geometry
parallelogram
ratio
AMC
AIME
rectangle
analytic geometry

In parallelogram $ ABCD$, point $ M$ is on $ \overline{AB}$ so that $ \frac{AM}{AB} \equal{} \frac{17}{1000}$ and point $ N$ is on $ \overline{AD}$ so that $ \frac{AN}{AD} \equal{} \frac{17}{2009}$. Let $ P$ be the point of intersection of $ \overline{AC}$ and $ \overline{MN}$. Find $ \frac{AC}{AP}$.

I used mass point geometry, it's much, much, much easier than coordinate bashing it. But, coordinate geometry does get the problem done, even if it takes a long time. Try putting it on a coordinate plane and solving the problem. Or if you know mass points, use it.

-rts2007

[hide="cheap"]
Assume that the parallelogram is a unit square with one vertex on the origin.  Then, the line passing through $ (0,17/2009)$ and $ (17/1000,0)$ is $ y\equal{}\frac{\minus{}1000}{2009}x\plus{}\frac{17}{2009}$.  This intersects the diagonal, $ y\equal{}x$, at $ (17/3009, 17/3009)$, so the ratio is $ \frac{3009}{17}\equal{}177$. [/hide]

Extend the line BC past B and MN past M so they intersect, and then use two similar triangles.

The answer that I got is 177.. (I think it is right..)

First, I draw the parallelogram w/ AB 1000, AM 17, AN 17 AD 2009, in a coordinate with

angle DAB = 30, then it comes out with M = (17,0), N = (17/2, 17(3)^0.5/2)

D = (2009/2, 2009 (3)^0.5 /2) then C = (4009/2, 2009 (3)^0.5 /2)

Then find the equation for AC and MN, set it up as same...

I got X value (which two lines meet) and compared it with x values of C,

=177/1= 177 (I am bad at explaining + this is really really bad way to solve this question..)

[quote="Walk Around The River"][hide="cheap"]
Assume that the parallelogram is a unit square with one vertex on the origin.  Then, the line passing through $ (0,17/2009)$ and $ (17/1000,0)$ is $ y \equal{} \frac { \minus{} 1000}{2009}x \plus{} \frac {17}{2009}$.  This intersects the diagonal, $ y \equal{} x$, at $ (17/3009, 17/3009)$, so the ratio is $ \frac {3009}{17} \equal{} 177$. [/hide][/quote]

Yep; on AIMEs you can do that since it has to work for all types of parallelograms

[hide]Or assume its a rectangle with $ AM \equal{} 17$, $ AN \equal{} 17$, $ AB \equal{} 1000$, and $ AD \equal{} 2009$. Find the equation of the line running from the origin to the point C then find the equation of the other line $ MN$. Solve for the point of intersection, then using a the distance formula, it reduces nicely to get the answer[/hide]

Extend $ MN$ past $ M$ and $ BC$ past $ B$ so they intersect at $ Q$. Let $ AN\equal{}17y.$ Then $ \frac{AM}{BM} \equal{} \frac{17}{983} \equal{} \frac{17y}{QB} \longrightarrow QB \equal{} 983y$. We also know that $ BC \equal{} 2009y,$ so $ QC \equal{} 2992y$. $ \frac{QC}{AN} \equal{} \frac{2992}{17} \equal{} \frac{PC}{AP}$, so $ \frac{AC}{AP} \equal{} \frac{2992\plus{}17}{17} \equal{} 177$.

[hide="using areas, and probably uglier"]
Let $ \frac{AC} {AP}\equal{}x$. Thus,
$ \frac{[APM]} {[ABC]}\equal{}\frac{AM} {AB}\frac{AP} {AC}\equal{}\frac{17} {1000x}$.
Similarly, $ \frac{[APN]} {[ADC]}\equal{}\frac{17} {2009x}$ and $ \frac{[AMN]} {[ABD]}\equal{}\frac{17^2} {2009*1000}$.
Since $ [ABC]\equal{}[ACD]\equal{}[ABD],$ we have $ \frac{[APM]\plus{}[APN]} {[ABD]}\equal{}\frac{17} {2009x}\plus{}\frac{17} {1000x}$,
$ \frac{[AMN]} {[ABD]}\equal{}\frac{17} {x}(\frac{1} {1000}\plus{}\frac{1} {2009})\equal{}\frac{17^2} {2009*1000}$.
We can solve for x and end up with $ x\equal{}\boxed{177}$.
[/hide]

Draw a similar triangle with vertex A such that this new triangle continues the sides AB and AD and is congruent to MAN, and the side opposite of A touches C. Obviously we have new similar triangles, and they are all also isosceles. We see that the side length of this big triangle is 3009, and the side length of the little is 17. The lines we are concerned with are corresponding parts of similar triangles, so they are congruent and proportional to 3009:17 for an answer of 177.

(Also, I have to say, this question was quite guessable.)

[quote="the future"]I assumed it was a rectangle with side lenghts 1000 and 2009.  As a result, I got massive numbers in my calculation (used distance formula to find the distances), as I had to find the square root of some number in the billions.  Luckily, I made no calculation mistake and got the answer.[/quote]

I did the same. You actually didn't need to find that massive square root - it canceled out at the end. At least, if you did what I thought you did and I remember correctly what I did....

[quote="buzzer11"]I assumed it was a rectangle.

[quote="Walk Around The River"][hide="cheap"]
Assume that the parallelogram is a unit square with one vertex on the origin.  Then, the line passing through $ (0,17/2009)$ and $ (17/1000,0)$ is $ y \equal{} \frac { \minus{} 1000}{2009}x \plus{} \frac {17}{2009}$.  This intersects the diagonal, $ y \equal{} x$, at $ (17/3009, 17/3009)$, so the ratio is $ \frac {3009}{17} \equal{} 177$. [/hide][/quote]

But I don't think it can be a unit square? The side lengths were given as 2009 and 1000.[/quote]
nah on the ratios.  I was thinking about unit sqaure instead of rectangle (which I went with), but I thought otherwise because unit squares would deal with fractions and I thought that I would be better off dealing with integers instead of fractions....I was wrong because the unit square's terms came out nicely.

[quote=vatatmaja]how do you just equate p and q components. This is literally assuming the parallelogram is a rhombus, which is exactly what I don't want. This is no different in its essence than using coordinates, can someone give another 'synthetic solution' where we don't assume any thing about the parallelogram...and no fake solving  :furious:[/quote]

There is no assumption. We set the line equations equal to each other, and can equate p and q components since p and q are [b]linearly independent[/b], which in 2 dimensions is just a fancy way of saying that p and q are [b]non-parallel[/b].

[quote=vatatmaja]Yes I think I have found a good solution.
Thanks.

Just 1 more question, after I extended MN to meet the extension of BC, the problem became rather simple and quite easy. What is the motivation to do this type of extension?[/quote]

The motivation is that when you extend a side length, you end up creating numerous similar triangles (always). The technique of extending appears quite commonly on the AMC and AIME. Therefore, if you ever have "almost" a similar triangle or "a part of" of a similar triangle, you should think of extending until you actually get a similar triangle.

Okay, sorry for the bump, but I kind of really need help.

I am just learning how to use mass points and this is like the ~2nd problem I did, so bear with my ignorance  :noo:

Well, even before looking at the solution, I tried something (obviously was wrong).

So basically, we know that $AN=17$ and $ND=1992$ right, so then with mass points, $D=\frac{17}{1992}$

Okay, then $AM=17$ and $MB=983$. So then $B=\frac{17}{983}$

Well, this is obviously wrong. I, being a very stupid person, did not realize that the diagonals bisected each other and therefore $B=D$.

However, why was my original logic wrong? I'm quite sure it's because the two ratios are not equal to each other, so we can't like equate them, but in another problem, someone did that and it turned out fine  :wacko:

So then, I read the solution, and I followed it up until the point where it said that $A$ had a mass of $2975$ and $O$ had a mass of $34$. But wait, that doesn't make sense, because $A$ can't be more than $O$?? Wut?

Please help a confused me.

Also, if anyone knows any handouts for mass points, I would be eternally grateful. I watched one youtube video and read the AoPS Wiki page and you can see where I ended up  :dry:

[quote=Archeon]Standard mass points do not work on this problem. You have to use [url=https://en.wikipedia.org/wiki/Mass_point_geometry#Splitting_masses]split masses.[/url][/quote]

Ohhh, so basically the mass is kind of variant for both sides :o and combined in the middle !

Thanks  :)

Wait, but the mass of A is still greater than the mass of O, which doesn't make sense? Also, how did they know that AO/AP was that ratio? Then wouldn't P have a mass of 2975+34?

@[b]submath[/b]: I've deleted your bump posts for the sake of thread continuity.  Also, obligatory not a mass point expert, but hopefully this should help.

[quote="submath"]Wait, but the mass of A is still greater than the mass of O, which doesn't make sense?[/quote]
Why does this not make sense?  The fact that the mass at $A$ is greater than the mass at $O$ means that point $P$ is much closer to $A$ than it is to $O$.  Intuitively, this [i]must[/i] be the case since $M$ and $N$ are also much closer to $A$ than they are to $B$ and $D$ respectively.

[quote="submath"]Also, how did they know that AO/AP was that ratio?[/quote]
This is because $\tfrac{AO}{AP} = 1 + \tfrac{OP}{AP}$, and this last ratio equals $\tfrac{2975}{34}$ by reading off the masses at points $A$ and $O$.

[quote="submath"]Then wouldn't P have a mass of 2975+34?[/quote]
Yes.  (Note that the mass at $P$ [i]must[/i] be greater than the masses of points $A$ and $O$; might this be what was confusing you?)

Thanks for your help dj!

I was confused that the mass of A was greater than the mass of O because isn't A+C=O, and then A can't then be greater than O?

Eh, that might be true if we were using mass points on the entire parallelogram $ABCD$, but in this case they're just being applied to triangle $ABD$, and so point $C$ isn't of interest to us.

In this problem I found it easiest to assume $ABCD$ was a rectangle with side lengths $AB=CD=1000$ and $BC=AD=2009,$ so the answer I got was $177$ via analytic geometry. I know this must be the correct answer because MAA doesn't tell us what type of parallelogram we are dealing with, so the answer must be the same for all parallelograms. But how do I prove that $\cfrac{AC}{PA}=177$ is true for [b]all[/b] parallelograms?

[hide=A Mass Points Solution]First, let the intersection of $\overline{BD}$ and $\overline{AC}$ be $X$. Since mass points works on triangles, not parallelograms, let's focus on $\triangle ABD$ and disregard $\triangle BDC$.

We know the ratio of $\frac{AM}{AB}=\frac{17}{1000}$ and $\frac{AN}{AD}=\frac{17}{2009}$. Then, the mass at $B$ is $17$, the mass at $D$ is $17$, and the mass at $A$ is $(1000-17)+(2009-17)=983+1992=2975$ (splitting masses). The mass at $X$ is $34$.

So, we know that $\frac{PX}{AP}=\frac{2975}{34}$. Adding the bottom to the top, $\frac{AX}{AP}=\frac{3009}{34}=\frac{177}2$, so finally, $\frac{AC}{AP}=\boxed{177}$. [/hide]

The degenerate parallelogram solution is really quick -- but is there an instance where taking a degenerate shape doesn't work??

[hide=Menelaus]Let $AM=AN=17.$ Extend $\overline{BC}$ past $B$ and $\overline{MN}$ past $M$, and call their intersection $F.$  Triangle similarity gives $FB=MB=983.$ Then from Menelaus, $$\frac{FB}{FC}\cdot\frac{CP}{PA}\cdot\frac{AM}{MB}=\frac{983}{2992}\cdot\frac{CP}{AP}\cdot\frac{17}{983}=1,$$ which after simplifying gives $\frac{CP}{AP}=176.$ $\frac{AC}{AP}=1+\frac{CP}{AP}=\fbox{177}$.[/hide]


Find the smallest integer k for which there is more than one non-decreasing sequence of positive integers a1, a2, a3, ... such that a9 = k and an+2 = an+1 + an.

Letting the first two integers be a and b, we get the formula for a_9 as 13a+21b, where b>a. (1,35)(22,22) are I believe the smallest pairs which will work. They give 13+35*21=22*13+22*21=748.

spoiler:

[hide]why must b > a? it says non decreasing, not increasing.... therefore a can equal 1 and b can equal 1 as well... the fibonacci sequence will work and we havent broken any rules, why can't k be 34?[/hide]

Yeah b can be greater than or equal to a. However, k=34 won't work as there has to be [b]two[/b] solutions for k.

wait, mystic, where did u get thsoe ordered pairs from?

well, two equivalent ordered pairs will have the form (a, b) and (a+21, b-13). In the second one, I want b-13:ge:a+21, giving b:ge:a+34. Since a has to be at least 1 (positive integer), I get the smallest b as 35.


[img]http://img466.imageshack.us/img466/9136/picvw2.png[/img]

How to find radius of this small (grey) circle?

Well, wlog let $r=1$ (you can scale this later).

The equation of the circle can be expressed in the form $(x-r)^{2}+(y-k)^{2}=r^{2}\implies x^{2}+y^{2}=2rx+2ky-k^{2}$

This equation has exactly one intersection with each of $x^{2}+y^{2}=1$ and with $x^{2}+y^{2}=2x$

[quote="Altheman"]if two circles are tangent, then their centers, and the point of tangency are collinear[/quote]

proof?

let L be the line connecting their centers

the circles are obviously symmetric about this line

hence if they intersect once, at P, that can be flipped over L, to P', to another intersection, for two intersections

however, this fails when P=P' i.e. P is on line L

[color=darkblue]Suppose w.l.o.g. that for the square $ABCD$ we have $AB=1$. Let $w(X,x)$ be the required circle exterior tangent to the circle $C(B,1)$, interior tangent to the circle $C(A,1)$ and tangent to the side $[AD]$. Then in the triangle $XAB$ we have $XA=1-x$, $XB=1+x$. Denote the point $E\in AB$, $XE\perp AB$. Observe that $EA=x$, $EB=1-x$ and $XE^{2}=XA^{2}-AE^{2}$ $\Longrightarrow$ $XE=\sqrt{1-2x}$. Therefore, from the Pitagora's theorem applied and in the right triangle $XEB$ obtain $(1-x)^{2}+(1-2x)=(1+x)^{2}$, i.e. $x=\frac{1}{6}$. In conclusion, for $AB=r$ $\Longrightarrow$ $\boxed{\ x=\frac{r}{6}\ }$.[/color]

[quote="The Zuton Force"]Draw the tangent to the circles at the point of tangency.

The lines connecting each circle's center to this point are both perpendicular to the line of tangency and both pass through the point of tangency and are thus the same line.

