geometry
search
trigonometry
circumcircle
vector
geometric transformation
dilation

[size=150][b]all my unsolved problems will be published here.I insisted to think about 3 problems per day.Some problem may be old but still worth seeking for different solving method.
I want your help if you can solve it.Thanks!      Geometry is full of fun![/b][/size]

Now [b]problem1[/b]

[quote="shoki"]it has been posted later,it's a well-known one.search...[/quote]

Really difficult to search
I have a idea of problem1
use sine or cos laws
but I haven't tried

[quote="tpjmroc"]all my unsolved problems will be published here.I insisted to think about 3 problems per day.Some problem may be old but still worth seeking for different solving method.
I want your help if you can solve it.Thanks!      Geometry is full of fun!

Now problem1[/quote]

This problem one was posted before in the past , but i can't find a reference i think it is some IMO

Better post 3 problem per week if you want that people get interested for your problems , that's using this forum efficiently

 :)

Problem 2:
Just prove that $ N$ is the orthocenter of $ \Delta OMA$.$ OA \perp AP,AP\parallel{}BD \implies OA \perp MN$.
it remains to prove $ AN \perp OP$.
Let $ S \equal{} CN \cap PA$,$ F \equal{} AB \cap CN,K \equal{} CP \cap AN$.we have $ \frac {DN}{AS} \equal{} \frac {CN}{CS},\frac {BN}{AS} \equal{} \frac {BF}{FB}$.

we have
$ \frac {DN}{AS} \equal{} \frac {BN}{AS} \implies \frac {CN}{CS} \equal{} \frac {NF}{FS}$.so the same is for $ P,C,K,B$(harmonic division).

so we must have $ PO \perp AN$.

⊿AFP∽⊿QFA  =>  AF^2=FP*FQ=FP*PG  =>  FP/PD=PE/PG  =>  ⊿FDP∽⊿EGP
=>  ∠GEB=∠EGP=∠FDP=∠AFD=∠AEM  =>  E,F,A,M are concyclic
Notice we have FPEA are concyclic  =>  EPFAM are concyclic
Analgously, we have PDMBG are concyclic  =>  ∠AMB=180-∠CAB-∠CBA =>  M,B,C,A are concyclic

[quote][b]Problem2[/b][/quote]
Since $ N$ is the midpoint of $ BD\|AP$, thus, $ (AD,AB,AN,AP)\equal{}\minus{}1$ or in other word, we have $ (AC,AB,AN,AP)\equal{}\minus{}1$, which implies that $ N$ and $ P$ are conjugated with respect to $ (O)$. Therefore, $ AN$ is the polar of $ P$ with respect to $ (O)$ $ \Longrightarrow$ $ AN\perp OP$ $ \equiv OM$. But we also have $ OA\perp DB\equiv MN$. As the result, $ N$ is the orthocenter of $ \triangle OAM$ $ \Longrightarrow$ $ ON\perp AM$, which completes our proof. $ \square$

Here are Problem 3,4,5.
I can solve problem 3 but not nice proof. :(
You can try these.

for problem4
you can see http://www.aoshoo.com/bbs1/dispbbs.asp?BoardID=43&ID=16478&skin=0
I have mentioned in 13L

it seems that there is sth wrong with Pr5.if $ CF$ cuts the circumcircle of $ OAE$ at $ G$ then ,as i know, $ OG$ is the diameter.
for Pr3, we can use vectors and use this equality $ \vec{OH}\equal{}\vec{OA}\plus{}\vec{OB}\plus{}\vec{OC}$.

[b]Problem 3[/b]
Since $ MNCB$ is concyclic quadrirateral, hence, $ MN$ is antiparallel with respect to $ BC$. But $ BC\|EF$, thus $ MN$ is antiparallel with respect to $ EF$, which implies that $ M,N,E,F$ are concyclic $ \Longrightarrow$ $ \overline {PE}.\overline {PF}$ $ \equal{}\overline {PM}.\overline {PN}$. Now denote $ (O_1)$, $ (O_2)$ respectively by the circumcircles $ \triangle AEF$ and $ \triangle AMN$. As the result, $ \mathcal {P}_{P/(O_1)}\equal{}\mathcal {P}_{P/(O_2)}$ $ \Longrightarrow$ $ AP$ is the radical axes of $ (O_1)$ and $ (O_2)$, which implies $ AP\perp O_1O_2$. But note that $ O_1$ and $ O_2$ respectively are the midpoints of $ AH$ and $ AO$, hence $ O_1O_2\|OH$ $ \Longrightarrow$ $ AP\perp OH$. Which completes our proof. $ \square$

[hide="Remark"]$ \bullet$ [b]Remark[/b] I will post another solution of this problem using Pole and Polar next time.[/hide]

I’m going to use harmonious quadrilateral property here:
Denote E is the midpoint of AP , F is the intersection of AN and BC
Using ceva thoerem ,we have AE,AB,PD are concurrent  =>  BD,CE,AF are concurrent at N
Which means P,B,F,C are harmonious division  => PB/PC=FB/FC
Notice the fact that if ABCG is a harmonious quadrilateral and F’=AG∩BC  then we have PB/PC=F’B/F’C   =>   F’=F
So extend AF to G ,from the property above => PG is the tangent line of ⊙O => OP⊥AN
which completes our proof

[quote="shoki"]it seems that there is sth wrong with Pr5.if $ CF$ cuts the circumcircle of $ OAE$ at $ G$ then ,as i know, $ OG$ is the diameter.
for Pr3, we can use vectors and use this equality $ \vec{OH} \equal{} \vec{OA} \plus{} \vec{OB} \plus{} \vec{OC}$.[/quote]
Why OG is the diameter?
I ensure that this problem is right
[b]If you want the Geometry Skechpad file ,download here![/b]
http://cid-5a3ba870f969e066.skydrive.live.com/browse.aspx/.Public?uc=2        (20090917，20090918)
Thanks to [b]mathVNpro[/b],Well done!
Use the method of antiparallel will be a fast way to solve this.I was awkward that I hadn't found this.
Good Work,[b]plane geometry[/b]! Using the[b] harmonious quadrilateral property [/b]can be a new way to me to handle this.
There is a [b]problem6[/b] that my classmates was confused and I suppose that was a good problem!

assume that $ \angle AOE\equal{}\alpha,\angle AGE\equal{}\frac{360^\circ\minus{}\alpha}{2}$.
as u mentioned $ AGEO$ is cyclic so we must have $ \angle  AGE\plus{}\angle AOE\equal{}\alpha\plus{}\frac{360^\circ\minus{}\alpha}{2}\equal{}180^\circ \implies 360^\circ \plus{} \alpha\equal{}360^\circ \implies \alpha\equal{}0^\circ$.contradiction.
am i missing sth?if yes,plz show me...

Let the intersection of $ AC$ and $ BD$ be $ P$.it's obvious that $ P \in OM$.Let the intersection of the circumcircles of $ OAD$ and $ OBC$ be $ K$.it's well-known that $ N,P,K$ are collinear and we have $ \angle NKO\equal{}90^\circ$.
from this we conclude that we must show $ N,O,K,M$ are concyclic.or in other words we must show that $ NP.PK\equal{}OP.PM$.but we have $ OP.PM\equal{}AP.PC$.so we must show that $ N,K,A,C$ are concyclic.which is obvious since
$ \angle NKC\equal{}\angle PKC\equal{}180^\circ \minus{} \angle PDC\equal{}180^\circ \minus{} \angle CAB\equal{}\angle CAN$.DONE!
Note that $ PKCD,PKAB$ are concyclics.

It is well-known that H is the orthocentr of ⊿OLN
Denote P the projection through N to OL    K the projection through O to NL  Q the intersection of LH and AB
In⊿LAB , AC,BD,LQ are concurrent at H => N,A,Q,B are harmonious division  =>  from the harmonious quadrilateral property I mentioned above => XAYB is a harmonious quadrilateral => XY⊥ON =>  NX^2=NY^2=NZ*NO=NC*ND=NH*NP  =>  PCDH are concyclic

now we have CDPH are concyclic  =>  ∠NPC=∠HDN=∠NAC  =>  PCNA are concyclic
PH*HN=HC*HA=OH*HK
Notice H is the radical center of the three circles  =>  OM*OH= HC*HA  =>  HK=HM
K=M

[quote="tpjmroc"][quote="shoki"]assume that[hide] $ \angle AOE \equal{} \alpha,\angle AGE \equal{} \frac {360^\circ \minus{} \alpha}{2}$.
as u mentioned $ AGEO$ is cyclic so we must have $ \angle AGE \plus{} \angle AOE \equal{} \alpha \plus{} \frac {360^\circ \minus{} \alpha}{2} \equal{} 180^\circ \implies 360^\circ \plus{} \alpha \equal{} 360^\circ \implies \alpha \equal{} 0^\circ$.contradiction.
am i missing sth?if yes,plz show me[/hide]...[/quote]
Owing to shoki ,I have found a mistake of pro5
The problem5 doesn't seem wrong of his body,but I have drawn the wrong picture.[b]So,I have updated the problem5[/b]. :oops: Sorry.[/quote]
∠DO1O=∠DOA=2∠DBA   =>  DA//OF  =>  ∠FOC=∠OAD  =>  ∠BGC=∠FOC=∠DAO=∠BEC  =>  G,E,B,C are concyclic  =>  ∠GEC=∠GBC=∠OFC  =>  GFHE are concyclic

On the onther hand ,OD=OB => in ⊙O1 we have ∠DGO=∠BGO  =>  ∠FGD=∠BGC  =>
∠FGD=∠BGC=∠BEC=∠DAO  =>  DGCA are concyclic

∠OEG=∠DEG+∠DEO=∠ODC+∠GBO=180-∠GDC=180-∠GAC  =>  AOEG are concyclic
Done！

A really tough and Strange Problem

$ P$ is outside the⊙$ O$.Two tangential lines and one secant line pass through $ P$.And the two tangential lines cut ⊙$ O$ at $ A$,$ B$.and the secant line intersects ⊙$ O$ at$ C$,$ D$
draw two segments $ DE$$ ,DF$ which satisfied$ DE$⊥$ PA$ and $ DF$⊥$ PB$.($ E$,$ F$ is the perpendicular foot)
Proof that:
$ AC*AD*PF \plus{} BC*BD*PE \equal{} AB^{2}*PD$

[b][size=100][color=DarkBlue]EQUIVALENT PROBLEM. - A cyclic quadrilateral $ ABCD$ is given, with circumcircle $ (O)$ and let $ (O_{1}),\ (O_{2})$ be, the circumcircles of the triangles $ \bigtriangleup AOC,\ \bigtriangleup BOD,$ intersect each other at points $ O$ ( the center of $ (O)$ ) and $ M.$ Prove that $ NM\perp MO,$ where $ N\equiv AB\cap CD.$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] - Let be the point $ H\equiv AC\cap BD\cap OM,$ as the radical center of the circles $ (O),\ (O_{1}),\ (O_{2}),$ as well.

As [b][size=100]plane geometry[/size][/b] said, it is well known that $ NO\perp LH$ $ ,(1)$ and $ LO\perp NH$ $ ,(2)$ where $ L\equiv AD\cap BC.$

( [b][size=100][color=DarkBlue]In every cyclic quadrilateral, convex or non convex, the line connecting the center of its circumcircle with the point of intersection of its diagonals, is perpendicular to the line connecting the points of intersection, of its opposite sidelines[/color][/size][/b]. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110887][b][color=red]Here[/color][/b][/url] for some proofs ).

From $ (1),\ (2),$ we conclude that the line segments $ LH,\ NH,$ are the polars of the points $ N,\ L$ respectively, with respect to the circle $ (O)$ as well.

( $ LH$ is the line through the point $ H\equiv AC\cap BD$ and perpendicular to $ NO$ and similarly, $ NH$ is the line segment through $ H$ and perpendicular to $ LO$ ).

$ \bullet$ It is easy to show that the line segment $ OB$ tangents to the circumcircle $ (O'),$ of the triangle $ \bigtriangleup BHM,$ because of $ \angle OMB \equal{} \angle ODB \equal{} \angle OBD$ and so, we have $ (OH)\cdot (OM) \equal{} (OB)^{2} \equal{} (OE)^{2} \equal{} (OF)^{2}$ $ ,(3)$ where $ E,\ F$ are the points of intersection of the circle $ (O),$ from the line segment $ OHM.$

From $ (3),$ based on the [b][size=100]Newton theorem,[/size][/b] we conclude that the points $ M,\ E,\ H,\ F,$ are in harmonic conjugation.

Because of now, the line segments $ LH,\ NH,$ as the polars of $ N,\ L$ respectively, with respect to $ (O),$ passe through the point $ H,$ we conclude that the points $ N,\ L,$ as their poles, lie on the line through the point $ M,$ as the harmonic conjugate of $ H$ with respect to $ E,\ F$ and perpendicular to $ OM,$ as the polar of $ H$ with respect to $ (O).$

Hence, we conclude that $ NM\perp MO$ and the proof is completed.

