geometry
IMO 1994
circumcircle
IMO
trigonometry
cyclic quadrilateral

Let $ ABC$ be an isosceles triangle with $ AB \equal{} AC$. $ M$ is the midpoint of $ BC$ and $ O$ is the point on the line $ AM$ such that $ OB$ is perpendicular to $ AB$. $ Q$ is an arbitrary point on $ BC$ different from $ B$ and $ C$. $ E$ lies on the line $ AB$ and $ F$ lies on the line $ AC$ such that $ E, Q, F$ are distinct and collinear. Prove that $ OQ$ is perpendicular to $ EF$ if and only if $ QE \equal{} QF$.

(*) if ${OQ}\perp{EF}$

we have that $OQCF$ is cyclic (because, by simmetry, ${OC}\perp{AC}$).

so, $\angle{OFE} = \angle{OCB} = \angle{OBC}$. ($OB = OC$)

see that $EQOB$ is cyclic, because $\angle{OBE}+ \angle{EQO} = 180$.

so, $\angle{OEQ} = \angle{OBC} = \angle{OFE} \Longrightarrow OE = OF \Longrightarrow QE = QF$.

(**) if $QE = QF$.

by menelaus, we have $\frac{EB . AC . QF}{AB . CF . QE} = 1$, but $AB = AC$ and $QE = QF$.

$\Longrightarrow CF = EB$. but, $OB = OC$, so, $\triangle{EBO} \equiv \triangle{OCF}$

$\Longrightarrow OE = OF \Longrightarrow {OQ}\perp{EF}$

and the problem is solved

This solution isn't very different from the nice E.Lopes's solution.

I note the reflection $E'$ of the point $E$ w.r.t. the point $B\ (BE=BE')$, i.e. the line

$BO$ is the bisector of the segment $EE'$. From the Menelaus' theorem applied to the

transversal $\overline {BQC}$ for  the triangle $AEF$ results the relation $\frac{BE}{BA}\cdot \frac{CA}{CF}\cdot \frac{QF}{QE}=1$,

i.e. $\frac{BE'}{CF}=\frac{QE}{QF}$. Thus, $QE=QF\Longleftrightarrow BE'=CF\Longleftrightarrow E'F\parallel BC\Longleftrightarrow$

$AO$ is the bisector of the segment $[E'F]\Longleftrightarrow$ the point $O$ is the circumcenter

of the triangle $E'EF\Longleftrightarrow OQ\perp EF$.

Here is a slightly different solution for when $ QE\equal{}QF$:
[hide="Solution"]
Let $ E'$ and $ F'$ be on $ AB$ and $ AC$ respectively such that $ E'F'\perp OQ$.  Then, by the first part of the problem, $ QE'\equal{}QF'$.  Hence, $ Q$ is the midpoint of $ EF$ and $ E'F'$, which means that $ EE'FF'$ is a parallelogram.  Unless $ E\equal{}E'$ and $ F\equal{}F'$, this is a contradiction since $ EE'$ and $ F'F$ meet at $ A$.  Therefore, $ E\equal{}E'$ and $ F\equal{}F'$, so $ OQ\perp EF$, as desired.  [/hide]

[quote="Nemion"]I have the same solution  as E.Lopes's[/quote]

Did you have to revive a 2 year old thread to just say this?

[quote="Virgil Nicula"]This solution isn't very different from the nice E.Lopes's solution.

I note the reflection $ E'$ of the point $ E$ w.r.t. the point $ B\ (BE \equal{} BE')$, i.e. the line

$ BO$ is the bisector of the segment $ EE'$. From the Menelaus' theorem applied to the

transversal $ \overline {BQC}$ for  the triangle $ AEF$ results the relation $ \frac {BE}{BA}\cdot \frac {CA}{CF}\cdot \frac {QF}{QE} \equal{} 1$,

i.e. $ \frac {BE'}{CF} \equal{} \frac {QE}{QF}$. Thus, $ QE \equal{} QF\Longleftrightarrow BE' \equal{} CF\Longleftrightarrow E'F\parallel BC\Longleftrightarrow$

$ AO$ is the bisector of the segment $ [E'F]\Longleftrightarrow$ the point $ O$ is the circumcenter

of the triangle $ E'EF\Longleftrightarrow OQ\perp EF$.[/quote]

really nice solution dear experienced virgil :clap2:

i will post my solution later! :wink:

[color=darkred][b][u]An easy extension.[/u][/b] Let $ w$ be the circumcircle of the triangle $ ABC$ . Let $ [AA']$ be the diameter of the circle $ w$ .

Consider three collinear points $ D\in (BC)$ , $ E\in AB$ , $ F\in AC$ . Prove that $ A'D\perp EF\iff\frac {A'E}{A'F} \equal{} \frac {AB}{AC}$ .[/color]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b] Let $ w$ be the circumcircle of the triangle $ ABC$ . Let $ [AO]$ be the diameter of the circle $ w$ .

Consider three collinear points $ Q\in (BC)$ , $ E\in AB$ , $ F\in AC$ . Prove that $ OQ\perp EF$ if and only if $ \frac {QE}{QF} \equal{} \frac {AB}{AC}$ .[/color][/quote]
nice extension but i don't think it's completly correct(we can't express $ \frac{QE}{QF}$ in terms of $ b$ and $ c$ only,I think)but i think if we change problem a little it would work .

[color=blue][b]Problem[/b].Let $ \omega$ be the circumcircle of the triangle $ ABC$ . Let $ AA'$ be the diameter of the circle $ \omega$ . Consider three collinear points $ D\in BC$ , $ E\in AC$ , $ F\in AF$ . Prove that $ A'D\perp EF$ if and only if $ \frac {A'E}{A'F} \equal{} \frac {A'C}{A'B}$[/color]

[i]Solution.[/i]Let $ A'D\perp EF$ we have $ A'B\perp AB$ therefore $ BDA'F$ is cyclic hence $ \angle DFA'\equal{}\angle DBA'\equal{}\angle A'AC$ hence $ AEA'F$ is cyclic
by similarity of Triangles $ EA'F$ and $ BA'C$ we have : $ \frac{A'F}{A'E}\equal{}\frac{A'B}{A'C}$ .
now consider $ \frac{A'F}{A'E}\equal{}\frac{A'B}{A'C}$  hence $ \frac{A'F}{A'B}\equal{}\frac{A'E}{A'C}$ and $ \angle A'CE\equal{}\angle A'BF\equal{}\frac{\pi}{2}$  hence $ \angle A'FA\equal{}\angle A'EC$ hence $ AEA'F$ is cyclic.
therefore $ \angle A'BC\equal{}\angle EAA'\equal{}\angle EFA'$ therefore $ BDA'F$ is cyclic then $ \angle FDA'\equal{}\frac{\pi}{2}$

For A different(own) solution, see [url=http://www.artofproblemsolving.com/blog/37910]here[/url]

Dear Mathlinkers,
for the initial problem, see
http://www.artofproblemsolving.com/Forum/index.php?f=47&t=330790.
Sincerely
Jean-Louis

For $QE=QF\Rightarrow OQ\perp EF$ i have a different approach, consider the triangle $AEF$, since the perpendiculars from $O$ to the sides $AE, EF, FA$ of these triangle are collinear, we have by the Simson lines that $AEOF$ is cyclic and then $\measuredangle{OEF}=\measuredangle{OAF}=\measuredangle{OAE}=\measuredangle{OFE}$, since $ABC$ is isosceles, then $OQ\perp EF$, qed.

In fact there is a spiral similarity with centre $O$ than sends $\triangle OBE$ to $\triangle OCF$, I feel very dim for not noticing this at first. Instead an almost equivalent solution by easy angle-chasing:

WLOG $E$ is on the segment $AB$ and $F$ is on the ray $AC$ extended beyond $C$.

First assume $OQ$ is perpendicular to $EF$, then $\angle OQE=\angle OBE=90^{\circ}$ then $OQEB$ is a cyclic quadrilateral (perhaps degenerate). So $\angle OBQ=\angle OEQ$. Next since $\angle OCF=90^{\circ}=\angle FQO$ so $QCOF$ is a cyclic quadrilateral which implies $\angle OQF=\angle QCO=\angle MCO=\angle MBO=\angle OEQ\implies\triangle OEF$ is isosceles. So $QE=QF$. For the next direction, use Menelaus on $AEF$ wrt $BQC$ to see $\triangle BEO$ is congruent to $\triangle OCF$. Then $OE=OF$ so $OQ$ is the median implies $OQ$ is the perp. bisector so $OQ\perp EF$.

i have a solution which is very similar to that of e.lopes

I have a solution which is not at all similar to that of e.lopes!

[hide="Solution"]

We consider it on the coordinate plane. Place $B$ at $(-1,0)$, $C$ at $(1,0)$, $A$ at $(0,m)$, and $M$ at $(0,0)$.

Clearly, $O$ is located at $\left(0,\frac{-1}{m}\right)$

Now, let $E$ be at $n, m(n+1)$, and $Q$ be at $(k,0)$.

Line $AC$ has equation $y=-m(x-1)$.

Line $EQ$ has equation $y=\frac{m(n+1)}{n-k}(x-k)$.

Their intersection is then $F$. First we find its x-coordinate:

$-m(x-1)=\frac{m(n+1)}{n-k}(x-k) \implies (x-1)(k-n)=(n+1)(x-k)$

$x(k-n-n-1)=k-n-kn-k$

$x_{F}=\frac{kn+n}{2n-1-k}$.

Since $y=-m(x-1)$, we have $y_{F}=-m\frac{n+kn-2n-1+k}{2n+1-k}=\frac{-m(k-1)(n+1)}{2n+1-k}$.

Now, consider the two conditions:

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

$OQ \perp EF \Longleftrightarrow OQ \perp QE \Longleftrightarrow \overrightarrow{QO} \cdot \overrightarrow {QE}=0$.

But $\overrightarrow {QO}=\left(-k,\frac{-1}{m}\right)$

and $\overrightarrow{QE}=(n-k,m(n+1))$

so this condition is equivalent to $-k(n-k)-(n+1)=0$ or $k^2-kn-n-1=0 \implies (k+1)(k-n-1)=0$ \implies $k=-1,n+1$.

$k=-1 \implies Q=B$ which we know is false, so $OQ \perp EF \Longleftrightarrow k=n+1$.

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

Our second condition is that $EQ=QF \Longleftrightarrow Q \text{ is the midpoint of } EF$

The x-coordinate of the midpoint of $EF$ is their average of their x-coordinates, so we must have that

$x_E+x_F=2x_Q$ or that

$n+\frac{n+kn}{2n+1-k}=2k \Longleftrightarrow (2n+1-k)+n(k+1)=2k(2n+1-k) \Longleftrightarrow n(2n+2)=2k(2n+1-k)$

Dividing by two and expanding, we have

$n^2+n=2kn+k-k^2 \Longleftrightarrow n^2-2kn+k^2+n-k=0 \Longleftrightarrow (n-k)(n-k+1)=0$

or $n=k, k-1$.

Meanwhile, the y-coordinates must also match, so

$m(n+1)-\frac{m(k-1)(n+1)}{2n+1-k}=0$.

So, either $n=-1$ or

$2n+1-k=k-1 \Longleftrightarrow 2n+2=2k \Longleftrightarrow n=k-1$.

If $n=-1$, then by the x-coordinate condition, $k=n,n+1$. If $k=n=-1$, $Q=B$, contradiction, so this condition is equivalent to $n=k-1$.

As the two conditions are equivalent to the same thing ($n=k-1$), they are therefore equivalent and we are done.[/hide]

[hide="I have a solution similar to my namesake!"]If $QE=QF$

Let $D$ be the point that belongs to $AB$ and $DF$ is parallel to $CB$. The triangle $EQB$ is similar to $EFD \cdot EF=2EQ$. Therefore $DF=2BQ, ED=2EB$.

Namely $EB=BD$.

So, when $OB$ perpendicular to $ED$, is the bisector. Therefore $OE=OD=OF$ and $OQ$ is bisector of $EF$, and therefore $EF$ is perpendicular to $OQ$.

Now suppose that $EF$ is perpendicular to $OQ$ with $E$ on the line $AB$ and $F$ on $AC$. Let $D$ be the point of $AB$ such that $DF$ is parallel to $BC$. The triangle $FDE$ is similar to $QBE$.

$OF=OD=OE'$ being $E'$ on $AB$. Thus, $OQ$ is bisector of $E'F$, so it must be $E'=E$.[/hide]

My solution:
First suppose $OQ\perp EF$ then $BQOE$ and $OQFC$ are both cyclic and thus  $\angle OEQ=\angle OBQ=\angle BAM=\angle MAC=\angle MCO=\angle OFQ$ implies $EQ=QF$ as desired.
Now suppose $EQ=QF$. Observe that as $\angle EBQ+\angle QCF =180^{\circ}$ and $\angle EQB=\angle DQF$ and $EQ=QF$, this means  $\Delta QBE$ and $\Delta QCF$ are both part of the ambiguous case. Hence $EB=FC$ & also $OB=OC$ So $\Delta EBO$ is congruent to $\Delta FCO$ implying  $OF=OE$.

I don't know why but the problem in the IMO Compendium is a bit different and more difficult:
N is an arbitrary point on the bisector of ∠BAC. P and O are points on the lines AB and AN, respectively, such that ∡ANP = 90◦ = ∡APO. Q is an arbitrary point on NP, and an arbitrary line through Q meets the lines AB and AC at E and F respectively. Prove that ∡OQE = 90◦ if and only if QE = QF .

