C).
    $$
    We have $\cos C$ bounded as $-1 \leq \cos C \leq 1$, and as $a + b - c > 0$ by the triangle inequality,
    $$
    (b - a)c \leq AD^2 - BC^2  \leq (b - a)(2b + 2a - c)\leq 2.
    $$
    If $b = a$, then this inequality holds.
    Now assume that $b > a$. We wish to satisfy
    $$
    (b - a)c \leq 2.
    $$
    However, as the sidelengths of $ABC$ are integers, we must have $b = 2$, $a = 1$ and $c = 1$ or $c = 2$.

    If $c = 1$, the triangle inequality $a + c > b \iff 2 > 2$ is not satisfied, so $ABC$ is not a triangle. Similarly, if $c = 2$, then we have equality of $(b - a)c = 2$, which implies that $\cos C = 1$ and $C = 0$, so $ABC$ is also not a triangle.
    Thus, if a triangle with integer sidelengths satisfies $\left| A D^{2}-B E^{2}\right| \leq 2,$ then $AC = BC$ as required.


choose 6 people out of 12 to sit into two columns of 3 seats each. if 3 of them only like to sit at the left hand side and 4 only like to sit at right and other 5 people willing to sit anywhere, how many ways of arrangements?

[hide]  Split this into 3 cases.  One case is where the choosey people get what they want.  The other case is where the seats on one side are taken by the non-choosers, but the other side is not not taken.  The last case is where the seats on both sides are taken.
  The first case has 3*2*5*4*4*3=1440 ways of happening.
  The second case has two subcases, the seats on the right being taken or the seats on the left being taken.  Each of those has two subacases, either one or two seats are taken.
  If 1 seat on the right was taken, there's 4*5*4*3*3*2=1440 ways of having a given seat on the right side being taken.  However, since there's 2 seats, there's 2880 ways of having 1 seat taken on the right side.
The chances of having 2 seats on the right taken is 3*2*5*4*3*2=720 ways.
  If 1 seat is taken from the left, there's 2*3*5*4*3*4*3=4320 possibilities.
  If 2 seats are taken from the left, there's 5*4*3*2*4*3=1440 possibilites.
  Therefore, the total possibilites for case 2 is 9360 ways.
  The third case has 2 subcases.  1 seat on both sides could be taken, or 1 seat on one side and two seats on another side is taken.  Actually, the second subcase is further divided into 2 subcases because it makes a difference whether the left or right side is fully taken.  No more subcases may exist because if the non-choosers filled up all the side seats, there's not enough of them left to fill up the middle seats.
  1 on each side: 5*4*3*2*3*4=1440 ways, but there's 4 ways to arrange the taking of one slot on each side, so that's 5760 ways.
  1 on one side and 2 on the other right side fully taken: 5*4*3*2*1*3=360 but there's two ways to arrange the side where only one seat is taken so that's 720 ways.
  1 on one side and 2 on the other left side fully taken: 5*4*3*2*1*4=480 but again, there's two ways to arrange the side where only one seat is taken so that's 960 ways.
  The third case has a total of 7440 ways.
  Adding it all up there's a total of 7440+9360+1440=18240 ways.[/hide]

im not too sure about your answer

because there is far more than just 3 cases.

so let there be 3L 5W and 4R ( where L is the people who want to sit left, W is where people sit wherever and R is where people sit right)

u can have 3L 3W, 3L 2W 1R, 3L 1W 3R, 3L 0W 3R..

and so on..

