trigonometry
limit
real analysis
real analysis solved

Well, you mean
\[\sum_{n\geq 0}3^n\sin^3\left(x\over 3^n \right),\]
right?

OK, use complex numbers. It is known that for real numbers $x$
\[e^{ix} = \cos x + i\sin x\]
so
\[e^{-ix} = \cos(-x) + i\sin(-x) = \cos x - i\sin x\]

Subtracting these two equations yields
\[\sin x = {e^{ix} - e^{-ix}\over 2i}\]

Thus
\[\sin^3 x = \left(e^{ix} - e^{-ix}\over 2i\right)^3 = -{e^{3ix} - e^{-3ix} - 3(e^{ix} - e^{-ix})\over 8i}\]
or
\[\sin^3 x = {1\over 4}(3\sin x - \sin(3x))\]

Well, one could derive this by simple trigonometric means (using the formula for $\sin(3x)$, fo instance), but I prefer doing calculations this way because you don't need to memorize trigonometric formulae. You only need to do the calculations.

Then the sum is equal to
\[\sum_{n\geq 0}3^n{1\over 4}\left(3\sin\left(x\over 3^n \right)-\sin\left(3x\over 3^n \right)\right) = {1\over 4}\left(\sum_{n\geq 0}3^{n+1}\sin\left(x\over 3^n \right) - \sum_{n\geq 0}3^n\sin\left(x\over 3^{n-1}\right)\right) = {1\over 4}\left(\sum_{n\geq 0}3^{n+1}\sin\left(x\over 3^n \right) - \sum_{n\geq 1}3^{n+1}\sin\left(x\over 3^n\right)\right) = 3\sin x\]

I hope my calculations are OK.

Hmm...

Actually, answer is $\frac{3x-\sin 3x}{4}$.  :D

and what is the problem with cyshine's solution?

I think it is here, should consider partial sum and you will see

${1\over 4}\left(\sum_{n\geq 0}3^{n+1}\sin\left(x\over 3^n \right) - \sum_{n\geq 1}3^{n+1}\sin\left(x\over 3^n\right)\right) = 3\sin x$

For the benefit of those who have been reading this and are still confused, I'll spell it out but hide it.

[hide="The explanation."]Using cyshine's trigonometric identity, we get that

$\sum_{n=0}^m3^n\sin^3\left(\frac{x}{3^n}\right) = \frac14\left(\sum_{n=0}^m 3^{n+1}\sin\left(\frac{x}{3^n}\right)-\sum_{n=1}^{m+1}3^{n+1}\sin\left( \frac{x}{3^n}\right)\right)$

$=\frac14\left(3\sin x-3^{m+2}\sin\left(\frac{x}{3^{m+1}}\right)\right).$

Now take the limit as $m\to\infty$. We have $\lim_{m\to\infty}3^{m+2}\sin\left(\frac{x}{3^{m+1}}\right) = 3x.$

Hence, $\sum_{n=0}^{\infty}3^n\sin^3\left(\frac{x}{3^n}\right) = \frac{3\sin x-3x}4.$

The orignal series is convergent. Where cyshine went wrong was that he represented this convergent series as the difference of two divergent series without noting that they were divergent.[/hide]

[quote="Myth"][quote="kunny"]My answer is $\frac{1}{4}(3\sin x-3x)$. :?[/quote]
Have you simply rewritten my answer?  :lol:  :lol:  :lol: (joke)[/quote]

No.You are wrong.

Sorry for bring this old topic to light, but I solved the initial problem, and then read the topic, but I still get as the answer

$S = \frac{1}{4}(3x-\sin(3x))$. Indeed, using the trigonometric identity $\sin^{3}x=\frac{1}{4}[3\sin x-\sin(3x)]$ I get

$\sum_{n=0}^{\infty}3^{n}\sin^{3}\left(\frac{x}{3^{n}}\right) =-\frac{1}{4}\sum_{n=0}^{\infty}\left(3^{n}\sin\left(\frac{x}{3^{n-1}}\right)-3^{n+1}\sin\left(\frac{x}{3^{n}}\right)\right)$
$=-\frac{1}{4}\sum_{n=0}^{\infty}(a_{n}-a_{n+1})$

So, we have here a telescopic series with the sum I posted above. What am I missing?

Thanks in advance, Carcul

[quote="Kent Merryfield"]
The orignal series is convergent. Where cyshine went wrong was that he represented this convergent series as the difference of two divergent series without noting that they were divergent.[/hide][/quote]

if i understand correctly this is why the telescoping method doesn't work.

But two divergent series can be sumed to yield a convergent series. For example

$\frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}$.

What are the partial sums of that telescoping series? Look closely before taking the limit.

I read the topic and I belive after my computation the answer is

$ S\equal{}\frac{3x\minus{}sin 3x}{4}.$


What is the number of centimeters in the diameter of a circle whose area is $100 \pi$ cm$^2$?

oh i see how i'm wrong
[hide]20, because 20/2 is 10 and 10^2 is 100.[/hide]

[quote="4everwise"]I got the answer in a different way.[/quote]
how did you do it?


geometry
incenter
circumcircle
projective geometry
geometry solved

Given the triangle $ABC.$ $I$ and $N$ are the incenter and the Nagel point of $ABC$, and $r$ is the in radius of $ABC.$ Prove that
\[IN=r \iff  a+b=3c \text{ or } b+c=3a \text{ or } c+a=3b\]

Please, tell what is Nagel point?
I can't remember all this names :blush:

let the outcircles touch the sides of the triangle in $A',B',C'$. then the nagel point is the intersection of $AA'$,$BB'$,$CC'$.

Let $x$, $y$ and $z$ be segments, on which incircle divides sides of the triangle.
Then $I$ is a center of gravity of $\frac{y+z}{2(x+y+z)}A$, $\frac{x+z}{2(x+y+z)}B$, $\frac{y+x}{2(x+y+z)}C$ and $N$ is a center of gravity of $\frac{x}{x+y+z}A$, $\frac{y}{x+y+z}B$ and $\frac{z}{x+y+z}C$.
Moreover, $r^2=\frac{xyz}{x+y+z}$.

We know Lagrange formula:
Let $P$ be center of gravity of $p_1A_1$, ..., $p_nA_n$ and $Q$ be center of gravity $q_1A_1$, ..., $q_nA_n$, where $p_1+...+p_n=q_1+...+q_n=1$. Then
\[PQ^2=-\sum_{i<j}(p_i-q_i)(p_j-q_j)A_iA_j^2.\]

Applying this formula to our configuration and after some transformations we obtain, that $IN^2=r^2$ is equivalent to
\[4(x+y+z)(x+y-z)(x-y+z)(-x+y+z)=0,\]
hence result !!!  :D  :D  :D

I have just found a beautiful solution for this problem  :)
Let the incircle (I) touches BC,CA,AB at X,Y,Z respectedly. XI, YI, ZI intersect (I) again at X',Y',Z'.
We all know that AX',BY',CZ' meet at K  ( the Nagel point )
So $K\in(I)$<->K=X' or K=Y' or K=Z'
And now the rest is very easy  :D

[b][EDIT: This is incredible! Stun and me posted the same solution with a difference of 6 minutes!][/b]

Well, I knew that such a solution would be posted, but actually there is a synthetic one. I already wanted to post it yesterday but was too tired - ice skating is arduous, especially for a complete beginner...

I start, as usual, with rewriting the problem:

[i]Prove that the Nagel point N of a triangle ABC lies on the incircle of triangle ABC if and only if b + c = 3a, c + a = 3b or a + b = 3c.[/i]

For the solution, I firstly note (as Peter Scholze did ;) ) that the Nagel point N of triangle ABC is the point of intersection of the lines AA', BB', CC', where A', B', C' are the points where the A-excircle, B-excircle, C-excircle of triangle ABC touch its sides BC, CA, AB. Well, that's just half of the truth. There is another important property of the Nagel point:

[b]Theorem 1.[/b] If the incircle of triangle ABC touches its sides BC, CA, AB at the points D, E, F, and if D', E', F' are the points diametrically opposite to the points D, E, F on this incircle, then these points D', E', F' lie on the lines AA', BB', CC', respectively.

This shows that the Nagel point N of triangle ABC could have also been defined as the point of intersection of the lines AD', BE', CF'.

I love hide tags, so you it will require a painful click to see my proof of Theorem 1 :D :

[hide="Proof of Theorem 1"]
[i]Proof of Theorem 1.[/i] In the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=5811 , or, more precisely, somewhere in the middle of my post there, I proved that the point D' lies on the line AA'. Similarly, the points E' and F' lie on the lines BB' and CC', and Theorem 1 is proven. (Yes, I am able to write short proofs: this one was just 2 lines long! :D )[/hide]

In the following, "arc EF", "arc FD" and "arc DE" will mean the arcs EF, FD and DE on the incircle of triangle ABC (where the arc EF, of course, doesn't contain the point D, and similarly the arcs FD and DE don't contain the points E and F, respectively). Here is another lemma, almost trivial (I also won't kill myself it turns out to be completely trivial):

[b]Theorem 2.[/b] The line BB' meets the arc FD at exactly one point, namely at the point E'.

My proof is boring, so I put it in a hider:

[hide="Proof of Theorem 2"]
[i]Proof of Theorem 2.[/i] At first we will show that the point E' lies on the arc FD. In fact, the point E' lies on the incircle of triangle ABC; thus, it lies on one of the arcs EF, FD and DE. Now, if it would lie on the arc DE, it would therefore lie inside the triangle DCE, and thus we would have $\measuredangle E^{\prime}EC < \measuredangle DEC$. But since the two tangents from a point to a circle are equal, we have CD = CE; thus, the triangle DCE is isosceles. Since a base angle of an isosceles triangle must always be < 90°, we thus have $\measuredangle DEC < 90^{\circ}$, so that $\measuredangle E^{\prime}EC < \measuredangle DEC < 90^{\circ}$. But on the other hand, since the point E' lies diametrically opposite to the point E on the incircle of triangle ABC, the point E' lies on the line EI, where I is the incenter of triangle ABC. But since a tangent to a circle is always perpendicular to the corresponding radius, we have $CA\perp EI$, and thus $\measuredangle E^{\prime}EC = 90^{\circ}$. This contradicts $\measuredangle E^{\prime}EC < 90^{\circ}$. Thus, the point E' cannot lie on the arc DE. Similarly, we show that the point E' cannot lie on the arc EF, and hence, the point E' must lie on the arc FD.

Theorem 1 shows that the point E' lies on the line BB'; thus, the point E' is a point of intersection of the line BB' with the arc FD. To prove Theorem 2, it now remains to show that this point E' is the only point of intersection of the line BB' with the arc FD. In other words, we have to prove that there are no two distinct points of intersection of the line BB' with the arc FD. In fact, assume that there are two distinct points of intersection of the line BB' with the arc FD. Call these points U and V, where BU < BV (we can always WLOG assume this). Since both points U and V lie on the arc FD, we have $\measuredangle FUD = \measuredangle FVD$. On the other hand, since both points U and V lie on the arc FD, they both lie inside the triangle FBD, and thus they both lie inside the segment BT, where T is the point of intersection of the segments BB' and FD.
Now, since BU < BV, we have $\measuredangle FUT < \measuredangle FVT$ and $\measuredangle DUT < \measuredangle DVT$, so that $\measuredangle FUD = \measuredangle FUT + \measuredangle DUT < \measuredangle FVT + \measuredangle DVT = \measuredangle FVD$, contradicting the equation $\measuredangle FUD = \measuredangle FVD$. Hence, there are no two distinct points of intersection of the line BB' with the arc FD. This completes the proof of Theorem 2.[/hide]

Our next lemma will be something more substantial:

[b]Theorem 3.[/b] We have BN = E'B' (with directed segments).

This is, by the way, a nice exercise to prove. My proof is again in a hider:

[hide="Proof of Theorem 3"]
[i]Proof.[/i] If $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC, then, since the point E is the point of tangency of the incircle of triangle ABC with the side CA and the point B' is the point of tangency of the B-excircle with the side CA, we have EA = s - a and CB' = s - a, so that CB' = EA. If M is the midpoint of the segment CA, then we have CM = MA. Thus, B'M = CM - CB' = MA - EA = ME, so that the point M is also the midpoint of the segment EB'. On the other hand, since the point E' is the point diametrically opposite to the point E on the incircle of triangle ABC, the center I of this incircle is the midpoint of the segment EE'. Altogether, in the triangle E'EB', the points I and M are the midpoints of the sides EE' and EB'. Thus, $IM=\frac{E^{\prime}B^{\prime}}{2}$, so that $E^{\prime}B^{\prime}=2\cdot IM$.

On the other hand, since the point M is the midpoint of the side CA of triangle ABC, the line BM is a median in triangle ABC, and thus it passes through the centroid G of triangle ABC and we also have BG : GM = 2. On the other hand, we know that the Nagel point N, the centroid G and the incenter I of triangle ABC are collinear and that we have NG : GI = 2. Hence, BG : GM = NG : GI, and it follows by Thales that BN || MI and that BN : IM = BG : GM. Thus, BN : IM = 2, and $BN = 2\cdot IM$. Comparing this with $E^{\prime}B^{\prime}=2\cdot IM$, we immediately get BN = E'B', and Theorem 3 is proven.[/hide]

The next step to the solution will be:

[b]Theorem 4.[/b] If $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC, then $\frac{NB^{\prime}}{E^{\prime}B^{\prime}}=\frac{s-b}{b}$.