$ \angle ABC=x,\angle OQB=y$
By Sine rule in $\Delta OQB$,
$\frac{BQ}{\sin(90-y+x)}=\frac{BO}{\sin y}\implies BQ=\frac{c\tan (90-x)\cos(x-y)}{\sin y}$
By Sine Rule in $\Delta OQC$,
$\frac{QC}{\sin(90-x-y)}=\frac{b\tan (90-x)}{\sin(180-y)}\implies QC=\frac{b\cot x \cos(x+y)}{\sin y}$
$\implies \frac{BQ}{CQ}=\frac{\frac{c\tan (90-x)\cos(x-y)}{\sin y}}{\frac{b\cot x \cos(x+y)}{\sin y}}$
$=\frac{c}{b}\cdot \frac{\cos(x-y)}{\cos(x+y)} =\frac{\cos(x-y)}{\cos(x+y)}$
====================
$OQ\perp EF\implies QE=QF$
====================
$\implies \angle QEB=y-x+90,\angle CFQ=x+y-90(\angle OQE=90)$
By Sine rule in $\Delta BEQ$
$\frac{\sin x}{QE}=\frac{\sin (y-x+90)}{BQ}$
By Sine rule in $\Delta QCF$
$\frac{\sin (180-x)}{QF}=\frac{\sin (x+y-90)}{QC}$
Dividing these two, we get,
$\frac{QF}{QE}=\frac{\frac{-\cos (x+y)}{QC}}{\frac{\cos(y-x)}{BQ}}=\frac{-\cos(x+y)}{\cos (y-x)}\cdot\frac{BQ}{QC}=\frac{-\cos(x+y)}{\cos (y-x)}\cdot \frac{\cos(x-y)}{\cos(x+y)}=1$
$\implies QE=QF$.
====================
$QE=QF\implies OQ\perp EF$
====================
Let $\angle EPQ=p$
By Sine Rule in $\Delta EBQ$,
$\frac{\sin(180-(p+x))}{BQ}=\frac{\sin x}{EQ}$
By Sine Rule in $\Delta FQC$
$\frac{\sin (x-p)}{QC}=\frac{\sin(180-x)}{QF}$
$\implies \frac{\sin (p+x)}{\sin (x-p)}\cdot \frac{QC}{BQ}=\frac{FQ}{EQ}$
$\implies \frac{\sin (p+x)}{\sin (x-p)}=\frac{BQ}{CQ}=\frac{\cos(x-y)}{\cos(x+y)}$
$\implies \frac{\sin (p+x)}{\sin (x-p)}=\frac{\cos(x-y)}{\cos(x+y)}$
After expanding, it is easy to get $p=90-y\implies \angle OQE=90-y+y=90\implies OQ\perp EF$

Indeed you trigo bashed it bravely.

But @elope's solution is easier and same to the one I got.

My solution:
By Perpendicularity lemma we have:
$OQ\perp ED$$\Longleftrightarrow$$OE^2-OF^2=QE^2-QF^2$. Since $OB\perp AB$ and $OC\perp AC$ so
$OF^2=OC^2+CF^2$ and $OE^2=OB^2+BE^2$ and $OB=OC$. So we have
$OE^2-OF^2=BE^2-CF^2$. From law of sine on($\angle EQB=x$,$\angle ABC=b$) $\triangle QBF$:$ BE=\frac{\sin x}{\sin b}QE$. Analogously $CF=\frac{\sin x}{\sin b}QF$. So
$OE^2-OF^2=BE^2-CF^2$ so
$OQ\perp ED$$\Longleftrightarrow$$OE^2-OF^2-BE^2+CF^2=0$$\Longleftrightarrow$ $(OE^2-OF^2)(\frac{sin^2(x)-sin^2(b)}{sin^2(b)}=0$ $ \Longleftrightarrow$ $QE=QF$(if $x=b$ or $x=180-b$ $EF$ is parallel to $AC$ or $AB$).

The feet of the perpendiculars from $O$ to the sides of $\triangle AEF$ are $B,Q,C$, and are collinear. By the converse of Simson, this means that $OEAF$ is cyclic. Therefore, $\angle FAO = \angle FEO$. Since
\[ \angle FAO = 90^\circ-\angle ACM = 90^\circ - (180^\circ-\angle BCO - 90^\circ) = \angle BCO ,\]we know that $\angle FEO = \angle BCO$. Now, we see that $\angle FEO = \angle EFO \iff EQ=QF$. Also,
\[ \angle BCO = \angle EFO \iff OFCQ \text{ cyclic } \iff \angle FGO = \angle OCF = 90^\circ.\]Therefore, $EQ=QF \iff \angle EQF = 90^\circ$.

Here's a (hopefully) different solution. First let's tackle the if $OQ \perp EF$ condition:
Note $O$ is the antipode to $A \implies \angle OBA =90^{\circ} \implies OQBE$ is cyclic. Similarly $OCFQ$ is cyclic
$\implies \angle OEQ = \angle OBC = \angle OCB = \angle OFQ \implies OF=OE$. As $OFE$ is isosceles and $OQ \perp EF \implies QE=QF$.

Now for the if $QE=QF$ condition:
Let $E'$ be the reflection of $E$ in $B$ and $F'$ the reflection of $F$ in $C$. First note $OE=OE'$, as $BE=BE'$ and $OB \perp EE'$.
First, we claim $BE=CF$. To see why, let's apply sine rule on triangles $BQE$ and $CQF$:
$\frac{BE}{\sin{\angle BQE}} = \frac{QE}{\sin{\angle QBE}}$ and $\frac{FC}{\sin{\angle FQC}} = \frac{QF}{\sin{\angle FCQ}}$.

Note that $\angle QBE=\angle OBE +\angle QBO=90^{\circ}+\angle OAC=180^{\circ}-\angle ACB =180^{\circ}-\angle FCQ \implies \sin{\angle FCQ}=\sin{\angle QBE}$.
As $\angle BQE =\angle FQC$ and $QE=QF \implies BE=CF$. Thus $AE \cdot AE'=AB^2-BE^2=AC^2-CF^2=AF \cdot AF'$, aka $AE'FF'$ is cyclic, with circumcenter $O$, as $OE=OE' \implies OE=OF$. As $Q$ is the midpoint of $EF$, we conclude $QO \perp EF$.

[b]Proving that if $OQ \perp EF$, $QE = QF$:[/b]
$BQOE$ is cyclic because $\angle EBO = \angle EQO = 90$. Therefore, $\angle QEO = \angle OBQ$. Since $OB^2 = AO^2 - AB^2$ and $OC^2 = OA^2 - AC^2 = OA^2 - AB^2$, $OC = OB$, so $\angle OBQ = QCO$. Since $\angle FQO + \angle FCO = 90 + 90 = 180$, $QOFC$ is cyclic with $\angle QCO = \angle QFO$. Putting this all together, $\angle QEO = \angle QFO$, so $EO = OF$. $EQ^2 = EO^2-QO^2$ and $QF^2 = FO^2 - QO^2 = EO^2 - QO^2$. Thus, $QE = QF$.

[b]Proving that if $QE = QF$, $OQ \perp EF$:[/b]
Note that $\angle EBO = \angle FCO = 90$. Since $OB^2 = AO^2 - AB^2$ and $OC^2 = OA^2 - AC^2 = OA^2 - AB^2$, $OC = OB$. Now, if $\angle BEQ = b-a$ and $\angle BQE =a$, then $\angle BQO = 90-b \Longleftrightarrow \angle QCO = 90-b \Longleftrightarrow \angle FCQ = b$. It’s also true that $\angle BQE = \angle FQC = a$. Therefore, by the Law of Sines, we have that:
$$\frac{QE}{\sin (180-b) } = \frac{BE}{\sin (a)} \Longleftrightarrow \frac{QE}{\sin (b) } = \frac{BE}{\sin (a)}$$
$$\frac{QF}{\sin (b)} = \frac{FC}{\sin (a)}$$
Therefore, $BE = FC$, and $\triangle BOE$ is congruent to $\triangle COF$, which implies that $EO = FO$, which in turn implies that $\triangle EQO$ and $\triangle FQO$ are congruent. This means that $OQ \perp EF$.

Firstly , if $OQ \perp{EF}$ ,
Clearly $\angle{OCF}=\frac{\pi}{2}$ , thus $OQCF$ is cyclic , hence $\angle{BFQ}=\angle{OFQ}=\angle{OCB}=\angle{OBC}=\angle{FBQ}$ wich implies that $QF=QB$ , and since $\angle{FBE}=\frac{\pi}{2}$ , thus Q is the circumcenter of $\triangle{BEF}$ , hence $QF=QE$.
Now if $QF=QE$ ,
similary $Q$ is the circumcenter of $\triangle{BEF}$ , thus $\angle{BEQ}=\angle{B}=\angle{C}=\angle{QCA}$ , hence $QCAE$ is cyclic , this implies that $\angle{EQB}=\angle{A}$ , since $ABOC$ is cyclic $\angle{A}=\pi-\angle{BOC}=\angle{COF}$ , thus $\angle{CQF}=\angle{EQB}=\angle{COF}$ , hence $CQOF$ is cyclic , $\longrightarrow \angle{OQF}=\angle{OCF}=\frac{\pi}{2}$ .

[hide = First direction]
We have been given that $OQ \perp EF$. It is sufficient to prove that $OEF$ is isosceles. Since $EQOB$ is a cyclic quadrilateral, we know that $\angle ACB = \angle ABC = \angle EOQ \implies \angle OEF = \tfrac{\pi}{2} - \angle ABC$. Therefore it is sufficient to prove that $\angle EFO = \tfrac{\pi}{2} - \angle ABC \implies \angle FOQ = \angle ABC \implies  QCOF$ is a cyclic quadrilateral. But $QCOF$ must be cyclic since $\angle OQF = \angle OCF = \tfrac{\pi}{2}$ (because $ABC$ is an isosceles triangle). Thus, $OEF$ is isosceles and we are done.
[/hide]

[hide = Second direction]
[b][color=#f00]Lemma. [/color][/b] In an isosceles triangle $ABC$, if $Q$ divides $EF$ into equal parts ($E \in AB, F \in AC, Q \in BC$), it follows that $BE = CF$.
Proof is trivial by menelaus.

We are given $EQ = QF$. Once again, it is sufficient to prove that $EOF$ is isosceles, i.e. we need to prove $OF = OE \implies CF^2 + OC^2 = BE^2 + OB^2 \implies CF^2 = BE^2 \implies CF = BE$ which is true by the lemma.

[/hide]

To show that "if QE = QF then (OQ) is perpendicular to (EF)", you can bash it quite easily with a repere. All points have some decent coordinates so it works fine.


vector
inequalities
calculus
calculus computations

how can i prove that given p(x),q(x), norms on a finite vector space $ V$, $ \exists \lambda, \mu>0$ such that $ \lambda q(x) \leq p(x) \leq \mu q(x) \ \forall x \in V$ iff $ \forall \epsilon >0, \ \exists \epsilon', \epsilon'' >0$ such that $ B_q(x, \epsilon') \subset B_p(x,\epsilon) \subset B_q(x,\epsilon'') \forall x\in V$, where $ B_p(x,\delta)\equal{}\{y \in V \ | \ p(x\minus{}y)<\delta \}$ ?

Suppose $ \| \cdot \|_1$ and $ \| \cdot \|_2$ are two norms on a finite vector space $ V$. Let $ \{ v_1, \cdots, v_n \}$ be a basis of $ V$ where $ n \equal{} \dim V$. Then define $ M_1 \equal{} \sup_{\| x \|_1 \equal{} 1} \| x \|_2$.

Notice that $ M_1$ is nonzero, since $ \| x \|_1 \equal{} 1$, then $ x \neq 0$, so $ \| x \|_2 > 0$. Also, if we write $ x \equal{} \sum_{i\equal{}1}^{n} a_i v_i$, then $ |a_i| \equal{} \frac{\| a_i v_i \|_1}{\|v_i\|_1} \leq \frac{\| x \|_1}{\| v_i \|_1} \equal{} \frac{1}{\| v_i\|_1}$, so we obtain an estimate of an upper bound $ \| x \|_2 \leq \sum_{i\equal{}1}^{n} |a_i|\|v_i\|_2 \leq \sum_{i\equal{}1}^{n} \frac{\|v_i\|_2}{\| v_i\|_1} < \infty$.

Then for any nonzero $ x \in \mathbb{R}^n$, we have

$ \| x \|_2 \equal{} \left\| \|x\|_1 \frac{x}{\|x\|_1} \right\|_2 \equal{} \|x\|_1 \left\| \frac{x}{\|x\|_1} \right\|_2 \leq M_1 \| x \|_1$.

This inequality also holds for $ x \equal{} 0$, so this is an absolute inequality. Interchanging the role of $ \| \cdot \|_1$ and $ \| \cdot \|_2$, we obtain $ \| x \|_1 \leq M_2 \| x \|_2$ for some positive $ M_2$. So we have proved

$ \frac{1}{M_1} \| x \|_2 \leq \| x \|_1 \leq M_2 \|x\|_2$.

Rest of the proposition is easy to prove... -o-


An equilateral triangles is drawn with side length A. A new equilateral triangle is formed by joining the midpoints of the first one. THen a third equilateral triangle is formed by joining the midpoints of the second; and so on forever. Find the limit of the sum of the perimeters of all the triangles thus drawn.

Hint: think infinate geometric series...

[hide]So if each is 1/4 the previous area

3A+A+.333A+.111A...=4.25A[/hide]

That'd work for area, but the question is about perimeter.

[hide]Each is 1/2 the Perimeter of the previous triangle.
3A/(1-1/2)= [b]6A ANS.[/b][/hide]


calculus
integration
trigonometry
invariant
algebra
polynomial
calculus computations

how would one even approach this problem?

my guess would be a way to factor the numerator and denominator into something like $u(x)^{2}+a^{2}$ and then trig sub.
this is because we have

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

yes, i do agree with that, however isn't factoring and simplifying the challenging part?

[quote="maokid7"]how were u able to factor the denominator?[/quote]

http://www.quickmath.com - in particular, [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=factor&s3=basic]the factor page[/url].

The next step: If we let $p(x)=x^{6}-4x^{5}+3x^{4}+4x^{3}-2x^{2}-2x+1$, then $p(1-x)=1-4x+3x^{2}+4x^{3}-2x^{4}-2x^{5}+x^{6}=x^{6}p(1/x)$. Setting $I=\int_{-\infty}^\infty \frac{x^{2}-2x+1}{p(x)}\,dx$ and making the substitution $t=1-x$, we get
$I=\int_{-\infty}^\infty \frac{t^{2}}{t^{6}p(\frac1t)}\,dt= \int_{-\infty}^\infty \frac{t^{-4}}{p(\frac1t)}\,dt$. Substituting again $t=\frac1y,dt=-\frac1{y^{2}}$, we get
$I=\int_{0}^{-\infty}\frac{y^{4}}{p(y)}\,\frac{-1}{y^{2}}dy+\int_{\infty}^{0}\frac{y^{4}}{p(y)}\,\frac{-1}{y^{2}}dy$
$I=\int_{-\infty}^\infty \frac{y^{2}}{p(y)}\,dy-\int_{-\infty}^\infty \frac{x^{2}}{p(x)}\,dx$.

Another round of the same operations (on the new form of $I$) gets us
$I=\int_{-\infty}^\infty \frac{t^{-4}-2t^{-5}+t^{-6}}{p(\frac1t)}\,dt$
$I=\int_{-\infty}^\infty \frac{x^{2}-2x^{3}+x^{4}}{p(x)}\,dx$

A third repetition would return us to the original form.

