Este pacat ca forumul din Romania nu este atat de solicitat...ar fi chair bine daca la interval de o saptamana sau cateva zile sa posteze cineva un set de 3 4 probleme frumoase...ca  pe urma sa se discute pe tema lor :D . O sa incerc sa postez eu acum un set.

Iata un set de 3 probleme care un elev de clasa a 6a ar avea cam aceleasi sanse sa le rezolve ca un elev de liceu:
[b]1.[/b] Fie patratul ABCD si E mijlocul lui BC.Punctul M din interiorul patratului are proprietatea ca <MAB=<MBC=<EMB=x.Aflati x.
[b]2.[/b] Fie triunghiul ABC isoscel(AB=AC), <A=80,M in interiorul triunghiului ABC astfel incat <MBC=30 si <MCB=10.Aflati <AMC.
[b]3.[/b] In triunghiul ABC, <B=15, <C=30, seprelungeste AB cu AD=AB.Sa se afle <ADC.

nimeni? prima este chiar simpla... :lol:

[b]1.[/b] [hide]Cum triunghiul MBE este isoscel rezulta ca $ BE\equal{}ME\equal{}a$.Cum $ \angle ABM\equal{}90\minus{}x$ si $ \angle BAM\equal{}x \equal{}> \angle ABM\equal{}90$.Fie N mijl lui AB=> NBEM e patrat de unde rezulta ca $ x\equal{}45$[/hide]

[b]2.[/b] [hide]Ducem AR_|_BC si fie Q intersectia dintre BM si AR, $ Q \in AR$ de unde QBC isoscel de unde $ \angle QCM\equal{}90\equal{}\angle ACQ$ $ \Delta AQC \equiv \Delta QCM$ de unde AC=MC => $ \angle AMC\equal{}70$[/hide]...va continua...tb sa ma duc la scoala[/url]

[quote]Cum $ \angle ABM\equal{}90\minus{}x$ si $ \angle BAM\equal{}x \equal{}> \angle ABM\equal{}90$[/quote]
Ar trebui sa fie $ \angle AMB\equal{}90$.

[quote="bike_me"][quote]Cum $ \angle ABM \equal{} 90 \minus{} x$ si $ \angle BAM \equal{} x \equal{} > \angle ABM \equal{} 90$[/quote]
Ar trebui sa fie $ \angle AMB \equal{} 90$.[/quote] Intradevar ....ai dreptate

Problema 1

In MBC,  mediana (ME) este jumatate din (BC), deci $\widehat{BMC}=90^\circ$. Dupa cum a demosntrat si Razvan  $\widehat{BMA}=90^\circ$, de unde A,M,C coliniare.
$BM\perp AC;\ BD\perp AC \Rightarrow $ B,M,D coliniare, de unde $x=\widehat{MBC}=45^\circ$.

Problema 2

Desi o solutie trigonometrica nu pare greu de dat, o sa prezint una sintetica momentant:
Fie $ {E}=BM\cap AC $ si F un punct  $\in Ext\triangle{ABC}$ a.i. F si C sunt in semiplane diferite determinate de dreapta AB si $\triangle{FAB}$ echilateral.
Din unghiuri, $BE=BA\ si\ BM=AE$.
Rezulta $\triangle{FBM}\equiv \triangle{BEA}\ (L.U.L.)\Rightarrow \widehat{MFA}=40^\circ$. Dar $FM=FA$, $\widehat{FMA}=70^\circ$. Deci, $\widehat{AME}=30^\circ$ de unde conchidem ca $\widehat{AME}=70^\circ$.

Problema 2, solutia trigonometrica

Fie $a=AB=AC$.
Din teorema sinusurilor in ABC, $BC=\frac{a\cdot\sin{80^\circ}}{\sin{50^\circ}}$

Din teorema sinusurilor in BMC, obtain  $MC=\frac{a\cdot\sin{80^\circ}\sin{30^\circ}}{\sin{40^\circ}\sin{50^\circ}}=a$

$MC=a=AC;\ \widehat{MCA}=40^\circ \Rightarrow \widehat{AMC}=70^\circ$

Problema 3

Fie $E$ mijlocul lui $BC$ si $F\in[CA$ a.i. $BF\perp AC$

Din unghiuri, $FE=FA$ si $\widehat{EFA}=30^\circ$ de unde$\widehat{AEF}=75^\circ$, deci $\widehat{AEB}=135^\circ$. Dar A este mijlocul lui BD, E este mijlocul lui BC, deci $AE\parallel CD$, de unde $\widehat{BCD}=\widehat{BEA}=135^\circ$.
$\widehat{DBC}=15^\circ; \widehat{BCD}=135^\circ \Rightarrow \widehat{CDA}=30^\circ$

Frumoase problemele, dar foarte cunoscute.

Pr. 2:
Notam $E\in AC\cap BM,$ luam $F$ pe $(ME$ asa ca $CF=CE$.
Nu credeti ca se poate aplica urmatoarea [b]problema ajutatoare[/b]?

[b][color=#4000FF]Pe latura $EM$, prelungita, a triunghiului isoscel $\Delta MEC$ cu $\widehat{MEC}=100^\circ$ se ia punctul $B$ asa ca $\overline{BM}=\overline{CM}-\overline{ME}$ ($\overline{BE}=\overline{CM}$), atunci $\widehat{MBC}=30^\circ, \widehat{BCM}=10^\circ$.[/color][/b]

[i]Demo:[/i]$\Delta CEF$ e un triunghi [b]Langley[/b] ($80^ \circ - 20^\circ - 80^\circ$). Se ia $P$ pe $\overline {CM}$ asa ca triunghiul $CFP$ sa fie echilateral, deci $P$ este circumcentrul $\Delta BCF$ si $BP=PC=PF$, deci $\Delta BMP$ este isoscel.

Asadar, revenind la problema, $BE=CM$, deci si $CM=AC$, iar triunghiul $ACM$ este isoscel, cu unghiul de la varf, $\angle ACM=40^\circ$. In concluzie, $\angle AMC=\angle MAC=70^\circ$, q.e.d.

[b]@Lawasu[/b]: fara a fi malitios, ma indoiesc ca ai facut aceasta legatura. Problemele simple au fost postate tocmai pentru a testa/dezvolta imaginatia creativa a pretendentilor la un post de olimpic.

Best regards,
sunken rock


Is it possible to have an obtuse angle in a quadrilateral adjacent to a right angle in the same quadrilateral with two opposite sides congruent. For example, say I have quadrilateral ABCD. and angle BCD is obtuse and CDA is a right angle and CD and DA are both congruent. Is this possible? I noticed that if the obtuse angle is 91 degrees,then the quadrilateral doesnt exist since the other two angles must be acute and they can't be acute when the other two angles are 91 and 90. I'm not sure if this has any relevance to what I'm trying to find out. Can you please explain?

Such a quadrilateral is possible, even with the obtuse angle 91 degrees. The other two angles do not need to be acute. For example, consider a square ABCD. Then moving point B away from point D on line BD will create a quadrilateral with angle BCD obtuse, CDA right, and CD congruent to DA.


calculus
integration
modular arithmetic
Divisibility Theory

Find all positive integers $n$ such that $n$ has exactly $6$ positive divisors $1<d_{1}<d_{2}<d_{3}<d_{4}<n$ and $1+n=5(d_{1}+d_{2}+d_{3}+d_{4})$.

Obviosly $ n\equal{}p^5$ is not solution, therefore $ n\equal{}pq^2\equal{}5(p\plus{}pq\plus{}q\plus{}q^2)\minus{}1$ or $ p(q^2\minus{}5q\minus{}5)\equal{}5q(q\plus{}1)\minus{}1$, therefore $ p>5,q>5$.
If $ q\equal{}7$ we get $ p\equal{}31$ (solution) if $ p\equal{}7$ we get contradition $ q(q\minus{}20)\equal{}17$. Because for $ q\ge 11$ we get $ p\le 7$, solution $ n\equal{}31*7^2$ is unique.

[hide="A100"]We know that $n=p^5$ or $pq^2$. If $n=p^5$, then $1+p^5=5(p+p^2+p^3+p^4) \Rightarrow p^5-5p^4-5p^3-5p^2-5p+1=0$, which obviously has no prime roots $p$. (RRT does the trick).

If $n=pq^2$, we have \[pq^2+1=5p+5q+5pq+5q^2 \Rightarrow p = \frac{5q^2+5q-1}{q^2-5q-5} = 5+\frac{30q+24}{q^2-5q-5}\]

Note that if $q=2, 3, 5$, $p$ is not integral. Otherwise, $p$ is positive. Let $f(x)=\frac{30q+24}{q^2-5q-5}$. Note that $f$ is decreasing from $5$ to infinity, and that $f(7)=26$ and $f(11)\approx 5.8$. If there is a solution, then $k \le 5$  Additionally, note that if $\frac{30q+24}{q^2-5q-5}  = k$ , then $kq^2-5kq-5=30q+24 \Rightarrow kq^2-(30+5k)q-(5k+24)=0$. If there are integral solution, $\Delta = 25k^2+300k+900+20k^2+96=3(15k^2+68k+300)$. If this is a perfect square, $15k^2+68k+300 \equiv 0 \pmod 3 \Rightarrow k \equiv 0 \pmod 3 \Rightarrow k = 3$. However, obviously there is no solution when $k=3$, so the only solution is $(p, q) = (31, 7)$, or $n=1519$.[/hide]


External Tangent Geometry Proof

Question :

Hello! I have a proof to complete but do not even know where to start:

Prove that the three pairs of common external tangents to three circles (no two of which are equal or concentric), taken two at a time, intersect in three collinear points.

Shobber,

I don't understand your reply.  Can you explain?

http://mathworld.wolfram.com/MongesCircleTheorem.html
Is that what you mean in the problem?

[quote="symmetry"]I'll go check out the site.[/quote]
Got it?  :)

yes, i was going to say that was just a theorem, but i could not remember what the name of it was

Thank you all.

Shobber:

Yes, I got it.


LaTeX
inequalities
triangle inequality
calculus
calculus computations

I spent 45 mins but couldn't get it working. Thanks [b]10000th user[/b] for the clerification. Is the proof fault???

Oh my god.  Clean up your $\LaTeX$.  Please.

Some more constructive criticism: Don't put a backslash before a dollar sign unless you actually intend to make a dollar sign. You want your dollar signs to end the LaTeX sequence, so you don't want backslashes.

You should always preview your posts before you submit them if there is a lot of code. Everyone hates to have to read poor typesetting.

[quote="mohd_ahsan"]Suppose that  $\sum_n{n}a_{n} \$ converges absolutely. Prove that$\sum_n{n}a_{n}\$ converges by showing that the sequence of partial sums $\S_{n}\$ is a Cauchy sequence.$Proof$By theorem$\\lim_{h\to infinity} \|a_{n} = 0\$
                          implies      $\|a_{n}< \in/n\$ for all$\n>N\$

                   Let  $\ m > n \$$\S_{n} = |a_{1}| + |a_{2}|+.........................+|a_{n} |\$
                 $\S_{m} =| a_{1}| + |a_{2}|+.........................+|a_{n}| + |a_{n+1}|+.............+|a_{m}|\$$\S_{m}-S_{n} = |a_{n+1}|+ |a_{n+2}|+................+|a_{m}|\$

$\|S_{m}-S_{n}| = ||a_{n+1}|+ |a_{n+2}|+................+|a_{m}||\$$\ = |a_{n+1}|+ |a_{n+2}|+................+|a_{m}|\$
                             $\\leq |a_{n}|+|a_{n}|+....................+|a_{n}|\$$\= (m-n)|a_{n}\$
                             $\<(m-n)\in/n\$$\<\in\$

i.e $\|S_{m}-S_{n}|<\in\$ Thus = |$\(S_{n})  \$ is cauchy and convergent[/quote]

 :rotfl: Imminent danger to browsers and your eyes. Below is my editing skills :D

[quote="mohd_ahsan"]Suppose that  $\sum_na_{n}$ converges absolutely. Prove that $\sum_na_{n}$ converges by showing that the sequence of partial sums $S_n$ is a Cauchy sequence.

[b]Proof:[/b] By theorem $\lim_{h\to\infty}|a_n|=0\Rightarrow |a_n|<\frac{\epsilon}n$ for all $n>N$.

Let  $m>n$.

$\begin{aligned} S_n&=|a_1|+|a_2|+\cdots+|a_n|\\ S_m&=|a_1|+|a_2|+\cdots+|a_n|+|a_{n+1}|+\cdots+|a_m|\\ S_m-S_n&=|a_{n+1}|+|a_{n+2}|+\cdots+|a_m|\\ |S_m-S_n|&=\left||a_{n+1}|+|a_{n+2}|+\cdots+|a_m|\right| \end{aligned}$

But,
$\begin{aligned} |a_{n+1}|+|a_{n+2}|+\cdots+|a_m|&\leq|a_n|+|a_n|+\cdots+|a_n|=(m-n)|a_n|\\ &<(m-n)\frac{\epsilon}n<\epsilon \end{aligned}$
$|S_m-S_n|<\epsilon$. Thus, $(S_n)$ is Cauchy and convergent[/quote]Well, I tried to make some sense out of it.

This problem is much easier than you're trying to make it. Your estimate is too generous, and it messes you up. Here's the problem: $m$ can be any amount larger than $n$. It can be huge compared to $n$. You can't say $(m-n)\frac{\epsilon}{n} < \epsilon$. Take $m = 3n$, for a mild example.

Also, note that your partial sums are $S_n = a_1 + a_2 + ... + a_n$. No absolute value. Unless I misunderstand the question, of course. But it sounds like you're supposed to prove that $\sum_n |a_n|$ convergent implies $\sum_n a_n$ convergent. You want to prove that $|S_m - S_n|$ is small. Hint: Use the triangle inequality, and the fact that $\sum_n |a_n|$ converges.