Oops, 2 columns of 3 and not 2 rows of 3.  I did 2 rows of 3.
[hide]OK, using sen's notation, 3L 3R has 3*2*1*4*3*2=144 ways.
3L 1W 2R has 3*2*1*5*4*3=360 ways if 1W 2R was the only arrangement, but 1R 1W 1R or 2R 1W exist as well, so that's 1080 ways.
3L 2W 1R has 3*2*1*5*4*4=480 ways, but there's ${3\choose 2}$ or 3 ways to arrange this so that's 1440 ways.
3L 3W has 3*2*1*5*4*3=360 ways.
Therefore, 3L has 2784 ways.
2L 3R 1W has 3*2*5*4*3*2=720 ways if 2L 3R 1W was the exact arrangement, but there's ${3\choose 1}$ arrangements, so there's 2160 arrangements.
2L 2W 2R has 3*2*5*4*4*3=1440 ways if 2L 2W and 2R was the exact arrangement, but there's ${3\choose 1}^2$ or 9 arrangements, so there's 12960 ways.
2L 3W 1R has 3*2*5*4*3*4=1440 ways if 2L 3W and 1R was the exact arrangement, but again there's ${3\choose 1}{3\choose 2}$ or 9 arrangements, so there's 12960 ways.
2L 4W has 3*2*5*4*3*2=720 ways if 2L 4W was the exact arrangement, but there's ${3\choose 1}$ or 3  arrangements, so there's 2160 ways.
2L has 30240 ways.
1L 3R 2W has 3*5*4*4*3*2=1440 ways if 1L 3R 2W was the exact arrangement, but there's ${3\choose 2}$ or 3 arrangements, so there's 4320 ways.
1L 2R 3W has 3*5*4*3*4*3=2160 ways if 1L 2R 3W was the exact arrangement, but there's ${3\choose 2}{3\choose 1}$ or 9 arrangements, so there's 19440 ways.
1L 1R 4W has 3*5*4*3*2*4=1440 ways if 1L 1R 4W was the exact arrangement, but there's ${3\choose 2}{3\choose 2}$ or 9 arrangements, so there's 12960 ways.
1L 5W has 3*5*4*3*2*1=360 ways if 1L 5W was the exact arrangement, but there's ${3\choose 2}$ or 3 arrangements, so there's 1080 ways.
1L has 37800 ways.
3W 3R has 5*4*3*4*3*2=1440 ways.
4W 2R has 5*4*3*2*4*3=1440 ways if 4W 2R was the exact arrangement, but there's ${3\choose 1}$ or 3 arrangements, so there's 4320 ways.
5W 1R has 5*4*3*2*1*4=480 ways if 5W 1R was the exact arrangement, but there's ${3\choose 2}$ or 3 arrangements, so there's 1440 ways.
0L has 7200 ways.
In total, there's 78024 ways.[/hide]
Is there a faster method?


Prove that:
$ \forall x \in [\minus{}1,1]: \,|ax^2\plus{}bx\plus{}c|\leq 1 \Rightarrow |cx^2\plus{}bx\plus{}a|\leq 1$

Seems wrong: $ (a,b,c) \equal{} (2,0,\minus{}1)$, and then $ x \equal{} 0$ in the conclusion. Maybe all of the coefficients are supposed to be nonnegative?


A person who was alive in 1900 died in 1950. The person's age at death was $\frac{1}{29}$ of the year of birth. What was the age of the person in 1910?

[hide]We can set up a system of equations:
$1950-x=y$ and $29x=y$
By substitution,
$29x=1950x$
$1950=30x$
$x=59$

Since the guy was 59 years old when he died, he was 19 in 1910.
$\boxed{19}$[/hide]

[quote="#H34N1"]I did this:
[hide]
Let $b$ be the year he was born.

Then, $(1950-b)\times 29=b$
$b=1885$

Therefore, he was $25$ years old.

What's wrong with mine?[/hide][/quote]
Your solution is correct.  laughinghead505, you messed up when you said that x=59.  x Should actually be 65.


function
algebra
functional equation
rational function

Sorry for this random comment guys, I've been trying this useless functional equation problem for half and hour, just to realize that there is no solution!

Find f: R-> R if f(x)f(y) - f(x+y) = x+y for all real x,y

Personally I don't think it deserves to be on the algebra section, if you're bored, try it.. otherwise, leave it alone, cuz it really is pointless..

[quote="jpark"]Just realized... We're writing Fermat 2 days before my birthday![/quote]

Hmm. I'm gonna be writing an SAT on my birthday  ;)

Aww not fair...I want a contest on my birthday too!!!!