Adding the three versions, $I=\int_{-\infty}^\infty \frac{x^{4}-2x^{3}+3x^{2}-2x+1}{x^{6}-4x^{5}+3x^{4}+4x^{3}-2x^{2}-2x+1}\, dx$, which is invariant under the substitution $x\mapsto \frac1{1-x}$. Chasing invariants look like a useful approach, but I haven't figured out the finish.

Let $K$ denote the desired integral. The denominator factors into two polynomials of degree 6, and the corresponding partial fraction form is :

 $ 2 K=\int_{0}^{+\infty}\frac{(x-1)^{2}}{x^{6}-4x^{5}+3x^{4}+4x^{3}-2x^{2}-2x+1} dx +\int_{0}^{+\infty}\frac{(x+1)^{2}}{x^{6}+4x^{5}+3x^{4}-4x^{3}-2x^{2}+2x+1} dx $

Substitution of $x=1/(u-1)$ in the second integral gives

$\int_{1}^{+\infty}\frac{u^{2}(u-1)^{2}}{u^{6}-4u^{5}+3u^{4}+4u^{3}-2u^{2}-2u+1} \ du$

Substitution of $x=(u-1)/u$ in the $\int_{0}^{1}$ portion of the first integral gives

$\int_{1}^{+\infty}\frac{u^{2}}{u^{6}-4u^{5}+3u^{4}+4u^{3}-2u^{2}-2u+1} \ du$

Adding these two, and leaving the $\int_{1}^{+\infty}$ portion of the first integral as is, we get

$2K=\int_{1}^{+\infty}\frac{x^{2}+(x-1)^{2}+x^{2}(x-1)^{2}}{x^{6}-4x^{5}+3x^{4}+4x^{3}-2x^{2}-2x+1} \ dx$

Let $p(x)=x^{3}-2x^{2}-x+1$ and $q(x)=x^{2}-x$. Note that the denominator of the integrand is equal to $p(x)^2 + q(x)^2$, while its numerator is equal to $q(x)p'(x)-q'(x)p(x)$
Therefore

$2K=\left[\arctan \frac{p(x)}{q(x)}\right]_{1}^{+\infty}=\pi$


[quote="mychrom"]If (a+b+c)(1/a +1/b +1/c)>10, than find a relation between a,b &c.[/quote]
This seems beyond moems  :huh:
[hide]$ab+bc+ac>10$ follows from $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ac}{abc}$[/hide]


As the question goes. I personally prefer blank paper, that way I don't sort of feel restricted. Plus it is clearer. :D

I prefer blank paper but I know most of my friends like lined paper. When I do my math homework on blank paper, they always ask "WHY IS THIS ON PRINTER PAPER?"

Ummmm....
You can't write proofs in your head....

I've always used lined paper since its easier to hand. The paper doesn't really bother me (unless its geometry)

My handwriting can slant after a period of time, so I prefer lined paper  :wink:

I like blank because I get OCD about my writing and the lines mixing. :)

Blank is better because it gives no limits on your line sizes.

I had to cram division with terms such as $ z^2 _2$, etc. into a college rule paper, before. Not fun.

i prefer blank paper. lined paper gets in the way when you have big equations (such as huge fractions that can't fit onto one line) or with geometric diagrams.

Very true. I also don't  like it when your square roots get lost in the lines. Or when the bars of your fractions get lost in the lines.

For formal proofs (like the AoPS challenge sets):
For working out the proofs (not writing it out, just formulating my ideas and stuff like that), I just use whatever paper is closest to me (if I write fractions and stuff like that on lined paper, I just use up 2 lines, or however many lines I need). When doing my final draft, I usually type it up.

For homework and stuff:
I use lined, because my math teacher makes us do our work in a certain way. So yeah.

Lined is better, because if you actually have good drawing skills for geo diagrams, its easy to do the line segment division cheating thing, if the answer is an integer. I don't write in the lines on lined paper when using it as scratch, but the lines let me plot out a grid which help make a good diagram. Graph paper does that too, but has too many lines so it messes up my brain when I try to read writing.

When it's possible, I use my head (and leave answer on my homework paper). However, I often get marked off for not showing my work as a result. When necessary, I use graph paper because I think you can pack more in it, and is good for drawing straight lines. But I prefer blank paper over line paper because my pencil can freely slide without the visual distraction of the lines (notably the margin).Blank paper is good scratch paper because it seems cleaner, which helps me concentrate.

I use blank paper for proofs whenever possible. Lined paper for homework, though since I'm taking linear algebra (the course sucks to be honest, it just has us do random computations with 5x5 matrices on calculators and the few proofs we actually do are half given to us and the half that's given has massive notational issues), the matrices take up a lot of space so I often ignore the lines. It helps when I'm writing words, though, since I have horrible handwriting.

Im not much fond of doing things on paper and firmly believe that if you can do it on paper, you might as well do it in your head :)

Blank paper is brighter white, and has more visual appeal to me.

I also like being able to utilize the space as i please (my writing tends to be ~1.2 spaced when I am doing math, rather than 1x or 2x)

I like using graphing paper (if its available, otherwise lined paper) for geometry, when they aren't a lot of square roots and fractions and other line-confusing things. But, for normal algebra blank

In @above's format, I use

Drawing - Blank
Homework - Lined
Notes - Lined
Calculations - Blank

wait i just found out this is a 2008 thread that was bumped


Here is the IMO shortlist problems 1993, I got it from internet.


inequalities calculus derivative function geometry algorithm email

Can experts in inequalities please solve the following question:

If x>=0, y>=0, x+2y<=20, 3x+y<=30, find the maximum value of x+y.

Is there an algebric method to solve this question? I was wondering if it is possible to solve by partial differentiation? :maybe:

Thank you very much! :P

[hide]To maximize the values of $x$ and $y$, we write $x+2y=20$ instead of $\le$ and $3x+y=30$ (again, instead of $\le$).

Solving, we get $x=8$ and $y=6$, so $x+y=14$.[/hide]

What if I want to find the maximum value of x-y, how can I solve it?

Thank you very much! :P

In general, to solve these types of questions (Maximising a function subject to given $\leq$ constraints) you draw a graph of the lines with the equals signs, here the equations $x+2y=20$ and $3x+y=30$.

The maximum value of your function (as long as it is linear in terms of $x$ and $y$) will occur at one of the vertices of the shape your graph forms. In this case, the vertices are

$(0,10)$, $(8,6)$, $(10,0)$

So the only things you need to try are these three pairs: the maximum is 10 when $x=10$, $y=0$.

This process is known as 'linear programming'- you can google it or something.

tim1234133, thank you very much! :P
I just wonder why will the maximum value of the linear function occur at one of the verticies? What is the theory behind this method? :huh:
If the function is not linear, is there any elementry method of solving it? :maybe:

Thank you very much! :P

The maximum of the function occurs at a vertex because the area that they enclose is convex.
If the function you want to maximise is (say) $2x+y$, then consider all the lines with equation $2x+y=k$. All these lines are parallel and as $k$ increases, the get further away from the origin. I'm not entirely sure how to prove from this that the maximum is at a vertex, but it does follow from the convexity of the enclosed area.

I do not know of a method for non-linear functions, but one does exist (I saw them mentioned in a textbook but it just said that it was much harder in this case).

If you solve a linear programing problem manually, try all the vertexes is ok. But if you would like to solve it in a systematic method, you can try Simplex algorithm.

Modeler, I am interested in the systematic approach, do you know the simplex algorithm? Can you please post it here? :oops:

Thank you very much! :P

[quote="toanhocmuonmau"]We have
\[x+y=\frac{2}{5}(x+2y)+\frac{1}{5}(3x+y) \le 14 \]
;)[/quote]

Sorry, I do not quite get how the 14 comes into the inequality, can toanhocmuonmau please explain to me?

Thank you very much! :P

[quote="puuhikki"][quote="schoolgirl"][quote="toanhocmuonmau"]We have
\[x+y=\frac{2}{5}(x+2y)+\frac{1}{5}(3x+y) \le 14 \]
;)[/quote]

Sorry, I do not quite get how the 14 comes into the inequality[/quote]
By the assumption you made, we have $x+2y\leq 20$ and $3x+y\leq 30$. Therefore $\frac{2}{5}(x+2y)+\frac{1}{5}(3x+y)\leq \frac{2}{5}\cdot 20+\frac{1}{5}\cdot 30=14$.[/quote]

puuhikki, thank you very much! :P
Does the inequality method works for x-y, and non-linear function?

Thank you very much! :P

[quote="toanhocmuonmau"]Ok, We have
\[3(x-y)=(3x+y)-4y \le 30 \]

\[\Rightarrow x-y \le 10 \]
Now, let $x=10,y=0$, we have $x-y=10$.
Hence
\[\max\{x-y\}=10 \]
;)[/quote]

toanhocmuonmau, thank you very much! :P
I just wonder if this method works for non-linear function?

Thank you very much! :P

[quote="guowenfei"]How to get in touch with you?I have a nice idea
Thank you very much!![/quote]

guowenfei, you can send me an email; my email is alicia.yau@googlemail.com

Thank you very much! :P


If I have (2x1 )Matrixs A, how can I find ortogonal Matrix to A?
Note my Mitrix includ complex nomber.

I assume you mean a column vector in $ \mathbb{C}^2$...

And then you solve the equation $ A \cdot \binom{c}{1} \equal{} 0$, and all column vectors of the form $ \binom{ck}{k}$ will work.

(i) Two non zero vectors, $ u$ and $ v$, are orthogonal if and only if  $ u\cdot v\equal{}0$.

(ii) Suppose that $ u$ and $ a\ne 0$ are both vectors in $ \mathbb{R}^2$ or $ \mathbb{R}^3$ then,

$ proj_a u\equal{}\frac{u\cdot a}{\|a\|^2}a$

and the vector component of $ u$ orthogonal to $ a$ is given by,

$ u\minus{}proj_a u\equal{}u\minus{}\frac{u\cdot a}{\|a\|^2}a$


Let $a, b, c>0$ such that $a+b+c=1.$ Prove that the following inequality holds:

                         \[ \frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq\frac{36abc}{13abc+1}. \]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=17078 and http://www.mathlinks.ro/Forum/viewtopic.php?t=1656 .

  darij


calculus integration function videos MATHCOUNTS

[quote="mewto55555"][quote="cognos599"][hide][hide][hide][/hide][/hide][/hide][/quote]

DUDE WTH U ALMOST GAVE ME CARPAL TUNNEL SYNDROME (not really i just quoted and found the middle). YOU PEOPLE FROM INDIANA ARE MESSED UP!!![/quote]
All you had to do was press "quote", then scroll to the middle, obviously, being retarded enough to click is out of the question. Oh wait this is Missouri nevermind you had a point there.

[quote="serialk11r"][quote="mewto55555"][quote="cognos599"][hide][hide][hide][/hide][/hide][/hide][/quote]

DUDE WTH U ALMOST GAVE ME CARPAL TUNNEL SYNDROME [u][b][size=200](not really i just quoted and found the middle)[/size][/b][/u]. YOU PEOPLE FROM INDIANA ARE MESSED UP!!![/quote]
All you had to do was press "quote", then scroll to the middle, obviously, being retarded enough to click is out of the question. Oh wait this is Missouri nevermind you had a point there.[/quote]

WOW SOMEONE NEEDS TO LEARN TO READ!!!!!

I don't think you get it, what I meant was it's so obvious you shouldn't mention it unless you're retarded. This is MO forum so that's expected of course.

[quote="cognos599"]dude serialk11er, what the heck, like mewto5555 said, learn to read and even if you *barely* can actually do it and don't call other people retarded if you are too retarded to either be able to read what has just been posted or actually do it.[/quote]

umm.. no offense cognos599, but why are you defending mewto55555 when you're the one that started this whole thing? What has MO done to IN? *SIGH* just when I thought the spamming was over...

[hide="Click if you want to know integral calculus"][hide="well, ok derivatives."][hide="ok only limits"]HE POSTED IT IN HIS SIG SO I DID IT!!![/hide][/hide][/hide]

lol so Indiana's spam leaked here. Is that why the spam's been decreasing on our forum, cognos?

huh? I was talking present tense. You're talking past tense. So you mean ccy? He was inactive on our forum..

fat people

have

fat

moderators

indiana

is

fat

etc etc etc

[quote="cognos599"][hide][hide][hide][/hide][/hide][/hide][/quote]
um what is the point of this just asking?

[quote="mathemagician1729"]wait i thought i saw you at the ku thing like in april and I am pretty sure you are a he.
EDIT: Why does everyone think we Kansans are rednecks and farmers. If you want to see that then head out west. Us eastern Kansans are (relatively) sophisticated compared to our western counterparts...[/quote]
Quite.

[quote="tinytim"]yay!!! :P....[/quote]
Let's both be happy together!!!!! :D

We're TWIN TELETUBBIES!!! :D  :D

you can display ... um stuff... on your belly tvs!

I believe those are called the [b]Teletubbies tellies[/b]

only fat kids watch teletubbies.

 seriously, television + tubbiness =/= good.

Missourians, lets talk in  secret number language so other states don't know what we're saying!

45 5 8 666 564 367 356 45445 431 6768 56 239 368036 3223 5556827 347 1447.

Yes but then I don't know what you are saying. ZOMG tinytim said 666 he's the devil!

you're supposed to pretend you know.... :|

$ \textbf{tellitubbies}$ $ \equal{}$ $ \boxed{\textbf{NOT REAL}}$ sorry for breaking yor heart...

the above statement is false, but my previous post is true.

[quote="pianoforte"]I do not pee on roads. The splatter effect would be enormous. Besides, you have enough flooding as it is.[/quote]

Just realized, this guy is a mod!

Wow.

You're making me bite my finger like Tinky Winky :D

[quote="mewto55555"]:o  Yeah and next you'll tell me the Easter Bunny grinds up children to make eggs.[/quote]

This is quite true. Have notice we are missing little Timmy.

Timmy? Little?

Little timmy= tinytim?!?!?!?!?!!?!?! :o


given $ p>0,q>0, p\plus{}q<1$ and sequence $ \left(a_n \right)_{n \in N}$ nonnegative  satisfying $ a_{n\plus{}2} \leq pa_{n\plus{}1}\plus{}qa_n, n\equal{}1,2, \cdots$  prove that $ (a_n)$ converges and find $ lim_{n \to \infty } a_n$

Fix $ n$.

If $ a_{n\plus{}1}\leq a_n$ then $ a_{n\plus{}2}\leq (p\plus{}q)a_n$.
If $ a_{n\plus{}1}\geq a_n$ then $ a_{n\plus{}2}\leq (p\plus{}q)a_{n\plus{}1}$.