Kostas Vittas.

thanks to [b]vittasko[/b],good proof.
I have thought on [b]Problem 7[/b].And I have solved it.Tomorrow [b]I will post my method on it.[/b] :D
Here is the [b]problem8[/b]

Let $ K \equal{} PA \cap QB$ and $ PA \cap (O_1) \equal{} S$.

we just have to prove $ \frac {QC}{CK} \equal{} \frac {QB}{KB}$(harmonic division).

which is equivalent to $ \frac {\sin \angle QAC}{\sin \angle CAK} \equal{} \frac {\sin \angle QAB}{\sin \angle PAB}$.

but we have $ \frac {\sin \angle QAC}{\sin \angle CAK} \equal{} \frac {AS}{AQ} \equal{} \frac {O_2O_1}{O_2Q}$.($ AO_1$ is the median).

and also we have $ \angle KAB \equal{} \angle BQP \equal{} \angle QO_1O_2$.and also we have $ \angle QAB \equal{} 180^\circ \minus{} \angle BPQ \equal{} 180^\circ \minus{} \angle O_2QO_1 \implies \sin \angle QAB \equal{} \sin \angle O_2QO_1$.

so we get $ \frac {\sin \angle QAB}{\sin \angle PAB} \equal{} \frac {\sin \angle O_2QO_1}{\sin \angle QO_1O_2} \equal{} \frac {O_1O_2}{QO_2} \equal{} \frac {\sin \angle QAC}{\sin \angle CAK}$

DONE!

Dear shoki.I appreciate your proof.But can you solve this using pure geometry?
Cheers!

Here is my geometric proof of problem 8.

It is easy to see $ \angle APQ \equal{} 90$, so $ \angle ABQ \equal{} 90$.  Then it follows that $ ABC$ and $ O_1QC$ are similar.  Let $ BQ$ intersect $ O_1O_2$ and $ PA$ at $ R,S$, respectively.  It is clear that $ O_1RC$ and $ ASC$ are similar.  Additionally, similarity factor is the same as that of $ ABC$ and $ O_1QC$ (for both it is $ O_1C: AC$).  So let the dilution $ D$ centered at $ C$ send $ A,B,S$ to $ O_1,Q,R$, respectively.  It follows that this dilution sends $ P$ to a point $ P^\prime$ on $ O_1O_2$.  Now $ \angle APC \equal{} \angle CP^\prime O_1 \equal{} \angle PP^\prime O_1$.  Also, $ \angle BPA \equal{} \angle O_1P^\prime Q$.  It is obvious that these two angles are equal (the triangle $ PP^\prime Q$ is iscoceles, with altitude $ P^\prime O_1$).

Dear Mathlinkers,
for problem 3 you can see my synthetic proof on
http://perso.orange.fr/jl.ayme   vol. 2 Les deux points de Schroeter p. 4.
Sincerely
Jean-Louis

Dear everyone.

here is my shoddy answer.

applying Pappus to (B,M,E) and (C,N,F), we get L,G,H=collinear.
thus L,G,H,K,O=collinear (Euler line). K=nine point circle center.
for cyclic quadrangle MENF, AP is perpendicular to KL (well known).
therefore AP is perpendicular to OH.

Dear everyone.

here is my shoddy answer.

$ PA\parallel O_{1}O_{2}$
$ PQ\bot O_{1}O_{2}\Longrightarrow PQ\bot PA$
$ \Longrightarrow\angle{ABQ} \equal{} 90^{\circ}$

$ O_{2}\in QA$, $ E \equal{} AP\cap (O_{1})$
$ E,O_{1},Q$=collinear
$ EQ\parallel BA$

$ (\infty,O_{1}/E,Q) \equal{} \minus{} 1 \equal{} A(\infty,O_{1}/E,Q) \equal{} (B,C/D,Q)$
$ \angle{DPQ} \equal{} 90^{\circ}$

$ \therefore\angle{BPD} \equal{} \angle{CPD}$

Very good solutions!
Dear mathlinkers from where can I learn drawing such pictures in pdf(Latex) or ...?

Perfect Answers!
I have found mistakes on problem7.So,I can't post my Proof.Sorry. :(
Here is Problem9 and 10            :P

[quote="77ant"]Dear tpjmroc.

please see April's awesome solution at http://www.mathlinks.ro/Forum/viewtopic.php?t=207440[/quote]

Thanks 77ant.I have read your solutions towards problem8,and it is great!
April's answer is very very very beautiful!~
Crazy man! :lol:

Dear tpjmroc

here is my shoddy answer.

$ (\infty,M/A,H) \equal{} \minus{} 1 \equal{} D(\infty,M/A,H) \equal{} (G,K/J,D)$
$ \therefore\square{GDKJ}$=harmonic

$ \Longrightarrow JJ,DD,GK$=concurrent
$ FE,JJ,DD$=concurrent (well known)

$ \therefore FE,JJ,DD,GK$=concurrent

$ (B,C/D,L) \equal{} \minus{} 1$
$ \angle{DKL} \equal{} 90^{\circ}$

$ \therefore\angle{BKD} \equal{} \angle{CKD}$

see my solution on http://www.aoshoo.com/bbs1/dispbbs.asp?BoardID=43&ID=16859&skin=0
It is true to both [b]incicrle and exincircle[/b]!

[b][color=red]IMPORTANT[/color]
 :oops: [color=darkred]Pro11 is a wrong problem for itself.So you can ignore it[/color].[/b]
[color=darkred]Now problem12[/color]
[size=150][b]Pro12[/b][/size]

[size=150]O, I are the circumcenter and the incenter of triangle ABC.D,E,F are the random points on the circumcircle.Try to show that it must exist such triangle DEF,which incenter is I.
[/size]

Strangely that I can't find the simple way.

well, :maybe: ,
$ OH^2 \equal{} 9R^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2) \implies a^2 \plus{} b^2 \plus{} c^2 \plus{} OH^2 \equal{} 9R^2$.
maybe u mean sth else...

[quote="shoki"]well, :maybe: ,
$ OH^2 \equal{} 9R^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2)$ [hide]$ \implies a^2 \plus{} b^2 \plus{} c^2 \plus{} OH^2 \equal{} 9R^2$.
maybe u mean sth else...[/hide][/quote]

[b]Thanks!I had realized the reason.
Appplying
$ OA^{2}\plus{}OB^{2}\plus{}OC^{2}\equal{}3OG^{2}\plus{}\frac{1}{3} (a^{2}\plus{}b^{2}\plus{}c^{2})$
and $ OH\equal{}3OG$
And that's OK.[/b]


Let triangle $ ABC$. Prove that
$ 15R^2\plus{}2R\sum\sqrt{R^2\plus{}b^2\plus{}c^2\minus{}a^2}\leq 3(a^2\plus{}b^2\plus{}c^2)$

[quote="Algadin"]Let triangle $ ABC$. Prove that
$ 15R^2 + 2R\sum\sqrt {R^2 + b^2 + c^2 - a^2}\leq 3(a^2 + b^2 + c^2)$[/quote]
But if $ \alpha\rightarrow180^{\circ}^-$ then $ R\rightarrow+\infty.$ :wink:

[quote="Algadin"]I'm sory. $ ABC$ is acute triangle[/quote]
I think, it's wrong too. Try $ a\equal{}b\equal{}5$ and $ c\equal{}7.$ :wink:


I need a singapore math book for grade 9, can you help me?

Do you know anyone in the United States? Because most of the math books I know of only ship to there.

But do you know in Singapore, which aspects they study in  Algebra and Geometry? Is there Combinatoric in their curriculum?


algebra
polynomial
calculus
superior algebra
superior algebra solved

Find and proove the smallest value of the polynomial

x^3(x^3+1)(x^3+2)(x^3+3)

I think the answer is -1 (I did it myself on my computer but

am unsure of the math that would get me to that response)

Step 1: Forget the cubes. $x\mapsto x^3$ has the entire real line in its image.

Now we want to find the absolute minimum of the fourth degree polynomial $x(x+1)(x+2)(x+3)$ in the real numbers.
We can rewrite this as $\left(y-\frac32\right)\cdot\left(y-\frac12\right)\cdot\left(y+\frac12\right) \cdot\left(y+\frac32\right)=\left(y^2-\frac94\right)\cdot\left(y^2-\frac14\right)$ where $y=x+\frac32.$

Let $z=y^2$. Our polynomial is now $\left(z-\frac94\right)\cdot(z-\frac14)=z^2-\frac52z+\frac9{16}= \left(z-\frac54\right)^2-1$ which clearly has its minimum at $z=\frac52$ and takes the value 1 there.

This could of course be done with calculus, but this problem was extremely symmetric.


calculus
integration
trigonometry
derivative
Galois Theory
calculus computations

Calculate:
$ I\equal{}\int_0^{\sqrt {3}} x^2\sqrt {1 \plus{} x^2} dx$
$ J\equal{}\int \frac{sinx}{x}dx$

$ \int_0^\infty \frac{\sin x}{x}dx\equal{}\frac{\pi}{2}$, but there is no anti-derivative (expressable in a finite combination of elementary functions).

hello
$ I\equal{}\int x^2\sqrt {1 \plus{} x^2} dx$
$ \equal{} x \frac{(x^2\plus{}1)^{\frac{3}{2}}}{3} \minus{} \frac{1}{3}\int (x^2\plus{}1)(\sqrt{x^2\plus{}1}) dx$
=$ x \frac{(x^2\plus{}1)^{\frac{3}{2}}}{3} \minus{} \frac{1}{3}\int x^2\sqrt{x^2\plus{}1} dx \minus{}\frac{1}{3}\int\sqrt{x^2\plus{}1}dx$
=$ x \frac{(x^2\plus{}1)^{\frac{3}{2}}}{3} \minus{} \frac{I}{3}\minus{}\frac{1}{3}[\frac{x\sqrt{x^2\plus{}1}}{2}\plus{}\frac{log(x\plus{}\sqrt{x^2\plus{}1}}{2}] \plus{}C$
$ \frac {4I}{3}\equal{}x \frac{(x^2\plus{}1)^{\frac{3}{2}}}{3}\minus{}\frac{1}{3}[\frac{x\sqrt{x^2\plus{}1}}{2}\plus{}\frac{log(x\plus{}\sqrt{x^2\plus{}1}}{2}] \plus{}C$
I hope now u can put the limits and get the value of $ I$.
thank u

hello, to the first integral setting $ x=\sinh(t)$ then we have $ \int\sinh^2(t)\cosh^2(t)\,dt=\frac{1}{4}\int\sinh^2(2t)\,dt$. We can calculate the last integral with integration by parts, we have
$ {\int\sinh(2t)\sinh(2t)\,dt=\frac{1}{2} \sinh(2t)\cosh(2t)-\int\cosh^2(2t)\,dt=\frac{1}{2} \sinh(2t)\cosh(2t)-\int1+\sinh^2(2t)\,dt}$. From here we get $ \int\sinh^2(2t)\,dt=\frac{1}{8}\sinh(4t)-\frac{t}{2}$. So we get the following result
$ \int_{0}^{\sqrt{3}}x^2\sqrt{x^2+1}\,dx=\frac{7\sqrt{3}}{4}-\frac{arsinh(\sqrt{3})}{8}$.
Sonnhard.

[quote="dulldrake"]
$ I \equal{} \int_0^{\sqrt {3}} x^2\sqrt {1 \plus{} x^2} dx$

[/quote]

$ x \equal{} \tan\theta \\
 \\
dx \equal{} \sec^2\theta d\theta \\
 \\
I \equal{} \int\tan^2\theta\sec^3\theta d\theta \\
 \\
 \\
\equal{} \int \left(\sec^3\theta \plus{} \sec^5\theta\right)d\theta \\
 \\
\text{Use Reduction Formula: } \\
 \\
I_n \equal{} \int\sec^n\theta d\theta \equal{} \frac {\sec^{n \minus{} 2}x\tan x}{n \minus{} 1} \plus{} \frac {n \minus{} 2}{n \minus{} 1}I_{n \minus{} 2}$

In the end, substitute $ \theta \equal{} \tan^{ \minus{} 1}x$ and evaluate the definite integral.
[url=http://archives.math.utk.edu/visual.calculus/4/recursion.2/]More Reduction Formulae[/url]

You could also use:

$ \int uv \equal{} u\int v \minus{} u'\int \int v \plus{}u''\int \int \int v\minus{}u'''\int \int \int \int v \plus{} ....$

Choose the $ u$ wisely, here if $ u\equal{}x^2$, it is interesting to note that the series terminates at $ u''' \equal{} 0$.

[quote="dulldrake"]
$ J \equal{} \int \frac {sinx}{x}dx$

[/quote]

You should read these two ([url=http://en.wikipedia.org/wiki/Differential_Galois_theory]1[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=264015]2[/url]) articles.


trigonometry
function
algebra
domain
Pythagorean Theorem
geometry

Some interesting piece of trig :

Discuss (without calculator) the graph of the function

f(x) = sin(arccos x).

[hide]

Draw a right triangle.  Let the hypotenuse=1.  Now, let the cos(theta)=x which means the bottom leg is x.    Now, use pythagorean theorem to find other leg.

x^2 + y^2=1 where y=the unknown leg

y^2=1-x^2

y=sqrt(1-x^2)

[/hide]

Hm ... what about the graph itself ?

[hide]sin(arccos x) = :sqrt:(1 - cos2(arccos x)) = :sqrt:(1 - x2)

Indeed. But what's the shape of the graph ? The domain of f ? ... ?

[/hide]

What about it?

It's symmetric about the y axis.  It is going to look similar to a very narrow parabola.  As x->1 or -1, y->infinity.  It has a minimum of 1.

Hm ... I expected something more. A hint :

[hide]The general form of the equation of a circle is

(x - a)2 + (y - b)2 = r2.

So rewrite the relation f(x) = :sqrt:(1 - x2). OK ?[/hide]

So ...

y = :sqrt:(1 - x2)
if and only if
y :ge: 0 and y2 = 1 - x2
if and only if
y :ge: 0 and x2 + y2 = 1.