BUWHAHAH!  Finals are over for me, I can breathe and do math for a week.  Yay!

[quote="JBL"]I had a 10 page paper due on my birthday this semester ...[/quote]Somehow this doesn't sound too appealing :D

No, I have to admit, it was not my most exciting brush with the academic world.  On the other hand, one year my brother had braces put on on his birthday, so I think I got the better deal of the two of us.

Ouch.. It seems like I'm not the only one..
Last year, I had to write CCC the day before my birthday..

By the way, this really is random..

[quote="Valentin Vornicu"]I wonder if there's anyone who took IMO on his birthday :D[/quote]
The famous example mentioned in other threads is Terry Tao who
turned 11,12 and 13.  I don't know if any of the Day 1 IMO sessions coincided
with the birthday but it was within a day or so in some of those years.

[quote="jpark"] useless functional equation problem ... [has] no solution!

Find f: R-> R if f(x)f(y) - f(x+y) = x+y for all real x,y [/quote]

There are very few functional equation (competition) problems that have
solutions that are both nontrivial (meaning the function is not constant or linear
or something similar) and unexpected (meaning that you can't immediately find
all solutions by, say, solving for the constant solutions). Usually the problem is
to show that the obvious or routinely derivable solutions are the only ones.

Exceptions to this are rare.  Does anybody remember any?

regardless of what you mean by "non-trivial" (hard to be guessed?) and unexpected, i can construct radom functional equations by hand...

did you look at the link i've provided above?  there are some good examples in there.

this one has an sort of "unexpected" and "non-trivial" solution
[quote]
Determine all continuous real functions f of a real variable for which
f(x + 2f(y)) = f(x) + y + f(y)

for all real x and y.
[/quote]

and this one is just bizarre, but it is my favourite of all time

[quote]
Suppose that f is a real-valued function defined on the reals for which

f(xy) + f(y - x) >= f(y + x)

for all real x and y.
i) Prove that f(x) >= 0 for all real x.
ii) determine f(x)
[/quote]
if you can guess the solution to the second problem, I salute you

[quote="bugzpodder"]construct random functional equations by hand... [/quote]
But will they have interesting functions as solutions?

It's also not easy to find equations satisfied by a given function.  It's not clear to me
how to do this for a given rational function of degree > 1 (other than trivialities like multiplying by denominator and taking N'th differences).

[quote] did you look at the link i've provided above? [/quote]
It has only the solutions, not the problems.
Thanks for the two that you posted.

There was a very bizarre functional equation from the mock olympiad at the winter camp..
Except I can't seem to find the problem at the moment..

If anyone has it, could you post it? I think it's either F1 or F2..

[quote="bugzpodder"][quote="jpark"]
Find f: R-> R if f(x)f(y) - f(x+y) = x+y for all real x,y
[/quote]

Ralph thought it was strange also:

http://www.student.math.uwaterloo.ca/~rfurmani/math/Functional_Equations.dir/Functional_Equations.pdf

I think i found two typos in this pdf, but i forgot what they were.  if you guys spot it, leave a msg here
(good 'ole' hexagon doesnt count)[/quote]

Those of you wondering what problems these are, go to http://www.nevada.edu/~baragar/functional.pdf .

Hehe, people at my school suck at playing the game invented by our honourable leader person, etc, etc, etc...

They play a modified version where you're allowed to talk and discuss the rules, and I crushed them with have a nice day rule.. *feeling evil*   :P


geometry
circumcircle
function
national olympiad

plase send us Problems of Asian Pasific Math olympiad 2008?

Algebrageometry!
APMO 2008 hasn't hold yet.
It will be held 10-11th march of this year!

The directions specifically say that the problems cannot be posted on the internet until they are up on the official website.

I think the website is:http://www.kms.or.kr/competitions/apmo/

Solutions posted!

[url]http://www.kms.or.kr/competitions/apmo/data/08APMO-SOL.pdf[/url]

Results pending...

I was told on that day after APMO that the results would be released in late April. I do hope it's not that late.

Australia is quick to get their results out:

http://www.amt.canberra.edu.au/apmo2008.html

I've posted the problems here.