Well, sadly, Duke is ranked #10. Who do you think can own the ACC this year?

maryland is apparently pretty good ?

duke sucks

state seems to be doing pretty ok considering

UNC isn't doing as well as I thought it would be

but it will still own everyone

Just curious, why didn't you include Wake, the other major NC team?

Also, why Maryland?

yeah I know MD does seem to be shaping up well (that is, after missing the NCAA tournament for a while, I think?)

also, glad to see UNC has pulled ahead in the poll

also nobody cares about ACC tournament we should get to the NITTY - GRITTY

Well Maryland is 6-0 at least, while Duke and UNC have lost  :mad:

So yeah...

and Wake Forest wasn't included because they suck

bah, UNC is mostly a young team, of course they're going to lose early

WHATEV, MAN

[quote="b-flat"]I of course, support Duke, but I do not think they will win  :(

We lost most of our good players, but you never know, we could do something incredible  :D[/quote]

OH MAN THIS IS COMPLETELY IN ACCORD WITH HOW THIS IS [quote="b-flat"]a place to talk about math in North Carolina.[/quote]

GA TEcH SHOULD WIN!
but to be honest i dunno anything about this sport

GO NC!

I think they should win because they have the strength and determination, as well as stamina

[quote="funcia"]wow i just watched duke beat g mason well i didn't watch the whole thing i am not sure[/quote]

I don't think Duke and G.Mason played. Did they?

Duke still has the NCAA to come back  :D

I was surprised NC State managed to get as far as they did, though.

[quote="MysticTerminator"]I wasn't, especially against a first-round opponent as bad as Duke and a semifinal opponent they swept in the regular season.

Duke's not going anywhere in the NCAA; heck, they might just lose to VCU (they're a good team).[/quote]

I agree. Duke stinks like always. :D

Yeah. I'm worried about the seeding in UNC's bracket though. Facing Texas in the sweet 16? Yeesh. Whose idea was that. Grumph.

Duke's bracket isn't all that tough.  I can see them getting fairly far.  I don't know about winning, but farther than they got in the ACC.

[quote="Valentin Vornicu"]Although nobody in the office seems to be sharing my opinion on this, I'm cheering for Duke :D[/quote]

DUDE NO WTF IS THIS

and duke's not getting past pittsburgh. I'm annoyed they got ranked so high and with a comparatively weak pod but HEY THEY'RE DUKE THEY SUCK OK

[quote="Valentin Vornicu"]Yeah that's pretty much what I've got from Dave this morning concerning this matter :)[/quote]

excellent.

i have UNC losing to Georgetown in elite eight, lol

I think duke can beat pitt, but ucla will be tough...

Also, congratulations on getting unbanned, Arnav. :wink:

huh I wasn't banned

y'all were just being stupid with your anti-UNC bias

so I left

but hey March Madness I AM BACK

dude duke is obviously not getting anywhere near the elite eight

this is a fact

are they getting to sweet 16?

no

are they getting to second round?

possibly

OH MY GOD I CALLED THIS BEAUTIFULLY

FIRST-ROUND LOSS TO VIRGINIA COMMONWEALTH ON PULL-UP JUMPER BY ERIC MAYNOR

DUKE SEASON:

SEASON-ENDING LOSSES TO MARYLAND, CAROLINA

FIRST-ROUND ACC TOURNAMENT LOSS TO 10TH SEEDED NCSU

FIRST-ROUND NCAA TOURNAMENT LOSS TO 11TH SEEDED VCU

Cant blame him for hating a horrible team.

har, har, har :cursing:

i don't expect UNC to make it to Final Four though (wait, they do have like a 60% chance, but who cares)


A map of Wyoming is drawn with a scale of $ \frac{1}{2} \mathrm{inch} \equal{} 1 \mathrm{mile}$. On that map, how long would the drawing of a road 10 miles long be? Express your answer as a decimal to the nearest tenth.

Probably no reason; just a glitch...

it should be 5

[quote=BOGTRO]Why is the answer 2.5....[/quote]

This problem is different from the one on For the win.  ;)

Ya the numbers and the scale were slightly different.


In rectangle $ ABCD$ with diagonals $ \overline{AC}$ and $ \overline{BD}$ intersecting
at $ E$, what is the ratio of the area of $ \triangle BEC$ to the area
of $ \triangle BDC$? Express your answer as a common fraction.

[asy]currentpen = fontsize(8pt); size(75); pair D = (0,0), C = (3,0), B = (3,2), A = (0,2), E = (3/2,1);
draw(A--B--C--D--cycle); draw(A--C); draw(D--B);
label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,N);[/asy]

[hide]Because ABCD is a rectangle, it is a parallelogram. parallelograms have diagonals that bisect each other (I forgot how to prove it, but they do), so EC is a median of BDC. Medians bisect the area of the triangle (and when you draw all three, they split the area into 6 triangles of equal area, and three are on one side of a given median, and 3 are on the other side. therefore, [BEC]=[ECD]=[b]1/2[/b][bdc][/hide]


inequalities
Euler
inequalities proposed

[quote="Ligouras"]Prove that

$ \frac {3(2R \minus{} r)}{r(4R^2 \plus{} 4rR \plus{} 3r^2)}\leq\frac {1}{h_a^2} \plus{} \frac {1}{h_b^2} \plus{} \frac {1}{h_c^2}\leq\frac {2R^2 \plus{} r^2}{r^3(16R \minus{} 5r)}$[/quote]
\[ \sum {\frac {1}{{h_a^2 }}} \equal{} \frac {{s^2 \minus{} 4Rr \minus{} r^2 }}{{2s^2 r^2 }}\]
$ \frac {3(2R \minus{} r)}{r(4R^2 \plus{} 4rR \plus{} 3r^2)}\leq\frac {1}{h_a^2} \plus{} \frac {1}{h_b^2} \plus{} \frac {1}{h_c^2}\leq\frac {2R^2 \plus{} r^2}{r^3(16R \minus{} 5r)}$

$ \Leftrightarrow \frac {3(2R \minus{} r)}{r(4R^2 \plus{} 4rR \plus{} 3r^2)}\leq \frac {{s^2 \minus{} 4Rr \minus{} r^2 }}{{2s^2 r^2 }} \leq\frac {2R^2 \plus{} r^2}{r^3(16R \minus{} 5r)}$

and use Gerretsen Inequality and Euler Inequality is easy to prove it...


In k-people association work (2^n - 1) comission (every no empty set members of association creates comission). In every comission have to choose chairman. It demands condition. If comission C is sum two comission A and B, chairman comission C will be chairman at least one of comission A, B. Mark number posibility of choice chairman. I know that result is k!, but I don't know how show it

This is an ongoing Polish competition problem -- please do not reply.


inequalities
function
inequalities proposed

If $a,b,c$ are positive numbers such that $a+b+c=3$, then

$\frac{\sqrt[3] a}{b+c}+\frac{\sqrt[3] b}{c+a}+\frac{\sqrt[3] c}{a+b}\geq \frac{3}{2}$.

This problem is one time posted by me, or something similar. By C-S inequality \[\left(\frac{a^{p}}{b+c}+\frac{b^{p}}{c+a}+\frac{c^{p}}{a+b}\right)\left(a^{2-p}(b+c)+b^{2-p}(c+a)+c^{2-p}(a+b)\right)\ge \left(a+b+c\right)^{2}.\] Now using the well-known problem, if $k\le 2$ then \[a^{k}(b+c)+b^{k}(c+a)+c^{k}(a+b) \le \max \left(6, 2(\frac{3}{2})^{k+1}\right).\] or you can see it here
http://www.mathlinks.ro/Forum/viewtopic.php?t=30732&start=20
Because \[2(\frac{3}{2})^{3-\frac{1}{3}}\approx 5.9 <6\] So we have done. Also, the best constant $p$ is \[(\frac{3}{2})^{3-p}=3\Leftrightarrow p\approx 0.30\] .

In my book, there is also a quite similar problem \[\frac{a}{1+b^{2}+c^{2}}+\frac{b}{1+c^{2}+a^{2}}+\frac{c}{1+a^{2}+b^{2}}\ge \frac{4}{5}\] if $a,b,c$ are non-negative real numbers and $a+b+c=1$.

We can also use the similar method to find the minimum of \[\frac{a}{r+b^{k}+c^{k}}+\frac{b}{r+c^{k}+a^{k}}+\frac{c}{r+a^{k}+b^{k}}\] where k is a positive real number smaller than $3$.

[quote="Flove"][quote="Marius Damian"]My solution:
We have to prove that \[\sum{\frac{\sqrt[3]{a}}{b+c}}\geq\frac{3}{2},\] equivalent with \[\sum\bigg({\frac{1}{\sqrt[3]{a^{2}}}\cdot\frac{a}{b+c}}\bigg)\geq\frac{3}{2}.\] From Chebyshev and Nesbitt, we have \[\sum\bigg({\frac{1}{\sqrt[3]{a^{2}}}\cdot\frac{a}{b+c}}\bigg)\geq\frac{1}{3}\bigg( \sum{\frac{1}{\sqrt[3]{a^{2}}}}\bigg)\bigg( \sum{\frac{a}{b+c}}\bigg)\geq\frac{1}{2}\bigg( \sum{\frac{1}{\sqrt[3]{a^{2}}}}\bigg).\] Using the convexity of the function $f: (0;3)\rightarrow(0;\infty), f(x)=\frac{1}{\sqrt[3]{x^{2}}},$ it follows \[\frac{1}{2}\bigg( \sum{\frac{1}{\sqrt[3]{a^{2}}}}\bigg)\geq\frac{3}{2}\cdot\frac{1}{\sqrt[3]{\big( \frac{a+b+c}{3}}\big)^{2}}=\frac{3}{2}.\] [/quote]

But you use Chebtshev's inequality in a wrong way.  :([/quote]

It is right! Big mistake! :blush:


On a four-by-four grid (see below), the points are 1 unit apart horizontally and vertically. Two distinct points P and Q are randomly selected from this grid. What is the probability that the distance between P and Q is less than 2.5 ? Express your answer as a common fraction.
          . . . .
          . . . .
          . . . .
          . . . .

[hide="ANSWER"]\frac{41}{60}[/hide]

Could someone please describe all of the lines that work?  Thanks.

There are three types of points

corner points
side non-corner points
middle points (not touching a side)

just do 1 case for each then multiply by the corresponding number

denominator is 16*15

Do not double the numerator to account for PQ and QP because that will count every line twice

[hide="Solution"]There are 16!/14! ways, or 16*15=240
Okay, the key here is the fact that the points are distinct.  Let's work with the corners first.   To the sides, there are 2 points per, and 1 on diagonal (not 2 because 2√2>2.5).  There are 4 corners for a total of 4*5=20 points.  However, we have to multiply by 2 because that would double the lines WITH DIFFERENT ENDPOINTS!  Now, as for the sidepoints, there are a total of 1 for every pair of adjacent points.  In addition, there are 2 towards the middle of the rectangle, and 1 for both diagonals for 2+2+1=5 lines.  (Note: We didn't count the lines to the corners because we have already counted those).  Hence, since there are a total of 8 sidepoints, for a total of 8*5=40 points.  We have overcounted though, specifically the line from each sidepoint to another sidepoint adjacent to the same corner, and also, the line from each sidepoint to the adjacent sidepoint.  We have only overcounted these once, so the total # of possible lines is from the sidepoints (and not to the corners) is 32.  We have to multiply by 2, to get 32*2=64.  Now, as for the middle, we have to stick to the four lone points since we have counted all other lines to them.  There are 4*3 of these lines (1 to each adjacent middle point and 1 to the diagonal), but we are overcounting 6 lines (each diagonal and each of the four sides) for a total of 6 sides*2=12 sides.  64+12+40=116.  However, we haven't counted 1 thing, the diagonal of a rectangle with side lengths 2 and 1.  There are 12 such rectangles, 12(2)=24 such diagonals, and 24(2)=48 lines.  Thus, there are 116+48=164 such lines.  164/240=41/60.
                                               . . . .
                                               . . . .
                                               . . . .
                                               . . . .
Note to "the future," you have to multiply by 2 because the lines PQ and QP are different if you switch the points P and Q because P and Q are distinct.[/hide]

[quote="Z-coli"]
Note to "the future," you have to multiply by 2 because the lines PQ and QP are different if you switch the points P and Q because P and Q are distinct.[/hide][/quote]

Umm no you dont have to multiply by 2 because both cases are already included (atleast in my solution).
If I pick a corner point as P and a middle point as Q, thats PQ

I can get PQ backwords by choosing the P as the middle point and Q as the corner point.
So no need to double
Notice for your solution you have "we must stick to the four middle points because all the other lines have already been counted".
Had you not doubled earlier you do not have to restrict that

My solution when I say use middle points, I mean its not just a middle point to middle point, its middle point to all points.

Basically,
4 corner points can link to 7 each (anywhere)
8 side non-corner points can link to 10 each (anywhere)
4 middle points can link to 14 each (anywhere)

Thats $ 4*7\plus{}8*10\plus{}4*14\equal{}164
164/(16*15)\equal{}41/60$ -same answer

Now I didnt have to double anything there and still got right answer
You doubled everything and had to worry about overcounts (which can sometimes be missed)

Heres another example to demonstrate what I am talking about

You said there was 1 pair that linked two adjacent vertices on the sides.  So yes in your solution you multiply by 2 because they can go backwords PQ or QP
But in mine, you do not because I am choosing each point individually (kind of hard to explain but if you just draw it you can see)

OK, OK, you're right.  Ours both worked, I had just thought you meant something different.  Very smart, though.


Let $ a;b;t;x>0$;$ a\plus{}b\equal{}2x$,$ t$ is a const.Prove that
$ \frac{x}{\frac{t^2}{4}\plus{}a^2}\plus{}\frac{x}{\frac{t^2}{4}\plus{}b^2} \leq \frac{2}{t}$

Unfortunately the proposed inequality is not true.

It can be disproved by simply substituting
a=3/2
b=1/2
x=1
t=1
into the proposed inequality


Alright, question and choices at the top.