Edit: Oops, didn't see Altheman's post...[/quote]

at first, i thought of that proof, but it relies on a shaky definition of tangency


geometry
circumcircle
trigonometry
perpendicular bisector
geometry solved

k, this is trivial:

We see that |AK|*|KF|=|AK|*(|AF|-|AK|) (this is true for acute and obtuse triangle ABC) =|BF|^2 * sin(X+A)/sin(2A) * ( sin(X)/sin(A) - sin(X+A)/sin(2A) ), where X = <ABP.

But |AJ|*|JE| = CE^2 * sin(X-A)/sin(2A) * ( sin(X)/sin(A) - sin(X-A)/sin(2A) ).

After some cancellations, it turns out both sides are just sin(X-A)sin(X+A), so we're done.

Sine Law and mindless trig.... :D

[quote="zhaoli"]Point $P$ is on the arc $BC$ (not including $A$) of the circumcircle of triangle $ABC$. $E$ is the intersection of $CP$ and $AB$; $F$ is the intersection of $BP$ and $AC$. The perpendicular bisector of $AC$ intersects $AB$ at $J$, and the perpendicular bisector of $AB$ intersects $AC$ at $K$. Prove that $\frac{CE^2}{BF^2} = \frac{AJ \cdot JE}{AK \cdot KF}$.[/quote]

Another easy angle chaser. We will use directed angles modulo 180°.

Since the point J lies on the perpendicular bisector of the segment AC, we have AJ = CJ, so the triangle AJC is isosceles, and thus < ACJ = < JAC. But by the sum of angles in triangle AJC, we have $\measuredangle CJA+\measuredangle JAC+\measuredangle ACJ=0^{\circ}$; in other words, $-\measuredangle CJA=\measuredangle JAC+\measuredangle ACJ=\measuredangle JAC+\measuredangle JAC=2\cdot\measuredangle JAC$, so that $\measuredangle CJA=-2\cdot\measuredangle JAC$. In other words, $\measuredangle CJE=-2\cdot\measuredangle JAC$. Similarly, $\measuredangle BKF=-2\cdot\measuredangle KAB$, so that $\measuredangle FKB=-\measuredangle BKF=-\left(-2\cdot\measuredangle KAB\right)=-2\cdot\left(-\measuredangle KAB\right)=-2\cdot\measuredangle JAC$. Hence, < CJE = < FKB.

Also, < CEJ = < (CP; AB) = < (CP; CA) - < (AB; CA) = < PCA - < BAK. Since the point P lies on the circumcircle of triangle ABC, we have < PCA = < PBA, and since the point K lies on the perpendicular bisector of the segment AB, we have KA = KB, so the triangle AKB is isosceles, and thus < BAK = < KBA. Thus, < CEJ = < PCA - < BAK = < PBA - < KBA = < FBK.

Since < CJE = < FKB and < CEJ = < FBK, the triangles CJE and FKB are directly similar. Thus, $\frac{CE}{CJ}=\frac{FB}{FK}$ and $\frac{CE}{JE}=\frac{FB}{KB}$, so that $\frac{CE^2}{CJ\cdot JE}=\frac{CE}{CJ}\cdot\frac{CE}{JE}=\frac{FB}{FK}\cdot\frac{FB}{KB}=\frac{FB^2}{FK\cdot KB}$. In other words, $\frac{CE^2}{FB^2}=\frac{CJ\cdot JE}{FK\cdot KB}$. Since AJ = CJ and KA = KB, this becomes $\frac{CE^2}{FB^2}=\frac{AJ\cdot JE}{FK\cdot KA}$. And the problem is solved.

  darij


[quote="worthawholebean"]The sides of a triangle have lengths of $ 15$, $ 20$, and $ 25$. Find the length of the shortest altitude.

$ \text{(A)}\ 6 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$[/quote]
[hide]These side lengths form a right triangle. Also the shortest altitude goes to the longest side length. The area of this right triangle is $ \frac{(15)(20)}{2}\equal{}150$.
The area of a triangle is base times height over two. Setting the altitude as x, the equation is $ \frac{25x}{2}\equal{}150$. Solving for x results in 12. So $ B$. [/hide]

[quote=worthawholebean]The sides of a triangle have lengths of $ 15$, $ 20$, and $ 25$. Find the length of the shortest altitude.

$ \text{(A)}\ 6 \qquad
\text{(B)}\ 12 \qquad
\text{(C)}\ 12.5 \qquad
\text{(D)}\ 13 \qquad
\text{(E)}\ 15$[/quote]

Sorry for bump, but would this be possible if it was not a right triangle with the given information? Or is there some sort of formula?

Well, you could always calculate the area of the triangle using Heron's Formula. Then, you would just do what ckck did.


geometry
trigonometry
angle bisector
area of a triangle
Heron\u0027s formula
trig identities
Law of Cosines

Let $a,b,c$ be the sides of a triangle. The area of the triangle is $(a-b+c)(a+b-c)$. Find the length of the angle bisector opposite to side $a$ in terms of $b$ and $c$.

if theta is the angle btwn b and c, we can find theta and then finding the length of the bisector shouldnt be too far.  Use heron's formula to get  $17a^2-17b^2+30bc-17c^2=0$, then use law of cos: $a^2=b^2+c^2-2bc\cos{\theta}$, and can solve for $\cos{\theta}=15/17$.

from here i can do it real messy, ill look for a better way later.  But using $\cos{\theta}=15/17$, you can solve for $a$ in terms of $b$ and $c$.  Then using the angle bisector thm, let the two sides cut be $x$ and $a-x$, then $x=ba/(c+b)$ from the angle bisector thm.  Then substitute for $a$ using law of cosines.  Then use stewarts theorem, and we can get it in terms of $b$ and $c$.

Assuming the notation a = BC, b = CA, c = AB, length of the internal bisector of the angle $\angle A$ is

$l_A = \sqrt{bc\left(1 - \frac{a^2}{(b + c)^2}\right)} = \frac{\sqrt{bc \left[(b + c)^2 - a^2\right]}}{b + c} = \frac{\sqrt{bc(b + c + a)(b + c - a)}}{b + c}$

Using Heron's formula for the triangle area and the given relation for the area,

$\triangle^2 = \frac{1}{16} (a + b + c)(b + c - a)(c + a - b)(a + b - c) = \frac{1}{16}(a + b + c)(b + c - a) \triangle$

$\triangle = \frac{1}{16} (a + b + c)(b + c - a)$

Substituting this into the angle bisector formula,

$l_A = \frac{\sqrt{bc \cdot 16 \triangle}}{b + c} = \frac{4 \sqrt{bc \cdot \triangle}}{b + c}$

Now we have to find the triangle area in terms of b, c. We have 2 relations for the triangle area in terms of a, b, c: the given one and the one deduced above. Hence, we can calculate a in terms of b, c and then eliminate a by substituting the result into either of these 2 formulas.

$\triangle = \frac{1}{16}(a + b + c)(b + c - a) = (c + a - b)(a + b - c)$

$\triangle = \frac{1}{16} \left[(b + c)^2 - a^2\right] = a^2 - (b - c)^2$

$17a^2 = (b + c)^2 + 16(b - c)^2 = 17(b^2 + c^2) + 2bc - 32bc = 17(b^2 + c^2) - 30bc$

$a^2 = b^2 + c^2 - \frac{30}{17} bc$

Substituting this to either of the above relations for the triangle area,

$\triangle = \frac{1}{16} \left[(b + c)^2 - a^2\right] = \frac{1}{16} \left[b^2 + 2bc + c^2 - \left(b^2 + c^2 - \frac{30}{17} bc\right)\right] =$

$= \frac{34 + 30}{16 \cdot 17} bc = \frac{64}{16 \cdot 17} bc = \frac{4}{17} bc$

or

$\triangle = a^2 - (b - c)^2 = b^2 + c^2 - \frac{30}{17} bc - (b^2 - 2bc + c^2) = \frac{34 - 30}{17} bc = \frac{4}{17} bc$

This serves as a check. Substituting this into the angle bisector formula in terms of $b, c, \triangle$, the final result is

$l_A = \frac{4 \sqrt{bc \cdot \triangle}}{b + c} = \frac{4 \sqrt{bc \cdot \frac{4}{17} bc}}{b + c} = \frac{8 bc}{(b + c) \sqrt{17}} = \frac{8 bc \sqrt{17}}{17(b + c)}$

This one appears also in Komal 1998, I think.

$S = (a - b + c)(a + b - c) = a^2 - b^2 - c^2 + 2bc$.
From the law of cosines $a^2 = b^2 + c^2 - 2bc\cos \phi$. Hence $S = 2bc(1 - \cos \phi)$.
On the other hand $\frac{1}{2}bc\sin \phi = 2bc(1 - \cos \phi)$ $\Longrightarrow$ $\sin \phi = 4(1 - \cos \phi)$(1).
Let $x$ be the length of the angle bisector. Then $\frac{1}{2}x\sin \frac{\phi}{2}(b + c) = 2bc(1 - \cos \phi)$.
So $x = \frac{bc}{b + c} \cdot \frac{4(1 - \cos \phi)}{\sin \frac{\phi}{2}}$.
From (1) we get $\tan \frac{\phi}{2} = \frac{1}{4}$ $\Longrightarrow$ $\frac{1 - \cos \phi}{1 + \cos \phi} = \frac{1}{16}$ $\Longrightarrow$ $\cos \phi = \frac{15}{17}$.
Finally $x = \frac{bc}{b + c} \cdot \frac{4(1 - \cos \phi)}{\sqrt{\frac{1 - \cos \phi}{2}}}$.
Substitute the value of $\cos \phi$ we get $x = \frac{bc}{b + c} \cdot \frac{8\sqrt{17}}{17}$.


A ferry trip departs from the ferry dock every 15 minutes.
Yesterday the first ferry trip left at 8 a.m. and the last ferry trip left
at 5 p.m. How many ferry trips left from the ferry dock yesterday?

[quote=pinkmuskrat]This was a period of $ 9$ hours, so in total there were $ 9 \cdot 4 \plus{} 1 \equal{} 37$ ferry docks, since we multiply by 4 an hour and add one to include the first dock, at 8:00.[/quote]

Your answer is correct, but you're supposed to add 1 to include the last one, not the first.

[quote=pinkmuskrat]This was a period of $ 9$ hours, so in total there were $ 9 \cdot 4 \plus{} 1 \equal{} 37$ ferry docks, since we multiply by 4 an hour and add one to include the first dock, at 8:00.[/quote]

That's how I would have done it. I was going to post before I saw your post.


Problem 1: $ 0\leq x, y, z\leq 2$ and $ x\plus{}y\plus{}z\equal{}3$. Find the maximum and minimum values of:
$ M \equal{} x^{4}\plus{}y^{4}\plus{}z^{4}\plus{} 12(1\minus{}x)(1\minus{}y)(1\minus{}z)$
Problem 2: Suppose that 4 real numbers a,b,c,d satisfy the conditions
$ a^{2}\plus{}b^{2}\equal{}c^{2}\plus{}d^{2}\equal{}4$ and $ ac \plus{} bd \equal{} 2$
Find the set of all possible values the number $ M \equal{} ab \plus{} cd$ can take
Problem 3: Find all integers x, y such that $ x^{2}\plus{}y^{2}\equal{}(2xy\plus{}1)^{2}$

hello, for 1) we get $ 3\le M \le 17$, the minimum will be obtained for $ x\equal{}y\equal{}z\equal{}1$ and the maximum for $ x\equal{}0,y\equal{}2,z\equal{}1$.
Sonnhard.

[hide="Solution to 3"]
$ x^{2}\plus{}y^{2}\equal{}(2xy\plus{}1)^{2}$
$ x^{2}\equal{}(2xy\plus{}1)^{2}\minus{}y^{2}$
$ x^{2}\equal{}(2xy\plus{}1\minus{}y)(2xy\plus{}1\plus{}y)$
$ 2xy\plus{}1\minus{}y\equal{}2xy\plus{}1\plus{}y$
$ y\equal{}0$
Then, $ x\equal{}1,\minus{}1$

Similarly, $ x\equal{}0$ and $ y\equal{}1,\minus{}1$[/hide]


I am stuck on the following:
(this is for a local contest)

1) Solve for $ x$: $ \frac{1}{x\minus{}a}\plus{}\frac{1}{x\plus{}a}\equal{}\frac{2}{x\minus{}1}$.
2) Find all zeroes of $ P(x) \equal{} 2x^{3}\plus{}3x^{2}\plus{}2x\plus{}3$.
3) A function $ f$ has the property that $ f(2\plus{}x) \equal{} f(2\minus{}x)$ for all $ x$. If $ f$ has exactly four real zeroes, find their sum.
4) Find all zeroes of the function $ f(x) \equal{} x^{3}\plus{}1$.
(I got the $ x \equal{}\minus{}1$ as a part of the answer).

Thanks!

EDIT: Thank you so much!!!

[hide="clue for number 1"]

CONJUGATE then diff of squares for the denominator and simplify

[/hide]

[hide="HINT For Four"]
DeMoivre.
[/hide]
[hide="Four"]
$ \cos\frac{\pi}{3}\plus{}i\sin\frac{\pi}{3}\equal{}\frac{1}{2}\plus{}\frac{i\sqrt{3}}{2}$
$ \cos\frac{3\pi}{3}\plus{}i\sin\frac{3\pi}{3}\equal{}\cos\pi \equal{}\minus{}1$
$ \cos\frac{5\pi}{3}\plus{}i\sin\frac{5\pi}{3}\equal{}\frac{1}{2}\minus{}\frac{i\sqrt{3}}{2}$.
[/hide]

[hide="1"] $ (x\minus{}1)( (x\minus{}a)\plus{}(x\plus{}a) ) \equal{} 2(x\minus{}a)(x\plus{}a)\Leftrightarrow x^{2}\minus{}x \equal{} x^{2}\minus{}a^{2}\Leftrightarrow\boxed{x \equal{} a^{2}}$ [/hide]
[hide="2"] $ P(x) \equal{} x^{2}(2x\plus{}3)\plus{}(2x\plus{}3) \equal{} (x^{2}\plus{}1)(2x\plus{}3) \equal{} 0\Leftrightarrow x \equal{}\boxed{\pm i,\minus{}\frac{3}{2}}$ [/hide]
[hide="3"] $ f(x) \equal{} f(4\minus{}x)$; hence pairs of the zeroes sum up to $ 4$ and all four zeroes must sum up to $ \boxed{8}$. [/hide]
[hide="4"] $ x^{3}\plus{}1 \equal{} (x\plus{}1)(x^{2}\minus{}x\plus{}1) \equal{} 0\Leftrightarrow x \equal{}\boxed{\minus{}1,\frac{1\pm i\sqrt{3}}{2}}$ [/hide]

[quote="t0rajir0u"][hide="1"] $ (x\minus{}1)( (x\minus{}a)\plus{}(x\plus{}a) ) \equal{} 2(x\minus{}a)(x\plus{}a)\Leftrightarrow x^{2}\minus{}x \equal{} x^{2}\minus{}a^{2}\Leftrightarrow\boxed{x \equal{} a^{2}}$ [/hide][/quote]

[hide="Necessary appendix"]For $ a\in\{0,\pm 1\}$ the equation has no solution. For all the other values of $ a$, the unique solution is $ x \equal{} a^{2}$.[/hide]


number theory
relatively prime
number theory proposed

Let a and b be positive relatively prime integers. Show that for every integer n>ab there exist positive integers x and y such that n=ax+by.

check out my post in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180036


My copy of the AoPS books have a copyright of 1993,1995.  Is the newest version, or are there newer versions?