$ 1.$ Let $ ABC$ be a triangle with $ \angle A < 60^\circ$. Let $ X$ and $ Y$ be the points on the sides $ AB$ and $ AC$, respectively, such that $ CA \plus{} AX \equal{} CB \plus{} BX$ and $ BA \plus{} AY \equal{} BC \plus{} CY$ . Let $ P$ be the point in the plane such that the lines $ PX$ and $ PY$ are perpendicular to $ AB$ and $ AC$, respectively. Prove that $ \angle BPC < 120^\circ$.

http://www.mathlinks.ro/viewtopic.php?t=195490

$ 2.$ Students in a class form groups each of which contains exactly three members such that any two distinct groups have at most one member in common. Prove that, when the class size is $ 46$, there is a set of $ 10$ students in which no group is properly contained.

http://www.mathlinks.ro/viewtopic.php?t=195492

$ 3.$ Let $ \Gamma$ be the circumcircle of a triangle $ ABC$. A circle passing through points $ A$ and $ C$ meets the sides $ BC$ and $ BA$ at $ D$ and $ E$, respectively. The lines $ AD$ and $ CE$ meet $ \Gamma$ again at $ G$ and $ H$, respectively. The tangent lines of $ \Gamma$ at $ A$ and $ C$ meet the line $ DE$ at $ L$ and $ M$, respectively. Prove that the lines $ LH$ and $ MG$ meet at $ \Gamma$.

http://www.mathlinks.ro/viewtopic.php?t=195491

$ 4.$ Consider the function $ f: \mathbb{N}_0\to\mathbb{N}_0$, where $ \mathbb{N}_0$ is the set of all non-negative
integers, defined by the following conditions :

    $ (i)$ $ f(0) \equal{} 0$; $ (ii)$ $ f(2n) \equal{} 2f(n)$ and $ (iii)$ $ f(2n \plus{} 1) \equal{} n \plus{} 2f(n)$ for all $ n\geq 0$.

$ (a)$ Determine the three sets $ L \equal{} \{ n | f(n) < f(n \plus{} 1) \}$, $ E \equal{} \{n | f(n) \equal{} f(n \plus{} 1) \}$, and $ G \equal{} \{n | f(n) > f(n \plus{} 1) \}$.
$ (b)$ For each $ k \geq 0$, find a formula for $ a_k \equal{} \max\{f(n) : 0 \leq n \leq 2^k\}$ in terms of $ k$.

http://www.mathlinks.ro/viewtopic.php?t=195495

$ 5.$ Let $ a, b, c$ be integers satisfying $ 0 < a < c \minus{} 1$ and $ 1 < b < c$. For each $ k$, $ 0\leq k \leq a$, Let $ r_k,0 \leq r_k < c$
be the remainder of $ kb$ when divided by $ c$. Prove that the two sets $ \{r_0, r_1, r_2, \cdots , r_a\}$ and $ \{0, 1, 2, \cdots , a\}$ are different.

http://www.mathlinks.ro/viewtopic.php?t=195496

[quote="Nemion"]How do you thing will the cut offs be?

Broze:?
Silver:?
Gold:?[/quote]

I think
Gold: 20
Silver: 15
Bronze: 10

the problems are as hard as last year's.

The results are online now. [url]http://www.kms.or.kr/Competitions/apmo/data/20th%20APMO.htm[/url]

Nemion: APMO gold!

Congratulations to all!!


[quote="kunny"]Maybe your problem is $\sqrt{x^{2}+2x+1}+\sqrt{x^{2}-2x+1}$, isn't it?[/quote]
no
is's
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/4/1/8/418c3505c40c0cc57dee21ed7b121fcd088e7400.gif[/img]

Sorry,we can't understand what your problem is. :(

[quote="bbbbbbbbbb"]define the value of expression
$P=\sqrt{x^{2}+x+1}+\sqrt{x^{2}-x+1}$  :roll:[/quote]
What is your question? :huh:

Maybe the problem is that find the minimum value of $P$.