If you poll though, please leave your age and education level, so I can use this for an assignment.

Thanks. :D

I dont think the question is very clear....

arent all polls opinion polls? ;)

No, some polls (e.g. "What is your annual income?", "Who did you vote for in the last presidential election?") are completely factual.

The question asks which factor an elected official should consider the most important when he/she asks a vote in Congress.

I'd say a mix between personal beliefs and what the region believes, since that's the whole point of a congressman, to represent his region.

I'm 14, and I've graduated from 8th grade (SO I'm in 9th grade right now).

Any of the first four is okay.  However, before being elected, the representative/candidate should make it abundantly clear to the voters which method he/she will use to make decisions.  If he/she will use the first method, he/she should also make his/her personal views abundantly clear to the voters.

Essentially, anything goes, as long as the voters know what they are voting on when the elect the representative.

Edit: Actually, the first two options are ideally the same thing, since the representative should be making his/her views known to the general population in advance of being elected.  And there should be enough variety of candidates and a sufficiently fine-tuned method of electing them that the candidate who wins will always be one whose personal views agree with those of the voters.  Unfortunately, neither the candidate making personal views fully known, nor the necessary variety of candidates, nor the fine-tuned election system is likely to happen anytime in the near future (at least not in the U.S.).

Age: 18
Education level: freshman in college

Popular belief of the intellectuals (and of course, great minds think alike :P).

From what I've seen of America, I've started to think that what's good for the country isn't just opinion. It really requires deep thought and consideration. Now of course that idea of listening to intellectuals isn't practical and would be quite controversial (What is an intellectual?).

Given our current legal system, I'd say the best would be consider all views, and just think, analyze, use logic. Do what seems best.

a congressman represents a state. he should thus represent the people's beliefs from the state he represents. usually a congressman's personal beliefs are similar to his people's, so there isn't much of a distinction between the first two options. but if there was a difference, i feel the congressman should vote the way the people feel
age: 17, education: senior HS

the congressman is an advocate of the state he or she represents.  therefore they should say what the majority of who they are representing believe in.

grade 7, age 12

i think the congressman should believe in anything he wants as long as it's reasonable.

age 13, grade 8

Good luck on your assignment! :thumbup:

I'd say the region he/she represents. After all, they are representing a region for a reason... so that everyone in the region doesn't have to vote on everything. And it would make sense for his/her personal beliefs to tend toward the majority belief in the region because that's why the region would elect him/her in the first place.

Age 16, Grade 11

I think they should vote for the region they represent.  Many people vote for the people that they believe will best represent them, and so the Congresspeople should respect this belief.  After all, Congress is just a way to get the input of the entire nation without having the entire nation vote.  Its point is not to pick out a couple people and feel how they feel about things.

I'm 13 and I'm in 8th grade.

The region that they represent because thats what they are there to do anyways right? (represent a group of people)

Age: 12 Grade: 7

I think the Congressman should vote based on his/her own personal beliefs, opinions, etc, but to get elected/re-elected, he/she must represent the interests of his/her district.  If Joe Blow represents a big widget-manufacuring area, and he is morally opposed to widgets, he should be able to vote for widget-control legislation, but then nobody would like him, and the voters would choose a more pro-widget candidate.

Wow, that was totally confusing.

17, (high school) senior

I marked other the congressman should vote in the way that best serves the people he/she represents.  The reason this is different from the others is that he/she(for the rest of this im just typing male pronouns because its easier) may compromise in some places for gains in others.   Democracy is about compromising and giving up somethings we want to keep and get the ones we really want.  For eexample in the case of the widget-manufacturing if there was a bill about something unrelated and the leaders of his party told him to vote for it and then they would help him  pass legeslation that would provide jobs and money for his constituents (relating to widgets) he should vote with his party.  If his party wants to ban widget production he should vote against his party.  He can vote based on his own opinions when neither his party or his constiuents demand that he vote one way or the other.  The one thing that is more or less irrelevant is the rest of the nations opinion since he doesn't represent the rest of the nation.


Hi,
   I was wondering if there are chemistry contests open to college (undergraduate) students, in the United States/North America.

Thanks

Well, you might be able to find some in school, and there's the SAT II


$Both \ systems \ is \ equivalent \ that \ \ Matrice \ with \ strings \ {a b c}, \ {a^{'} b^{'} c^{'}}, \ {a^{\"} b^{\"} c^{\"}} \ is \ ortogonal.$


calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

Find :

\[
\displaystyle\int \frac {\ln^3(\sqrt{x}+1)} {\cos^2(\sqrt[3] {x}-\ln x)}\,dx
\]

Is it possible to do it by hand?

Or, better, would anybody with any level of sanity do it by hand when you could just plug it into a computer integrator.

If anybody solved this by hand I'm very impressed.


Nat Sprint 2002, #12:

The hour hand of a clock is 6 inches lonng and the minute hand is 8 inhes long. What is the ratio of the distance traveled by the hour hand to the distance traveled by the minute hand between noon and 3 PM?

Billy

[hide="My answer"]1/16
The hour hand travels 1/4 of a circle.  The diameter of it's circle is 12, so it travels 3$\pi$.
The minute hand travels 3 circles.  It's diameter is 16, so it travels 48$\pi$.
The ratio is:
$\frac{3\pi}{48\pi}=\frac{3}{48}=\frac{1}{16}$

:D
[/hide]
:P

Nothing more to say here, other than the fact that the formula for this sort of problem is $\frac{h}{12m}$, where $h$ is the length of the hour hand and $m$ is the length of the minute hand.


linear algebra
matrix
vector
linear algebra unsolved

I am new to linear algebra.
While solving a problem, I came up with this one. Please help me to solve it. Thank you

[b]A is a singular matrix. Prove that there exists a matrix X that satisfy AX=0. Find X. [/b]

You should add the condition $ X \neq O_{n \times n}$, since for such X, AX = O is always verified. Let $ X_1, X_2, ..., X_n$ be the columns of X. Then AX = O is equivalent to

$ A [X_1 \,\, X_2 \,\, ... \,\, X_n] \equal{} O_{n \times n}$

$ < \equal{} > [AX_1 \,\, AX_2 \,\, ... \,\, AX_n] \equal{} O_{n \times n}$

$ < \equal{} > A X_i \equal{} O_{n \times 1}$, for i = 1,2,...,n.

But these are systems of linear equations which, since A is singular, have infinitely many (non-trivial) solutions. Pick n those solutions and put them as the columns of X: then we have AX = O with $ X \neq O_{n \times n}$. $ \Diamond$

Exercise. Under the same conditions, show that there is a matrix $ X \neq O_{n \times n}$ such that $ XA \equal{} O_{n \times n}$.

Thank you for your solution
I think the exercise can be solved with the same strategy.

Let $ X_{1},X_{2},...,X_{n}$ be the rows of $ X$

$ XA\equal{}0 \\ <\equal{}>
\left[ {\begin{array}{cc}
 X_{1} \\
 X_{2} \\
 .... \\
 X_{3}  \\
 \end{array} } \right]
A \equal{} O_{n \times n} \\
<\equal{}>
\left[ {\begin{array}{cc}
 X_{1}A \\
 X_{2}A \\
 .... \\
 X_{3}A  \\
 \end{array} } \right]\equal{}O_{n \times n} \\
<\equal{}> X_{i}A \equal{} O_{1 \times n}$,for i=1,2,...,n.

$ X_{i}A\equal{}O_{1 \times n}$ is just as $ A^{T}X_{i}\equal{}O_{n \times 1}$
because $ A$ is singular, then $ A^{T}$ must be singular. As you wrote, the linear equations have infinitely many non-trivial solutions. Just pick n of them as the rows of $ X$, then we receive a matrix $ X \neq 0$, such that $ XA\equal{}0$

Yes, that's right. Now, if A is non-singular is there any matrix X such that AX = O?

There exists $ X, X \neq 0_{n \times n}$ and $ AX\equal{}0$ (1)

if A is invertible, then there exist $ A^{\minus{}1}$ such that $ A^{\minus{}1}A\equal{}I$
so $ A^{\minus{}1}AX\equal{}X ( \neq 0_{n \times n})$
but from (1) we have $ A^{\minus{}1}(AX)\equal{}A^{\minus{}1}0_{n \times n}\equal{}0$
so (1) must be false.

[u]Proof[/u]. You can also define the determinant as the constant $ \det \hat{A}$ such that $ \wedge^{\dim V}\hat{A}^N\equal{}(\det \hat{A})\hat{1}_V$, where $ V$ is a vector space with some basis $ \mathcal{B}\equal{}\{e_i\}$ and dimension equal to the number of rows of $ A$.  Suppose $ A$ is the matrix of $ \hat{A}$ w.r.t. $ \mathcal{B}$.  Then $ \bigwedge_{i\equal{}1}^{\dim V}\hat{A}e_i\equal{}\det\hat{A}\wedge_{i\equal{}1}^{\dim V}e_i\equal{}0$, Therefore, the columns of $ A$ are linearly dependent.  The conclusion follows. $ \Box$

[u]Remark[/u].  Defining $ \hat{A}\equal{}\sum_{i,j}A_{ij}e_i\otimes e_j^*\in V\otimes V*\cong \text{End}(V)$, we have $ A$ as the matrix of $ \hat{A}$

Boxedexe, what is the meaning of the symbol $ \bigwedge_{i\equal{}1}^{\dim V}$?


[quote]how many ways are there to make exactly 42 cents?

you can choose from 16 coins. --4 quarters, 4 nickels, 4 dimes, 4 pennies[/quote]
[hide]$31\ \text{ways}$[/hide]

[quote="13375P34K43V312"]Can you give a solution as well?[/quote]
[hide="Solution"]I made a list. :rotfl:[/hide]

nice one i like pie

[quote]how many ways are there to make exactly 42 cents?

you can choose from 16 coins. --4 quarters, 4 nickels, 4 dimes, 4 pennies[/quote]

so, the most quarters u can use is 1, nicels: 4, pennies: 4, and dimes: 4

q/d/n/p
1/1/1/2
1/0/2/2
0/4/0/2
0/3/2/2
0/2/4/2                   5 ways

and there are no more because the greatest you can make with nickels and pennies is 24 cents and below 2 dimes and 1 quarter, u can not make the remaining 18 cents
-jorian


These are some really hard problems i have come up against and i need some help with them. could you please post solutions ASAP as i have a deadline to show the students i tutor how to do them. Funny how i cannot solve them and that is why they had no clue:P
anyways thankyou for your help:)

[b]1. [/b]

Professor Pipkin has engineered two new species of bacteria, blue and red, that behave in the following way: after one minute every blue bacterium turns into $6$ red ones, and every red bacterium turns into $1$ blue and $1$ red bacterium into a test-tube, so after one minute there were $1$ blue bacterium and $7$ red one; and after two minutes there were $7$ blue bacteria and $13$ red ones. [b]Prove that after $n$ minutes there will be

$3/5 * 3^n + 2/5 * (-2)^n$
blue bacteria in the test-tube.[/b]

[b]2.[/b]
The vertices [i]A[/i] and [i]C[/i] of a triangle [i]ABC[/i] lie on the circumference of a circle. The circle cuts the sides [i]AB[/i] and [i]CB[/i] at points [i]X[/i] and [i]Y[/i] respectively. If [i]AC = XC[/i] and $<angle> YXC = 40 degrees$, [b]Find the size of $<angle> ABC$[/b]

[b]3.[/b]
 In triangle [i]ABC[/i], [i]D[/i] and [i]F[/i] are points on [i]BC[/i], and [i]E[/i] is a point on [i]AB[/i] such that [i]AD[/i] is parallel to [i]EF[/i], $<angle>CEF$ = $<angle>ACB$, [i]AD[/i]=$15$ m, [i]EF[/i] = $8$ m and [i]BF[/i] = $8$ m. [b]Find the length of BC.[/b]

Thanks again!
Keshin

it has something to do with setting up a recursive equation to find the amounts, let R_n be the number of reds in the tube after an n minutes, and let B_n be the number of blues at minute n, so using the given info, we get
$6*B_n + R_n = R_{n+1}$, and $R_n = B_{n+1}$
and then substituting the second into the first, $6*B_n + B_{n+1} = B_{n+2}$
then converting recursive to explict equations (you have to know the procedure, look at
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=51794]www.artofproblemsolving.com/Forum/viewtopic.php?t=51794[/url])
6 + x = x^2
0 = x^2 - x - 6
(x-3)(x+2) = 0
so the form must be $B_n = a*(3)^n + b* (-2)^n$ for some constants a and b, then find a few values of b_n and solve the system to get the answer, this is a pretty difficult problem, i am suprised that a school would have something like this

number 2
this one is not that hard, if you draw it out, the basic idea that you have to do is that since you get those 2 congruent sides, you have an isosoles triangle, and you have one of those angle and the 40 degree angle intercept the same arc, therefore they are congruent, then you can use the 180 degrees in a triangle to find AXC, and use the fact that an inscribed angle is half the arc it intercepts for a bunch of stuff, and then you can find all the arcs, except XC, and subtract from 360 to get arc XC, then use the formula for 2 secants to find the angle

number 3
this can all be done with similar triangles, namely $ADB \sim EFB$ and $CEF \sim ACD$, i dont have time to type all of the solution, but it is easy (use the parallel lines)

No.3
Sorry If Im dumb but
[hide="3"]
$23$
If this is right Ill post a sol
[/hide]

Hello, the first problem looks very simple: Let after n minutes,  $R_n$ be number of Red bacteria  and $B_n$   be number of Blue ones.
You say:

$R_n = 6 B_{n-1} + R_{n-1} $  and
$B_n = R_{n-1}$

Which  you can write as
$R_n = 6 R_{n-2} + R_{n-1} $ ..  But you are interested in Blue only so for Blue:
$B_{n+1} =  6 B_{n-1}+B_{n} $     (remember  $R_n = B_{m+1}$

This is standard recursive eauation, the corrosponding characterstic  equation is $x^2 =  x+6$   roots are $3,-2$

So general solution is: $B_n= a 3^n + b (-2)^n$   we are given $B_0 = 1$ and $B_1 =1 $ so
$a,b$ can be calculated  as when n=0  ; a+b= 1
and n=1   ; 3a- 2b = 1

so a=3/5 and b=2/5.
Hope this helps.

