geometry
incenter
trigonometry
Triangle
IMO
IMO 1998

I haven't thought about a generalization, but here's a solution to this problem.

The assertion is equivalent to $BI^2>BS\cdot BR$. It's easy to see that triangles $BSK,BMR$ are similar (in this order of the vertices). We only need to employ a simple angle chase. From here we derive $\frac{BS}{BK}=\frac{BM}{BR}\Rightarrow BS\cdot BR=BM\cdot BK=BM^2$, and since in the right triangle $BMI$ $BM$ is a leg, while $BI$ is the hypothenuse, we get $BI>BM\Rightarrow BI^2>BM^2=BS\cdot BR$, Q.E.D.

[quote="grobber"]I haven't thought about a generalization, but here's a solution to this problem.
[/quote]

What a beautiful proof!!

The generalization which I was talking of is that the result holds even if the line through B in not parallel to MK.

Regardless RS//MK, it holds that AS,CR and MK concur at the pole of RS wrt c(ABC). Given that this point is the orthocenter of the traingle RIS and is inside of RIS, we have the claim.

[quote="marko avila"]i  really  dont get your  proof  please elaborate  thanks   :)[/quote]

First of all I have to correct the mistyping of points of tangency.  As K lies on AB and so on I am going to prove that AS, RC and KL concurr at one point wich is the pole of RS wrt c(ABC).
I don't remember anymore my previous proof but I con give the following argument.

If T and U are the points of tangency from R to c, let's call P and Q the intersection of KL, UT and TL, UM respectively. From Pascal theorem to the hexagon  KULMTT we have that P, Q and R are collinear. Since, as is well known,  R, Q and C are collinear, then RC cuts KL at P.

As P = KL/\UT, then P is the pole of RB. As R, P and C are collinears then the respectiv polars are concurrent. This means that RB, TU and ML concurr at S. As R lies on KM also A lies on TU.

Since, as is easy to prove, P is the orthocenter of RIS and P is between I and the foot of I on RS, we have the claim.

Let me know if now it is clear enough .

Very nice your proof, Grobber ! I liked specially the characterization of an acute angle, i.e.
[b]Lemma.[/b] $\triangle ABC,\ D\in BC,\ AD\perp BC\ : \ A<90^{\circ}\Longleftrightarrow AD^2>DB\cdot DC\ .$
[u]Proof.[/u] $h_a=AD=c\sin B=b\sin C,\ DB=c\left|\cos B\right|,\ DC=b\left|\cos C\right|\ .$ Therefore,
$AD^2>DB\cdot DC\Longleftrightarrow$$bc\sin B\sin C>bc\left|\cos B\cos C\right|\Longleftrightarrow \sin B\sin C>\left|\cos B\cos C\right|\ .$
$\cos B\cos C\le 0\Longrightarrow B\ge 90^{\circ}\ \vee\ C\ge 90^{\circ}\Longrightarrow A<90^{\circ}\ ;$
$\cos B\cos C>0\Longrightarrow \cos (B+C)<0\Longrightarrow \cos A>0\Longrightarrow A<90^{\circ}\ .$

I proved easily the relation $\boxed {\ BR\cdot BS=(p-b)^2\ }$ and from the well-known relation
$BI^2=\frac{ac(p-b)}{p}$ results $BI^2>BR\cdot BS\Longleftrightarrow \frac{ac(p-b)}{p}>(p-b)^2\Longleftrightarrow$
$ac>p(p-b)\Longleftrightarrow 4ac>(a+c)^2-b^2\Longleftrightarrow b^2>(a-c)^2\Longleftrightarrow b>\left|a-c\right|\ .$

[u]Remark.[/u] If $X\in RS\cap AC$, then the division $(Y,R,B,S)$ is harmonically and the point $I$ belongs to the exterior of the circle with the diameter $RS\ !$

It's easy to see that the triangles $RKB$ and $LSB$ are similar so : $\frac{BK}{BS}=\frac{BR}{BL}=\frac{RK}{SL}=\frac{KM}{LM}$ (last equality by Thales theorem's).
Now let $\alpha= <BIS$ and $\beta=<BIR$ then :
\[ \tan(\alpha)=\frac{BS}{BI}=\frac{BK.LM}{KM.BI}=\frac{BK}{BI}\frac{LM}{KM}\] and by the sin law and the fact that \[<KLM=<ILM+<ILK=\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}-\frac{A}{2}\] and by the same way \[<MKL=\frac{\pi}{2}-\frac{C}{2}\] therefore :
\[ \tan(\alpha)=\frac{\cos(\frac{B}{2})\cos(\frac{C}{2})}{\cos(\frac{A}{2})}\] and by the same way \[\tan(\beta)=\frac{\cos(\frac{B}{2})\cos(\frac{A}{2})}{\cos(\frac{C}{2})}\]
so \[ \tan(\alpha+\beta)=\frac{\cos(\frac{B}{2})(\cos^2(\frac{C}{2})+\cos^2(\frac{C}{2}))}{\cos^2(\frac{A}{2})\cos^2(\frac{C}{2})\sin^2(\frac{B}{2})} >0\]
and finally $RIS < \frac{\pi}{2}$

[u][b]Outlines of my bash[/b][/u]

Apply the law of sines in $\triangle{RKB},\triangle{BLS}$ and express $BR,RK,BS,SL$ in terms of $r$ where $r$ is the inradius.Then apply the law of cosines in $\triangle{IKR},\triangle{ILS}$ to get $IR^2,IS^2$.Also calculate $RS=BR+BS$ in terms of $r$.Then show that $IR^2+IS^2-RS^2 >0$(it wont be too difficult as one would get rid of $r^2$ and only the trigonometrical terms would remain).Thus $cos(\angle{RIS}) >0$ and we are done.

A similar proof to sprmnt21 above can be done with just polar transformation:

Applying a polar transformation centered at the incircle, $t$ will be mapped to its pole $T$ on $KL$ (since this is the polar of $A$). With the given parallel condition this point would be the midpoint of $KL$, but that doesn't really matter. Now the polar of $R$ goes through $T$, as does the polar of $S$, and because they are polars, these are perpendicular to $RI$,$RS$ respectively, giving that $T$ is the orthocenter of $RIS$, done.

Any time an angle centered on the center of a circle, especially the incenter, is a part of a problem, poles and polars may have a use!

[quote=sprmnt21]Let $I$ be the incenter of triangle $ABC$. Let $K,L$ and $M$ be the points of tangency of the incircle of $ABC$ with $AB,BC$ and $CA$, respectively. The line $t$ passes through $B$ and is parallel to $KL$. The lines $MK$ and $ML$ intersect $t$ at the points $R$ and $S$. Prove that $\angle RIS$ is acute.[/quote]

Quirky solution :P

Let $P$ be the midpoint of $KL$.

[b]Lemma.[/b] $\angle APC>90^{\circ}$

[i]Proof.[/i] Let $N$ be the midpoint of $AC$. Then $$|\overrightarrow{PN}|=\left|\tfrac{1}{2}\left(\overrightarrow{AK}+\overrightarrow{CL}\right)\right| \le \tfrac{1}{2}\left(|AK|+|CL|\right)=\tfrac{1}{2}\left(|AM|+|CM|\right)=\tfrac{1}{2}|AC|$$ so $P$ lies inside the circle with center $N$ and radius $\tfrac{1}{2}|AC|$, proving the claim. $\blacksquare$

Now observe that $\overline{AM}$ is the polar of $R$ and $\overline{CM}$ is the polar of $S$ in the incircle of $\triangle ABC$. Thus, $IR \perp PA$ and $IS \perp PC$. Moreover, we see that $\angle API<90^{\circ}$ and $\angle IPC<90^{\circ}$ since $AKLC$ is convex and $P$ lies on segment $KL$. Thus, $\angle RIS=180^{\circ}-\angle APC$ and the conclusion follows. $\blacksquare$

Angle chasing gives $SBKM$ and $RBLM$ cyclic. Because $\angle SKM + \angle RLM = \angle SBM + \angle RBM = \pi$, $SK$ and $RL$ intersect at a point $Y$ on the incircle. This means $AS$ is the polar of $R$ and $CR$ is the polar of $S$, so $X:=AS\cap CR$ should be the orthocenter of $\triangle RIS$. By Pappus, $X$ lies on $\overline{KL}$ and because $IX\perp RS\parallel KL$, $X$ is in fact the midpoint of $KL$. Therefore $X$ lies strictly inside $\triangle RIS$, and $\angle RIS$ is acute.

[b][color=#BF9000]Claim 1:[/color][/b] $H$ is the orthocenter of $RSI$ if and only if $BI \cdot BX = RB \cdot BS$.
Trivial by considering the reflection of the orthocenter over the foot of $I$ to $RS$.

[b][color=#BF9000]Claim 2: [/color][/b]$BI \perp RS$.
Note that $KILB$ is cyclic, since $KI \perp BK$ and $IL \perp BL$. Therefore, $\angle KBI = \angle IBL = \angle LKI = \angle KLI$, which implies that $\angle BKL = \angle KLB = 90 - \angle IKL$ from which it follows that $BI \perp KL$. Since $KL$ is parallel to $RS$, $RS \perp BI$.

[b][color=#BF9000]Claim 3: [/color][/b] $\triangle BRK \sim \triangle BLS$.
Note that since $BI \perp RS$ and $\angle KBI = \angle IBL$, $\angle RBK = \angle LBS$. Since $AKIM$ is cyclic:
$\angle IAM = \angle MKI$ from which $\angle RKB = 90 - \angle BAI$. In the same vein, $\angle BLS = 90 - \angle ICB$. Since $KRB = 180 - \angle RKB - \angle RBK = 180 - (90 - \angle BAI) - (90 - \angle KBI) = \angle BAI + \angle KBI$, which by symmetry is equal to $\angle LSB$. The desired similarity follows.

[b][color=#BF9000]Claim 4: [/color][/b] The midpoint of $KL$ is the orthocenter of $\triangle RSI$.
Note that the midpoint of $KL$ (call it $X$) is also the intersection of $BI$ and $KL$. Let $b$ be the length of $AC$ and let $r$ be the inradius of $ABC$. By claim $3$, we know:
$$\frac{RB}{KB} = \frac{BL}{BS} \Longleftrightarrow BL^2 = RB \cdot BS \Longleftrightarrow (s-b)^2 = RB \cdot BS$$
It is easy to see that $BX = (s-b) \cos (KBI)$ and $XI = r \cos (KBI)$ from which we get that:
$$BI \cdot BX = ((s-b)\cos (KBI) + r \cos (KBI))(s-b)\cos (KBI) = (s-b)^2 - (s-b)^2\sin ^2 (KBI) + r(s-b)\cos^2(KBI)$$
Note that $$\tan (KBI) = \frac{r}{s-b}$$ from which we see that:
$$BI \cdot BX = RB \cdot BS = (s-b)^2$$
and by claim $1$ we win.

Solution with [b]awang11[/b].

Let $P$ be the midpoint of $MK$. By point-line duality about the incircle, we have $\angle APC=180^\circ-\angle RIS$. By Pappus on $ALC$ and $RBS$, $AS\cap CR=P'$ lies on $MK$. Let $P_\infty$ be the point at infinity along line $MK$. Note that $MK\perp IB\perp RS$. Observe that
\[(P',P_\infty;M,K)\stackrel{C}{=} (R,P_\infty;CM\cap RS,B)\stackrel{M}{=}(L,MK\cap AC;C,A)=-1,\]
with the last result following from the Ceva harmonic division configuration. Thus $P'$ is the midpoint of $MK$, so we have $\angle RPS=180^\circ-\angle RIS$. Since $P$ is closer than $I$ to line $RS$, the result follows.

Wow, I actually liked this problem a lot.

[color=#ff9a00][b]Solution 1.[/b][/color]

Note that $RS$ is actually external angle bisector of $\angle ABC$. Trivial angle chase yields that $KBSM$ and $RBLM$ are cyclic.

[b]Claim. [/b] $AS\perp IR$.
[i]Proof.[/i]
Let $(AKIM)$ and $(IBS)$ intersect at $P$. By the radical axis on $(AKIM)$, $(IBS)$ and $(KBSM)$, we obtain that $KM$, $BS$ and $IP$ are concurrent at $R$. The claim follows.

Now all we need to do is to verify that line $AS$ lies between the lines $AI$ and $AR$.
Let $F=AI\cap ML$, then $$\measuredangle IFL=\measuredangle AFM=\measuredangle CML+\measuredangle IAC=90^\circ-0.5(\measuredangle A+\measuredangle C)=0.5\measuredangle B=\measuredangle IBL,$$
thus $F$ lies on the circle with diameter $BI$. As $RS$ is tangent to $(BI)$, then $AI$ must intersect $ML$ before it intersects $RS$. Also $ML$ intersects side $BC$, and therefore cannot intersect $BC$ beyond $B$ of the line $BC$. Hence $AS$ lies inside $\angle IAB$ and therefore inside $\angle IAR$ as $MK$ intersects side $AB$, and therefore cannot intersect $AB$ beyond $B$ of the line $AB$. We do exactly the same with $C$ and obtain that orthocentre of $\triangle IRS$ lies inside the triangle, hence $\angle RIS$ is acute.

[color=#ff9a00][b]Solution 2.[/b][/color] [which I found during the write up of the first solution.]
Note that $RS$ is actually external angle bisector of $\angle ABC$. Trivial angle chase yields that $KBSM$ and $RBLM$ are cyclic.
Hence as $RK\cdot RM$ is the power of $R$ wrt $(KLM)$ and $SL\cdot SM$ is the power of $S$ wrt $(KLM)$, we get that
$$RS^2=RB\cdot RS+SB\cdot SR=RK\cdot RM+SL\cdot SM=RI^2-r^2+SI^2-r^2<RI^2+SI^2,$$
where $r$ denotes the inradius of $ABC$. By law of cosines, we have $RI^2+SI^2>RS^2=RI^2+SI^2-2\cos{\angle RIS} \cdot RI\cdot SI\implies \cos{\angle RIS}>0\implies \angle RIS$ is acute.

Let $X=BI\cap MK$. It is obvious that $X$ lies between $B$ and $I$.

[b]Claim:[/b] $X$ is the orthocenter of $\triangle RIS$.

[i]Proof.[/i] Clearly $BI\perp RS$. It is sufficient to prove that $BR\cdot BS=BI\cdot BX$ by [i]PoP[/i]. Notice the angles: \[ \angle BRM=\angle KML=\angle KLC=\angle LKC=\angle SKB \] Similarly $\angle RMB=\angle KSB$. That means $\triangle BRM\sim \triangle BKS$. So
    \begin{align*}
        \frac{BR}{BM}=\frac{BK}{BS}\Longrightarrow BR\cdot BS=BM^2=BI\cdot BX
    \end{align*}
hence $X$ is the orthcenter. $\square$

Since the orthocenter $X$ lies between $B$ and $I$ therefore the angle $\angle RIS$ must be acute.