If so, let $A\left(-\frac{1}{2},\ \frac{\sqrt{3}}{2}\right),\ B\left(\frac{1}{2},\ \frac{\sqrt{3}}{2}\right)$ and $Q(x,\ 0)$, for $A'\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$,

we have  $P=\overline{QA}+\overline{QB}\geq \overline{A'B}=2$, the equality holds $x=0.$


geometry
circumcircle
trigonometry
parallelogram
angle bisector
geometry proposed

Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$.
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$.
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.

[quote="Nima Ahmadi Pour"]Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$.
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,BC$ at $M,N,L$.
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.[/quote]

I think you meant that it touches $AB$ at $L$...

Anyway, let $BI_a$ meet the circumcircle of $\triangle{ABC}$ at $K$. Then $K$ is the midpoint of the arc $\widehat{ABC}$ , thus $OK\perp{AC}$. Since $I_aN\perp{AC}$ and $OK=I_aN$ we conclude that $KI_a||ON$, but $LM\perp{KI_a}$ $\Longleftrightarrow$ $LM\perp{ON}$.

Using similar arguments we obtain the conclusion.

Sorry, I just made it correct.
Anyway, thanks for the notification.

I thought of
$R=r_a \rightarrow R=p tg \frac{A}{2} \rightarrow |OM|^2-|ON|^2=|LM|^2-|LN|^2$ ...

[b]An interesting remark.[/b] Denote the projection $D$ of the vertex $A$ on the opposite side $[BC]$. Then $\boxed {R=r_a\Longleftrightarrow D\in OI}\ .$

this problem is way too easy for IRAN tst.
a solution very similar to sailor's but even simpler:
M on BC, N on AC, L on AB. Suffices to prove OL perpendicular to MN (because then the similar argument that OM perpendicular to LN finishes off the problem neatly). MN is perpendicular to IC with I = excentre of A-excircle. So it suffices to prove a stronger result, i.e. ICOL is an iscoceles trapezium. OC = IL (since R=r_a), so it suffices to prove <OCI=<LIC. Simple angle chasing yields if <A = 2a,etc; then <OCI=<OCB+<ICB=90-2a+90-c=2b+c; <LIC=<LIB+<BIC = b + (b+c) = 2b+c, so we are done.

hey virgil nicula, do you have a good proof of the 'interesting remark' about D,I,O collinear when R= r_a? i cant do it without side-bashing or trig-bashing.

I have a solution:
  Let X be the excentre opposite A and D be the point (#A) where AX intersect (ABC). Because OD // MX ( perpendicular BC),
OD=MX so ODXM is a parallelogram,AX perpendicular LN so OM perpen dicular LN.
 Let the internal angle bisector of BAC intersect (ABC) at E, EBX = 90, EA = EC. Let T the point of intersection of BX and(ABC)  so TA = TC and TE passes O.Let B' the foot of the perpendicular on AC from T. TB' passes O
 Because OT // NX and OT= NX so TONX is a parallelogram so TX // ON hence ON perpendicular NM
Hence O is the ortheocentre of MNL

Let $I_A$ be the $A$-excenter ; lines $I_AA,I_AB,I_AC$ meet $\odot(ABC)$ again at points $A',B',C'$, respectively. It isn't hard to note that:
[list]
[*] $I_A$ is the orthocenter of $\triangle A'B'C'$.
[*] corresponding sides of $\triangle A'B'C'$ and $\triangle MNL$ are parallel to each other and the two triangles have opposite orientations.
[/list]
Using the fact that $\triangle A'B'C'$ and $\triangle MNL$ have equal circumradius, we obtain there is a point $S$ such that reflection in $S$ (which we denote by $\mathcal R$) swaps $\{A',M\},\{B',N\},\{C',L\}$. Then as $O,I_A$ are corresponding cirucmenters of $\triangle A'B'C,\triangle MNL$, so $\mathcal R$ swaps  $\{O,I_A\}$. Since $I_A$ was the orthocenter of $\triangle A'B'C'$, hence $O$ is the orthocenter of $\triangle MNL$. $\blacksquare$