Added later: Sorry did not see,  some have already answered it, hope it is helpful anyway..

For 2, if nothing else: (let < stands for angle)  and <XAC is called  $A$
Now:   <BYX = 180 -<xyc  = <XAC  = $A$

Since  <YXC=40,  and  <CXA= <XAC = $A$  we have  <YXA = 40+ $A$
Hence required angle is 40. (Because <BYX =$A$ and  <BYX+<CBA = <YXC)

I can't resist to post  a "clever" (some  may call it "cheating"  but  I call it "physicist" way to do math) to do above. If you are tutoring others,  you might like to pass  this kind of "tool" in their tool box. -- (May be we can inspire some future physicist!)

Let us say <BAC is a rigth-angle   (I know, it is not given, but no one is saying it is [b]not [/b]a right angle, .. so why not, after all we live in a free country :).

Now, in this (limiting) case, AC is diameter of the circle, and A and X lies  at the same point (how convinient - AC is indeed equal to CX :))

Now of course, angle AYC is  right angle too (being in a semi-circle and all that..))
And of course  <YAC (I mean YXC ! - same thing) = 40
So YAB= 50 and so ,YBA (or CBA)  is 40 degrees!

Okay so BAC  is not given as 90 degrees,  but if we assume the person who posed the problem knows his math,  (s)he would see to it that  answer is same no matter what the angle  BAC is :)

In any case, it does not hurt to try simple (particular) case to  simplify any given  problem, at the minimum it will validate your  answer.

Hope this is helpful.

but making assumptions like that is bad because it is likely that this person will only be doing math in school, and on tests in school, you have to show all of your work, and you cannot do things like that

[quote]but making assumptions like that is [b]bad[/b][/quote]
Not really, in fact , IMO, some of the brightest students I have seen,  use this. Nothing [i][b]bad[/b][/i] about "creative" thinking, it is, agian  I think,  and like to teach that to my students  - even at university level-  one of  many  strong "tools"   which people use in problem solving.

As long as you are [b]aware[/b]  (and clear in your writing)  that you are making an assumption (or taking a particular case- of a more general case), it is okay, in fact  many times,  it is a strong  method  which can prove  "negativity" of a theorem.  .. And believe me, many times it gives you ideas on how to tackle the more general case. Sure, in an exam,  which is "formal", you may  not get full marks,   but in practice:

A - If you are clear enough to think  up a "creative" solution, you are all but likely to find the 'rotuine' solution without any difficulty.
B- By now you can  [i]double check[/i] your answer (and thus avoid making silly mistakes)
C - If the  coach/teacher e is anything like me,  (s)he would certainly choose you for a math team ! :)
Just my thoughts - Not everyone has to agree with it. :)

PS -  To give you an example, it took me  only a few seconds   to solve the  3rd problem above. Not because I have seen this problme before, but rather I just "assumed" EF (and thus AD) is perpendicular to BC and I equated the tangents of two angles to get  ( tan FEC = (7+x)/8 and tan BCA = 15/x ) and I got x=8.

Of course , it took only a few more seconds to 'extend' that idea to 'finding similar triangles' .. precisely  because now I knew which triangles to look for.
Hope this is useful. ..

[quote="Gyan"]For the last question, traingle EFC is similar triangle  CDA  (actual congruent but lets not spoil  one's fun!)

So    $\frac {FC}{FE} = \frac {DA}{DC} $ Let  $DC=x$  and we get $\frac {7+x} 8 = \frac {15} x $
which gives $x=8 $ (ignoring  x=-15, of course..)  So $BC=23$[/quote]
Yes! So I was right! :D

[quote="zeus_three"]Induction (with steps n=1, and n=2) will yield the solution.

By the way, you shouldn't be asking for help with Competition questions (noether program)[/quote]

Moderator: If this indeed a Competition problem, please delete the thread, and warn the postor. Thanks.  (You may also like to inform the competition coordinator - if possible)


number theory
greatest common divisor
number theory unsolved

For $ n \equal{} 2$ and $ gcd\left(p,q\right) \equal{} 1$, it is known that the largest number that cannot be written as $ px \plus{} qy$ ($ p,q,x,y\in\mathbb{N}$) is $ pq \minus{} p \minus{} q$.  I know the proof where you show that $ pq \minus{} p \minus{} q$ cannot be written as such but $ pq \minus{} p \minus{} q \plus{} k$ can be for all $ k\geqslant 1$.  However, I am having trouble with the following alternative proof (attached as a JPG file):

It seems to me mysterious why the author writes $ p \equal{} px \plus{} qy$ instead of $ c \equal{} px \plus{} qy$ for arbitrary $ c$ (why write $ p$ as such?).  If it were arbitrary $ c$, then clearly since $ px\plus{}qy\equal{}\gcd\left(p,q\right)\equal{}1$ for some $ x,y$, we can find $ p\left(xc\right)\plus{}q\left(yc\right)\equal{}c$.

Also, how do we know that such a number is representable as $ px \plus{} qy$ precisely when $ 0\leqslant x\leqslant q \minus{} 1$ and $ 0\leqslant y$?

Looks like a typo.  That $ p$ should just be another variable.

OK that's what I thought.  Now, for $ 0\leqslant x\leqslant q \minus{} 1$ why is it decidedly apparent that $ c\in\mathbb{N}$ is not representable when $ y < 0$ and is representable when $ 0\leqslant y$?

$ c$'s representation takes the form $ px \plus{} qy$ for some non-negative (not positive; your original problem is slightly wrong) integers $ x, y$.  If we know that $ c$ takes the form $ px' \plus{} qy'$ for some integers $ x', y'$, then we can describe all possible solutions to $ c \equal{} pa \plus{} qb$ in integers $ a, b$, and if $ c$ has a representation in the form $ px \plus{} qy$ in non-negative integers $ x, y$, it must be among these solutions, and can be isolated by constraining $ x$ to its smallest non-negative value.

If $ x,y\in \mathbb{N}$ then the number $ pq > pq \minus{} p \minus{} q$ must be expressed in the form $ px\plus{}qy$. But then $ q|x$ and $ p|y$ so $ pq \equal{} px \plus{} qy\geq pq \plus{} qp \equal{} 2pq$ - contradiction. In matter of fact the largest number that can be expressed in the form $ px \plus{} qy$ $ (x,y\in \mathbb{N})$ is exactly $ pq$ :P


analytic geometry
graphing lines
slope
geometry
geometric transformation
reflection
AMC

In the Cartesian Plane, triangle ABC is defined as A(0,0), B(40,0),
C(40,9). ABC is then reflected about the hypotenuse. POint D is the image of B. FInd the slope AD.

Um, no
Remember what happens a reflection: [i]the distance of the original point to the line of reflection is the same as the distance from the image to the line of relection.[/i]
Obviously, in an isosceles right triangle, when u reflect a vertex, u end up with a square, since the diagonals are perpendicular to each other.
So what u need to do is, from P, draw a line perpendicular to the hypotenuse, and then work from there.

Do you see?

[hide]if the angle a line makes with the x-axis is :theta:, then the slope of that line will be tan :theta:. Let us let :theta: in this case be CAB. We wish to find tan 2:theta:. By the double-angle identity, this is equal to 2tan:theta:/(1-tan:^2::theta:).  We know tan:theta: is 9/40, so you can plug that in.[/hide]

[quote]We wish to find tan2 :theta: [/quote]

But how did you know that from the start?[/quote]

[hide]

Slope of hypotenuse=9/40

Segment BD has to be perpendicular to AC

BD has a slope of -40/9

[/hide]

err
im sorry, no
i believe ur reasoning is incorrect


function
calculus
derivative
calculus computations

The function y = x^4 + bx^2 + 8x + 1 has a horizontal tangent and a point of inflection for the same value of x. What must be the value of b?

A. -1
B. 4
C. 1
D. 6
E. -6

[quote="Prime factorization"]Well in order for that to happen, both the first and second derivative of y must equal 0. Thus:

$ y' \equal{} 4x^{3} \plus{} 2bx \plus{} 8 \equal{} 0$
$ y'' \equal{} 12x^{2} \plus{} 2b \equal{} 0$
meaning
$ 4x^{3} \plus{} 2bx \plus{} 8 \equal{} 12x^{2} \plus{} 2b$

From there I'm not 100% sure what to do actually  :lol:[/quote]

From the two equations...
[hide]
The second derivative equation implies $ b\equal{}\minus{}6x^2$. Multiplying the equation for $ y^\prime$ by -3, we then get
\[ 0\equal{}2bx\minus{}6bx\minus{}24\Rightarrow bx\equal{}\minus{}6\]
Now multiply $ b\equal{}\minus{}6x^2$ by $ b^2$ to get $ b^3\equal{}(\minus{}6)^3$. So $ b\equal{}\boxed{\minus{}6}$.

Of course, given the multiple choice, you could brute force each of the options.  :P
[/hide]


analytic geometry
geometry
circumcircle
conics
parabola
parameterization
perpendicular bisector

Given a curve $C:y=x^2-\frac{5}{4}$ on $xy$ plane.Let $p,q$ be the $x$-coordinates of two points $P,Q$ on $C$ respectively such that $p-q=1, p\neq q$ and let $R$ be the intersection point of two tangent lines of $C$ at $P,Q$.Find the locus of the center of the circle passing through three points $P,Q,R$.

Your problem will immediately follow from the following more general result:

[color=blue][b]Problem. [/b]Given the curve $C:y = x^2  - \frac{5}{4}$  on the xy plane. Let p, q be the x-coordinates of two points P, Q on C such that  $p - q = k, k \ne 0$, and let R be the intersection point of two tangent lines of C at P, Q. The locus of the circumcircle of $\triangle PQR$  is the cubic
\[8x^2  - \left( {y + 1} \right)\left( {2y + 1 - k^2 } \right)^2  = 0\][/color][i]Proof.[/i]  (The boring calculation are done with the help of [i]Mathematica[/i] !)  The equations of the tangents lines to the parabola C at the points Q, P are respectively :

$[1] \qquad 2q\left( {x - q} \right) - y = \frac{5}{4} - q^2   $

$[2] \qquad  \left( {2q + 2k} \right)\left( {x - q - k} \right) - y = \frac{5}{4} - \left( {q + k} \right)^2 $

To find the coordinates of the point R let’s solve the system

$
\left\{ \begin{array}{l}
 2q\left( {x - q} \right) - y = \frac{5}{4} - q^2  \\  \\
 \left( {2q + 2k} \right)\left( {x - q - k} \right) - y = \frac{5}{4} - \left( {q + k} \right)^2  \\  \\
 \end{array} \right.{\rm{   }} \Rightarrow {\rm{   }}\left\{ \begin{array}{l}
 x = \frac{1}{2}\left( {2q + k} \right) \\  \\
 y = \frac{1}{4}\left( {4q^2  + 4qk - 5} \right) \\  \\
 \end{array} \right.
$

The perpendicular bisector of PR, QR are respectively:

$[3] \qquad 4x + 8\left( {q + k} \right)y = 8q^3  + 20q^2 k + \left( {16k^2  - 6} \right)q + 4k^3  - 7k$

$[4] \qquad 4x + 8qy = 8q^3  + 4kq^2  - 6q + k$

Solving the system given by [3], [4] we get the circumcenter of  PQR

$[5] \qquad \left\{ \begin{array}{l}
 x =  - \frac{1}{4}\left( {8q^3  + 12q^2 k + 4qk^2  - 2q - k} \right) \\  \\
 y = \frac{1}{2}\left( {4q^2  + 4qk + k^2  - 2} \right) \\  \\
 \end{array} \right.$

The [5] are the parametric equations of the locus. By eliminating the parameter q we get the Cartesian equation
\[8x^2  - \left( {y + 1} \right)\left( {2y + 1 - k^2 } \right)^2  = 0 \qquad \blacksquare\]

[color=blue][b]Initial problem.[/b][/color] If k=1 the the equations of the locus is :

\[2x^2  - y^2 \left( {y + 1} \right) = 0\]

Wao! Fantastic! Thank you for your reply[b], Leon[/b]. :)

Your solution is perfect!

kunny


induction inequalities algebra function domain

The following more general inequality holds by induction and the proof is identical:  let $ a_1, a_2, ... a_n$ be reals greater than or equal to $ \minus{}1$ and all of the same sign.  Then

$ (1 \plus{} a_1)(1 \plus{} a_2)...(1 \plus{} a_n) \ge 1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n.$

The desired result follows by letting $ a_i \equal{} x$.

Actually, isn't that a special case of the [url=http://planetmath.org/encyclopedia/BonferroniInequalities.html]Bonferroni inequalities[/url] ($ k \equal{} 1$), with a bit of extension of the domain?

[quote="JasonSullivan"][b]By induction:[/b]
Prove Bernoulli’s Inequality which states that
(1 + x)n ≥ 1 + nx for x ≥ −1 and n ∈ N.[/quote]
No one corrected this from the OP.

It's (1 + x)^n >= 1 + nx for every integer n >= 0 and every
real number x > -1.


Hi

I just read in a textbook that f:R-->R where f(x) = x^3 is a onto function. I feels it is not correct because every element in co-domain (i,e Set of Real numbers) doesn't have a corresponding element in domain of the function. Can somebody please confirm if I am correct?

[quote="vijay_singh"]because every element in co-domain (i,e Set of Real numbers) doesn't have a corresponding element in domain of the function. [/quote]
Why not?  The inverse function is given by $ f(x) \equal{} \sqrt[3]{x}$ and is defined for all reals.  Have you looked at a graph of this function?