My copy (volume 2) says fifth edition, 2003, and I think that this is the most recent edition.


For a positive integer $M$, if there exist integers $a$, $b$, $c$ and $d$ so that:
\[ M \leq a < b \leq c < d \leq M+49, \qquad ad=bc \]
then we call $M$ a GOOD number, if not then $M$ is BAD. Please find the greatest GOOD number and the smallest BAD number.

First, one observation: If all four numbers $a,b,c,d$ can be equal,
there is no point for solving this problem, so I think there is a
mistake in a statement of the problem. I suppose that it's $M\leq a<b\leq c<d\leq M+49$.

I found the greatest GOOD number, and I am pretty sure that all
numbers not greater then that are GOOD but i can't prove that (for
now).

Suppose that $b=a+k_1,c=a+k_2$ and $d=a+k_3$ where $0<k_1\leq k_2<k_3<50$. Then from $ad=bc$ we get $k_3-k_2-k_1=\frac{k_1k_2}a$.
So $50>k_3>k_1+k_2$, so $max\{k_1+k_2\}=48 \Rightarrow max\{k_1k_2\}=24^2$. The last one implies that $a\leq24^2$, so if
$M\geq 24^2+1$ then $a\geq M\geq 24^2+1$ contradiction!

If $M=24^2$ there is a solution: $a=24^2,b=c=24^2+24,d=24^2+49$, so
greatest GOOD number is $24^2$. For now, this is it, I'll be back
soon :)

bye

Well, I guess you're wrong.
I assume the condition is $a<b \leq c <d$. If $ad=bc$ we may find $m,n,p,q$ with $a=mn, b=mp, c=nq, d=pq$ then $m<q, n<p$ so $(p-1)(q-1)\geq mn=a$, and the new system $(p-1)(q-1), (p-1)p, (q-1)q, pq$ also satisfies the condition.

Thus we have come to the following characterization:

A number $M$ is bad if and only if $pq\leq M+49$ implies $(p-1)(q-1)<M$.
Now if $M<18^2$ then $M$ id good, because if we let $(k-1)^2<M+49 \leq k^2$ and $r\leq \sqrt{2k-1}$ with $k^2-r^2\leq M+49<k^2-(r-1)^2$ we take $p=k+r, q=k-r$ yielding $(k-1)^2-r^2 \leq M-1 =m+49-50 <k^2-(r-1)^2-50$ so $k+r \geq 25$ which implies $k \geq 19$.

Now we have some work to do (I mean $p=15, q=22$ make all numbers from $330$ to $343$ good and so on) till we get the desired number. I think it's 470 or smth close.

I really don't understand what you want to say. I said (and proved)
that greatest GOOD number is $24^2=576$ and I found one $a,b,c,d$
which are fulfilling given condition, so obviously is not, like you
said,"470, or something close".

Alright, let's give this ROBUST problem its due diligence ;)
[quote]For a positive integer $M$, if there exists integers $a$, $b$, $c$, $d$ such that $ad=bc$ and \[ M \le a < b \le c < d \le M + 49 \] then we say that $M$ is [i]good[/i], otherwise we say that $M$ is [i]bad[/i]. Determine the greatest good number and the smallest bad number.[/quote]

The largest good number is $24^2=576$ and the smallest bad number is $443$. We prove this through a series of several claims.

[b]Lemma[/b]:  	The number $M$ is good if and only if: 	there exist integers $p$ and $q$ for which $(p+1)(q+1) \le M + 49$ 	but $pq \ge M$.

[i]Proof.[/i]  	First, assume $M$ is good. 	Then given $ad=bc$ we set $a=wx$, $d=yz$, $b=wy$, $c=xz$. 	Then $a < b$ implies $x < y$, and $b < d$ implies $w < z$. 	Then $M \le a \le wx \le (z-1)(y-1)$ so we just take 	$p = z-1$ and $q = y-1$.

For the converse direction, if WLOG $p \le q$ then take 	$(w,x,y,z) = (p,q,q+1,p+1)$ to get $a$, $b$, $c$, $d$. $\blacksquare$

It is now easy to determine the largest good integer.

[b]Lemma[/b]:  	The largest good integer is $576 = 24^2$.

[i]Proof.[/i]  	To see $576$ is good, take $p=q=24$ in the lemma above. 	 	Conversely, assume $M$ is good, so $p$ and $q$ exist as described above. 	Then we must have $p+q+1 \le 49$ hence $p+q \le 48$. 	So $M \le pq \le 24^2 = 576$. $\blacksquare$

It remains to determine the smallest bad integer. First, we contend that:

[b]Lemma[/b]:  	Every integer $M \le 288$ is good.

[i]Proof.[/i]  	Note that there is some multiple of $13$ in $\{M+37, M+38, \dots, M+49\}$, 	call it $K$. 	We claim that we can take $q = 12$, and $p = K/13-1$. 	Indeed, 	\[ pq = \frac{12}{13}K - 12 		\ge \frac{12}{13} (M+37) - 12 		= M + \frac{12 \cdot 24 - M}{13} \ge M \] 	as desired. $\blacksquare$

[b]Lemma[/b]:  	Every integer $287 \le M \le 442$ is good.

[i]Proof.[/i]  	Note that any pair $(p,q)$ of integers is a witness 	to all $pq - \delta \le M \le pq$ being prime, 	where $\delta = 48-p-q$. 	In light of this, we construct the following 24 cases: 	\[ 	\begin{array}{cccc} 		p \cdot q & pq & \delta & pq-\delta \\ \hline 		15 \cdot 20 & 300 & 13 & 287 \\ 		14 \cdot 22 & 308 & 12 & 296 \\ 		15 \cdot 21 & 315 & 12 & 303 \\ 		18 \cdot 18 & 324 & 12 & 312 \\ \hline 		15 \cdot 22 & 330 & 11 & 319 \\ 		18 \cdot 19 & 342 & 11 & 331 \\ \hline 		14 \cdot 25 & 350 &  9 & 341 \\ 		19 \cdot 19 & 361 & 10 & 351 \\ \hline 		14 \cdot 26 & 364 &  8 & 356 \\ 		17 \cdot 22 & 374 &  9 & 365 \\ 		19 \cdot 20 & 380 &  9 & 371 \\ \hline 		16 \cdot 24 & 384 &  8 & 376 \\ 		13 \cdot 30 & 390 &  5 & 385 \\ 		18 \cdot 22 & 396 &  8 & 388 \\ 		20 \cdot 20 & 400 &  8 & 392 \\ \hline 		17 \cdot 24 & 408 &  7 & 401 \\ 		18 \cdot 23 & 414 &  7 & 407 \\ 		16 \cdot 26 & 416 &  6 & 410 \\ 		20 \cdot 21 & 420 &  7 & 413 \\ \hline 		17 \cdot 25 & 425 &  6 & 419 \\ 		18 \cdot 24 & 432 &  6 & 426 \\ 		15 \cdot 29 & 435 &  4 & 431 \\ 		21 \cdot 21 & 441 &  6 & 435 \\ \hline 		17 \cdot 26 & 442 &  5 & 437 	\end{array} 	\] 	For readability we have placed dividing lines 	under any pair with $|p-q| \le 1$. 	Since the intervals $[pq-\delta, pq]$ cover $[287, 442]$, 	the lemma is proved. $\blacksquare$

[b]Lemma[/b]:  	The number $M = 443$ is bad.

[i]Proof.[/i]  	Assume for contradiction $pq$ exists, 	meaning $pq \ge 443$ and $(p+1)(q+1) \le 492$. 	Then $pq \le 491 - (p+q)$. 	Now $p+q \ge 2\sqrt{443} \implies p+q \ge 43$, 	hence $pq \le 448$.

Hence $K = pq \in [443, 448]$. 	We now compute the factorization of each $K$ with $p+q$ minimal 	(in other words $p$ and $q$ as close as possible). 	\begin{align*} 		443 &= 1 \cdot 442 \\ 		444 &= 12 \cdot 37 \\ 		445 &= 5 \cdot 89 \\ 		446 &= 2 \cdot 233 \\ 		447 &= 3 \cdot 149 \\ 		448 &= 16 \cdot 28 	\end{align*} 	All of these fail the inequality $(p+1)(q+1) \le 492$, so $443$ is bad. $\blacksquare$

The greatest good number is $576$ and the smallest bad number is $443$. \\

Claim: $M$ is good iff there exist positive integers $x \le y$ (WLOG) such that $xy \ge M$ and $(x+1) (y+1) \le M + 49$ (or in other words $(x+1) (y+1) - 49 \le M \le xy$).
Proof: The if direction is true by considering $a = xy, b = (x+1) y, c = x(y+1)$, and $d = (x+1)(y+1)$. Now we prove the only if direction.

Suppose such $x,y$ do not exist, but we had $a, b, c, d$ where $ad = bc$ and $M \le a < b \le c < d \le M + 49$. By Four Number Lemma, there exist positive integers $p,q,r,s$ with $pq = a, rs = d, pr = b, qs = c$. We must have $q < r$ and $p < s$. Therefore $rs \ge (p+1) (q+1)$, so $pq \ge M$ and $(p+1) (q+1) \le M + 49$, as desired. $\square$

Clearly $576$ is good by setting $x = y = 24$. If $M$ is good, then $xy \ge M$ and $(x+1) (y+1) \le M + 49$ for some $x,y$, so $x + y  \le 48$, meaning the maximal value for $xy$ is $24\cdot 24 = 576$, hence all good numbers are at most $576$, so $576$ is the largest good number. Now it remains to prove that $443$ is the smallest bad number.

Claim: All $M \le 290$ are good.
Proof: Let $10y$ denote the smallest multiple of $10$ at least $M$. Setting $x = 10$ and $y = y$ gives that if $11(y+1)  - 49 \le M \le 10y$ implies $M$ is good. Let $M = 10y - c $, where $0\le c \le 9$. We have $11 y - 38 \le 10y - c$, so $y \le (38 - c)$. Since $c$ is at most $9$, any $y \le 29$ works, meaning any $M \le 290$ is good. $\square$

Now we prove that every integer in $[291,442]$ is good.

For $[291,300]$ take $(x,y) = (15,20)$.
For $[301,311]$ take $(x,y) = (13,24)$.
For $[312,324]$ take $(x,y) = (18,18)$.
For $[325,330]$ take $(x,y) = (15,22)$.
For $[331,340]$ take $(x,y) = (17,20)$.
For $[341,350]$ take $(x,y) = (14,25)$.
For $[351,361]$ take $(x,y) = (19,19)$.
For $[362,368]$ take $(x,y) = (16,23)$.
For $[369,378]$ take $(x,y) = (18,21)$.
For $[379,384]$ take $(x,y) = (16,24)$.
For $[385,390]$ take $(x,y) = (15,26)$.
For $391$ take $(x,y) = (15,28)$.
For $[392,400]$ take $(x,y) = (20,20)$.
For $[401,408]$ take $(x,y) = (17,24)$.
For $[409,414]$ take $(x,y) = (18,23)$.
For $[415,420]$ take $(x,y) = (20,21)$.
For $[421,425]$ take $(x,y) = (17,25)$.
For $[426,432]$ take $(x,y) = (18,24)$.
For $[433,434]$ take $(x,y) = (14,31)$.
For $[435,441]$ take $(x,y) = (21,21)$.
For $442$ take $(x,y) = (17,26)$.

Therefore every positive integer less than $443$ is good. It remains to show that $443$ is bad. Suppose there existed $x,y$ where $(x+1)(y+1) - 49 \le 443 \le xy$. Let $xy = 443 + c$. We must have $xy + x + y - 48 \le xy - c \le xy$, so $x  + y \le 48 - c \le 48$.

However, since $xy \ge 443$, we have $x + y \ge 2\sqrt{xy} > 42$, so $c < 6$. Now it remains to disprove $c\in [0,5]$.