Again I hide the proof from the curious eyes of Myth :D (no, rather from the bored eyes of Myth):

[hide="Proof of Theorem 4"]
[i]Proof.[/i] Since BN = E'B' by Theorem 3, it is enough to prove $\frac{NB^{\prime}}{BN}=\frac{s-b}{b}$.
But actually, we know that the Nagel point N is the point of intersection of the lines AA', BB', CC'; thus, by the van Aubel theorem, $\frac{BN}{NB^{\prime}}=\frac{BC^{\prime}}{C^{\prime}A}+\frac{BA^{\prime}}{A^{\prime}C}$. Now, since the points C' and A' are the points where the C-excircle and A-excircle of triangle ABC touch the sides AB and BC, we have BC' = s - a, C'A = s - b, BA' = s - c and A'C = s - b, so that

$\frac{BN}{NB^{\prime}}=\frac{s-a}{s-b}+\frac{s-c}{s-b}=\frac{\left(s-a\right)+\left(s-c\right)}{s-b}$.

But (s - a) + (s - c) = 2s - (a + c) = (a + b + c) - (a + c) = b, and thus $\frac{BN}{NB^{\prime}} = \frac{b}{s-b}$, so that $\frac{NB^{\prime}}{BN}=\frac{s-b}{b}$. This proves Theorem 4.[/hide]

Now, to solve the problem, it is enough to prove two assertions:

[b]Assertion 1.[/b] If the Nagel point N of triangle ABC lies on the incircle of triangle ABC, then b + c = 3a, c + a = 3b or a + b = 3c.

[b]Assertion 2.[/b] If b + c = 3a, c + a = 3b or a + b = 3c, then the Nagel point N of triangle ABC lies on the incircle of triangle ABC.

[i]Proof of Assertion 1.[/i] Since the Nagel point N of triangle ABC lies on the incircle of triangle ABC, it must lie on one of the arcs EF, FD and DE on this incircle. WLOG assume that the point N lies on the arc FD. Thus, the point N is a point of intersection of the line BB' with the arc FD. But after Theorem 2, there is only one point of intersection of the line BB' with the arc FD, and this point is the point E'. Hence, N = E'. This yields NB' = E'B', and $\frac{NB^{\prime}}{E^{\prime}B^{\prime}}=1$. But after Theorem 4, we have $\frac{NB^{\prime}}{E^{\prime}B^{\prime}}=\frac{s-b}{b}$. Thus, $\frac{s-b}{b}=1$. This implies s - b = b and s = 2b. But $s=\frac{a+b+c}{2}$; thus, $\frac{a+b+c}{2}=2b$, so that a + b + c = 4b. Now we subtract b and get c + a = 3b. Similarly, in the case when the point N lies on the arc EF, we find b + c = 3a, and in the case when the point N lies on the arc DE, we find a + b = 3c. This proves Assertion 1.

The [i]proof of Assertion 2[/i] can be easily found by reversing the arguments we used for proving Assertion 1; it is even simpler than the proof of Assertion 1 since we don't have to apply Theorem 2.

  Darij

[quote="darij grinberg"]

Again I hide the proof from the curious eyes of Myth :D (no, rather from the bored eyes of Myth):

  Darij[/quote]

That is very smart of you, Darij,  :lol: Just don't write a message to Myth not at the beginning or at the end of a post but somewhere inbetween. So you can make sure he won't read it.  :P

[quote="Myth"]Of course, didn't read his proof, because no one appreciates my quick one :([/quote]

Myth, we all love you.  :love: As most of your proofs are excellent we cannot admire all of them in a reply.  :lol:

[quote="Stun"]Given the triangle ABC. I and N are the incenter and the Nagel point of ABC, r is the in radius of ABC. Prove that IN=r <-> a+b=3c
or b+c=3a
or c+a=3b[/quote]

I think these properties are equivalent to this one : the centre of gravity $G$ of triangle $ABC$ lies on $DE$, on $EF$ or on $FD$ !  :D

Hi :)
I have a question on Darij' proof of theorem 1.
Here is what it states:

[quote]Let the A-excircle of triangle ABC touch BC at X'. Then, with the notation s = (a+b+c)/2, we have BX = s-b and CX' = s-b.

[/quote]

I'm sorry to say I can't get why CX' = s-b is true.

Could you give me a hand with this?

Ok, this must be true that you understand better when you ask, for I have already got the thing. :)

So, I'll post my proof if nobody does it.
The statement is as follows : $b+c = 3a$ $\Leftrightarrow$ $G$ lies on $EF$, where $D$, $E$ and $F$ are the points of tangency of the incircle with the sides $BC$, $CA$ and $AB$ respectively.

Here we use the following lemma : Let $X$ be the intersection of $DI$ with $EF$, then the line $AX$ bissects $BC$.

Proof : Let $Y$ the point of intersection of $EF$ with the parallel line to $BC$ through $A$. The polar line of $X$ with respect to $(I)$ is perpendicular to $PI$, so it's parallel to $BC$. Since $X$ lies on the polar line of $A$, the polar line of $X$ runs through $A$, so it's $AY$. Since $Y$ lies on the polar lines of $A$ and $X$, its polar line is $AX$. Therefore $Y, X, E, F$ are harmonic points, and thus $AY, AX, AE, AF$ is a harmonic line batch. This means exactly that $AX$ bissects $BC$.

So, $G$ lies on $AX$. Moreover $G$ lies on $IN$ where $N$ is the Nagel point. So $G$ is the intersection of $AX$ with $IN$. We know (because this is the subject of this topic  :D ) that $N$ lies on $DI$ if and only if $b+c = 3a$, so $G$ lies on $EF$ if and only if $b+c = 3a$, and when it's the case, $G$ = $X$, qed.

Benjamin

[quote="benjamin"] [color=indigo]$D,\,E$ and $F$ are the points of tangency of the incircle with the sides $BC,\,CA$ and $AB$ respectively of triangle $ABC.$ Let $X$ be the intersection of $DI$ with $EF,$ then the line $AX$ bissects $BC.$[/color]
[/quote]
[b] Proof.[/b] Denote: $\delta$ be the parallel line to $BC$ through $A;\,Y=EF\cap \delta,\,S=ID\cap\delta,\,M=AX\cap BC$
We have: $S,\,E,\,F$ lie on cicle with diameter $AI$ and $IE=IF$ so $SI$ is the bisector of angle $SEF\quad (1)$
On the otherhand: $SY\perp SI\Longrightarrow SY$ is the external bisector of angle $SEF\quad (2)$
From $(1),\,(2)\Longrightarrow (EFYX)=-1\Longrightarrow A(EFYX)=-1$
Otherwise: $BC\parallel AY$
Hence: $MB=MC$

Now an exercise which was proposed to France last year :
Let $ABC$ a triangle, $I$ its incenter and $D$, $E$, $F$ the points of tangency of its incircle with sides $BC$, $CA$, $AB$. $BI$ and $CI$ meet $PQ$ at $K$ and $L$ respectively. Let $\gamma$ the circumcircle of $IKL$. Show that $IKL$ and $(I)$ are tangent if and only if $b+c = 3a$


group theory
abstract algebra
pigeonhole principle
combinatorics unsolved
combinatorics

An [i]$ n\minus{}m$ society[/i] is a group of $ n$ girls and $ m$ boys. Prove that there exists numbers $ n_0$ and $ m_0$ such that every [i]$ n_0\minus{}m_0$ society[/i] contains a subgroup of five boys and five girls with the following property: either all of the boys know all of the girls or none of the boys knows none of the girls.

I assume "none of the boys knows none of the girls" should be "all of the boys know none of the girls". Otherwise, consider the society where all boys know none of the girls. Then, the problem statement is false.

[hide]

Consider n_0 = 10 and $ m_0 \equal{} 8 \binom{10}{5} \plus{} 1.$

We will prove any $ n_0\minus{}m_0$ society contains a subgroup of five boys and five girls with the following property: either all of the boys know all of the girls or none of the boys knows none of the girls. Call a boy girl-knowing if he knows at least 5 of the 10 girls. If a boy is not girl-knowing, then he knows less than 5 girls, so he doesn't know more than 5 girls.

Consider the number of girl-knowing boys and the number of boys who are not girl-knowing. By pigeonhole, one of the numbers is at least $ 4 \binom{10}{5} \plus{} 1.$

The case where the number of girl-knowing boys is at least $ 4 \binom{10}{5} \plus{} 1:$ For each boy, he must know at least 5 girls. For each boy, assign a set of 5 distinct girls who he knows. There are $ \binom{10}{5}$ possible sets and there are at least $ 4 \binom{10}{5} \plus{} 1$ boys. By pigeonhole, there is a set of 5 girls which is assigned to at least 5 boys. Choose these 5 boys and the 5 girls in the set. They satisfy the conditions.

The case where the number of non-girl-knowing boys is at least $ 4 \binom{10}{5} \plus{} 1$ is similiar.

[/hide]


Four different points lie on a circle. Mark the midpoints between each pair of adjacent points on the arc and connect the opposite midpoints. Show that the two lines cross at right angles.
 :D

It looks like you are a new member, so you probably don't have the feel of the difficulty levels of the forums yet.  Proofs like this tend to wind up in the High School forums, so I'll move this one.

Nothing yet? There's actually a quite simple solution.

If you already know the solution, why do you still put a "?" on it?

The measure of an angle between two chords is equal to the half of the sum of the two arcs that it faces. In this problem, the sum of the arcs is $180^o$, so $\text{Angle}=90^o$.


[quote="math92"]I entered a pie eating contest at the country fair. I was determined to win and went into training for 6 days. Each day I ate 4 more pies than the day before. I ate 150 pies while in training. How many pies did I eat each day?[/quote]
[hide]
x=1st day
x+4=2nd day
x+8=3rd
x+12=4th
x+16=5th
x+20=6th

6x+60=150
6x=90
x=15

so those days, you ate

1st=15
2nd=19
3rd=23
4th=27
5th=31
6th=35
[/hide]


A certain rectangle has a perimeter of 34 inches and an area of 60 square inches. What is the number of inches in a diagonal of the rectangle?

5-12 works, and its diagonal by Pythag is [b]13[/b].

(Mathcounts always has integer answers (most of the time, esp. when they do not ask for simplest radical answer))


Given a right triangle $ABC$,three regular triangles $ BCP_1,CAP_2,ABP_3 $ are formed outside each side.

Let $G_1,G_2$ and $G_3$ be barycentres of $\triangle{BCP_1,\triangle{CAP_2}}$ and $\triangle{ABP_3}$ respectively.

Find the minimum value of the folllowing ratio.

\[\frac{|G_1P_2|^2+|G_2P_3|^2+|G_3P_1|^2}{|\triangle{ABC}|}\]

Here is my solution.It is rather simple.
Let $\angle{CAB}=\alpha,\angle{CBA}=\beta,AC=b$ and $BC=a$.
It is obviouse that the points $G_{1},C$ and $P_{2}$ are collinear.Using this we can calculate
${G_{1}P_{2}}^{2}$ in terms of $a$ and $b$.So ${G_{1}P_{2}}^{2}=\frac{{a}^{2}}{3}+{b}^{2}+2ab\cdot\frac{\sqrt{3}}{3}$.Applying the cosines rule for the triangles $G_{2}BP_{3}$ and $G_{3}AP_{1}$,and using that $\cos({90+\alpha})=-\sin{\alpha}$ we obtain
${G_{2}P_{3}}^{2}={a}^{2}+\frac{4{b}^{2}}{3}+2ab\cdot\frac{\sqrt{3}}{3}$ ,
${G_{3}P_{1}}^{2}=\frac{4{a}^{2}}{3}+\frac{{b}^{2}}{3}+2ab\cdot\frac{\sqrt{3}}{3} .$
$[ABC]=\frac{a\cdot{b}}{2}.$
Using our equations and grouping we get $\frac{{G_{1}P_{2}}^{2}+{G_{2}P_{3}}^{2}+{G_{3}P_{1}}^{2}}{[ABC]}=\frac{16}{3}\*(\frac{a}{b}+\frac{b}{a})+4\sqrt{3}$.From AM-GM we get that the minimum value of the given expression is
$\frac{32+12\sqrt{3}}{3}$.

Thank you for your reply, Armo.

You almost could sove! Regrettably you are wrong in calculation $G_3P_1^2$.

Please try again!

My solution is exactly same as [b]Armo's[/b].

kunny

Thanks kunny.I have already edited it.Is my answer right? :P


are we allowed to talk about ICTM scores and questions?

From past years? I think so. I don't really see a problem with it, but who knows?

lol that was my old account. it was from about 1 year ago and i forgot the password.

:rotfl:  :rotfl:  :rotfl: i was so stupid back then....just thinking about it...but then...in a few nore years.....i'll feel the same way...

true, i feel that way about myself too...

i've never taken the ictm is it hard? :blush:

i didnt make it to state last year but this year i was 34th at state. i hope to make nats next year :roll:

i got 39/40 in pre-algebra last year, too! (even though i was in seventh grade :oops: ) i said i was taking algebra on my sheet, but my math teacher yelled at me and signed me up for pre. i made a careless mistake on some triangle problem. i got 64  :(  in state last year. i was 30th in sixth grade, though! i don't know what happened to me in 7th grade...