[quote="t0rajir0u"][quote="vijay_singh"]because every element in co-domain (i,e Set of Real numbers) doesn't have a corresponding element in domain of the function. [/quote]
Why not?  The inverse function is given by $ f(x) \equal{} \sqrt [3]{x}$ and is defined for all reals.  Have you looked at a graph of this function?[/quote]

Thanks, I get it now. The reason I was getting confused because the textbook also said f:R-->R where f(x)=x^2 is not a onto function. And I get it now, it is because range of the f(x)= X^2 is all positive real numbers.

But thanks for responding so quickly, didn't expect so prompt reply :-).

Vijay


linear algebra
matrix
linear algebra unsolved

Let $a,b,c$ be complex numbers and $A,B\in\mathcal l M_n(C)$ such that $(a+b)AB+cI_n=aBA$. Prove that $AB-BA$ is nilpotent.

when $a=0$,$b=0$,$a+b=0$ &$b=-2a$ it's easy to solve the problem so let's suppose the case where none of these conditions is fullfiled:
for $\lambda \in C$
$(a+b)[(AB-BA)-\lambda I_n]=-[(c+\lambda(a+b))I_n+bBA]$
and $a[(AB-BA)-\lambda I_n]=-[(c+\lambda a)I_n+bAB]$
Thus if $\lambda$is an eigenvalue of $AB-BA$ then ${\lambda a}/{a+b}$ is an eigenvalue of $AB-BA$ too ,as we've got$|a/{a+b}|\neq 1$ thus this way if $\lambda \neq 0$ we're gonna have infinite eigenvalues for the matrix $AB-BA$ which is obviously wrong .Then all the eigenvalues of $AB-BA$ are null and this way we prove that $AB-BA$ is nilpotent :).


If $ \frac{x}{y}\equal{} 2$ and $ \frac{z}{x}\equal{} 4$, what is the value of $ \frac{z}{y}$?

$ \frac{z}{y} \equal{} \frac{x}{y} \times \frac{z}{x} \equal{} 2 \times 4 \equal{} \boxed{8}$.


[hide="#1"]Note that $ \cos(a\plus{}b)\equal{}\cos a \cos b\minus{}\sin a \sin b$.

$ \cos a\equal{}\cos (\minus{}a)$ and $ \sin a\equal{}\minus{}\sin (\minus{}a)$ since cosine is an even function and sine is an odd function.

Hence, $ \cos 83 \cos 17\plus{}\sin 83 \sin 17\equal{}\cos (83\minus{}17)\equal{}\cos 66$ or $ \cos (17\minus{}83)\equal{}\cos 294$.

Answer: $ \cos 66$ or $ \cos 294$[/hide]

[quote="GautamR"]express in exact (radical, pi, or fraction; NO DECIMAL) form.  Use sum, difference, product, or half-angle formulas.

1.  cos83cos17 + sin83sin17
  I got to cos66, but I cant figure out how to get that in exact form

2.  cos((1/2)arctan(12/5)) Angle in Quadrant I

Thanks for your help![/quote]

I think with number $ 2$ you can convert arctan into arccos and change $ \frac{12}{5}$ to $ \frac{5}{13}$. then multiply $ \frac{5}{13}*\frac{1}{2}\equal{}\frac{5}{26}$ and then since $ \frac{5}{26}$ is in the domain of arccos, the cos and arccos functions will cancel each other out, leaving you with $ \boxed{\frac{5}{26}}$.

[quote="FantasyLover"][hide="#1"]Note that $ \cos(a \plus{} b) \equal{} \cos a \cos b \minus{} \sin a \sin b$.

$ \cos a \equal{} \cos ( \minus{} a)$ and $ \sin a \equal{} \minus{} \sin ( \minus{} a)$ since cosine is an even function and sine is an odd function.

Hence, $ \cos 83 \cos 17 \plus{} \sin 83 \sin 17 \equal{} \cos (83 \minus{} 17) \equal{} \cos 66$ or $ \cos (17 \minus{} 83) \equal{} \cos 294$.

Answer: $ \cos 66$ or $ \cos 294$[/hide][/quote]

cos66=cos294

he did say he wanted an exact answer...

try doing $ cos(2\theta)$ a few times to get a nice number, then doing angle addition.

$ cos(2*66))\equal{}cos(132)$
$ cos(2*132)\equal{}cos(264)$

Aha!
$ cos(264)\equal{}cos(294\minus{}30)$
$ cos(264)\equal{}cos(66\minus{}30)$

Plug the formula in a few times and then solve.

to convince yourself on number 2, what you can do is draw a right triangle that satisfies the inner part of the question

[hide]
we are given that we want cos of arctan 12/5
we can draw a triangle with an angle of arctan 12/5 by drawing the triangle with the legs 12 and 5
from this hypotenuse is 13

then, we use cosine

[/hide]

for 1, are you sure the question is right?  cos 66 does not seem doable with a lot of work or a calculator[/hide]

im sure                                     .

[quote="rd5493"]
$ cos(264) \equal{} cos(294 \minus{} 30)$
$ cos(264) \equal{} cos(66 \minus{} 30)$
[/quote]

This isn't true, but you've got good point.
Let $ \cos66^\circ\equal{}x$, then $ \sin66^\circ\equal{}\sqrt{1\minus{}x^2}$

$ \cos132^\circ\equal{}2x^2\minus{}1$
$ \cos264^\circ\equal{}2(2x^2\minus{}1)^2\minus{}1\equal{}8x^4\minus{}8x^2\minus{}3$
$ \cos264^\circ\equal{}x\cos30^\circ\minus{}\sqrt{1\minus{}x^2}\sin30^\circ$

$ 16x^4\minus{}16x^2\minus{}6\equal{}\sqrt3x\minus{}\sqrt{1\minus{}x^2}$

Now you can solve a one variable equation...

well, it seems like it may get into really high powers...maybe like 8 or 16?!?!?!?!

I suk at trig, IDK how to do it either

I have computed that the exact value for $ \cos 66^\circ$ is:
\[ \frac 18 \left(\sqrt 3 \plus{} \sqrt {15} \minus{} \sqrt {10 \minus{} 2 \sqrt 5}\right)
\]
Edit:
[quote="Bugi"]$ \cos264^\circ \equal{} 2(2x^{2} \minus{} 1)^{2} \minus{} 1 \equal{} 8x^{4} \minus{} 8x^{2}\color{red} \plus{} 1$[/quote]
Ah-ha! Your faulty step lies here! :D

Edit 2:
[hide="Solution to #2"]
Drawing a 5-12-13 right triangle, if $ \tan \theta \equal{} \frac {12}{5}$, then it follows that $ \cos\theta \equal{} \frac {5}{13}$ and then $ \cos \left(\frac 12 \theta\right) \equal{} \sqrt {\frac {1 \plus{} \cos\theta}2} \equal{} \sqrt {\frac {1 \plus{} \frac {5}{13}}2} \equal{} \sqrt {\frac {9}{13}} \equal{} \frac 3{\sqrt {13}}$.[/hide]


algebra
polynomial
analytic geometry
function
LaTeX
quadratics
floor function

[b]The problem is :[/b]

\[ Find\ the\ no.\ of\ distint\ terms\ in\  \\
\\(x_1)\ (x_1\ +\ x_2\ )\ (.....)\ (x_1\ +\ x_2\ +\... +\ x_n\ )\ \]

[b]According to me the reqd answer is equal to the the no. of integer solns of [/b]
\[ e_1\ +\ e_2\ +\ ....\ +\ e_n\ =\ n!\ \\
\\s.t.\ e_1\ \geq e_2\ \geq ....\ \geq e_n\ \]

[b]Is this right? If yes, how do you solve the last eqn. to find the no. of integer solns.[/b]

no it is not right.what you are claiming it is actually too large-your answer is the partition number of $n!$ and that is a
 v large number.it is true that there are a total of $n!$ terms in the expansion in all but then there are many terms that regroup on account of being the same.
what instead you can do is the following: the number of distinct terms is counted by the number of distinct sequences of degrees $(a_1,a_2,\ldots,a_n)$ where $a_i$ is the degree of $X_i$.now note that $a_n\leq 1,a_{n-1}\leq 2$ and so on.but further something else is true:we actually also have $a_n+a_{n-1}+\cdots+a_{n-k}\leq n-k+1\ \forall k\leq n$.thi is so since there are only $(n-k+1)$ terms that contain all of the variables $X_n,X_{n-1},\ldots,X_{n-k}$.
thus the sequences we wish to include are those(and only those!) which satisfy the above inequalities.counting these from back to front we want the number of $n$-tuples $(b_1,b_2.\ldots,b_n)(b_i=a_{n-i+1})$ that satisfy,$\sum_{i=1}^{k} b_i \leq k\forall k$.but these simply correspond to the 'paths' from $(0,0)$ to $(n,n)$ that never go above the line $y=x$, and so the number of terms is the Catalan number $\frac{\binom{2n}{n}}{n+1}$.

I don't know anything about catalan numbers.Can yo elaborate.Anyways,can one solve the eqn I have given and what would be its applications?

by solve the equation if you mean get a 'closed form' formula for it in terms of polynomials and other things-no.the partition numbers are the number of ways you can write a number as sum(s) of smaller numbers.for eg: 4=3+1=2+2=2+1+1=1+1+1+1,so that you can always arrange these in deceasing order.one can get some asymptotic formulas for the partition numbers(as was done by the legendary Ramanujan with Hardy) but there is no closed form formula.
the catalan numbers are just $\frac{\binom{2n}{n}}{n+1}$.they count several things and are easily as important combinatorial number sequences as the binomial coeffs.some of the things they count are the number of binary bracketings of n letters (Catalan's problem), the solution to the ballot problem, the number of trivalent planted planar trees, the number of states possible in an n-flexagon, the number of ways of forming an n-fold exponential, the number of rooted planar binary trees with n internal nodes, the number of rooted plane bushes with n graph edges, the number of extended binary trees with n internal nodes, the number of mountains which can be drawn with n upstrokes and n downstrokes and so on and on! just google on catalan numbers for more info.

I realise now that in my original question \[n\ \] should replace \[ n\ !\ \] and what was wanted was not the no.of solns but the different type of solns.Its very confusing so leave it.The book which has this problem has given the same soln as you have and I am satisfied.

But I have another question regarding the ubiquitous Catalan numbers.I read somewhere that they also appear in finding the no. of paths from (0,0) to (n,n) on the coordinate plane which do not cross y =x . I've interpreted it in a certain way and want to know if i'm correct.(it's all mine - not from a book I know)

I have associated every horizontal diplacement with 0 and every vertical displacement with 1. let each displacement (0 or 1) be denoted by Xi.

therefore
\[ X_{i}\ =\ 0\ or\ 1\ \\
\\The\ required\ no.\ of\ paths\ is\ equal\ to\ the\ number\ of\ solns\ of\ the\ eqn\ \\
\\ X_1\ +\ X_2\ +\ ....\ X_{2n}\ =\ n\ s.t.\ \\
\\ X_1\ +\ X_2\ +\ ....\ X_r\ \leq  [\ \frac{r}{2} ]\ \ \ \forall\  r\  \leq 2n\ \\
\\ where\ [ \ ]\ denotes\ the\ greatest\ integer\ fuction\  \]

I hope this interpretation of catalan nos. is correct.Also how does one solve this last set of eqns. and ineqns. using a purely alegbraic method i.e. a method which does not involve interpreting this eqn. in reference to another problem(even if my interpretation above is wrong).I wold also like to know in which problem did the catalan nos. first appear?Also can you recommend to me some widely available , popular and excellent Combinatorics books which cover all theory till graduate level and also have good problems and exercises?

[b]THANKS!!![/b]

(is there any symbol in latex for greatest integer fn.Also many people have told me to write latex enclosed in $ symbols to write in the same line but so far I have been unsuccessful in interpreting them.can you post me an example code illustrating how to write latex and normal text in same line.)

[quote="seshadri"]you don't need $x_1+x_2+\cdots+x_r\leq \left[\frac{r}{2}\right]$; it should be $r$ instead.
An easier way to think about these is to think of a number of paths from (0,0) to (2n,0) with each step being +1 or -1.and you want to count the number of paths which don't cross the X-axis(or rather the paths which stay always above(below) the x-axis).
 a great book in Enumerative Combinatorics is the classic by Richard Stanley.[/quote]

[b]won't "r": be a huge overstatement[/b]. r sort of refers to the [b]steps of the binary sequence[/b] I constructed (consisting of 0's and 1's) . If you look at this binary sequence , the sum indeed comes out to be less than greatest integer fn of (r/2) (i have tested it for lower values of n - it's correct)(i suppose you didn't keep in mind the r = odd cases !!)

Also you haven't elaborated a 'purely algebraic ' soln. of the above eqn.You've simply given another interpretation of the same original problem (something which I have already done) Can you illustrate such an algebraic soln?!

I suppose the book you are referring isnt available in India.I got Alan tucker's Applied Combinatorics.What do you make of that.Has any other other given the interpretation I have given of catalan nos.??

Also , can you kindly answer the latex problem I have posted in my previous post.It's getting a bit frustrating not able to write latex in the same line as your normal post!!

well, it seems to me that what you understand by paths not crossing that line refers to paths both above $\it{and}$ below that line, whereas what we generally do hile counting the so called Dyck paths is consider only those entirely above or entirely below.

an algebraic way of calculating the catalan numbers is to count these as the number of paths from (0,0) to (2n,0) with each step being one to he right and either a step up or a step down. then if $C_n$ denotes the number of such good paths, then we can get the following recurrence relation:
  $C_n=\sum_k C_{k-1}C_{n-k}$.Then if we form the generating function $F(x)=\sum_{k=0}^{\infty} C_nx^k(C_0=1)$ then the above relation gives us $F(x)-1=xF(x)^2$ solving back(as a quadratic) for $F(x)$ from this gives $F(x)=\frac{1-\sqrt{1-4x}}{2x}$.from this one can get back $C_n$  as the coeff of $x^n$ in its series expansion.
 stanley's book is titled Enumerative Combinatorics(2 volumes,though the second vol is more advanced material) and is easily oen of the best books available in this subject.it should posibly be available in some libraries.
finally about latex: one way to learn latex is to use the latex help in this forum.or next time just quote someone when you want to make a post;that way you will see their latex code and you can pick  up things from that directly.
for integer value you can use\lfloor,\rfloor or \lceil,\rceil to give the following symbols in latex:
$\lfloor,\rfloor,\lceil,\rceil$.


find $ f: Z\minus{}>Z$ such that:
$ f(a^3\plus{}b^3\plus{}c^3)\equal{}f(a)^3\plus{}f(b)^3\plus{}f(c)^3$

[quote="chien than"]It is a problem in Vietnam TST 2005
See at http://www.mathlinks.ro/viewtopic.php?search_id=982674035&t=14908[/quote]
yes,chien than! :lol:


quadratics
number theory
prime numbers
number theory open

Are there infinite many prime numbers of the form $5n+2$, with $n$ an integer?