[hide=Diagram][img width=50]https://media.discordapp.net/attachments/986049049834713139/996623358995210250/Screen_Shot_2022-07-12_at_8.45.15_PM.png?width=1104&height=1026[/img][/hide]

Let $D$ and $E$ be $CI\cap ML$ and $AI\cap MK$ respectively. They are also midpoints of $ML,MK$ respectively. Since $\angle IDS=\angle IBS=90^\circ$ and $\angle IER=\angle IBR=90^\circ$ we have $IDBS,IBRE$ cyclic. Also, we have $ILBK$ cyclic.

We have \[\angle SIR=\angle SMR+\angle MSI+\angle MRI=\angle LMK+\angle DBI+\angle EBI=\angle LMK+\angle DBE.\] To prove that this is less than $90^\circ$ we only need to show that double of that is less than $180^\circ.$ Conveniently, \[2(\angle LMK+\angle DBE)=\angle LIK+2\angle DBE=180^\circ-\angle LBK+2\angle DBE\] so it suffices to show that $\angle LBK\ge 2\angle DBE.$ Well, to do this, we dissect $\angle DBE$ into $\angle DBM+\angle EBM$ and we show that $\angle LBD>\angle DBM$ and $\angle KBE>\angle EBM.$ Note that $\angle MKB,\angle MLB$ are obtuse.

We claim that this is a general lemma: let $XYZ$ be an obtuse triangle with $\angle X$ obtuse and $P$ be midpoint on $XY$ then $\angle XZP=\angle YZP.$

Clearly, $ZY>ZX$ and that is the most important part. By sine law, \[\frac{\sin(\angle XZP)}{XP}=\frac{\sin(\angle XPZ)}{XZ}>\frac{\sin(\angle YPZ)}{YZ}=\frac{\sin(\angle YZP)}{YP}.\] Since $XP=YP$ we have $\sin(\angle XZP)>\sin(\angle YZP).$ $\angle X$ is obtuse so both these angles are clearly acute. Thus, we may conclude that $\angle XZP>\angle YZP$ so our proof is complete.

Won't post a solution but this question can also be solved by barycentric coordinates with reference triangle $\triangle RSM$ (after noting that $MB$ is a symmedian you end up with an inequality equivalent to $1>\sin RSM$ which obviously holds).

We show this by showing that the orthocenter lies inside of triangles RIS. We have KRB=MKL=MLC=SLB and similarly RKB=LSB. Then by sim triangles we have BS*BR=BL^2, and letting X be midpoint of RS we have BX=BLcos(b/2) and BI=BLcos(b/2) hence by transitive BI*BX=BS*BR. It’s easily derived that in a triangle ABC by pop BD*DC=AD*DA’=AD*HA’ where D altitude from A, H orthocenter, and A’ reflection of H over D onto circumcircle. Then since B is the foot of altitude (IX perp. KL parallel to RS), we conclude that X is the orthocenter of RIS, to which the problem becomes evident since KML is acute so X is inside MRS, while the line thru B is on the other side of KL than X. (can someone confirm this is rigorous? i was thinking abt proving that the cosine of that angle positive but idk how)

I've gotten much much worse at geometry, but this should work right?

The main claim is that $D=IB\cap KL$ is the orthocenter of RIS, which finishes (since it's closer to RS the external angle bisector of ABC than I is). Very easy to deduce (using well known angles to chase) that RBLM, SBKM are cyclic quads, so SKM+RLM=SBM+RBM=180 so $KS\cap RL=P$ lies on the incircle. Finally, by Brokard $KL\cap YM$, R, S, I is an autopolar orthocentric system, so D is the orthocenter as claimed.

The quintessential geometric inequality: so clean to do correctly, so easy to overcomplicate. Here's the nice solution, which took me much too long.

We will show that $RI^2+IS^2 > RS^2$. In particular, we have the following claim:

[b][color=#f00]Claim.[/b][/color] $MLSB$ is cyclic.

[i]Proof.[/i] $\measuredangle MBS = \measuredangle BMK = \measuredangle MLS$. $\blacksquare$

Similarly, $KBRL$ is cyclic. Hence $$RI^2 + IS^2 > RM \cdot RL + SK \cdot SL = RB \cdot RS + SB \cdot RS = RS^2$$ where we use the deep estimate $r>0$.


Let x,y,z>0

    prove that

   (a²+ab+b²)(b²+bc+c²)(c²+ca+a²)>=(ab+bc+ca)*3

From Holder Inequality
we have $ {((ab\plus{}b^2\plus{}a^2)(b^2\plus{}bc\plus{}c^2)(a^2\plus{}c^2\plus{}ca))}^{1/3} >\equal{} ab\plus{}bc\plus{}ca$.


Prove that in any nonisoscelles trapezoid $ ABCD$ , $ AB$ || $ CD$,that $ \frac {AC^2 \minus{} BD^2}{AD^2 \minus{} BC^2} \equal{} \frac {AB \plus{} CD}{AB \minus{} CD}$

[quote="balan razvan"]Prove that in any nonisoscelles trapezoid $ ABCD$ that $ \frac {AC^2 \minus{} BD^2}{AD^2 \minus{} BC^2} \equal{} \frac {AB \plus{} CD}{AB \minus{} CD}$[/quote]
Try $ BC\parallel{}AD,$ $ AB\equal{}\sqrt2,$ $ BC\equal{}CD\equal{}1,$ $ AD\equal{}2.$
Then $ BD\equal{}\sqrt2,$ $ AC\equal{}\sqrt5$ and $ \frac {AC^2 \minus{} BD^2}{AD^2 \minus{} BC^2} \neq\frac {AB \plus{} CD}{AB \minus{} CD}.$ :wink:

[hide]
Let $ M$ be the foot of the perpendicular from $ A$ to $ CD$ and let $ N$ be the foot of the perpendicular from $ B$ to $ CD$.  Notice that $ AM\equal{}BN$, $ MN\equal{}AB$, and $ MD\plus{}NC\equal{}CD\minus{}AB$.  We have

$ \frac{AC^2\minus{}BD^2}{AD^2\minus{}BD^2}\equal{}\frac{AM^2\plus{}MC^2\minus{}(BN^2\plus{}ND^2)}{AM^2\plus{}MD^2\minus{}(BN^2\plus{}NC^2)}$

$ \equal{}\frac{MC^2\minus{}ND^2}{MD^2\minus{}NC^2}$

$ \equal{}\frac{(MC\minus{}ND)(MC\plus{}ND)}{(MD\minus{}NC)(MD\plus{}NC)}$

$ \equal{}\frac{(NC\plus{}MN\minus{}(MD\plus{}MN))(MD\plus{}MN\plus{}NC\plus{}MN)}{(MD\minus{}NC)(MD\plus{}NC)}$

$ \equal{}\frac{(NC\minus{}MD)(CD\plus{}AB)}{(MD\minus{}NC)(CD\minus{}AB)}$

$ \equal{}\frac{AB\plus{}CD}{AB\minus{}CD}$ [/hide]

[quote="cosinator"][hide]... $ MD \plus{} NC \equal{} CD \minus{} AB$ ... [/quote]
If $ D\in (MN)$ and $ N\in (DC)$ , i.e. the order is $ M<D<N<C$ , where " $ X\ <\ Y$ " means " $ X$ is at left of $ Y$ " , the upper relation is [b]not[/b] truly !


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integration
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Study the convergence of $ ({\sin (n)})^{n}$

Do you mean $ \sin(n^n)$ or $ (\sin n)^n$ ?

suppose it convergent.
then any subsequance will convergent too.
to prove divergent of your sequance it's enough to show that sequance $ \sin 2^{n^{2}}$ divergent, after now I have no idea  :(

Shall we post it once every 16 months? :P
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2125608574&t=90103[/url]

Let's study the convergence of $ \int_0^{\sqrt{n}}\sin^n x\,dx$ instead. Here $ n$ is an even integer.

OK, that's $ \left(\sqrt {n} \plus{} O(1)\right)\frac {1\cdot3\cdot5\cdots(2n \minus{} 1)} {2\cdot4\cdot6\cdots(2n)} \plus{} o(1).$

This comes from writing $ \left[0,\sqrt {n}\right] \equal{} \left[0,\frac {k\pi}2\right]\cup\left[\frac {k\pi}2,\sqrt {n}\right],$ noting that $ \sqrt {n} \minus{} \frac {k\pi}2$ is bounded, and noting that the integral of $ \sin^nx$ on any sequence of intervals of bounded length tends to zero by the DCT.

Writing that product as $ \frac {(2n)!}{2^{2n}(n!)^2}$ and using Stirling, I get that the sequence tends to $ \frac {1}{\sqrt {\pi}}.$

[color=green][Edit: off by a factor of $ \sqrt{2}.$ Fixed below.][/color]

I did not remember the reduction formula for $ \int \sin^nx\,dx$, so I used Euler's formula instead: $ \sin^n x \equal{} \frac {1}{(2i)^n}(e^{ix} \minus{} e^{ \minus{} ix})^n$, then binomial and Stirling. But this is what Maple 11 thinks of this limit:
[img]http://www.math.tamu.edu/~lkovalev/files/limit-1.jpg[/img]
[img]http://www.math.tamu.edu/~lkovalev/files/limit-2.jpg[/img]
The last output reminded me of [url=http://xkcd.com/10/]this xkcd comic[/url].

OK, $ \frac1{\sqrt{\pi}}\approx 0.564$ and $ \sqrt{\frac{2}{\pi}}\approx 0.797.$

I won't swear that I didn't slip a factor of $ 2$ (or $ \sqrt{2}$) somewhere.

But the biggest problem is that $ \sqrt{n}$ is not a "nice" endpoint for the interval of integration; the reduction formula calculation I was quoting only works nicely for intervals consisting of whole numbers of quadrants. So I had to split off a piece. The piece I split off does go to zero, but does so agonizingly slowly, and in a very non-monotone fashion. $ n\equal{}50$ is surely nowhere near big enough for any reasonable accuracy.

Er ...
[quote="Maple"]$ \dots,32300,\dots,\minus{}0.278715\ 10^7,\dots$[/quote]
Say what? How can the integral of a positive function no larger than $ 1$ on an integral no larger than $ [0,5]$ be that large? Or negative? The problem doesn't look weird enough to me to cause total algorithmic insanity.

My calculations gave $ \sqrt {2/\pi}$. I have no idea where the Maple answers came from. Today it changed its mind and gave a different answer to the same command:
[img]http://www.math.tamu.edu/~lkovalev/files/limit-3.jpg[/img]
However, the nonsensical answers again occur at $ k\equal{}20,22$.  :huh:

By the way, the integral came up last night when a student dropped in to ask if $ \lim_{n\to\infty}\int_0^{\infty}\sin^n x\,dx \equal{} 0$, triggering a discussion of double limits.

in my post I considered sequance $ \sin (n^{n})$...

I found my mistake: I was was using $ 2n$ to mean $ n$ but not consistently. Here's my earlier post rewritten with $ n\equal{}2k$ and the mistake fixed.

That's $ \left(\sqrt {2k} \plus{} O(1)\right)\frac {1\cdot3\cdot5\cdots(2k \minus{} 1)} {2\cdot4\cdot6\cdots(2k)} \plus{} o(1).$

This comes from writing $ \left[0,\sqrt {2k}\right] \equal{} \left[0,\frac {m\pi}2\right]\cup\left[\frac {m\pi}2,\sqrt {2k}\right],$ noting that $ \sqrt {2k} \minus{} \frac {m\pi}2$ is bounded, and noting that the integral of $ \sin^{2k}x$ on any sequence of intervals of bounded length tends to zero by the DCT.

Writing that product as $ \frac {(2k)!}{2^{2k}(k!)^2}$ and using Stirling, I get that the sequence tends to $ \sqrt{\frac {2}{\pi}}.$

The output of Maple command evalf(int(sin(x)^44,x=0..sqrt(44))); turned out to be unstable. It usually gives a number about $ \minus{}3\cdot 10^6$, but sometimes it returns $ 0$. I tried to fix this problem by increasing the number of digits in floating-point numbers. The default is $ 10$. This is what I got by setting Digits:=...
[quote]Digits | Integral
11 | 526375
12 | -115999.902851
13 | 5100
14 | -1000
15 | -80.10
16 | -2.57757
17 | 1.0618235439984632
18 | .663655331878393240
19 | .761532
20 | .7506646
[/quote]
These outputs are not stable either: they change from one try to another. The latter numbers appear reasonable, which is actually a bigger problem: if I had Digits set to $ 18$ and got $ 0.6636$ as an answer, I would have no way of knowing that the answer is wrong.

In contrast to the above, evalf(Int(sin(x)^44,x=0..sqrt(44))); always returns the correct answer $ 0.7514952184$. What is going on? The difference between [b]int [/b]and [b]Int [/b]is that [b]int [/b]tries to compute the integral using the Fundamental Theorem of Calculus, which leads to a messy formula with some large coefficients in front of cosines and sines. Apparently, [b]evalf [/b]has issues with that formula because of subtraction of large and nearly equal terms.

The command [b]Int [/b]does not do anything to the integral: it's just recorded for future use. In this case [b]evalf[/b] performs numeric integration. It's easier to evaluate $ \sin(x)^{44}$ at several partition points than to evaluate its antiderivative at the endpoints $ 0$ and $ \sqrt{44}$.

But of course, we don't need to know Calculus now that we have computers. :mad:


Ten freshly laundered socks (two each of five different colors) lie in a heap.  If a toddler chooses socks at random to create five sock balls, then on average how many of the five balls do we expect to consist of a matched pair of socks?

Please show all work.

Thanks
Neil

Are there no solvers?

I am thinking we can do this direct, but if anyone has a clever solution, please post.

I have a very long computation. If P(n) is the probability of getting n pairs, the answer is obviously $ P(1) \cdot 1 \plus{} P(2) \cdot 2 \plus{} P(3) \cdot 3 \plus{} P(4) \cdot 4 \plus{} P(5) \cdot 5$
Think of the pairs as a string of ten letters. There is $ \frac {10!}{2!2!2!2!2!} \equal{} K$ total possibilities. We compute
$ P(5) \equal{} \frac {5!}{K}$.
$ P(4) \equal{} 0$.
$ P(3) \equal{} \frac {{5 \choose 3}^2 \cdot 6 \cdot 4}{K}$.
$ P(2) \equal{} \frac {{5 \choose 2}^2 \cdot 2 \cdot \frac {6!}{2!2!2!}}{K} \minus{} P(3) \minus{} P(5)$
$ P(1) \equal{} \frac {5 \cdot 5 \cdot \frac {8!}{2!2!2!2!}}{K} \minus{} P(2) \minus{} P(3) \minus{} P(5)$
So if I have everything corect, this is how you do it.
On my calculator, I get [hide]$ \frac {698}{945}$[/hide]

Hmm...the answer is 5/9.