In both cases we get recursion to lower $ n$. In the end of iteration process we get that $ a_{n\plus{}2}\leq (p\plus{}q)^{\theta_1 (n)}a_1$ or $ a_{n\plus{}2}\leq (p\plus{}q)^{\theta_2 (n)}a_2$ (it depends on $ n,\;a_1,\;a_2$), where $ \lim_{n\to\infty}\theta_1(n)\equal{}\lim_{n\to\infty}\theta_2(n)\equal{}\infty$. And because $ p\plus{}q<1$ we get $ \lim_{n\to\infty}a_n\equal{}0$.


[quote="4everwise"]If the sum of two numbers is 1 and their product is 1, then the sum of their cubes is:

$\text{(A)} \ 2 \qquad \text{(B)} \ -2-\frac{3i\sqrt{3}}{4} \qquad \text{(C)} \ 0 \qquad \text{(D)} \ -\frac{3i\sqrt{3}}{4} \qquad \text{(E)} \ -2$[/quote]

[hide](E) -2
$a+b=1$
$ab=1$
$a^3+b^3=(a+b)^3-3ab(a+b)=1-3$
$a^3+b^3=-2$[/hide]

[hide="solution"]$x+y=1$ and $xy=1$ so $x^2+2xy+y^2=1$ which means $x^2+y^2=-1$. $x^3+y^3=(x+y)(x^2-xy+y^2)$ substitute the values you know and you should get $(-1-1)=-2$. $\boxed{E}$[/hide]


trigonometry
analytic geometry
calculus
derivative
complex numbers
trig identities
Law of Cosines

Three points $A(-1,0),B(\cos \theta,\sin \theta),C(\cos 2\theta,\sin 2\theta) $ lie in on a coordinate plane.

Let $0\leqq\theta \leqq\pi$, Find the maximum and minimum value of $AC+BC$.

Using complex numbers,

$A = -1$
$B = e^{i \theta} = \cos \theta + i \sin \theta$
$C =  e^{2i\theta} = \cos{2\theta} + i \sin{2\theta}$

$AC = |C - A| = |\cos{2\theta} + i \sin{2\theta} + 1| =$
$= \sqrt{1 + 2 \cos{2\theta} + \cos^2{2\theta} + \sin^2{2\theta}} = \sqrt{2 + 2 \cos{2\theta}} = 2 \cos \theta$

$BC = |C - B| = |\cos{2\theta} + i \sin \theta - \cos \theta - i \sin \theta| =$
$= \sqrt{\cos^2{2\theta} - 2 \cos{2\theta} \cos \theta + \cos^2{\theta} + \sin^2{2\theta} - 2 \sin{2\theta} \sin \theta + \sin^2{\theta}} =$

$= \sqrt{2(1 - \cos{2\theta} \cos \theta - \sin{2\theta} \sin \theta)} = \sqrt{2[1 - \cos \theta (\cos^2{\theta} - \sin^2{\theta}) - 2 \sin^2{\theta} \cos \theta]} =$
$= \sqrt{2(1 - 2 \cos^3{\theta} + \cos \theta + 2 \cos^3{\theta} - 2 \cos \theta)} = \sqrt{2(1 - \cos \theta)} = 2 \sin{\frac \theta 2}$

$AC + BC = 2\left( \cos \theta + \sin{\frac \theta 2} \right)$

For minimum or maximum $AC + BC$:

$\frac{d(AC + BC)}{d\theta} =  2\left( -\sin \theta + \frac 1 2 \cos{\frac \theta 2} \right) = 0$

$\frac 1 2 \cos{\frac \theta 2} -  2 \sin{\frac \theta 2} \cos{\frac \theta 2} = 0$

Hence, either $\cos{\frac \theta 2} = 0$, $\theta = \pi$, or $\sin{\frac \theta 2} = \frac 1 4$, $\theta \doteq 0.16\ \pi$.

$\frac{d^2(AC + BC)}{d\theta^2} =  2 \left(-\cos \theta - \frac 1 4 \sin{\frac \theta 2} \right)$

For $\theta = \pi$, the 2nd derivative is equal to

$2 \left(-\cos \pi - \frac 1 4 \sin{\frac \pi 2} \right) = \frac 3 2 > 0$

and we have a minimum. For $\sin{\frac \theta 2} = \frac 1 4$, the 2nd derivative is equal to

$2 \left(2 \sin^2{\frac \theta 2} - 1 - \frac 1 4 \sin{\frac \theta 2} \right) = -\frac{15}{8} < 0$

and we have a maximum.

Needed a TI-89 to do this... :(

[hide="My inelegant solution."]A, B, and C are all on the unit circle (let's call the center O).  We can apply the Law of Cosines to triangles BOC and AOC to get
\[BC=\sqrt{1^2+1^2-2\cos(\theta)}=\sqrt{2-2\cos(\theta)}\]
and
\[AC=\sqrt{1^2+1^2-2\cos(\pi-2\theta)}=\sqrt{2+2\cos(2\theta)}\]
so the expression we wish to minimize and maximize is
\[\sqrt{2-\cos(\theta)}+\sqrt{2+\cos(2\theta)}\]
Using a TI-89, I could determine that the minimum is $\sqrt2$, attained when $\textstyle{\theta=\frac{\pi}{2}}$, and the maximum is 4, attained when $\theta=\pi$.

Can anyone optimize the above expression without a TI-89?  There has to be a way to do so...[/hide]

Edit: found a completely different, very simple way to determine the maximum (but not the minimum)...

[hide="Here it is."]BC and AC are both chords of the unit circle.  Both are maximized when they are diameters; that is, when they have length 2.  It is possible to maximize both at once by making $\theta=\pi$, since AC and BC will then both be diameters.  Thus, the maximum is 2+2=4.[/hide]

Thank you for your replies,yetti and dts.

To [b]yetti[/b]:

You should remark the condition $0\leqq\theta  \leqq\pi$.Then you have $AC=2|\cos \theta|$. ;)

To [b]dts[/b]:

For maximum value,your answer is right.Especially your geometric solution is wonderful. :)

To all of you:

You can use double-angle and half-angle formula of triangular function.

kunny


calculus
integration
trigonometry
calculus computations

Evaluate

\[ \frac{1}{\displaystyle \int_0^{\frac{\pi}{2}} \cos ^{2005}x\ \sin {2007x}\ dx}. \]

Thank you for your reply as usual.
Regrettably your answer is incorrect, anik_andrew. :(

kunny

I am frisking ! Denote $I_{n,m}=\int^{\frac{\pi}{2}}_0 \cos ^n x\sin mx\ dx$, where $\{m,n\}\subset N.$ Then, for any $n\in N$ and $m\ge 4$ there is the following recurrent relation

$\boxed {\ I_{n,m}=2I_{n+2,m-2}-I_{n,m-2}+I_{n+1,m-1}-I_{n+1,m-3}\ }\ !$

Oh! problem is very nice. Thank! Thank!  :roll:


Exercise 5.6. Show that the product of any three consecutive integers has
6 as a factor. (How many cases should you use here?)

If I have $ (n\minus{}1)n(n\plus{}1)$, expanding I get $ n^{3}\minus{}n$, it shows no indication of being divisible by 6? Help plz

[hide="Method 1"] We want to prove that a product of three consecutive integers is divisible by $ 2$ and by $ 3$.  Say we are talking about $ n\minus{}1, n, n\plus{}1$.

[b]Case:[/b]  $ n$ is even.  Then $ 2 | n | (n\minus{}1)n(n\plus{}1)$.

[b]Case:[/b]  $ n$ is odd.  Then $ 2 | n\minus{}1 | (n\minus{}1)n(n\plus{}1)$.

Now we want to decide whether the expression is divisible by $ 3$.

[b]Case:[/b]  $ n\equiv 0\bmod 3$.  Then $ 3 | n | (n\minus{}1)n(n\plus{}1)$.

[b]Case:[/b]  $ n\equiv 1\bmod 3$.  Then $ 3 | n\minus{}1 | (n\minus{}1)n(n\plus{}1)$.

[b]Case:[/b]  $ n\equiv 2\bmod 3$.  Then $ 3 | n\plus{}1 | (n\minus{}1)n(n\plus{}1)$. [/hide]
[hide="Method 2"] $ n\equiv n^{2}\equiv n^{3}\bmod 2\implies 2 | n^{3}\minus{}n$
$ n\equiv n^{3}\bmod 3\Leftrightarrow 3 | n^{3}\minus{}n$ (by Fermat's Little Theorem) [/hide]
[hide="Method 3"] $ {n\plus{}1\choose 3}$ is always an integer. [/hide]
[b]Generalization 1:[/b]  $ k!$ divides a product of $ k$ consecutive integers.

[b]Exercise:[/b]  What is the largest positive integer that divides $ n^{4}(n\plus{}1)^{3}(n\plus{}2)^{2}(n\plus{}3)$ for all positive integers $ n$?


limit
LaTeX
logarithms
calculus
calculus computations

HINT:
Notice that the limit above is in the form of f(x)^g(x) where f(x)->1 and g(x) tends to infinity i.e.
the limit is in the form 1^ (infinity ).
Such type of limits transfer into this form:
lim(x->a)f(x)^g(x)=
 e^(lim x->a){ g(x)*h(x)}
where h(x)=f(x)-1.

do the problem in this way ,u will get the answer. :)

hello, by the rule of L'Hospital we get ${{ \lim_{x \to \infty}(e^x+x)^{\frac{1}{x}}=e^{\lim_{x \to \infty}\frac{\ln(e^x+x)}{x}}}=e^{\lim_{x \to \infty}\frac{e^x+1}{e^x+x}}}=e^1=e$
Sonnhard.

hey Dr.
Dude,x is tending to zero not to infinity .

Thus my answer is e^2.
Kindly check it.

$ \lim_{x \to 0} (e^x\plus{}x)^{\frac{1}{x}} \equal{} \lim_{x \to 0} \left[e^x \left(1 \plus{} \frac{x}{e^x}\right)\right]^{1/x} \equal{} e \cdot \lim_{x \to 0} \left[ \left(1 \plus{} \frac{x}{e^x}\right)^{\frac{e^x}{x}} \right] ^{\frac{1}{e^x}} \equal{} e \cdot e^1 \equal{} e^2.$

hello, ohh, this was a mistake, it must be ${{ \lim_{x \to 0}(e^x+x)^{\frac{1}{x}}=e^{\lim_{x \to 0}\frac{\ln(e^x+x)}{x}}}=e^{\lim_{x \to 0}\frac{e^x+1}{e^x+x}}}=e^2$
I was a little bit tired in this morning. Thank you for your hint, kabi.
Sonnhard.

It is fine Dr.
Sometimes Dr. also get tired.Go and take rest. :lol:

hello, that's clear, i'm sitting here at my writing table, watching a dvd and preparing a class examination for my students and doing math all the time, so i have no time to get rest.
Sonnhard.

Dr.
I can't install Latex from this site as speed of my net is not coming in good speed now days.I hope it will be all right by Monday.I will install it that day only.
Well for  doing so much work ,that u have mentioned , One can't take rest.
Ur students must be proud of getting such a hardworking teacher.
 Good work Dr.
:clap2:

[quote="Carcul"]$ \lim_{x \to 0} (e^x \plus{} x)^{\frac {1}{x}} \equal{} \lim_{x \to 0} \left[e^x \left(1 \plus{} \frac {x}{e^x}\right)\right]^{1/x} \equal{} e \cdot \lim_{x \to 0} \left[ \left(1 \plus{} \frac {x}{e^x}\right)^{\frac {e^x}{x}} \right] ^{\frac {1}{e^x}} \equal{} e \cdot e^1 \equal{} e^2.$[/quote]

I think $ \lim_{x \to 0} (e^x \plus{} x)^{\frac {1}{x}} \equal{} e$

Thank you for your post.

Thanham, it's been shown already that the limit is, in fact, $ e^2$.

hi thanhnam2902
answer is e^2 only.
It will be better if u just post ur solution , may be we can clarify it.

how about without any differential (l'hopital/taylor) methods

$ \\
L \equal{} \displaystyle\lim_{x\to0} (e^x\plus{}x)^{1/x}
\\
\\
\ln{L} \equal{} \displaystyle\lim_{x\to0} \frac{\ln{(e^x\plus{}x)}}{x} \cdot \frac{x}{e^x\minus{}1} \equal{} \frac{\ln{(e^x\plus{}x)}}{e^x\minus{}1}
\\
\\
u \equal{} e^x\plus{}x \to \ln{L} \equal{} \displaystyle\lim_{(x,u)\to(0,1)} \frac{\ln{u}}{u\minus{}x\minus{}1}
\\
\\
\displaystyle \frac{1}{\ln{L}} \equal{} \displaystyle\lim_{(x,u)\to(0,1)} \frac{u\minus{}x\minus{}1}{\ln{u}} \equal{} \frac{u\minus{}1}{\ln{u}} \minus{} \frac{x}{\ln{(e^x\plus{}x)}} \equal{} 1 \minus{} \frac{1}{\ln{L}}
\\
\\
\displaystyle\frac{2}{\ln{L}} \equal{} 1 \to L \equal{} e^2$

Yes! the true is $ e^2$. Thank you very much.


search
number theory
prime numbers
number theory proposed

$ P$ is a set of positive prime numbers with at least $ 3$ elements such that for any proper finite subset $ M$ of $ P$, any prime divisor of

$ \prod_{p \in M}p \minus{} 1$ is also in $ P$.

Show that $ P$ is the set of all primes.


hey I have a problem:

Determine all p prime numbers such that p+2,p+4,p^2+2 and p^2+4 are simultaneously primes.

If p=3k+1 then p+2 is not prime and if p=3k+2 then p+4 is not prime so the only solution is p=3.

as  simple as that!
:D


I was hoping that some people could help me with some math problems.
[b]
D. Write the following equations in vertex form. State the vertex and the axis of symmetry:
[/b]

2. y = x^2 - 10x + 35

[b]E. Graph each function. State (a) the vertex, (b) the equation of the axis of symmetry, and (c) the maximum or minimum value:[/b]

1. y = -(x-3)^2 + 1

2. y = (x + 1)^2 - 3

3. y = 3(x - 2)^2

4. y = 1/2(x + 3)^2

[b]F. In the equation y = a(x-k)^2 + h[/b]

1. What does the "a" do to the graph of the equation y = x^2?

2. What does the "k" do to the graph of the equation y = x^2?

3. What does the "h" do to the graph of the equation y = x^2?

[b]G. In the equation y = -2(x + 3)^2 + 4[/b]

1. What does the "-2" do to the graph of the equation y = x^2?

2. What does the "3" do to the graph of the equation y = x^2?

3. What does the "4" do to the graph of the equation y = x^2?

4. If the "-2 was changed to "1/2", how would the graph change?

5. Graph the first equation stated.

[b]H. Define the following:[/b]

1. Degree of an Equation -

2. Quadratic Equation -

3. Vertex Form of a Quadratic Equation -

If anyone could help me solve these or even teach me the method of how to do these problems I would GREATLY appreciate it.