[quote="math888"]1. What is the sum of all the possible 4 digit numbers using the digits 1, 2, 3, 4.
(I just want an easy way to do it without calculating tons of stuff)[/quote]
Note that each number is in each place 6 times. So, our sum is

$ 6(1111\plus{}2222\plus{}3333\plus{}4444)\equal{}6666(1\plus{}2\plus{}3\plus{}4)\equal{}66660$.

math888, can each digit only be used once in a number, or as many times as you want?


calculus integration geometry trigonometry function LaTeX limit

Using calculus techniques, is there a proof for the area under the curve of si(x) (or sinx / x) between the bounds zero and infinite. (the answer happens to equal pi / 2).

Well the integral $ \int\frac{\sin x}{x}dx$ cant be evaluated as introductory functions but the integral $ \int^{\infty}_{0}\frac{\sin x}{x}dx$ can be evaluated(For example we cant evaluate $ \int\exp\left(\minus{}x^{2}\right)dx$ but $ \int^{\infty}_{0}\exp\left(\minus{}x^{2}\right)dx \equal{}\frac{\sqrt{\pi}}{2}$), because its [hide="convergent"]for the convergence of $ \int^{\infty}_{0}\frac{\sin x}{x}dx$ we'll prove $ \int^{\infty}_{\frac{\pi}{2}}\frac{\sin x}{x}dx$ is convergent,
we have:
$ \int^{\infty}_{\frac{\pi}{2}}\frac{\sin x}{x}dx \equal{}\left[\minus{}\frac{\cos x}{x}\right]^{\infty}_{\frac{\pi}{2}}\minus{}\int^{\infty}_{\frac{\pi}{2}}\frac{\cos x}{x^{2}}dx \equal{}\minus{}\int^{\infty}_{\frac{\pi}{2}}\frac{\cos x}{x^{2}}dx$

but the RHS integral is convergent because $ \int^{\infty}_{\frac{\pi}{2}}\left|\frac{\cos x}{x^{2}}\right|dx\leq\int^{\infty}_{\frac{\pi}{2}}\frac{1}{x^{2}}dx$[/hide]

So what is the proof for the area under si(x)= pi/2 for  0<x<infinite

And i believe  you meant exp(-x^2) when you were giving that example (sorry for being pedantic, im a physics student as well you see).

Are you asking how to calculate $ \int_{0}^{\infty}\frac{\sin x}{x}\; dx$? Then this would be an answer:

[Preliminary]
$ 1.\;\;\int_{0}^{\infty}e^{-sx}\; dx =\frac{1}{s}$
$ 2.\;\;\int_{0}^{\infty}\sin x e^{-sx}\; dx =\frac{1}{1+s^{2}}$

If you are familiar with laplace transform, then you would know these two equations. Also,

$ 3.\;\;\int_{0}^{\infty}\frac{1}{1+x^{2}}\; dx =\frac{\pi}{2}$

Now let's begin the proof.

\begin{eqnarray*}\int_{0}^{\infty}\frac{\sin x}{x}\; dx & = &\int_{0}^{\infty}\sin x\left(\int_{0}^{\infty}e^{-tx}\; dt\right)\; dx\\ & = &\int_{0}^{\infty}\int_{0}^{\infty}\sin x e^{-tx}\; dt\; dx\\ & = &\int_{0}^{\infty}\int_{0}^{\infty}\sin x e^{-tx}\; dx\; dt\quad\quad\text{(change the order of integration)}\\ & = &\int_{0}^{\infty}\frac{1}{1+t^{2}}\; dt\\ & = &\frac{\pi}{2}\end{eqnarray*}

p.s. Why you don't use latex?

That's not a justified interchange- the integral's not absolutely convergent.
There are ways to get around this, including integrating by parts.

I'm sorry... I knew that this is not a justified method, but I thought it is 'intuitively simple'. I think this solution would be more rigorous than previous one:

[hide="Solution 2"]Let $ I(s) =\mathscr{L}_{t}\left\{\frac{\sin t}{t}\right\}(s) =\int_{0}^{\infty}\frac{\sin t}{t}e^{-st}\; dt$.