Expected Value when there are seperate events is the sum of the expected valuee of each of the events, even when the events are dependent on each other.

[hide]
Consider each ball of sock seperately. The probability that the socks are the same is 1/9. Therefore, the sum of all the expected values is 5/9. So the answer is 5/9.
[/hide]

[quote="pythag011"]Expected Value when there are seperate events is the sum of the expected valuee of each of the events, even when the events are dependent on each other.[/quote]
I am confused by this. Is it that expected value is the most likely value, and I found the average value, which are different? He said "on average what would you expect?". Does this mean expected value or average value?

By brutest-of-brute force, I obtain 5/9. I'm not sure if I agree with [b]pythag[/b]'s method, but I believe the answer is correct.

Define $ f \; : \; \mathbb{N}^{10} \to \{ 0,1,2,3,4,5 \}$ as follows, in Mathematica:
[code]f[n_]:= Which[n[[1]] == n[[2]], 1, True, 0] +  Which[n[[3]] == n[[4]], 1, True, 0] +  Which[n[[5]] == n[[6]], 1, True, 0] + Which[n[[7]] == n[[8]], 1, True, 0] +  Which[n[[9]] == n[[10]], 1, True, 0][/code]
In English, this means if the first element equals the second element, add 1. Then, if the third element equals the fourth element, add 1. etc, up to element 9 and 10.

Define the set p, which is the set of distinct permutations of {1,1,2,2,3,3,4,4,5,5}
[code]p = Permutations[{1, 1, 2, 2, 3, 3, 4, 4, 5, 5}];[/code]

For each permutation in p, add the value of f(p), and divide by the number of permutations in p
[code]Total[Table[f[p[[n]]], {n, 1, Length[p]}]]/Length[p][/code]

Result:
[code]5/9[/code]

After reading [url]http://en.wikipedia.org/wiki/Expected_value[/url] it appears I am right as long as my calculations are right.

pythag's solution looks fine to me.  As he said, he is using the principle that if $ X$ and $ Y$ are random variables, then $ E(X \plus{} Y) \equal{} E(X) \plus{} E(Y)$.  Here E is expected value, which is the same thing as average value.

Here is an article that Dave Patrick and Richard Rusczyk wrote on the subject:
[url]http://www.usamts.org/Forms/ExpectedValue.pdf[/url].  See their Problem 5 for a related example.  (They mentioned me in the last problem!)


inequalities
function
calculus
three variable inequality

A non-elegant solution, but which doest not use the hypothesis that the numbers are greater than 3/4
Let f(x) = x/(1+x^2).

we have f'(x) = (1-x)/(1+x^2)^2 and f''(x) = 2x(x^2 - 3)/(1+x^2)^3.
Then f is increasing on [0,1] and decreasing on [1,+oo[, moreover f is concave on [0,sqrt(3)].

Wlog, we may assume that a =< b =< c.

- If c >= 1 : let's fix b, and then a = 1 - b - c.
Moreover, since 2a =< a+b = 1-c =< 0, we have -3/4 =< a =< 0.
Let's consider the LHS as a function of c, say S(c), whose derivative is S'(c) = f'(c) + (-b-c)(1-b-c)/(something)^2 = f'(c) + (a-1)(a+1)/(something)^2 =< 0.
Thus, S is a non-increasing finction in c, from which we deduce that S(c) =< S(1) = 1/2, since if c = 1 then a = -b.

From now on, we assume that c < 1. Then b > 0.

- If a >= 0, since f is concave on [0,sqrt(3)], Jensen's inequality leads to : S =< 3*S( (a+b+c)/3 ) = 3*S(1/3) = 9/10.

- If a =< 0 :
We use that f is concave, but only on b and c, which gives S =< f(a) + 2*f((a+b)/2) = f(a) + 2*f( (1-a)/2 ).
Then (ugly...) we clear the denominators, and expand, which leads to
9a^4 - 68a^3 + 114a^2 -28a + 85 >= 0 which holds since a =< 0.

Pierre.

How one could alternatively handle the second case: ($a \leq 0 \leq b \leq c$)

Let $f : \mathbb R \to \mathbb R$, $f(x) = \frac{x}{x^2+1}$. Then $f$ is odd, increasing on $\left[ 0,1 \right]$, decreasing on $\left[ 1,\infty \right)$, concave on $\left[ 0,\sqrt3 \right]$. Hence, $f(x) \leq f(1) = \frac12$, for all $x$.

Thus, we want to prove that $f(x)+f(y)-f(z) \leq \frac9{10}$ when $x,y,z \geq 0, \, x+y-z=1$.

We have $f(x)+f(y) \leq 2 f(1) = 1$. Let $u<v$ be numbers s.t. $f(u)=f(v)=\frac1{10}$, i.e. $u,v=5 \pm 2 \sqrt 6$.

If $z \in (u,v)$, then $-f(z) < - \frac1{10}$, which, combined with $f(x)+f(y) \leq 2 f(1) = 1$, yields the desired inequality.

If $z>v>9$, then $x+y=z+1>10$, so at least one of $x,y$ is $>5$, WLOG $x$. We have that $f(x) < f(5) = \frac5{26}$. Consequently, $f(x)+f(y)-f(z) \leq f(x)+f(y) \leq \frac5{26} + \frac12 = \frac{18}{26} < \frac9{10}$.

If $z<u<0.11$, then $x+y=z+1<1.11$, so $x,y,z \in \left[ 0,\sqrt 3 \right]$. By Jensen's inequality we get that $f(x)+f(y)-f(z) \leq \mathbf{ f(x)+f(y) \leq 2 \cdot f \left( \frac{x+y}2 \right) } < 2 \cdot f(.555) < 0.9$.

One line solution : The inequality is equivalent to

$(a-\frac{1}{3})^2(a+\frac{9}{12})\geq 0$           :D

[quote=harazi]The following one is very nice and not at all easy:
 Prove that if $a,b,c > -3/4$ and $a+b+c= 1$ then
$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+ \frac{c}{c^2+1}  \leq \frac{9}{10}$$.[/quote]

$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+ \frac{c}{c^2+1}  \leq \frac{9}{10}$$
$$\iff \sum_{a,b,c} \frac{3}{10} - \frac{a}{a^2+1} \ge 0$$
$$\iff \sum_{a,b,c} \frac{3a^2-10 a + 3}{a^2+1} \ge 0$$

Without loss of generality assume $a \ge b \ge c$
Then $\implies 3a^2 + 3 -10a \le 3b^2 + 3 - 10b \le 3c^2 +3 - 10c$ and $\frac{1}{a^2+1} \le \frac{1}{b^2+1} \le \frac{1}{c^2+1}$

So by Tchebychev's inequality
$$\iff \sum_{a,b,c} \frac{3a^2-10 a + 3}{a^2+1} \ge \frac{1}{3}(\sum_{a,b,c} 3a^2-10a+3)(\sum_{a,b,c} \frac{1}{a^2+1})$$

But since $a^2+b^2+c^2 \ge \frac{1}{3}$
$$\implies \frac{1}{3}(\sum_{a,b,c} 3a^2-10a+3)(\sum_{a,b,c} \frac{1}{a^2+1}) \ge 0$$

Hence proved.

And I don't think we need $a,b,c \ge \frac{-3}{4}$ anywhere here as [b]arqady[/b] mentions above :)

Edit : But to show $3a^2 + 3 -10a \le 3b^2 + 3 - 10b \le 3c^2 +3 - 10c$, we do need $a,b,c \ge \frac{-7}{3}$

[quote=harazi]The following one is very nice and not at all easy:
  Prove that if a,b,c are greater than -3/4 and add up to 1 then
a/(a^2+1)+b/(b^2+1)+c/(c^2+1)<=9/10.[/quote]

The condition $a,b,c\ge\frac{-3}{4}$ in fact validates the following proof:
\[ \cfrac{x}{x^2+1}\le\cfrac{72x+6}{100}\Leftrightarrow (3x-1)^2(4x-3)\ge 0 \]
Thus
\[\sum_{cyc}\cfrac{a}{a^2+1}\le\cfrac{72(x+y+z)+18}{100}=\cfrac{9}{10}. \]

[quote=arqady][quote=harazi]The following one is very nice and not at all easy:
  Prove that if $a,b,c$ are greater than$-\frac{3}{4}$ and add up to 1 then
$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq\frac{9}{10}.$$[/quote]
This inequality is true for all reals $a$, $b$ and $c$ such that $a+b+c=1$.[/quote]

$\sum_{cyc} \frac{2a}{a^2+1}\le \frac{9}{5} \ \ \Leftarrow \sum_{cyc} \frac{(1-a)^2}{a^2+1}\ge \frac{6}{5} \ \ \Leftarrow \sum_{cyc} \frac{(b+c)^2}{a^2+1}\ge \frac{6}{5}$

$\Leftarrow (a^2+b^2+c^2+1)\sum_{cyc} \frac{(b+c)^2}{a^2+1}\ge \frac{6}{5}(a^2+b^2+c^2+1)$

$\Leftarrow \sum_{cyc}(b+c)^2+\sum_{cyc} \frac{(b+c)^2(b^2+c^2)}{a^2+1}\ge \frac{6}{5}(a^2+b^2+c^2+1)$

$\Leftarrow \sum_{cyc} \frac{(b+c)^2(b^2+c^2)}{a^2+1}\ge \frac{1}{5}(a^2+b^2+c^2+1)$

$\Leftarrow \sum_{cyc} \frac{(b+c)^4}{a^2+1}\ge \frac{2}{5}(a^2+b^2+c^2+1)$

$\Leftarrow \frac{(a^2+b^2+c^2+1)^2}{a^2+b^2+c^2+3}\ge \frac{2}{5}(a^2+b^2+c^2+1)$

$\Leftarrow 5(a^2+b^2+c^2+1)\ge 2(a^2+b^2+c^2+3) \Leftarrow 3(a^2+b^2+c^2)\ge 1  \Leftarrow \sum_{cyc} (a-b)^2 \ge 0$

[quote=utkarshgupta]

Without loss of generality assume $a \ge b \ge c$
Then $\implies 3a^2 + 3 -10a \le 3b^2 + 3 - 10b \le 3c^2 +3 - 10c$ and $\frac{1}{a^2+1} \le \frac{1}{b^2+1} \le \frac{1}{c^2+1}$
[/quote]
It's wrong for real variables.

[quote=mudok]$\sum_{cyc} \frac{2a}{a^2+1}\le \frac{9}{5} \ \ \Leftarrow \sum_{cyc} \frac{(1-a)^2}{a^2+1}\ge \frac{6}{5} \ \ \Leftarrow \sum_{cyc} \frac{(b+c)^2}{a^2+1}\ge \frac{6}{5}$[/quote]
From idea of [b]mudok [/b]we can write inequality as
\[\sum \frac{(b+c)^2}{a^2+(a+b+c)^2}\geqslant \frac{6}{5}.\]
Using Cauchy-Schwarz Inequality, we have
\[\sum \frac{(b+c)^2}{a^2+(a+b+c)^2} \geqslant \frac{\displaystyle \left[\sum (b+c)^2\right]^2}{\displaystyle \sum \left[a^2+(a+b+c)^2\right](b+c)^2},\]
and
\[\left[\sum (b+c)^2\right]^2 - \frac65 \sum \left[a^2+(a+b+c)^2\right](b+c)^2 = \frac15  \sum \left[(2a+2b+c)^2+3c^2\right](a-b)^2 \geqslant 0.\]


x,y,z are nonnegative reals with sum 1
prove that
$ 2<\equal{} sum of (1\minus{}x^2)^2<\equal{}(1\plus{}x)(1\plus{}y)(1\plus{}z)$

[quote="ishfaq420haque"]x,y,z are nonnegative reals with sum 1
prove that
$ 2 \leq \sum_{cyc}(1 \minus{} x^2)^2 \leq (1 \plus{} x)(1 \plus{} y)(1 \plus{} z)$[/quote]
$ \sum_{cyc}(1 \minus{} x^2)^2 \leq (1 \plus{} x)(1 \plus{} y)(1 \plus{} z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x\plus{}y)^2(x\plus{}y\plus{}2z)^2\leq(x\plus{}y\plus{}z)\prod_{cyc}(x\plus{}y\plus{}2z)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3y\plus{}x^3z\minus{}2x^2yz)\geq0.$
$ 2 \leq \sum_{cyc}(1 \minus{} x^2)^2\Leftrightarrow\sum_{cyc}(x\plus{}y)^2(x\plus{}y\plus{}2z)^2\geq2(x\plus{}y\plus{}z)^4\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2x^2y^2\plus{}8x^2yz)\geq0.$
Done!

thanks for the solution but is there any solution without such type of multiplications??? :blush:


trigonometry
floor function
ceiling function
function
algebra
polynomial
geometry

At my high school we give problem sets to the students in analysis or above. I thought that it would be a good idea to post a few of the problem sets that we have done so far.
They are not that hard, but there is a great deal of variability in problem difficulty since there is a lot of variability in terms of talent on the team.
Here are the first 3  :wink:
I will not be making more JV Problem Sets because that duty has been relegated to other captains.

Just remember, I could make these problems a lot harder, but i have to think about putting a few easier ones for the freshmen and sophomores.

To download: right-click the links and select "save target as." Also, if the link is being blocked as a pop-up, hold "ctrl" and then click the link.

(There is a maximum number of attachments per post. The limit is 3. I will post the next 3 in another post)

Internet Explorer can't download it and my Firefox isn't working. :|

have you tried right clicking and selecting "save target as"?

also, if it is being blocked as a pop-up, hold ctrl and click the link.

The "Save Target As..." thing worked. Thank you.

You're welcome  :)

Let me know what you think about the problem sets.

Isn't the geometric mean the $\sqrt[n]{a\cdot b\cdot ...}$ where $n$ is the amount of numbers?

Yeah. If you have $n$ real numbers $x_{1}, x_{2}, \ldots, x_{n}$, their geometric mean, GE, is $(x_{1}x_{2}\ldots x_{n})^{1/n}$.

Oh, never mind. In the problem it says the square root of $a$ and $b$ only, not generally.

EDIT: Too bad I'm only in 8th grade. Haven't learned sine and cosine yet. Would've had fun solving those problems. :(

Well a lot of them don't need trigonometry.

Here is an easier one then  :) (no trig.)

Now I know that $\lfloor\ \rfloor$ means floor. :rotfl:

What about $\lceil\ \rceil?$

$f(x)=\lfloor x \rfloor$ is the floor function, or greatest integer function.

the other one is the cieling function, or the least integer function

[quote="Padmavathi"]Hi
Nice problems...i dwnlded all..they're fun!
Which school is this from, and which grade?[/quote]
It says at the top of the documents: Columbus High School - Junior Varsity.

which ones are you guys stuck on? i could give you hints

*tries to crack knuckles but fails again*

Darn!