Thanks,
Chris. :)

[color=red][this part of post removed by mod on request.][/color]

I'm not going to do your take-home test for you but,

Vertex Form of a Quadratic Equation is $ a(x \minus{} h)^2 \equal{} y \minus{} k$.
The vertex of this quadratic equation is $ (h,k)$.

Here is an example of how to put a quadratic equation in vertex form.
Take $ y \equal{} 2x^2 \minus{} 8x \plus{} 3$.
First, factor out the 2 from the terms with an x: $ y \equal{} 2(x^2 \minus{} 4x) \plus{} 3$
Next, recognize that $ (x \minus{} h)^2$ is in the form $ x^2 \minus{} 2hx \plus{} h^2$. So in this case, $ h \equal{} 2$ and thus, $ h^2 \equal{} 4$ so you must add 4 inside the parentheses.
Since everything inside the parentheses is being multiplied by 2, you are really adding 8 and must add 8 on the left side as well to compensate.
$ y \plus{} 8 \equal{} 2(x^2 \minus{} 4x \plus{} 4) \plus{} 3$.
Now, rewrite $ x^2 \minus{} 4x \plus{} 4$ as $ (x \minus{} 2)^2$ and subtract 3 from both sides.
$ y \plus{} 5 \equal{} 2(x \minus{} 2)^2$. You now have the quadratic in vertex form and see that the vertex is located at $ (2, \minus{} 5)$.

Also for future reference, problems of this level should go into the High School Basics forum.


Solve this equation:
$ x^5\equal{}(x\plus{}1)^2$

The best I can do for you. [url]http://www.wolframalpha.com[/url]


I am only posting solutions...

[hide="#5"]Multiply by $ xy$.

Then you get
$ y - x = - 7xy$
$ y + x = 3xy$
Eliminate $ x$ by adding them together.
So $ 2y = - 4xy$
Divide each side by $ 2y$
That should give you $ x = - \frac12$
Then you substitude to get $ y$ which is $ \frac15$, and add them up to get $ \boxed{ - \frac3{10}}$[/hide]

I figured #9 out.. :lol:
[hide="#9"] We know that $ \frac {20!}{2^n}$ should give us an positive integer.

So what I did was  $ 20! = 20\times19\times 18\times \cdots \times 1$, but only thing matters is the even numbers.

$ 20 = 2^2\cdot5$
$ 18 = 2^2\cdot3^2$
$ 16 = 2^4$
$ 14 = 2^7$
$ 12 = 2^2\cdot3$
$ 10 = 2\cdot5$
$ 8 = 2^3$
$ 6 = 2\cdot3$
$ 4 = 2^2$
$ 2 = 2^1$

Then you count up $ 2^x$ to get $ \boxed{18}$
[/hide]

[hide="#14"] You just have to know the divisibillity rule...
It's,in this case, $ A - 4 + 6 - 8 + B - 0 + 5$ has to be multiples of 11.
So that simplifies into $ A - 1 + B$, we can perceive that $ A + B$ should equal to $ \boxed{12}$[/hide]
----------------------------------------------------------------------------------------------------------
From here, I solved it, so answers may not be right...

[hide="#18"]Multiply each terms by $ \frac14$

Then you get $ \frac {13}{10}$,$ \frac {17}{13}$,$ \frac {21}{16}$,$ \frac {25}{38}$, respectively.

So the smallest fraction is $ \boxed{\frac {25}{76}}$
[/hide]

[hide="#19"]#19 was pretty easy and obvious...$ \boxed{\frac13}$[/hide]

[hide="#20"]There are 5 parts, and half has 6.
So it's $ \frac56\times\frac12 = \frac5{12}$
Multiply it by 100 and round it up to get $ \boxed{42}$(%)[/hide]

[hide="#23"]Plug and Chug..
$ 3\diamond4 = 1$

$ 1\diamond5 = \boxed{\frac { - 3}{2}}$[/hide]

[hide="#24"]Multiply by 100 on each terms and simplify as much as you can.

That would get you $ \frac {1255\cdot61}{622\cdot 41}$.which also can be look at it like this..

$ \rightarrow \frac {1255}{622}\cdot\frac {61}{41}$
The first fraction is close to $ 2$ and the second one is close to $ 1$, so $ 2 + 1 = \boxed{3}$[/hide]

[hide="#25"]I rounded up $ \sqrt {2}$ as $ 1.4$
$ 1 + 1.4 = 2.4$
So the range should be $ 0 < x^{\frac13} < 2.4$ which also can be $ \rightarrow 0 < x < 2.4^3$
$ 2.4^3 = 15.324$

So, $ 0 < x < 15.324$ and $ x$ is an integer, $ \boxed{15}$ integers satisfiy the inequality.[/hide]

[hide="#26"]Simplify to get $ \frac {1}{6\cdot5\cdot3\cdot2\cdot1} + \frac {1}{5\cdot4}$

$ \longrightarrow \frac {1}{180} + \frac1{20} = \frac1{180} + \frac9{180} = \frac {10}{180} = \boxed{\frac {1}{18}}$[/hide]

[hide="#27"]You need to know how to put numbers in algebraric form.

It was given that $ A + B = 12$ and the reversed number was 54 greater than the original.

[u]Equations:[/u]

$ 10B + A - (10A + B) = 54\longrightarrow \ 9B - 9A = 54$

$ \begin{array}{c} 9B + 9A = 108 \
 9B - 9A = 54 \end{array}$

Add them to get $ 18B = 162\rightarrow B = 9$

[u]Substitude[/u]= $ A + 9 = 12\rightarrow A = 3$

Thus, the product would be $ 9\times 3 = \boxed{27}$[/hide]

[hide="#28"]Notice that 5 liked only strawberry. So we can subtract those 5 people from 22 students.
So 17 students left.

Draw a Venn Diagram(way easier) or make an equation..

Then the answer should be $ \boxed{2}$[/hide]

I will try to solve more..

If you guys have a solution to the problem that I didn't post, please post..

#15:  You didn't find enough decimal places on $ \sqrt2$

#18:  I don't follow your line of thought.  why multiply by $ \frac{1}{4}$?

[quote="danb80"]Note:  These questions can be found on the materials page of the the JMS MATHCOUNTS web page (94 state)[/quote]

I noticed that, but I want to look at it later... because the temptation's too great for me... :lol:

[quote="danb80"]Note:  These questions can be found on the materials page of the the JMS MATHCOUNTS web page (94 state)[/quote]
When I went to the Jenks MATHCOUNTS webpage yesterday, and typed in the password it loaded the site but only displayed:
"Sorry, the page you requested was not found."
I checked twice again this morning but it still didn't load.I'm wondering is it my computer's fault or is the webpage faulty? Maybe I put the wrong password ?If so could you pm me ,danb80, with the right one? PLEASE?  :help:

[quote="danb80"]#15:  You didn't find enough decimal places on $ \sqrt2$

#18:  I don't follow your line of thought.  why multiply by $ \frac {1}{4}$?[/quote]

On #15, [hide]then it should be 14[/hide]..

On #18, I found my mistake.
[hide]You multiply those by $ 4$ to make fractions in order to break them down into smaller cases..
If I am right, the answer should be $ \frac {13}{40}$ Is it right??[/hide]


algebra
polynomial
induction
algebra unsolved

If f(x)+f(1/x)=f(x)f(1/x),then find f(x) with a proof. :)

f(x) is a polynomial function.

There is a missing hypothesis, otherwise there an infinite number of solutions...

Pierre.

let f(x) = a0 + a1x + a2x^2 + ... an-1x^(n-1) + anx^n   (where n belongs to whole numbers)

given (a0 + a1x + a2x^2 + ... + anx^n)(a0 + a1(1/x) + a2(1/x)^2 + ... + an(1/x)^n)
 = (ana0)x^n + (ana1+ an-1a0)x^(n-1) + (ana2 + an-1a1 + an-2a0)x^n-2 .. + (anak+an-1ak-1+an-2ak2+
      .... +an-ka0)x^(n-k) +.... + (an^2 + an-1^2+...a0^2)

equating coffecients for x^n we have ana0=an. Since f is an n degree polynomial an cannot be 0
hence a0 = 1 ........... I

equating coefficents for x^(n-1) we have ana1+an-1a0 = an-1
=> ana1 + an-1 = an-1 from I
=> ana1 = 0. Since an is not 0, we must have a1 = 0 ........... II

equating coeff. of xn-2 we and using I & II above we get a2 = 0

now assume we have equated all coefficients from x^n  to x^(n-k+1) and found a1, a2,..,ak-1 to be 0
then equating coeff of xn-k we observe
anak + an-1ak-1 + an-2ak-2+...+an-k+1a1 + an-ka0 = an-k
=> anak +0+0+0+...+0+an-ka0 = an-k   (assumed a1 to ak-1 are all 0)
therefore anak + an-k = an-k   (a0=1)
=>anak=0
therefore ak=0   (an cannot be 0)
hence if a0 to ak-1 are all 0 then ak must be 0, so bu induction and equating all coefficients for non zero powers of x we obtain a1 to an-1 as 0.

equate coeff of x^0 and substitue values of a0 to an-1:
a0^2 + a1^2+...+ an^2 = 2a0
=>an^2 = 1
hence an= +-1

so f(x) = 1+-x^n for all n belonging to whole numbers.
more precisely
f(x) = 1,2,1+-x^n for n € N

 :blush: Writing this was a pain. Can someone please help me with writing equations with appropriate symbols and sub/superscripts, Pleaseeeeeeeeeeeeeeeeeee.[/code][/list]

Let $ g(x)\equal{}f(x)\minus{}1$, then the equation becomes $ g(x)g(\frac{1}{x})\equal{}1$
Now let $ g(x)\equal{}a_nx^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\dots  \plus{}a_1 x\plus{}a_0$,where $ a_n \neq 0$
we have $ (a_n x^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\dots  \plus{}a_1 x\plus{}a_0)(a_0x^n\plus{}a_1x^{n\minus{}1}\plus{} \dots \plus{}a_{n\minus{}1}x\plus{}a_n)\equal{}x^n$
As $ x$ tends to $ 0$, we have $ a_0*a_n\equal{}0$, so $ a_0\equal{}0$
Then we have $ (a_n x^{n\minus{}1}\plus{}a_{n\minus{}1}x^{n\minus{}2}\plus{}\dots  \plus{}a_1)(a_0x^n\plus{}a_1x^{n\minus{}1}\plus{} \dots \plus{}a_{n\minus{}1}x\plus{}a_n)$ and thus $ a_1\equal{}0$, similarly we have $ a_2,\dots ,a_{n\minus{}1}\equal{}0$,Now the equation becomes $ a_n^2\equal{}1$, $ a_n\equal{}\mp 1$,
It's then easy to check that the only possible polynomials $ g(x)\equal{}\pm x^n$, i.e. $ f(x)\equal{}1 \pm x^n$ for $ n$ being any non-negative integers satisfies the conditions.


Euler
number theory proposed
number theory

For any positive integer $n$, let $w\left(n\right)$ denote the number of different prime divisors of the number $n$. (For instance, $w\left(12\right)=2$.) Show that there exist infinitely many positive integers $n$ such that $w\left(n\right)<w\left(n+1\right)<w\left(n+2\right)$.

let's put $n=2^m$ then it's obvious that $w(n)<w(n+1)$ so now we have to show that $w(2^m+1)<w(2^m+2)=w(2^{m-1}+1)+1$
holds for infinitely many $m$
suppose the contrary i.e. that for all big enough $m\geq m_0$
we have $w(2^m+1)> w(2^{m-1}+1)$ let's write $w(2^m+1)=f(m)$ then $f(m_0+1)>f(m_0)>1$ etc. so
$f(m)\geq m+1-m_0$ for all $m>m_0$
but this can not hold true as $\prod_1^n p_i >2^{i-a}+1$ for some $n$ large enough

[quote="ZetaX"]For $n=2^m$ it isn't obvious that $w(n) < w(n+1)$. Indeed, this is the hardest part to handle with when using that approach.[/quote]
true, sorry for my rush
but its true when m is not a power of two...
still I believe that this approach may yieald a solution!

We will try to find infinite number of triples of the form $2^k, 2^k+1, 2^k+2$,
which satisfy $w(2^k)<w(2^k+1)< w(2^k+2)$.

Assume contradiction which implies $w(2^k+1)> w(2^{k-1}+1)$.
Remember Euler statement $2^k+1$ is prime if $k=2^t$.
Then take  numbers from
$a_1 =2^{2^t}+1, a_2 =2^{2^t+1}+1,..., a_{2^t} =2^{2^{t+1}}+1$
From assumed contradiction we have that

$w(a_1)\geq 1, w(a_2)\geq 2,...w(a_{2^t})\geq 2^t$,

but from $w(a_{2^t})\geq 2^t$ we get that $a_{2^t} \geq p_1 p_2...p_{2^t}> 4^{2^t} ~2^{2^{t+1}}+1= a_{2^t}$ (I hope you understood this is true for $t$ enough large $3\cdot5\cdot7\cdot 11=1055>4^5$)

Thus in the interval $[2^{2^t}+1, 2^{2^{t+1}}+1]$ there exist $k$ such that
$w(2^k+1)\leq w(2^{k-1}+1)$ and $2^{k-1}+1$ is not a prime, that’s all I think.

Same as above: why is $w(2^a+1) >1$ in all your cases¿ $m$ doesn't need to be prime to fulfill $w(m)=1$!

Sorry, anyway easy assume $2^t+1=p^u$

1 case $u$  even, than one from $p^{u/2}-1$ or $p^{u/2}+1$ is divisible just by $2$,
hence $p = 3$ and $u=2$.

2 case $u$ - odd, $p-1 | p^u-1$ and $p-1 | 2^t$  thus $p=2^q+1$

putting in the first one we get that $q=n$ or $2^t+1=p$, this case is discussed.

[quote="bae"][quote="prowler"]
Remember Euler statement $2^k+1$ is prime if $k=2^t$.
[/quote]

Are you kidding?[/quote]

No, he isn't...
Although his "is" should be replaced with a "can be" and the "if" with an "iff". :)

Sorry for my English, but you understood what I meant.
If $2^k+1$ is prime number - $k$ should be of the form $2^l$.