If $c=0$, then $xy = 443$. To check that $443$ is prime, we see it is $420 + 23$, so it can't be divisible by $2,3,5,7$. It is also $143 + 300$, so it isn't divisible by $11$ or $13$, and it is $323  + 120$, so it isn't divisible by $17$ or $19$. If $443$ was composite, then it would have a prime factor under $\sqrt{443} < 22$, but we proved this is impossible. Thus, we must have $(x,y) = (1,443)$ and permutations, so $x + y = 444 > 48$, absurd.
If $c = 1$, then $xy = 444$ and $x + y \le  48$. However, one of $x,y$ must be a multiple of $37$ since $37 \mid 444$ and $37$ is prime. If the multiple of $37$ is over $74$, then it's obviously impossible. Thus, one of $x,y$ is equal to $37$, so the other one must be $12$, meaning $x + y = 49 > 48$, absurd.
If $c = 2$, then $xy = 445$ and $x + y \le  48 - c = 46$. Then since $\frac{445}{5} = 89$ is prime, one of $x,y$ is a multiple of $89$, so $x + y > 89$, absurd.
If $c = 3$, then $xy = 446$, and $x  + y \le 45$. We see that $\frac{446}{2} = 223$ is prime because it is $2\cdot 3\cdot 5\cdot 7 + 13$, $11\cdot 20 + 3$, $13\cdot 17 + 2$ so it can't be a multiple of any prime under $\sqrt{223}$. Therefore, one of $x,y$ is a multiple of $223$, so $x + y \ge 223$, absurd.
If $c = 4$, then $xy = 447$ and $x  + y \le 44$. Since $\frac{447}{3} = 149$ is prime, one of $x,y$ is a multiple of $149$, so $x + y \ge 149$, absurd.
If $ c= 5$, then $xy = 448$ and $x + y \le 43$. Clearly $7\mid 448$, so one of $(x,y)$ is a multiple of $7$. If one of $(x,y)$ is $7$ or $14$, then either $x + y = 7 + 64$ or $14 + 32$, both of which exceed $43$. Since $21\nmid 448$, neither of $x,y$ can equal $21$. Now if one of $x,y$ is $28$, then $x + y = 28 + 16 = 44 > 43$. Otherwise, one of $x,y$ is a multiple of $7$ over $42$ (since $35$ and $42$ don't divide $448$), so $x + y > 43$, absurd.


floor function
ceiling function
algorithm
calculus
calculus computations

Let $ x_0 \equal{} a$, $ x_1 \equal{} b$ be real numbers, and for $ n \geq 2$ set $ x_n \equal{} |x_{n \minus{} 1} \minus{} x_{n \minus{} 2}|$.  Find conditions on $ a, b$ so that the sequence $ (x_n)$ converges.  If $ (x_n)$ does not converge, what is its end behavior?

case 0: $ a \equal{} b$, then $ x_2 \equal{} 0, x_3 \equal{} |b|, x_4 \equal{} |b|, x_5 \equal{} 0, ...$
$ x_{3k} \equal{} |b|, x_{3k \plus{} 1} \equal{} |b|, x_{3k \plus{} 2} \equal{} 0,...$
converges iff $ a \equal{} b \equal{} 0$
case 1: $ |a| < b$
$ x_2 \equal{} b \minus{} a, \to x_3 \equal{} |b \minus{} b \minus{} a| \equal{} |a|, \to x_4 \equal{} b \minus{} |a| \to x_5 \equal{} |a|, ...$
$ x_{2k} \equal{} b \minus{} |a|, x_{2k \plus{} 1} \equal{} |a|,...$
converges iff $ |a| \equal{} b \minus{} |a| \to b \equal{} 2|a|$
case 2:$ a > b > 0\to (N \plus{} 1)b > a > Nb$ for some natural N
the sequence will start off like
$ x_{3k \plus{} 1} \equal{} b, x_{3k \plus{} 2} \equal{} a \minus{} (2k \plus{} 1)b, x_{3k \plus{} 3} \equal{} a \minus{} (2k \plus{} 2)b$
 N odd $ \to x_{3K \minus{} 1} \equal{} a \minus{} Nb \equal{} a', x_{3K} \equal{} (N \plus{} 1)b \minus{} a \equal{} b'$ converges iff  (a,b)=(a',b') converges
 N even $ \to x_{3K} \equal{} a \minus{} Nb, x_{3K \plus{} 1} \equal{} b, x_{3K \plus{} 2} \equal{} b', x_{3K \plus{} 3} \equal{} a'$ converges iff (a,b)=(b',a') converges
note $ a' \equal{} a_j \minus{} \lfloor \frac {a}{b} \rfloor b > 0, b' \equal{} \lceil \frac {a}{b} \rceil b \minus{} a > 0$

then $ (a,b)$ induces a sequence $ (A_i,B_i)$ by
$ (A_{i \plus{} 1},B_{i \plus{} 1}) \equal{} (A_i',B_i')1_{\lfloor \frac {A_i}{B_i} \rfloor \equal{} 2k \plus{} 1} \plus{} (B_i',A_i')1_{\lfloor \frac {A_i}{B_i} \rfloor \equal{} 2k}$
which will either
(i) continue forever, when $ A_i > B_i, \forall i$
(ii) stop after finitely many steps $ (A_M, B_M)$, and $ B_M \ge A_M > 0$ which converges iff $ B_M \equal{} 2A_M$

it looks really ugly to get a general form of either of these two..?
ps i know i still havent done the negative parts cases

case 3: $ a < b < 0$
$ x_2 \equal{} b \minus{} a, x_3 \equal{} \minus{} a, x_4 \equal{} \minus{} b,$ (a,b) converges iff (-a, -b) does, which is case 2

case 4: $ b < min (0,a)$
$ x_2 \equal{} a \minus{} b, x_3 \equal{} a \minus{} 2b, x_4 \equal{} \minus{} b, x_5 \equal{} a \minus{} b$ (a,b) converges iff (-b,a-b) does
which is an earlier case depending on sign of a

case 5: $ a < 0 < b\le |a|$
$ x_2 \equal{} b \minus{} a, x_3 \equal{} \minus{} a, x_4 \equal{} b$ converges iff (-a,b) does which is case 2

these are allpossible cases, so if we get case 2 we are done
let $ \gamma_j \equal{} \frac {A_j}{B_j}$

$ \gamma_m \equal{} r > 0 \to rB_m \equal{} A_m\to \gamma_{m \minus{} 1} \minus{} \lfloor \gamma_{m \minus{} 1} \rfloor \equal{} \frac {r}{r \plus{} 1}$

$ \to \gamma_{m \minus{} 1} \equal{} 2k_{m \minus{} 1} \plus{} 1 \pm \frac {r}{1 \plus{} r}, k_{m \minus{} 1} \ge 0$

so type (ii) solutions would be numbers of the form
 $ 2k_1 \pm \frac {1}{2k_2 \pm \frac {1}{3}}$, or $ 2k_1 \pm \frac {1}{2k_2 \pm \frac {1}{2k_3 \pm \frac {1}{3}}}$

for finitely many natural numbers $ k_i \ge 0$

I haven't quite tried to figure out everything you've done, but already in case 1 you had an error:[quote="kenn4000"]case 1: $ |a| < b$
$ x_2 \equal{} b \minus{} a, \to x_3 \equal{} |b \minus{} b \minus{} a| \equal{} |a|, \to x_4 \equal{} b \minus{} |a| \to x_5 \equal{} |a|, ...$
$ x_{2k} \equal{} b \minus{} |a|, x_{2k \plus{} 1} \equal{} |a|,...$
converges iff $ |a| \equal{} b \minus{} |a| \to b \equal{} 2|a|$[/quote] If $ x_2 \equal{} b \minus{} a$ and $ x_3 \equal{} |a|$ then $ x_4 \equal{} |b \minus{} a \minus{} |a\parallel{}$ which could be any of a variety of things but almost certainly isn't going to be what you wrote ($ b \minus{} |a|$).  (In fact, they are equal exactly when $ a \equal{} 0$ or $ a \equal{} \frac {2b}{3}$.)

Another way to have caught this error: if $ b \equal{} 2|a|$ then we quickly end up in your "case 0": if $ b \equal{} 2a \geq 0$ our sequence is $ a, 2a, a, a, 0, \ldots$, while if $ b \equal{} \minus{} 2a \geq 0$ our sequence is $ \minus{} \frac {b}{2}, b, \frac {3b}{2}, \frac {b}{2}, b, \frac {b}{2}, \frac {b}{2}, 0, \ldots$.

More broadly, the positive versus negative division isn't really significant: regardless of the signs of $ x_0$ and $ x_1$, the sequence thereafter will have all nonnegative terms, so you could just as well focus on the "both initial terms are positive" case.

[hide="Hint"]Consider those sequences that contain 0 separately -- which sequences eventually contain 0?[/hide]

aha gotta love when your base case has the problem so the fun stuff is guilty by association.

Nice problem!  Thought I'd revive it with an additional [hide="hint:"]If $ \frac {a}{b}\in \mathbb{Q}$, the problem is closely related to [url=http://en.wikipedia.org/wiki/Euclidean_algorithm]Euclid's algorithm[/url].[/hide]


calculus
integration
logarithms
trigonometry
function
calculus computations

[hide="Number 1"]$ \int\frac{x \ln x}{\sqrt{1\minus{}x^2}}\,dx \equal{} \minus{}\frac{1}{2}\left(2 \ln x \sqrt{1\minus{}x^2} \minus{} 2\int\frac{\sqrt{1\minus{}x^2}}{x}\,dx\right)$

$ \equal{} \int\frac{\cos ^2t}{\sin t}\,dt \minus{} \ln x \sqrt{1\minus{}x^2} \equal{} \int\frac{1\minus{}\sin^2t}{\sin t} \minus{} \ln x \sqrt{1\minus{}x^2}$

$ \equal{} \ln\left|\frac{1\minus{}\cos t}{\sin t}\right| \plus{} \cos t \minus{} \ln x \sqrt{1\minus{}x^2}$

$ \equal{} \ln\left|\frac{1 \minus{} \sqrt{1\minus{}x^2}}{x}\right| \plus{} (1 \minus{} \ln x) \sqrt{1\minus{}x^2} \plus{} C$[/hide]

Very good solution! What about the second one?

The second one is wrong... the corect statement is:   $ \int \ \frac {1}{x\sqrt {x^{4} \plus{} x^{2} \plus{} 1}}\,dx$

$ I\equal{}\int\frac{dx}{x\sqrt{x^{4}\plus{}x^{2}\plus{}1}}\equal{}\minus{}\int\frac{udu}{\sqrt{u^{4}\plus{}u^{2}\plus{}1}}\equal{}\minus{}\frac{1}{2}\int\frac{d(u^{2}\plus{}\frac{1}{2})}{\sqrt{(u^{2}\plus{}\frac{1}{2})^{2}\plus{}(\frac{\sqrt{3}}{2})^{2}}}$
$ I\equal{}\minus{}\frac{1}{2}\int\frac{d(\frac{\sqrt{3}}{2}\tan\theta)}{\sqrt{(\frac{\sqrt{3}}{2}\tan\theta)^{2}\plus{}(\frac{\sqrt{3}}{2})^{2}}}\equal{}\minus{}\frac{1}{2}\int\sec\theta d\theta\equal{}\frac{1}{2}\ln |\sec\theta \minus{}\tan\theta |\plus{}C$
All you need to do is substitute it back...(hope I've made no mistake) :D

[quote="Carcul"][hide=Number 1]$ \int\frac {x \ln x}{\sqrt {1 \minus{} x^2}}\,dx \equal{} \minus{} \frac {1}{2}\left(2 \ln x \sqrt {1 \minus{} x^2} \minus{} 2\int\frac {\sqrt {1 \minus{} x^2}}{x}\,dx\right)$[/quote]
I don't understand this passage. Can you be more clear? Is it integration by parts? If it is, which are the two functions f and g?

[quote="GioMott"]
I don't understand this passage. Can you be more clear? Is it integration by parts? If it is, which are the two functions f and g?[/quote]

$ \int\frac {x \ln x}{\sqrt {1 \minus{} x^2}}\,dx$
$ \equal{} \minus{} \frac12 \int\frac {\ln x}{\sqrt {1 \minus{} x^2}}\,d(1 \minus{} x^2)$
$ \equal{} \minus{} \frac12(\ln x (2\sqrt {1 \minus{} x^2}) \minus{} \int 2\sqrt {1 \minus{} x^2} d(\ln x))$
$ \equal{} \minus{} \frac {1}{2}\left(2 \ln x \sqrt {1 \minus{} x^2} \minus{} 2\int\frac {\sqrt {1 \minus{} x^2}}{x}\,dx\right)$

[quote="borislav_mirchev"]what is the substitution you made? u=?[/quote]

$ x \equal{} \frac1{u}$

Thanks to isomorphism for the explanation. Now, how can I find that
$ \int \frac{1}{sint}\, dt\equal{}\ln\left|\frac {1 \minus{} \cos t}{\sin t}\right|$?
I can easily verify that $ \frac{d}{dt}\ln\left|\frac {1 \minus{} \cos t}{\sin t}\right|\equal{}\frac{1}{sint}$, but I don't know how to calculate the integral. Any hints?

$ sint \equal{} 2sin\frac {t}{2}cos\frac {t}{2}$
$ I \equal{} \int\frac {sin^2\frac {t}{2} \plus{} cos^2\frac {t}{2}}{2sin\frac {t}{2}cos\frac {t}{2}}dt$ :)

Alternately, $ \int \frac{1}{\sin t} \, dt \equal{} \int \frac{\sin t}{\sin^2 t} \, dt \equal{} \int \frac{\sin t}{1 \minus{} \cos^2 t} \, dt \equal{} ...$.


probability
inequalities
geometry
ratio
analytic geometry
calculus
integration

The question:
Independently of each other, two numbers are selected at random from among the positive real numbers less than 2, with all numbers equally likely to be selected.  What is the probability that the sum of the squares of the two selected numbers is less than 4.

This is what I did:
x an y: the two numbers to be selected
2>x>0 and 2>y>0
I need to find the probability of this being true: x^2+y^2<4

Graphing these inequalities, I get a square of side length 2, with its lower left vertex on the origin, in the first quadrant, with a quarter-circle of radius 2, with its center at the origin, within the square.  So, all the possible (x,y) are within the square, and all the (x,y) that make x^2+y^2<4 true are within the quarter circle.  Then, I kind of guessed, and said I could represent the infinite number of points within the square with its area, and the infinite number of points within the quarter circle with its area.  The ratio of the area of the quarter circle to the area of the square would be the answer: pi/4.  This answer is correct.  However, I don't completely understand how a ratio of two sets, each with an infinite number of elements (the points within the different regions) can yield a number: pi/4.  Can anyone explain this?  Also, I don’t fully understand how the finite areas of the two regions can represent an infinite number of points: (x,y).  Could someone please explain this too?

I tried doing this another way:
x an y: the two numbers to be selected
2>x>0 and 2>y>0
4>x^2>0 and 4>y^2>0
So: 8> x^2+y^2>0.  This would represent all the possible values of x^2+ y^2
I’m looking for the probability that x^2+y^2<4 is true.
Graphing the two previous inequalities yields two quarter circles with centers at the origin, one with radius of 2, and the other with radius of 2sqrt(2).  The ratio of the area of the smaller to the larger would be the probability that x^2+y^2<4 is true.  However, this yields ½, which is wrong.  What did I do wrong?

Probability is so hard because it's often so unintuitive.

The problem in the second method you used is saying incorrectly that you can pick $ 8 > x^2 \plus{} y^2 > 0$. While this is true whenever $ 2 > x > 0$ and $ 2 > y > 0$, and it does represent all possible values of $ x^2 \plus{} y^2$, some cases where this is true does not correspond to valid values of x and y, for example x = 2.1 and y = 0.
Then the quarter-circle with $ 2\sqrt{2}$ you got contains a lot of points (x, y) which can never be picked using the random prodecure mentioned, and the probability is incorrect.

As for the paradox of dividing two infinite sets and getting a finite, precise number, I don't know how to address. But I would suggest that the area and the size of the set are almost completely unrelated; for instance, the Cantor set is infinite but doesn't even contain any kind of segment. At the same time, dividing infinity by infinity is undefined, and, well, infinity behaves very strangely.