I think that I can only make comments up here.

I can't figure out 2, because I forgot how to graph circles.  :noo:
I'm not that good at trig, so I can't do 4. At this rate, I can't do #6.

And the pattern works. I can't do #6 because I'm no good at trig.

I can't do #7, because I'm not good at these types of problems.

#11 is impossible, because there are only three equations and there are four variables. There must be a fourth equation so it can be solvable.

#12 is another trig problem.

Absolutely no clue on #14 or #15

I'll now do the second problem set.

[hide="the very first problem set"][hide="#1"]First, we know that the arithmetic mean is equal to the geometric mean. That means $\frac{x+y}{2}=\sqrt{xy}$. We can square both sides to get $\frac{x^{2}+2xy+y^{2}}{4}=xy$ and then we can simplify.
$\frac{x^{2}+y^{2}}{4}+\frac{xy}{2}=xy$
$\frac{x^{2}+y^{2}}{4}=\frac{xy}{2}$
$x^{2}+y^{2}=2xy$
$(x-1)(y-1)=0$
That can only be true if [b]x or y is one.[/b][/hide]

[hide="#3"]Powers of i run in patterns of four: i,-1,-i,1. Therefore, the equation also runs in patterns of four: 0, -2, 0,2. Therefore, the whole equation is equal to $i^{2005}+\frac{1}{i^{2005}}+i^{2006}+\frac{1}{i^{2006}}$, which is equal to [b]-2[/b][/hide]

[hide="#5"]We can simplify the second equation into 6x+6y+x+y=63=7x+7y=7(x+y)
and further into
x+y=9.
We can square that to get
$x^{2}+2(6)+y^{2}=81$
$\boxed{x^{2}+y^{2}=69}$[/hide]

[hide="#8"]We can use the binomial theorem:
$(x+y)^{n}=x^{n}\binom{n}{0}+x^{n-1}\binom{n}{1}+\cdot\cdot\cdot+y^{n}\binom{n}{n}$
And we plug in 1 for x and y.
$\boxed{2^{n}=\binom{n}{0}+\binom{n}{1}+\cdot\cdot\cdot+\binom{n}{n}}$[/hide]

[hide="#9"]We remember the theorem and see that 2(2x+2)=(x+1)(x+2)=4(x+1)
Therefore, [b]x=2[/b][/hide]

[hide="#10"]We reduce to get [b]b/a[/b][/hide]

[hide="#13"]We can add 5's until we reach 104. We can add a ton more 5's to get 1004. Therefore, 1001=f(1004)=f(999)=f(84)

Therefore, [b]f(84)=1001[/b][/hide][/hide]

nutz_for2.718281828

#1: Check your final factorization.

#9: You applied the theorem wrong. Make sure you look at the picture.

#10: Check your cancellation.

#11: Not impossible to determine $w$. Combining equations in a unique way gives you a 4th equation.

#13: Your answer to this problem is incorrect.

Good job on the other problems! Don't be discouraged by the comments I left, just rework some of the problems with more careful reasoning. Here are some hints for the last two problems:

#14: How do you derive the formula for the sum of a standard geometric series? Try using the same approach on this series and see what happens.

#15: Finding the answer to this question requires strong algebraic and number theoretic reasoning. I would suggest looking at it as a number theory problem.

Set 1, problem 10 has a GRAVE error. It should've been $\binom{n}{k}=\binom{n}{n-k}$

problem set 2: #7 careful on this problem, if f(x)=0, then it is undefined, yet it possesses the desired properties, also if f(22) is a root, the value is undefined, you need to restrict the degree of the polynomial

#9 on the jv set is unclear because it does not state wheter or not $a,b,c$ are relatively prime

[quote="drunner2007"]thanks. i'll make the appropriate changes.  :wink:

but on #7 doesn't the wording imply that those are the only roots? the presence of the word "the" should signal that those are the only roots, right?[/quote]

it is still unclear, what if there are double roots? it is easier to just day a polynomial degree n and list the n roots

[quote="drunner2007"]At my high school we give problem sets to the students in analysis or above. I thought that it would be a good idea to post a few of the problem sets that we have done so far.
They are not that hard, but there is a great deal of variability in problem difficulty since there is a lot of variability in terms of talent on the team.
Here are the first 2  :wink:

Just remember, I could make these problems a lot harder, but i have to think about putting a few easier ones for the freshmen and sophomores.

To download: right-click and select "save target as." Also, if the link is being blocked as a pop-up, hold "ctrl" and then click the link.[/quote]

where am I supposed to click?  :maybe:

Ok,
I reattached everything with the corrections, and included a new PSet #3. Have fun with this one (its harder than the first 2 in my opinion).
Also, let me know of any mistakes or comments.

deimos, right-click the link.

altheman, I see what you're saying. I changed the problem.

I should be updating every week.


algebra
polynomial
complex analysis
complex analysis unsolved

[quote="transseries"]Multiply it out, put everything on one side, you get a polynomial of degree 4.  Can you factor it?[/quote]

I don't know if it's a kind of hint or help?

Try to make this factorization more beautiful. Multiply both sides by $ 2 \times 4 \times 6 \times 8$. Then it will be $ 8(3z \plus{} 1)6(4z \plus{} 1)4(6z \plus{} 1)2(12z \plus{} 1) \equal{} (24z \plus{} 8)(24z \plus{} 6)(24z \plus{} 4)(24z \plus{} 2) \equal{} 2(384)$. Put $ a \equal{} 24z \plus{} 2$ so we will have $ a(a \plus{} 2)(a \plus{} 4)(a \plus{} 6) \equal{} (a^2 \plus{} 6a)(a^2 \plus{} 6a \plus{} 8) \equal{} 768$ and again put $ b \equal{} a^2 \plus{} 6a$ and so $ b^2 \plus{} 8b \minus{} 768 \equal{} 0$. Solve it for $ b, a$ and $ z$ resp.


How do you prove $ C_m^n\equal{}\frac{n!}{m!(n\minus{}m)!}$ is an integer?
[hide="My way"]By proving $ C_m^{n\plus{}1}\equal{}C_{m\minus{}1}^n\plus{}C_m^n$ and $ C_0^n\equal{}C_n^n\equal{}1$. Then use the induction, we get the result.[/hide]

it's an integer because it counts the number of ways to do something -- namely, choose $ n$ objects from $ m$ given ones.

[hide="Another Way"]Write the binomial coefficient as $ \frac{n(n\minus{}1)(n\minus{}2)\cdots (n\minus{}m\plus{}1)}{m(m\minus{}1)(m\minus{}2)\cdots 1}$.  We count the number of factors of some prime $ p$ in both the numerator and denominator by adding the number of terms divisible by $ p$, the number of terms divisible by $ p^{2}$, the number of terms divisible by $ p^{3}$, etc.  Suppose there are $ n$ terms divisible by $ p^{k}$ in the denominator.  Then $ m\ge n\cdot p^{k}$.  But the terms in the numerator are $ m$ consecutive integers.  Since $ m\ge n\cdot p^{k}$, we can take $ n\cdot p^{k}$ consecutive terms, $ n$ of which are divisible by $ p^{k}$.  Therefore there are at least $ n$ terms divisible by $ p^{k}$ in the numerator, and we're done.[/hide]

Similarly, the multinomial $ \frac{(a_1 \plus{} a_2 \plus{} ... \plus{} a_n)!}{a_1! a_2! ... a_n!}$ is an integer (there are arguments in both matt's and pleurestique's styles).


Ooh, ooh, pick me!

The practice was sort of silly. They seem unable to make good problems and therefore just give ugly ones.

EDIT:
[quote="redcomet46"]Meh...[/quote]because she refuses to post.

[quote="jrshoch"]I support the revival

Whoever that kid is with the long red hair obviously is smart if he solved all those power round questions so quickly, but I think he has to work on the teamwork part... he could've explained his ideas to another kid and not have had to write two pages of proof by himself in twenty minutes[/quote]The power round at the actual competition should go much better. since we will have stuff for people to do instead of just 4 questions.

YAY 1000 POSTS lol

Honestly, this is just to get something happening in here.

ARML was definitely fun.

Edit: WHAT

YOU POSTED

WOAH

Oh, uh... So it appears I'm the new mod in team. Y'all tread lightly around here, y'hear?

[quote="Ihatepie"]hey. I'm going to high school next year.[/quote]Nice. How was Mathcounts?

[quote="Ihatepie"]made nats.

I messed up a bit, so I got 79th place. Ct got 19th.[/quote]You should come to ARML next year.

[quote="Ihatepie"]how do I join?[/quote]I'll send you the info next February. If I forget, ask the FCML people.

I'm still in contact with the other ct state mc team, should I contact them about this?

I'll tell Lucian to come myself. Just tell the others to come next February or whatever.

Does anyone know how exactly the Connecticut ARML A team is going to get our calculators? There was that post on the eboard that talked about meeting up, but it has been a while and there are no signs of plans. They really should just mail them to us. I'd be more than happy to just pay the $5 shipping.

B is supposedly going to get them in the mail at some point. He has not yet received them.

ARML-American Regions Math League

Yikes, that was a huge bump...


Find all continuous functions $f: \mathbb{R}\to\mathbb{R}$ such that \[f(xyz)=f(x)f(y)f(z)\;\forall x,y,z \in\mathbb{R}.\]

$x=y=z=0$ ,therefore $f(0) = f^{3}(0)$, so there are 2 cases:
$I. f(0)=0$, so now taking $x=y=z=1$, we have another 2 subcases :
a)$f(1)=0$, but then taking $y=z=1$, we have $f(x)=0$.
b)$f(1)=1$, and now taking $z=1$, we have that $f(xy)=f(x)f(y)$, but this is cauchy, so the solutions are $f(x) =x^{c}$, or $f(x) = |x|^{c}$, or $f(x)=0$

$II.f(0)=1$, but now we take $y=z=0$, therefore $f(x)=f(0)=1$.

so the solutions are: $f(x)=0, f(x)=1, f(x)=x^{c}, f(x)=|x|^{c}$.

1.You are wrong at  a point ( guess where :P ), because $f(x)=x^{c}$ clearily works.
2. Take $z=1$ and define the function $g(x)=f(x)\cdot f(1)$. The rest is standard. If anybody wants, I will post my solution.

continuous solutions are $f\equiv 0$, $f\equiv 1$, $f\equiv-1$
and $f(x)=\{\begin{array}{ll}0&\text{for }x=0\\ \epsilon x^{c}&\text{for }x>0\\ \delta |x|^{c}&\text{for }x<0 \end{array}$
for some $c\in\mathbb{R}_{+}$ and $\epsilon,\delta\in\{\pm 1\}$.

yes, i edited now, i did a mistake when i wrote the solutions to $f(xy) = f(x)f(y)$


floor function
combinatorics unsolved
combinatorics

Hi everyone,
Prove that $p(n)\ge 2^{ \lfloor \sqrt n \rfloor}$  for $n\ge 2$

$p(n)$ stands for the number of partitions of $n$
sorry for that.

Note that $p(n)$ is non-decreasing, because any partition of $n$ can become a partition of $n+1$ by adding an extra $1$. Then it's enough to prove the problem for $n$ a perfect square, say $k^2$. We want to prove $p(k^2) \geq 2^k$. But for this bound, we can easily give $2^k$ distinct partitions of $k^2$: take $k+1$ slots, one with a $1$, one with a $2$, ..., one with a $k$ and an empty slot. You can either take or leave the number in each slot; the final slot will be $k^2-S$, with $S$ the sum of the chosen slots. It's easy to see that the $2^k$ possible partitions are all different, and they're all legal, since $k^2-S \geq k^2-\frac{k(k+1)}{2}=\frac{k(k-1)}{2}>k$ for $k \geq 4$. The smaller cases can be done by hand.


Let $ x;y;z\ge 0$
Prove that
$ (x\plus{}y\plus{}z)^6\ge (x^2\plus{}yz)(y^2\plus{}xz)(z^2\plus{}xy)$

[quote="tranvanluan"]Let $ x;y;z\ge 0$
Prove that
$ (x \plus{} y \plus{} z)^6\ge (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)$[/quote]
$ (x \plus{} y \plus{} z)^6\geq64(x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)\geq (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy).$

[quote="arqady"]    $ (x \plus{} y \plus{} z)^6\geq64(x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy)\geq (x^2 \plus{} yz)(y^2 \plus{} xz)(z^2 \plus{} xy).$[/quote]
Explain your steps, arqady. :wink:

$ (x \plus{} y \plus{} z)^2 \ge x^2 \plus{} (y \plus{} z)^2 \ge x^2 \plus{} 4yz \ge x^2 \plus{} yz$ etc.

or

$ 4(x \plus{} y \plus{} z)^2 \equal{} [(x \plus{} y) \plus{} (y \plus{} z) \plus{} (z \plus{} x)]^2 \ge \sum (x \plus{} y)^2$
$ \equal{} 2 (\sum x^2 \plus{} yz) \ge 6 \sqrt [3]{\prod (x^2 \plus{} yz)}$

$ \Rightarrow (x \plus{} y \plus{} z)^2 \ge \frac {3}{2} \sqrt [3]{\prod (x^2 \plus{} yz)}$

$ \Rightarrow (x \plus{} y \plus{} z)^6 \ge \frac {27}{8} (x^2 \plus{} yz)(y^2 \plus{} zx)(z^2 \plus{} xy)$


Hi all mathlinkers,
I'm a newcomer.
All about myself is that I love maths but find it hard to solve problems.  :o
I like to know results of IMO team of India & Bangladesh.
Can anyone help,guys?

Thank you. :jump:
Will u say how2 make my own page?

Do you mean how to create a blog?
Go to the "profile" and there will be an option "Create a blog" :wink:


i was wondering if anyone had any neat ideas on the problems and just wanted to discuss them (if that is allowed?)

People can still email or fax solutions until midnight.

no....until 11:59 October 3rd....i think

Please do not discuss the problems until next week.


In programa de olimpiada de anul acesta, se cere teorema lui Jarnik referitoare la proprietatea lui Darboux a raportului a două funcţii derivate. Are cineva idee in ce consta?

I think it is the following theorem.

[b]Theorem.[/b] Suppose $f$ and $g$ are differentiable on $[a,b]$ and $g'(x)\neq 0$ on $[a,b]$. Then ration $\frac{f'(x)}{g'(x)}$, where $x\in [a,b]$, takes all intermediate values between $\frac{f'(a)}{g'(a)}$ and $\frac{f'(b)}{g'(b)}$.