[quote="cognos599"][quote="indianamath"]there's pre-alg, alg, geo, alg II, and cumulative. i got 39/40 on the pre-alg last year in 6th grade. i got 47th in the state in MATHOUNTS last year.[/quote]

in our school you cannot take contests that are ahead of your grade level. so i couldn't take pre last year even though i was in pre.(it is strange,we do pre-algebra in both 6th and 7th grade)[/quote]

yes you can!

as was my case

i got 21 on the algebra II test last year, with like 4 people participating from my class.. i don't even want to know what place i got, but i remember it was one point above the average or something  :oops:

EDIT: yeah, and taking the comprehensive test this year.. printed out all past tests :)

lol when you guys were in 7th grade (to mwp), i was in 6th grade ^_-, and there were alot of new 6th graders who kinda owned -_-... so theres ur answer ^^... and then more people suddenly came (whoaaa)

can you take ICTM comprehensive even though you are only in Algebra II?  cause theres no point in me taking the algebra II test..

no, you can't do that. you have to take the test that corresponds with the course you are taking now (at least that's what my teacher said when i asked her)...

O well, I signed up for Comprehensive..I dont think they would care that much because Im "challenging" myself... :rotfl:

[quote="mwpl11"]i tried signing up for the algebra test last year (i was in pre algebra) and my teacher ended up changing it on my form (without telling me until like a little bit before the test). oh, well, you can try it. good luck with that[/quote]

But arent your teachers those people with ego problems? (that you mentioned)

yes. but i don't see how that really relates? they're just like "you messed up on your form. you said that ur doing algebra, so i changed it to say pre-algebra."
    and i'm just like " i didnt mess up on my form. i wanted to take algebra."

    "you can't take algebra."

    "why not?"

    "because you aren't involved in an algebra course."

    "? oookaaay?"

i asked again this year and my teacher now said no also. oh well, it's not like i'd do well in comprehensive anyways  :lol: ....

[quote="mwpl11"]are you even from indiana?[/quote]
Why would I visit the "Indiana" forum if I weren't?

Yea..the ICTM problems are stupid-just look at the 2005 comprehensive test-I can brute force half of them..

[quote="cognos599"]have either of you ever gotten a perfect score on ICTM. if not DONT SAY SOMTHING IS STUPID UNLESS YOU HAVE GOTTEN  A PERFECT SCORE ON IT.[/quote]
I think the pre-algebra, algebra, and the geometry tests are stupid :D

and how exactly did you take those tests? u weren't even in indiana when you were taking pre algebra!!

because I've taken a few as practice and I've looked at all of them?

besides, the pre-algebra and algebra tests seem to have a lot of simplifying/solving equations going on.. i learned that a long time ago.

yeah, and this same arrogant kid failed to compute 100 - 2 correctly

it still nags me.

[quote="cognos599"]have either of you ever gotten a perfect score on ICTM. if not DONT SAY SOMTHING IS STUPID UNLESS YOU HAVE GOTTEN  A PERFECT SCORE ON IT.[/quote]
Dude, no need for caps. It makes you sound angry, therefore repulsive. Yes, I did get perfect, therefore the test is stupid if [i]I[/i] can get perfect, along with 20 others.

i got 39/40 last year. i couldn't do it because my number theory sucked.

omg are you andrew keiser???????????

jk i know who you are


Determine the number of integers $n \ge 2$ for which the congruence \[x^{25}\equiv x \; \pmod{n}\] is true for all integers $x$.

Suppose that $n$ is such number and:
$n = p_1^{\alpha_1}p_2^{\alpha_2}...p_m^{\alpha_m}$
so for $i=1,2,...,m$ we have:
$x^{25} \equiv x \mod{p_i^{\alpha_i}}$
By plugging $x=p_i$ we can easily see that $\alpha_i=1$. Now, for every prime $p_i$ there exist a primitive root mod $p_i$. Call it $g_i$. Since order of $g_i$ is $p_i-1$ and we have $x^{24} \equiv 1 \mod{p_i}$ it follows that $p_i-1|24$. Now it's only a matter of simple calculating.

Isnt this Bulgaria 1995 Mathematical Olympiad ?

It could be, I don't have access to these sources. Can anyone confirm this?

Thanks for the report by the way. :)

Yes it is confirmed .
See this ......this is what I could get from here http://www.math.bas.bg/bcmi/


floor function
inequalities
calculus
calculus computations

Let $ a>1$ and $ u_n$ is the greatest integer not exceeding $ a^n$.
show that $ u_{n\plus{}1}/u_n\rightarrow a$.

as $ u_n\equal{}\lfloor a^n\rfloor$, the following inequality holds: $ a^n\minus{}1<u_n\leq a^n$ (*)
because of (*) is $ \frac{u_{n\plus{}1}}{u_n}\leq\frac{a^{n\plus{}1}}{u_n}<\frac{a^{n\plus{}1}}{a^n\minus{}1}\equal{}\frac{a^n\cdot a}{a^n(1\minus{}1/a^n)}\rightarrow a$ (1)
and $ \frac{u_{n\plus{}1}}{u_n}\geq\frac{u_{n\plus{}1}}{a^n}>\frac{a^{n\plus{}1}\minus{}1}{a^n}\equal{}a\minus{}\frac{1}{a^n}\rightarrow a$ (2)
according to (1) and (2) $ \frac{u_{n\plus{}1}}{u_n}$ is bigger than a series converging to $ a$ and smaller than a series converging to $ a$, therefore $ \frac{u_{n\plus{}1}}{u_n}\rightarrow a$


You have to pay $\$$ 10 to a cashier using 2 and 1$\$$ notes. (You have any number of 2 and 1 $\$$ notes available.) You place the notes on the counter in a row, e.g., 2 2 1 1 2 2 Placing the same notes in a different order is to be counted as different. For example the above notes could be permuted as 2 2 1 2 1 2, and it is considered different. (But interchanging the first and the second 2 $\$$ notes would not make it different.) You could have also placed a different set of notes such as 2 1 1 1 2 1 2 which is different. In how many different ways can you pay 10$\$$?[/code]

Hmmm could you use a generating function to do this?


conics
parabola
analytic geometry
algebra
polynomial
calculus
calculus computations

Which point on the parabola $ y\equal{}x^2$ is nearest to $ (1,0)$? Find the coordinates to two decimals. [Hint: Minimize the square of the distance- this avoids square roots.]

I thought about the difference quotient, but I don't know how to start. What does the hint means?

The hint means exactly what it says.  The distance of a point $ (x, x^2)$ from the point $ (1, 0)$ is given by

$ d \equal{} \sqrt{ (x \minus{} 1)^2 \plus{} (x^2 \minus{} 0)^2 }$

If we try to minimize $ d$ by differentiating and setting to zero, we'll run into some tedious algebra.  But since $ d$ is non-negative, minimizing $ d$ is equivalent to minimizing the [b]square[/b] of the distance, which is given by

$ d^2 \equal{} (x \minus{} 1)^2 \plus{} (x^2 \minus{} 0)^2 \equal{} x^4 \plus{} x^2 \minus{} 2x \plus{} 1$

which is just a polynomial, and much easier to differentiate and minimize.


The integers from 1 to 10, inclusive, are partitioned into two sets of five elements each. Find the probability that 1 and 2 are in the same set.

[hide]After either 1 or two is sent to a set, there is a 4/9 chance of the other being chosen for the same set.

The answer is 4/9


Here's how you play this game: First you pick which team you think is going to win the championship (Lakers or Magic), then you predict a score for each upcoming game( for now game 1), then you pick which player on that team will be the top scorer for the upcoming game.

I say Lakers will win the whole thing, the first game score will be 103-97 lakers on top, and that Kobe Bryant will be the leading scorer.

:dry:

you frontrunner...

i am going to go with the Magic winning the entire thing

they will win game 1 97-94

and, obviously, since htey have learned well, dwight howard will be the leading scorer

GO LAKERS!!!   Lakers are gonna win the whole thing but they are going to be surprised in the first game and lose.

Game 1:   114-97 Magic win.  Rashard Lewis is the high scorer

Hey! I'm from around L.A.! I really like the Lakers!

Anyways, I also really like the magic and I especially like Rashard Lewis and Hedo Turkgolu because they are amazing 3-point shooters. I'm not a crazy fan of Dwight Howard, but I do think he is good.


calculus
integration
logarithms
derivative
analytic geometry
trigonometry
real analysis

Let $R$ be the region inside the unit circle and above the line $x+y=1$. Calculate:

$\int\int_{R}\frac{1}{\ln^{2}x+\ln^{2}y}\frac{dx dy}{xy}$.

[quote="sylow_theory"]You can find the anti-derivative for,

$\frac{1}{\ln^{2}x+\ln^{2}y}\cdot \frac{1}{xy}$
By using $u=\ln x$[/quote]

Yes, but that doesn't lead to an easily evaluated integrand in y.  Try it and you'll see.  The answer is $\frac{\pi}{2}\log 2$, but I do not yet know how to prove it.

Atomicwedgie's answer is correct. If someone gets completely stuck, here is a small
[hide="Hint"]Actually, it makes sense to make the natural change of variable $u=\ln x$, $v=\ln y$, then to switch to the polar coordinates $u=r\cos\theta$, $v=r\sin\theta$ and to integrate with respect to $\theta$ first. The trick on this way is not to try to come up with any [b]explicit[/b] formula for the intergal with respect to $\theta$: there is none. ;) [/hide]
Most likely, it is not the only way to do it but this way requires almost no computation and brings up one useful integral.


geometry
trapezoid
projective geometry
geometry proposed

Let $ ABCD$ be a [fixed] trapezoid with parallel sides $ AB$, $ CD$. Let $ K\in AB$, $ L\in DC$ such that $ AK: KB=CL: LD=t$. Let $ X\in AL\cap KD$ and $ Y\in BL\cap KC$. Prove that as $ t$ varies through positive real numbers, $ XY$ passes through a fixed point.

It’s easy problem, because the point $ L$ on $ CD,$ is the symmetric point of $ L',$ with respect to the midpoint of the segment $ CD,$

where $ L'\equiv CD\cap SK$ and $ S\equiv AD\cap BC$ $ ($ $ \frac{AK}{KB}= \frac{DL'}{L'C}= \frac{CL}{LD}$ $ ).$

So, we have the collinearity of the points $ L,$ $ P\equiv AC\cap BD$ and $ K$ $ ($ easy to prove $ ).$

Hence, apply the [b][size=100]Pappus theorem[/size][/b], we conclude that the segment line $ XY,$ passes through the point $ P$ and the proof is completed.

Kostas Vittas.

Yes, nice solution; that is essentially the solution that I was looking for.

Let $ AC\cap BD=Z$, $ KZ\cap CD=L'$. Then through triangle similarity, $ AK: CL'=KZ: ZL'$ and $ BK: L'D=KZ: ZL'$. Hence $ AK: CL'=BK: L'D\implies L'D: CL'=BK: AK$. Hence $ L\equiv L'$.

Then we apply pascal to $ ALBDKC$; $ AL\cap KC=X$, $ LB\cap CA=Y$, and $ BD\cap CA=Z$, so $ XYZ$ is a line. Since $ Z$ is fixed, the result follows.

I guess that the problem is pretty simple, but it is nice. Problem proposing not only helps me solidify theorems and concepts, but it helps me learn to apply them.


[quote="indybar"]100 prisoners are about to be executed. The judge gives each of them a hat of color either black or white. Each person cannot see his own hat but can see all other hats. The judge ask, one-by-one, which color the hat is. If the prisoner gets it correct, he survive. If not, he is executed. The prisoners are not allowed to communicate with each other, but can see when others answer the question. What is the strategy to save most number of people?

(ex: first person looks and 2nd and 3rd and answer white if the color are the same and black otherwise. The first person will die but 2nd and 3rd will survive, and so on. 66 people will be saved.)[/quote]

[hide="fun game"]
1st person says white if parity of white hats he sees is odd, black if parity of white hats he sees is even. From this, the 2nd person and so on can determine the color of their own hat by looking at the parity of the other hats he sees. So 99 people survive (100 if the 1st guy is lucky).[/hide]

[quote="krustyteklown"]hmm, they were not allowed to communicate before hand tho, are they each gonna figure out the parity strategy

i think this problem was on the BAMM 2005[/quote]

Well, the problem asks for a strategy, and it doesn't say the prisoners weren't allowed to communicate beforehand - obviously, if there is strategy involved, the prisoners must all know about it or it won't work.  :)


given *x^2+*x+*=0

two player are playing a game.player 1 will pick three different number (not a zero).player 2 will subsitute each of the three numbers into each of  the * in the equation in any order he likes.if the equation has two distinct roots,then player 2 will win.if not,player 1 will win.Proof that player 2 can always win.[/list][/code]

That's an algebra problem and definitely not Open. So I will move it.

And I find it ridiculous that this is formulated as a game. Sounds more like "A and B play a game: A wins if the Riemann hypothesis is true. Show that A can win."...  :wink:


The side length of a cube is doubled. The volume of the new cube is how many times the volume of the original cube?

The ratio of the volumes is the cube of the ratio of the side lengths, so we have $ 2^3\equal{}\boxed8$.

If you wanted to check your answer, you could also substitute "easy" numbers. For instance, let the original cube have side length 1 and the new cube have side length 2. Clearly, the answer is $ 2^3 \div 1^3 \equal{} \boxed{8}$.

