Yeah, 9 and -9
Because

9^2 is 81

and -9^2 is also 81.

[quote="iwolf"]:idea:
how many values are there for the square root of 81?[/quote]

[quote="Z-coli"]Note to Elio(n)-your solution wasn't correct in how you wrote it.  When you write the sqrt(81), you are implying that it is positive.[/quote]

The original problem implies that it's positive. I think a better worded question would be: how many solutions are there to the problem $ x^2 \minus{} 9 \equal{} 0$?

Z- coli was taliking about my latex . OK  :wink:

[quote="lotsofmath"]For a positive square root, there are always 2 possibilities, so the answer is [b]2[/b].[/quote]

 :wink: The square root of any number has two answers.

i've heard somewhere that when you find the square root of a certain number such as 81, you are saying that you want to find the positive answer, which would be 9 in this case. it's true that you can square negative 9 and get 81 too, but it's just that i think i've heard somewhere that when you say you are finding the square root of a certain number, you are finding the positive value.

$ sqrt{81}\equal{}9$ and $ \minus{}sqrt{81}\equal{}\minus{}9$

Put a back slash (\) in front of the Latex commands.

:wink: The square root of a number always has two values.  However, the [i]principal[/i] square root of a number is the postive, real one.

This is completely unrelated to the OP but to AstroPhys, it finally it me what the joke of your "location" is: it is an allusion to the Heisenberg Uncertainty Principle. :oops:

JRav , yes, it is an allusion to the Heisenberg Uncertainty Principle.  :lol:

[quote="iwolf"]:idea:
how many values are there for the square root of 81?[/quote]

There is only one value: $ \sqrt{81} \equal{} 9$.

You'll often see the solution to an algebraic equation like:

$ x^2 \equal{} 81$

$ \therefore x \equal{}\pm\sqrt{81} \equal{}\pm 9$.

In such an expression $ x$ could be positive or negative; you don't know which, but the square root itself is positive.

Here's another way to look at it:

$ x^2 \equal{} 81\\ \sqrt{x^2} \equal{}\sqrt{81}\\ \sqrt{x^2} \equal{} \left|x\right|\\ \sqrt{81} \equal{} 9$

$ \Rightarrow \left|x\right| \equal{} 9\\ x \equal{}\pm \left|x\right|\\ \therefore x \equal{}\pm 9$.


How much time do they give on each part of the state math contest?

I assume you are talking about the Comprehensive?

It is 90 minutes total, but I don't remember exactly how it is broken up (there may not be any break at all).

There is no break. And then there's the sprint round, which I hate. Every time I do one of those, I'm stuck doing the whole thing, which is about the most exhaustive thing you could imagine.

$Score = A*[number\;correct]+B*[number\;incorrect]$ where $A,B$ are constants, usually with $A > 0 \ge B$.

Usually also $A = 10$, $B =-4$ i think...

i feel your pain, jb.  those things can be hard, especially given all the potential for dumb errors.

The score on the state comprehensive is computed as:

4(number of multiple choice correct) +
1(number of multiple choice left blank) +
5(number of integer response correct)

Typically, the contest has 20 multiple choice and 10-15 integer response questions.  The maximum possible score varies depending on the number of questions.


Let a triangle ABC. The incircle touches BC,CA,AB at D,E,F. DA',EB',FC' are the bisector of triangle DEF (A' on EF, B' on FD, C' on DE). Denote M,N,P,Q,R,S be the mipoints of segments AA',BC,BB',CA,CC',AB. Prove that MN, PQ, RS are concurrent.

It is a direct application of the [b][size=100]Cevian Nests Theoerem[/size][/b], coming from the past (I met it first time, in a Hellenic publication of 1971)

Please see at: http://www.mathlinks.ro/Forum/viewtopic.php?t=6579

[b][size=100]EDIT[/size][/b]: Sorry dear all my friends, because of I posted my reply in a wrong topic. Unfortunately, I can't delete it.

It is recpect to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=228464]Concurrence with incircle[/url], also of [b][size=100]mr.danh[/size][/b].

Kostas Vittas.

[quote="jayme"]Dear Mathlinkers,
I saw this result in a book of "Döttl J., Neue merkwürdige Punkte des Dreiecks (1886)" n° 39.
Sincerely
Jean-Louis[/quote]
Really?  :(
It seems that I am not successful in finding out new problems.
Dear Jean-Louis, can you introduce about the proof in the book you mentioned? (I hope that it is nicer than my solution).
Thanks in advance.