Finite areas represent infinite points because areas are two-dimensional while points are zero-dimensional. I guess it's kind of inherent in the definition.
Still, I think we need a third opinion.

Yea, because individual poitns have no area, a finite area (or even nonexistent) is capable of containing infinite points.

So what is the correct way to solve this problem? (without resorting to 3-D coordinates plane & integrals if possible)

The first method is correct.  Uniformly choosing two independent coordinates in $ [0, 2]$ is the same as uniformly choosing a point in the square $ [0, 2] \times [0, 2]$, and what "uniform" means in this context is precisely that the probability of a set of outcomes is equal to the area of that set.


logarithms
calculus
calculus computations

(i)  Given the convergent series $ \sum{a_{n}}$ with nonnegative terms, determine the set of $ p$ such that $ \sum \sqrt{a_{n}}n^{-p}$ must converge.

(ii)  Decide whether the absolute convergence of $ \sum a_{n}$, where no $ a_{n}=-1$, implies the absolute convergence of $ \sum a_{n}/(1+a_{n})$.

For (i), I can show that it will converge for $ p > \frac{1}{2}$ and diverge for $ p<\frac{1}{2}$, but I am not sure about $ p=\frac{1}{2}$.

For $ p > \frac{1}{2}$, since $ \sum a_{n}< \infty$ and $ a_{n}\geq 0$ for all $ n$, we must have for sufficiently large $ n$ that $ a_{n}< \frac{1}{n}$ which implies $ \sqrt{a_{n}}< \frac{1}{\sqrt n}\rightarrow \sum \frac{ \sqrt{a_{n}}}{n^{p}}< \sum \frac{1}{n^{p+\frac{1}{2}}}< \infty$.

For $ p<\frac{1}{2}$ it is sufficient to consider series like $ a_{n}=\frac{1}{n^{1+\epsilon}}$ to see the divergence.

For (ii), we can (similar to (i)) write for sufficiently large $ n$:
$ \frac{|a_{n}|}{|1+a_{n}|}< \frac{|a_{n}|}{1-\frac{1}{n}}<2 |a_{n}|$ and we know $ \sum 2|a_{n}|$ converges absolutely.

that's right.  For $ p = 1/2$, it will not necessarily converge.  You can't use $ \sum 1/n$ as the counterexample, because that diverges, but you can use something that's pretty close... maybe $ \sum \frac{1}{n \log^{2}n}$  :)


inequalities
complex numbers
algebra proposed
algebra

$P(z)=R(z) + iQ(z)$ where $R(z) \in \mathbb{R}[X]$, $Q(z) \in \mathbb{R}[X]$.
Suppose that every complex root $z_0$ of $P$ satisfies  $Im(z_0)>0$.
Does it follow that $R(z)$ and $Q(z)$ have all their roots in $\mathbb{R}$

Very beautiful problem and evil examinator. I searched a little bit for contreexamples, but coudln't find and I guessed it must be true. And it is true. Suppose for example that $R(z)=0$ . Then of course $R(c(z))=0$ ( here $c(z)$ is the complex conjugate of $z$) and thus $ P(c(z))=i Q(c(z))$. But on the other hand we have $ c(P(z))=-i Q(c(z))$. It follows that $|P(c(z))|=|P(z)|$. Now, write $ P(z)=(z-z_1)...(z-z_n)$ and observe that if $ Imz>0$ then we have $ | c(z)-z_i|>|z-z_i|$ for all $i$ (or the inverse inequality, I don't have a pen now to check it) anbd multiplying them we find the desired contradiction. Hope it's good.


Let a1 in Z+ such that:  an=[((3a(n-1)))/2]+1.  (n >=2)
Is there exist a1 satisfy that:
        an is even number  for n=1,2...10^5
        a(10^5+1) is odd number

If I'm reading this right, $a_n=\frac{3a_{n-1}}{2}+1$, so that $b_n=a_n+2$ satisfies $b_n=\frac{3}{2}b_{n-1}$, and hence $b_n=\left(\frac{3}{2}\right)^{n-1}b_1$. But $b_n$ is an even integer if and only if $a_n$ is. If we let $b_1=2^nr+2^{n-1}$, we get $b_i$ even for $1\le i\le n-1$, but $b_n$ odd. In particular, $b_n$ (and hence $a_n$) is an even integer for $1\le n\le10^5$ but odd for $n=10^5+1$ if and only if $a_1=2^{10^5+1}r+2^{10^5}-2$ (for $r\in\mathbb{N}$).


trigonometry inequalities calculus derivative

Find $\max_{x\in\mathbb{R}}{ \frac{\sin x\cos x}{1+\sin x+\cos x}}$.

[hide="not sure if this is right"]
Since $(sinx + cosx)^2 = 1+2sinxcosx$,

$sinx + cosx = \sqrt{1+2sinxcosx}$. (Since we want the maximum, take the positive root).

If we let $sinxcosx = k$, the original equation becomes,$\frac{k}{1+\sqrt{1+2k}}$.

Rationalizing the denominator and simplifying leads to $\frac{\sqrt{1+2k}-1}{2}$.

Since the maximum of $sinxcosx = \frac{1}{2}$, the maximum value of the expression is
$\frac{\sqrt{2}-1}{2}$.   Btw, can we use inequalities such as AM-GM for trig equations?
[/hide]

Using calculus:

[hide]

Take the derivative to get $\frac{(1+ \sin x+ \cos x)(\cos^2x - \sin^2x)}{(1+\sin x + \cos x)^2}$, so we want

$\sin x + \cos x = -1$ or $\cos^2x - \sin^2x = 0$.

Reject the first because it gives a 0 denominator.  The second gives us $x=\frac{(2k+1)\pi}{4}$, for $k$ an integer.  Plugging back into the original equation, we get $-\frac{1}{2}$ and $\frac{\sqrt{2}-1}{2}$.  The second is the maximum.

 [/hide]

can you see why my sol doesn't work?...

[hide]

clearly we can consider only $0 \leq x \leq \frac{\pi}{2}$, so $\sin x \geq 0$ and $\cos x \geq 0$.

Then we can apply GM-QM to the numerator:

$\sqrt{\sin x \cos x} \leq \sqrt{\frac{\sin^2x + \cos^2x}{2}} \implies \sin x \cos x \leq \frac{1}{2}$

and AM-QM to the denominator

$\frac{1+\sin x +\cos x}{3} \leq \sqrt{\frac{1+\sin^2x+ \cos^2x}{3}} \implies 1+\sin x +\cos x \leq 3\sqrt{\frac{2}{3}}$

dividing the two inequalities

$\frac{\sin x \cos x}{1+\sin x +\cos x} \leq \frac{1}{6 \sqrt{\frac{2}{3}}}$
[/hide]

I can solve it now.
[hide]
$f(x)=\frac{\sin x\cos x}{1+(\sin x+\cos x)}\cdot\frac{1-(\sin x+\cos x)}{1-(\sin x+\cos x)}$
$=\frac{\sin x\cos x(1-\sin x-\cos x)}{1-(\sin x+\cos x)^2}$
$=\frac{\sin x\cos x(1-\sin x-\cos x)}{-2\sin x\cos x}$
$=-\frac12 [1-(\sin x+\cos x)]$
$=-\frac12 \left[1-\sqrt2\sin\left(x+\frac{\pi}{2}\right)\right]$
$\leq -\frac12(1-\sqrt2)=\frac{\sqrt2-1}2$[/hide]

[quote]1+sinx+cosx $\leq3\sqrt{\frac{2}{3}}$
[/quote]

1+sinx+cosx $\leq3\sqrt{\frac{2}{3}}$
$\implies$ $\frac{1}{1+\sin x +\cos x}\geq\frac{1}{3\sqrt{\frac{2}{3}}}$

 :D  :lol:  :roll:  :oops:  :)

[quote="hydro"]
Then we can apply GM-QM to the numerator:

$\sqrt{\sin x \cos x} \leq \sqrt{\frac{\sin^2x + \cos^2x}{2}} \implies \sin x \cos x \leq \frac{1}{2}$

and AM-QM to the denominator

$\frac{1+\sin x +\cos x}{3} \leq \sqrt{\frac{1+\sin^2x+ \cos^2x}{3}} \implies 1+\sin x +\cos x \leq 3\sqrt{\frac{2}{3}}$

dividing the two inequalities

$\frac{\sin x \cos x}{1+\sin x +\cos x} \leq \frac{1}{6 \sqrt{\frac{2}{3}}}$
[/quote]
Hydro, let 0<a<X<b, 0<c<Y<d, what do you think the ranges of X-Y and X/Y are?


I am having a very easy problem, yet I am not able to solve it. I have a signal which contains parts in which the function looks a lot like a parable. I want to be able to detect these parts of the signal.
How would I approach this problem
[img]http://www.plotti.com/download/signal.jpg[/img]

In the image above you see my signal and the very specific peak in the second half of the signal.

The solution to this problem should give me a good hint whenever i encounter such parabolic peaks, but should should neglect all the other peaks. Talking of parables:

a*x^2+c where a should be able to have different values from 0.3 - 2 for example.

What is the format of the data you're working with - is it just a list of values sampled at some fixed rate? If so, how long is the data (how many samples) you want to analyze?


I was wondering if people have the Chinese, Russian, Romanian, Indian or Turkish Olympiad problems? I really want to get some more problems to do.

Have you gone through everything in "Resources"?!?


let a,b,c be integers such that a+b+c divides $ a^2\plus{}b^2\plus{}c^2$.Prove that there are infinitely many positive integres n such that a+b+c divides $ a^n\plus{}b^n\plus{}c^n$
sorry I am learning English very bad

[quote="Rust"]Let $ \sigma_1 \equal{} a \plus{} b \plus{} c,\sigma_2 \equal{} ab \plus{} bc \plus{} ca,\sigma_3 \equal{} abc,A_n \equal{} a^n \plus{} b^n \plus{} c^n$, then
$ A_{n \plus{} 3} \equal{} \sigma_1A_{n \plus{} 2} \minus{} \sigma_2A_{n \plus{} 1} \plus{} \sigma_3A_n$.
From $ \sigma_1|A_2 \equal{} \sigma_1^2 \minus{} 2\sigma_2$ we get $ \sigma_1|2\sigma_2$.
If $ \sigma_1|\sigma_2$, then $ \sigma_1|A_{3k \plus{} 2} \ \forall k\in N$.
If $ \sigma_1|2\sigma_2, \ \sigma_1\not |\sigma_2$, then $ 2|A_3\to 2|A_{3k}$, therefore
$ \sigma_1|A_{3k \plus{} 2} \equal{} \sigma_1 A_{3k \plus{} 1} \minus{} 2\sigma_2\frac {A_{3k}}{2} \plus{} \sigma_3 A_{3k \minus{} 1}$.[/quote]

What's the main line here? i cannot see it. For example, are there an argument $ \epsilon\geq A_{n \plus{} 1} \minus{} A_{n}$ so either we have a contradiction or $ A_{n \plus{} 1} \equal{} A_{n}$; we are continouing with the second one, so if we have only one $ A_{i}$ for some $ i\in Z$ then we have also there are infinitely many $ A_{n}$ satisfying the condition we have said. Note that, the initial $ i$ can be for example as $ i \equal{} 2$.


geometry
geometric transformation
real analysis
real analysis theorems

Can someone recommend me good books where Fourier Analysis is taught?

That depends on the level. At an undergraduate (no measure theory) level, I'm partial to the book [i]A First Course in Fourier Analysis[/i] by David Kammler. It was published by Prentice Hall, but it seems to have gone out of print already.

For a book just on Fourier series, there's [i]Fourier Series[/i] by Georgi P. Tolstov. The English translation was published by Dover - it's an old book and has been around for a long time, but with Dover, it might still be available. (And maybe the Russian original is still out there.)

If you do know some measure theory, there's the book by Yitzhak Katznelson, also on Dover. Hmm - I can't find my copy of Katznelson at the moment, so I'll have to guess about the title. It's something like "An Introduction to Euclidean Harmonic Analsyis."


Lockers are numbered with consecutive positive integers beginning with 1 (it would be fun to be locker #1--my school starts at 1000 or something). The digit 2 is used exactly 90 times (who counts these things, anyway?) What is the number of the last locker?

I got
[hide]237[/hide]Is that right? I lost track of what I was doing.

I need an explanation, please.

I posted this because the explanation on mathcounts.saab.org was really bad. They listed out all the possibilites.

And no, the correct answer is
[hide]235[/hide]

My explanation's about the same: :oops:
You find 20 2's between 1 and 100, so between 101 and 199 there's also 20 2's, which brings the total up to 40.From 200, just count 1 2 for every number then add either 1 2 or 2 2's for numbers like 212 and 222 until the total goes up to 235. It's not a great solution...

[hide]1-100 has it 20 times. Same for 100-199. Now comes the tricky part. With 50 twos left you go to 250. Minus five for each 2 as a units digit to get 245. Subract 10 for the 220s to get [b][u]235[/u][/b][/hide][/hide]
EDIT: POST 100 WOOT


How many sets $S$ which have $k$ elements and its elements in $\{1,2,3,..,n\}$ ? ($k<n$) Its elements are not necessarry different

Not really, because it says that the elements NEED NOT be different. The correct answer is ${{n+k-1}\choose n}$

[quote="Farenhajt"]${{n+k-1}\choose n}$[/quote]
Yes, and these are known as "combinations with repetition". ;)


Prove that there exists infinitely many primes of form x^2+1.

Bomb

You are so funny, hohohoho  :lol: Not!!!!

What d'ya mean, Im joker. I am very very serious, its never been solved to date, only proved Hasse closer. That's why its here, to be provveeeeddd.

BOMB :D

No. This is an open question, so it has to be in the open question section!  :mad:

Pierre.


Express $ 0.1\overline{7}$ as a common fraction.

Let $ 0.1\overline{7} \equal{} x$. Multiply both sides by 100 to get:

$ 17.\overline{7} \equal{} 100x$

Keep in mind that $ 0.\overline{x} \equal{} \frac{x}{9}$.

$ 17 \frac{7}{9} \equal{} 100x$

$ \frac{160}{9} \equal{} 100x$

$ \frac{8}{45} \equal{} x$

Thus that is our answer.

If we let that be x, then multiplying by 10 gives:

$ 10x\equal{}1.7\overline{7}$
$ \text{ }1x\equal{}0.1\overline{7}$

And subtraction gets rid of all the repeating decimals:

$ 9x\equal{}1.6$, so $ x\equal{}\frac{1.6}{9}\equal{}\frac{16}{90}\equal{}\frac{8}{45}$

EDIT: beaten to it.

My method.

Split the decimals into $ 0.1$ and $ 0.0\overline{7}$.