The formula for the volume of a cube is $ s^3$ Letting the two sides be $ s_1$ and $ s_2$, $ \frac{s_2^3}{s_1^3}\equal{}\left(\frac{s_1}{s_2}\right)^3$


limit
function
logarithms
calculus
calculus computations

"Exponentiating by a non-rational complex number" is a "function" which takes infinitely many values, so you'd probably better explain what you think your question means.

what i mean is

$\imath\uparrow\uparrow 1\;=\;\imath\;=\;0\;+\;\imath$

$\imath\uparrow\uparrow 2\;=\;\imath^\imath\;\left( =\;e^{-\frac{\pi}{2}}\right) \;=\;0.208\;+\;0\;\imath$

$\imath\uparrow\uparrow 3\;=\;\imath^{\imath^{\imath}}\;=\;0.947\;+\;0.321\;\imath$

$\imath\uparrow\uparrow 4\;=\;\imath^{\imath^{\imath^{\imath}}}\;=\;0.05\;+\;0.602\;\imath$

show that the values eventually [u]spiral[/u] (as $n$ increases indefinitely) in the complex $z-plane$
and eventually settles down at the value specified.    :D

you may choose to give the equation of the spiral if you prefer, something like:

$r\;=\;r_{0}\;+\;\alpha \left( e^{-\beta (\theta-\theta_{0})}\right) \;\;\;\;\;\quad where\;\;\; \alpha,\beta\;\in\;\mathbb{R}^{+}$

$i^{i}= \left(e^{\pi i/2+2\pi n i}\right)^{i}=\left \{e^{(2n-1/2)\pi}\right\}_{n \in \bf N}$.  Raising any non-zero complex number to a non-rational complex power results in an infinite family of values.  If you want to pick a single one of these values in general, you need to explain how you're doing it.

ok, since $\ln\;z$ is a multi-valued function, as in $\imath^\imath\;=\;e^{\imath\ln\imath}$,
you can choose $[0,2\pi)$ as your principal branch...  :rotfl:

just use some definitions of tetration.

$z^{z^{z^{.^{.^{.}}}}}=-\frac{\mathrm{W}(-\ln{z})}{\ln{z}}$

${\frac{2 i}{\pi }W\left(-\frac{1}{2}i \pi \right) }$

err.. but does not make much sense give a numerical solution... since we have calculators.

if you were asking for a non-numerical solutions, thats it :)


linear algebra
matrix
probability
parameterization
algebra
system of equations

Hello all,

I am deparing me with a problem or a doubt about the calculation of probabilities in one matrix.

I have 3 events (A,B,C)

Percentage of probability of each event: A - 2.3% B - 10% C - 15%

Conditions:
- A imposes B
- A imposes C

My Matrix:
A B C
0 0 0
0 0 1
0 1 0
0 1 1
1 1 1

Goal: The sum of all lines of matrix must be 100%, the sum of lines of each event must be equal to the percentage of probability given before the matrix generation.

Right now I am changing the percentage of each impose event (A) and I left the imposed event (B,C) with the original percentage.
Thus, I have this changed percentages:

P(A) = P(A|(B*C)) = P(A) / P(B)*P(C) = 1.53 (153%) or could consider 100%
P(B) = P(B) = 0.1 (10%)
P(C) = P(C) = 0.15 (15%)

If I made the calculations with this values, I will obtain 100% for all lines and for all events, but i will obtain negative values because the probability of (A) its bigger than 100%, so (~A)<0.
If i consider 100% instead of 153% I will obtain wrong values for (A) event.

The point is, I am not sure if it is possible to calculate this situation obtaining positive and correct values and I would like to know if anybody have some kind of help or some tips that can help me.

Thank you, and Best Regards.

[quote="JoaoPaixao"]P(A) = P(A|(B*C)) = P(A) / P(B)*P(C) = 1.53 (153%) or could consider 100%[/quote]

Well, it is surely the case that either this calculation is wrong or the situation is impossible: if the rule and data were correct (and I have no idea what rule you were trying to apply) then we would have an event with $> 100\%$ probability, which is nonsense.  I suggest you not try to use whatever rule you were using.

Here is how I, in my naive way, would go about this problem: associate probabilities $p_{1}, \ldots, p_{4}$ with the four rows of your matrix 001, 010, 011, 111.  (The probability associated with the row 000 will just have to be whatever is left over.)  Then you must solve a system of equations in order to find $p_{1}, p_{2}, p_{3}, p_{4}$ given what you know about the likelyhoods of A, B, C.  In fact, I think you will find that there is an infinite, one-parameter family of solutions.


Prove that:$a^{4}+b^{4}+c^{4}+abc \geq a^{3}+b^{3}+c^{3}$
where $a;b;c$ are non-negative and satisfying $a+b+c=2$

If non of $a,b,c$ is $0$ we get homogen inequality $[4,0,0]+\frac{[2,1,1]}2\geq [3,1,0]$ which is true by Shure theorem (it's can be even sharper if you add $\frac{[4,0,0]}{2}$) to the right side.

Let $c=0$, and $a,b>0$.

Then it's reduced to $a+b=2$ and $a^{4}+b^{4}\geq \frac{(a+b)(a^{3}+b^{3})}{2}$, which is true by Muirhead.

Let $b=c=0$ and $a>0$

Then is reduced to $a=2$ where simple check can prove inequality.


function
limit
algebra
functional equation

Find all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(x+y)+f(x-y)=2f(x)f(y)\wedge \lim_{x\rightarrow\infty}f(x)=0. \]
(Without using known results regarding d'Alembert's functional equation, of course.)

[quote="boxedexe"]
(Without using known results regarding d'Alembert's functional equation, of course.)[/quote]
What do you mean by this?
[hide="You mean this?"]
Let's say $ f$ twice differentiable (I know, I shouldnt do that, I am merely asking a question).

Define, $ u(x,y) = f(x+y)+f(x-y)$
Then, $ u_{xx}(x,y) = f''(x+y)+f''(x-y)$
And, $ u_{yy}(x,y) = f''(x+y)+f''(x-y)$

This gives us the partial differential equation,
$ \frac{\partial^{2}u}{\partial x^{2}}= \frac{\partial^{2}u}{\partial y^{2}}$
This is the [url=http://en.wikipedia.org/wiki/Wave_equation]wave equation[/url].

Now we can solve this using [url=http://en.wikipedia.org/wiki/D%27Alembert%27s_formula]d'Alembert's formula[/url] (at least partially). Is this what you mean?

[/hide]

Er...
[hide]
$ y=x-k$
\[ \implies f(k)=2f(x)f(x-k)-f(2x-k)\]
Take the limit of both sides as $ x$ goes to $ \infty$. The right hand side has to go to $ 0$, by the given condition, so this implies that $ f(k)=0$ for all $ k \in \mathbb{R}$...[/hide]

@ Sylow_theory: That is definitely not what I meant.  The functional equation I wrote above is called d'Alembert's [i]functional[/i] equation, not d'Alembert's formula.  K81o7's answer is correct.

Let $ x=y$ and rearrange to obtain $ f(0)=2[f(x)]^{2}-f(2x)$.  Take the limit of both sides as $ x\rightarrow\infty$ to find $ f(0)=0$.  Set $ y=0$ to get $ f(x)\equiv 0\ \forall x\in\mathbb{R}$.


There are two tuba players.  They are playing a one -note warm up piece. They both play the same note.  Tuba player A has a 3/4 chance of getting the note right.  Tuba player B has a 4/5 chance of getting the note right.    What is the chance that they both play the right note?

We multiply the individual probabilities to get
\[ \frac{3}{4} \cdot \frac{4}{5}\equal{}\boxed{\frac{3}{5}}.\]

[quote="123456789"]We multiply the individual probabilities to get
\[ \frac {3}{4} \cdot \frac {4}{5} \equal{} \boxed{\frac {3}{5}}.
\]
[/quote]
Sorry this not right, try again please.

[quote="math154"]How is it not correct? :huh: The probabilities are independent, right...?[/quote]
[hide]Well it is more logic than probability.  There is something in the question that makes that answer incorrect.[/hide]

So one has to use the fact that they are guaranteed to play the same note? In that case, the question is contradictory, as the probabilities surely have to be the same of getting the right note if they always play in unison.

[hide="Logic"]
1/4 that only A doesn't do it right, 1/5 that only B doesn't do it right, 1/20 that they both don't do it right. Subtract from 1 to get 1/2.[/hide]

Is the probability of "getting the note right" the probability given that the two play the same note or independent of that fact?

If math154's interpretation is right, the question would be a little better posed as "if they both play the same note, what's the probability that they played the correct one?"

[hide]
In which case it's $ \frac{\frac{3}{5}}{\frac{13}{20}} \equal{} \frac{12}{13}$
[/hide]

Bugi, you overcounted. The chance that only B doesn't do it right is, for example, 3/20. You'll eventually reach the 3/5 answer again.

However, more clarification would still be needed--it is possible that Tuba A could only play the right note and, say, D, and Tuba B could only play the right note and something like E, in which case the probability would be $ 1$.

[quote="ffao"]
Bugi, you overcounted. The chance that only B doesn't do it right is, for example, 3/20. You'll eventually reach the 3/5 answer again.[/quote]

Yes,yes,sorry it should be:
[hide]
only A fails: 1/5
only B fails: 3/20
both fail: 1/20
both well: 12/20=3/5[/hide]

[quote="ffao"]If math154's interpretation is right, the question would be a little better posed as "if they both play the same note, what's the probability that they played the correct one?"

[hide]
In which case it's $ \frac {\frac {3}{5}}{\frac {13}{20}} \equal{} \frac {12}{13}$
[/hide]

Bugi, you overcounted. The chance that only B doesn't do it right is, for example, 3/20. You'll eventually reach the 3/5 answer again.[/quote]
[hide]
You are right.
They both have to play the same note.  Therefor, either two things can happen, either the both get it wrong or both get it right.
There is only one way for them to both get it wrong.
There are 12 ways to get it right.
Hence, 12/13
I guess both of you are right, though, the question would be better posed as ffao said and math154 implied.
It is supposed to be as tricky as possible though. :) [/hide]


inequalities
geometry
inradius
geometry proposed

Given triangle $ABC$ inradius $r$ we known the inequality
$s=\frac{a+b+c}{2}\ge 3\sqrt{3}{r}$
Prove the inequality with weight
Forall $x,y,z>0$
$\frac{x}{y+z}a+\frac{y}{z+x}b+\frac{z}{x+y}c\ge 3\sqrt{3}r$

Actually you can strengthen the RHS up to $\sqrt{3r(4R+r)}$.  :)

That's right treegoner, my inequality is base on the algebraic inequality posted in crux
$\frac{x}{y+z}(b+c)+\frac{y}{z+x}(c+a)+\frac{z}{x+y}(a+b)\ge 3\frac{ab+bc+ca}{a+b+c},\forall x,y,z,a,b,c>0$
If $b+c\rightarrow a,c+a\rightarrow b,a+b\rightarrow c$ then
$3\frac{ab+bc+ca}{a+b+c}=\frac{3r(4R+r)}{s}\ge 3\sqrt{3}r$
Thus you made it stronger
$\frac{x}{y+z}(b+c)+\frac{y}{z+x}(c+a)+\frac{z}{x+y}(a+b)\ge \sqrt{3(ab+bc+ca)}\ge 3\frac{ab+bc+ca}{a+b+c},\forall x,y,z,a,b,c>0$
It is a nice algebraic inequality. :)

An open question
$3\sqrt{3}r\le\frac{x}{y+z}(b+c)+\frac{y}{z+x}(c+a)+\frac{z}{x+y}(a+b)\le \frac{3\sqrt{3}}{2}R$ forall $x,y,z,a,b,c>0$
Surely?


Let M be the set of all primes, and let M be a subset of P with M > 2, such that for any proper subset A of M, the number \pip is in Ap - 1 is in M. Prove that M = P.

The problem you posted here was given at a Romanian selection in 2003. A solution can be found on ajorza's website (you ca find the address in the memberlist of this forum). You'll have to download the files from there.

I do have a solution, but it's rather long because this problem was a tricky one: notice that one would be tempted to take  {p is in M} p - 1 but it says "proper subset" so we can't do that. It's not such a big problem, but it makes the solution longer. I didn't participate in that test, btw.


Amy can mow 600 square yards of grass in $1\frac{1}{2}$ hours.  At this rate, how many minutes would it take her to mow 600 square feet?

[quote="G-UNIT"][hide]

1800ft=1.5hours

divide by 3

600ft  =[b].5 hours or 30 mins[/b] [/hide][/quote]

That's wrong...

[hide]1 sq yd=9 sq ft

1.5 hours=90 minutes
90/9=10.
[b]10[/b] minutes.[/hide]

o, i wasnt sure because in Canada we use the metric system.......

i think the metric system is alot better than the US customary system...you just have to multiply or divide everything by 10 to find another measure.


what integer x satisfies $ (2x)^x\equal{}2985984$


In an acute triangle $ ABC$,the longest altitude $ AH$ has the same length as the median $ BM$.Prove that $ \hat{ABC} \equal{} 60$

Please, check the statement in your problem.

In my drawing with as $ AH$ the longest altitude and $ AH \equal{} BM,$ where $ M$ is the midpoint of $ AC,$ all the angles of the acute triangle $ \bigtriangleup ABC,$ are different than $ \ 60^{o}.$

Kostas Vittas.

This problem is not true! It's only true if ABC is equilateral
Construct K on BC such that MK$ \perp$BC. Then MK=1/2AH=1/2BM. So $ \hat{MBK}$=30.
If $ \hat{B}$=60 then $ \bigtriangleup$ABC is equilateral


Don't know which ones those were, please tell me.