[b]Proof.[/b] Note that $g'(a)$ and $g'(b)$ have the same sign. Indeed, otherwise there is $c\in [a,b]$ s.t. $g'(c)=0$, since $g'$ has the Darboux property.
Let $\lambda$ be an arbitrary intermediate value between $\frac{f'(a)}{g'(a)}$ and $\frac{f'(b)}{g'(b)}$. Consider the function $F(x)=f(x)-\lambda g(x)$. We have $F'(a)=g'(a)\left(\frac{f'(a)}{g'(a)}-\lambda\right)$ and $F'(b)=g'(b)\left(\frac{f'(b)}{g'(b)}-\lambda\right)$, thus $F'(a)$ and $F'(b)$ have different signs. It follows that there is $c\in [a,b]$ s.t. $F'(c)=0$ (due to Darboux property for $F'$). But then for the point $c$ we obtain $f'(c)-\lambda g'(c)=0$, so $\frac{f'(c)}{g'(c)}=\lambda$. QED.


LaTeX
number theory proposed
number theory

Hi to all
a nice and not hard problem :
prove every integer can be written in this form:
((a²+b²+c²+d²+e²)/(1+abcde)),a,b,c,d,e∈N

 :)

Casue you are Amaetur and new commer, You must learn somthings.
[b]1st[/b].Name of Topics must say what problem  is about in best way .

Read this :
http://www.mathlinks.ro/viewtopic.php?t=135915
:p

[b]2nd.[/b] You must learn $ \text{\LaTeX}$ , for learning read here: http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14262

Your Problem in Latex form :
$ \forall t\in \mathbb{N}\ \ \exists a,b,c,d,e\in \mathbb{N}$ that $ t=\frac{a^2+b^2+c^2+d^2+e^2}{1+abcde}$


Dena and Shibani each took three math tests. The median of Dena's three scores is the same as Shibani's. Their lowest scores are also the same. However, Dena's mean is 2 points higher than Shibani's mean? How many points higher is Dena's best test score than Shibani's?

Call Dena's scores $ D_1, D_2,$ and $ D_3$.
Call Shibani's scores $ S_1, S_2,$ and $ S_3$.
The problem gives us that $ D_1 \equal{} S_1$. Also, since the median (middle number) is the same, $ D_2 \equal{} S_2$. Thus, we must find $ D_3 \minus{} S_3$. We know that the difference of the averages was $ 2$ in Dena's favor. Since there are $ 3$ tests, and the other $ 2$ were the same scores, the answer is $ 2 \times 3 \equal{} \boxed{6}$.


trigonometry
invariant
algebra unsolved
algebra

For a finite set $ X$ of positive integers, let $ \Sigma(X) \equal{} \sum_{x \in X} \arctan \frac{1}{x}.$ Given a finite set $ S$ of positive integers for which $ \Sigma(S) < \frac{\pi}{2},$ show that there exists at least one finite set $ T$ of positive integers for which $ S \subset T$ and $ \Sigma(S) \equal{} \frac{\pi}{2}.$

[i]Kevin Buzzard, United Kingdom[/i]

[quote="orl"]For a finite set $ X$ of positive integers, let $ \Sigma(X) \equal{} \sum_{x \in X} \arctan \frac {1}{x}.$ Given a finite set $ S$ of positive integers for which $ \Sigma(S) < \frac {\pi}{2},$ show that there exists at least one finite set $ T$ of positive integers for which $ S \subset T$ and $ \Sigma(S) \equal{} \frac {\pi}{2}.$[/quote]

Quite nice problem.

$ \tan(\Sigma(S))$ is a rational number and we are looking for a complementary set $ U$ such that $ \tan(\Sigma(U))$ $ \equal{}\frac{1}{\tan(\Sigma(S))}$. Then $ T\equal{}S\cup U$ would be an answer.

So it would be enough to show that, given any positive rational number $ \frac ab$ (with $ \gcd(a,b)\equal{}1$), we can always find a set $ U$ such that $ \tan(\Sigma(U))\equal{}\frac ab$

If $ a\equal{}1$, then $ U\equal{}\{b\}$ is the answer.
If $ a>1$, let then $ n\equal{}1\plus{}[\frac ba]$ and $ x\equal{}\frac{an\minus{}b}{bn\plus{}a}$. If we find a set $ V$ such that $ \tan(\Sigma(V))\equal{}x$, then $ \tan(\Sigma(V\cup\{n\}))\equal{}\frac ab$ and we got the answer.
It remains just to see that $ n\equal{}1\plus{}[\frac ba]$ and $ a>1$ $ \implies$ $ n\minus{}1<\frac ba < n$ $ \implies$ $ 0 < an\minus{}b < a$
And so the numerator of $ x$ is less than $ a$ and we are sure we can repeat the process a finite number of times to get the result.

The end.

[quote="pco"][quote="orl"]For a finite set $ X$ of positive integers, let $ \Sigma(X) \equal{} \sum_{x \in X} \arctan \frac {1}{x}.$ Given a finite set $ S$ of positive integers for which $ \Sigma(S) < \frac {\pi}{2},$ show that there exists at least one finite set $ T$ of positive integers for which $ S \subset T$ and $ \Sigma(S) \equal{} \frac {\pi}{2}.$[/quote]

Quite nice problem.

$ \tan(\Sigma(S))$ is a rational number and we are looking for a complementary set $ U$ such that $ \tan(\Sigma(U))$ $ \equal{} \frac {1}{\tan(\Sigma(S))}$. Then $ T \equal{} S\cup U$ would be an answer.

So it would be enough to show that, given any positive rational number $ \frac ab$ (with $ \gcd(a,b) \equal{} 1$), we can always find a set $ U$ such that $ \tan(\Sigma(U)) \equal{} \frac ab$

If $ a \equal{} 1$, then $ U \equal{} \{b\}$ is the answer.
If $ a > 1$, let then $ n \equal{} 1 \plus{} [\frac ba]$ and $ x \equal{} \frac {an \minus{} b}{bn \plus{} a}$. If we find a set $ V$ such that $ \tan(\Sigma(V)) \equal{} x$, then $ \tan(\Sigma(V\cup\{n\})) \equal{} \frac ab$ and we got the answer.
It remains just to see that $ n \equal{} 1 \plus{} [\frac ba]$ and $ a > 1$ $ \implies$ $ n \minus{} 1 < \frac ba < n$ $ \implies$ $ 0 < an \minus{} b < a$
And so the numerator of $ x$ is less than $ a$ and we are sure we can repeat the process a finite number of times to get the result.

The end.[/quote]

No. Elements in set need to be distinct. You only proved this problem for list.

[quote="greentreeroad"]No. Elements in set need to be distinct. You only proved this problem for list.[/quote]

You are perfectly right and I thank you for having noticed this.

This can be rather easily solved using the fact that $ \arctan(\frac 1p)\equal{}\arctan(\frac 1{p\plus{}1})\plus{}\arctan(\frac 1{p^2\plus{}p\plus{}1})$

So we can replace in $ U$ any "bad" element $ p$ by the couple $ \{p\plus{}1,p^2\plus{}p\plus{}1\}$ so to transform the list in set (with $ U\cap S\equal{}\emptyset$). This process of replacement need to be conducted with some caution in order to avoid creating infinitely many double values, but it's rather easy to find such a process.

[quote="pco"]So we can replace in $ U$ any "bad" element $ p$ by the couple $ \{p \plus{} 1,p^2 \plus{} p \plus{} 1\}$ so to transform the list in set (with $ U\cap S \equal{} \emptyset$). This process of replacement need to be conducted with some caution in order to avoid creating infinitely many double values, but it's rather easy to find such a process.[/quote]

How does this process work precisely?
What are the invariants?

It is not obvious to me.

We can modify pco's proof in the following way: say the sum we have is arctan(a/b), then you want to come up with some more denominators such that the arctan sum becomes arctan(b/a). We can keep subtracting the largest possible fraction 1/n<b/a that is not in S. Clearly, the arctan(b'/a') value left will always have b'/a'>0 (or 0, in which we're done), and we basically keep going until we've cleared all the elements of S (we're subtracting smaller fractions than all the ones in S). It's not too hard to show that (1) we won't subtract the same fraction twice and (2) we'll eventually hit a point where all the fractions we take are such that 1/n<b/a<1/(n-1). From here, the numerators will decrease.

(We'll need the observation that arctan(1/n)~1/n)

For the sake of completeness...

[quote="test20"]How does this process work precisely?
What are the invariants?[/quote]It's basically the same as the proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=266774]here[/url], except you do $n\to\{n+1,n^2+n+1\}$ instead of $n\to\{n+1,n^2\}$.

BTW, the same thing (with $n\to\{n+1,n(n+1)\}$) works for Egyptian fraction representations as well (but so does Catalyst's greedy algorithm).

pco your "easy process" is actually very hard and not at all trivial

Best thing for this problem is Catalyst's

Copied from my solution in Math StackExchange.
(I indeed know that this is pretty much a rewording of pco's proof)

I will prove a stronger statement - the RMM problem itself.
It generalizes the two integers $x,y$ to a set of positive integers.

First, for convenience, we define $f(x)=\arctan x$.

Also, for a set $X$, we define $g(X)=\sum_{x \in X} f(\frac{1}{x})$.

Now, say we are given a set $X$, where $g(X)=f(\frac{a}{b}) < \frac{\pi}{2}$ and $(a,b)=1$.
It suffices to find a set $Y$, such that $X \cap Y = \emptyset$ and $g(Y)=f(\frac{b}{a})$, since we will have $$g(X \cup Y)=g(X)+g(Y)=f(\frac{a}{b})+f(\frac{b}{a})=\frac{\pi}{2}$$

Disregard the fact that we need distinct positive integers, for now.

Now if $b=1$, we are done since we can just let $Y=\{a\}$ to finish the problem.

Assume that $b \ge 2$. Now set an integer $l \in Y$ which will be determined later.
We have $$g(Y-\{l\})=g(Y)-f(\frac{1}{l}) = f(\frac{b}{a})-f(\frac{1}{l})=f(\frac{bl-a}{al+b})$$
We want $0<bl-a<b$, which motivates us to set $l=1+\lfloor \frac{a}{b} \rfloor$, since $a$ cannot be a divisor of $b$.
Now the numerator strictly decreases, since $\frac{a}{b}<l<1+\frac{a}{b}$. Therefore, after performing this finitely, we will have the numerator equal to $1$, i.e. $b=1$, a case we already had a look on.

[b]NOTE: If this is uncomfortable, use strong induction on the numerator.[/b]

[quote="Induction Proof"]The base case $b=1$ is the same.

Assume that there is a set $P$ such that $g(P)=f(\frac{x}{y})$ if $x<b$.
Since $0<bl-a<b$, by induction hypothesis, we have a set $P$ such that $g(P)=f(\frac{bl-a}{al+b})$.
Add the number $l$ to find $g(P \cup \{l\})=f(\frac{b}{a})$, which completes the induction.[/quote]

We have determined set $Y$, but we don't know that it has a repeating number.

Perform the map $t \mapsto \{t+1,t^2+t+1\}$ on the elements of $Y$.

Since the functions $p(x)=x+1$ and $q(x)=x^2+x+1$ are not bounded above, we can perform it finitely to replace the repeating numbers.

To make this argument more rigorous, order the elements by size and start from the smallest number. When encountered with a repeating number(s), keep performing the map on the resulting numbers until the numbers that result from this map are larger than any other element.

Since the number of the elements of the original set is finite, the number of times that this map needs to be used is finite as well, so the resulting set is finite. We are done. $\blacksquare$


modular arithmetic
quadratics
number theory
relatively prime
combinatorics unsolved
combinatorics

Let $p\ge5$ be a prime and let $r$ be the number of ways of placing $p$ checkers on a $p\times p$ checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that $r$ is divisible by $p^5$. Here, we assume that all the checkers are identical.

Considering the number of all possible arrangements, it will be $C_{p^{2}}^{p}$. We have $p$ "bad" arrangements.
So we must prove that:
$C_{p^{2}}^{p}-p \vdots p^{5} \Leftrightarrow$
$\frac{p^{2}\cdot(p^{2}-1)\cdot\ldots(p^{2}-p+1)}{p!}-p \vdots p^{5}\Leftrightarrow$
$\frac{p^{2}\cdot(p^{2}-1)\cdot\ldots(p^{2}-p+1)-p\cdot p!}{p!}$ $\vdots$ $p^{5}\Leftrightarrow$
$\frac{p(p^{2}-1)(p^{2}-2)\ldots (p^{2}-(p-1))-p!}{(p-1)!}$ $\vdots$ $p^{5}$. Since $p$ is a prime number, we ignore $(p-1)!$. Dividing by $p$, we must prove that:
$(p^{2}-1)\ldots (p^{2}-p+1)-(p-1)!$ $\vdots$ $p^{4}$.
In the process of opening the brakets, if we select 2 $p^{2}$'s, then the whole product will be divisible by $p^{4}$.
So, we must prove that:
$p^{2}(\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\ldots \frac{(p-1)!}{p-1})+(-1)^{p-1}(p-1)!-(p-1)!$ $\vdots$ $p^{4}$.
Since $(-1)^{p-1}=1$, then it suffices to prove that:
$\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\ldots \frac{(p-1)!}{p-1}$ $\vdots$ $p^{2}\Leftrightarrow$
$(p-1)!\cdot(1+\frac{1}{2}+\ldots+\frac{1}{p-1})$ $\vdots$ $p^{2}$
The last follows immediately from Wolstenholme's Theorem.

I would just like to comment on the official solution for this problem.  Apparently, everything that Freemind wrote above is worth 3 marks, and the organisers had expected the contestants to prove Wolstenholme's Theorem for the next 4 marks, without realising that it was a well-known theorem.  As a result, according to the official guidelines, anybody who quotes Wolstenholme's Theorem in addition to whatever is written above is awarded 7 marks, whereas those who do not know that it is a well-known result are awarded 3. :mad:

That is a little strange.
Actually, up till yesterday, i didn't know that it is a result of Wolstenholme. I just met this in another problem, so i thought it should be known problem, and starting looking for it. I discovered in a book that it was actually a theorem.
:)

I know the Wolstenholme's  theorem.

but i can't und how freemind brought that fractional form from that product form? can anybody clr me?

mahbub, Wolstenholme states that the numerator of $1+\frac{1}{2}+\ldots+\frac{1}{p-1}$ is divisible by $p^{2}$. Since $p$ is a prime number, the denominator will not be divisible by $p$. Let $1+\frac{1}{2}+\ldots+\frac{1}{p-1}=\frac{M}{N}$. Then $M=p^{2}m$, $N=(p-1)!$. Multiplying by $(p-1)!$ we get $mp^{2}$. I didn't put the condition that $(M,N)=1$, because ii isn't necessary. That's why we may take simply $N=(p-1)!$.

[quote="shyong"][quote="freemind"]So, we must prove that:
$p^{2}(\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\ldots \frac{(p-1)!}{p-1})+(-1)^{p-1}(p-1)!-(p-1)!$ $\vdots$ $p^{5}$[/quote]

I think mahbub wants to know how you arrive to this step instead of Wolstenholme Theorem :P[/quote]
Yes, I did a mechanical mistake :( . I edited already. I wrote we must prove that is divisible by $p^{4}$, but in the relation, I wrote $p^{5}$. :)
I think everything is clear now.