Now x2 + y2 = 1 represents the unit circle.
(Take a = b = 0, r = 1 in the above hint.)
Since y :ge: 0 we get only half of the unit circle.
So the graph is a semicircle.

And you wrote some strange things DaSweepa :

-When x -> 1 or x -> -1, y -> infinity ? How that ?
-It has a minimum of 1 ? Probably you mean a maximum.

I was thinking it was the 1/sqrt(1-x^2) for some reason.  Don't ask me why cause I have no idea.  :oops:  :oops:


calculus
integration
trigonometry
linear algebra
matrix
calculus computations

hope i have not asked this before....  :maybe:

show

$ \int_{0}^{\infty}\;\;\frac{\sin^{2k}\;x}{x^{2k}}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{\pi}{2}\cdot\frac{1}{(2k\minus{}1)!}\cdot\left\langle\begin{matrix}2k\minus{}1\\ k\minus{}1\end{matrix}\right\rangle}$

what are the numbers $ \boxed{\left\langle\begin{matrix}n\\ m\end{matrix}\right\rangle}$ typically called ? :D

[quote="sylow_theory"]What is that symbol supposed to mean? Combinations?[/quote]

sylow, it is a symbol that refers to a certain type of numbers, just like
$ \textbf C_{n}^{m}\;\;\equal{}\;\;\dbinom{n}{m}$ represents combinations,
and $ \textbf s_{n}^{m}\;\;\equal{}\;\;\left[\begin{matrix}n\\ m\end{matrix}\right]$ represents Stirling numbers of the 1st kind,
and $ \textbf S_{n}^{m}\;\;\equal{}\;\;\left\{\begin{matrix}n\\ m\end{matrix}\right\}$ represents Stirling numbers of the 2nd kind, etc.

the numbers $ \boxed{\left\langle\begin{matrix}n\\ m\end{matrix}\right\rangle}$ are fairly well known...  :D


Thorough Ted, a construction engineer, correctly computed that the maximum safe load of a bridge being planned would be 1000(99-70 (suare root of) 2)tons.

Speedy Sam, the safety supervisor, was asked to design a sign for motorists saying how much weight the bridge could safely hold. He took thorough Ted's expression, used 1.4 as an approximation for (saure root of)2, and created the sign based on his calculations. But on the day the bridge opened to traffic, it collasped under a load less than a tenth of the weight shown on Sppedy Sam's sign.

Speedy Sam told the city council that he had simply used Ted's figures. Thorough Ted has been over and over his figures and doesn't see how they could be wrong. Write a clear explanation for the city counctil why the bridge collasped.

(IF ANYONE CAN HELP)>>please do  :D

$ 99 \minus{} 70 \sqrt {2}$ is very small.  Calculate it.  (It should be about $ 0.005$.)  Why is this a problem?

cuz i have noooo idea what i'm doing....um...do you think you can explain it in more detail. please  :P
This is a important assignment...and i dont know what to do... :|

No, no, I meant "why is it a problem that $ 99 \minus{} 70 \sqrt{2}$ is very small?"  Have you done the calculation?  Have you also calculated $ 99 \minus{} 70 (1.4)$?  Did you go on to multiply by $ 1000$?  What is the error?

There is a large difference between $ 1000(99\minus{}70\sqrt2)$ and $ 1000(99\minus{}70 \times 1.4)$.  The first number is about $ 5.05$.  However, the second is $ 1000$.  And since $ 1000 > 5.05$, the bridge collapsed.


Let $ ABC$ be a triangle and $ A'B'C'$ be a Cevian triangle of point $ P$. $ PA',BC$ intersect circumcircle $ (A'B'C')$ at $ \{A',A_1\}$ and $ \{A',A_2\}$, resp. $ A_1A_2$ intersect $ B'C'$ at $ H$. Tangent of $ (A'B'C')$ at $ A_2$ intersects $ B'C',AB,AC,AH$ at $ I,J,K,L$,resp. $ IA$ intersect $ A_1A_2,A_2B',A_2C'$ at $ P,M,N$, resp. Prove that $ NJ,KL,MP$ are concurrent.

Small mistake, thanks, I fix it

Let $ ABC$ be a triangle and $ A'B'C'$ be a Cevian triangle of point $ P$. $ PA',BC$ intersect circumcircle $ (A'B'C')$ at $ \{A',A_1\}$ and $ \{A',A_2\}$, resp. $ A_1A_2$ intersect $ B'C'$ at $ H$. Tangent of $ (A'B'C')$ at $ A_2$ intersects $ B'C',AB,AC,AH$ at $ I,J,K,L$,resp. $ IA$ intersect $ A_1A_2,A_2B',A_2C'$ at $ I,M,N$, resp. Prove that $ NJ,KL,MI$ are concurrent.


$ Calculate 2008C0 \plus{} 2008C4 \plus{} 2008C8 \plus{} ... \plus{} 2008C(4k) \plus{} ... \plus{} 2008C2008$
how do you calculate this??? its on my homework for one of my classes...

[hide="LaTeX tip"]you can write combinations as

[code]$\binom{a}{b}$[/code]

which gives

$ \binom{a}{b}$.[/hide]

[hide="For problem"]Basically use the fourth roots of unity and the binomial theorem.[/hide]

[hide="Solution"]
[b]Theorem[/b] (Roots of Unity Filter). Let $ P(x)\equal{}\sum a_kx^k$.
Let $ \omega$ be an $ n^{\text{th}}$ root of unity.
Then:
$ \frac{1}{n}\sum_{k\equal{}0}^{n\minus{}1}P(\omega^k)\equal{}\sum_{n|k}a_k$.
[i]Proof.[/i] I'll refer you to this blog post: http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=245909

---

Using this, $ \sum_{k\equal{}0}^{502}\binom{2008}{4k}\equal{}\frac{1}{4}\sum_{k\equal{}0}^{3}P(\omega^k)$, where $ P(x)\equal{}(x\plus{}1)^{2008}$ and $ \omega$ is a 4th root of unity.

Our answer is, then, $ \frac{2^{2008}\plus{}(1\plus{}i)^{2008}\plus{}(1\minus{}i)^{2008}}{4}$.

Noting that $ 1\plus{}i\equal{}\sqrt{2}\cdot\text{cis}\;\frac{\pi}{4}$, and applying DeMoivre's, we eventually get:

$ \frac{2^{2008}\plus{}2^{1006}}{4}\equal{}\boxed{2^{2006}\plus{}2^{1004}}$.
[/hide]


solve the following inequality:

sum((a^2+bc)/(b+c))>=sum(a); a,b,c >0

PS: I appologize for the past "signature" I had!

 :(

heya!!!! here is my solution:

[b]Solution[/b]:

you write the ineq like so: sum((a^2+bc)/(b+c)-a)>=0 which is equivalent to: sum ((a-b)(a-c)/(b+c))>=0 which gives us: sum ((a-b)((a-c)/(b+c)- (b-c)/(c+a))>=0 which gives in the end that:
sum ((a-b)^2*(a+b)/((b+c)(c+a))>=0 which is true!!!
we have equality when a=b=c!!!

cheers! :D :D :D

[b]Another solution:[/b]

use that (a^2+bc)/(b+c)+(b^2+ca)/(c+a)>=a+b  and the other simmilar ineqs!!!

and for (a^2+bc)/(b+c)+(b^2+ca)/(c+a)>=a+b you get to (a-b)^2(a+b)>=0 which is true... and ineq is if a=b=c.

cheers! :D :D :D


Consider a distribution with pdf $ f(x;\theta)\equal{}\theta{x}^{\theta\minus{}1}$ if $ 0<x<1$ and zero otherwise.

(a) Based on a random sample of size n=1, find the most powerful test of $ H_0: \theta\equal{}1$ against $ H_a: \theta \equal{} 2$ with $ \alpha\equal{}0.05$

(b) Compute the power of the test in (a) for the alternative $ \theta\equal{}2$

(c) Derive the most powerful test for the hypotheses in (a) based on a random sample of size n

I wonder if you had heard of [url=http://en.wikipedia.org/wiki/Neyman%E2%80%93Pearson_lemma]Neyman-Pearson lemma[/url].
Because if yes, (a) and (b) are easy, (c) is just a bit harder.


Congrats to the 36 USAMO qualifiers from MA.  An impressive number which includes 1 7th grader, 3 8th graders, 5 9th graders, 11 10th graders, and 8 each in 11th and 12th grade.

Best of luck to you all!!

Mr Frost

Yes, this year, I did not fail AMC 10 with a 112.5 :D (I got 150).

According to an analysis done by jb05 on Facebook, Massachusetts is second only to New Hampshire in USAMO qualifiers by State Population. New Hampshire has ~12 per million, MA ~5.5, and the next highest is Maine with ~3. It should be noted that New Hampshire's qualifiers all come from Exeter, which contains very few actual New Hampshire residents. So nice job, everyone.

(Although it also should be noted that Andover has a lot of non-permanent residents as well, though this still puts us at about 4.5 if we take away the Deerfield kid as well).

But this isn't enough! My goal for this state is 8 MOPpers. Whether that group contains me or not...we'll see.

[quote="perfect628"][quote="Yongyi781"](I got 150).[/quote]
Ugh seriously how many times are you going to post that you got 150 on the AMC 10? I must have seen this like 15 times.

Anyway, I enjoy my ID number this year. Disliked it last year...I keep getting very low IDs...[/quote]

Hmm, actually, doing a search with "150" by author "Yongyi781" only yields 3 results which contain me saying I got a 150.

I only said I got 150 here just in case anyone thought I actually got 112.5 this year (take it out and you'll see the ambiguity).

You could have just said "I made it this year."

Representing the oddly-weak senior class!  :lol:

Why hasn't the central-location for all MA test-takers idea been repeated? I think it's a good idea to have everyone meet up and mingle before/after sweating over proofs.

I was never aware that there was such an idea.

Personally, a little peace is good for "sweating over proofs"


For what value of $ a$ is there a right triangle with sides $ a \plus{} 1$, $ 6a$, and $ 6a \plus{} 1$?

Naturally think that $ a \plus{} 1$ is probably smaller than $ 6a$, and, of course, $ 6a < 6a \plus{} 1$. Thus, set up an equation and solve.

$ (a \plus{} 1)^2 \plus{} (6a)^2 \equal{} (6a \plus{} 1)^2$
$ a^2 \plus{} 2a \plus{} 1 \plus{} 36a^2 \equal{} 36a^2 \plus{} 12a \plus{} 1$
$ a^2 \plus{} 2a \plus{} 1 \equal{} 12a \plus{} 1$
$ a^2 \plus{} 2a \equal{} 12a$
$ a^2 \equal{} 10a$
$ a \equal{} 0$ or $ a \equal{} 10$

$ 0$ would not work because a non-degenerate triangle can't have any side length of $ 0$. (In this case, if $ a \equal{} 0$, then $ 6a \equal{} 6(0) \equal{} 0$.) Thus, $ a$ must equal $ \fbox{10}$.


Evaluate:  $\sum_{k=1}^{\infty}\frac{1}{F_{k}}$, where $F_{n}$ is a Fibonacci sequence

This is the reciprocal Fibonacci constant.  There is no known closed formula.   Opportunity to enter the history books, then!


This forum deserves a topic on this:

[size=200][color=red]Happy New Year and Welcome to the Year of the Rooster!!![/color][/size]

 [hide="Happyness and laughs to everyone"]:D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl: :D :rotfl:[/hide]

OK, maybe my post is considered overusage of emoticons but hey, NO ONE REPLIED TO THIS TOPIC?!?!? :!:  :?:  :o
I'm Chinese and I'm not even born in China or doesn't know much about Chinese stuff, and I celebrate Chinese New Year. I know a lot of the users in here are Chinese, and yet not a single Chinese user (other than me) have replied here. I'm very disappointed  :(  for a moment like this. Where are you Chinese people!!!!!!!???????
 :mad:  :mad:  :mad:

Chinese users are on Earth.  :D
I overlooked this post, sorry..   :blush:
Happy Chinese New year every one! :D  :D

Happy chinese new year

There was some firecracker in my area

Yes, [b]Happy New Year[/b] (or [b]Spring Festival[/b] as it is now called in China)!!!!!!!!!! :D

HAPPY NEW YEAR!!!!! Even though I've already posted a post on another one about chinese new year...
[size=200]
新年快?， ?大家。。。。一年比一年?明???[/size]

Wait, are any of you even the Chinese ones??

what do u mean by are any of you chinese ones? I dont' get it...   :? but I"m chinese and proud of it....  :D

[quote="Moubinool"]There was some firecracker in my area[/quote]
Too much firecracker here at 0 o'clock on 9th... Very Noisy...

im proud of being a chinese too..haha

happy new year guys,
get some good chicks in chick year.  :D


#14:

Let a sequence be defined as follows: a(sub1)=3, a(sub2)=3, and for n>=2.

a(sub n+1)*a(sub n-1)=[a(sub n)]^2+2007. Find the largest integer less than or equal to

[a(sub 2007)]^2+[a(sub 2006)^2
____________________________
a(sub 2007)*a(sub 2006)

NOTE: the line is a fraction bar.


If there is enough people today at 7:00 EST, I'll do a math chat today.

If you want to do it, let me know by posting here.

Thanks.

sure, what time is it in AL, I'm terribly bad with the time thing

Wait, I've never been to a math chat b4, what do we do there, what do we do it through? AIM?

You can invite yourself into the room "Math Chat" or if you don't know how to do that IM me ( my sn is BaseballJoe2002 ).