I am trying to load that right now, but I have a dial up modem, and it is not working well at all.

Okay, finally got the problems.

2.
[hide]2 colors. Color A and Color B. Remember to draw in the last vertex on the cube.  Label one vertex A.  All next to it must be B.  Any touching those must be A.  Continue, and you can finish all with only As and Bs.
[/hide]

4.
[hide]3. Since all are [b]less than ten[/b] and triangle=(5/2)hexagon, triangle=5 and hexagon=2. For the last equation, 10=trapazoid + 2 >> trapazoid=8.  For first equation, 5 + square =8 >> square=3.
[/hide]

5.
[hide]8. In general n+(n+2)+(n+4)+(n+6)=4n+12. Since n is odd, 4n= 4 mod 8. 12=4 mod 8. So, 4n+12 = 0 mod 8, and the sum must be divisible by 8. 1+3+5+7=16 and 3+5+7+9=24, so it cannot be a larger number.  [/hide]

8.
[hide]657. First, Chantel gets 1. Then, John gets 1002. Now there are 1002.  Next, Mike gets 501. Now there are 501.  Chantel gets 1 and John gets 250. Now there are 250.  Mike gets 125. He now has 626.  Now there are 125. Chantel gets 1.  John gets 62. Now there are 62.  Mike gets 31. He now has 657.  Chantel gets 1. John gets 15. Now there are 15. Chantel gets 1 and John gets 7. Chantel gets 1 and John gets 3. Chantel gets 1 and John gets 1. Chantel gets the last one. Mike ended up with 657.[/hide]

[hide="2"]Remember, these are VERTICES. If you unfold them, you get a grid.
And with grids, you get checkerboards! So 2 colors[/hide]

[hide="4"]THis is just substitution.
In 3rd Equation, plug in 1st equation for the pent.
Plug that into the 2nd equation (left side)
You get tri+squ=4hex=trap
3+5=4*2=8
5+5=2+8 (2tri=trap+hex)
squ=3
>>3<<[/hide]

[hide="5"]Just list out the sums of 4 consec. odds:
16, 24, 32,...
They're all mult. of 8
>>8<<[/hide]

[hide="8"]Make a table: (At the end of round...)
1->1002 left, mike=0
2->501 left, mike=501
3->250 left, mike=501
4->125 left, mike=626
5->62 left, mike=626
6->31 left, mike=657
7->15 left, mike=657
and from then on, It just stays odd.
>>657<<(I think)[/hide]

Good. That means perfect target score! :)  :)  :)  :lol:  :lol:  :lol:

[color=indigo]I made a really stupid mistake on #2 and somehow got [/color][color=green]4 :( [/color][color=indigo]. Usually I'm good at target but I missed this one and some other one.[/color]

I dunno how, but I missed the one that had the multiples of 8, and put FOUR, because I added 21, 23, 25, and 27 to to get 100 =/

i missed the same problem the same way! wait, does target count the same amount as sprint? this guy in our district missed like 3 on target and got like 1 or 2

[quote="theaznkungfu"]# 2   was a countdown problem a couple year back, so our whole team knew the answer right away :lol:[/quote]

There was a countdown problem very similar this year also.

Targets are worth 2 pts. per ques.  Sprint are worth 1.

Yeah the targets were easy compared to last year.. I got 'em all right... but don't expect anything so easy for the states.. I mean last years' weren't exactly a walk in the park... still for this year's target round, you either knew the problem or screwed the entire thing up... and of course there were the stupid errors... I think three or four of them were either recognizing a pattern or seeing simple algebra...


function
modular arithmetic
algorithm
number theory
Euclidean algorithm

Well, I was wondering how you could find the multiplicative inverse of a number in some specific mod. For example, say if you needed to find what, uh, say $ 47^{999}$ was in (mod 100). From the phi function, you reduce it to $ 47^{39}\pmod{100}$, but to find that, you need the multiplicative inverse.

Help?

Finding the inverse of $ a \bmod b$ is equivalent to solving the linear Diophantine $ ax \plus{} by \equal{} 1$.  (The inverse is then $ x$.)  This is done with [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]the extended Euclidean algorithm[/url].

Demonstration:

$ 100 \equal{} 2 \cdot 47 \plus{} 6$
$ 47 \equal{} 7 \cdot 6 \plus{} 5$
$ 6 \equal{} 1 \cdot 5 \plus{} 1$.

This gives us $ 1 \equal{} 6 \minus{} 5 \equal{} 6 \minus{} (47 \minus{} 7 \cdot 6) \equal{} 8 \cdot 6 \minus{} 47 \equal{} 8 \cdot (100 \minus{} 2 \cdot 47) \minus{} 47 \equal{} 8 \cdot 100 \minus{} 17 \cdot 47$.  We therefore have $ 47^{\minus{}1} \equiv \minus{}17 \equiv \boxed{83} \bmod 100$.

By way of verification, $ 47 \cdot 83 \equal{} 3901$.

A sometimes faster way is to look for finding the inverse of $ a^b \pmod{n}$ is to assume the solution is of the form $ a^c$ and then looking at $ a^b*a^c \equal{} 1 \equal{} a^{\phi(n)}$.  Then $ b\plus{}c \equal{} \phi{n}$ so $ c \equal{} \phi(n) \minus{} b$.  Thus the inverse is $ a^{\phi(n)\minus{}b}$.

In your case this is particularly nice since $ \phi(100) \equal{} 40$:

$ 47^{39}*47^c \equiv 1 \equiv 47^{40}$ so $ c\equal{}1$ so $ 47$ is the inverse you're looking for.

And indeed, this is the same as $ 83^{39}$, but its way easier to work with.

xpmath wants to calculate $ 47^{39} \bmod 100$, not its inverse.  That requires finding $ 47^{ \minus{} 1}$.  (And note that we can replace $ \varphi$ by the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url]; here $ \lambda(100) \equal{} 20$.)

[b]t0rajir0u[/b]'s method is a special case of [url=http://en.wikipedia.org/wiki/Bezout%27s_identity]Bézout's identity[/url].


in the second row:2
in the third row:2
in the fifth row:2

so in the 2^x+1 row 2 odds if x is an integer.

2 odds.

[quote="math92"]In Pascal’s Triangle, how many odd numbers will there be in the 129th row?[/quote]

[hide]

As Hamster noted, the top row is row 0 but ordinally it is the 1st row. Baring that in mind, the coefficients of the 129th row are given by: $\binom{128}{n} | n\in\{0,1,2,3\dots , 128\}$

Except for $\binom{128}{0} = 1$ and $\binom{128}{128}=1$, $\binom{128}{n}\equiv 0\mod{2}$ (That means all of them are even.)

Therefore, 2 is how many odd numbers there are in the 129th row of Pascal's Triangle.
[/hide]

[quote="Hamster1800"]hold on! there's a little thing about this problem which need to be clarified. The top row of pascals triangle is defined as row 0, so is the second row 11 or 121?[/quote]
I think that row 129 was supposed to mean the 129th row.  In mathcounts there was a pascals triangle problem that defined the fourth number in row 8 as the 8th row.

I found this cool thing about Pascals triangle:
If instead of writing the numbers, you fill in a black dot where an ever number should be and leave the odd numbers blank,  it will shoe Sierpinski's triangle


Obviously MathCamp considers much more than how many questions you solved on the qualifying quiz to determine admission, but I'm curious to know the distribution of how many questions were answered (at least by AoPSers).

I solved all 10, and wrote them out somewhat roughly on a piece of paper, but decided not to apply...

[quote="cosinator"]I wrote up solutions to all 10, but I think only 7 or 8 are fully correct.[/quote]

[quote="iostream.h"]ha, i procrastinated a bit.  i wrote something down for all of them, but i only think i have good, rigorous solutions to 8 of them.

hopefully the threshold for graduating seniors isn't too high.[/quote]

I'm just an alumni, not a staff member, but unless you completely forgot the rest of the application, 7-8 problems correct should easily get you in.

I solved everything but 5a, and my solution to 6 is a bit unrigorous.

I think I had decent solutions for everything except 5b and 8.

I solved seven, though it appears that one is incorrect.

i think i have rigorous solutions to 5. Decent solutions to other 2.

I solved 1 through 9, and made progress on 10.

I just realized that this sort of gives away information to people who are still working on the problems, so let's continue this via PM (and edit your post too).

Out of curiosity, does anyone know when they usually tell us if we got in?

Last year (when I applied and did not get in :-( ), it was a few weeks after the deadline (I have no idea when the deadline was, but I got my answer on 2007-05-03). I remember reading that it was earlier than usual, though. My guess would be late May (around 2007-05-20, probably).

Last year, the deadline was April 26, according to the transcript of last year's math jam. So we might even get results this week.

[color=darkblue]I solved all problems except problems 5 and 10.  I have attempted both of them and checked the 8 I did solve for fallacies; there are none.  I did, however, prove problem #2 using a GNU Fortran program to the number [/color][color=darkred]*******[/color][color=darkblue] and tackling the tail analytically.  My friends have told me that my personal statement is a killer; do you think that I will get in? (There is the chance that my academic recommender was lazy and didn't send a letter in; that would preclude me from attending.)[/color]

[quote="E^(pi*i)=-1"]
I'm just an alumni, not a staff member, but unless you completely forgot the rest of the application, 7-8 problems correct should easily get you in.
[/quote]

So yeah, probably. I am not even an alumnus, but I imagine that recommendations are important. However, I wouldn't be too certain there are no fallacies in your argument, not so much because I doubt your ability (I don't know you; you may be a genius), but once you come up with something, you tend to think it's correct.

I got all 10, though I had to resort to some not-so-mathematical methods for number 10, as in I used a calculator at one point.

I turned in solutions for all problems except #8. I must have really missed something, but I just could not think of a method to approach the problem. My answer on #5 is a little shaky, but it must have done alright because I got in. I solved #10 on a gut feeling and it turned out to be correct which amazed me. I derived a lot of equations for that problem to use as constraints. I think I spent the longest time on #7 because I misread the question. I wish they would have let me know which questions I got right  :(


Stie cineva vreun link/o idee de rezolvare a acestei probleme la nivel de cam clasa VIIIa?
$\forall n>1,a_1, a_2,$...$,a_{2n-1} \in \mathbb{Z}$,$\exists i_1,i_2,$...$,i_n \in \overline{1,2n-1}$ a.î. $a_{i_1}+$...$+a_{i_n} \equiv 0(\mod n)$?

problema este cunoscuta si nu este deloc usoara...
un link util : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=14797[/url]


Kind of lame...i give a $ \frac{3}{10}$.

I don't know how much credit to give your ranking.

How much did you get out of it?

I need specifics from you.

Well I understand the joke...but I just don't think the punchline is very strong.


inequalities trigonometry geometry LaTeX

If $ AD,BE,CF$ are the angle bisectors of $ \angle A,\angle B,\angle C$ of $ \triangle ABC$ respectively, Prove that
$ (a\plus{}b\plus{}c)\sqrt{3}\ge2(AD\plus{}BE\plus{}CF)$

Let $ BC\equal{}a,CA\equal{}b,AB\equal{}c,AD\equal{}l_a,BE\equal{}l_b,CF\equal{}l_c$
We have: $ l_a\equal{}\frac{2\sqrt{bcp(p\minus{}a)}}{b\plus{}c}\le\sqrt{p(p\minus{}a)}$
Likewise, $ l_a\plus{}l_b\plus{}l_c\le\sqrt{p(p\minus{}a)}\plus{}\sqrt{p(p\minus{}b)}\plus{}\sqrt{p(p\minus{}c)}\le\sqrt{3p^2}$
Therefore, we have $ 2(l_a\plus{}l_b\plus{}l_c)\le\sqrt3(a\plus{}b\plus{}c)$ as desired

Kyoshiro's solution is correct. :coolspeak: But I have a challenge..........
[hide=challenge]
Let the bisector of $ \angle C$ of a triangle meet $ AB$ at $ D$. Show that $ CD^2<AC\cdot BC$            :lol:
NB: usage of some trigonometry is needed for my solution

[quote="Potla"]Kyoshiro's solution is correct. :coolspeak: But I have a challenge..........
[hide=challenge]
Let the bisector of $ \angle C$ of a triangle meet $ AB$ at $ D$. Show that $ CD^2 < AC\cdot BC$            :lol:
NB: usage of some trigonometry is needed for my solution[/quote]
Applying AM-GM, we get  $ 4p(p\minus{}c)<(p\plus{}p\minus{}c)^2\equal{}(a\plus{}b)^2$ (the equality doesn't occur)
So that, we have $ l_c^2\equal{}\frac{4abp(p\minus{}c)}{(a\plus{}b)^2}<ab$ as desired.
:lol:

How did you get this?
 $ l_a\equal{}\frac{2\sqrt{bcp(p\minus{}a)}}{b\plus{}c}$

[quote="IW@IT"]How did you get this?
 $ l_a \equal{} \frac {2\sqrt {bcp(p \minus{} a)}}{b \plus{} c}$[/quote]

Try using [url=http://en.wikipedia.org/wiki/Stewart's_theorem]Stewart's theorem[/url] on angle bisectors.