$ I$ is contiouous on $ [0,\infty)$ and is differentiable on $ (0,\infty)$ with $ I'(s) =-\int_{0}^{\infty}\sin t e^{-st}\; dt$.

Computing $ I'(s)$ gives $ -\frac{1}{1+s^{2}}$. This means that $ I(s) = C-\arctan s$ for some $ C$.

To evaluate $ C$, observe $ C-\frac{\pi}{2}=\lim_{s\to\infty}I(s) = 0$. Then we have $ C =\frac{\pi}{2}$.

Therefore, $ I(s) =\frac{\pi}{2}-\arctan s$ and $ I(0) =\frac{\pi}{2}$.[/hide]

No, that's essentially the same. To justify bringing the integral inside, you need something like Fubini's theorem.

The integration by parts trick:

$ \int_{0}^{\infty}\frac{\sin x}{x}\,dx \equal{}\left[\frac{1\minus{}\cos x}{x}\right]_{0}^{\infty}\plus{}\int_{0}^{\infty}\frac{1\minus{}\cos x}{x^{2}}\,dx \equal{}\int_{0}^{\infty}\frac{1\minus{}\cos x}{x^{2}}\,dx$ (In the first term, the limit at each endpoint is zero).

Now $ \frac{1}{x^{2}}\equal{}\int_{0}^{\infty}te^{\minus{}tx}\,dt$, so
$ \int_{0}^{\infty}\frac{1\minus{}\cos x}{x^{2}}\,dx \equal{}\int_{0}^{\infty}\int_{0}^{\infty}(1\minus{}\cos x)\cdot te^{\minus{}tx}\,dt\,dx$. Everything's positive, so we can reverse the order (Tonelli's theorem):
$ \int_{0}^{\infty}\int_{0}^{\infty}te^{\minus{}tx}\minus{}t\cos xe^{\minus{}tx}\,dx\,dt \equal{}\int_{0}^{\infty}\left[\minus{}e^{\minus{}tx}\minus{}\left(\frac{t}{1\plus{}t^{2}}\sin x e^{\minus{}tx}\minus{}\frac{t^{2}}{1\plus{}t^{2}}\cos x e^{\minus{}tx}\right)\right]_{x \equal{} 0}^{x \equal{}\infty}\,dt$
$ \equal{}\int_{0}^{\infty}\frac{1}{1\plus{}t^{2}}\,dt \equal{}\frac{\pi}{2}$.

Can i have a proof for the preliminary parts number 2, also how did jmerry go from tonellis theorem to his last line of working. Other then that thanks guys, ive learnt a lot.

jmerry, I can't understand why my second solution is essentially same as the first one. Once we establish that $ I(s)$ has the derivative $ \minus{}\int_{0}^{\infty}\sin t\ e^{\minus{}st}\; dt$ and $ \lim_{s\to\infty}I(s) \equal{} 0$, then we don't need to evaluate a double integral at all. We only have to determine $ I(s)$ using the contition $ I'(s) \equal{}\minus{}\frac{1}{1\plus{}s^{2}}$ and $ I(\infty) \equal{} 0$.

Xile, you may use integration by parts (or complex functions, or else) to obtain the general formula $ \int\sin ax\ e^{bx}\ dx\; \equal{}\;\frac{e^{bx}(b\sin ax\minus{}a\cos ax)}{a^{2}\plus{}b^{2}}$ where $ a^{2}\plus{}b^{2}> 0$.
Then just evaluate $ \int_{0}^{x}\sin t\ e^{\minus{}st}\ dt$ and take limit $ x\to\infty$.

jmerry, i think you have a small mistake on your second last line of working, the curved brackets need a negative out the front. Cheers sos for that proof.

[quote="Xile"]jmerry, i think you have a small mistake...[/quote]
Fixed, along with another typo.

[quote="sos440"]Once we establish that I(s) has the derivative...[/quote]
That's the part you need something like Fubini for. Bringing the derivative inside the integral isn't automatic.