[quote="Arne"]This problem has been discussed a few times already. It appeared also in Brazil in 1995, I think.[/quote]

Brazil 1995 was http://www.mathlinks.ro/Forum/viewtopic.php?t=2005 which is different.

Anyway, it is problem from longlist 1979 if I remember correctly - proposal of Yugoslavia

[quote="ZetaX"]Yes your approach works (it's the same as mine and some others who solved it), but there are exceptions from $m$ being a power of $2$, namely $3^2=2^3+1$ (but thats the only one indeed).[/quote]

@All: how can we show that ?

davron

See this : http://www.mathlinks.ro/Forum/viewtopic.php?t=57843 . :)

but i am asking : why the solution is unique?


vector
real analysis
Functional Analysis
advanced fields
advanced fields unsolved

Does anyone have interesting applications of Zorn's lemma? I know it is used to show that every vector space has a basis, that there are maximal ideals in rings with identity, and it is also used in the proof of the Hahn-Banach theorem.

There is more, e.g. it is used in general to prove the existence of (real) algebraic closures or ultrafilters (thus also of spherical closures or surreal numbers, where for the latters we don't really need it) and for the compactness of products of compact spaces (Tychonoff).

It's also used to prove that there exist lebesgue-non-measurable sets.

[quote="ZetaX"][quote="julien_santini"]This is actually equiavalent to Tychonoff's theorem, isn't it ?[/quote]
What is equivalent¿[/quote]

Zorn's lemma

There are very many applications throughout mathematics, check out
http://en.wikipedia.org/wiki/Zorn's_lemma
http://en.wikipedia.org/wiki/Axiom_of_choice


function
number theory unsolved
number theory

Find all positive integer $ n$ such that $ n|(x^2 \minus{} 1)$ for every $ x\in\mathbb Z$ and $ gcd(x,n)\equal{}1$

[quote="Taidoka-X"]Problem is to find all positive integers n such that Phi(n)=2. It is easy to see that only solutions are 2,3,4 and 6[/quote]
You are wrong. (Firstly $ \phi(2) \equal{} 1$, but that's beside the point :P)
e.g. $ n \equal{} 8$, works perfectly well.

Right solution:
First we notice that if $ p$ is a primedivisor of $ n$ then $ p \le 3$.
Assume there exists a prime $ p > 3$ such that $ p \mid n$, then there exists $ m$ such $ m \equiv 2 \bmod p$ and $ (m,n) \equal{} 1$.
Then $ p \mid n \mid m^2 \minus{}1$. So since $ m \equiv 2 \bmod p$ we get $ p \mid 3$. Contradiction since we assumed $ p > 3$.
Clearly it doesn't work with $ 2^4 \mid m$, and $ 3^2 \mid m$. So the only solutions are the divisors of $ 2^3 \cdot 3\equal{}24$. There are $ 8$ of them:
$ n \in \{1,2,3,4,6,8,12,24\}$.

What we really want is the positive integers such that $ \lambda(n) | 2$, where $ \lambda$ denotes the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url], since that's the exponent of the group.

Equivalent with this problem: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=106668#106668


Drop a perpendicular from $ O$ to $ N$, the midpoint of $ AB$. This perpendicular meets $ DC$ at its midpoint, $ M$.

Let each side of the squaere have length $ 2x$ and let $ OM \equal{} y$.

$ \tan 15 \equal{} \frac {y}{x}$

$ y \equal{} x \tan 15$

$ ON \equal{} x(2 \minus{} \tan 15)$.

Now $ \tan 15 \equal{} \tan (45 \minus{} 30) \equal{} \frac {\tan 45 \minus{} \tan 30}{1 \plus{} \tan45 \tan30} \equal{} \frac {1 \minus{} \frac {1}{\sqrt {3}}}{1 \plus{} \frac {1}{\sqrt {3}}}$

$ \equal{} 2 \minus{} \sqrt {3}$

Therefore $ ON \equal{} x \sqrt {3}$

So triangle $ AON$ is 30-60-90 and by symmetry, triangle  $ AOB$ is equilateral.

[quote="AndrewTom"] $ \tan (45 \minus{} 30) \equal{} \frac {\tan 45 \minus{} \tan 30}{1 \minus{} \tan45 \tan30} \equal{} \frac {1 \minus{} \frac {1}{\sqrt {3}}}{1 \plus{} \frac {1}{\sqrt {3}}}$
[/quote]
I think you meant $ \tan (45 \minus{} 30) \equal{} \frac {\tan 45 \minus{} \tan 30}{1 \plus{} \tan45 \tan30}$

Yes Complex_Ninja, I did.
Thanks, I've edited it.

Is there a way to do this without trignometry?

[quote="ThetaPi"]Consider a square $ ABCD$. A point $ O$ is in the interior of $ ABCD$ such that $ \angle ODC \equal{} \angle OCD \equal{} 15^{\circ}$. Prove that the $ \triangle OBA$ (with $ OA$ and $ OB$ drawn) is equilateral. [hide="Solution"] Let $ O'$ be a point in the interior of $ ABCD$ such that $ \triangle O'BA$ is equilateral. Because $ O'A \equal{} AB \equal{} AD$, $ \angle DAO' \equal{} 30^{\circ}$ and $ \angle ADO' \equal{} 75^{\circ}$. From here, we prove that $ \angle O'DC \equal{} \angle O'CD \equal{} 15^{\circ}$. Therefore $ O'$ is $ O$. QED. [/hide]

Your solutions? Maybe someone can come out with a more 'direct' approach?[/quote]I'm doubting about the validity of this solution... it's not quite logically correct. Any thoughts anyone?


induction
integration
vector
function
linear algebra
matrix
calculus

Given that $g(x,0)=0$ and that \[ D_w[g(x,w)] = - \lambda g(x,w) + \lambda g(x-1,w)  \] for $x=1, 2, 3, \cdots$. If $g(0,w)= e^{-\lambda w}$, show, by mathematical induction, that \[ g(x,w) = \frac{(\lambda w)^x e^{- \lambda w}}{x!} , \qquad x = 1, 2 , 3, \cdots  \]

This is not a homework problem, please write your solution.  :)

This is simply a first-order linear differential equation, which we can solve by the usual multiplying factor technique.

Suppose that $g(x-1,w)=\frac{(\lambda w)^{x-1}e^{-\lambda w}}{(k-1)^!}.$

(This is the inductive hypothesis, and is given to be true when $x=1.$)

Write $y=y(w)=g(x,w).$ We have the differential equation

$y'=-\lambda y+\lambda\cdot\frac{(\lambda w)^{x-1}e^{-\lambda w}}{(k-1)^!}$

$y'+\lambda y=\frac{\lambda^x w^{x-1}e^{-\lambda w}}{(k-1)^!}$

The multiplying factor is $e^{\int \lambda\,dw}=e^{\lambda w}.$

$e^{\lambda w}y'+\lambda e^{\lambda w}=e^{\lambda w}\frac{\lambda^x w^{x-1}e^{-\lambda w}}{(k-1)^!}$

$\left(e^{\lambda w}y\right)'=\frac{\lambda^xw^{x-1}}{(k-1)!}$. Integrate both sides of this to get

$e^{\lambda w}y=\frac{(\lambda w)^x}{k!}+C$

We have the initial condition $y(0)=0.$ From that we determine that $C=0.$

Multiply both sides by $e^{-\lambda w}$ and we have the result.

--

Another point of view would be to regard $g(w)$ as an infinite-dimensional vector function of $w$ which satisfies the following differential equation:

$g'(w)=\begin{\bmatrix}-\lambda&0&0&\cdots\\ \lambda&-\lambda&0&\cdots\\ 0&\lambda&-\lambda&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{bmatrix}g(w);\,g(0)=\begin{bmatrix} 1\\0\\0 \\ \vdots\end{bmatrix}$


function
trigonometry
special factorizations

Find the greatest value of function

$ f(x)\equal{}(sinx)^{2}\plus{}cosx$

[quote="Invariant"]Find the greatest value of function

$ f(x) = (sinx)^{2} + cosx$[/quote]

[hide]
Completing the square

\begin{align*} f(x) & = {\sin x }^2 + {\cos x} \\
& = 1 - {\cos x }^2 + \cos x \\
& = - [{\cos x }^2 - \cos x - 1] \\
& = - [(\cos x - \frac {1}{2})^2 - \frac {5}{4}] \\
& = \frac {5}{4} - (\cos x - \frac {1}{2})^2 \\
\end{align*}

Therefore the maximum is $ \frac {5}{4}$ when $ \cos x = \frac {1}{2}$ or $ x = \frac {\pi}{3}$ amongst others
[/hide]


We have 100 marbles, with 20 White + 20 Red + 30 Black + 30 Green
Suppose we chose 60 marbles at random WITH replacement. Simulate the random numbers of the number of marbles for each color.

Here is my approach:

First, we generate a random number of a hypergeometric with 20 White and 80 non-White balls, from the 60 balls.
Then, given that we know the number of White balls (call it Xw), we generate a random number of hyperg with 20 Red and 80 non-Red balls, from the (60-Xw) balls.
Then, given that we know Xw and Xr, generate a r.n. of hyperg with 30 Black and 70 non-Black balls, from the (60 - Xw - Xr) balls.
Then, finally, Xg = 60 - Xw - Xr - Xb

Finally, we perform all these steps several times to complete the simulation.

Is that a good approach to solve the problem? I have written the codes on S-Plus, and I realized that the numbers of green balls are always almost twice of the numbers of black balls, which is unusual (should likely be equal instead). I believe that there shouldn't be any error in my codes for S-Plus....

Can we also use binomial to solve this problem??

[quote]
First, we generate a random number of a hypergeometric with 20 White and 80 non-White balls, from the 60 balls.
[/quote]
What exactly do you mean by this and what does this give you? :? I do not see how it is relevant to the problem in question: hypergeometric normally relates to choices [b]without[/b] replacement. Also the next steps are clearly wrong: once you determined the number of white balls, the conditional distribution requires considering 20 red and [b]60[/b] non-red balls (whites are no longer allowed), and so on. During the last step you should have 30-30, not 30-70. I would go for binomial distribution with 60 trials and probability of success $ \frac{20}{100}$ to determine the number $ W$ of white balls. Then use the binomial with $ 60\minus{}W$ trials and probability of success $ \frac{20}{80}$ to find the number $ R$ of reds, then the binomial with $ 60\minus{}W\minus{}R$ trials and probability of success $ \frac{30}{60}$ to find the number $ B$ of blacks and finally put $ G\equal{}60\minus{}W\minus{}R\minus{}B$.

[quote]hypergeometric normally relates to choices without replacement.[/quote]

Why is that true? I am confused with the notions of 'with' and 'without' replacement.

Can you also explain why using binomial makes sense here? I follow the steps that you have written here, but I am not too convinced why binomial is preferred over hypergeometric.

"[i]With replacement[/i]" means that you draw a ball and then put it back. "[i]Without replacement[/i]" means that you draw a ball and throw it away, so the urn after the first draw is not the same as it was in the beginning. Hypergeometric relates to "without replacement" more or less by definition (it is the distribution that arose exactly in this (or equivalent) problem). So, if the condition was "without replacement", hypergeometric would be the right choice. As it is, you have binomial that describes a sequence of independent Bernoulli trials with fixed probability of success and that's exactly what is going on for drawings with replacement: the probability to draw, say, a white ball does not depend on what you drew before that and always stays at $ 1/5$. For drawing without replacement, it changes with each drawing and hypergeometric takes this into account. If you want to see the difference, consider just 2 balls: 1 white and 1 black in the urn and draw 2 balls with and without replacement. When you draw with replacement, you have a chance of $ 1/4$ to get 2 whites, $ 1/2$ to get 1 white and 1 black and $ 1/4$ to get 2 blacks. Without replacement, you always get 1 white and 1 black (because you drew out all the balls in the urn). Now write your simulations for this simple case and see what they give you with binomial and with hypergeometric.


AMC
AMC 12
college
AIME
USA(J)MO
USAMO
Berkeley Math Circle

Dear Mr. Dunbar,

I took AMC 12A at Berkeley math circle and AMC 12B at my school, and then took the AIME at Berkeley math circle.  I was wondering if the AMC would be able to recognize my AMC 12 scores as belonging to the same person, even though I took the test at 2 different locations.  (It turns out that I scored higher on the AMC 12B, and would like my index to be calculated using that score.)

I would like to take the USAMO at Berkeley math circle, but when printing the certificates, I was wondering if the name of my high school could also be included.

Thank you.

Benjamin Yao

Please don't bump threads back up to the top.

Also, if you have a question for the AMC, your best bet is to contact the AMC directly, rather than posting here.  Contact info is on the [url=http://www.unl.edu/amc]AMC website[/url].


I'n sure that there exists one constant $k$ st

\[ 3(a^4+b^4+c^4)+2abc(a+b+c)+k(a^3b+b^3c+c^3a) \ge 0 \]

Using expanding square? :D Sure  :lol:

[quote="red3"]i have found a way by which we can prove [size=200][color=#FF0000][b]all[/b][/color][/size] cyclic-3v-ineq in 4th degree :D  :D  :D,

and i used it proved [size=200]many hard ineq[/size] :lol:[/quote]

Can you solve this inequality with your method:

$\sum{}x^4-xyz\cdot \sum{}x \ge \frac{\sqrt{21}-3}{2}\cdot \left( \sum{}x^2y^2-\sum{}x^3y \right) $

Also, please find cases of equality.