[quote="Kyoshiro"]
[quote="potla"]
Kyoshiro's solution is correct.  :coolspeak:  But I have a challenge..........
[hide="challenge"]
Let the bisector of $ \angle C$ of a triangle meet $ AB$ at $ D$ . Show that $ CD^2 < AC.BC$  :lol:
NB: usage of some trigonometry is needed for my solution [/quote]
Applying AM-GM, we get $ 4p(p - c) < (p + p - c)^2 = (a + b)^2$(the equality doesn't occur),
So that, we have $ l_c^2 = \frac {4abp(p - c)}{(a + b)^2} < ab$ as desired.
 :lol: [/quote]
Wow! You have found a very good solution! I did not see it, but did it in a bit different way.
[hide]We have $ \Delta ABC = \Delta ADC + Delta BDC$ where $ \Delta$ denotes area of the triangles.
Therefore $ \frac {1}{2} AC\cdot BC\cdot sin\angle ACB = \frac {1}{2}\cdot AC\cdot AD\cdot sin\angle ACD + \frac {1}{2}\cdot BC\cdot CD\cdot sin\angle DCB$.......................(1)

CD is the bisector of [u]/  [/u]ACB, hence
$ \angle ACD = \frac {\angle ACB}{2} = \angle DCB$.

using this (1) gives
$ AC\cdot BC\cdot sin\angle ACB = (AC + BC)CD\cdot sin\frac {\angle ACB}{2}$ ............(2)

We know that sinA=2sinA/2.cosA/2.

Using this we can write (2) in the form of
$ 2AC\cdot BC\cdot sin\frac {\angle ACB}{2}\cdot cos\frac {\angle ACB}{2} = (AC + BC)CD sin\frac {\angle ACB}{2}$.

or, $ \frac {1}{CD}\cdot cos\frac {\angle ACB}{2} = \frac {1}{2}(\frac {1}{AC} + \frac {1}{BC})$.................(3)

As [u]/   [/u]ACB/2 is an acute angle 0<cos[u]/  [/u]ACB/2 <1

This gives:
$ \frac {1}{CD} > \frac {1}{2}(\frac {1}{AC} + \frac {1}{BC}) \ge (\frac {1}{AC}\cdot\frac {1}{BC})^{\frac {1}{2}}$
from the AM GM ineq.

Hence $ CD^2 < AC\cdot BC$ is proved

As you see it is too long and too tiring to write in Latex, I have not written some easy parts;still I did my best or else it would be called an "ununderstandable or undecodable post"    ......

 :lol: :spider:
[/hide][/hide]


The set of all solutions of the system

\[ x\plus{}y \le 3\]
\[ 2x\plus{}y \ge 2\]
\[ x \ge 0\]
\[ y \ge 0\]

is a trapezoidal region. Find the number of units in the length of the longest side. Express your answer in simplest radical form.

ugh gotta learn asymptote

Anyways, a good start would be to graph the lines.
  [geogebra]80bbdb4524de08d48497f14f7c46c4d851cd6a9d[/geogebra]

The "trapezoidal (not really)" region is the section in the first quadrant. Obviously, the longest side is the hypotenuse of the 45-45-90 triangle. Its length is ${ \boxed{3\sqrt2}}$.

huh? is there a way to do it without a graph?
thx


conics
superior algebra
superior algebra solved

Show that $X^2 + Y^2 - Z^2 \in \overline{K}[X,Y,Z]$ is irreducible, where $\overline{K}$ denotes an algebraically closed field of characteristic $0$.

Since the field is algebraically closed, it is enough to prove that $X^2+Y^2+Z^2$ is irreducible.
If it was reducible, then it would split as a product of two linear factors: $X^2+Y^2+Z^2=(aX+bY+cZ)(AX+BY+CZ)$.
This gives the system: $aA=bB=cC=1$ and $aB+Ab=aC+Ac=bC+Bc=0$ which implies $A^2+B^2=B^2+C^2=C^2+A^2=0$. From this we get $A=B=C=0$ which is impossible since $aA=bB=cC=1$.

Some other very quick proofs are:
1. $X^2+Y^2+Z^2$ is the only smooth conic in $\mathbb P^2_{\mathbb C}$.
2. The circle is not the union of two lines.


analytic geometry
ratio
combinatorics unsolved
combinatorics

What maximal number of colors can we use to color the squares of a $ 2n\times 2n$ square board so that each row or column contains squares of at most $ n$ colors?

[hide="Is this right?"]I think it's n^2. Take some square with coordinates (x,y). There must be some square with the same color with coordinates (x,a) and another square with coordinates (b,y). It is only necessary to color the square with coordinates (b,a) that same color to satisfy the conditions. This can be done with any color and any initial square. So there are at least 4 squares of the same color for each color, and since there are 4n^2 squares, there are n^2 colors minimum.
If we color the grid as a checkerboard, with each color occupying a 2x2 square, we see that this is achievable.[/hide]

Uh less squares/color=more colors...that is the maximum (or what I think it is).
I can rigorize it if anyone wants.

I think your answer is right, but your proof isn't.  In particular,
[quote="serialk11r"]Take some square with coordinates (x,y). There must be some square with the same color with coordinates (x,a) and another square with coordinates (b,y).[/quote] is just wrong.  I'd like to see what you think the rigorized version is.

Haha I know it's wrong, it was just my path of thinking, that's all.
Okay so:
[hide]
If there are only n colors for 2n squares, then there must be at least 2 squares for each color in each row or column.
Suppose that some square with coordinates (x,y) is colored with a color that does not appear at least 4 times. We know that there exists squares (x,b) and (a,y) with a and b not equal to x and y, respectively, that are colored with the same color as square (x,y). We know that there must be some square on the same column as (a,y) that is the same color as the previous squares indicated. However, this square is not one of (x,y), (a,y), or (y,b), since that square is on column a, and cannot be (a,y) by definition, we have another square outside of the first 3, and 3+1=4, a contradiction.
We see that it is possible to use n^2 colors, if we arrange all the squares with the same color into 2x2 squares respectively.
[/hide]

[quote="serialk11r"]
If there are only n colors for 2n squares, then there must be at least 2 squares for [b]each[/b] color in each row or column.[/quote]

I think this is wrong...

Maybe my argument isn't rigorous.
Consider the sum of the ratios: $ \frac {numberofdifferent colors}{n}$ for each row and column. Each of the terms is $ \leq 1$ so the whole sum is $ \leq 4n$. This sum can be seen in the following way: is the sum over each color of the # of different rows and different columns in which the color appears over $ n$. In each row and column the average of appearances is $ \geq 2$ (this means that for each row the average (per color) of the number of columns with it (average of the averages) will be $ \geq 2$) then the average of appearances  is at least $ 4$. If we had $ n^2 \plus{} 1$ or more colors. Then the considered sum $ \geq (n^2 \plus{} 1)4 > 4n^2$ contradiction.

[quote="Nemion"]. This sum can be seen in the following way: is the sum over each color of the # of different rows and different columns in which the color appears over $ n$. [/quote]

Please explain this....

The sum over the number of different colors that appear in each row and column (sum over all rows and columns) is the same as the sum over the number of different rows and columns in which each color appears (the sum of all these over all colors).

Sorry for being so vague


inequalities
probability
function
integration
conics
parabola
linear algebra

[color=orange][size=150]Open Competition for University Students[/size]
of The Faculty of Mechanics and Mathematics of Kyiv National Taras Shevchenko University[/color]

3.03.2004
[color=indigo]Problems for 1-2 years students[/color]

[b]1.[/b] Prove that for every positive integer $n$ an inequality $\dfrac1{3!}+\dfrac3{4!}+\ldots+\dfrac{2n-1}{(n+2)!}<\frac12$ holds.
(R. Ushakov)

[b]2.[/b] One cell is erased from the $2\times n$ table in arbitrary way. Find the probability of the following event: It is possible to cover without overlappings the rest of the table with figures $\begin{picture}(17, 17) \put(0, 0){\line(1, 0){16}}\put(0, 8){\line(1, 0){16}}\put(0, 0){\line(0, 1){16}}\put(8, 0){\line(0, 1){16}}\put(0, 16){\line(1, 0){8}}\put(16, 0){\line(0, 1){8}}\end{picture}$ (with any orientation).
(A. Kukush)

[b]3.[/b] For any continuous convex at $[0, 1]$ function $f$ prove the inequality $\dfrac25\int_{0}^{1}f(x)\, dx+\dfrac23\int_{0}^{3/5}f(x)\, dx\ge \int_{0}^{4/5}f(x)\, dx.$
(A. Prymak) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=572963#572963][color=olive]link to discussion/solution[/color][/url]

[b]4.[/b] Find all odd continuous functions $f: \mathbb{R}\to\mathbb{R}$ such that the identity $f(f(x))=x$ holds for every real $x$.
(A. Kukush)

[b]5.[/b] Construct with the compass and the ruler the circle of maximal radius which lies inside the given parabola and touches to it in its vertice.
(Zh. Chernousova)

[b]6.[/b] Let $A$,  $B$,  $C$,  $D$ be (not necessarily square) real matrices such that $A^{T}=BCD$, $B^{T}=CDA$, $C^{T}=DAB$, $D^{T}=ABC$. For $S=ABCD$ prove that  $S^{3}=S$.
(V. Brayman) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=566609#566609][color=olive]link to discussion/solution[/color][/url]

[b]7.[/b] Denote by  $A_{n}$ the maximum of the determinant of an $n\times n$ matrix with entries $\pm1$. Does there exist a finite limit $\lim\limits_{n\to\infty}\sqrt[n]{A_{n}}$?
(A. Bondarenko) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=565034#565034][color=olive]link to discussion/solution[/color][/url]

[b]8.[/b] Let $\{x_{n}, \, n\ge1\}$ be a sequence of positive numbers which contains at least two distinct elements. Prove or disprove that
$\liminf\limits_{n\to\infty}(x_{1}+\dots+x_{n}-n\sqrt[n]{x_{1}\dots x_{n}})>0.$
(D. Mitin)

[b]9.[/b] Some permutation of entries of a matrix maps any non-singular $n\times n$ matrix into non-singular one and maps identity matrix into itself. Prove that this permutation preserves the determinant of the matrix.
(V. Brayman) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=577789#577789][color=olive]link to discussion/solution[/color][/url]

[b]10.[/b] A rectangle with side lengths  $a_{0}, b_{0}$ is dissected into smaller rectangulars with side lengths $a_{k}, b_{k},\, 1\le k\le n$. The sides of smaller rectangulars are parallel to the corresponding sides of the big rectangular. Prove that $\mid \sin a_{0}\sin b_{0}\mid \le \sum_{k=1}^{n}\mid \sin a_{k}\sin b_{k}\mid .$
(Zsolt P\'ales, Hungary)[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=561683#561683][color=olive] link to discussion/solution[/color][/url]

[color=indigo]Problems for 3-4 years students[/color]

[b]1.[/b] Let random variable $\xi$ be distributed as $\mid \gamma\mid^\alpha$, $\alpha\in\mathbb{R}$, where $\gamma$ is standard normal variable. For which $\alpha$ there exists an expectation $M\xi$?
(U. Mishura)

[b]2.[/b] One cell is erased from the $2\times n$ table in arbitrary way. Find the probability of the following event: It is possible to cover without overlappings the rest of the table with figures $\begin{picture}(17, 17) \put(0, 0){\line(1, 0){16}}\put(0, 8){\line(1, 0){16}}\put(0, 0){\line(0, 1){16}}\put(8, 0){\line(0, 1){16}}\put(0, 16){\line(1, 0){8}}\put(16, 0){\line(0, 1){8}}\end{picture}$ (with any orientation).
(A. Kukush)

[b]3.[/b] Normed space $Y$ is said to be strongly normed if for every $y_{1},y_{2}\in Y$ the equality $\| y_{1}\| =\|y_{2}\| =\left\| \tfrac{y_{1}+y_{2}}2 \right\|$ implies $y_{1}=y_{2}$. Let $X$ be normed space, let $G$ be subspace of $X$ and let the adjoint space $X^{*}$ be strongly normed. Prove that for every functional from $G^{*}$ there exists the unique extension $X^{*}$ which preserves the norm.
(O. Nesterenko) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=581487#581487][color=olive]link to discussion/solution[/color][/url]

[b]4.[/b] Let $R(z)=\tfrac{z^{2}}2-z+\ln(1+z)$, $z\in\mathbb{C}$, $z\ne-1$. Prove that for every real $x$ the following inequality holds: $\mid R(ix)\mid \le\tfrac{|x|^{3}}3$.
("ln" means the value of logarithm from the branch with $\ln\,1=0.$)
(D. Mitin) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=581480#581480][color=olive]link to discussion/solution[/color][/url]

[b]5.[/b] Let $A$, $B$, $C$, $D$ be (not necessarily square) real matrices such that $A^{T}=BCD$, $B^{T}=CDA$, $C^{T}=DAB$, $D^{T}=ABC$. For $S=ABCD$ prove that $S^{2}=S$.
[i]Remark: for 1-2 years students it is proposed to prove that $S^{3}=S$.[/i]
(V. Brayman) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=566609#566609][color=olive]link to discussion/solution[/color][/url]

[b]6.[/b] Let $e$ be nonzero vector in $\mathbb{R}^{2}$. Construct a non-singular matrix $A\in\mathbb{R}^{2\times2}$ such that for $f_{d}(x): =\|A(x+d)\|^{2}$, $x,d\in\mathbb{R}^{2}$ the set of pairs
$M: =\{(x,y): \ f_{e}(x)=1,\ f_{-e}(y)=1$ and there exist real numbers $\lambda, \mu$ such that $(x,y)$ is a stationary point of Lagrange function $F(x,y): =\|x-y\|^{2}+\lambda f_{e}(x)+\mu f_{-e}(y)\}$ contains at least 8 pairs of points.
(A. Kukush)