Sorry, I'm waiting your proofs, dear Mathlinkers.


A number is called [i]capicua[/i] if when it is written in decimal notation, it can be read equal from left to right as from right to left; for example: $8, 23432, 6446$. Let $x_1<x_2<\cdots<x_i<x_{i+1},\cdots$ be the sequence of all capicua numbers. For each $i$ define $y_i=x_{i+1}-x_i$. How many distinct primes contains the set $\{y_1,y_2, \ldots\}$?

$0 < 1 < \cdot\cdot\cdot < 9 < 11 < \cdot\cdot\cdot < 99 < \cdot\cdot\cdot n$.
From this the possible values of $y$ are $1, 2, 11$.
If $x_{i}$ has $n$ digits($n \geq 3$), then $x_{i+1}$ ends with the same digit as $x_{i}$, i.e it's divisible by 10, or not. In this case the two consecutive [i]capicua[/i] can be of the form $x_{i} = a9...9a$, $x_{i+1} = b0...0b$ with $b = a + 1$ and $y_{i} = 11$. If $x_{i+1}$ has one more digit than $x_{i}$, they are of the form: $x_{i} = 9a...a9$, $x_{i+1} = 1b...b1$ with $y_{i} = 2$. Hence the set contains two distinct primes.

Given a capicua number, its easy to see that the most frequent way to find the next one is by changing the middle slot or slots by 1, for example if you had $12321$ the next capicua number is $12421$, if you had $257752$ then the next one would be $258852$ so on and so forth. Now, if it is of the form $99....99$ then the next one will be $100...01$ or if the middle spot cant go any further such is the case for $48799784$ for example, then you would have to change the slots closest to the middle spots and start over, in this particular case $48800884$, when you do that, the difference of those capicua numbers is just $11\cdot 10^2$. This process would repeat again for the 2nd pair of slots you changed, and they could eventually hit $9$ in which case if $x=$number of slots "untouched" to the right, you would have to add $11\cdot 10^x$ or $1\cdot 10^x$ if there is only 1 middle slot (i.e. if the capicua has an odd number of digits),this is just because we are repeating the case of the middle slots but we have to consider the "unused" decimal places  of the number. Hence we have the only primes are $2$, $11$, since if any difference ends in $0$ it cant be a prime.


symmetry
geometry
geometric transformation
rotation
function
reflection
integration

[size=134][b]Problem 1[/b] (8 points)[/size]
A meson is composed of two quarks with rather complicated interaction between them. To learn the physics inside the meson high energy electrons are usually employed to collide inelastically with the mesons. The process is very difficult to interpret, and to grasp the essential elements of it the Parton model was developed. The model assumes that the incoming electron collides elastically with only part of the meson, i. e., with only one quark. The quark then passes its energy to its parent meson. The essence of the model can be illustrated with the following drastically simplified model. The electron of mass $m_{1}$ and kinetic energy $E$ collides elastically with a quark of mass $m_{2}$ in a meson. The other quark has mass $m_{3}$. The two quarks are connected by a weightless spring. Before the collision the two quarks are at rest and the spring is at its natural length. All motions are along one dimension only, and ignore all relativistic effects. Find the internal energy of the meson as represented by the vibration energy of the spring, and the kinetic energy of the meson as a whole after the collision.

[size=134][b]Problem 2[/b] (8 points)[/size]
A uniform iron pulley of total mass $M_{1}$ and radius $R$ is fixed above the ground. It can spin freely about its center. Additional weights in the form of many small magnets with total mass $M_{2}$ can be attached to the pulley. A long thread is winding around its edge, and at its end hangs a weight of mass $M_{3}$ at height $H$ from the ground that can be released suddenly. Assume that there is no slip between the thread and the pulley.

(a) How should the magnets be placed with circular symmetry on the pulley such that the final speed of $M_{3}$ is minimum when it reaches the ground?

(b) Find the minimum speed.

[size=134][b]Problem 3[/b] (10 points)[/size]
A cylindrical vessel of radius $R$ is filled with water (density $\rho$) up to height $H$.

(a) Calculate the force of water exerting on a narrow vertical strip of width $d\, (\ll R<H)$ of the cylinder wall. (3 points)

(b) The vessel is then placed on a platform spinning at angular speed $\omega\, (<\sqrt{gh}/R)$. The rotation axis coincides with the central axis of the vessel. The water spins at the same angular speed as the platform. Find the shape of the water surface, and the extra force of water exerting on the strip due to spinning. ($g$ is the gravity acceleration.) (7 points)

[size=134][b]Problem 4[/b] (12 points)[/size]
A cylinder of fixed volume $V$ is partitioned into left and right compartments separated by a movable piston of negligible mass. The left and right compartments are filled with $n_{\text L}$ and $n_{\text R}$ moles of the same ideal gas, respectively, at initial temperatures $T_{\text L}$ and $T_{\text R}$.