[quote="MathUniverse"]

Can you solve this inequality with your method:

$\sum{}x^4-xyz\cdot \sum{}x \ge \frac{\sqrt{21}-3}{2}\cdot \left( \sum{}x^2y^2-\sum{}x^3y \right) $

Also, please find cases of equality.[/quote]

of course i can prove it ,but it is hard to get the case of equality because i  am weak in solving system
i will try it later

[quote="Vasc"][quote="arqady"]Let $a,$ $b,$ $c$ are real numbers, such that $a^3b+b^3c+c^3a=0.$
Prove that $3(a^4+b^4+c^4)+2abc(a+b+c)\geq0.$[/quote]
This is the best:

$a^4+b^4+c^4+(\sqrt 3-1)abc(a+b+c)\ge 0.$

[hide]See
http://www.emis.de/journals/JIPAM/article1123.html?sid=1123[/hide][/quote]

yes.and i have proved this best one .but my way is very easy  :D

oh... :(  :( ,this is exactly what i found........

you are really great in ineq ,Vasc :)

[quote="MathUniverse"][quote="Vasc"][quote="arqady"]Let $a,$ $b,$ $c$ are real numbers, such that $a^3b+b^3c+c^3a=0.$
Prove that $3(a^4+b^4+c^4)+2abc(a+b+c)\geq0.$[/quote]
This is the best:

$a^4+b^4+c^4+(\sqrt 3-1)abc(a+b+c)\ge 0.$

[hide]See
http://www.emis.de/journals/JIPAM/article1123.html?sid=1123[/hide][/quote]

Mr. Vasile, I've seen this result before and I'm really interested how did you find this sum of squares... Did you use computer program or you have some beautiful method?[/quote]

i think he found by his articles' last chapter.

orther ralated inequalities,but i prefer x=\sum a^4-\sum a^2bc,2y=\sum a^3(b+c)-2\sum a^2bc, z=\sum a^2b^2-\sum a^2bc,by which we will only have 3 varible :)

[quote="Vasc"][quote="arqady"]Let $a,$ $b,$ $c$ are real numbers, such that $a^3b+b^3c+c^3a=0.$
Prove that $3(a^4+b^4+c^4)+2abc(a+b+c)\geq0.$[/quote]
This is the best:

$a^4+b^4+c^4+(\sqrt 3-1)abc(a+b+c)\ge 0.$

[hide]See
http://www.emis.de/journals/JIPAM/article1123.html?sid=1123[/hide][/quote]
My new CS technique solves this too. :lol:

I will present all these techniques in the English version and the second edition of Vietnamese version (which will be re-published in this year) of the CS book. Please wait for them.

it must be very useful because they all have a special case of equality :lol:

[quote="red3"]in fact :[quote="fjwxcsl"]$T_{148}.$Let
${a,b,c\in\mathbb{R},\frac{-v-\sqrt{12(k+1)-3v^2}}{2}}\le{u}\le{\frac{-v+\sqrt{12(k+1)-3v^2}}{2}},$

$-2(k+1)\le{v}\le{2(k+1),k\ge{-1},}$

prove that

$\sum_{cyc}{a^4}+k\sum_{cyc}{a^2b^2}+u\sum_{cyc}{a^3b}+v\sum_{cyc}{ab^3}-(1+k+u+v)\sum_{cyc}{a^2bc}\ge0.$[/quote]

my way can also prove this general one  :) ,and your ineq is a special case of it[/quote]

sorry,$-2\sqrt{k+1}\le{v}\le{2\sqrt{k+1}}.$

[quote="red3"]i have found a way by which we can prove [size=200][color=#FF0000][b]all[/b][/color][/size] cyclic-3v-ineq in 4th degree :D  :D  :D,

and i used it proved [size=200]many hard ineq[/size] :lol:[/quote]

red3 can you show your method??

Thanks in advance.

[quote=arqady]Let $a,$ $b,$ $c$ are real numbers, such that $a^3b+b^3c+c^3a=0.$
Prove that $3(a^4+b^4+c^4)+2abc(a+b+c)\geq0.$[/quote]

[img]http://s1.sinaimg.cn/middle/006ptkjAzy76JhAoGw810&690[/img]

[quote=thuanz123]Similar one
Let $a,b,c$ are real numbers.Prove:
$$3(a^4+b^4+c^4)+(ab+bc+ca)^2 \ge 6(a^3b+b^3c+c^3a)$$[/quote]
We have
\[3\sum a^4 + \left(\sum bc\right)^2 - 6 \sum a^3b = \frac 12 \sum (a^2-2b^2+c^2+2bc-2ca)^2.\]

[quote=Vasc]This is the best:
$a^4+b^4+c^4+(\sqrt 3-1)abc(a+b+c)\ge 0.$
[hide]See
http://www.emis.de/journals/JIPAM/article1123.html?sid=1123[/hide][/quote]
The following inequality stronger is true
\[a^4+b^4+c^4+(\sqrt 3-1)abc(a+b+c) \geqslant \frac{1}{\sqrt{3}}(ab+bc+ca)^2.\]

[quote=Nguyenhuyen_AG][quote=Vasc]This is the best:
$a^4+b^4+c^4+(\sqrt 3-1)abc(a+b+c)\ge 0.$
[hide]See
http://www.emis.de/journals/JIPAM/article1123.html?sid=1123[/hide][/quote]
The following inequality stronger is true
\[a^4+b^4+c^4+(\sqrt 3-1)abc(a+b+c) \geqslant \sqrt{3}(a^2b^2+b^2c^2+c^2a^2).\][/quote]

1,1,-1

[quote=luofangxiang]
1,1,-1[/quote]
I fixed.


geometry
geometric transformation
rotation
trigonometry
linear algebra
matrix
calculus

A few days ago in geometry class, we went over trig rotations using matrices, namely

\[ \begin{bmatrix}\sin x & \cos x\\
\cos x & \minus{}\sin x
\end{bmatrix}
\begin{bmatrix}x\\
y
\end{bmatrix}
\equal{}
\begin{bmatrix}x'\\
y'
\end{bmatrix}\]

I noticed that the second entry of the first row was the derivative of the first entry ($ \frac{d}{dx} \sin x \equal{} \cos x$), and the same with the second row ($ \frac{d}{dx} \cos x \equal{} \minus{} \sin x$).

I was wondering if the entries in the 2x2 matrix had any correlation to one another, or if it was merely coincidence?

The rotation matrix as you have written it is the [url=http://en.wikipedia.org/wiki/Wronskian]Wronskian[/url] of the functions $ \sin x, \cos x$, although generally we write your second row first.  (The matrix you gave is not a rotation by $ x$ counterclockwise.)


Solve in rational numbers the equation : $4x^{2}-y^{2}=36$

[hide]If $x$ and $y$ are rational then so are $2x\pm y$. Let $2x-y=\frac{6p}{q}$ for any coprime $p$ and $q$.
From the system $2x-y=\frac{6p}{q}, \ \ 2x+y=\frac{6q}{p}$ all the solutions could be found.[/hide]


A particular positive three-digit integer is divisible by 5.  The integer also is divisible by 11.  The sum of the three digits of the integer is 13.  What is the integer?

Let the number be $ xyz$.  Then the number must be a multiple of 55.  So the choices are 110, 165, 220, 275, 330, 385, 440, 495, 550, 605, 660, 715,... and $ x\plus{}y\plus{}z\equal{}13$ from which it is clear that 715 is what we want.

Another method: Obviously it ends in 5 or 0.

If it ends in 0:

we have a number of the form $ xx0$ Since x is an integer, the sum is $ 2x$ an even number (and 13 is odd!) so the number cannot end in 0!

If ite ends in 5:

The divisibility rule for 11 is to alternately add and subtract digits in the number, and if you get a multiple of 11, the original number is a multiple of 11! Thus we see

$ xy5$

WE can set up two equations here

$ x\plus{}y\plus{}5\equal{}13$ and $ x\minus{}y\plus{}5\equal{}11$
Simplifying we get

$ x\plus{}y\equal{}8$
$ x\minus{}y\equal{}6$

We can quickly see x=7 and y=1 so our number is $ 715$


Given positive integer $ m \geq 17$, $ 2m$ contestants participate in a circular competition. In each round, we devide the $ 2m$ contestants into $ m$ groups, and the two contestants in one group play against each other. The groups are re-divided in the next round. The contestants compete for $ 2m\minus{}1$ rounds so that each contestant has played a game with all the $ 2m\minus{}1$ players. Find the least possible positive integer $ n$, so that there exists a valid competition and after $ n$ rounds, for any $ 4$ contestants, non of them has played with the others or there have been at least $ 2$ games played within those $ 4$.

I'm not sure but I think the answer is $m-1$ for even $m$ and $m$ for odd $m$.

I think that the answer is always $m-1$.

First let's construct this.

For even $m$, label the people $A_1, A_2, A_3, \cdots, A_m, B_1, B_2, \cdots, B_m$ arbitrarily. For $1 \le i < m$, on day $i$, $A_j$ will play $A_{i + j}$ for each $1 \le j \le m$, where indices are modulo $m$. For day $m$, $A_j$ will play $B_j$ for $1 \le j \le m.$ For the last $m-1$ days, we simply schedule so that the $A_j$'s play each other once and the $B_j$'s play each other once.

For odd $m$, label the people in the same way, and schedule games in the same way for the first $m-1$ days. For the $(m-1+i)$th day, for $1 \le i \le m$, we will have $A_i$ play $B_i$. Then, we pair up the remaining players so that $A_x$ plays $A_{2i - x}$, $B_x$ plays $B_{2i - x}$, for $x \neq i$, where indices are again modulo $m.$

In this way, we've shown that $m-1$ always works.

Suppose now that $n \le m-2$ worked. Consider the graph where we connect two people iff they have played in the first $n$ days. This graph is clearly $n-$regular.

Consider the largest set $S$ of people, none of whom have played each other. Suppose $|S| = k$. By Turan's Theorem, it's easy to see that $k \ge 3.$

Note that if anybody else plays at most $k-2$ of the people in $S$, then we can find a $4-$tuple with exactly one match played among them. Hence, suppose that everyone else plays at least $k-1$ of the people in $S$. This implies that, because each of the people in $S$ plays $n$ others, there are at least $\frac{nk}{k-1}$ other people. Hence $k + \frac{nk}{k-1} \ge 2m.$ After simplifying, this is equivalent to $k^2 - (2m+1-n)k + 2n \ge 0.$

As $n \le m-2$, this implies that $k \ge 2m-n.$ This implies that since there are at most $n$ people not in $S$, we have $|S| = 2m-n$ and everybody in $S$ must play all of the people not in $S$. However, this means that anybody not in $S$ has played each of the $2m-n > n$ people in $S$, contradiction.

So $n \le m-2$ do not work, and the answer is $\boxed{m-1}.$

$\square$


trigonometry
real analysis
real analysis unsolved

Does

$\sum_{n=1}^{\infty}\frac{(\frac{2}{3}+\frac{1}{3}\sin{n})^{n}}{n}$

converge?

Can you prove it?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]here[/url]. It's one of the classic threads of the forum.

By the way, the answer is yes.


Prove
\[ \sum_{k\equal{}1}^n\left[\log_k\frac{n}{x}\right]\equal{}\left[\frac{n}{x}\right],\: \forall n\in\mathbb{N}^*,k\in\mathbb{N}^*\setminus\{1\},\: x\in\mathbb{N}^*, x\not\equiv 0\mod k\]

[quote="cckek"]Prove
\[ \sum_{k \equal{} 1}^n\left[\log_k\frac {n}{x}\right] \equal{} \left[\frac {n}{x}\right],\: \forall n\in\mathbb{N}^*,k\in\mathbb{N}^*\setminus\{1\},\: x\in\mathbb{N}^*, x\not\equiv 0\mod k
\]
[/quote]

There seems to be a contradiction between the $ k\equal{}1$ in the "$ \sum_{k\equal{}1}$" and the statement $ k\in\mathbb{N}^*\setminus\{1\}$

$ \sum_{k\equal{}2}^n[log_k\frac nx ]\equal{}[\frac nx]$ is wrong too.

My fault. Here is the right text:
Prove:

\[ \sum_{x\equal{}1}^n\left[\log_k\frac{n}{x}\right]\equal{}\left[\frac{n}{k}\right],\: \forall n\in\mathbb{N}^*,k\in\mathbb{N}^*\setminus\{1\},\: x\in\mathbb{N}^*, x\not\equiv 0\mod k\]

[quote="cckek"]Prove:
\[ \sum_{x \equal{} 1}^n\left[\log_k\frac {n}{x}\right] \equal{} \left[\frac {n}{k}\right],\: \forall n\in\mathbb{N}^*,k\in\mathbb{N}^*\setminus\{1\},\: x\in\mathbb{N}^*, x\not\equiv 0\mod k
\]
[/quote]
$ [log_k\frac nx ] \equal{} m$ if $ \frac {n}{k^{m \plus{} 1}} < x\le \frac {n}{k^m}$.
Therefore $ \sum_{x \equal{} 1}^n [log_k \frac nx ] \equal{} \sum_{m \equal{} 1}^{\infty} [\frac {n}{k^m}] \ge [\frac nk ]$.
For example $ n \equal{} 16,k \equal{} 2$ give $ [log_2\ 16] \plus{} log_2 8 \plus{} [log_2\frac {16}{3}] \plus{} [log_2\frac {16}{4}] \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 0... \plus{} 0 \equal{} 15$
and $ [\frac {16}{2}] \plus{} [\frac {16}{2^2} \plus{} ... \equal{} 15 > 8$.

Ok!
\[ \sum_{1\le x\le n, k\not |x}[log_k\frac nx ]\equal{}\sum_{m\equal{}1}^{\infty}mf(\frac{n}{k^m},k),\]
were
\[ f(y,k)\equal{}\#\{x|\frac yk <x\le y, \ k\not |x\}\equal{}\#\{x|\frac yk <x\le y\}\minus{}\#\{\frac{y}{k^2}<x\le \frac yx \}\equal{}[y]\minus{}2[\frac yk]\plus{}[\frac{y}{k^2}]\equal{}g(y,k)\minus{}g(\frac yk ,k), \ g(y,k)\equal{}[y]\minus{}[\frac yk ].\]
It give $ \sum_m mf(\frac{n}{k^m},k)\equal{}\sum_{m\equal{}1}^{\infty } g(\frac{n}{k^m},k)\equal{}[\frac nk ].$


calculus
integration
function
algebra
domain
real analysis
real analysis unsolved

Let $ f$ and $ g$ be continuous functions on a domain $ D$ that contain [0,1]. Suppose that $ \int_0^1 f(x) \, dx \equal{} \int_0^1 g(x) \, dx \equal{} 1$. Prove that there exist $ c,d\in [0,1]$ such that
\[ \int_c^d f(x)\, dx \equal{} \int_c^d g(x) dx \equal{} \frac {1}{2}
\]

edited: $ a$ should be $ 1$.

I doubt there is a very elementary solution (without Brouwer & Co). Anyway, if I remember well, it has been discussed on the forum.

How would you solve this using Brouwer & Co?  (Are you referring to the Brouwer fixed point theorem?)


In trying to prove the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6538]problem posted here[/url], I created two silmilar problems as follows, not sure original or not, but they are pretty easy:

If $a, b, c$ are positive numbers with $a + b + c = 1$, then prove

(1) $\displaystyle \frac{bc}{bc+a}+\frac{ca}{ca+b}+\frac{ab}{ab+c}\geq \frac{3}{4}$

(2) $\displaystyle \frac{a}{bc+a}+\frac{b}{ca+b}+\frac{c}{ab+c}\leq \frac{9}{4}$

Yeah, it's kind of the same...