Turn them into fractions: $ \frac1{10} \plus{} \frac{7}{90}$

$ \frac9{90} \plus{} \frac7{90} \equal{} \frac{16}{90} \equal{} \boxed{\frac8{45}}$


$ x^x \plus{} y^y \equal{} 31$

$ x^y \plus{} y^x \equal{} 17$

Solve it..

And please post a general solution for this

[hide]The only way I can see to solve this is to break both parts up into perfect squares or cubes.
$ 31 \equal{} 27 \plus{} 4$
$ 17 \equal{} 9 \plus{} 8$

So its easy to see that $ (2, 3)$, and $ (3, 2)$ are solutions. [/hide]


Here is a question from a test I recently I just took. My teacher and I disagree on the answer.

"A printer printed a card with black ink on blue paper and an envelope with blue ink on blue paper. If the colors were chosen at random from a stock of 7 inks and 6 papers, what was the probobility of the given result?"

He thinks it is 1/21. I think it is 1/1764. He says that the reason that it is 1/21 is that there are two seperate events, but he cannot explain this any further, and I am not convinved.

Can anyone help me on this?

Thanks.

I'm not sure if this is correct (i'm not the best person to ask), but I got 1/1764:

black ink on blue: $\frac{1}{7} * \frac{1}{6}$
blue ink on blue: $\frac{1}{7} * \frac{1}{6}$
then multiply $\frac{1}{42} * \frac{1}{42}$
to get $\frac{1}{1764}$
This is true only if there was no limited amount of ink or paper; ie the ink and paper in the last problem were not depleted.  You should wait for someone else to confirm though...
Since the events were seperate they would be multiplied by each other: ie, there is a total of 7*6*7*6 1764 different possible combinations to be made with two events, and that is one of the events...

Thats what I get as well, and I think the word stock makes it seem that there are multiple of each color. However, my teacher took the problem to 3 other math teachers and they all got the same answer   :huh: (1/21).

for each event there is a $\frac{1}{42}$ chance of each occuring. The question now is do we add or multiply the two probabilities together?

This web page may help you decide

http://ep.llnl.gov/bep/math/9/SDouble.html

This is the option I chose (I may be wrong):

 The probability that two mutually exclusive events will have different occurrences is twice the product of the probability one will have the occurrence times the probability the other will not have the occurrence.

The chances of both the card and the envelope of having the requested combination of colors is $2 \displaystyle(\frac{1}{42})(\frac{41}{42}) = \frac{41}{882}$

The option you chose has this example.

"For example, the probability when two dice are thrown that only one will be a 'four' is found as follows"

This is how the second probobility (in the example) ends up being 5/6. I do not think that this is what probobility we are looking for.

However, I could be wrong.

are you sure that is the question stated exactly as it was on the test, you answer is correct for the problem that you gave, but in any event, you teacher is wrong, your teacher did this: $\frac{1}{6*7}+\frac{1}{6*7}\frac{1}{42}$

the reason why that is wrong is because to have both, you have to multiply the probabilities, these are independant events, so you multiply the probabilities, not add, also perhaps the question was what is the probability of at least 1 of those events occuring? this would justify the adding, but it is still wrong though because it does not take into account the fact that the 2 events could occur at the same time, and the way they are doing it, they double count that situation


topology
rotation
advanced fields
advanced fields unsolved

Find an example of a topological group $ G$ with continuous action on a topological space $ X$ such that the natural map $ G/G_x \to Gx$ is not open (note: it is always continuous and bijective).

$ G \equal{} \mathbb{Z}$, $ X \equal{} S^1$, the action $ G \to Aut(X)$ induced by $ 1 \mapsto (z \mapsto z  \exp(2 \pi i \theta))$ where $ \theta$ is irrational. so we rotate $ S^1$ by an irrational angle. then $ G/G_x \equal{} G$ is discrete, whereas $ Gx$ is not.

however, if $ G$ is compact and $ X$ is hausdorff, $ G/G_x \to Gx$ is a homeomorphism.


[quote="Ulanbek_Kyzylorda KTL"]Let f,g: R >R be functions like that so [b]f(g(x))=g(f(x))=-x[/b] for any x is element of R
a) prove that f and g are odd functions
b) Make an example of these two functions [b]f[/b] isn't equal to [b]g[/b]
Please solve it[/quote]

a) :
$ g(f(g(x)))\equal{}g(u)$ where $ u\equal{}f(g(x))\equal{}\minus{}x$ and so $ g(f(g(x)))\equal{}g(\minus{}x)$
$ g(f(g(x)))\equal{}g(f(v))\equal{}\minus{}v$ where $ v\equal{}g(x)$ and so $ g(f(g(x)))\equal{}\minus{}g(x)$
So $ g(\minus{}x)\equal{}\minus{}g(x)$ and $ g(x)$ is an odd function.

Same computation with $ f(g(f(x)))$ shows that $ f(x)$ is an odd function.

b)
Choose $ f(x)\equal{}2x$ and $ g(x)\equal{}\minus{}\frac x2$


Given a quadrilateral $ABCD$, point $K$ and $L$ is obtained as the intersection of ray $AB$ and $AD$, respectively, with the circle centered at A whose radius is the geometric mean of the sides of the quadrilateral starting at vertex A. Similarly, points $M$ and $N$ are obtained as the intersections of rays $CB$ and $CD$, respectively, with the circle centered at $C$ whose radius is the geometric mean of the sides starting at vertex $C$. Find the condition which guarantees that $KL=MN$.

Denote the quadrilateral sides a = AB, b = BC, c = CD, d = DA and its internal angles $\alpha = \angle A, \beta = \angle B, \gamma = \angle C, \delta = \angle D$. The triangles $\triangle AKL, \triangle CMN$ are isosceles with the sides $AK = AL = \sqrt{da}$, $CM = CN = \sqrt{bc}$ and angles $\angle KAL = \alpha, \angle MCN = \gamma$. Thus $KL = 2 \sqrt{da} \sin \frac \alpha 2$, $MN = 2 \sqrt{bc} \sin \frac \gamma 2$. The condition KL = MN then means

$2\sqrt{da} \sin \frac \alpha 2 = 2\sqrt{bc} \sin \frac \gamma 2$

$4da \sin^2 \frac \alpha 2 = 4bc \sin^2 \frac \gamma 2$

$2da (1 - \cos \alpha) = 2bc (1 - \cos \gamma)$

Denote p = BD one of the quadrilateral diagonals. Using the cosine theorem for the triangles $\triangle ABD, \triangle CBD$, we have

$p^2 = d^2 + a^2 - 2da \cos \alpha = b^2 + c^2 - 2bc \cos \gamma$

$2da \cos \alpha = d^2 + a^2 - p^2,\ \ 2bc \cos \gamma = b^2 + c^2 - p^2$

Substituting this to the last formula for the condition KL = MN,

$2da - (d^2 + a^2 - p^2) = 2bc - (b^2 + c^2 - p^2)$

The diagonal p disappears from the equation and we obtain

$d^2 - 2da + a^2 = b^2 - 2bc + c^2$

$(d - a)^2 = (b - c)^2$

$d - a = c - b$ [color=white].[/color] or [color=white].[/color] $d - a = b - c$

$b + d = a + c$ [color=white].[/color] or [color=white].[/color] $b - d = c - a$

The 1st condition is the well known sufficient and necessary condition for the sides of a tangential quadrilateral with an incircle: The sums of the opposite sides are equal. The 2nd possible condition is a similar condition for a tangential quadrilateral with an excircle: The differences of the opposite sides are equal.


show that the function f(t) = A Cos(at+z)  + B Cos(bt+x) + C Cos(ct+y) is periodic when  c/a and b/a are rational numbers.

Just to make it easier for people to solve.

$f(t) = A \cos (at+z) + B  \cos (bt+x) + C \cos( ct+y)$ is periodic when $\frac {c}{a}$ and $\frac {b}{a}$ are rational numbers. And oh yeah, show this. :)

Function f is periodic iff $\exists T \in \mathbb{R}:f\left( {t + T} \right) = f\left( t \right) \Leftrightarrow $
$ \Leftrightarrow A\cos \left( {at + aT + z} \right) + B\cos \left( {bt + bT + x} \right) + C\cos \left( {ct + cT + y} \right) = $
$ = A\cos \left( {at + z} \right) + B\cos \left( {bt + x} \right) + C\cos \left( {ct + y} \right)$

But $\frac{c}{a},\frac{b}{a} \in \mathbb{Q} \Rightarrow \exists r,p,q,w \in \mathbb{Z}$ such that  $\frac{c}{a} = \frac{r}{p}$  and  $\frac{b}{a} = \frac{q}{w}$

If  $\exists k_1 ,k_2 ,k_3  \in \mathbb{Z}$ such that  $aT = 2\pi k_1 ,bT = 2\pi k_2 ,cT = 2\pi k_3 $
 then  f is periodic.
Then $k_1  = pw,k_2  = qp,k_3  = rw$  verifay  first identity and f is periodic.


sorry to bother you guys

should i start with Engel, Problem solving through problems, or treasures?

You can find many topics about "which book(s) are the best to start with".

[quote="TheHeinrich"]sorry to bother you guys

should i start with [b]Engel, Problem solving through problems, or treasures[/b]?[/quote]

May i ask what these are???


Keiko tosses one penny and Ephraim tosses two pennies. What is the probability that Ephraim gets the same number of heads as Keiko gets?

A. 1/4  B. 3/8   C. 1/2   D. 2/3  E. 3/4.

How would you do this problem?

Divide the outcomes into cases

[u]Case I: Keiko gets 1 head [/u]
The probability of Keiko getting 1 head is
$\frac{1}{2}$
The probability of Ephraim getting 1 head is
$\binom{2}{1}(\frac{1}{2})^{2}=\frac{1}{2}$

[u]Case II: Keiko gets 0 heads[/u]
The probability of Keiko getting 0 heads is
$\frac{1}{2}$
The probability of Ephraim getting 0 heads is
$(\frac{1}{2})^{2}=\frac{1}{4}$

Therefore, the probability of Ephraim receiving the same number of heads as Keiko is
$(\frac{1}{2})(\frac{1}{2})+(\frac{1}{2})(\frac{1}{4})=\frac{3}{8}$

[quote="NeverOddOrEven"]Divide the outcomes into cases

[u]Case I: Keiko gets 1 head [/u]
The probability of Keiko getting 1 head is
$\frac{1}{2}$
The probability of Ephraim getting 1 head is
$\binom{2}{1}(\frac{1}{2})^{2}=\frac{1}{2}$

[u]Case II: Keiko gets 0 heads[/u]
The probability of Keiko getting 1 head is
$\frac{1}{2}$
The probability of Ephraim getting 0 heads is
$(\frac{1}{2})^{2}=\frac{1}{4}$

Therefore, the probability of Ephraim receiving the same number of heads as Keiko is
$(\frac{1}{2})(\frac{1}{2})+(\frac{1}{2})(\frac{1}{4})=\frac{3}{8}$[/quote]

you put keiko getting 1 head chance is 1/2 under keiko gets 0 heads. ..  :)

wow I got the answer in a different way, yours is easier


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ R$ be a commutative ring in which $ a^2\equal{}0$ only if $ a\equal{}0$. Show that if $ q(x)\in R[x]$ is a zero-divisor in $ R[x]$, then, if $ q(x)\equal{}a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots \plus{}a_{0}$ there is an element $ b\neq 0$ in $ R$ such that $ ba_{0}\equal{}ba_{1}\equal{}\cdots \equal{} ba_{n}\equal{}0$.

Idea so far:
Let there be a polynomial $ p(x)\equal{}b_{m}x^{m}\plus{}\cdots\plus{}b_{0}$ such that $ q(x)p(x)\equal{}0$. This implies that $ b_{m}a_{n}\equal{}0$ and $ b_{0}a_{0}\equal{}0$. Notice that $ \text{deg } b_{m}q(x) \equal{} n \minus{} 1 < \text{deg } q(x) \equal{} n$.

Your idea is good. Is there a polynomial $ p(x)$ like the one you found that has minimal degree? what must its degree be?

The problem without the assumption that $ a^2\equal{}0$ implies $ a\equal{}0$ is still true, try to prove this.

[quote="TomciO"]The problem without the assumption that $ a^2 \equal{} 0$ implies $ a \equal{} 0$ is still true, try to prove this.[/quote]

This general formulation is sometimes called McCoy's theorem. You can find a proof in my sci.math post here
http://google.com/groups?selm=y8z3c6fwpme.fsf%40nestle.csail.mit.edu


Every pair of communities in a couny are linked directly by exactly one mode of transportation: bus, train, or airplane.

*All three modes are used in the county
*no community is served by all three modes
*no three communities are linked pairwise by the same mode.

Prove that five communities cannot be linked in this manner.

Casework works nicely:  if there exists a mode of transportation $X$ and a city $A$ such that $X$ connects $A$ to $B, C, D$, consider the modes connecting $B, C$ and $D$ to each other.  If there is no such pair, you can show that there must be at least 4 edges of each color, a fairly immediate contradiction.


Er... This seems to easy an answer for this forum, but is it just similar triangles?


I am having problems using commands that I created to display special characters in the example environment.  The special characters display properly, but apparently everything following a special character is treated as part of one word.  This causes problems with the gloss aligning properly.  Does anyone have any suggestions for how to fix this problem?  Here is a minimal example:

[quote]\documentclass[11pt]{amsart}
\usepackage{geometry}                % See geometry.pdf to learn the layout options. There are lots.
\geometry{letterpaper}                   % ... or a4paper or a5paper or ...
%\geometry{landscape}                % Activate for for rotated page geometry
%\usepackage[parfill]{parskip}    % Activate to begin paragraphs with an empty line rather than an indent
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\usepackage[utf8x]{inputenc}
\newcommand{\R}{\unichar{"1E5B}} %This is the command for r underdot
\newcommand{\M}{\unichar{"1E43}} %This is the command for m underdot
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}
\usepackage{cgloss4e}
\usepackage{gb4e}

\begin{document}

\begin{exe}
\ex \gll indro v\R tra\M\ jayati \\
Indra.nom Vrtra.acc defeat.pres.3s\\ %`defeat' should line up under `jayati'
\glt `Indra defeats Vrtra.'
\end{exe}

\end{document}  [/quote]

Commands need to be ended so use \R{} instead of {}. You would do the same with \LaTeX{} otherwise the space after would be ignored. Using \ as you've done after \M is an alternative.


A group consist of $n$ tourists. Among every $3$ of them there are $2$ which are not familiar. For every partition of the tourists in $2$ buses you can find $2$ tourists that are in the same bus and are familiar with each other. Prove that is a tourist familiar to at most $\frac{2n}{5}$ tourists.