This is random but this new smiley is cool  :D ... :P

Announcement!

Math Chat will start in 20 minutes, which is 7:20 to 7:30.

The reason?

I have to get something to eat now.

Thanks. :D

can you post the transcript for today's chat.  Thx


[hide] If an integer leaves a remainder of $ 1$ when divided by both $ 3$ and $ 5$, then it will leave a remainder of $ 1$ when divided by $ 15$ (I think this is CRT, but not exactly sure). Also, note that $ 15$ leaves a remainder of $ 1$ when divided by $ 7$. Therefore, in order to obtain a remainderof $ 5$, we should multiply by $ 4$ and then add $ 1$. We thus have $ 15(4)\plus{}1\equal{}\boxed{61}$. [/hide]

First, I would list out all the factors of seven, up to 7*9, since seven is the number of greatest value.

7, 14, 21, 28, 35, 42, 49, 56, 63,

Then, I would add five to each since, when divided by seven, it leaves a remainder of five.

12, 12, 26, 33, 40, 47, 54, 61, 68

Know you just have to divide them my three and five and you will find that the number you are looking for is 61.
61/5=12r1 61/3=20r1 and 61/7= 8r5

[quote="AndrewTom"]What is the smallest positive integer which leaves remainders of $ 1, 1$ and $ 5$ when divided by $ 3, 5$ and $ 7$ respectively?[/quote]

61.  61/3=20r1, 61/5=12r1, 61/7=8r5


[quote="Rust"]Only x=0.
Consider $x>0$ and $x<0$.[/quote]

in general, how do you find roots (non-zero) for an eqn which has two forms of x?

ok maybe the qn isn't very good since e^x is always >1 if x > 0

how about... x + e^x = 2

or should i change it again...

x + e^x = 2.148721271

i know x = 0.5 is a solution (because i set the qn), but how do i find it the right way?

@ahgohgoh:
Please read carefuly
[quote="Rust"]$x+e^{x}=a$ had only unique solution, because $f'(x)=(x+e^{x})'=1+e^{x}>0$.[/quote]

[quote="Rust"]$x+e^{x}=a$ had only unique solution, because $f'(x)=(x+e^{x})'=1+e^{x}>0$.[/quote]

why would $f'(x)=(x+e^{x})'=1+e^{x}>0$ show that it has unique solution? and what's the implication of it having a unique solution? How do we find that unique solution?

[quote="Rust"]You solve $f(x)=a, f(x)=x+e^{x}$. If $f(x)=a$ had at least to solution $f(x) =f(y)=a,x<y$, then must be $x<z<y$, suth that $f'(z)=0$. But $f'(z)=1+e^{z}>1>0$. Therefore $f(x)=a$ had no more one solution. We have $f(-\infty )=-\infty , \ f(\infty )=\infty$, therefore f -surjective and had one solution $\forall a\in R$.[/quote]

thanks for showing that the root is unique...

but how do i find the unique root for ans that is not 0?

can it be calculated? or do we need a supercomputer?

You cannot find the root in explicit form. You can find it approximately using binary search.


integration
geometry
calculus
calculus computations

A city is built around a circular lake that has a radius of 1 mile. The population density of the city is f(r) people per square mile, where r is the distance from the center of the lake in miles. What is the expression that gives the number of people who lives within 1 mile of the lake?

$\int_1^2{f(r)}$, right?
Or you might need to use the circle area: $\pi (\int_1^2{f(r)})^2$

Neither of those is correct, me@home.

Imagine a very narrow annulus at a distance $r$ from the center and with width $dr.$ This strip has area $2\pi r\,dr$ and population density $f(r).$ Hence it has population $2\pi rf(r)\,dr.$

The answer is $\int_1^22\pi rf(r)\,dr.$

Units of measurement can be a help. $f(r)$ has the units $\frac{\text{persons}} {(\text{mile})^2}.$ $r$ is in miles.

You should definitely include $dr$ (or d(whatever)) in your proposed formulas. Note that $dx$ has the same units of measurement as $x,$ and doing the integral is then just taking a sum - the sum of things with the same units of measurement has the same units.

Your first effort was $\int_1^2f(r)\,dr.$ Let's check the units for that:

$\frac{\text{persons}} {(\text{mile})^2}\cdot\text{miles}=\frac{\text{persons}} {\text{mile}}.$ That's not right.

Your second effort was $\pi\left(\int_1^2f(r)\,dr\right)^2.$ The units for this would be

$\left(\frac{\text{persons}} {\text{mile}}\right)^2.$ That's not right, either.

By contrast, $\int_1^22\pi rf(r)\,dr$ gives you

$\text{miles}\cdot \frac{\text{persons}} {(\text{mile})^2}\cdot \text{miles} = \text{persons}.$

That is the correct units. Being the correct units doesn't prove that it's the right formula, but it's a start.

I dont understand how you know to intergrate from 1 to 2
on the area how do you know the area of the strip is 2(pi)(r)dr

Ok, think of a circle of radius 2, then a smaller concentric circle of radius 1 representing the lake.  Then the city is an annulus (ring), with it's inner most circle being 1 unit from the center and it's outer most circle being 2.  Then imagine an infinitely thin strip of the annulus, which would be a circle centered at the center of the lake with a radius of $r$ and a circumference of $2\pi r$.  The reason we use a cirucmference is because the area of an infinitely thin annulus would be the circumference of a circle (with the same radius).  Therfore, the desired area is given by $2\pi\int_1^2 rf(r) \ dr$.

yes. I definitely messed up stupidely when I first looked at this problem...
then the next day in calculus class, I saw the exact same problem on a multiple choice test and the result was trivial, explained in the same way as Kent Merryfield. What a coincidence!


I have a question about the problem series classes offered. Do the AMC-10 & Mathcounts problem series classes contain the same problems as last session?

Thanks (:

Yes, they do.

If at any point we decide to write entirely new problem series for a testing level, we'll make it clear that it's a new set of problems.  We're currently in the works for an AIME B Problem Series, but that will be the first problem series with two tracks of distinct problems.


Rohit I am surprised at you my friend. Even I was able to solve this problem and you call this high standards????? :O:O

Let There exist a $ d$   for which

$ 2d - 1 = n^{2}$
$ 5d-1 = k ^{2}$ ---- 1
$ 13d- 1 = z^{2}$ ---- 2

$ 2d -1 = n^{2} \implies n^{2}$is odd
$ \implies n$ is odd

All odd perfect squares are $ 1 \pmod( 4)$

$ 2d - 1 \equiv 1 \pmod 4$
$ \implies 2d \equiv 2 \pmod 4$
$ \implies d \equiv 1 \pmod 4$
$ \implies d$ is odd

Since d is odd
$ 5d - 1$ is even $ \implies k$ is even
$ 13d - 1$ even $ \implies z$ is even

Subtract 2 and 1, We get
$ 8d = (z-k) (z+k)$
Let $ z'$ and $ k'$ be $ \frac{z}{2}$ and $ \frac{k}{2}$ resp since $ z$ and $ k$ are even

$ 2d = (z'-k')(z'+k')$

We know that $ d$ is odd. So that means only one of the two terms on the RHS can be even, but thats a contradiction cause $ a+b$ and $ a-b$ always have same parity.

Contradicted.

Hence the Proof

P.S I know this isn't very rigorous, but meh!

[size=150][color=blue]MOD EDIT:I $ \text{\LaTeX}$ ified that[/color][/size] :ewpu:

Who said this was tough :huh: . I said always that this was easy for an imo problem if u didnt get that impression im sorry to have not been a bit more clearer  :) . Also even i got the sum and was shocked duly that this was an imo sum even though of 1986. So i decided to give the sum  :D . And ofc i expected everyone to solve this sum easily and had hence prefixed the tag easy so mister shreyas u neednt get excited or tense @ all (i was just fuelling you up so that i could see u doing some math :P)

shre u should have atleast tried latexing it... :huh:
u r 385 posts old :)

Sorry da! I still have small age. But you know that latexing really takes my time and already I'm getting screwed here for spending too much time online. :(

But thanks for latexing it. Sure rohit, I never got excited mind you. Just happy that I solved an IMO problem but I know that my method was slicker than yours and you really feeled too much. :D:D Ha!


LaTeX
search
number theory unsolved
number theory

prove that  for any  integer $n$ there $a,b,c$ is integer st $n=a^{2} +b^{2}-c^{2}$

[mod.: $\LaTeX$-ed]

Thank ZetaX verymuch.

Your TeX-code is not far away from being perfect, just don't set those "\" (what did they stand for¿) and put $\$$ around the formulas :)

To the problem:
I think I posted a solution to this one before, but I don't want to search for it now, so here it goes ;)

By considering cases it's not hard to show that every $k \not \equiv 2 \mod 4$ can be written as difference of two squares, namely $k=b^2-c^2$.
Also, for any $n$ there is a $k \not \equiv 2 \mod 4$ such that $n-k$ is a perfect square, more exactly, we can choose $k=n$ or $k=n-1$, at least one will not be $\equiv 2 \mod 4$.
But then $n-k=a^2$, thus $n=a^2+k=a^2+b^2-c^2$, done.

But now try a harder problem: prove that every integer $n$ can be written as $n=a^2+b^2+c^3$ with integers $a,b,c$.
(note that by changing the sign of $c$, this looks very similar to the above)

Can't we just pick $a = 1, b^2 - c^2 = n-1$ in the case $n \equiv 2 \bmod 4$?  I think that would be simpler :)

This is what I did (but hided very good) ;)


I am not an adept of calculs  :lol:  but
[hide]$\frac{88+89+90+100+x}{5}=91,4\\367+x=457\Rightarrow x=90$
Therefore he got  90%[/hide]

[quote="math92"]In math class, you have taken five tests. Your average grade on those tests is a
91.4%. Your first four test grades were 88%, 90%, 100%, and 89%. What was your fifth
test score?[/quote]

90


I am individualy studying chemistry with a book that has no answers.,

okay so...

Is adding milk and fruit to a bowl of cereal a chemical change or a physical change?

When the milk and fruit are added, the only thing that happens is that the cereal and fruit absorb some of the milk and become soggy. No new species are produced, so this would be a physical change.

yes it should be physical because the chemical make up of the ingredients do not change


How do you box a number in LaTeX?  I have a huge wall of text in my USAMTS, and I want to box the important parts of the solution.

\boxed{}

$ 2\plus{}2\equal{}\boxed{2^2}$

The answer is $ \boxed{\left[\frac{(1\plus{}x)^2\minus{}x}{e^x\plus{}e^{\minus{}x}}\right]^{12}}$


function
search
Putnam
superior algebra
superior algebra unsolved

Prove that the identity is the only increasing automorphism of the group of positive rational numbers.

[quote="ZetaX"]The following stronger problem was already discussed (and solved):

Let $ f:\mathbb N\to\mathbb R$ be a multiplicative function and non-decreasing. Then $ f(n) \equal{} n^{a}$ for some $ a$[/quote]
Can you give me a link?Thank you(I don't how to search it) :blush:

http://www.mathlinks.ro/Forum/viewtopic.php?p=381479#381479
http://www.mathlinks.ro/Forum/viewtopic.php?p=179524#179524
(I just searched for "multiplicative increasing" :D)

Thank you,But how can we prove for every irrational number $ \alpha$,$ n^{\alpha}$ is not rational for some natural number n?

ZetaX has indeed given a precious argument, but I really think there are still some arguments, which are not trivial, for Zhaobin's question. There was a Putnam problem if we replace rational by integers, but I think this variant is more subtle.


Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$. Prove that
\[ a^2\plus{}b^2\plus{}c^2\plus{}6(ab\plus{}bc\plus{}ca) \minus{}21 \ge 20 (a\minus{}1)(b\minus{}1)(c\minus{}1).\]

P/s: From this inequality, the following inequality (of Hanjingjun)
\[ \sum \frac{34a^2\plus{}11}{4a^2\plus{}2a\plus{}3} \ge 15\]
follows immediately. :)

I tried $ a\equal{}\frac {x}{y}$ . $ b\equal{}\frac{y}{z}$ , $ c\equal{}\frac{z}{x}$. But still i couldn't solve it.
can you give me just a little hint?

[quote="shaaam"]I tried $ a \equal{} \frac {x}{y}$ . $ b \equal{} \frac {y}{z}$ , $ c \equal{} \frac {z}{x}$. But still i couldn't solve it.
can you give me just a little hint?[/quote]
I used Mixing Variables, my friend. :)


vector
linear algebra
matrix
superior algebra
superior algebra solved

Hello everyone,

I've encountered this question in one of the previous years linear algebra (1) tests:

Let v1 = (1, i, i), v2 = (2i, 1, 1), v3 = (2-2i, -1+2i, -1+2i) be vectors in C^3 (that is supposed to be C with a three on the right-top :-)).
A basis for V = span{v1,v2,v3} is:

1. {(1,i,i),(0,i,-i)}
2. {(1,0,0),(0,1,1)}
3. {(1,i,i),(0,0,i)}
4. {(1,0,0),(0,1,0),(0,0,1)}

I reduced the corresponding matrix to row-echelon form and got {(1,0,0),(0,1,1)} as a result (answer 2). But there's a theorem that states that if a group of vectors is linearly independent and contains the same number of vectors as the basis, then it is a basis itself. So it seems to me answers 1 and 3 are correct aswell, as I find them linearlly independent. There's supposed to be just one right answer. How can this be?