(1)  [tex]\displaystyle \sum \frac{bc}{bc+a} = \sum \frac{bc}{bc+a(a+b+c)} = \sum \frac{bc}{(a+b)(a+c)} = \displaystyle\frac{\sum bc(b+c)}{(a+b)(b+c)(c+a)}[/tex]

So the inequality we wish to prove becomes

[tex]\displaystyle 4 \sum (b^2c+bc^2) \geq 3(a+b)(b+c)(c+a)[/tex]

After expanding and simplifying,

[tex]\displaystyle \sum (b^2c+bc^2) \geq 6abc[/tex], which holds by AM-GM.

(2) is similar, [tex]\displaystyle \sum \frac{a}{bc+a} = \sum \frac{a(a+b+c)}{bc+a(a+b+c)} = \sum \frac{a(a+b+c)}{(a+b)(a+c)} = \frac {\sum a(b+c)(a+b+c)}{(a+b)(b+c)(c+a)}[/tex]

So we want:

[tex]4 \sum a(b+c)(a+b+c) \leq 9(a+b)(b+c)(c+a)[/tex]

After simplification,

[tex] 6abc \leq \sum (a^2b+ab^2)[/tex], again by AM-GM.

Here is my proof:

We have

$\displaystyle \frac{bc}{bc+a}+\frac{ca}{ca+b}+\frac{ab}{ab+c}=\frac{\left( bc\right) ^{2}}{b^{2}c^{2}+abc}+\frac{\left( ca\right) ^{2}}{c^{2}a^{2}+abc}+\frac{\left( ab\right) ^{2}}{a^{2}b^{2}+abc}$.

Now we can apply the Cauchy-Schwarz inequality in the Engel form to get

$\displaystyle \frac{\left( bc\right) ^{2}}{b^{2}c^{2}+abc}+\frac{\left( ca\right) ^{2}}{c^{2}a^{2}+abc}+\frac{\left( ab\right) ^{2}}{a^{2}b^{2}+abc}\geq \frac{\left( bc+ca+ab\right) ^{2}}{\left( b^{2}c^{2}+abc\right) +\left( c^{2}a^{2}+abc\right) +\left( a^{2}b^{2}+abc\right) }$.

Thus,

$\displaystyle \frac{bc}{bc+a}+\frac{ca}{ca+b}+\frac{ab}{ab+c}\geq \frac{\left( bc+ca+ab\right) ^{2}}{\left( b^{2}c^{2}+abc\right) +\left( c^{2}a^{2}+abc\right) +\left( a^{2}b^{2}+abc\right) }=\frac{\left( bc+ca+ab\right) ^{2}}{b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3abc}$.

Now we will prove that

$\displaystyle \frac{\left( bc+ca+ab\right) ^{2}}{b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3abc}\geq \frac{3}{4}$.

In fact, this is equivalent to

$4\left( bc+ca+ab\right) ^{2}\geq 3\left( b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3abc\right)$,

i. e. to

$4\left( bc+ca+ab\right) ^{2}\geq 3\left( b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3abc\left( a+b+c\right) \right)$

(since a + b + c = 1). ASA (after some algebra), this turns out to be equivalent to

$b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}\geq a^{2}bc+b^{2}ca+c^{2}ab$,

i. e. to

$\left( ca-ab\right) ^{2}+\left( ab-bc\right) ^{2}+\left( bc-ca\right) ^{2}\geq 0$,

what is clearly true.

Hence we have shown

$\displaystyle \frac{\left( bc+ca+ab\right) ^{2}}{b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3abc}\geq \frac{3}{4}$.

Together with

$\displaystyle \frac{bc}{bc+a}+\frac{ca}{ca+b}+\frac{ab}{ab+c}\geq \frac{\left( bc+ca+ab\right) ^{2}}{b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}+3abc}$,

this yields

$\displaystyle \frac{bc}{bc+a}+\frac{ca}{ca+b}+\frac{ab}{ab+c}\geq \frac{3}{4}$.

So (1) is proven. Now, subtracting 3 on both sides, we get

$\displaystyle \left( -\frac{a}{bc+a}\right) +\left( -\frac{b}{ca+b}\right) +\left( -\frac{c}{ab+c}\right) \geq -\frac{9}{4}$,

establishing (2).

  Darij


to only  :blush: [hide] :cong 3/2 vao` 2 ve'
dua ve` dong` bac (a^3+b^3+c^3=a^3+b^3+c^3/abc)  sau do' chuan hoa' a+b+c=3
roi` dung` tre bu sep'
c/m tong cua 1/a^2+(3-a)^2 =<3/5
djch ra ho tau cay'[/hide]

[quote="Math pro"]to only  :blush: :add $ \frac {3}{2}$ to 2 HS
we have $ (a^3 \plus{} b^3 \plus{} c^3 \equal{} a^3 \plus{} b^3 \plus{} c^3/abc)$  and setup $ a\plus{}b\plus{}c\equal{}3$
Then use Chebyshev
Prove that $ \sum \frac {1}{a^2} \plus{} (3 \minus{} a)^2 \le \frac {3}{5}$
[/quote]
Oh my friend;Are U mistake? I have already given $ abc \equal{} 1$. why do U setup $ a \plus{} b \plus{} c \equal{} 3$ ?

[quote="Math pro"]no mistake my friend!
[hide] khi dua ve` dong` bac roi` thj` xet' bai` toan' rieng bjet c/m LHS =< 1/5(a^3+b^3+c^3/abc+3) +3/2
luc' do' thj` chuan hoa' duoc[/hide][/quote]
Can U post your detail proof ?

[quote="onlylove_math"]Let$ a;b;c \ge\ \frac {1}{\sqrt {3} }$  and $ abc \equal{} 1$. Prove that:
$ \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \plus{} \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \le\ \frac {1}{5}(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3)$[/quote]
It is really easy  onlylove_math,howerver,your problem is not strong.Here is my proof (I think your way is diffrent from mine)
We will prove $ \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \plus{} \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \le\ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{5abc}$
ther $ a,b,c\geq 0$,and$ a\plus{}b\plus{}c\equal{}1$ (It is very easy for you to find this ineq be similar to yours)
$ LHS\equal{}\sum \frac{c(1\minus{}c)}{c^2\plus{}(1\minus{}c)^2} \le\ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{5abc}$
We have $ \frac{c(1\minus{}c)}{c^2\plus{}(1\minus{}c)^2} \leq 1,08 c\plus{}0,04$ hence $ LHS \leq 1,2$
$ 1,2\leq \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{5abc}$<=> $ a^3 \plus{} b^3 \plus{} c^3 \geq  3abc$(true since AMGM)
We have done
With my proof we dont need $ a;b;c \ge\ \frac {1}{\sqrt {3} }$
Post your proof,detail

[quote="Allnames"]
ther $ a,b,c\geq 0$,and$ a + b + c = 1$ (It is very easy for you to find this ineq be similar to yours)
$ LHS = \sum \frac {c(1 - c)}{c^2 + (1 - c)^2} \le\ \frac {a^3 + b^3 + c^3 + 3abc}{5abc}$
We have $ \frac {c(1 - c)}{c^2 + (1 - c)^2} \leq 1,08 c + 0,04$ hence $ LHS \leq 1,2$
$ 1,2\leq \frac {a^3 + b^3 + c^3 + 3abc}{5abc}$<=> $ a^3 + b^3 + c^3 \geq 3abc$(true since AMGM)
Post your proof,detail[/quote]
Can U detail it?
[quote="onlylove_math"]Let$ a;b;c \ge\ \frac {1}{\sqrt {3} }$  and $ abc = 1$. Prove that:
$ \frac {c(a + b)}{c^2 + (a + b)^2} + \frac {a(b + c)}{a^2 + (b + c)^2} + \frac {b(c + a)}{b^2 + (c + a)^2} \le\ \frac {1}{5}(a^3 + b^3 + c^3 + 3)$[/quote]
Here is my solution;
$ LHS = \sum \frac {\frac {a + b}{c}}{1 + (\frac {a + b}{c})^2}$
to examine  $ f(t) = \frac {t}{1 + t^2}$
We easy prove $ f"(t) < 0$with $ a;b;c \ge\ \frac {1}{\sqrt {3} }$
Follow Jensen we have:
$ LHS = f(\frac {a + b}{c}) + f(\frac {b + c}{a}) + f(\frac {c + a}{b}) \le\  3f({\frac{\frac {a + b}{c} + \frac {b + c}{a} + \frac {c + a}{b}}{3})}$
And we have $ RHS = \frac {1}{5}(a^3 + b^3 + c^3 + 3)$
$ = \frac {1}{5}\frac {a^3 + b^3 + c^3 + 3abc}{abc} \ge\ \frac {1}{5}\frac {\sum ab(a + b)}{abc}$
$ RHS \ge\ 3f({\frac{\frac {a + b}{c} + \frac {b + c}{a} + \frac {c + a}{b}}{3})}$
We have done!
To be honest; my ineq develop from ineq Jensen

[quote="Math pro"][quote="Allnames"][quote="onlylove_math"]Let$ a;b;c \ge\ \frac {1}{\sqrt {3} }$  and $ abc \equal{} 1$. Prove that:
$ \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \plus{} \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \le\ \frac {1}{5}(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3)$[/quote]
It is really easy  onlylove_math,howerver,your problem is not strong.Here is my proof (I think your way is diffrent from mine)
We will prove $ \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \plus{} \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \le\ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{5abc}$
ther $ a,b,c\geq 0$,and$ a \plus{} b \plus{} c \equal{} 1$ (It is very easy for you to find this ineq be similar to yours)
$ LHS \equal{} \sum \frac {c(1 \minus{} c)}{c^2 \plus{} (1 \minus{} c)^2} \le\ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{5abc}$
We have $ \frac {c(1 \minus{} c)}{c^2 \plus{} (1 \minus{} c)^2} \leq 1,08 c \plus{} 0,04$ hence $ LHS \leq 1,2$
$ 1,2\leq \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{5abc}$<=> $ a^3 \plus{} b^3 \plus{} c^3 \geq 3abc$(true since AMGM)
We have done
With my proof we dont need $ a;b;c \ge\ \frac {1}{\sqrt {3} }$
Post your proof,detail[/quote]
to all names[hide]thằng khốn Thiện béo mj hiểu cách làm của tau mà còn giả hỏi đồ mất dậy[/hide][/quote]
[hide="to mathpro"]Bạn nên  xem lại, cách bạn nói trên với cách mình  # nhau 1 trời 1 vực đó .Ba,j thử post rõ cách mà dùng Trebysev đó coi,mình nghĩ là ko thể :rotfl: [/hide]
To onlylovemath.My proof is quite clearly,Which line that you cant understand,tell me?


Find all positive integers n such that n^3+3^3 is aperfect square

I think it's never.

[hide]

u(n) = n^3 + 3^3 = (n+3)*(n - 3n + 3)

Let p a prime number which divides u(n), square. If p divides

(n+3) and (n - 3n + 3) then p=3.

If 3 divides u(n) then n=3N, u(n) = 3^3 (N^3+1) = 9*A. But then 3

would divide A which leads to a contradiction.

So we must have (n+3) square and (n - 3n + 3) square.

Looking with the trusty "modulo 4"   , n+3 is square for n=1 or 2 mod 4,

n - 3n + 3 is square for n=0 or 3 mod 4.

So u(n) is never a square.    [/hide]

[quote="JBL"]Actually, that problem recurs throughout your proof, and I think it ruins the mod-4ness of it.[/quote]

As far as I can see, no :

n+3 mod 4 :
0: 3
1: 0
2: 1
3: 2

n-3n+9 mod 4 :
0: 1
1: 3
2: 3
3: 1

By the way, it does not work for n-3n+3 mod 4 (it was really a typo in the whole text) :
0: 3
1: 1
2: 1
3: 3


That's different for every college and every professor.

[quote="kool_dudy"]At college, what books do they generally use in first year?

(In all subjects, I mean English, Math, 3 Sciences, etc.)[/quote]

Hmmm, confuted is right that this is decided at least by each college, and possibly by more than one professor in each college. What the first-year course is in math varies GREATLY from college to college, but in most cases that would be a multivariable calculus course for math majors at a strong college, and a single-variable calculus course for most people studying "hard" subjects.

I rather doubt that most colleges have their first-year students take three different science classes in the first year. The first-year honors biology book that I studied back in the 1970s is surely obsolete by now, and I think the Halliday-Resnick first-year physics book, which was once quite dominant, is not so dominant any more.

What does anyone else have to say about this?

To find out about a PARTICULAR college, land on its Web site, look for descriptions of courses offered, and see whether the Web syllabuses for courses at that college include descriptions of what textbooks are used. That's how I've found out what textbooks are (sometimes) used at Harvard, for instance.

Hope this helps!

Thank you everyone.  I was taking your example, TK, at Harvard and I have found some textbooks that they recommend during their courses.


Find the equation of the parabola whose vertex is $ (\minus{}4,3)$ and whose focus is at $ (4,8)$.

I hope there's a much easier way.

[color=darkblue]You see the picture. I will post the solution soon.[/color]

[color=darkblue]$ \bullet$ $ S$ is midpoint of $ PF$ , because $ S( \minus{} 4;3)$ and $ F(4;8)$ $ \Longrightarrow P( \minus{} 12; \minus{} 2)$

$ \bullet$ $ \Delta$ pass point $ P$ and $ \Delta \perp SF \Longrightarrow$ the equation of line $ \Delta: 8(x \plus{} 12) \plus{} 5(y \plus{} 2) \equal{} 0$

$ \Longleftrightarrow 8x \plus{} 5y \plus{} 106 \equal{} 0$

$ \bullet$ Point $ M(x;y) \in (P) \Longleftrightarrow d(M,\Delta) \equal{} MF$

$ \Longleftrightarrow \frac {|8x \plus{} 5y \plus{} 106|}{\sqrt {8^2 \plus{} 5^2}} \equal{} \sqrt {(x \minus{} 4)^2 \plus{} (y \minus{} 8)^2}$

So easy you get the equation of the Parabola $ (P)$[/color]

[color=darkblue]Some same problem:

1) Find the equation of the parabola whose vertex is $ S(\minus{}2; 5)$ and $ \Delta : 3x \plus{}y \minus{} 6 \equal{} 0$

2) Find the equation of the parabola whose focus is at $ F(5; 3)$ and $ \Delta : x\minus{}2y \plus{}3 \equal{} 0$. [/color]