[b]7.[/b] Some permutation of entries of a matrix maps any non-singular $n\times n$ matrix into non-singular one and maps identity matrix into itself. Prove that this permutation preserves the determinant of the matrix.
(V. Brayman) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=577789#577789][color=olive]link to discussion/solution[/color][/url]

[b]8.[/b] The croupier and two players are playing the following game. The croupier chooses an integer in the interval $[1,2004]$ with uniform probability. The players guess the integer in turn. After each guess the croupier informs them whether the chosen integer is bigger or smaller or has just been guessed. The player who guesses the integer first wins. Prove that both players have a strategy such that their chances of winning are at least $\frac{1}{2}.$
 (S. Shklyar)

[b]9.[/b] A rectangle with side lengths  $a_{0}, b_{0}$ is dissected into smaller rectangulars with side lengths $a_{k}, b_{k},\, 1\le k\le n$. The sides of smaller rectangulars are parallel to the corresponding sides of the big rectangular. Prove that $\mid \sin a_{0}\sin b_{0}\mid \le \sum_{k=1}^{n}\mid \sin a_{k}\sin b_{k}\mid .$
(Zsolt P\'ales, Hungary)[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=561683#561683][color=olive] link to discussion/solution[/color][/url]

[b]10.[/b] Does there exist a sequence $\{x_{n},\,n\ge1\}$ of vectors from $l_{2}$ with unit norm which satisfy $(x_{n},x_{m})<-\frac1{2004}$ for $n\ne m$, $n,m\in\mathbb{N}$?
(A. Bondarenko) [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=581481#581481][color=olive] link to discussion/solution[/color][/url]

Part 1-2 years students

[hide="solution 1"]

$\sum_{1}^{\infty} \frac{2n-1}{(n+2)!}=\sum_{1}^{\infty} \frac{(2n+4)-5}{(n+2)!}=1-3( \sum_{3}^{\infty} \frac{1}{n!})=1-3(e-2.5)=8.5-3e <0.4$

[/hide]

[hide="solution 2"]
If $3\not| n-2$ the probability is zero. (trivial)

if $n=3k+2$ the probability is $\frac{2k+2}{3k+2}$

Proof:

Divide the triangle in one $2*2$square and $k$ $3*2$rectangles , if the erased cell is in the $2*2$, it is solved.
( this is possible iff the erased cell isn't in the $3j^{th}$ column for some $j$)
If not, it isn't possible.

[/hide]

[hide="solution 4"]

$f$ is clearly injective and $f(0)=0$.

Now we look to $f(a)$ for $a>0$ :

$f(a)>a$ means $f(f(a))>a$ by injectivity.
$0<f(a)<a$ means $f(f(a))<f(a)<a$ by the conditions.
$-a<f(a)<0$ means $f(-a)<a$v and $f(f(a))$ is between $f(0)$ and $f(-a)$, hence it isn't $a$.
$f(a)<-a$ and $f(-a)>a$ means $f(f(a)) > f(-a)>a$ by the conditions.

Hence $f(a)= \pm a$.
Now by injectivity and continuity we have this holds over the whole domain.

it is easy to check the functions $f(x)=-x$ and $f(x)=x$ satisfy the question and hence are the only two solutions.

[/hide]

[hide="solution 5 ( probably not complete)"]

Firsts if $y=ax^2$ is the parabola after moving it to the center. $(a>0)$

We say the center of the circle is $(0,y)$.
Then $x^2+(y-ax^2)^2 \ge y^2$ or $a^2x^4+(1-2ay)x^2 \ge 0$ this will mean $1-2ay \ge 0.$
Hence $y \le \frac{1}{a}.$

If we are given the normal coordinate axes:

we can find the vertex of the parabola ( parallel line cutting parabola in $2$ points, perpendicular line through the midpoints of those $2$ points..)
Then we can draw the coordinate axes, so we have the parabola $y=ax^2$ in our new cartesion coordinatze system.

Now we can draw a bissector intersecting the parabola in $( \frac 1a, \frac 1a )$ and then the projection on the y-axis gives our desired center of the circle.

If not: all we need to find is the vertex ( if it is given, it is similar solved) but when we just have the parabola drawn on an empty page, it seems to be yet a crucial idea to find it.

[/hide]

part for 3-4 year students

[hide="solution 2"]

see the part for 1-2 year students

[/hide]

[hide="solution 8"]

We will solve it for the general case :
The croupier chooses an integer in the interval $[1,N]$ with uniform probability.

We prove the following facts [b] $(IH)$[/b]:

If $N=2k$ for some natural $k$, the probability of winning is $\frac{1}{2}$ when both players play with optimal strategy.
if $N=2k+1$, the first player can make he win with a chance of $\frac{k+1}{2k+1}$ and the second with a probability of $\frac k {k+1}$.

[b]$(IB)$[/b]

For $N=1,2$ this is trivial.

Induction step:

It is proven for $N=1,2 \cdots, 2k$.
We will prove it for $N=2k+1$.

If the first player chose an odd number $t$.
By the induction hypothese  he has a probability to win of $\frac{t-1}{2k+1}* \frac{1}{2}+\frac{1}{2k+1}+\frac{2k+1-t}{2k+1}* \frac12 = \frac {1+k} {2k+1}$
If the first number is an even number $t$,
he wins with a chance of $\frac{t-1}{2k+1}* \frac{\frac t2 -1}{t-1}+\frac{1}{2k+1}+\frac{2k+1-t}{2k+1}* \frac{ k-\frac t2}{2k+1-t}= \frac{k}{2k+1}$

hence if he plays optimal by chosing an odd $t$, the given claims are proven for $2k+1.$

Similar we can prove now for $2k+2$ the probability to win
 is $\frac 12$ ( which doesn't depend on his first move).

By complete induction, the IH holds for every $N$ and the Original problem is solved as well.
[/hide]


How many prime numbers are there less than 300?


Prove that $\log_{a}\frac{c}{b}+\log_{b}\frac{a}{c}+\log_{c}\frac{b}{a}>0$ for $1<a<b<c.$

This problem doesn't make sense to me.  What are we proving?

EDIT: Problem fixed.

[hide="Solution"]
Let $x = \ln a, y = \ln b, z = \ln c$ are positive reals with $z > y > x$.  We wish to show

$\frac{z-y}{x}+\frac{x-z}{y}+\frac{y-x}{z}> 0$

Let $a = \frac{y}{x}, b = \frac{z}{y}$ be positive reals $>1$.  Then the problem can be rewritten

$ab-a+\frac{1}{a}-b+\frac{1}{b}-\frac{1}{ab}> 0$
$(a-1)(b-1)+(1-\frac{1}{a})(1-\frac{1}{b}) > 0$

Which is true (all of the factors involved are positive).  QED. [/hide]

That's right, t0rajir0u.

You can prove directly as follows.

After setting as you did, $\frac{z-y}{x}+\frac{x-z}{y}+\frac{y-x}{z}=\cdots=\frac{(y-x)(z-x)(z-y)}{xyz}>0.$


The numerator of a fraction is 3 less than the denominator. If the numerator is tripled and the denominator is increased by 7, the value of the resulting fraction is 3/2. Find the original fraction.


Which of the following atoms is never found with more than octet of electrons around it: C,P,S,Br? Why?

small atoms cannot have too many ligands (bonds, pairs of electrons, etc). I have an outline on the reasons of this somewhere, but a few of them are here:

"crowding" effect - there is just not enough physical space for the ligands

too few orbitals - hybridization allows for bonding; with s and p mixing, only four ligands may be present (sp3). for higher coordination, d orbitals must be introduced for sp3d, sp3d2 and so on.

electron pair repulsion - remember coulomb's law: like repels like. since there are too many ligands, there are too many electron pairs close to each other. thus, the pairs repel each other: a thermodynamically unstable and unfavorable configuration.

A fact that one should know is d orbitals are energetically available to bond for atoms larger than Si; in other words, as soon as the 3p orbitals are filled with one electron each, the d orbitals are available.

To answer the question, C almost always forms 4 bonds. The other atoms can have extended octets. For example, P can form PCl3 and PCl5, S can form SF6, and Br can form BrO4- anion.

[quote="JRav"]I haven't taken chemistry in over a year but I believe the answer is carbon.  When carbon bonds, it needs to have a complete octet.  I'm pretty sure sulfur and phosporus can, but I'm not too sure about bromine.[/quote]

Funny thing is quite a while ago I posted a problem concerning structure of a class of compounds called hyperlithiated compounds (e.g. $ \ce{CLi6}$). Carbocation chemistry has also revealed the existence of pentavalent carbon intermediates such as $ \ce{CH5\plus{}}$.


Given a triangulation of a polygon, call two vertices adjacent if they are connected by an edge of a triangle. What is the maximum number of colors we need to use in order to guarantee that no two adjacent vertices have the same color?

Followup: find an algorithm to color such a polygon.

[quote="Phelpedo"]Given a triangulation of a polygon, call two vertices adjacent if they are connected by an edge of a triangle. What is the [b]maximum [/b]number of colors we need to use in order to guarantee that no two adjacent vertices have the same color?

Followup: find an algorithm to color such a polygon.[/quote]Isn't this supposed to be 'minimum'? Also, am I correct in understanding that a triangulation of a polygon is where you draw all possible line segments connecting two points?

A triangulation is just any partition of a surface into triangles.

I should clarify- what is the maximum number of colors that is [b]needed[/b], i.e. the minimum number of colors that will suffice for any situation.

Coloring the vertices is the same as coloring the faces of its dual. That is the four color theorem, for which I don't think there is a nice proof. If you have a solution in mind, can you post a hint?

Edit: Never mind, this is just a specific case of the four color theorem, since the vertices of the dual are of degree three. So the upper bound is four. I guess if you can find a polygon that cannot be colored with three, you're done.

[hide="hint"]
Actually, 3 colors always suffice  :wink:  There is a nice proof (try induction)
[/hide]


The “pigeon” version of the pigeonhole principle states that if there are h holes and p pigeons in the holes and h < p, then there must be at least two pigeons in one hole. That is, if there is a mapping between two finite sets of unequal size, then at least one element in the smaller set must be paired with more than one element in the larger set.

The extended pigeonhole principle says that if p > nh for some integer n, then at least one hole will contain n + 1 pigeons. Informally, the most even distribution of pigeons assures this result and larger populations within a hole are possible (note that some specific holes may be empty—we are only making a claim about the existence of a more crowded hole).

[quote="neelnal"]What is this thing people are refering to as pigeonhole?

Is it a real world math concept? or an AOPS saying?[/quote]

It's just a rather interesting name for a theorem. If you have x+1 objects and you want to put them in x boxes, somewhere there will be two objects in one box. The "pigeon" part is probably because the original problem was about pigeons and pigeonholes.

What "original problem" are you talking about? This is one of the simplest useful concepts in proofs. It must have been around for millenia, as it is actually "common sense" (I found myself using it in daily life once before I learned it)
Actually, on the wikipedia article I read a while back it said something like pidgeonholes hold ink in the desks in England so it is common... or something like that.
It is not just an AoPS saying


1:实数$a_1,a_2\cdots a_k$ 满足 $a_1+a_2+\cdots+a_k=0$.求证:
\[ \max_{1 \le i \le k} (a_i^2)\leq \frac{k}{3}\cdot[(a_1-a_2)^2+(a_2-a_3)^2+\cdots+(a_{k-1}-a_k)^2] \]
2:正整数$a_1,a_2 \cdots a_{2006}$满足$\frac{a_1}{a_2},\frac{a_2}{a_3} \cdots \frac{a_{2005}}{a_{2006}}$两两不同，求这$2006$个数中至少出现多少个不同的正整数？

3:正整数$m$，$n$，$k$满足$mn=k^2+k+3$，求证方程$x^2+11y^2=4m$及$x^2+11y^2=4n$至少有一个有奇数解。

4:在Rt△ABC中，∠ACB=90°，其内切圆O分别于边BC、AC、AB相切于点D、E、F。连AD与内切圆O相交于点P，连BP、CP，若∠BPC=90°。求证AE+AP=PD。

5:实数列$\{a_n\}$满足：$a_1=1/2, a_{k+1}=-a_k+\frac{1}{2-a_k},k \in N_+$。证明:
\[ \left( \frac{n}{2(a_1+a_2+\cdots+a_n)}-1 \right)^n \leq \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n\left(\frac{1}{a_1}-1\right)\cdots\left(\frac{1}{a_n}-1\right) \]
6:设$X$是一个$56$元集合，求最小的正整数$n$，使得对$X$的任意$15$个子集，只要他们中任何$7$个的并集的元素个数都不少于$n$，则这$15$个子集中一定存在$3$个，他们的交集非空。