Thanks for reading so far. My final exam in linear algebra is in two days, so I hope to get that out of the way before it :-)

-- trmason

The vectors in 1 and 3 may not span V. They may span another subspace of dimension 2


quadratics
analytic geometry
graphing lines
slope
algebra

If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show that:

(a) r^2(1 + m^2) = b^2

(b) The point of tangency is the point (-(r^2m/b), (r^2/b))

(c) The tangent line is perpendicular to the line having the center of the circle and the point of tangency.

1. Put $ y\equal{}mx\plus{}c$ in the eqn $ x^2 \plus{} y^2 \equal{} r^2$ and solve the quadratic equation with discriminant = 0
    If you put D = 0 you get the result .
2. Solve for the resulting x and y in 1 .
3.Hint
[hide]
For 2 curves to be perpendicular product of slopes of the 2 curves = -1
[/hide]
 :)


:rotfl:  :pilot:  :play_ball:
hello plz prove this inequality


1.  Find two unit vectors in $ R^2$ that are perpendicular to $ ( \minus{} 3,4)$

[hide="1."]Do you know the formula for angles between vectors?  If $ \textbf{a},\textbf{b}$ are vectors, then
$ \frac{\textbf{a}\cdot\textbf{b}}{\left|\textbf{a}\right|\left|\textbf{a}\right|}\equal{}\cos{\theta}$.
If the vectors are perpendicular, then $ \cos{\theta}\equal{}0$ so multiplying through, we get that $ \textbf{a}\cdot\textbf{b}\equal{}0\equal{}(\minus{}3,4)\cdot (x,y)\Rightarrow \minus{}3x\plus{}4y\equal{}0$.  Now you just need to find a solution to this equation, say $ (4,3)$.  Of course, there are others.  Note that this is not a unit vector so you'll have to normalize it, and I'll leave that to you.
[/hide]

It looks like you're trying to get us to do your homework for you.  If you're going to post homework problems, make it clear that you've attempted the problem or explain what you don't get.


Supose y is proportional to x and y = -8 when x=3. Find y when x=-6

$x: y \implies3: -8$

$\implies -6: y = 3: -8 \implies y=16$

Proportional means that

$y=kx$, for some $k \in \mathbb{Z}^{+}$

Then $-8 = 3k$, $k = -\frac{8}{3}$.

then $y=-\frac{8x}{3} \Rightarrow y = -\frac{8 \cdot -6}{3} = -16$

sorry in my proofs i put arrows which dont realli mean anything. they js mean a new line i guess. it suppose to mean $a\implies b$ statement a implies statement b. but i dont adhere to that strict definition.


calculus
integration
inequalities
logarithms
trigonometry
function
real analysis

Prove the following inequality.

\[\int^{\frac{\pi}{4}}_{ -\frac{\pi}{4}} \ln (1-\sin \theta)d\theta <\frac{1}{4}\left(1-\frac{\pi}{2}\right)\]

Since the integral is over an interval symmetric about the origin, the only part that matters is the even part of the function, namely

$\frac{\ln(1-\sin\theta)+\ln(1+\sin\theta)}2=\frac12\ln(1-\sin^2\theta).$

Hence, $\int^{\frac{\pi}{4}}_{ -\frac{\pi}{4}} \ln (1-\sin \theta)d\theta =\int_0^{\frac{\pi}4}\ln(1-\sin^2\theta)d\theta> \int_0^{\frac{\pi}4}-\sin^2\theta d\theta.$

Computation of the last integral gives the answer.

Yes.You are right,Kent Merryfield. :) Thank you!

kunny


Find the minimum value of\[ (u-v)^2+\left(\sqrt{2-u^2}-\frac{9}{v}\right)^2 \]for $0<u<\sqrt{2}$ and $v>0$

[quote="luimichael"]Using Calculus, the miniumum value is found to be 8.
Is there any other method?[/quote]
I have said in my previous post.
This problem is to find the shortest distance between the circle $x^2+y^2=2$ and the hyperbola $xy=9$, which is clearly $2\sqrt{2}$. Sqaure it we get $8$.

[quote="shobber"][quote="luimichael"]Using Calculus, the miniumum value is found to be 8.
Is there any other method?[/quote]
I have said in my previous post.
This problem is to find the shortest distance between the circle $ x^2 \plus{} y^2 \equal{} 2$ and the hyperbola $ xy \equal{} 9$, which is clearly $ 2\sqrt {2}$. Sqaure it we get $ 8$.[/quote]

I don't get it.  Why is it clearly $ 2\sqrt {2}$????  And how is it related to the original equations???

hello, i have also found that the minimum is $ 8$ for $ u\equal{}1$ and $ v\equal{}3$.
Sonnhard.

[quote="sexynsmartjenny"]And how is it related to the original equations?[/quote]

$ (u, \; \sqrt {2 \minus{} u^2} )$ is a point on the upper half of the circle $ x^2 \plus{} y^2 \equal{} 2$

$ \left( v, \; \frac {9}{v} \right)$ is a point on the hyperbola $ xy \equal{} 9$

The square of the distance between these two points is $ (u \minus{} v)^2 \plus{} \left( \sqrt {2 \minus{} u^2} \minus{} \frac {9}{v} \right)^2$, by the distance formula, which is identical to the expression we want to minimize.

[quote="sexynsmartjenny"]Why is it clearly $ 2\sqrt {2}$?[/quote]

Graph the circle and the hyperbola. You'll see that the shortest path from the circle to the hyperbola is along the line $ x \equal{} y$.

This intersects the circle at $ (1,1)$ and the hyperbola at $ (3,3)$, so the minimum distance is the distance between these points, or $ 2 \sqrt {2}$

Then, the minimum square distance, which is what we want, is $ (2 \sqrt {2})^2 \equal{} 8$

[quote=smartjenny][quote="TZF"]
Graph the circle and the hyperbola. You'll see that the shortest path from the circle to the hyperbola is along the line $ x \equal{} y$.[/quote]

How do you see the shortest path must lie on the line$ x\equal{}y$?[/quote]

You are effectively minimizing the distance from the point on the hyperbola to the origin because you can just subtract $\sqrt{2}$


ratio
geometric sequence
geometric series

Let $ A$ be the sum of the first $ 10$ terms of a geometric sequence, and let $ B$ be the sum of the $ 11$th through $ 20$th term.  Obtain the sum of the $ 21$st through $ 30$th term.  Assume that $ A \neq 0$.

Given a geometric sequence whose sum of the first $ 10$ terms is $ 2$, and whose sum from the $ 11$th through $ 30$th term is $ 12$, obtain the sum from the $ 31$st through $ 60$th term.

The sum of the first ten terms is given by $ A\equal{}x\frac{1\minus{}r^{10}}{1\minus{}r}$, where $ r$ is the common ratio, and $ x$ is whatever the constant is.  Similarly, the formula for the first 20 terms is $ x\frac{1\minus{}r^{20}}{1\minus{}r}$.  Hence $ B\equal{}x\left(\frac{1\minus{}r^{20}\minus{}1\plus{}r^{10}}{1\minus{}r}\right)\equal{}x\frac{r^{10}\minus{}r^{20}}{1\minus{}r}$.  You can similarly obtain the next 10 terms.

[quote="JRav"]The sum of the first ten terms is given by $ A \equal{} x\frac {1 \minus{} r^{10}}{1 \minus{} r}$, where $ r$ is the common ratio, and $ x$ is whatever the constant is.  Similarly, the formula for the first 20 terms is $ x\frac {1 \minus{} r^{20}}{1 \minus{} r}$.  Hence $ B \equal{} x\left(\frac {1 \minus{} r^{20} \minus{} 1 \plus{} r^{10}}{1 \minus{} r}\right) \equal{} x\frac {r^{10} \minus{} r^{20}}{1 \minus{} r}$.  You can similarly obtain the next 10 terms.[/quote]

So basically, we perform the arithmetic as follows:

$ S_{30} \minus{} S_{20} \implies \dfrac{a(r^{30} \minus{} 1)}{r \minus{} 1} \minus{} A \minus{} B$ and then we plug in the values of $ A$ and $ B$.  I see, that makes sense.

For number two, I get a 30th degree equation when I try to solve for $ r$.  (Where $ r$ is the common ratio of the geometric sequence.)

Any hints on it?  Am I doing it the wrong way?

This is the equation I am getting:
$ r^{30} \minus{} 7r^{10} \plus{} 6 \equal{} 0$

Letting $ r^{10} \equal{} k$, we still have
$ k^3 \minus{} 7k \plus{} 6 \equal{} 0$

One solution is $ k \equal{} 1$, and another is $ k \equal{} 2$.  However, I am pretty sure both these values contradict the geometric series.

[quote="AdithyaGanesh"]
For number two, I get a 30th degree equation when I try to solve for $ r$.  (Where $ r$ is the common ratio of the geometric sequence.)

Any hints on it?  Am I doing it the wrong way?

This is the equation I am getting:
$ r^{30} \minus{} 7r^{10} \plus{} 6 \equal{} 0$

Letting $ r^{10} \equal{} k$, we still have
$ k^3 \minus{} 7k \plus{} 6 \equal{} 0$

One solution is $ k \equal{} 1$, and another is $ k \equal{} 2$.  However, I am pretty sure both these values contradict the geometric series.[/quote]

[hide="Hint"]
Let the geometric series have first term $ t_{1}$ and common ratio $ r$.
Then, we are given:

$ t_{1}\frac{1\minus{}r^{10}}{1\minus{}r}\equal{}2$     $ (1)$

$ t_{1}\frac{r^{10}\minus{}r^{30}}{1\minus{}r}\equal{}12$     $ (2)$

Just manipulate these equations for the answer. I do not think you set up your equations correctly...

[/hide]

[hide="Solution"]
We want the sum from the $ 31$st through the $ 60$th term, or $ t_{1}\frac{r^{30}\minus{}r^{60}}{1\minus{}r}\equal{}t_{1}\frac{r^{30}(1\minus{}r^{30})}{1\minus{}r}$.

Divide $ (2)$ by $ (1)$ to get $ \frac{r^{10}\minus{}r^{30}}{1\minus{}r^{10}}\equal{}r^{20}\plus{}r^{10}\equal{}6$.

Solving for $ r$, we get $ r^{10}\equal{} 2$, so $ r^{30}\equal{} 8$.

Add $ (1)$ and $ (2)$ to get $ t_{1}\frac{1\minus{}r^{30}}{1\minus{}r}\equal{}14$.

Multiply to get $ t_{1}\frac{r^{30}(1\minus{}r^{30})}{1\minus{}r}\equal{}14*8\equal{}\boxed{112}$.  Alternatively, to solve for the sum, you can solve for $ t_{1}$ and $ r$.
[/hide]


geometry
circumcircle
inequalities
trigonometry
geometry unsolved

let the $M$ be a point in a triangle $ABC$ . let $x, y, z$  be the distances between $M$ and $ A ,B ,C$. and
$u , v , w , $ be the distances between $M$ and $BC, CA , AB$. and $a , b , c $ be the lenghts of sides of triangle
and $S$ its area and $R , r$ be its circumradius and inradius. show that we have these inequalities:

$1$) $ax+by+cz \geq 4S$

$2$) $x+y+z \geq 2(u+v+w)$

$3$) $xu+yv+zw \geq 2(uv+vw+wu)$

$4$) $2( \frac1x+\frac1y+\frac1z) \leq \frac1u+\frac1v+\frac1w$

$5$) $xyz \geq \frac R{2r}(u+v)(v+w)(w+u)$

$6$) $xyz \geq \frac{4R}ruvw$

$7$) $xy+yz+zx \geq \frac{2R}r(uv+vw+wu)$

if someone prove the numbers $4 , 6 $ then simply we can prove $7$

[img]http://img.photobucket.com/albums/v218/hudrea/ineggeojpg.jpg[/img]

Prove that:
$x\geq\frac{bv+cw}a\qquad (*)$
and
$x\geq\frac{cv+bw}a\qquad (**)$
[u][b]Solution:[/b][/u]
$\left( * \right) \Leftrightarrow x \ge \frac{{bv + cw}}{a} \Leftrightarrow \frac{{B'C'}}{{\sin A}} \ge \frac{{bv + cw}}{a} \Leftrightarrow $
$ \Leftrightarrow \frac{{B'C'}}{{\sin A}} \ge \frac{{2Rv\sin B + 2Rw\sin C}}{{2R\sin A}} \Leftrightarrow $
$ \Leftrightarrow B'C' \ge v\sin B + w\sin C \Leftrightarrow $
$ \Leftrightarrow v^2  + w^2  + 2vw\cos A \ge v^2 \sin ^2 B + w^2 \sin ^2 C + 2vw\sin B\sin C \Leftrightarrow $
$ \Leftrightarrow v^2 \cos ^2 B + w^2 \cos ^2 C - 2vw\cos B\cos C \ge 0 \Leftrightarrow $
$ \Leftrightarrow \left( {v\cos B - w\cos C} \right)^2  \ge 0$
that is true.