Yes by Dirichlets Theorem on primes in arithmethic progression.
It says that for any coprime integers $a,m$ there are $\infty$ many primes $p \equiv a \mod m$.
There are elemetary proves for some special cases like $a \equiv 1 \mod m$ or $m=2,3,4,6,8$, but I don't know any for the case $a=2, m=5$ (using the law of quadratic reciprocity it should be possible to prove that there are at least $\infty$ many $\equiv 2,3 \mod 5$).

Ok, thanks!
Now I propose a bit more difficult problem :)
Are there $\infty$ many prime numbers of the form $n^2-n+1$? (I think it is an unsolved problem, right?)


I've been noticing some people going to a User on AoPS' userpage and creating a mini-biography on them. While that's definitely good faith, userpages are not for biographies ;) It is generally considered inproper etiquette on Wikipedia to edit someone else's userpage unless it contains things like personal attacks, or was vandalized, or you're fixing a link or something.

I don't think it's been done often.  I cautioned one user against it (it might have been your user page), but I don't think it's been rampant.  Either way, we should be sure to add that to the netiquette page.  I would do it myself, but as of now, I'm on vacation for a few days.


I know that some schools consider AMC and AIME scores when looking at applications. If you're applying, do you know if schools will also consider that year's AMC and AIME scores, even though they wont be in until after your application has been sent? (I'm banking on getting a good first round USAMTS score to add onto my resume, as I haven't really been heavy in the math competitions until recently.)

The AIME takes place around the time regular decision gets mailed out, so somehow I doubt it.  I don't even know if the AMC is scored early enough to affect the decision, and I also don't know that it matters that much.  It's more important that you've participated, I think, than that you've scored well (phenomenal scores aside).

As for the USAMTS, I have no idea what the schedule for that is.

Nevertheless, there's no harm in sending the scores anyway.

It doesn't hurt to send them in if you get deferred or waitlisted.

I sent my AMC12 score in to MIT, along with a short cover letter about how I wanted to go there. I can only guess what effect it would have had, because they lost my humanities rec-letter.

well it's not likely your AIME/AMC stuff will help out because it is so late.
USAMTS's first round result should be out in November I think, but I don't think most college even know about it.


1.  Two roots of the real quartic [tex]x^4 - 18x^3 + ax^2 + 200x - 1984 = 0 [/tex]have product -32. Find a.

[hide]1984 factors to 2^6*31

if the equation's roots are m,n,o,p then mn=-32 and op=62

if the equation is (x-m)(x-n)(x-o)(x-p), then expanded, that looks like:

x^4-(m+n+o+p)x^3+(mn+mo+mp+no+np+op)x^2-(mno+mnp+nop)x+mnop

so we get 18=m+n+o+p

a=mn+mo+mp+no+np+op

-200=mno+mnp+mop+nop

and as we found before, mnop=-1984, thus yeilding to mn=-32 and op=62

substitute into the equations, making a=mn+mo+mp+no+np+op

a=62-32+mo+mp+no+np or

a-30=(m+n)(o+p)

-200=mno+mnp+mop+nop

-200=62m+62n-32o-32p

-200=62(m+n)-32(o+p)     look familiar?

make m+n=x and o+p=y

x+y=18

xy=a-30

62x-32y=-200

just use x+y=18 and 62x-32y=-200 for now. it's just a system of 2 equations which yields x=4 and y=14

so, a-30=56 and a=86[/hide]

Rep, I think you made a little error here:

Rep123max wrote:[hide]Now we can go into the third equation, and substituting what we know for ab and cd we have 62c+62d-32a-32b=-200. This is equivalent to 62(c+d)-32(a+b)=-200.[/hide]

I think you meant [hide]62a+62b-32c-32d=-200[/hide]

I could be wrong though, because it's pretty early in the morning and I'm playing a game right now.

[quote="Rep123max"]Ya, I did. It still doesnt effect the answer though (I dont think).[/quote]
yeah, the answer's still the same either way

That was really really easy USAMO problem, but I think that year must have a really hard #4 or #5.

[quote=sugar_rush]lol this could be an AIME #8[/quote]

fr tho

maybe even an aime 5 lol

WoW this is a 16 year old thread. Nice job @benchkodakshark for bumping it

[quote=benchkodakshark][quote=eagles2018]Nice bump. I just solved it; does $a=86$? This should be a 7 on the AIME[/quote]

but its a usamo problem and aime isnt proofs[/quote]

True, but it still could have been used as an AIME problem. The problem itself doesn't require you to write a proof; it could easily appear in a short answer test.

The lack of latex in those solutions in 2004 makes me a bit uncomfortable

[quote=OlympusHero][quote=ilovepizza2020]This is the classic olyhero flexing vietas problem.[/quote]

I am not very happy about this comment :maybe:

I didn't even use Vieta :P[/quote]

Hmm then what did you do to solve it?

Undetermined Coefficients  :winner_first:

[quote=benchkodakshark]thanks @kante314[/quote]

Like it was sarcasm, but 86 is the answer


[quote="arp"]Show that for odd prime $p$, if $2^{m}$ is not congruent to $1$  $\mod p$ (how do you write not congruent symbol?), then
$\sum_{n=0}^{p-1}n^{m}\equiv 0 \mod p$[/quote]
If $(p-1)\not m$, then $\sum_{n=0}^{p-1}n^{m}\equiv 0 \mod p$.
For example $2^{3}\equiv 1 \mod 7$, but $\sum_{n=0}^{6}n^{3}=3^{2}7^{2}\equiv 0 \mod 7$.

[quote="arp"]Show that for odd prime $p$, if $2^{m}$ is not congruent to $1$  $\mod p$ (how do you write not congruent symbol?), then
$\sum_{n=0}^{p-1}n^{m}\equiv 0 \mod p$[/quote]
An derectly solution$(2^{m}-1)\sum_{n=0}^{p-1}n^{m}\equiv \sum_{n=0}^{p-1}{2n}^{m}-\sum_{n=0}^{p-1}n^{m}\equiv \sum_{j=2n}{j}^{m}-\sum_{n=0}^{p-1}n^{m}\equiv 0$
becuase$2^{m}-1\not \equiv 0$ then $\sum_{n=0}^{p-1}n^{m}\equiv 0$
quote my post ,you will find that how to input $\not \equiv$ :)

[quote="arp"](how do you write not congruent symbol?)[/quote]

\equiv : $\equiv$
\not\equiv: $\not\equiv$


[color=blue]What's your favorite beautiful word in various meaning?[/color]

Confuzzled :!:
Means Confused and Puzzled COMBINED.  ==>  :ewpu:


Denote by $S$ the set of all primes $p$ such that the decimal representation of $\frac{1}{p}$ has the fundamental period of divisible by $3$. For every $p \in S$ such that $\frac{1}{p}$ has the fundamental period $3r$ one may write \[\frac{1}{p}= 0.a_{1}a_{2}\cdots a_{3r}a_{1}a_{2}\cdots a_{3r}\cdots,\] where $r=r(p)$. For every $p \in S$ and every integer $k \ge 1$ define \[f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2r(p)}.\] [list=a] [*] Prove that $S$ is finite. [*] Find the highest value of $f(k, p)$ for $k \ge 1$ and $p \in S$.[/list]


Find all polynomials with real coefficients, for which the equality
\[ P(2P(x)) \equal{} 2P(P(x)) \plus{} 2(P(x))^{2}\]
holds for any real number $ x$.

Let $ P(x) \equal{} x^n \plus{} a_1x^{n \minus{} 1} \plus{} \dots \plus{} a_n$

$ \Rightarrow P(2P(x)) \equal{} 2P(P(x)) \plus{} 2(P(x))^{2} \Leftrightarrow 2^nx^{n^2} \plus{} \cdots \equal{} 4x^{2n} \plus{} \cdots$
$ \Rightarrow n \equal{} 2\Leftrightarrow P(x) \equal{} x^2 \plus{} ax \plus{} b$

$ 4(x^2 \plus{} ax \plus{} b)^2 \plus{} 4a(x^2 \plus{} ax \plus{} b) \plus{} 2b \equal{}$ $ 2(x^2 \plus{} ax \plus{} b)^2 \plus{} 4a(x^2 \plus{} ax \plus{} b) \plus{} 2b \plus{} 2(x^2 \plus{} ax \plus{} b)^2$  :ninja:

How about the cases for $ P(x)\equal{}a_0 x^n \plus{} a_1 x^{n\minus{}1}\plus{} \cdots\plus{}a_n$ with $ a_0 \neq 1$?

[quote="BG Yoda"]$ 2^nx^{2n} \plus{} \cdots \equal{} 4x^{2n} \plus{} \cdots$[/quote]
$ deg(P(2P(x)))\equal{}n^2$ and not $ 2n$

let$ P(x) \equal{} a_n x^n \plus{} a_{n \minus{} 1} x^{n \minus{} 1} \plus{} \cdots \plus{} a_0$,
let $ P(2P(x)) \minus{} 2P(P(x)) \equal{} g(x)$,
when $ deg(P(x))\ge3$,i.e. $ a_n\neq0$
$ \deg(P(2P(x))),\deg(P(P(x)) \equal{} n^2 > deg(P(x))^2 \equal{} 2n$
but $ P(2P(x)) \minus{} 2P(P(x)) \equal{} 2(P(x))^{2}$
$ \therefore deg(g(x)) \equal{} 2n$
the coefficient of $ x^{n^2}$ in $ g(x)$ is $ (2^n \minus{} 2)a_{n}^{n \plus{} 1}$,
$ \therefore (2^n \minus{} 2)a_{n}^{n \plus{} 1} \equal{} 0, a_{n} \equal{} 0$which is contradictory to $ a_n\neq0$
when $ \deg(P(x)) \equal{} 2$,
then$ 2(a_2 \minus{} 1)[P(x)]^2 \equal{} a_0,\forall x\in R$,
$ \therefore a_2 \equal{} 1, a_0 \equal{} 0$,
$ P(x) \equal{} x^2 \plus{} a_1x$ where a_1 is any real number
when $ \deg(P(x)) \equal{} 1$,
then$ 2(a_{1}x \plus{} a_0)^2 \equal{} \minus{} a_0,\forall x\in R$
which is impossible.
therefore there is only one solution  $ P(x) \equal{} x^2 \plus{} ax,\forall a \in R$

If $ | \{ P(x) : x \in \mathbb{R} \} | < \infty$ then $ P(x) \equal{} c \forall x$. Substitute into the equation, one has
\[ c \equal{} 2c \plus{} 2c^2 \iff c \equal{} 0 \vee c \equal{} \minus{}\frac 12\]

Now, in the other case, since $ P(x)$ consumes infinitely many values, it must be the case that
\[ P(2y) \equal{} 2P(y) \plus{} 2y^2 \forall y \in R\]
Let $ Q(x) \equal{} P(x) \minus{} x^2$. It turns out to be $ Q(2x) \equal{} 2Q(x) \forall x$.
The only possible case are $ Q(x) \equiv 0$ or $ Q(x) \equiv \alpha x$ where $ \alpha \not\equal{} 0$. These two cases could be merged to $ Q(x) \equiv \alpha x$.
Therefore,

\[ P(x) \equal{} \begin{cases}
0 \\
\minus{} \frac 1 2 \\
x^2 \plus{} \alpha x
\end{cases}\]

[quote=AlcumusGuy]I know this is a major revive, but I was doing this problem today, and I appear to be getting something different. Can someone check my solution please?

[hide="Solution"]
Let $P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ with $a_n \ne 0$ and all $a_i \in \mathbb{R}$ so that it is a polynomial of degree $n$. We start by considering the leading term of each expression:

$P(2P(x)) = P(2(a_nx^n + ...)) = a_n(2a_nx^n + ...)^n + ... = a_n(2^na_n^n(x^n)^n + ...) + ... = 2^na_n^{n+1}x^{n^2} + ...$

$2P(P(x)) = 2P(a_nx^n + ...) = 2[a_n(a_nx^n + ...)^n + ...] = 2a_n(a_n^n(x^n)^n + ...) + ... = 2a_n^{n+1}x^{n^2} + ...$

$2(P(x))^2 = 2(a_nx^n + ...)^2 = 2(a_n^2(x^n)^2 + ...) = 2a_n^2x^{2n} + ...$

For the two sides to be always equal, the coefficients of each term must be equal. In particular, the leading coefficients must be equal. On the left side (expanded), we have a polynomial of degree $n^2$. On the right side, we have the sum of a polynomial of degree $n^2$ and a polynomial of degree $2n$, so the combined degree will be $\max(n^2, 2n)$.

If $n>2$, then $n^2 > 2n$, so the expression $2(P(x))^2$ does not contribute to the leading coefficient of the expanded polynomial on the right side. Comparing leading coefficients, we have $2^na_n^{n+1} = 2a_n^{n+1}$. We assumed in our original definition that $a_n \ne 0$, so we can divide both sides by $a_n^{n+1}$ to get $2^n  = 2$. The only solution to this is $n = 1$, but we assumed that $n>2$, hence we have a contradiction, so this case is not possible.