(a) What are the volumes of the left and right compartments $V_{\text L}$ and $V_{\text R}$ initially when the system is at equilibrium? (2 points)

(b) Suppose heat can be exchanged between the two compartments but there is no heat exchange between the cylinder and outside world. The amount of heat flow per unit time $\frac{\Delta Q_{\text L\to \text R}}{\Delta t}$ from left to right compartment is proportional to the temperature difference between the two compartments, i.e. $\frac{\Delta Q_{\text L\to \text R}}{\Delta t}=k(T_{\text L}-T_{\text R})$. Find the temperature difference, $T_{\text L}-T_{\text R}$ as a function of time. (8 points)

(c) Find out how $V_{\text L}$ and $V_{\text R}$ changes with time. (2 points)

[size=134][b]Problem 5[/b] (12 points)[/size]
A planar boundary at $y = 0$ separates two non-magnetic media with refractive indices $n_{1}$ (real) and $n_{2}$ (can be complex). An electromagnetic wave $\vec{\mathbf E}_{\text I}=E_{0}\vec{\mathbf x}_{0}\exp\{i(k_{1}y-\omega t)\}$ is incident onto the boundary, and its reflected wave and transmitted wave are given by $\vec{\mathbf E}_{\text R}= E_{\text r}\vec{\mathbf x}_{0}\exp\{i(-k_{1}y-\omega t)\}$ and $\vec{\mathbf E}_{\text T}= E_{\text t}\vec{\mathbf x}_{0}\exp \{i(k_{2}y-\omega t)\}$, respectively.

(a) Use the boundary conditions of electric and magnetic fields to find $r=E_{\text r}/ E_{0}$ and $tr=E_{\text t}/E_{0}$ in terms of $n_{1}$ and $n_{2}$. (8 points)

(b) For $n_{1}= 1, n_{2}= 2.0i$, where $i=\sqrt{-1}$, find the reflection $R=|r|^{2}$. (1 points)
(c) For $n_{1}= 1, n_{2}= 1.0+2.0i$, find the phase shift of the reflected wave relative to the incidence wave. (3 points)

Hey, this is a nice collection of problems, I didn't know that this Olympiad existed... Who organizes it? Anyway:

[b]1.[/b] There are two parts of the solution: solving the electron-quark collision and solving the oscillating system. Conservation of energy and momentum gives:
\[E = E_{e}+E_{q}\quad \text{and}\quad p = p_{q}-p_{e}.\]
Indices $e$ and $q$ represent the electron and the quark immediately after the collision. The first equation gives
\[m_{1}v^{2}= m_{1}v_{e}^{2}+m_{2}v_{q}^{2}\Rightarrow m_{1}(v-v_{e})(v+v_{e}) = \frac{p_{q}^{2}}{m_{2}}.\]
From the 2nd equation we get
\[(v-v_{e}) p_{q}= \frac{p_{q}^{2}}{m_{2}}.\]

One solution is $p_{q}= 0$, which is the situation immediately before the collision. Supposing $p_{q}\neq 0$ we get
\[v_{e}= v-\frac{p_{q}}{m_{2}}\Rightarrow p-p_{e}= \frac{m_{1}}{m_{2}}p_{q}.\]
Now we can get
\[p_{q}= \frac{2pm_{2}}{m_{1}+m_{2}}= \frac{2m_{2}\sqrt{2m_{1}E}}{m_{1}+m_{2}},\]
\[E_{q}= \frac{4m_{1}m_{2}E}{(m_{1}+m_{2})^{2}},\]
and
\[v_{q}= \frac{2\sqrt{2m_{1}E}}{m_{1}+m_{2}}.\]

The momentum of the entire meson after the collision is $p_{q}$, and the speed of its center of mass is

\[v_{c}= \frac{p_{q}}{m_{2}+m_{3}}= \frac{2m_{2}\sqrt{2m_{1}E}}{(m_{1}+m_{2})(m_{2}+m_{3})}.\]

The velocity of the first quark in the center-of-mass system is the initial speed of the oscillations, and the energy of these oscillations is the kinetic energy of that quark in the C-system:

\[U = \frac{m_{2}(v_{q}-v_{c})^{2}}{2}= \frac{4m_{1}m_{2}m_{3}^{2}E}{(m_{1}+m_{2})^{2}(m_{2}+m_{3})^{2}}.\]