-------------------------------------------------------------------------------------------------------------------
[color=red]1.)[/color]
It is possible to prove:
$x \ge \frac{{bv + cw}}{a}$

then we have:
$\left. \begin{array}{l}
 ax \ge bv + cw \\
 by \ge cw + au \\
 cz \ge au + bv \\
 \end{array} \right\} \Rightarrow ax + by + cz \ge 2(au + bv + cw) = 4S$
--------------------------------------------------------------------------------------------------------------------
[color=red]2.)[/color]
$\text{the proof of Erd}\ddot{o}\text{s-Mordell inequality follows immediately}$
From:
$\left. \begin{array}{l}
 x \ge \frac{{cv + bw}}{a} \\
 y \ge \frac{{aw + cu}}{b} \\
 z \ge \frac{{bu + av}}{c} \\
 \end{array} \right\} \Rightarrow x + y + z \ge u\left( {\frac{b}{c} + \frac{c}{b}} \right) + v\left( {\frac{c}{a} + \frac{a}{c}} \right) + w\left( {\frac{a}{b} + \frac{b}{a}} \right) \Rightarrow $
$ \Rightarrow x + y + z \ge 2\left( {u + v + w} \right)$
[color=red]Erdos-Mordell inequality[/color]
---------------------------------------------------------------------------------------------------------------------
[color=red]3.)[/color]
From:
$x \ge \frac{{bv + cw}}{a}$
we have:
$\left. \begin{array}{l}
 ux \ge \frac{b}{a}uv + \frac{c}{a}uw \\
 vy \ge \frac{c}{b}vw + \frac{a}{b}vu \\
 wz \ge \frac{a}{c}wu + \frac{b}{c}wv \\
 \end{array} \right\} \Rightarrow ux + vy + wz \ge uv\left( {\frac{a}{b} + \frac{b}{a}} \right) + vw\left( {\frac{b}{c} + \frac{c}{b}} \right) + wu\left( {\frac{c}{a} + \frac{a}{c}} \right) \Rightarrow $
$ \Rightarrow ux + vy + wz \ge 2\left( {uv + vw + wu} \right)$
----------------------------------------------------------------------------------------------------------------------


inequalities
function
geometry
geometric transformation
reflection
Cauchy Inequality

Solve following system

2^x + 2^y = 4
3^x + 3^y = 6

Namdung

My teacher , this is my answer :
Difines : 2^x=a , 2^y=b , c= ln2/ln3 then we have
     a+b=4   and    a^c + b^c = 6
uses the expansion of cauchy inequality : a^c + b^c  \geq 2*((a+b)/2) <sup>c/sup> =6 . then we have a=b=2 . therefore x=y=1 .

I hope my solution can bring in some light:

Multiply both equations like this:

6(2 x  + 2 y ) = 4(3 x  + 3 y )

Then isolate the two variables (like separation of variables):

3 * 2 x  - 2 * 3 x  = 2 * 3 y  - 3 * 2 y

Now divide both sides by 6 to get

2 x-1  - 3 x-1  = 3 y-1  - 2 y-1

In this form we can actually see that LHS is the negative of the RHS so let f(t) = 3 t - 2 t

Clearly f(t) is an increasing function for t \geq 0 and only has one zero when t = 0.

So f(x-1) + f(y-1) = 0 ==> the only solution for x \geq 1 and y \geq 1 is x=y=1.

This result similarly holds for x,y \leq 1 since we can look at the function g(t) = 2 t - 3 t for t \leq 0

Lastly suppose WLOG that x \leq 1 \leq y. Realize that g is the image reflection of f with respect to the origin, so -f(-t) = g(t) and and again we get that x=y=1.

Suggestions and comments are extremely welcomed.


geometry
incenter
inradius
parallelogram
trigonometry
circumcircle
geometric transformation

Dear Mathlinkers,
let ABC be a triangle, A'B'C' the orthic triangle of ABC, DEF the contact triangle of ABC, I the incenter of ABC, A" the meetpoint of AI and B'C', B", C" cyclically...
Prove synthetically that A'A", B'B" and C'C" are concurrent at the orthocenter of DEF.
Sincerely
Jean-Louis

Here is an outline of the proof:

[b]Lemma 1[/b]. Let $ P_1$, $ P_2$, $ P_3$ denote the incenters of triangles $ AB'C'$, $ BA'C'$, and $ CA'B'$, respectively.  Then these points also act as the orthocenters of triangles $ AEF$, $ BDF$, and $ CDE$, respectively.  Furthermore, $ P_1 E \equal{} P_1 F \equal{} P_2 D \equal{} P_2 F \equal{} P_3 D \equal{} P_3 E \equal{} r$, where $ r$ is the inradius of triangle $ ABC$.

[b]Lemma 2[/b]. Quadrilateral $ P_1 XDI$ is a parallelogram, where $ X$ is the orthocenter of triangle $ DEF$.

[b]Lemma 3[/b]. Let $ T_1 T_2 T_3$ denote an arbitrary triangle, and let $ U_1$, $ U_2$, $ U_3$ be points on sides $ T_2 T_3$, $ T_1 T_3$, and $ T_1 T_2$, such that $ T_2 U_1: U_1 T_3 \equal{} c/b$, $ T_3 U_2: U_2 T_1 \equal{} a/c$, $ T_1 U_3: U_3 T_2 \equal{} b/a$, where $ a$, $ b$, and $ c$ are the side lengths of triangle $ ABC$.  Then $ T_1 U_1$, $ T_2 U_2$, $ T_3 U_3$ concur at a point $ V$, and $ T_1 V: VU_1 \equal{} (b \plus{} c)/a$.

[b]Lemma 4[/b]. Triangles $ AB'C'$ and $ ABC$ are similar.  (This is because $ AC' \equal{} b \cos A$ and $ AB' \equal{} c \cos A$.)

Applying Lemmas 3 and 4 to triangle $ AB'C'$, we get $ AP_1: P_1 A'' \equal{} (b \plus{} c)/a$.

By angle bisector theorem, $ C'A'': A''B' \equal{} b/c$, so by Lemma 3, $ A'A''$, $ B'B''$, and $ C'C''$ concur at some point $ X'$, such that $ A'X': X'A'' \equal{} (b \plus{} c)/a$.  Hence, $ P_1 X' \parallel AA'$.  Furthermore, $ AA'/r \equal{} (a \plus{} b \plus{} c)/a$, so $ P_1 X' \equal{} r$.  Therefore, $ X'$ coincides with $ X$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/0c2289870dfd981b4bff59503a0d36a2.png[/img]

$ (P)$ is circumcircle of the right angled kite $ AEIF,$ with diameter $ AI.$ $ AI$ cuts $ BC$ at $ X$ and $ (Q)$ is a circle with diameter $ AX,$ cutting $ CA, AB$ at $ E', F'.$ $ \triangle AB'C' \sim \triangle ABC$ with coefficient $ \cos A$ $ \Longrightarrow$ $ AA'' \equal{} AX \cos A$ $ \Longrightarrow$ $ A''$ is orthocenter of the isosceles $ \triangle AE'F'$ and reflection of $ X$ in $ E'F'.$ Let $ J$ be orthocenter of the isosceles $ \triangle AEF,$ the reflection of $ I$ in $ EF,$ and $ K$ orthocenter of $ \triangle DEF,$ then $ DK \equal{} IJ.$ Circles $ (P), (Q)$ are similar with center $ A$ and coefficient $ \frac {DA'}{XA'} \equal{} \frac {AI}{AX} \equal{} \frac {IJ}{XA''} \equal{} \frac {DK}{XA''}.$ As $ DK \parallel XA''$ (both $ \perp EF$), $ \triangle A'XA'' \sim \triangle A'DK$ are centrally similar by SAS with center $ A'$ and the same similarity coefficient, hence $ K \in A'A''.$ Similarly, $ K \in B'B'', K \in C'C''.$

Nice problem!
Can anyone post solution using barycentric coordinates? I'm not good at barycentric coordinates.

Dear Mathlinkers,
a synthetic proof of this result originated has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis


write an explicit formula for the nth term of each geometric sequence:

250, 100, 40, 16......

[hide]
$ \frac {100}{250} \equal{} \frac {40}{100} \equal{} \frac {2}{5}$
So the nth term is $ \left(\frac {2}{5}\right)(n \minus{} 1)$

Meh recursion.
[/hide]


geometry
geometric transformation
homothety
circumcircle
geometry proposed

Construct a triangle $ ABC$ given $ b\minus{}c$,the median $ m_a$ and the internal bisector $ w_a$

[hide="horrendously ugly bash"]

using stewart's, we can find the following formulas for the lengths of an angles bisector and median in a triangle AB=c, BC=a, AC=b:

$ m_a \equal{} \sqrt {\frac {2b^2 \plus{} 2c^2 \minus{} a^2}{4}}$

$ w_a \equal{} \sqrt {ab(1 \minus{} \frac {c^2}{(a \plus{} b)^2}}$

then using these two equations with $ b \minus{} c$ we can solve for a,b,c i think[/hide]

[hide="alternatively, just an idea:"]

In a triangle with incircle I, let I touch BC at D and let M be diametrically across D on the incircle. Let N be intersection of AD and BC. we can find the ND = b -c using the homothety that takes the excircle across A to the incricle and equal tangents[/hide]

edit: [hide] also note that (if b > c) BN = DC from equal tangents stuff, so that one end of the median should intersect the segment ND at its midpoint[/hide]

Length $w_a$ can  be expressed as $ {w_a}^2 \equal{} AB \cdot AC \minus{} BV \cdot CV,$ where $ V$ is the foot of the A-angle bisector. Since $BV \cdot CV$ is the power of $V$ WRT the semicircle $(M)$ with diameter $\overline{BC},$ we rewrite ${AV}^2 \equal{} AB \cdot AC \minus{} \frac {_1}{^4}BC^2 + VM^2.$

$\Longrightarrow  {AV}^2\equal{}\frac{_1}{^2}(AC^2\plus{}AB^2)\minus{}\frac{_1}{^2}(AC\minus{}AB)^2\minus{} \frac {_1}{^4}BC^2 + VM^2 \ \ (1)$

By Apollonius theorem ${AM}^2 \equal{} \frac {_1}{^2}(AB^2 \plus{} AC^2) \minus{} \frac {_1}{^4}BC^2 \ \ (2)$

Combining $(1)$ and $(2)$ gives $ (AC \minus{} AB)^2 \equal{} 2({AM}^2 \minus{} {AV}^2 \plus{} VM^2),$

which permits to construct $\triangle AMV$ with three known sides, then it remains to determinate the vertices $B$ and $ C.$ Note that circle with center $ M$ and radius $ \frac {_1}{^2}|AC \minus{} AB|$ cuts $ AV$ at the orthogonal projections of $B,C$ onto $\overline{AV}.$ Hence, this gives the construction of $B,C$ completing the triangle $\triangle ABC.$

nice solution!

Is there a nice geometry proof of how to get $ bc\minus{}BW \times WC \equal{} w_a$?

Ray $ AV$ cuts circumcircle $(O)$  at $ P.$ From $\triangle ABV \sim \triangle APC$ we get

$ AP \cdot AV \equal{}AB \cdot AC \Longrightarrow (AV\plus{}PV)AV\equal{}AB \cdot AC$

But $ AV \cdot VP\equal{}BV \cdot CV \Longrightarrow AV^2=AB \cdot AC-BV \cdot CV$


modular arithmetic
number theory
Diophantine equation

Hi,

I don't understand the solution to this in the wiki: http://www.artofproblemsolving.com/Wiki/index.php/2000_AMC_12_Problems/Problem_16

Here's the problem:

A checkerboard of 13 rows and 17 columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered 1,2,\ldots,17, the second row 18,19,\ldots,34, and so on down the board. If the board is renumbered so that the left column, top to bottom, is 1,2,\ldots,13,, the second column 14,15,\ldots,26 and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).

Explain of the solution in there, or an alternate solution are welcome.

Thank you! :)

[hide="Solution"]For the first numbering, the number in $ (x,y)$ (the $ x^\text{th}$ row and $ y^\text{th}$ column) is $ 13(x-1)+y$, since the number increases by 13 each row we go down and 1 each column we go across. Similarly, for the second numbering, the number in $ (x,y)$ is $ 17(y-1)+x$.

To find squares that have the same numbering in both systems, we equate the expressions: $ 13(x-1)+y=17(y-1)+x$. Simplifying gives $ 4y=3x+1$. Since $ x$ and $ y$ are integers, $ 3x+1\equiv0\pmod4$, so $ 3x\equiv3\pmod4$, or $ x\equiv1\pmod4$. Since $ 1\leq x\leq17$, $ x\in\{1,5,9,13,17\}$. Since $ y$ increases by 3 for every increase of 4 in $ x$, and when $ x$ is 1 $ y$ is 1, we know $ y\in\{1,4,7,10,13\}$ (because $ 1\leq y\leq13$).

Using the first ordering to determine the number in these squares, the number in $ (x,y)$ is $ 13(x-1)+y$. Call the five possible values of $ x$ $ x_1,x_2,\ldots,x_5$ and the five possible values of $ y$ $ y_1,y_2,\ldots,y_5$. Thus, the sum of the values in the squares is
\begin{eqnarray*}S&=&13(x_1-1)+y_1+13(x_2-1)+y_2+\cdots+13(x_5-1)+y_5\\
&=&13(x_1+x_2+\cdots+x_5-5)+y_1+y_2+\cdots+y_5\\
&=&13(9\cdot5-5)+7\cdot5\\
&=&\boxed{555}\end{eqnarray*}

This is basically the solution in the wiki, except for the computation at the end. Let me know if you have any more questions.[/hide]

Explaining the solution given:
Basically they just constructed an equation to find the value of any square.

They are saying that for arrangement I, the value of the square $ (x, y)$ in row $ x$ and column $ y$ is $ 17(x\minus{}1) \plus{} y$.  With similar logic, the value of the square $ (x, y)$ in row $ x$ and column $ y$ is $ 13(y\minus{}1) \plus{} x$.

We now set the two equations equal:
$ 17x\minus{}17\plus{}y\equal{}13y\minus{}13\plus{}x \implies 16x\minus{}12y\equal{}4 \implies 4x\minus{}3y\equal{}1$.