[quote="freemind"]${p(p^{2}-1)(p^{2}-2)\ldots (p^{2}-(p-1))-p!}{(p-1)!}$ $\vdots$ $p^{5}$.

$p^{2}(\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\ldots \frac{(p-1)!}{p-1})+(-1)^{p-1}(p-1)!-(p-1)!$ $\vdots$ $p^{4}$.
[/quote]

How does the second line comes from the first line? I mean by which transformation u got the second line?

[I hope u can understand my Q now ;) )

How many marks do i get, if I proved half of the Wolstenholme's Theorem. I proved that $p|\sum_{i=1}^{p-1}\frac{1}i$ but I couldn't prove it for $p^2$

Sorry, but thats not half of it... I would say it is not even a part of it... However, about marks, i dont really know...

Have you read the first posts!! They finished the problem using Wolstenholme's Theorem, I did the same, but i didn't knew it was a known result. But i didn´t stop at that point, I did a little bit more than that. So it really is a part of it.

[quote]I proved that  $p|\sum_{i=1}^{p-1}\frac{1}{i}$ but I couldn't prove it for $p^2$[/quote]

What I meant is that it is not even half of Wolstenholme's theorem, in fact showing that $p$ divides it is trivial while for $p^2$ it is not trivial at all!!! So maybe the problem ended that way, but if you didnt prove Wolstenholme or neither mentioned you wont get many marks!!! Check the marking scheme. I personally checked part of your test and i can tell you wont get lot of marks here.

BTW A theorem is a theorem, you cannot talk about "half" of a theorem, in my opinion.

No more spam.

[quote="silouan"][quote="freemind"], we must prove that:
$(p^{2}-1)\ldots (p^{2}-p+1)-(p-1)!$ $\vdots$ $p^{4}$.

So, we must prove that:
$p^{2}(\frac{(p-1)!}{1}+\frac{(p-1)!}{2}+\ldots \frac{(p-1)!}{p-1})+(-1)^{p-1}(p-1)!-(p-1)!$ $\vdots$ $p^{4}$.
.[/quote]

Could you explain how you reach to the second  from the first  ?? I  can't see this[/quote]
When opening the braktes, if you choose 2 factors $p^{2}$, then their product with other $p-3$ terms, will be divisible by $p^{2}\cdot p^{2}=p^{4}$.
So we must consider, when opening the brakets, only the terms containing one $p^{2}$, or no$p^{2}$.
There is only one term containing no $p^{2}$. That is $(-1)(-2)\ldots (-(p-1))=(-1)^{p-1}\cdot(p-1)!$.
If we choose one $p^{2}$ from some bracket, say from $p^{2}-k$, then it will be multiplied by $(-1)(-2)\ldots(-(k-1))(-(k+1))\ldots(p-1)=(-1)^{p-2}\frac{(p-1)!}{k}$. When considering the divisibility, the minus doesn't matter, since $(-1)^{p-1}(p-1)!-(p-1)!=0$

Thank you nice explanation and nice solution . ;)

Via a counting argument,
\[r=\binom{p^2}{p}-p\]
We now want:

 \begin{align*} \frac{p^2\left(p^2-1\right)\cdots \left(p^2-p+1\right)}{p\left(p-1\right)!}\equiv p\pmod{p^5}\\\frac{\left(p^2-1\right)\cdots \left(p^2-p+1\right)}{\left(p-1\right)!}\equiv 1\pmod{p^4} \\  \left(p^2-1\right)\cdots \left(p^2-p+1\right)-\left(p-1\right)!\equiv 0\pmod{p^4} \\ p^2\left(-1\right)^{p-2}\left(\sum_{i=1}^{p-1}\frac{(p-1)!}{i}\right)+\left(-1\right)^{p-1}\left(p-1\right)!-\left(p-1\right)! \equiv 0\pmod{p^4}\\ -p^2\left(p-1\right)!\sum_{i=1}^{p-1}\frac{1}{i}\equiv 0\pmod{p^4}\end{align*}

Which is true by Wolstenholme’s theorem.

How do you do mods with fractions? How does it work? For example, can you add fractions in mods, like $\frac{1}{2}+\frac{1}{3}\equiv\frac{5}{6} (mod ...)$ or something?

Once $m$ is coprime with $n$, it is invertible modulo $n$, so there exists its inverse, denoted $m^{-1}$ modulo $n$.
To write $m^{-1} = \dfrac {1}{m}$ is just an issue of notation, thus allowed. As for your other question, yes, say you have $2^{-1}+3^{-1} \equiv a \pmod{n}$, where $\gcd(n,6)=1$. Multiply by $6$ to get $5=2+3 \equiv 6a \pmod{n}$. Since $6$ is invertible, we can also write  $5\cdot 6^{-1} \equiv a \pmod{n}$, i.e. $2^{-1}+3^{-1} \equiv 5\cdot 6^{-1} \pmod{n}$.

So technically, $m^{-1}$ or $\frac{1}{m}$ are integers $mod  n$ for $m,n$ relatively prime? Digressing a little, this means that $(\frac{a}{b})^2\equiv (a\cdot b^{-1})^2 \equiv 2(mod  n)$ implies 2 is a quadratic residue of $n$? since $\frac{a}{b}$ is an integer?

Because $r={{p^2}\choose{p}}-p$ we can easily reduce it to $p^4 | \displaystyle\prod_{i=1}^{p-1} (p^2-i) - \displaystyle\prod_{i=1}^{p-1} (i) $. Expanding the first product out, it reduces to $p^2 | \displaystyle\sum_{i=1}^{p-1} \displaystyle\frac{(p-1)!}{i}$ which is equivalent to the Wolstenholme Lemma.

It suffices to show that $\dbinom{p^2}{p} \equiv p\pmod{p^5}$
Expanding it,we get \begin{align*}
\dfrac{p^2(p^2-1)\cdots (p^2-p+1)}{p!} &\equiv p\pmod{p^5}\\
(p^2-1)\cdots (p^2-p+1) &\equiv (p-1)!\pmod{p^4}\\
(p-1)! - p^2(p-1)!\left(1^{-1}+2^{-1}+\cdots +(p-1)^{-1}\right) &\equiv (p-1)! \pmod{p^4}\\
1^{-1}+2^{-1}+\cdots +(p-1)^{-1} &\equiv 0 \pmod{p^2}
\end{align*} which is just Wolstenholme's Theorem.$\blacksquare$


\[ \ x,y,z>0;x\plus{}y\plus{}z\equal{}1 .\]
Prove that :\[ \ 1\plus{}5(xy\plus{}yz\plus{}zx)\geq 9(x\plus{}y)(y\plus{}z)(z\plus{}x)\]
 :D  :D

[quote="mai quoc thang"]
\[ \ x,y,z > 0;x \plus{} y \plus{} z \equal{} 1 .
\]
Prove that :
\[ \ 1 \plus{} 5(xy \plus{} yz \plus{} zx)\geq 9(x \plus{} y)(y \plus{} z)(z \plus{} x)
\]
:D  :D[/quote]
 $ 1 \plus{} 5(xy \plus{} yz \plus{} zx)\geq 9(x \plus{} y)(y \plus{} z)(z \plus{} x)$

$ \leftrightarrow 1 \plus{} 5(xy \plus{} yz \plus{} zx)\geq 9(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$

$ \leftrightarrow 1 \geq 4(xy \plus{} yz \plus{} zx) \minus{} 9xyz$

postulate $ x \equal{} min(x,y,z) \rightarrow x\le \frac {1}{3}$

put $ f(x,y,z) \equal{} 4(xy \plus{} yz \plus{} zx) \minus{} 9xyz$

$ f(x,\frac {y \plus{} z}{2},\frac {y \plus{} z}{2}) \minus{} f(x,y,z) \equal{} (y \minus{} z)^2(1 \minus{} \frac {9}{4}x)\ge 0$   $ (x\le \frac {1}{3})$

$ \rightarrow f(x,\frac {y \plus{} z}{2},\frac {y \plus{} z}{2})\ge f(x,y,z)$

we need to prove $ f(1 \minus{} 2t,t,t)\le 1$ (with $ t \equal{} \frac {b \plus{} c}{2}$)

$ \leftrightarrow 4[2t(1 \minus{} 2t) \plus{} t^2] \minus{} 9(1 \minus{} 2t)t^2\le 1$ (easy:D a variable )

[quote="bunhiacovski"][quote="mai quoc thang"]
\[ \ x,y,z > 0;x \plus{} y \plus{} z \equal{} 1 .
\]
Prove that :
\[ \ 1 \plus{} 5(xy \plus{} yz \plus{} zx)\geq 9(x \plus{} y)(y \plus{} z)(z \plus{} x)
\]
:D  :D[/quote]
 $ 1 \plus{} 5(xy \plus{} yz \plus{} zx)\geq 9(x \plus{} y)(y \plus{} z)(z \plus{} x)$

$ \leftrightarrow 1 \plus{} 5(xy \plus{} yz \plus{} zx)\geq 9(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$

$ \leftrightarrow 1 \geq 4(xy \plus{} yz \plus{} zx) \minus{} 9xyz$

[/quote]
oh,is a Schur Ineq
$ 9xyz\ge (x\plus{}y\plus{}z)(4(xy\plus{}yz\plus{}zx)\minus{}(x\plus{}y\plus{}z)^2)\equal{}4(xy\plus{}vz\plus{}zx)\minus{}1$


My favorite is Toad (I know, it's not that obvious) and Bullet Bike

Anyone try Zip Zip a lot? What's it like?

I've played that game once and i like yoshi and i forget the vehicle.

Something like Cucumber.

Toad and the Cheep Charger :)
Favorite course: Rainbow Road (it's hard AND fun :))

It's a remake of the GBA track right? I remember my old neighbor playing with her Mario Kart... there was a Rainbow Road on there I think; it was easier ("-_-)

[quote="spaceguy524"]Yoshi and Sugarscoot.

Favorite course is also Rainbow Road.[/quote]

I HATE Rainbow Road. And Sherbet Land, Ghost Valley, and Bowser Castle (Lightning Cup)

Also, why sugarscoot? I use MachBike.

[quote="Math Champion"]Luigi and w/e the fastest car is.[/quote]

Toad/Bullet bike pwns. But my second choice is Yoshi or Mii and Mach Bike.

Toad, regular bike.
how do you unlock stuff?

Attempt to beat the game in Grand Prix mode...
 :P

I've beat everything for 50cc for grand prix and still not unlocking stuff.

dude spear PWNS on Luigi circuit, almost all my friends can beat the expert ghost with that

[quote="spaceguy524"][quote="AwesomeToad"]Really fast bike, okayish sort of bad drift, pwns on time trials where you don't need lots of drift[/quote]

Small, medium, or large?

How do you increase your driver's rankings?

I just beat all the grand prixes (all=1st place).[/quote]

Big.

To increase drivers rankings, do all the cups again get 1st on almost all, well win by a sizable amount, don't fall, use your items wisely... blahbalh blah

Good luck :)

Hey... don't you need 12 fast staffs for that or something?

btw, you know internet? How do people get like 3 stars or 2 stars, or one big star, or a silver wheel, or other stuff in their names?

and what's driver's rankings?  :stink:

:wallbash:
I just noticed that SonyWii is incorrect... it is nintendo wii.... wow (at myself)

[quote="AwesomeToad"]Just pwn on all the cups and don't screw up much. Gah--don't be toad on mushroom gorge, found that out the hard way[/quote]

Why not?

Ok toad is sort of pwn hahaha :)
but I got rammed by heavy guys on the mushrooms (waluigi, funky/kong, wario,...)

ARRGHHH :mad:  :mad:
POW and lightning PWN on Mushroom gorge

wait you guys... how do people get 3 stars or 2 or 1 star on their internet names in wifi?!

:(

Rankings are based on:

1) How many times you finish 1st
2) Amount of time spent in 1st
3) Item hits (hitting others with items)
4) Item hits received (getting hit with items)

[quote="spaceguy524"][quote="AwesomeToad"]1) Getting out 3 times in a row
Description: I was going on the shortcut on warios gold mine. Using Flame Runner/Funky Kong

I stopped to avoid a gold cart, then went forward, since the acceleration sort of sucked, I fell short of the other side.

I got back in 11th. Then a STAR rammed me off the track. I got back on and got run over by a GOLD cart

2) My friend got caterquacked 3 times in a row teehee hee (DK/Flame Runner)
I distracted him by **accident**, then he ran into a duck. He got back on the track, then got run over again. Then he got back on and got rammed into a caterquck

Boy he was mad[/quote]

For your friend, what track?

I hate blue spiny shells. I go from 1st place to 7th place.[/quote]

Peach Beach, GCN. Where the ducks are :P

I hate spiny shells too. And red shells if not blocked. 2nd to 11th???


Prove that for every positive inetger $n$ there exist a unique ordered pair $(a,b)$ of positive integers such that
$n=\frac{(a+b-1)(a+b-2)}{2}+a$

Obviously, there exists a unique positive integer m, such that

$1+2+\ldots+(m-1)<n\leqslant 1+2+\ldots+(m-1)+m$.

Then let $a = n-(1+2+\ldots+(m-1)) = n-\frac{m(m-1)}{2}$ and $b = (1+2+\ldots+(m-1)+m)-n+1=\frac{m(m+1)}{2}-n+1$. Here $a$ and $b$ are positive integers and $m=a+b-1$, that's why

$n = \frac{m(m-1)}{2}+a=\frac{(a+b-1)(a+b-2)}{2}+a$.


calculus
integration
function
geometry
rectangle
calculus computations

evaluate the integral
$\int_{0}^{1}(\sqrt[2006]{1-x^{2007}}-\sqrt[2007]{1-x^{2006}})$
where
$\sqrt[n]{x}$
stands the n-th root of x

1) You forgot the $dx$.
2) This does not belong here; it belongs in the Calculus section.

Here is how I would solve this problem.

More generally we have that:

$\int_{0}^{1}\sqrt[a]{1-x^{b}}dx=\frac{1}{b}\beta(\frac{1}{a}+1,\frac{1}{b})$,

where $\beta$ is the Beta function and $a$ and $b$ are positive real numbers.
 The integral can be evaluated via the substitution $x^{b}=y$.
Therefore your integral is equal to zero, since

$\frac{1}{2007}\beta(1+\frac{1}{2006},\frac{1}{2007})-\frac{1}{2006}\beta(1+\frac{1}{2007},\frac{1}{2006})=0$.

It is much easier than that: just notice that for any continuous decreasing bijection $f : [0,1]\to [0,1]$, we have $\int_{0}^{1}f=\int_{0}^{1}f^{-1}$.

How do you prove that result? If $f$ were differentiable we could use integration by parts, but what about the general case?