Next we consider $n < 2$. If $n = 1$, then $n^2 < 2n$, so the left side will have a degree (1)  less than that of the right side (2), so this case is not possible. If $n = 0$, then $P(x) = y$, for some constant $y$. Substituting this into the original polynomial equation yields $y = 2y + 2y^2$. Rearranging gives $2y^2 + y = 0$, so $y = 0$ or $y = -\frac{1}{2}$. We can easily check to see that both work, so $\boxed{P(x) = 0}$ and $\boxed{P(x) = -\frac{1}{2}}$ are possible solutions.

Finally, we take the case $n = 2$. Then comparing leading coefficients gives us $4a_2^3 = 2a_2^3 + 2a_2^2$. Rearranging and dividing out a factor of $2$ gives $a_2^3 - a_2^2 = 0$, or $a_2^2(a_2 - 1) = 0$. Since we must have $a_2 \ne 0$, the only solution is $a_2 = 1$. Hence our polynomial takes the form $P(x) = x^2 + bx + c$. We now substitute this into our original polynomial equation the following:

\[P(2(x^2 + bx + c)) = 2P(x^2 + bx + c) + 2(x^2 + bx + c)^2\]
\[(2(x^2 + bx + c))^2 + b[2(x^2 + bx + c)] + c = 2[(x^2 + bx +c)^2 + b(x^2 + bx + c) + c] + 2(x^2 + bx + c)^2\]
\[4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c = 2(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c + 2(x^2 + bx + c)^2\]
\[4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c = 4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c\]

Both sides are identically equal, so all polynomials of the form $\boxed{P(x) = x^2 + bx + c}$ for any $b, c \in \mathbb{R}$ also work.

[hide="Sidenote"]
Is there a notation that I can use for leading coefficients so that I don't have to keep writing $...$?
[/hide]
[/hide][/quote]

there is error at these lines:
\[(2(x^2 + bx + c))^2 + b[2(x^2 + bx + c)] + c = 2[(x^2 + bx +c)^2 + b(x^2 + bx + c) + c] + 2(x^2 + bx + c)^2\]
\[4(x^2 + bx + c)^2 + 2b(x^2 + bx + c) + c = 2(x^2 + bx + c)^2 + 2b(x^2 + bx + c) +   >2c<      + 2(x^2 + bx + c)^2\]
so c=0 and answer will $\boxed{P(x) = x^2 + bx}$


induction
quadratics
number theory unsolved
number theory

The sequence $ (a_n)$ for $ n > 0$ is defined by $ a_1 \equal{} 1,a_2 \equal{} 3$ and $ a_n \equal{} 4a_{n \minus{} 1} \minus{} a_{n \minus{} 2}$ for $ n > 2$
Prove that each term of this sequence is of the form $ a^2 \plus{} 2b^2$ for some natural numbers $ a$ and $ b$

Can you check your formulas? This fails already at the next term: $ a_3 \equal{} 4\cdot 2 \minus{} 1 \equal{} 7$ but $ 7$ is not of the form $ a^2\plus{}2b^2$.

[quote="randomgraph"]Can you check your formulas? This fails already at the next term: $ a_3 \equal{} 4\cdot 2 \minus{} 1 \equal{} 7$ but $ 7$ is not of the form $ a^2 \plus{} 2b^2$.[/quote]

I had edited it

Here's one way to go about this - not necessarily the smartest, but pretty straightforward so you can see "how to solve it" and not just the solution itself.

First thing you can do is calculate some values of this: for $ n\equal{}1,2,3,4,5,6$ you get

$ a_n \equal{} 1, 3, 11, 41, 153, 571$

Now try to write each of these as $ a^2\plus{}2b^2$. When a positive integer can be written in this way, there's typically only a few ways to do it, so we can reasonably expect to be able to find "the" way of writing $ a_n$ in this way so that the pattern, if there is one, will reveal itself. So:

$ 1 \equal{} 1^2 \plus{} 2\cdot 0^2$

$ 3 \equal{} 1^2 \plus{}2\cdot 1^2$

$ 11 \equal{} 3^2 \plus{} 2\cdot 1^2$

$ 41 \equal{} 3^2\plus{}2\cdot 4^2$

$ 153 \equal{} 11^2 \plus{} 2\cdot 4^2$ (to be honest, there are more options here, but this one stands out as part of the pattern)

$ 571 \equal{} 11^2 \plus{} 2\cdot 15^2$

at this point the pattern is almost clear: the $ a$ in $ a^2\plus{}2b^2$ looks just like the sequence $ a_n$ itself, taken two at a time. The $ b$ seems less obvious, but one may notice that it satisfies exactly the same recursion relation as the $ a_n$, only with different starting values ($ 1$ and $ 4$ instead of $ 1$ and $ 3$). So we can make the following educated guess: Let $ b_n$ be defined by $ b_1\equal{}1$, $ b_2\equal{}4$ and $ b_{n\plus{}1} \equal{} 4b_n \minus{} b_{n\minus{}1}$. It'll be convenient to further define $ b_0\equal{}0$ which makes the recursion work even at step $ n\equal{}1$. Then (we guess)

1. $ a_{2n} \equal{} a_n^2 \plus{} 2b_n^2$ for $ n \equal{} 1,2,3,\ldots$

2. $ a_{2n\plus{}1} \equal{} a_{n\plus{}1}^2 \plus{} 2b_n^2$ for $ n \equal{} 0,1,2,3,\ldots$

Now we have a concrete guess to prove, which is much easier than working with some unknown $ a$ and $ b$.

(BTW: one may also notice that $ b_n \equal{} a_1 \plus{} \cdots \plus{}a_n$. This is cute, but we won't use it below).

Proving equations 1. and 2. can be done by induction, but it's a little bit messy. Instead, we'll take an even more brute force approach: find an explicit formula for $ a_n$ and $ b_n$, in the same manner as this is done for the Fibonacci sequence, and use it to prove 1. and 2.

Finding the formulas is routine: we optimistically write $ a_n \equal{} x^n$, observe that the recurrence relation becomes $ x^{n\plus{}1} \equal{} 4x^n \minus{} x^{n\minus{}1}$ which is just $ x^2 \equal{} 4x\minus{}1$, find the two roots $ x_1$ and $ x_2$, and write $ a_n \equal{} \alpha x_1^n \plus{} \beta x_2^n$ for some appropriate $ \alpha$ and $ \beta$ chosen so as to make this work for $ a_1$ and $ a_2$. It's actually easier to work with $ a_0$ and $ a_1$, and we observe that setting $ a_0 \equal{} 1$ makes the recursion work. The result, by sheer routine calculation, is

$ a_n \equal{} \frac{3\minus{}\sqrt{3}}{6}(2\plus{}\sqrt{3})^n \plus{} \frac{3\plus{}\sqrt{3}}{6}(2\minus{}\sqrt{3})^n$

Similarly,

$ b_n \equal{} \frac{\sqrt{3}}{6}(2\plus{}\sqrt{3})^n \minus{} \frac{\sqrt{3}}{6}(2\minus{}\sqrt{3})^n$

and now it's just a matter of a little bit more computation to calculate

$ a_n^2 \equal{} \frac{2\minus{}\sqrt{3}}{6}(2\plus{}\sqrt{3})^{2n} \plus{} \frac{2\plus{}\sqrt{3}}{6}(2\minus{}\sqrt{3})^{2n} \plus{} \frac{1}{3}$

$ 2b_n^2 \equal{} \frac{1}{6}(2\plus{}\sqrt{3})^{2n} \plus{} \frac{1}{6}(2\minus{}\sqrt{3})^{2n} \minus{} \frac{1}{3}$

and indeed, $ a_n^2 \plus{} 2b_n^2$ is none other than $ a_{2n}$. This proves 1., and 2. can be done in exactly the same way.

One final note: when a sequence of numbers is expressible as the values of some quadratic form, it is not necessarily the case that the variables of the form are defined by some simple formula, as they are here. To take a totally trivial example, every prime congruent to $ 1$ mod $ 4$ has the form $ a^2\plus{}b^2$, but finding a pattern in those $ a$'s and $ b$'s is quite hard. So the approach taken here is risky - it may not work in all cases. However when it does it turns the problem to something quite routine.


integration
logarithms
calculus
limit
advanced fields
advanced fields unsolved

Let $0<p<\infty$. Prove that
\[
\int_{0}^\infty\frac{dx}{x^{p/2}(1+|\ln x|)^p}<\infty
\]
if and only if $p=2$.

I have my own solution but I want to see other solutions.  :)

For $p>2$, the integral converges on $[a,+\infty)$ because $\frac{1}{x^{p/2}(1+\ln x)^p} < \frac{1}{x^{p/2}}$. But if diverges on $(0,a]$ because we can find an $\varepsilon>0$ s.t. $\frac{p}{2}>1+\varepsilon$, then $\lim_{x\to 0} \frac{x^{p/2-\varepsilon}}{x^{p/2}(1-\ln x)^p}=\infty$ and $\int_0^a \frac{dx}{x^{p/2-\varepsilon}}$ diverges.

For $p<2$, the integral converges on $(0,a]$ because $\frac{1}{x^{p/2}(1-\ln x)^p} < \frac{1}{x^{p/2}}$. But if diverges on $[a,\infty)$ because we can find an $\varepsilon>0$ s.t. $\frac{p}{2}<1-\varepsilon$, then $\lim_{x\to 0} \frac{x^{p/2+\varepsilon}}{x^{p/2}(1+\ln x)^p}=\infty$ and $\int_a^\infty \frac{dx}{x^{p/2+\varepsilon}}$ diverges.

If $p=2$, then the integral becomes $\int_0^\infty \frac{dx}{x(1+|\ln x|)^2}=\int_0^\infty \frac{d(\ln x)}{(1+|\ln x|)^2}=\int_0^1 \frac{d(\ln x)}{(1-\ln x)^2} +\int_1^\infty \frac{d(\ln x)}{(1+\ln x)^2}=2\int_0^\infty \frac{dt}{(1+t)^2} = 2$


If $ a @ b$ is defined as $ a @ b$ = $ 3a \minus{} 3b$, what is the value of $ 3 @ 5$?

just plug in the values and solve

$ a@b\equal{}3a\minus{}3b$
$ 3@5\equal{}3(3)\minus{}3(5)$
$ 3@5\equal{}9\minus{}15$
$ 3@5\equal{}\minus{}6$

our answer is $ \boxed{\minus{}6}$

another solution

$ a@b\equal{}3a\minus{}3b\equal{}3(a\minus{}b)$.
Thus, $ 3@5 \equal{} 3(3\minus{}5) \equal{} 3 \cdot (\minus{}2) \equal{} \boxed{\minus{}6}$.


Ok so I have these 3 logic problems that I can't solve for the life of me. I've been doin fine in my logic class until I hit these problems. Here is the question:

Write a proof for each of the following arguments. You may use the Rules of Inference.

>= horseshoe/negation  *= dot/conjunction

1)
1. ~A > [~B > (C > B)]
   2. A v ~B
   3. ~A                    /~C

2)
1. (P * Q) > (R v S)
   2. (~R v L) > P
   3. (~R v M) > Q
   4. ~R * T                /S

3)
1. ~(~A * ~B) > C
   2. D > (B v A)           /D > C

4)
1. (L > M) * (N > M)
   2. L v P
   3. N v ~P                /M

I know it's a lot but any help is appreciated. I am really struggling with this crap and it's killing me.

Please use $\LaTeX$.
[hide]1.
$\neg A\rightarrow\left(\neg B\rightarrow\left(C\rightarrow B\right)\right)$
$A\vee\neg B$
$\neg A/\neg C$

2.
$\left(P\wedge Q\right)\rightarrow\left(R\vee S\right)$
$\left(\neg R\vee L\right)\rightarrow P$
$\left(\neg R\vee M\right)\rightarrow Q$
$\neg R\wedge T/S$

3.
$\neg\left(\neg A\wedge\neg B\right)\rightarrow C$
$D\rightarrow\left(B\vee A\right)/D\rightarrow C$

4.
$\left(L\rightarrow M\right)\wedge\left(N\rightarrow M\right)$
$L\vee P$
$N\vee\neg P/M$[/hide]

I think that's four problems you need help with, not 3.
What laws of inference are you allowed to use? There are many different axiom systems, so it wouldn't be any help if I used stuff you haven't encountered.
What kind of proof are you looking for? Two-column, truth tables, something written out?

:?  I appologize, you're right it is 4. I wrote that when I was very tired. Anyways, yeah sorry for the lack of info.

The answers would need to be two column. You're trying to get the conclusion by itelf again through the other premises.

And the laws you can use are as follows:

MP, MT, HS, DS, CD, Simp, Conj, Add, DM, Com, Assoc, Dist, DN, Trans, Impl, Equiv, Exp, Taut.
Not sure if thats all the rules... as you can probably tell this stuff is very new to me.

hope that helps. Message me if you need anymore info, or if I'm just stupid and keep not giving you enough.

thanks.

Ok so I got 1-3. It's the 4th one that is giving me the most trouble. Can anyone help a brotha out?

The best I could do is assume $\neg M$ and derive a contradiction.
$\neg M$ Assumption
$L\rightarrow M$ Definition of $\wedge$
$\neg L$ Modus Tollens
$P$ Disjunctive Inference
$N\rightarrow M$ Definition of $\wedge$
$\neg N$ Modus Tollens
$\neg P$ Disjunctive Inference
Since assuming $\neg M$ leads to a contradiction ($P\wedge\neg P$), $\neg M$ must be false, which, in turn, shows that $M$ must be true.


[quote="gabriel ponce"]Write all number from 1 to 20 using only four digts 1 and the math operations![/quote]
you mean, for example for 1 it would be
$1=1\times 1\times 1\times 1\times$? :maybe:

[quote="gabriel ponce"]Write all number from 1 to 20 using only four digts 1 and the math operations![/quote]
This is more like a puzzle. Anyways, let me try a few. I presume we have to use exactly four $1's$ each time and no other digits.