The pairs of integers that fulfill the Diophantine equation are $ (1,1),(4,5),(7,9),(10,13),(13,17)$, and we just plug the values into any of the equations, add them up, and get $ \fbox{555}$.


calculus
integration
trigonometry
inequalities
calculus computations

For a constant $a,$ let $f(x)=ax\sin x+x+\frac{\pi}{2}.$ Find the range of $a$ such that $\int_{0}^{\pi}\{f'(x)\}^{2}\ dx \geq f\left(\frac{\pi}{2}\right).$

hello kunny, we have
$ f(x) \equal{} ax\sin(x) \plus{} x \plus{} \frac {\pi}{2}$
from here we get
$ f'(x) \equal{} a\sin(x) \plus{} ax\cos(x) \plus{} 1$
and our integral is
$ \int_{0}^{\pi}(a\sin(x) \plus{} ax\cos(x) \plus{} 1)^2\,dx \equal{} \pi \plus{} \frac {a^2\pi}{4} \plus{} \frac {a^2\pi^3}{6}$
further
$ f(\frac {\pi}{2}) \equal{} \frac {a\pi}{2} \plus{} \pi$
so we have to solve the following inequality
$ \frac {a^2\pi}{4} \plus{} \frac {a^2\pi^3}{6}\geq \frac {a\pi}{2}$
and a computation gives us
$ \minus{} \infty < a\le 0 \cup \frac {6}{3 \plus{} 2\pi^2}\le a < \infty$
Sonnhard.


The code

draw(circle((0,0),1));

produces
[asy]draw(circle((0,0),1));[/asy]

If I replace circle by Circle, everything works:
[asy]draw(Circle((0,0),1));[/asy]
What's up? (both are legitimate Asymptote commands; the only difference is that circle is slightly less accurate than Circle) :?

Also cse5 commands seem to be completely disabled :(

In fact I think only asymptote images that have been rendered before will render now. When I try to copy an existing asy code, the image shows up correctly, but if I change anything at all it doesn't work. For example, I copy fedja's code and change something minor

[code][asy]
draw(Circle((0,0),1));
[/asy]
[asy]
draw(Circle((0,0),1.0));
[/asy]
[/code]

[asy]draw(Circle((0,0),1));[/asy]
[asy]draw(Circle((0,0),1.0));[/asy]

Thanks for reviving Asymptote, Valentin!  :first: By the way, if not a secret, which version of it does AoPS use? (or you always upgrade to the latest one?)

All right problem should be now permanently solved (there was a path issue involved to the asymptote executable).


1)Find three different two-digit prime numbers such that the average of all three primes is also prime and the average of any two of the primes is also prime.
2)What is the sum of all the digits needed to write each counting number from 0 through 1,000,000? For example, the sum of all the digits needed to write each counting number from 18 through 23 would be 1+8+1+9+2++2+1+2+2+3=33.

 :rotfl:  :rotfl:
 :rotfl:

Please make one post per problem.  The first of these problems is much harder than the second and belongs in a different forum.  Thanks.

[quote="fireemblem13"][quote="nonie"]Is there an easy way to do the second one??[/quote]

easier than what?

i would just for every digit: ones, tens, hundreds..., multiply 45 by as many numbers that included that digit for every digit. it's kind of long though

BTW, i lost my glasses so I have to lean towards the computer to read what's on the monitor and that's kinda uncomfortable.[/quote]

Sorry about your glasses.

About the problem, there's an easier and combinatoric way.... just think about it!

Hello this is my first post. I attemped to solve the (b) part. I hope it is correct. I am new here and I would like someone to advice if I write correct. What I mean is whether I right too mush details in my solutions or if I nedd to write in more detailed and that sort of things.

[hide]b) Let x=1+2+3+4+5+6+7+8+9
The sum of the digits From 0 to 9 = 1+2+3+...+9=x
-From 10 to 19 = 1+0+1+1+1+2+...+1+9=10*1+x
-Form 20 to 29 = 20+x and so on
=> From 0 to 99 = x+(10+x)+(20+x)+... (90+x)=10x+10(1+2+3+...+9)=20x
-From 100 to 199 = 1+1+1+...+1+ (From 0 to 99) =100+20x
-From 200 to 299 = 200+20x and so on
=> From 0 to 999 = 20x+(100+20x)+(200+20x)+... (900+20x) = 200x+100x=300x
=> From 0 to 9,999 = 10(300x)+1000x=4000x
=> From 0 to 99,999 = 40000x+10000=50000x
=> From 0 to 999,999 = 500000x+100000=600000x
=> From 0 to 1,000,000 = 600000(45)+1=[b]27000001
[/b][/hide]

[quote="bedakilla"]Ugh I hate to do the second problem but why hasn't anyone solved the first problem yet??
 :huh:
Here's answer to the first problem  ;)  I might be wrong (I don't think thought  :) )
[hide]

Answer is $11, 47, 71$ because
1. They are all Primes I think... :P.
2.if you average $11$ and $47$ you get $29$ which is also prime (I think).
3.if you average $11$ and $71$ you get $41$ which is prime.
4.if you average $47$ and $71$ you get $59$ which is prime.

5.And finally average of $11,47,71$ is $43$ which is also prime.
[/hide]
First one wasn't THAT hard  ;)

(if I got the answer wrong I'd be very embarrassed from posting this reply :? )[/quote]

Do you have a solution?? Guess-and-Check all 1330 possible combinations?

[quote="lotrgreengrapes7926"] Guess-and-Check all 1330 possible combinations?[/quote]

Obviously not


Let a,b,c be positive  real numbers such that $ a\plus{}b\plus{}c \ge \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$.  Prove that:  $ a\plus{}b\plus{}c \ge  \frac{1}{a\plus{}b\plus{}c}\plus{} \frac{2}{abc}$ .

$ A\equal{}a\plus{}b\plus{}c \ge \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}$

From Andreescu we get that;
$ A\ge \frac{9}{a\plus{}b\plus{}c}$

So  $ a\plus{}b\plus{}c \ge \frac{3}{a\plus{}b\plus{}c}\plus{} \frac{2}{abc}$  $ \Leftrightarrow$

     $ \frac{9}{a\plus{}b\plus{}c} \ge \frac{3}{a\plus{}b\plus{}c}\plus{} \frac{2}{abc}$  $ \Leftrightarrow$

$ 3abc\ge a\plus{}b\plus{}c$  $ (1)$

But the condition is also equivalent to;
$ a\plus{}b\plus{}c\ge \frac{ab\plus{}bc\plus{}ca}{abc}$
From Chebychev;  $ ab \plus{}bc\plus{}ca\ge\frac{(a\plus{}b\plus{}c)^2}{3}$  so;

$ a\plus{}b\plus{}c\ge\frac{(a\plus{}b\plus{}c)^2}{3abc}$  $ \Leftrightarrow$
$ 3abc\ge a\plus{}b\plus{}c$

Which means that $ (1)$ holds and thus the proof is complete.

Are you sure cause $ ab \plus{}bc\plus{}ca\ge\frac{(a\plus{}b\plus{}c)^2}{3}$ is real?

From Chebychev;

$ \frac{ab \plus{}bc\plus{}ca}{3}\ge \frac{a \plus{}b\plus{}c}{3}\frac{b\plus{}c\plus{}a}{3}$ $ \Leftrightarrow$

$ ab\plus{}bc\plus{}ca\ge\frac{(a\plus{}b\plus{}c)^2}{3}$

Follows directly from the theorem.

One sequence is increasingand one is decreasing.And Chebyshev Inequality become $ n\sum_{cyc}a_ib_i\leq\sum_{cyc}a_i\sum_{cyc}b_i$. Or  $ ab \plus{}bc\plus{}ca\leq\frac{(a\plus{}b\plus{}c)^2}{3}$  because  $ 3(ab\plus{}bc\plus{}ca)\leq a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca)$  and  $ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca$ (true...). Is ok?

Well i considered the triplets $ (a,b,c)$ and $ (a,b,c)$ and because of symmetry we can assume that $ a\ge b \ge c$ and again $ a\ge b\ge c$ and because they are both decreasing the inequality becomes;

$ \frac {1}{n}\sum_{cyc}a_ib_i\ge(\frac {1}{n}\sum_{cyc}a_i)(\frac {1}{n}\sum_{cyc}b_i)$

In order to apply it to $ ab \plus{} bc \plus{} ca$. That's how i justify my result, don't know if i missed anything obvious.

EDIT: I have realized the error, even tho Chebychev led me otherwise :oops: . Thanks for your observation friend, but now it is really late, i will reconsider my proof tomorrow.

THe inequality is indeed very weak.
We know by AM GM that $ (a\plus{}b\plus{}c)(1/a \plus{} 1/b \plus{} 1/c) \ge 9$ and hence $ a\plus{}b\plus{}c \ge 3$ and now we try to prove that $ 3\ge RHS$ of the given inequality which is very easy.

what do you mean by the inequality is very weak?is the problem easy or is the inequality not true for some real numbers?

[quote="baleanu"]Let a,b,c be positive  real numbers such that $ a \plus{} b \plus{} c \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$.  Prove that:  $ a \plus{} b \plus{} c \ge \frac {3}{a \plus{} b \plus{} c} \plus{} \frac {2}{abc}$ .[/quote]
This inequality very nice :)
Since $ a\plus{}b\plus{}c \ge \frac{1}{a} \plus{}\frac{1}{b}\plus{} \frac{1}{c}$.
We have: $ a\plus{}b\plus{}c \ge \frac{9}{a\plus{}b\plus{}c} \equal{}> a\plus{}b\plus{}c \ge 3$  (1)
and$ abc(a\plus{}b\plus{}c) \ge ab\plus{}bc\plus{}ca$
On the orthe hand $ ab\plus{}bc\plus{}ca \ge \sqrt{3(abc(a\plus{}b\plus{}c)}$
So, $ abc(a\plus{}b\plus{}c) \ge \sqrt{3abc(a\plus{}b\plus{}c)} \equal{}> abc(a\plus{}b\plus{}c) \ge 3$  (2)
Since (1) we have:$ a\plus{}b\plus{}c \minus{}\frac{3}{a\plus{}b\plus{}c} \ge \frac{2(a\plus{}b\plus{}c)}{3}$
I will must prove that:
$ \frac{2(a\plus{}b\plus{}c)}{3} \ge \frac{2}{abc}$
$ <\equal{}> abc(a\plus{}b\plus{}c) \ge 3$
Which is obvious true since (2)  :)

[quote="Agr_94_Math"]THe inequality is indeed very weak.
We know by AM GM that $ (a \plus{} b \plus{} c)(1/a \plus{} 1/b \plus{} 1/c) \ge 9$ and hence $ a \plus{} b \plus{} c \ge 3$ and now we try to prove that $ 3\ge RHS$ of the given inequality which is very easy.[/quote]
you can post solution :oops:

[quote="Papalakis"]From Chebychev;

$ \frac {ab \plus{} bc \plus{} ca}{3}\ge \frac {a \plus{} b \plus{} c}{3}\frac {b \plus{} c \plus{} a}{3}$ $ \Leftrightarrow$

$ ab \plus{} bc \plus{} ca\ge\frac {(a \plus{} b \plus{} c)^2}{3}$

Follows directly from the theorem.[/quote]
What ?  :huh:
How does that follow from Chebychev?
Try the numbers $ a\equal{}1,b\equal{}c\equal{}2$. Then $ ab\plus{}bc\plus{}ca \equal{} 6$ and $ \frac{(a\plus{}b\plus{}c)^2}{3} \equal{} \frac{25}{3} > 6$.... In fact we have $ ab \plus{} bc \plus{} ca\le\frac {(a \plus{} b \plus{} c)^2}{3}$

[quote="geniusbliss"][quote="Mathias_DK"][quote="baleanu"]Let a,b,c be positive  real numbers such that $ a \plus{} b \plus{} c \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$.  Prove that:  $ a \plus{} b \plus{} c \ge \frac {1}{a \plus{} b \plus{} c} \plus{} \frac {2}{abc}$ .[/quote]
Using the condition we have to prove:
$ \frac {\sum_{cyc} ab}{abc} \ge \frac {3}{\sum_{cyc} a} \plus{} \frac {2\sum_{cyc} a}{\sum_{cyc} ab}$. Now $ \left(\sum_{cyc} ab \right)^2 \ge 3abc \left( \sum_{cyc} a \right )$ is wellknown, so we have:
$ \frac {3\sum_{cyc} a}{\sum_{cyc} ab} \ge \frac {3}{\sum_{cyc} a} \plus{} \frac {2\sum_{cyc} a}{\sum_{cyc} ab} \iff \sum_{cyc} a^2 \ge \sum_{cyc} ab$, and we are done :)[/quote]
 i am sorry but i dont quite follow the first step mathias_dk
can u elaborate plz?[/quote]
Of course. By the condition we have $ a \plus{} b \plus{} c \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {ab \plus{} bc \plus{} ca}{abc}$. And we also have $ \frac {2}{abc} \le \frac {a \plus{} b \plus{} c}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \cdot \frac {2}{abc} \equal{} \frac {2(a \plus{} b \plus{} c)}{ab \plus{} bc \plus{} ca}$ :)