第二題是46還是47呢？47 一定可以，但46我寫不出那個排列

真是不容易啊，CMO才刚考完3个小时楼主就把题目发出来了，那我就来写答案了。
第一题：用$n$代替$k$
        令$b_i=a_{i+1}-a_i$,设$\max_{1 \le i \le n} (a_i^2)=a_t^2$，
        则$\displaystyle \sum_{i=1}^n a_i=0 \iff na_1+\sum_{i=1}^{n-1} (n-i)b_i=0 \iff a_1=-\sum_{i=1}^{n-1} \frac{(n-i)b_i}{n}$
           原不等式等价于$\displaystyle a_t^2 \leq \sum_{i=1}^{n-1} b_i^2$
           那么，由Cauchy不等式，$\displaystyle \max_{1 \le i \le k} (a_i^2)=a_t^2=%Error. "bracket" is a bad command.
( \sum_{i=1}^{t-1} \frac{ib_i}{n}-\sum_{i=t}^{n-1} \frac{(n-i)b_i}{n} %Error. "bracket" is a bad command.
)^2 \leq ( \sum_{i=1}^{t-1} \frac{i^2}{n^2}+\sum_{i=t}^{n-1} \frac{(n-i)^2}{n^2})(\sum_{i=1}^{n-1} b_i^2)$
           易证明$\displaystyle \sum_{i=1}^{t-1} \frac{i^2}{n^2}+\sum_{i=t}^{n-1} \frac{(n-i)^2}{n^2} \leq \frac{n}{3}$

第二题：答案是46。
        46是非常容易得到的，就不写了，主要写一下构造。
        构造采用归纳法，证明这样一个构造存在：对任意正整数$k$,存在正整数序列$%Error. "bracket" is a bad command.
( a_i %Error. "bracket" is a bad command.
)_{i=1}^{k(k-1)+2}$使得$a_1=a_{k(k-1)+2}=2$且符合题设的那个条件，并且序列$%Error. "bracket" is a bad command.
( a_i %Error. "bracket" is a bad command.
)_{i=1}^{k(k-1)+2}$中恰有$k$个不同的正整数$2=p_1<p_2<\cdots<p_k$,这里$2=p_1<p_2<\cdots<p_k$是前$k$个素数。
       归纳过渡如下：从$k$到$k+1$，令$a_{k(k+1)+2}=2,a_{k(k+1)+2-2i}=p_{i+1}$这里$1 \leq i \leq k-1$,
最后令$a_{k(k-1)+3}=a_{k(k-1)+5}=\cdots=a_{k(k+1)+1}=p_{k+1}$

第五题：设$\displaystyle \left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^n=M$
            先证明$1-\frac{\sqrt{2}}{2} \leq M$
            再用$Jensen$不等式证明$\displaystyle (\frac{1}{M}-1)^n \leq \left(\frac{1}{a_1}-1\right)\cdots\left(\frac{1}{a_n}-1\right)$
            接下来的不等式非常显然，在这里就不写了。
第六题：非常麻烦，晚上再来写，先说一声，答案是41。

继续下午写的答案
第六题：设$X=\left\{ 1,2,3,\cdots,56 \right\}$,15个集合记作$A_1,A_2,\cdots,A_15$.
     证明分三步完成：$1)$证明$n\geq41$
                             $2)$证明$n_{min}\leq42$
                             $3)$证明$n_{min}\leq41$
                             这样就可以得到$n_{min}=41$
$1)$这只需对$n=40$给出一个反例即可。
定义15个集合如下：$C_i=\left\{8i-7,8i-6,\cdots,8i\right\},i=1,2,\cdots,7$
                            $D_i=\left\{k,k\equiv i(mod8)\right\},i=1,2,\cdots,8$
验证过程由大家自己完成。

$2)$这里，采用反证法证明$n=42$的情形。
将$\displaystyle (i;A_{j_1},A_{j_2},A_{j_3},A_{j_4},A_{j_5},A_{j_6},A_{j_7}),1 \leq j_1<j_2<j_3<j_4<j_5<j_6<j_7 \leq 15$
配为一对当且仅当$i$不属于$\displaystyle \bigcup_{k=1}^7 A_{j_k}$,记这种对子的个数为$M$
一方面,从$i$来看,每个i至多属于$A_1,\cdots,A_{15}$中的$13$个,产生至少$C_{13}^7$个对子,有$M \geq 56C_{13}^7$
另一方面,从七元组$(A_{j_1},A_{j_2},A_{j_3},A_{j_4},A_{j_5},A_{j_6},A_{j_7})$来看,每个七元组的并至少有$42$个元素,故至多能产生$14$个对子,故有$M \leq 14C_{15}^7$
综合上面两个式子,矛盾!
故反设不成立,有$n_{min}< 43$(我不知道为什么这上面显示不出正确的小于等于号了)

$3)$仍采用反证法
由于这时所有的元素的重数和不超过$112$,故必定有一个集合中至多只有$7$个元素.
不妨设$7\geq k=A_{15}$,$A_{15}=\left\{1,2,3,\cdots,k\right\}$
将$\displaystyle (i;A_{j_1},A_{j_2},A_{j_3},A_{j_4},A_{j_5},A_{j_6}),1 \leq j_1<j_2<j_3<j_4<j_5<j_6\leq14$
配为一对当且仅当$k+1\leq i\leq 56$且$i$不属于$\displaystyle \bigcup_{k=1}^6 A_{j_k}$,记者中对子总数为$L$.
一方面,从来$i$看,每个i至多属于$A_1,\cdots,A_{14}$中的12个,产生至少$C_{12}^6$个对子,有$L\geq (56-k)C_{12}^6$
另一方面,从六元组来看,每个六元组的并至少有$41-k$(为什么?请大家自己思考)个元素,故至多能产生$15$个对子,故有$L\leq 15C_{14}^6$
综合上面两个式子,矛盾!
故反设不成立,有$n_{min}<42$
至此,全命题证毕.

问一声了，有没有人自愿把这个题目和答案翻译一下的？

[quote="Forward"]提供一些有关冬令营的消息：第三题全冬令营只有一个人做出来，这个人叫邓煜，来自深圳高级中学。
不知这里有没有人认识？
另，
冬令营的最佳题评选结果如下：
一等奖：第六题；
二等奖：第三、五题。[/quote]
在aoshoo.com 看了他第三題的解答，真是精彩

[quote="Forward"]提供一些有关冬令营的消息：第三题全冬令营只有一个人做出来，这个人叫邓煜，来自深圳高级中学。
不知这里有没有人认识？
[/quote]
这是个牛人，去年（高一）就进了集训队，今年说不定可以进国家队。

就他现在的水平，肯定进国家队的

[quote="Soarer"][quote="zhaobin"][quote="libra_gold"]就他现在的水平，肯定进国家队的[/quote]
呵呵，几年前有个方家聪，现在又有了一个看来中国的牛人是成周期出现呀 :)[/quote]
付云浩呢？[/quote]
付云浩，好象是高二开始才进集训队吧，他们两个好象从高一就开始了。
另外好象付云浩的数论现在超强，满分是很正常

[quote="Forward"]第三题：我没做出来,只能给一个初步的结果,易于得到的是当$m,n > 1$时可以设$m$的所有素因子$p$均有$\displaystyle (\frac{p}{11})=1$,接下来可以证明这样一个结论：设可以表示的所有数组成集合$A$,那么由$a$属于$A$及$b$属于$A$可以推出$ab$属于$A$,
最后需要证明对任意素数$p$满足$\displaystyle (\frac{p}{11})=1$均有$p$属于$A$.
很遗憾,最后这一点我不知道怎么证明.
第四题：此题出得极烂，解析就可以了。[/quote]
可以证明当m的所有素因子p都满足(p/11)=1时有奇数解.
m,n而比有一个满足此条件.
我个人认为此题极烂，因为完全是x^2+y^2=n
的拓广，证法可完全照搬其证法。

[quote="zhaobin"][/quote][quote="Soarer"][quote="zhaobin"][quote="libra_gold"]就他现在的水平，肯定进国家队的[/quote]
呵呵，几年前有个方家聪，现在又有了一个看来中国的牛人是成周期出现呀 :)[/quote]
付云浩呢？[/quote][quote="zhaobin"]
付云浩，好象是高二开始才进集训队吧，他们两个好象从高一就开始了。
另外好象付云浩的数论现在超强，满分是很正常[/quote]
康嘉引初三就进了。

请问一下有这次的获奖情况吗？

[quote="dingdongdog"][/quote][quote="zhaobin"][/quote][quote="Soarer"][quote="zhaobin"][quote="libra_gold"]就他现在的水平，肯定进国家队的[/quote]
呵呵，几年前有个方家聪，现在又有了一个看来中国的牛人是成周期出现呀 :)[/quote]
付云浩呢？[/quote][quote="zhaobin"]
付云浩，好象是高二开始才进集训队吧，他们两个好象从高一就开始了。
另外好象付云浩的数论现在超强，满分是很正常[/quote][quote="dingdongdog"]
康嘉引初三就进了。[/quote]

真牛..

真是羡慕啊~~~这次好想进冬令营，可惜联赛考得太差了。
自己做了一下，完全做出1-4题，但是5题没做出，6题只想到一个反例。
我们省听说考得很差，呜呜呜~~~

[quote="libra_gold"][quote="Forward"][quote="libra_gold"]这样看来，每题好像都很简单的样子（除了第三题），我81分，侥幸进了集训队~ :blush:[/quote]
问一下，你是陈祖维还是姜子麟阿？认识一下。[/quote]

我是上海队的队员~ :ninja:[/quote]
恭喜下啦 :)

[quote="libra_gold"][quote="Forward"][quote="libra_gold"]这样看来，每题好像都很简单的样子（除了第三题），我81分，侥幸进了集训队~ :blush:[/quote]
问一下，你是陈祖维还是姜子麟阿？认识一下。[/quote]

我是上海队的队员~ :ninja:[/quote]
那好了，姜子麟，我们在集训队会见面的。我叫汪哲楠，湖北队的。

你们好,我也进集训队了,吉林东北师大附中金龙

我就没那么幸运了。。5555555555,才72分.........看来真得好好学组合了
不过还好，还有一年的机会。

[quote="jin"]是啊 我们都高三了[/quote]
可是你高二就进集训队了啊

就是说我去年在集训队时考得很差

楼上的也姓金吗？？
我也姓金的啊，交个朋友如何？
偶的QQ:8371447 MSN:whtllxr@gmail.com

[quote="liudaveliu"]where can I get past cmo questions?[/quote]
mathlinks上好像就有啊

呵呵，我是福建的
今年没进冬令营，不过我才高二，明年还有机会


I think that's something called a Diophantine equation but I don't know how to solve  those things(I don't have AoPS Vol. 1 or Intro to NT :().  But I'm pretty sure there are other ways than guess and check.

[url]http://www.geometer.org/mathcircles/diophantine.pdf[/url]
awesome site if your learning the diophantine equations for the first time

You could always google it or wiki it.  :wink:

[quote="Cliu0301"]1. Let 3x + 5y = 90. What is the number of ordered pairs (x,y) that could satisfy this equation?

Anyway to solve this other than guess and check?[/quote]

[i]They don't have to be integers, right?[/i]
If not, this is how I think we solve it.
3x+5y=90.
3x=90-5y.
Dividing by 3, we get x=(90-5y)/3.
Now, using our common number theory sense, we know that (90-5y) is divisible by 3.
90-5y=0.
90-5y=3.
90-5y=6.
.
.
.
90-5y=90.

You know that for each equation there is a different value for y for each of the above equations, there are...
((90-0)/3)+1=31 values for y.
For each value of y there is a different value of x, so there are 31 ordered pairs?


Let ABCD be quadrilateral. AD cuts BC at E, AB cuts CD at F. Prove that the midpoints of AC,BD,EF form a line.

That is Gaus`s line.

Maybe spelling is not good. :oops:

[quote]That is Gaus`s line. [/quote]

Please can you give us some reference!

See this: http://www.gogeometry.com/geometry/newton_theorem_line_gauss.htm

In fact, it`s Newton-Gauss line. :)

For - probably - a new proof, see: http://www.mathlinks.ro/viewtopic.php?t=202546.

Best regards,
sunken rock

Dear Mathlinkers,
see:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=514534195&t=176253
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=514534195&t=79751
Sincerely
Jean-Louis


I am a contractor and oftem deal with final contract numbers that have a predetermined % markup on them.
This may be a really silly question but I am puzzled on how to figure this out with a calculator in hand.

I am trying to determine the cost of the labor and material before they are marked up with 15%. If I know that the final contract dollar amount is $\$$ 4,800.00, and there is a 15% markup on this number, what was the cost of the labor and materials?
Is there an easy way to figure this out with a caculator?
Every time I take (4800-15%) I get 4080. Than when I take (4080+15%) I get 4692. I just don't get it.

Thanks for taking a look at this for me. I apologize if this is a really silly question.

John

No problem, Jwatch.

Think of it this way, you want to "undo" the markup to find the solution.  What operation does a 15% markup correspond to?  What operation can you use to undo it?

For example, (this is an easy example, I'm just using it to get the point across) if I knew that there was a 10 dollar markup on something that wound up costing 50, and I wanted to find the original cost, I want to undo the 10 dollars that were added in, so I subtract 10 from 50 to end up with 40 dollars.

 [hide] Try marking up something worth 100 by 15%, what do  you need to do to the new value (115) to get back to 100? [/hide]

Yeah, the markup is applied to the original cost, not to the final amount.  $\$4800$ is 115% of the cost, and the cost is 100%.  Thus, it's a ratio of $\frac{100}{115}$, which is not the same as $\frac{85}{100}$, like you were doing...


geometry
analytic geometry
geometry proposed

given $ {P\in C_1 : y^2\equal{}2x}$; and points $ {B,C \in y\minus{}axis}$ such that :$ {C_2: (x\minus{}1)^2\plus{}y^2\equal{}1}$ is inscribed in $ {\triangle PBC}$ find the mininum area of $ {\triangle PBC}$..
 :wink:

The point P has x coordinate 4 and the minimum area is 8