Take the obvious graph with tourists as vertices and edges between familiar people. The first condition says that the graph has no triangles; the second says that it's not bipartite, so it must have a cycle of odd length ($>3$, of course).

Take an odd cycle of minimal length, $T_1T_2 \ldots T_{2k+1}$. It is clear that this subgraph can't have any edges $T_iT_j$ except $T_1T_2, \ldots, T_{2k+1}T_1$, as otherwise there would be an odd cycle of smaller length. Also, note that if a vertex $X$ outside this subgraph has two edges $XT_i, XT_j$ then $T_i \ldots T_jX$ and $T_iXT_j \ldots T_{i-1}$ are cycles, one of which has odd length, and the odd cycle is of length $<2k+1$ unless $j=i+2$. However, if there are at least $3$ edges $XT_i, XT_j, XT_m$, then two of $i, j, m$ are at distance $\neq 2$ apart, so that we can find a cycle of smaller length, absurd. Then any $X$ must have at most two edges $XT_i, XT_j$. The sum of the degrees of the $T_i$ is then $\leq 2(n-2k-1)+2(2k+1)=2n$, so one of them has degree $\leq \frac{2n}{2k+1}$. Since $2k+1>3$, it is at least $5$, thus the result.

Why does the second condition say that the graph is not bipartite?

Because for every partition of the vertices in $2$ sets there is an edge between $2$ vertices in the same set.


The perimeter of an isosceles right triangle is $ 2p$. Its area is:
$ \textbf{(A)}\ (2\plus{}\sqrt{2})p \qquad\textbf{(B)}\ (2\minus{}\sqrt{2})p \qquad\textbf{(C)}\ (3\minus{}2\sqrt{2})p^2\\
\textbf{(D)}\ (1\minus{}2\sqrt{2})p^2 \qquad\textbf{(E)}\ (3\plus{}2\sqrt{2})p^2$

hello
using perimeter , the sides of triangle are $ \frac{2p}{2\plus{}\sqrt{2}},\frac{2p}{2\plus{}\sqrt{2}},\frac{2p\sqrt{2}}{2\plus{}\sqrt{2}}$
Area=$ (3\minus{}2\sqrt{2})p^2$
thank u

[hide="Click for solution"]
Let $ x$ be one of the equal sides of the triangle.  $ 2p\equal{}2x\plus{}x\sqrt{2} \implies x\equal{}\frac{2p}{2\plus{}\sqrt{2}} \implies$  $ A\equal{}\frac{1}{2}x^2\equal{}\frac{1}{2} \cdot \frac{4p^2}{6\plus{}4\sqrt{2}} \cdot \frac{6\minus{}4\sqrt{2}}{6\minus{}4\sqrt{2}}\equal{}p^2(3\minus{}2\sqrt{2})$, or $ \boxed{\textbf{(E)}}$.
[/hide]


MATHCOUNTS
articles
inequalities
function
absolute value

I'm going to add some problems daily but please do not post your answers to these yet.

Save these for later when you attended the math chat and then learned something from it and then look at it.

The topic is Algebra. I have changed the wording to prevent any troubles from directly posting the national mathcounts problems. But I did put the year and the number.

1. Let's generalize $x$ as $\frac {1}{2}$ and $y$ as $\frac {1}{3}$. Then compute the value of:

$\left(\frac {1}{x} + \frac {1}{y}\right) + \frac {x}{y}$

From 1983 #14

2. When General McHenry bought the article from Starbucks, the price was 100 dollars. During the sale, the price was decreased by 20%. If the rate of sales tax is 6%, calculate how many dollars would General McHenry have to pay for his article from Starbucks. Express your answer in ab.cd dollars, where a,b,c,d represent the tens, unit, and two decimal digits.

From 1983 #27

3. It is given that for $a,b \in \mathbb {R}$ integers: (This just represent the real numbers that are integers)

$\frac {7}{3} = \frac {a+b}{a-b}$ with $0 \leq ab \leq 10$. Then what is the value of $ab$ in this problem?

From 1983 #31

4. Let's denote two integers $r$ and $s$ such that $r+s = 10$ and $2r-s = 11$. Then find the sum of the squares of $r$ and $s$, in another word, what is $r^2+s^2$?

From 1983 #32

5. If $1<p$, which expression would be ALWAYS (excluding when $p=1$ by restriction) greatest number?

$(p+1)(p-1), (p^2+p), (p+1)^2$

From 1983 #39

6. The operation $\bullet$ is defined as $U \bullet = U^2-1$. Calculate:

$10(3 \bullet) - 5(4 \bullet)$.

From 1992 #9

7. $k(\zeta) = |\zeta - 1|$ such that $k(\zeta) < 3$ for some numbers. Let's call this number $K$. What is the value of $K$? (This just means for $K$ numbers, the inequality holds true)

From 1992 #13

I forgot the sources on the followings.

8. Geometric mean is given as:

$({x_1 \cdot x_2 \cdot x_3 .... x_n})^{\frac {1}{n}}$

If the geometric mean of 81,243,and 729 equals $j^k$, find $j+k$.

9. What real number is not found on the function $j(z)$ defined as $j(z) = \frac {1}{z} - 2$?

10. $\varphi = 50!$. How many zeros does $\varphi$ end when expressed in standard base ten notation?

11. We're given the fact that:

$7x^2+3x^2y+14xy-2xy^2 = 61$ and $7y^2-3x^2y+2xy^2 = 2$.

What is the absolute value of the difference of the two variables?

12. A super-engined car circles the Mars in 48 hours. A super-dooper-engined car, which travels from the oppostite direction and at the same altitude, passes super-engined car in every 16 hours. It takes $\gamma$ hours for super-dooper-engined car to circle the earth. Evaluate the absolute value of $\gamma$.

The more problems will be posted. I'm thinking posting around 30 problems. :)

[quote]11. We're given the fact that:

7x^2+3x^2y+14xy-2xy^2 = 61 and 7y^2-3x^2y+2xy^2 = 2.

What is the absolute value of the difference of the two variables? [/quote]
[hide]Add the 2 equations and we get $7x^2+14xy+7y^2=63$. Divide both sides by 7 to get $x^2+2xy+y^2=9$. This factorizes into $(x+y)^2$, so x+y=3.[/hide]

I'm not sure of the difference though, hopefully this will help.

Billy

[quote]1. Let's generalize  1/2 as x and 1/3 as y . Then compute the value of:
 $(\frac {1}{x} + \frac {1}{y}) + \frac {x}{y}$[/quote]

[hide]Within the parentheses, you immediately have 1/(1/2)+1/(1/3). Simplifying, you get 2+3=5. Now you have to solve for the x/y. You have (1/2)/(1/3) Sorry I'm a bit lazy and don't want to use tex... so that simplifies out to 3/2. When you add 5 to 3/2, you get 13/2.[/hide]

[quote]1. Let's generalize  1/2 as x and 1/3 as y . Then compute the value of:
 $(\frac {1}{x} + \frac {1}{y}) + \frac {x}{y}$[/quote]
[hide]The $(\frac {1}{x} + \frac {1}{y}$ simplifys to $\frac{x+y}{xy}$, and setting a common denominator, we get $\frac{x+y+x^2}{xy}$. Then substituting in, we get $\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{\frac{1}{6}}$. That is 13/2
[/hide]

Billy


The conning tower of a ship is about 20 feet above sea level. How far can one see from the conning tower?

Use the fact that the radius of the Earth is 3960 miles.

same principle as [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127376[/url]

Thanks for your quick reply and math tip.


induction
combinatorics unsolved
combinatorics

A positive integer N0 is written on the blackboard.
Two players, A and B, play the following game.
A erases N0 and writes N1 \in {N0 - 1, [N0/3]} instead.
B erases N1 and writes N2 \in {N1 - 1, [N1/3]} instead.
...
They continue playing until 0 appears on the blackboard.
The player who managed first to write 0 wins the game.
Which player has a winning strategy when

(a) N0 = 120?
(b) N0 = (32002 - 1)/2?
(c) N0 = (32002 + 1)/2?

It is easy to prove (by induction) N0=11..13 is a good case for Player 1. Then we show N0=11..103 and N0=11..123 are good cases for Player 2.

You are right. It is not very difficult problem. Let's think about arbitrary N0. It's very interesting.

Namdung

Namdung!
Tell us about common case. I can't catch thought.

The main idea:
Denote f(n) = 1 if the first player wins, = 0 if otherwise then we have
f(n) = 1 - f(n-1)f([n/3])

Study this sequence we can find some patern. Let's try first. I will post some my results later. (I do not receive the final result, but quite general, for example, we can say who wins if N0=2004)

Namdung

Namdung! It is time to tell us your results ;) After 1 year...

Thank you!


hello !  :)
find all $ f: \mathbb{R}\to\mathbb{R}$ such that:

$ (\forall x\in{\mathbb{IR}})$ $ (fofofof)(x)\equal{}(fofof)(x)\plus{}2x$

[hide="My solution"]this pb can be solved easily by sequences.
so: Let $ (x_{n})$ a sequence such that: $ x_{0}\equal{}x$ and $ f(x_{n})\equal{}x_{n\plus{}1}$ so we get:
$ x_{n\plus{}4}\equal{}x_{n\plus{}3}\plus{}2x_{n}$,and by solving the equation: $ x^{4}\minus{}x^{3}\minus{}2\equal{}0$ we get:
$ x\equal{}\minus{}1$ or $ x\equal{}a\approx1,5436890126921$ then:
$ x_{n}\equal{}a^{n}\alpha\plus{}(\minus{}1)^{n}\beta$ with $ x\equal{}\alpha\plus{}\beta$.
$ f(x)\equal{}a\alpha\minus{}\beta$.
since $ (fofofof)(x)\equal{}(fofof)(x)\plus{}2x$ we have: $ a^{4}\alpha\plus{}\beta\equal{}\beta\plus{}\alpha(a^{3}\plus{}2)$ so we conclude that $ \alpha\equal{}0$ then : $ f(x)\equal{}\minus{}x$.  [/hide]

I want to have others solutions to this problem.

thanx :)

[quote="reason"]hello !  :)
find all $ f: \mathbb{R}\to\mathbb{R}$ such that:

$ (\forall x\in{\mathbb{IR}})$ $ (fofofof)(x) \equal{} (fofof)(x) \plus{} 2x$

[hide="My solution"]this pb can be solved easily by sequences.
so: Let $ (x_{n})$ a sequence such that: $ x_{0} \equal{} x$ and $ f(x_{n}) \equal{} x_{n \plus{} 1}$ so we get:
$ x_{n \plus{} 4} \equal{} x_{n \plus{} 3} \plus{} 2x_{n}$,and by solving the equation: $ x^{4} \minus{} x^{3} \minus{} 2 \equal{} 0$ we get:
$ x \equal{} \minus{} 1$ or $ x \equal{} a\approx1,5436890126921$ then:
$ x_{n} \equal{} a^{n}\alpha \plus{} ( \minus{} 1)^{n}\beta$ with $ x \equal{} \alpha \plus{} \beta$.
$ f(x) \equal{} a\alpha \minus{} \beta$.
since $ (fofofof)(x) \equal{} (fofof)(x) \plus{} 2x$ we have: $ a^{4}\alpha \plus{} \beta \equal{} \beta \plus{} \alpha(a^{3} \plus{} 2)$ so we conclude that $ \alpha \equal{} 0$ then : $ f(x) \equal{} \minus{} x$.  [/hide]

I want to have others solutions to this problem.

thanx :)[/quote]

This solution is wrong.
Solution to $ x_{n \plus{} 4} \equal{} x_{n \plus{} 3} \plus{} 2x_{n}$ is not $ x_{n} \equal{} a^{n}\alpha \plus{} ( \minus{} 1)^{n}\beta$

You should include the two complex roots of the equation $ x^{4} \minus{} x^{3} \minus{} 2 \equal{} 0$


function
inequalities
algebra proposed
algebra

Does there exist a function  $ f\colon \mathbb{R}\to\mathbb{R}$ such that for all x the following inequality holds
$ f(x)\plus{}f(y)\plus{}f(z)\plus{}sin(x\plus{}y\plus{}z)<1$

_______________________________________
Azerbaijan Land of the Fire  :rotfl:

[quote="Pirkuliyev Rovsen"]Does there exist a function  $ f\colon \mathbb{R}\to\mathbb{R}$ such that for all x the following inequality holds
$ f(x) \plus{} f(y) \plus{} f(z) \plus{} sin(x \plus{} y \plus{} z) < 1$ [/quote]

Yes, it does : $ f(x)\equal{}\minus{}1$


Solve for x: :)
$ \sqrt x  \plus{} \sqrt[4]{{x(1 \minus{} x)}} \plus{} \sqrt[4]{{(1 \minus{} x)^3 }} \equal{} \sqrt {1 \minus{} x}  \plus{} \sqrt[4]{{x^3 }} \plus{} \sqrt[4]{{x^2 (1 \minus{} x)}}$ :wallbash:

hello, your equation has the roots $ x_1\equal{}0$, $ x_2 \approx 0.67719$ and $ x_3\equal{}1$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, your equation has the roots $ x_1 \equal{} 0$, $ x_2 \approx 0.67719$ and $ x_3 \equal{} 1$.
Sonnhard.[/quote]
Can you find  $ x_2 \approx 0.67719$ exactly :)

Are you sure the second summand on the right wasn't meant to be $ \sqrt[4]{x(1\minus{}x)^2}$?

[quote="JBL"]Are you sure the second summand on the right wasn't meant to be $ \sqrt [4]{x(1 \minus{} x)^2}$?[/quote]
Yeah, I'm also doubt about that but it's a problem in my book and it doesn't have the solution.However, I think the problem is right because it has some nice roots. :roll:

[quote="Dr Sonnhard Graubner"]hello, your equation has the roots $ x_1 \equal{} 0$, $ x_2 \approx 0.67719$ and $ x_3 \equal{} 1$.
Sonnhard.[/quote]
please post ur solution

[quote="JBL"]Sorry, by "right" I meant "left."  (In this case, the equation would be more symmetric between $ x$ and $ 1 \minus{} x$ and would have [i]all[/i] "nice" roots rather than one root that doesn't seem to have a better expression than the root of some high-degree polynomial.)  Books have been known to have typos.

murgi, I would think it self-evident that Sonhard went and plugged the equation into some numerical solver.[/quote]
Yes, That is what I'm doubt about.I think it should be symetric but in fact, it has 0 and 1 are the roots.Thus, I don't know that it's wrong or right? :)


geometry
circumcircle
power of a point
radical axis
geometry proposed

Let a triangle ABC  and its circumcircle $C_{k}$ .The interior and exterior bisector of $\widehat{BAC}$ intersect BC at  D,E respectively. Let a circle  $C_{1}$ of diameter DE which intersect $C_{k}$  at R which is different than A and AR intersects $BC$ at K. The tangent of $C_{k}$ at  A intersect  BC at  L. Now let a circle $C_{2}$ which passes through  K,L. Prove that when the  $C_{2}$ varies ,the radical axis of $C_{1}$ and $C_{2}$ passes through the midpoint $M$ of  BC.