Take a Riemann lower sum for the integral of $f,$ and associate that with the graph of a nonincreasing step function. Draw in the vertical segments needed to make that graph a connected set. This set bounds a region which is a finite union of rectangles, hence having a well-defined area. Now reverse the roles of the variables; this set becomes the graph of nonincreasing step function; the area "below" it is the same as the previously computed area, and that area serves as a Riemann lower sum for the integral of $f^{-1}.$ We can do exactly the same thing with upper sums. We conclude that any number which is a lower sum for the integral of $f$ is a lower sum for the integral of $f^{-1},$ and the same for upper sums. Hence $f$ and $f^{-1}$ have equal Riemann integrals.

If $f$ were an increasing bijection from $[0,1]$ to $[0,1]$ and we tried the same argument, we'd be turning upper sums into lower sums and vice versa, and we would wind up concluding in that case that

$\int_{0}^{1}f+\int_{0}^{1}f^{-1}=1.$

Here is another way to see fedja's result, I think.  Let $ x$ and $ y$ be independent random numbers chosen uniformly from 0 to 1.  Then $ \int_{0}^{1}f$ is equal to $ \Pr[y \le f(x)]$.      Similarly $ \int_{0}^{1}f^{-1}$ is equal to $ \Pr[x \le f^{-1}(y)]$.  But $ y \le f(x)$ is equivalent to $ x \le f^{-1}(y)$.


geometry
perimeter
rectangle
Pythagorean Theorem

If the perimeter of a rectangle is $ p$ and its diagonal is $ d$, the difference between the length and width of the rectangle is:

$ \textbf{(A)}\ \frac {\sqrt {8d^2 \minus{} p^2}}{2} \qquad\textbf{(B)}\ \frac {\sqrt {8d^2 \plus{} p^2}}{2} \qquad\textbf{(C)}\ \frac {\sqrt {6d^2 \minus{} p^2}}{2}$
$ \textbf{(D)}\ \frac {\sqrt {6d^2 \plus{} p^2}}{2} \qquad\textbf{(E)}\ \frac {8d^2 \minus{} p^2}{4}$

[hide]

Let the length of the rectangle be $ l$ and the width $ w$. By the Pythagorean Theorem, we have $ l^2\plus{}w^2\equal{}d^2$. Also, we have $ l\plus{}w\equal{}\frac{p}{2}$. If we square this, then we have $ d^2\equal{}(l\plus{}w)^2\minus{}2lw\equal{}\frac{p^2}{4}\minus{}2lw$, so $ 2lw\equal{}\frac{p^2}{4}\minus{}d^2$. Finally, we have $ l\minus{}w\equal{}\sqrt{l^2\plus{}w^2\minus{}2lw}\equal{}\sqrt{d^2\minus{}\frac{p^2}{4}\plus{}d^2}\equal{}\frac{8d^2\minus{}p^2}{2}$, so $ \textbf{(A)}$

[/hide]


probability
geometry
parallelogram
trapezoid
analytic geometry
rectangle
symmetry

I'll give only a hint for only number 1 as I don't have time. BALLS AND URNS.

Er, no.

A better hint: Consider two dice.

For two, its $ 7$ (therefore you have to be at the second green spot in Monopoly to have the highest chance of getting Boardwalk).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=583825456&t=233798]#6[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=551993461&t=227762]#3[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=624556475&t=263880]#4[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1118811821&t=217885]#5[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=42781649&t=266170]#7[/url]

Hint: [b][size=200][color=red]SEARCH![/color][/size][/b]

@AIME- True that: I forgot about $ 0$ and that there were $ 48$ numbers! I was in a rush so I don't know what I was thinking.

[quote="k00lperson"]1) When four standard six- sided dice are rolled, what is the most likely sum of the four numbers?

2) What is the positive difference between the probability of a fair coin landing heads up exactly 2 times out of 3 flips and the probability of a fair coin landing heads up 3 times out of 3 flips?

3) In parallelogram ABCD, AB = 16 cm, DA = 3 root 2cm, and sides AB and DA form a 45- degree interior angle. In isosceles trapezoid WXYZ with WX is not equal to YZ, segment WX is the longest parallel side and has length 16 cm, and two interior angles each have a measure of 45 degrees. Trapezoid WXYZ has the same area as parallelogram ABCD. What is the length in centimeters of segment YZ?

4) The lines 2x + y = 6 and x +y = 10 together with the x - axis and y- axis are drawn on a coordinate grid to form quadrilateral ABCD. What is the area of quadrilateral ABCD?

5) Two vertices of a rectangle are ( -3,1) and (-3,-3). The line with equation x =1 is a line of symmetry for this rectangle. How many square units are in the area of a rectangle?

6) Mrs. Read can knit one pair of children's mittens with a ball of yarn six inches in diameter. How many pairs of identical mittens can she knit with a ball of yarn twelve inches in diameter?

7)Two trains are headed directly towards each other at rates of 80 mph and 120 mph. How many miles apart are they 1 minute before impact? Express your as a  mixed.[/quote]

Are you trying on these? It seems as if every problem you come upon, you tell us to solve it. Next time, please tell us where you have gotten to make sure you aren't just throwing out problems. I have no problem with you asking questions, but just us giving you the whole solutions to everything.

For instance: 3 cans of tomatoes cost 1.68. How much does 10 cans of tomatoes cost excluding tax?

In this problem, you could say I alread got the equation: $ 3x\equal{}1.68$ and divided all sides by 3 and got $ x\equal{}.56$, but what do I do after that?
 :wink:

[quote="jonathanchou711"][quote="k00lperson"]1) When four standard six- sided dice are rolled, what is the most likely sum of the four numbers?

2) What is the positive difference between the probability of a fair coin landing heads up exactly 2 times out of 3 flips and the probability of a fair coin landing heads up 3 times out of 3 flips?

3) In parallelogram ABCD, AB = 16 cm, DA = 3 root 2cm, and sides AB and DA form a 45- degree interior angle. In isosceles trapezoid WXYZ with WX is not equal to YZ, segment WX is the longest parallel side and has length 16 cm, and two interior angles each have a measure of 45 degrees. Trapezoid WXYZ has the same area as parallelogram ABCD. What is the length in centimeters of segment YZ?

4) The lines 2x + y = 6 and x +y = 10 together with the x - axis and y- axis are drawn on a coordinate grid to form quadrilateral ABCD. What is the area of quadrilateral ABCD?

5) Two vertices of a rectangle are ( -3,1) and (-3,-3). The line with equation x =1 is a line of symmetry for this rectangle. How many square units are in the area of a rectangle?

6) Mrs. Read can knit one pair of children's mittens with a ball of yarn six inches in diameter. How many pairs of identical mittens can she knit with a ball of yarn twelve inches in diameter?

7)Two trains are headed directly towards each other at rates of 80 mph and 120 mph. How many miles apart are they 1 minute before impact? Express your as a  mixed.[/quote]

Are you trying on these? It seems as if every problem you come upon, you tell us to solve it. Next time, please tell us where you have gotten to make sure you aren't just throwing out problems. I have no problem with you asking questions, but just us giving you the whole solutions to everything.

For instance: 3 cans of tomatoes cost 1.68. How much does 10 cans of tomatoes cost excluding tax?

In this problem, you could say I alread got the equation: $ 3x \equal{} 1.68$ and divided all sides by 3 and got $ x \equal{} .56$, but what do I do after that?
 :wink:[/quote]
yeah, these problems shouldn't be that hard if u think about them. Besides, instead of letting everyone else answering ur questions for u, u should answer them urself. that way u truly understand the solution

I think these may be from Saab MATHCOUNTS.


geometry
circumcircle
trigonometry
geometric transformation
reflection
exterior angle
angle bisector

In triangle ABC, AB>AC. The exterior angle bisector of A intersect the circumcircle in E, through E construct $EF \perp AB$. Show that $2AF=AB-AC$. (1989 China League)

Trigo's power again  :D
[hide]Let $\angle BAC = x$.

Then $AE = \frac{AF}{\sin \frac{x}{2}}$, $EC = 2R\sin \left(90+\frac{x}{2}\right)$, $BC = 2R\sin x$, and $BE = 2R\cos \frac{x}{2}$.

So by Ptolemy's theorem $AB\cdot 2R\cos \frac{x}{2}= \frac{AF}{\sin \frac{x}{2}}\cdot 2R\sin x+AC\cdot 2R\cos \frac{x}{2}$.

From which $(AB-AC)\cos \frac{x}{2}= \frac{AF}{\sin \frac{x}{2}}$, and finally $AB-AC = 2AF$.[/hide]

Without trig it is still doable.  :lol:

Let $D$ be a point on $FB$ such that $FD=FA$. Let $ED$ meet the circle at $G$, connect $AG$ and $EC$. Then $\angle{ACE}=\angle{AGD}$, $\angle{ADG}=180^{o}-\angle{ADE}= 180^{o}-\angle{EAH}=\angle{EAC}$($H$ is on the extension of $CA$), then $\triangle{ADG}\sim \triangle{EAC}$. Notice that $BC=AG$, we have $AD= \frac{AE \cdot BC}{EC}$, therefore $2AF = \frac{AE \cdot BC}{EC}$.

Connect $EB$, Use Ptolemy in quadrilateral in $AEBC$: \[AB \cdot EC = AE \cdot BC+BE \cdot AC\] Thus $AE \cdot BC = AB \cdot EC-BE \cdot AC$, hence $2AF= \frac{AB \cdot EC-BE \cdot AC}{EC}=AB-AC$.

[color=darkblue][u]It is O.K. Shobber, the your proposed problem is nicely and easily, just for this section ! [/u]

[b]The first proof.[/b] Suppose $c>b$. Denote $L\in BC\cap AE$. Therefore: $\boxed{AF=AE\sin \frac{A}{2}}\ ;\ \triangle ABE\sim \triangle ALC$ $\Longrightarrow \boxed{AE\cdot AL=bc}\ .$
$\boxed{2bc\sin\frac{A}{2}=(c-b)\cdot AL}\Longrightarrow \boxed{2\cdot AF=c-b}$ (the last relation is the product of the previous three relations).

[b]The second proof.[/b] Suppose $c>b$. Denote : the reflection $M$ of the point $E$ w.r.t. the circumcenter $O$ of the triangle $ABC$ ;
the projection $P$ of the point $M$ to the line $AB$. From a well-known property $BP=AF$ and the relations $BP=MB\cos \left(B+\frac{A}{2}\right)$,
$MB=2R\sin\frac{A}{2}$ results $AF=2R\sin \frac{A}{2}\cos\left(B+\frac{A}{2}\right)$ $\Longrightarrow$ $AF=R(\sin C-\sin B)$ $\Longrightarrow$ $\boxed{\ 2\cdot AF=c-b\ }\ .$[/color]

Let $F$ be the projection of $E$ onto $AC$; since $AE$ is the bisector of $\angle PAF$, we have $EP=EF\ (\ 1\ )$, consequently $AP=AF\ (\ 2 )$. $CE=CE, (1)$ and $\angle BFE=\angle CPE$ give $\triangle CPE\cong\triangle BFE$, consequently $CP=BF$; witn $(2)$ we get the required relation.

Best regards,
sunken rock

Dear Mathlinkers,
this problem is based on the broken chord theorem coming from Archimede.
Sincerely
Jean-Louis


modular arithmetic
quadratics
function
number theory
relatively prime

[hide="Solution?"]If $ x\not\equiv 0\pmod{2}$, then $ x\equiv 1\pmod{2}$.  This means that $ x$ must be an odd integer.  Thus, $ x\equiv 1, 3, 5, \text{or } 7\pmod{8}$.  Thus, $ x^2\equiv 1^2\equiv 3^2\equiv 9\equiv 5^2\equiv 25\equiv 7^2\equiv 49\equiv 1\pmod{8}$.

Is this what you were asking for?
[/hide]

Take it like this:

Obviously $ (8a\plus{}b)^2\equiv b^2\pmod 8$. So, to find the residue of square of some number modulo 8, we need to find residue of square of residue of that number modulo 8.

The residues of $ 0,1,2,3,4,5,6,7$ are $ 0,1,4,1,0,1,4,1$, so $ 0,1,4$ are quadratic residues modulo 8.
As odd numbers give odd residues modulo 8, they all must be $ 1$.

[quote="limac"]Hi,

I am not completely sure why/how $ x \not\equiv 0 \pmod{2}$ can be written as $ x^2 \equiv 1 \pmod{8}.$ I mean, what if $ x \equal{} 2,4,6,$ etc?

Thank you for the help.[/quote]

$ x$ cant be 2,4,6 etc. because $ x \not\equiv 0 \mod 2$ and 2,4,6... are equivalent to 0 mod 2.

If x is not divisible by 2, then x is odd. Like jxl said, if x is odd, then $ x \equiv 1,3,5,7 \mod 8$, $ x^2 \equiv 1^2 \equal{} 1 \equal{} \boxed{1},3^2 \equal{} 9 \equal{} \boxed{1},5^2 \equal{} 25 \equal{} \boxed{1},7^2 \equal{} 49 \equal{} \boxed{1} \mod 8$

Oh, I didn't realize that the number had to be odd. Thank you very much for the help. :)

Alternatively, the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url] yields $ \lambda (2^3) \equal{} 2^1 \equal{} 2$, so $ x^2 \equiv 1 \pmod{8}$ for all integers $ x$ such that $ \gcd{(x,8)} \equal{} 1$. All odd numbers are clearly relatively prime to 8, so it holds.


ok[hide]
$\frac{9}{150}$ are 8th graders
$\frac{6}{200}$ are 7th graders
$\frac{15}{350}$ percent total
final percent is $4.5$
this is prob wrong as there's no way it's inbetween 6 and 3
[/hide]
-jorian

Somewhere you got a little confused, I think. [hide="solution"] You can just assume that there are 100 7th graders and 150 8th graders. Taking the proper percentages, we find that there are 3 7th graders and 9 8th graders. So we have $\frac{12}{250}= \boxed{4.8 \text{ percent}}$ [/hide]

jhredsox would be correct if there are the same number of seventh graders as eighth graders.
[hide="I agree with humblemastermind"]Let there be $E$ eighth graders and $S$ seventh graders. In mathcounts, there are ${\frac{6}{100}E}$ eighth graders and  ${\frac{3}{100}S}$ seventh graders. So we want to find:
${\frac{\frac{6}{100}E+\frac{3}{100}S}{E+S}}$

We already know: $E={\frac{3S}{2}}$. Substituting,
${\frac{\frac{6}{100}({\frac{3S}{2}})+\frac{3}{100}S}{{\frac{3S}{2}}+S}}$

Canceling the $S$:
${\frac{\frac{6}{100}({\frac{3}{2}})+\frac{3}{100}}{{\frac{3}{2}}+1}}$

${\frac{\frac{9}{100}+\frac{3}{100}}{{\frac{5}{2}}}}$

${\frac{\frac{12}{100}}{{\frac{5}{2}}}}$

${\frac{\frac{3}{25}}{{\frac{5}{2}}}}$

${\frac{6}{125}}$

$4.8 \%$[/hide]

You guys are correct. Oh, and you're not allowed to use a calculator on this.