$1 = 1^{111}$
$2 = 1\cdot 1+1\cdot 1$
$3 = 1\cdot 1+1+1$
$4 = 1+1+1+1$
$5 =$
$6 =$
$7 =$
$8 =$
$9 = 11-(1+1)$
$10 = 11^{1}-1$
$11 = 1\cdot 11^{1}$
$12 = 11^{1}+1$
$13 = 11+(1+1)$
$14 =$
$15 =$
$16 =$
$17 =$
$18 =$
$19 =$
$20 =$


Factor Completely.

$(x^2-y^2)(x^2+2xy)+(x^2-y^2)(y^2)$

Im thinking along the lines of $(x^2-y^2)(x^2+2xy+y^2)$ Im not sure at all if this is right, or complete.

[quote="TFO_ForsakenGod"]Factor Completely. $(x^2-y^2)(x^2+2xy)+(x^2-y^2)(y^2)$[/quote]
I assume you are working in $\mathbb{Q}$. In that case we have
\begin{eqnarray*} (x^2-y^2)(x^2+2xy)+(x^2-y^2)(y^2)&=&(x^2-y^2)(x^2+2xy+y^2)\\ &=&(x-y)(x+y)(x+y)^2\\ &=&(x-y)(x+y)^3. \end{eqnarray*}

I have 1 more problem.

The area of a square is 16 centimeters. Find the value of x.

It is shown that one side of the square is $x-4$.

Naturally, I do this....

Because $x-4$ is one side of the square......

I do this $(x-4)^2$  because a side times a side would be the area.....
so $(x-4)^2=16$
$(x-4) (x-4)=16$
$(0-4) (0-4)=16$
$(-4)(-4)=16$

Where did I go wrong?

[hide]Why do you make $x=0$?
Since $(x-4)^2=16$, you can square root both sides to get $x-4=4$ (if you get $x-4=-4$, x can't be 0 otherwise you don't have a square) and then $x=8$. [/hide]

[hide="hard way"]$x^2-8x+16=16$
$x^2-8x=0$
$(x-8)(x)=0$
$x=8$  You can't have a negative side length!   :D
Check my work,$(8-4)(8-4)=16$[/hide]

or

[hide="easy way"]$(x-4)^2=16$
take the square root
$(x-4)=4$
$x=8$

This way I find much faster when working with squares.  Hope I have helped![/hide]


[quote="t0rajir0u"][quote="brady00n"]In permutations, order matters, but lets say you're selected a president, vice president, and secretary from a group of 6 people.  It doesn't matter the order you choose the three positions, as long as the same people are chosen, right?  But that's contradicting the definition...  :?[/quote]

Let's say you've got a "President" box, a "Vice President" box, and a "Secretary" box.  The boxes are distinct.  Candidates A, B, C in the boxes (in that order) is different from candidates B, A, C in that order, is different from candidates A, B, D, etc.

If you were doing combinations, you would have three boxes.  They would not be distinguishable.  Candidates A, B, C in three boxes is the same as candidates B, A, C in three boxes.[/quote]

Wow, that's an awesome explanation!  I completely understand now.  :)

For that president problem, you would use 6C3 then permute them to find the possible number of combinations right?

[quote="nateharman1234"]yea 6C3 x 3!

however the 3! cancels out from the bottom so you are just left with 6 x 5 x 4 which is some ridiculous number i don't want to multiply out[/quote]

$6\times5\times4$ is only $120$. :P

Here's how I remember: there are ALWAYS more permutations!!! (or at least as many i guess).

[quote="lotrgreengrapes7926"][quote="chess64"]Permutations are just $P(n,r)=\frac{n!}{r!}$. For combinations order doesn't matter so you divide by $(n-r)!$ to get $C(n,r)=_{n}C_{r}=\binom{n}{r}=\frac{n!}{r!(n-r)!}$.

Make sure you understand where these formulas come from. If you don't, just ask. :)[/quote] Isn't permutations $_{n}P_{r}=\frac{n!}{(n-r)!}$? :?[/quote]

oops im dumb.

[quote="ShoeFactory"][quote="lotrgreengrapes7926"][quote="chess64"]Permutations are just $P(n,r)=\frac{n!}{r!}$. For combinations order doesn't matter so you divide by $(n-r)!$ to get $C(n,r)=_{n}C_{r}=\binom{n}{r}=\frac{n!}{r!(n-r)!}$.

Make sure you understand where these formulas come from. If you don't, just ask. :)[/quote] Isn't permutations $_{n}P_{r}=\frac{n!}{(n-r)!}$? :?[/quote]

They're the same thing! Look at it again and you'll realize; its a basic identity.[/quote]

Uh, they're not.  chess64 got it wrong.  ;)

[quote="ShoeFactory"][quote="lotrgreengrapes7926"][quote="chess64"]Permutations are just $P(n,r)=\frac{n!}{r!}$. For combinations order doesn't matter so you divide by $(n-r)!$ to get $C(n,r)=_{n}C_{r}=\binom{n}{r}=\frac{n!}{r!(n-r)!}$.

Make sure you understand where these formulas come from. If you don't, just ask. :)[/quote] Isn't permutations $_{n}P_{r}=\frac{n!}{(n-r)!}$? :?[/quote]

They're the same thing! Look at it again and you'll realize; its a basic identity.[/quote]

you might be thinking of $\binom{n}{r} = \binom{n}{n-r}$

[quote="xxreddevilzxx"][/quote][quote="ShoeFactory"][quote="lotrgreengrapes7926"][quote="chess64"]Permutations are just $P(n,r)=\frac{n!}{r!}$. For combinations order doesn't matter so you divide by $(n-r)!$ to get $C(n,r)=_{n}C_{r}=\binom{n}{r}=\frac{n!}{r!(n-r)!}$.

Make sure you understand where these formulas come from. If you don't, just ask. :)[/quote] Isn't permutations $_{n}P_{r}=\frac{n!}{(n-r)!}$? :?[/quote]

They're the same thing! Look at it again and you'll realize; its a basic identity.[/quote][quote="xxreddevilzxx"]

you might be thinking of $\binom{n}{r} = \binom{n}{n-r}$[/quote]

$_nP_r=\frac{n!}{(n-r)!}$

You can see this upon expansion. I believe there is some confusion regarding the identity. The identity is for combinations, not permutations. If you ever get confused, just write the two expressions as factorials and it should be clear which two are equal.

Oh sh.... that's what I meant. Yeah, I'm better with doing problems than stating formulas  I guess. :P


induction
modular arithmetic
algebra
polynomial
number theory unsolved
number theory

This problem looks identical to the one Valentin used for the [url=http://www.mathlinks.ro/Forum/contest/2nd%20Edition/2nd%20edition%20-%20round%202.pdf]Mathlinks contests 2.2[/url].

[quote="andre.l"]Consider the sequence $(a_n)_{n\in \mathbb{N}}$ with $a_0=a_1=a_2=a_3=1$ and $a_na_{n-4}=a_{n-1}a_{n-3} + a^2_{n-2}$. Prove that all the terms of this sequence are integer numbers.[/quote]

Why dont we use the method of Induction ?

Davron

Well, the solution for Fábio Dias Moreira is by induction itself. But it is not so easy...
I will put only some ideas of this solution...

[b]Lemma:[/b]
$GCD(a_k,a_{k-1})=GCD(a_k,a_{k-2})=GCD(a_k,a_{k-3})=1$.

And, after this lemma, it is suffices to show that $a_{n-3}a_{n-1}+a_{n-2}^2 \equiv 0 (mod a_{n-4})$.
Now the problem is purely brutal-force  :D .

P.S.: This is almost the same solution of Toumaf...

[quote="Johann Peter Dirichlet"]This is called a "Somos-4" sequence. There exists a combinatorial proof but I cannot remember it at now :([/quote]

This is an attachment about the problem itself. It is so interesting... :P

Here is a solution using induction.

By induction one can easily show $a_n,a_{n+1}, a_{n+2},a_{n+3}$ are pairwise coprime for any $n$.
Now, we will proceed by induction. It suffices to show $a_{n-4}|(a_{n-1}a_{n-3} + a_{n-2}^2)$.
The base cases are trivial.
Now, because ${a_{n-8}}$ is an integer we know $a_{n-4}|(a_{n-5}a_{n-7} + a_{n-6}^2)$
Now, remark that $a_{n-1}a_{n-5} = a_{n-2}a_{n-4} + a_{n-3}^2$, thus $a_{n-1}a_{n-5} \equiv a_{n-3}^2 \pmod{a_{n-4}}$.
We also know that $a_{n-2}a_{n-6} = a_{n-3}a_{n-5} + a_{n-4}^2$ so $a_{n-2}a_{n-6} \equiv a_{n-3}a_{n-5} \pmod{a_{n-4}}$.
Lastly, we can find $a_{n-3}a_{n-7} \equiv a_{n-5}^2 \pmod{a_{n-4}}$

So now our method is to "eliminate" variables. Denote $a,b,c,d,e,f$ as $a_{n-1}, a_{n-2}, a_{n-3}, a_{n-5}, a_{n-6}, a_{n-7}$ modulo $a_{n-4}$.
Now we know:
$ad = c^2$
$be = cd$
$cf = d^2$
$df + e^2 = 0$
and we want to show $ac + b^2 = 0$
Using the first equation, we derive $a = \frac{c^2}{d}$.
Thus we want $\frac{c^3}{d} + b^2 = 0 \Longleftrightarrow c^3 + db^2 = 0$
Using the second equation we kill $b$. We have $b = \frac{cd}{e}$ so we want:
$c^3 + d\left (\frac{cd}{e} \right)^2 = 0 \Longleftrightarrow e^2c^3 + d^3c^2 = 0 \Longleftrightarrow e^2c  + d^3 = 0$
And we know:
$cf = d^2$
$df + e^2 = 0$
We kill $c$ since $c = \frac{d^2}{f}$ so we want:
$\frac{d^2e^2}{f} + d^3 = 0 \Longleftrightarrow e^2 + fd = 0$
However, we know this already so we're done!

[quote="Johann Peter Dirichlet"]I really want the combinatorial proof. It is very hard to find?[/quote]

Google "Somos-4 graphs". There is also a very nice technique involving the so-called Laurent polynomials and cluster algebras (yes, I'm name-dropping, so you can look for this stuff; I confess I don't know exactly what all that stuff is).


conics
hyperbola
calculus
derivative
Pythagorean Theorem
geometry

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

[hide="hint"] Pythagorean theorem. [/hide]

[quote="knexpert"][hide="Solution"]The distance is minimized when the point on the circle is $(1,1)$ and the point on the hyperbola is $(3,3)$.[/hide][/quote]

Could you elaborate? Thanks  :)

'course you could use second partial derivatives to find stationary points and whether they're minima or maxima, but that's probably not the best method.

[quote="knexpert"][hide="Solution"]
Instead of using the Pythagorean Theorem, let's use the distance formula. We can square root the expression to find the minimum. The two points we are measuring the distance between are $(u,\sqrt{2-u^{2}})$ and $(v,\frac{9}{v})$. These are just the parameterized equations of two curves, a half-circle with a radius of $\sqrt{2}$ and a rectangular hyperbola. The distance is minimized when the point on the circle is $(1,1)$ and the point on the hyperbola is $(3,3)$. The distance between these points is $\sqrt{8}$. The minimum is this distance squared, so the minimum is $\boxed{8}$.[/hide][/quote]

I consider difference formula the same thing as the Pythagorean theorem   :wink:  However, I don't think this is rigorous proof that those two points provide the minimum value.

Nothing can be more rigourous than this

line of shortest distance is along the common normal to the two curves xy=9 and x^2+y^2=2

Equation of normal to xy=9 at(x,y) is

Y-y=((x^2)/9)(X-x)

If it is normal to the circle then it should pass through the centre of the circle i.e. (0,0)

y=(x^3)/9

solving with xy=9

we get x=3,y=3

(3,3) is that point on xy=9 and also on the normal

Distance of (3,3) from (0,0) is 3sqrt.(2) and now subtracting the radius from 3sqrt.(2),we get the shortes tdistance between the two curves which is 2.sqrt(2)

therefore the required answer is obtained by squaring the shortest distance obtained .Hence the answer is 8.
[/quote]

[quote="pankajsinha"]line of shortest distance is along the common normal to the two curves [/quote]

Is this always true?

Oops forgot to mention that it is true for non-intersecting curves

I don't think you can say "Nothing can be more rigorous than ... if you don't prove the statement in question (that Kurt Godel quoted).


Let $ f(x)\equal{}x^3\plus{}x\plus{}1$. Prove that there exists an integer such that $ 3^n\ (n\in\mathbb{N })$ divides only one among  $ f(1),\ f(2),\ \cdots \ ,\ f(3^n)$.

[quote="kunny"]Let $ f(x) \equal{} x^3 \plus{} x \plus{} 1$. Prove that there exists an integer such that $ 3^n\ (n\in\mathbb{N })$ divides only one among  $ f(1),\ f(2),\ \cdots \ ,\ f(3^n)$.[/quote]
$ n \equal{} 1$?   :roll:

I have edit the context as follows.

[color=blue]Let $ f(x) \equal{} x^3 \plus{} x \plus{} 1$. Prove that there exists some integer only one such that $ 3^n\ (n\in\mathbb{N })$ divides $ f(1),\ f(2),\ \cdots \ ,\ f(3^n)$.[/color]

Obicious from Helsen lemma.
Let $ f(x)\equiv 0 (\mod p)$ has r solution $ \{x_1,...,x_r\}$ and
$ \gcd(f'(x_i),p) \equal{} 1$
 then the equation
$ f'(x)\equiv 0 (\mod p^k)$ has r solution $ \in [1,p^k]$


I just upgraded Adobe reader from 8.1.2 to 9.0.0. Now TeXnic center is unable to display the output for PDF LaTeX. How do I fix this?

Build > Define Output Profiles > Viewer tab. Change the path of the executable ie acrord32.exe. You may only need to change 8 to 9 but if not then use the browse button [...] to find acrord32.exe.