[quote="Mathias_DK"][/quote][quote="geniusbliss"][quote="Mathias_DK"][quote="baleanu"]Let a,b,c be positive  real numbers such that $ a + b + c \ge \frac {1}{a} + \frac {1}{b} + \frac {1}{c}$.  Prove that:  $ a + b + c \ge \frac {1}{a + b + c} + \frac {2}{abc}$ .[/quote]
Using the condition we have to prove:
$ \frac {\sum_{cyc} ab}{abc} \ge \frac {3}{\sum_{cyc} a} + \frac {2\sum_{cyc} a}{\sum_{cyc} ab}$. Now $ \left(\sum_{cyc} ab \right)^2 \ge 3abc \left( \sum_{cyc} a \right )$ is wellknown, so we have:
$ \frac {3\sum_{cyc} a}{\sum_{cyc} ab} \ge \frac {3}{\sum_{cyc} a} + \frac {2\sum_{cyc} a}{\sum_{cyc} ab} \iff \sum_{cyc} a^2 \ge \sum_{cyc} ab$, and we are done :)[/quote]
 i am sorry but i dont quite follow the first step mathias_dk
can u elaborate plz?[/quote][quote="Mathias_DK"]
Of course. By the condition we have $ a + b + c \ge \frac {1}{a} + \frac {1}{b} + \frac {1}{c} = \frac {ab + bc + ca}{abc}$. And we also have $ \frac {2}{abc} \le \frac {a + b + c}{\frac {1}{a} + \frac {1}{b} + \frac {1}{c}} \cdot \frac {2}{abc} = \frac {2(a + b + c)}{ab + bc + ca}$ :)[/quote]

er..sorry mathias i was a fool to ask
p.s. nice solution

[quote="geniusbliss"][quote="Agr_94_Math"]THe inequality is indeed very weak.
We know by AM GM that $ (a \plus{} b \plus{} c)(1/a \plus{} 1/b \plus{} 1/c) \ge 9$ and hence $ a \plus{} b \plus{} c \ge 3$ and now we try to prove that $ 3\ge RHS$ of the given inequality which is very easy.[/quote]

to prove $ a \plus{} b \plus{} c \ge \frac {3}{a \plus{} b \plus{} c} \plus{} \frac {2}{abc}$
since $ a \plus{} b \plus{} c \ge 3$
substituting the expression in the $ RHS$ with a lesser value in the denominator to make the fraction bigger,
we get
$ 3 \ge \frac {3}{3} \plus{} \frac {2}{abc}$
or,to prove$ abc \ge 1$
which can be done by proceeding similarly as by quykhtn-qa1
so am i right?  :P[/quote]
Why $ abc \ge 1$ :wink:

er.. i am sry i dint seee the loophole thanx for noticing quykhtn-qa1!
yes  $ abc \ge 1$ is not neccesary


Which digit will appear in the 534th place after the decimal point in the decimal representation of $ \frac{5}{13}$?

Well, you would have to first find the decimal representation, which is:

$ 0.\bar{384615}$.

Since $ 534 \text{ mod} 6$ is $ 0$, the answer is $ \boxed{5}$.

Dang, beaten to it.  And how do you make the bar over the whole entire thing?

I used:

[code]\bar{384615}[/code]

What am I doing wrong?

I don't think there is a way to get the bar over the number.

@Jongao

In the future, can you please explain in more detail, like ernie did? Thanks.

Sorry if this counts as spam, I don't think it does.

ernie, for questions about latex, plz post those in the latex section of site and general information
however, since this was quick and concerning the prob (sorta) then i guess it's ok

[quote="Jongao"]the pattern is .384615..., so the answer must be 5.[/quote]
Can't you find a pattern?

That is the pattern. The decimal goes on like 0.384615384615384615...


The anniversay AHSME is a great source of problems, but there appear to be two different versions of it:

http://www.unl.edu/amc/a-activities/a7-problems/a7-1-archives/1999-pa/anniver.pdf
http://www.math.uncc.edu/~hbreiter/AHSME/anniver.pdf

Not only are some of the questions different, but they disagree on the answers of their 1960 and 1959 problems (which are exactly the same in both versions). I don't have the skill to verify which one is correct, so could someone tell me which one I should trust more? Thanks!

[quote="EFuzzy"]...could someone tell me which one I should trust more? Thanks![/quote]
Based on the 1959 and 1960 problems, I suggest you trust Professor Reiter's version.


AMC
AMC 8
AMC 10
AMC 12
AIME
USA(J)MO
USAMO

I was wondering what people are aiming to achieve. First, post whether you are in elementary school (grades 1-5), middle school (grades 6-8), high school (grades 9-12), university, or other. Then, post your goals.

I am in middle school. My goals are:

MC Nats: Top 150
Math Kangaroo: 110
Purple Comet: 10 (individual)
AMC8: 24
AMC10: 100
AMC12: 70

I'm in high school.

AMC 12: 150

AIME: 13+

USAMO: 21

Mandelbrot: Make first leaderboard tier

ARML: help team win again

Middle School(8th grader next year)

My goals:

[u]MC[/u]
School: 46/46

Chapter: CD

State: Make Nats

Nats: Top 50

[u]AMC[/u]
AMC8:25/25

AMC10: Don't fail(<120)

AIME: 5

USAMO(With the very small chance I'll make it): A positive score

As for every other competition(list below), my goal is to not fail, the criteria of which im to lazy to post.
Other Competitions:
Purple Comet
ASMA
SIGMA
NEML
IMLEM

High School (yay, skipping freshman year. Gonna be a sophomore!) :D

Goals:

AMC 10: 150
AMC 12: 130+
AIME: 9+
USAJMO: 10+
USAMO: 5+, hopefully make CGMO and Blue MOP if possible
Math Prize for Girls: Top 10 (and therefore win some money)
ARML: 7+ Individual, Make Chicago Team A
Mandelbrot: 1st tier leaderboard, I guess?
NSML (North Suburban Math League, local math contest): Perfect score on all meets, win orals contests
HMMT: idk if I'll do it
USAMTS: 70+, or hopefully silver
ICTM: 1st place state, win orals

going into 9th

MC- awww not doing it anymore
AMC10- 134
AMC12- 105
AIME- 5
Mandelbrot- make 3rd tier national level
UNM PNM- 1st in division, top 10 overall, 46+
ARML- 3
iTest- 80
Purple Comet HS- 14
Mandelbrot Team- 21
USAMTS- actually submit my solutions!

8th grade next year

See my sig.

FYI I won't do AMC 8 (boring).

Well if you want to see my goals, they are in my BLOG. My BLOG also includes many cool things like problems and quotes of the day. Visit my BLOG.

\shameless plug

Anyways, here are my goals (going to 8th)

OUTRAGEOUSLY IMPOSSIBLE GOALS-
Make Mathcounts Nationals Countdown
Win Mathcounts Nationals (hey, a guy can dream right?)
Make USAMO
Make the International Mathematical Olymp.........I wont even go there.
Get a 150 on the AMC 10a/b
Get a 150 on AMC 12 a/b

NORMAL GOALS-
Make it to Mathcounts Nationals and place in top 50 (last year FAILED states)
Make AIME and score decent (5+) (last year 3  :( )
Get a 138+ on AMC 10a/b (last year 120)
Get a 100+ on AMC 12 a/b (didnt take it last year)
Get a perfect MOEMS (last year 25 :lol: )
Get a perfect on AMC 8 (last year 24.... :mad:)

@ PoP: Dont aim for 24 on the AMC 8! Im sure if you studied Mathcounts tests over the summer, you can get a perfect with your eyes closed!

[quote="Thunder365"]

@ PoP: Dont aim for 24 on the AMC 8! Im sure if you studied Mathcounts tests over the summer, you can get a perfect with your eyes closed![/quote]

except for calculation errors :wacko:  :oops_sign:  :ewpu:

Ninth grade goals:
AMC 10: 150 (made in 8th grade)
AMC 12: 141+ (closest was 133.5)
AIME: 10+ (closest was 8)
USAMO: + (closest was 0)
USAMTS: 75+ (first year!)
HMMT: trophy
PUMaC: Make final on something.
NYSML: 10
ARML: 6+ (closest was 4)
Mandelbrot: Convince club to take it.

Y is the NYSML goals so much higher than the ARML goal?  Aren't they like the same thing?

NYSML is easier.

Been getting one away from perfect for THREE YEARS, and my best ARML score was a 4.

[quote="1=2"]Ninth grade goals:
Mandelbrot: Convince club to take it.[/quote]

 :rotfl:  :rotfl:  :rotfl:

Eh, I'll fail all my goals again, but whatever. (When I mean a perfect, I add more points)

AMC8 - 26 [24 argh argh argh]
AMC10 - 156 [150 :)]
AMC12 - 144 [138 what the]
AIME - 12 [10. failish]
USAMO - MOP [ow]
Mathcounts - 47 (school), 47 (chapter), 47 (state), CD (nats) [44 (what!?), 46, 45, fail]
ARML - 10 [9, notfail]
SMT - combined score 21 (max = 20) [18. um]
Purple Comet - Perfect + 1

AMC 8: 25
AMC 10: 150
AMC 12: 135-ish
AIME: 10+
USAJMO (if qualified): 30+? I don't know the exact difficulty, but if they really are like hard AIME problems I'll be fine,
USAMO (if qualified): 2+ :P

School: 46 (win)
Chapter: 45 (win)
State: 46 (top 3)
Nats: 43 (make countdown, do not fail too much)

ARML: 7

USAMTS: 70+

To james4l: If Silas isn't going to be at Stanford next year, who's going to run SMT?

No idea. Although I'm pretty sure Stanford still has enough math students to sustain the program, right?

Well, seeing as the USAJMO has never been administered as of now, what do you expect? You should probably practice with ending AIME problems/some JBMO problems (I believe that some are easier are than USAMO problems, although I personally find them more or less at USAMO #1 level  :P )/easier USAMO problems...

Also, it would've been better to ask that question in the AMC forum. Perhaps you would have seen [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691]this[/url] sticky...

I'm in middle school. Goals;

AMC 8: 23+
AMC 10: 20+
AMC 12: 18+
Qualify for AIME
AIME: 5+
Finish reading;
AoPS Books

hey guys I have this urge to share personal math competition goals with all of you, without even a pretense of stating them except the apparent but by mutual agreement unstated one of showing off that I'm amazing enough at math to have relatively high goals.

@game.guy: AMC10 and AMC12 are out of 150 not 25. You probably meant number of questions right, but what about blanks?

darn, fwolth's post describes me (and probably everyone else who posted) perfectly.

hm uh anyway

MATHCOUNTS: go to florida, as an assistant coach or something. i dunno. and make sure ed atleast gets top 30.
AMC10: 135+ (eh, failed last year... i didn't even make AIME ;__;)
AMC12: 120+ (blah i'll probably only take the AMC10 though cause our school doesn't offer the B test, but i'll try to take both... somehow)
AIME: 8+
USAJMO: positive score
USAMO (okay, i'd be super surprised if i even made this): positive score
ARML: convince my parents to drive me to lehigh (which is like 70 miles from my house), and get at least a 4+and not be on the last team
USAMTS: hm/bronze, and atleast a 15+ on this round
PUMAC: set up a team somehow, and get top 10 in atleast one thing (B division of course :P)

SAT: 2401
ACT: 37
AP: 6
what other standardized tests are there?

shoot i just noticed the last post was a few weeks ago.

[quote="Sephiroth"]how is this a discussion of games?[/quote]

simple, the game of math.

7th grade, middle school, skipped fifth

[b]AMC8[/b]: Get 22/25 or higher, highest I got was 20/25, 6th grade  :|
[b]MATHCOUNTS[/b]: Get into primary team and score 3rd or higher out of the 4 members
[b]AMC10[/b]: Move on
[b]NSF Math Bee[/b]: Pwn everybody :D
[b]USAMTS[/b]: Get at least 5 Questions Right out of All 20
[b]Algebra 1 Math Class[/b]: Nothing more than boredom  :)

if this helps you figure out how i operate...  :rotfl:

6th grade, middle school

AMC8: 23+

MathCounts: get to nats]

MOEMS: get at least 20

ICAE: get to class IIB2

6th grade horrendously impossible goals
MC: Nats Countdown
AMC 8: Ace it
AMC 10: 144
AIME: 8 / Make USAJMO???

Middle School

Yeah, I am going to be an odd ball and say something completely different and say...

1. OM(Odyssey of the Mind)- Make into world and be in the top 20 (considering it's world...)
2. Become a theoretical physicist

6th grade, still elementary school in our district unlike some others.

AMC 8 - Get a score of 20/25 at the least.
MATHCOUNTS - Try to get into a team or individual and go to state
AMC 10 - (From AMC 8) Try to get at least 15 correct.
AoPS - Do well in Intro to Geometry in the summer
School - Do well on test today.  :o

Extra - Get 5 stars by the end of March. :P

[quote="Scamper"]
6th grade, still elementary school in our district unlike some other[/quote]
Hi five! Same in our district :D

Is this a huge revival of a long dead thread?

BTW: Goals: Try to make Countdown in MC State.

Apparently so

I'm going into 8th grade

My goals:

AMC 8: 25
AMC 10: 100
Mathcounts: 40+ at school and chapter.


Find all integer numbers $a$ and $b$, for which there exist polymonial $W(x)$ (with integer coefficients), such that polymonial $P(x)=(x^{2}+ax+b)W(x)$ has coefficients equal to $1$ or $-1$

This is problem A1, IMO Shortlist 2005. I don't know if it's appropriate to post solutions at this time.
By the way, the proposer is Dr. Laurentiu Panaitopol, Romania.

I direct discussion here:

http://www.mathlinks.ro/Forum/viewtopic.php?p=475370#p475370