Here's a trick that allows us to compute ${\frac{6}{125}}$ without a calculator:

${\frac{6}{125}}$
$={\frac{6 \cdot 8}{125 \cdot 8}}$
$={\frac{48}{1000}}$
$=4.8\%$

Do something like this when faced with a denominator such as $25, 50, 125, 250, 500$.

[hide]I just pretended that there were 100 7th graders and 150 8th graders so there were 3 7th grade participants and 9 8th grade participants, so 12 total out of 250 makes 48 out of 1000 or 4.8 out of 100; [b]4.8%[/b][/hide]

[hide="solution"]Let's assume that there are 2x 7th graders and  3x 8th graders. Then there are 5x total. 6% of 3x=9x/50. 3% of 2x=3x/50. That means 6x/25 participate in mathcounts.6x/25=6/125 of 5x. (6/5)/25=(24/5)/100. 24/5=4.8

4.8[/hide]

[quote="vishalarul"]jhredsox would be correct if there are the same number of seventh graders as eighth graders.
[/quote]

where'd i go wrong?

there's something wrong with $\frac{15}{350}$ as it actually DOESN'T come out to 4.5, it's more like ~ 4.2
-jorian

Um, this needs to be exact. In a problem like this, you can assume WLOTMG that there are 1500 8th graders and 1000 7th graders and work from there :)

umm...so where wasn't i exact, im confused

-jorian

[quote="Ignite168"]You went wrong because $\frac{150}{200}\not=\frac{3}{2}$ which is your 8th to 7th graders wasn't the one they stated.[/quote]

oh thank you very much, i forgot about that

-jorian

[quote="Walk Around The River"][quote="indianamath"]wasn't this a MC problem? like 2002 state or something[/quote]

yes, hence the reason it was posted in the "MATHCOUNTS" forum.  :P  :rotfl:[/quote]

NOt all problems in the Mathcounts Forum are MathCounts Problems.

So, it doesn't have to be whatever it is.

And not all math counts problems are posted in Math Counts. :wink:

[quote="davidli"]6% of eighth graders and 3% of seventh graders participate in Mathcounts. There are 1.5 times as many 8th graders as 7th. What percent of the 7th and 8th graders, taken together, participate in Mathcounts?[/quote]
[hide]
there are 2.5x the amount of 7th graders in all. so it is 3% of 2/5 + 6% of 3/5, which is 6/5% + 18/5%, which is
$\frac{24}{5}percent$

[/hide]


$a,b,c>0$
If $a+b+c=6$ and $a^2+b^2+c^2=14$ find the minimum price of
$A=a^2b+b^2c+c^2a$

thank you very much, friend silouan!!

But are you sure it is right?? I proved that $A>\frac{121}{6}$ (?!) :?

Well
we easily find that $ab+bc+ca=11$

Now from BCS ineq we get $(a+b+c)A\geq(ab+bc+ca)^2=121$

so $A\geq\frac{121}{6}$
equality holds iff  $\frac{a^2b}{b}=\frac{b^2c}{c}=\frac{c^2a}{a}$
                     iff  $a=b=c$ since $a,b,c>0$
                     iff  $a=b=c=2$ and  $a=b=c=\sqrt{\frac{14}{3}}$ (impossible)
so the ineq is strict.....

do you have something else in mind??
 :)

Sorry I have not examine when the equality hold and I make mistake. I will try to ''fix''the ineq


modular arithmetic
number theory proposed
number theory

Let $ n$ be a natural number, and $ n \equal{} 2^{2007}k\plus{}1$, such that $ k$ is an odd number. Prove that \[ n\not|2^{n\minus{}1}\plus{}1\]

$ n|2^{n\minus{}1}\plus{}1$ if and only if n=1.

[quote]Let $ n$ be a natural number, and $ n \equal{} 2^{2007}k \plus{} 1$, such that $ k$ is an odd number. Prove that
\[ n\not|2^{n \minus{} 1} \plus{} 1\]
[/quote]
Assume that $ n\mid 2^{n \minus{} 1} \plus{} 1 \equal{} \left (2^k \right )^{2^{2007}} \plus{} 1$. Let $ p$ be an arbitrary prime divisor of $ n$ (Of course, $ p$ must not less than $ 3$) and set $ 2^k \equal{} t$. We have $ t^{2^{2007}}\equiv \minus{} 1 \pmod {p}$ $ \Longrightarrow$ $ t^{2^{2008}}\equiv 1 \pmod {p}$. Let $ d \equal{} \text {ord}_{p}(t)$. It is followed that $ d\mid 2^{2008}$ $ \Longrightarrow$ $ d \equal{} 2^{2008}$. Thus $ 2^{2008} \equal{} d\mid p \minus{} 1$, i.e. $ p\equiv 1 \pmod {2^{2008}}$. Therefore:
\[ n\equiv 1 \pmod {2^{2008}} \Longrightarrow 2^{2007}k\equiv 0 \pmod {2^{2008}}\Longrightarrow 2\mid k.\]
wich leads to the contradiction. The result of the problem is lead as follow.

[hide]OOps. Sorry for if my solution's idea is the same to any posters before mine. It is accidential, no on purpose[/hide]


search
ratio
factorial
geometric series
calculus
calculus computations

hi!!

we have $ e\equal{}\sum_{k\equal{}0}^{\infty}\frac{1}{k!}$

prove that $ e$ is an irrational number.

thanx.

Assume the contrary, $ e \equal{} \frac {p}{q}$, $ p\in\mathbb{Z},q\in\mathbb{N}$. Now, using the remainder in Lagrange form, we have:

$ e \equal{} 1 \plus{} 1 \plus{} \frac {1}{2} \plus{} \cdots \plus{} \frac {1}{n!} \plus{} \frac {1}{(n \plus{} 1)!}e^\theta$

Obviously, the series expansion holds for every $ n$, so we can take $ n > q$.

Multiplying both sides with $ n!$ gives:

$ n!\frac {p}{q} \equal{} 2n! \plus{} \frac {n!}{2} \plus{} \cdots \plus{} \frac {n!}{n!} \plus{} \frac {e^\theta}{n \plus{} 1}$. Since $ n > q$, all fractions in the equation are integers except for the $ \frac {e^\theta}{n \plus{} 1}$ which isn't integer, since  $ 0 < \theta < 1$, so we obtain contradiction.

P.S. maybe the search option would bring you some other proofs...

[b]COPIED WORD TO WORD FROM THE BELOW GIVEN LINK[/b]
Consider now INFINITE SUM expression of the number 'e':
e = e^1 = 1 +1/1! + 1/2! + ... + 1/n! + ...
Assert: 'e' is rational, that is it can be written as a ratio, denoted: N/D
where the numerator 'N' and the denominator 'D' are positive integers
greater than unity.
     The product of D factorial, D!, and 'e' is:
D!*e = D! + D!/1! + D!/2! + D!/3! + ... + D!/D! + other terms denoted by R,
the remainder.
     If 'e' is rational, then D!*e is also rational, since D! is a positive
integer.
     Examine the other terms of the product D!*R:
D!*R = [D!/(D+1)! + D!/(D+2)! + D!/(D+3)! +...]
     So R = [1/(D+1) + 1/(D+1)(D+2) +...]
since in each ratio of factorials the first D! in the numerator and
denomenator cancel.
     So, R = [1/(D+1) + 1/(D+1)(D+2) +...] < [ 1/(D+1) + 1/(D+1)(D+1) +...]
which is obtained by substituting (D+1) for each term in the sum of
reciprocals in the expression for R. This is true because (D+j) < (D+1) for
any term for j>1.
     Factoring out the first common term: 1/(D+1) gives:
D!*R = 1/(D+1) * [ 1 + 1/(D+1) + 1/(D+1)^2 + 1/(D+1)^3 + ... + 1/(D+1)^j +
...]
     Now the term in the brackets [....] is just the geometric series
expansion of the form: [1+ x + x^2 + x^3 + ...] where x = 1/(D+1). This
INFINITE SERIES sums to: 1/(1 - x ).
     So: R<1/(D+1) * [ 1/{1 - (1/(D+1)}] = 1/(D+1) * [ (D+1)/D ] = 1/D.
     Summarizing: 0 < R < 1/ D. From the original definition of R,  both R
and D are positive integers. This is a contradiction. Hence the original
assertion that
'e' is rational is false. Thus 'e' is irrational.
[b]SOURCE :http://www.newton.dep.anl.gov/askasci/math99/math99119.htm  [/b]

[quote="Dr Sonnhard Graubner"]hello, here you can find another proof:
http://cazelais.disted.camosun.bc.ca/250/e-irrational.pdf
Sonnhard.[/quote]

thank you Dr Sonnhard Graubner  :)

there is an other solution :

We prove that:

$ (\exists t ) 0<t<1$ such that:

$ e=u_{q}+\frac{t}{qq!}$ then we prove that:

$ q!e=\sum_{k=0}^{q}\frac{q!}{k!}+\frac{t}{q}$

so since $ \sum_{k=0}^{q}\frac{q!}{k!}$ $ \in{\mathbb{N^{*}}}$ and $ q!e$ also.

$ \frac{t}{q}$ must be $ \in{\mathbb{N^{*}}}$ but $ 0<t<1$ Contradiction!

All of these proofs are essentially identical. Anyone know a proof that is substantially different?


What is the sum of the first nine prime numbers?


Find the area enclosed by the following function:

$ |x| \plus{} |x\minus{}y| \equal{} 4$

Is there a general approach to solving enclosed area problems where $ x$ and $ y$ are both in absolute values?

Would I consider solving for y?

I don't I think I would, because I don't prefer to work with absolute value quantities
(or their equivalents).

However, if I were to do it, I would try this:

$ |x| \plus{} |x \minus{} y| \equal{} 4$

$ |x \minus{} y| \equal{} 4 \minus{} |x|$

This is equivalent to:

$ \sqrt {(x \minus{} y)^2} \equal{} 4 \minus{} |x|$

$ [\sqrt {(x \minus{} y)^2}]^2 \equal{} (4 \minus{} |x|)^2$

$ (x \minus{} y)^2 \equal{} (4 \minus{} |x|)^2$

$ x \minus{} y \equal{} \pm(4 \minus{} |x|)$

$ \minus{} y \equal{} \minus{} x \pm(4 \minus{} |x|)$

$ y \equal{} x \pm(4 \minus{} |x|)$

=========================================================================
The portion of the graph where $ x \ge 0$  gives:

$ y \equal{} x \pm(4 \minus{} x)$.  This splits into $ y \equal{} x \plus{} 4 \minus{} x$, which gives $ y \equal{} 4$, and another section
$ y \equal{} x \minus{} (4 \minus{} x) \equal{} x \minus{} 4 \plus{} x$, which gives $ y \equal{} 2x \minus{} 4$.

The portion of the graph where $ x < 0$ gives:

$ y \equal{} x\pm(4 \minus{} ( \minus{} x))$.  This is the same as $ y \equal{} x \pm(4 \plus{} x)$.  This splits into $ y \equal{} x \plus{} 4 \plus{} x$,
which gives $ y \equal{} 2x \plus{} 4$, and another section $ y \equal{} x \minus{} (4 \plus{} x) \equal{} x \minus{} 4 \minus{} x$, which gives $ y \equal{} \minus{} 4$.

[quote="Stank"][quote="Arrange your tan"][quote="redcomet46"]Find the area enclosed by the following >> function <<:

$ |x| \plus{} |x \minus{} y| \equal{} 4$

Is there a general approach to solving enclosed area problems where $ x$ and $ y$ are both in absolute values?[/quote]

It is not a function; it is a relation.

[hide]
Four parts:

--------------------

x + (x - y) = 4
2x - y = 4
y = 2x - 4

==============

x + (y - x) = 4
y = 4

==============

-x + (x - y) = 4
-y = 4
y = -4

==============

-x + (y - x) = 4
-2x + y = 4
y = 2x + 4

==============

The intersection of these lines consist of the line segments from
(0, 4) to (4, 4) to (0, -4) to (-4, -4) to (0, 4).

This is a parallelogram.  Its area is base times height.

The base is 4 units.  The height is 32 units.

The area is (4 units)(8 units) = 32 square units.
  [/hide][/quote]

I believe the term is "implicitly defined function"[/quote]

Then the mathematics in use is inconsistent itself and should NOT
even be using the word "function" to describe it as it assigns
at least two y-values to at least one x-value.  As another example,
a website described $ x^2 \plus{} y^2 \minus{} 1 \equal{} 0$ as an "implicitly defined
function."  A circle is a relation which happens not to be a function,
so, again, to use a term "function," as part of a longer term to
describe something, of which it is not, is inconsistent.


[b]Tom picks a polynomial [i]p[/i] with nonnegative integer coefficients. Sally claims that she can ask Tom just two values of [i]p[/i] and then tell him all the coefficients. She asks for p(1) and p(p(1) + 1). For example, suppose p(1) = 10 and p(11) = 46, 610. What is the polynomial, and how did she know it?[/b]

[hide="Solution"] $p(1)$, the sum of the coefficients, is guaranteed to be greater than or equal to each individual coefficient.  $p(p(1)+1)$, expressed in base $p(1)+1$, has digits which must, therefore, be the coefficients.

For example, in base $11$ we have

$46610_{10}= 32023_{11}$ and therefore

$p(x) = \boxed{ 3x^{4}+2x^{3}+2x+3 }$ [/hide]


inequalities
number theory proposed
number theory

[quote="Stifler"]Denote by $ d(n)$ the number of positive divisors of $ n$. Suppose that $ r$ and $ s$ are positive integers with the property that $ d(ks)\geq d(kr)$ for each $ k\in\mathbb{N}$. Prove that $ r$ divides $ s$.[/quote]

Suppose there exists a prime $ p$ such that $ v_p(r)>v_p(s)$
Let then $ A\equal{}\{$ prime $ q\ne p$ such that $ q|rs\}$

If $ A\equal{}\emptyset$, then $ r\equal{}p^{v_p(r)}$ and $ s\equal{}p^{v_s(r)}$ and $ d(s)<d(r)$ and so contradiction

If $ A\ne\emptyset$, let $ k_n\equal{}\prod_{q\in A}q^n$. We get :

$ \frac{d(k_ns)}{d(k_nr)}\equal{}\frac{v_p(s)\plus{}1}{v_p(r)\plus{}1}\prod_{q\in A}\frac{v_q(s)\plus{}n\plus{}1}{v_q(r)\plus{}n\plus{}1}$

Setting $ n\to \plus{}\infty$, $ \frac{d(k_ns)}{d(k_nr)}\to\frac{v_p(s)\plus{}1}{v_p(r)\plus{}1}<1$, which is impossible.

Hence the result.