I think that according to your problem, we have $A=K$.

[quote="Albanian Eagle"]maybe $K$ is the intersection of$AR$and $BC$?[/quote]

Yes you are right ,I made a typo. Now I corrected and the problem wait for you  :lol:

Note that the circle $C_{1}$ is the locus of point $X$ sch that $\frac{XB}{XC}= \frac{c}{b}$. Thus $\frac{KB}{KC}=-\frac{c^{2}}{b^{2}}$. On the other hand, since $\frac{LB}{LC}= \frac{c^{2}}{b^{2}}$, we have $(BCKL) =-1$.
Since $(BCDE) =-1$, it follows that $MB^{2}= MC^{2}= MD.ME = MK.ML$. And we are done.


[hide]i got 988 by guess and check[/hide]

[quote="Timothy"]What is the greatest three-digit number which is a multiple of 13?[/quote]

NOT using guess and check:

[hide]We know that 1300 is a multiple of 13 and so is 130.

Subtract 130 from 1300:

$1300-130=1170$

That must be a multiple of 13.

Subtract 130 from 1170:

$1170-130=1040$

We know that 1040 is a multiple of 13 now.

Now we keep on subtracting 13 now.

$1040-13=1027$

$1027-13=1014$

$1014-13=1001$

$1001-13=\boxed{988}$

[/hide]

[hide]
I used guess and check for this one.
76*13=988[/hide]


This problem was from a monthly problem set we had to do. I already solved it and sent in my solution, and I'm interested in seeing other solutions... mine seemed a bit longish, I think there may be a shorter way to do it. Anyway, here it is:

Prove that the inequality
$\frac{x+y}{x^2-xy+y^2} \le \frac{2\sqrt{2}}{\sqrt{x^2+y^2}}$
holds for all real numbers $x$ and $y$, not both equal to $0$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=14342 .

  darij


The four-digit number `25AB' is divisible by nine, with A the tens digit and B the units digit. How many different such four-digit numbers could `25AB' represent?

a+b must be congruent to 2 mod 9. For each value of b there is one value for a except if b is 2 when a can be 0 or 9. Therefore the answer is 10+1=11.


I'm wondering whether I should bother getting AoPS 1, or whether I should just simply get AoPS 2? (books)

Info about me:
Grade: 9
I can usually get 120 or more on AMC 12
I can usually get at least an 8 on AIME

NOTE: I am still looking for feedback!

Well, you have to ask yourself "Do I know most of what's covered in the Table of Contents?"

[url=http://www.artofproblemsolving.com/Books/AoPS_B_TOC.php?page_id=1]Table of Contents![/url]

Given that you can usually get 8 or more on the AIME, I can only guess that Volume 1 would probably mostly be a review.

Well, just by looking at the TOC of Volume 1, I seem to know pretty much all of the concepts. Looks like I should buy Volume 2.

I notice that it seems to coincide with AoPS courses (and I may take some of them).


linear algebra
matrix
calculus
calculus computations

Given a system of differential equation with two variables and constant coefficients, find a resolvent matrix R(t,t0).

Example of system of differential equation with two variables and constant coefficients:
y1(t)' = a*y1(t)+b*y2(t)
y2(t)' = c*y1(t)+d*y2(t)

Please help..I have no idea what is a resolvent matrix is..I could not find a satisfying result on google nor book..

When searching for mathematical definitions, try wikipedia. It doesn't seem to have its own page, but this seems to be the definition you were looking for:
[quote="Wikipedia"]The matrix function $ (A\minus{}sI)^{\minus{}1}$ is called the [b]resolvent matrix[/b] of the differential operator $ \frac{d}{dt}\minus{}A$.[/quote]
$ A$ here is $ \begin{bmatrix}a&b\\c&d\end{bmatrix}$.

I thought of giving the same answer but it is clearly incompatible with the notation $ R(t,t_0)$ :?

actually the equation is given in following form:

y1' = ay1
y1' = by2
y2' = cy1
y2' = dy2

will it make difference?

I don't get what t and t0 means in R(t,t0)...

Yes, it will make a difference. What you wrote now as the equations is a complete nonsense. So I start to suspect that $ R(t,t_0)$ belongs to the same category.

my mistake! it says following "two" systems..so one with

y1'=ay1
y2'=by2

and one with

y1'=cy2
y2'=dy1

well....what would be the resolvent matrix..? or am I interpreting the problem wrong..

I should have gone to class..


Find all the solutiones of the next equation in the positive integers:

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

$3^{x}=2^{x}y+1 \Rightarrow 2^{x}|3^{x}-1$. Of course $x=1$ gives a solution. If $x>1$ then $4|2^{x}|3^{x}-1$ and $x$ must be even since $3^{x}-1 \equiv (-1)^{x}-1 (mod\ 4)$. Let $x=2^{a}b$, where $b$ is odd. Then $3^{x}-1 = 3^{2^{a}b}-1 = ((3^{b})^{2}-1)((3^{b})^{2}+1)((3^{b})^{2^{2}}+1)((3^{b})^{2^{3}}+1)\ldots((3^{b})^{2^{a-1}}+1)$
For every $k\geq1 (3^{b})^{2^{k}}+1 \equiv 1+1=2 (mod\ 4)$, so $(3^{b})^{2^{k}}+1$ is divisible by exactly one 2 $\Rightarrow ((3^{b})^{2}+1)((3^{b})^{2^{2}}+1)\ldots((3^{b})^{2^{a-1}}+1)$ is divisible by exactly $a-1$ 2's. We also see that if $b$ is odd then $b=2c+1$ for some natural $c$, so $(3^{b})^{2}-1=3^{4c+2}-1=81^{c}\cdot9-1\equiv 1\cdot9-1 =8 (mod\ 16)$. That's why this term is divisible by exactly three 2's, and the whole product (equal $3^{x}-1$) is divisible by exactly $a+2$ 2's.

Now we have that if $2^{x}|3^{x}-1$, then $x\leq a+2 \Leftrightarrow 2^{a}b \leq a+2$. If $a=1$, then $2b\leq3 \Rightarrow b=1$, $x=2$; if $a=2$ then $4b\leq4 \Rightarrow b=1$, $x=4$. If $a\geq3$, then $2^{a}b\geq 2^{a}\geq a+2$, so we have no solution. Thus the only solutions can be $x=1,2,4$ and we easily check that these three numbers satisfy given condition; all the solutions to the original equation are $(x,y)\in\{(1,1),(2,2),(4,5)\}$

Thanks!!
Your solution is very good!!
This is a problem of the selectiv exam in Mexico!!

 :P


I practice whistle for at least half an hour every day . . . plus all my other istruments -- so around an hour. Every day, depending on whether it's a pipes day or a flute day . . . or an "other" day for the ones I just play around with and am not very good at. :P

On average, around 30 minutes. Yeah, I'm a slacker :P . But today, I practiced around two hours.

About 1 hour on piano and 30 to an hour on percussion.  I have a friend in college who does 6 to eight a day though.

[quote="miraculouspostmaster"]  I have a friend in college who does 6 to eight a day though.[/quote]

I am assuming he is a music major though... :D

About 35 minutes on piano, and 30 on viola when I remember to bring it home from school  :D

I probably should practice more

[quote="shobber"]my answer would be: $0$![/quote]Same here!

[quote="shobber"]my answer would be: $0$![/quote]

Well put! Me too! :D (although my violin teacher seems to think that that is the VERY minimum! :? )

My teacher isn't too strict about practicing...but my old teacher insisted on an hour a day at least.  he said that he used to practice 3-4 hours a day at my age...
At this point i practice about an hour a day, so i would be in between those two categories.  But i have orchestra at school 3 times a week for 45 minutes each time, so i said 1-2 hours a day.

I am now practicing 1-2 hours a day, due to the immense amount of material I have to go through.  I'm working 3 solo pieces, scales in 13 different rhythm/style/bowings, 4 etudes in 9 styles each, and whatever school and youth orchestra music I have.

$0!$ would be one wouldn't it? :D

I dont practice much.. whenever I feel like it.. which is bad I know. ;)

I practice around 2 hours a day on the violin.  Does anyone else play the violin.

I practice piano around 30 minutes a day. I play violin but I almost never practice.

[quote="smartazn"][quote="mathemagician1729"][quote="smartazn"]hm...I sometimes practice violin half-thinking about math problems (I outlined an entire proof in my head before I finished the first movement of my concerto :D)

anyone else do that? (<---seems like a bad habit, but most of the time practice is just mindless repetition, UNLESS you're prepping for a recital/competition)[/quote]
I don't know about you, but for most people that would be a dangerous habit. For a violin at least, real progress during practice comes from the mental labor of constantly finding your mistakes and attempting to fix them. Leopold Auer strongly believed that one should only practice with absolute concentration because otherwise the student is merely perfecting his or her own faults.[/quote]

Well, it's just during things such as scales and etudes that I kinda wander off, since I only need half of my concentration to catch intonation or articulation errors.

I "practice with absolute concentration" for pieces since there's just too much to watch for.[/quote]

I am the opposite: I wander on pieces and concentrate on scales and etudes. Sometimes I think I think more about other stuff more than the piece I'm playing! :rotfl:

I am supposed to practice 2-3 hours, but during the focused school year, it's more like 1-2 hours. In summer, I slack more.
 :noo:

45 min for piano...15 f9r my flute...if i remember it...lke 10 for guitar lol

I practice around an hour and a half a day on piano.  I wish I had more time though.  I could play piano all day.

I technically could if it was raining and I could play around and I had no good books and if I was in a pianistic mood :)
Oh, that a mans reach could exceed his grasp, or what's a heaven for? ~Robert Browning

I'm [i]supposed [/i]to practice cello three hours and piano one, but, normally, I just do an hour each. ;)

Trombone, 45 min and a little bit of Electric Base

:P an hour of piano every day.  I'm a slacker.

My violin practicing time is near $0$ in the summer (which is bad), but I practice more during the school year.  :laugh:

About half an hour to two hours of piano a day.

My vibrato is a little bit wonky now when I sing!

Hi, I voted here a long time ago but now I try to play piano for 2 (generally more because my technique sucks) hours every day that I can.

I also try to sing 30 minutes by myself at least 2 times per week, I was really bad with this but I have allstate soon so yeah...

I'm so bad at practicing...I really should be doing that right now.


please post the sources correctly, as described in the source menu.
for one change this is Jarnik 2004, problem 4, but maybe it has another source than that  :maybe: and you can find the solution here http://www.mathlinks.ro/viewtopic.php?t=4959


For any positive integer $n$ , let $s(n)$ denote the number of ordered pairs $(x,y)$ of positive integers for which $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$ . Determine the set of positive integers for which $s(n) = 5$

Let'a have a go for this one :

Equation is rewritten $n(x+y) = xy$ or $x = n + n^2/(y-n)$ with, wlog, $x \leq y$

For which it is deduced that $2n \leq y \leq n^2+n$ and $y-n \mid n^2$

The problem is then equivalent to finding all integers $n$ such as there are exactly 5 divisors of $n^2$ between $n$ and $n^2$.

It's not the case if n is prime or if n has more than 3 prime divisors or if n is divisible by a square.

Only solutions are $n = p1 * p2$ with p1 and p2 prime.

$s(n)$ is defined to be the number of [b]ordered[/b] such pairs, so your result is incorect. Considering this fact $n=p^2$ where $p$ is an arbitrary prime.

Rewrite the equation as $ (x\minus{}n)(y\minus{}n)\equal{}n^2$. So $ s(n)$ denote the number of divisors of $ n^2$. If $ s(n)\equal{}5$, then $ n^2$ is in form of $ p^4$, where $ p$ is a prime number. Thus, $ n$ is a square of prime.

Using the Technique from PSS,
we can say $x=n+p$ and $y=n+q$.
then, $\frac{1}{n+p} + \frac{1}{n+q} = \frac{1}{n}$
or $n^2=pq$
Thus, the number of order pairs is the same as the number of divisors of $n^2$, thus the answer is $n=s^2$ where $s$ is a prime number.

We have,
$1/x+1/y=1/n$
$x+y/xy=1/n$
$xy-nx-ny+n^2=n^2$
$ (x-n)(y-n)=n^2$
So,
$n=p^2$,where p is a prime number the above equation gets satisfied .
$n^2=p^4$
$p^4=1.p^4,p^4.1,p^2.p^2,p^3.p,p.p^3$
So, it has 5 ordered pairs$ (x,y)$ such that $s(n)=5$.

\[\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{n} \; \Rightarrow \; x+y=\dfrac{x y}{n}\]
Seeing the fact, both x and y are integers implies $x+y$ should be integer which implies $n \mid xy$. Now the problem becomes trivial, we just need to find a category of $n's$ for which 5 ordered pairs exist such that their product is divisible by $n$. So this means a number who have 5 divisors. So if $n=p_1^{q_1}p_2^{q_2}p_3^{q_3}...p_n^{q_n}$ then Number of divisors is given by$(\sigma_n(0)=(q_1+1)(q_2+1)...(q_n+1))$. Since 5 is a prime ($5 \cdot 1$), $5=(q_1+1)(q_2+1)$. This implies one of $q_i$ is 0 and other is $4$. Hence all n which are power of 4 of a prime satisfies the given condition.\\
$n \in \{16,81,625,...,p^4\}$ where p is a prime.

Multiply the equation by $xyn$ and apply SFFT to get $(x-n)(y-n)=n^2$. In order for $s(n)=5$, we must have that $n^2$ has $5$ factors. This is only possible if $n^2=p^4$, for some prime $p$. Hence, $s(n)=5$ when $n=p^2$ for a prime number $p$.

\begin{align*}
\dfrac{1}{x} + \dfrac{1}{y} &= \dfrac{1}{n}\\
xn+yn &= xy\\
0 &= xy - xn - yn\\
n^2 &= (x-n)(y-n)
\end{align*}

$s(n)=5$ indicate that $n^2$ has $5$ factors as each factor correspond to a unique $x$ and a corresponding $y$.
Thus we conclude $n=p^2$ for some prime $p$.