The total kinetic energy of the meson is

\[T = E_{q}-U = \frac{m_{2}(v_{q}-v_{c})^{2}}{2}= \frac{4m_{1}m_{2}^{2}(m_{2}+2m_{3}) E}{(m_{1}+m_{2})^{2}(m_{2}+m_{3})^{2}}.\]

[b]2.[/b] Let the speed of the weight when it hits the ground be $v$. Conservation of energy gives
\[M_{3}gH = \frac{M_{3}v^{2}}{2}+\frac{I\omega^{2}}{2},\]
where $I$ is the moment of inertia of the pulley and $\omega$ is the angular velocity of the pulley, given by the non-slipping condition as $\omega = v/R$. The velocity $v$ is now given by
\[v^{2}= \frac{2M_{3}gH}{M_{3}+I/R^{2}}.\]

Clearly, the speed is minimized when the moment of inertia of the pulley is maximized, ie. when the magnets are positioned on the rim of the pulley. Then we would have
\[I = \frac{M_{1}R^{2}}{2}+M_{2}R^{2},\]
which results in the final speed of
\[v = \sqrt{\frac{4M_{3}gH}{M_{1}+2M_{2}+2M_{3}}}.\]

[b]3.[/b] When the water isn't moving, the force on the strip of wall is
\[F = \int_{0}^{H}\rho gd h\text dh = \frac{\rho gdH^{2}}{2}. \]
Now, let $y = f(x)$ denote the 'profile' of the spinning water, with the $y$-axis coinciding with the central axis of the system and with $y = 0$ at the bottom of the cylinder. Considering the forces an elementary water the surface, we can obtain
\[\omega^{2}x \text dm\cos\phi = g\text dm\sin\phi, \]
which comes from equating the components of the centrifugal and the gravitational force. Here $\phi$ is the slope of the curve $y(x)$, and we can see that
\[\tan\phi = \frac{\omega^{2}}{g}x. \]
Since $\text dy/\text dx \equiv \tan\phi$, by separating variables and integrating we get
\[y = \frac{\omega^{2}}{2g}x^{2}+C, \]
where $C$ is a constant of integration. As we can see, the profile of the water is parabolic, and the surface of the water is a rotational paraboloid.

The force acting on the strip of the cylinder wall is given with the same equation as before, but the limit of integration is now $y(R)$ rather than $H$. We can find $C$ (and thus $y(R)$) by equating the volumes of water. The volume of the rotating water is given by
\[V = \int_{R^{2}\pi}y\text dS = \int_{0}^{R}\left( \frac{\omega^{2}}{2g}x^{2}+C\right) 2x\pi\text dx = \frac{\omega^{2}\pi R^{4}}{4g}+\pi C R^{2}. \]
We also have that $V = R^{2}\pi H$, so we can find that $C = H-\frac{\omega^{2}R^{2}}{4g}$.

[According to the given boundary for $\omega$, the $4g$ in the denominator is a mistake... I'm not sure whether I've made a mistake I keep overlooking, or there was a lapsus calami in the problem.]

Now we have that
\[F' = \int_{0}^{y(R)}\rho gd h\text dh = \frac{\rho gd}{2}\left( H+\frac{\omega^{2}R^{2}}{4g}\right)^{2}. \]
This is a pain to compare to $F$ unless $\omega$ is small enough, so I'll leave it like this.

Ok, since nobody else posted anything I looked at the solutions. It appears that I've correctly solved 2. and 3. and made a mistake somewhere in the first problem, although the idea is ok. However, the expression for the speed of the quark after the collision I found in the solution can't be correct, and I can't find any mistake in my equations, so there may be hope yet that my solution is correct (yeah, right). Anyway, there is a mistake in the official solution  :roll:

Good job, Djole. :thumbup:

[quote="Djole"]... Anyway, there is a mistake in the official solution  :roll:[/quote]
Do you mean the official solution the following document?
[url]http://hkpho.phys.ust.hk/Resources/Pan%20Pearl%20River%20Delta%20Physics%20Olympiad%202006%20Answers.doc[/url]
Before posting the questions, I just had the problems set but the answers. (I hadn't searched for it.)

Yes, that's the one. Their expression for $v_{q}$ can't be right  :maybe:

I think so. But my friend got the same final answer as official. (Actually, I'm not good at physics and english, and so I can't express his idea to you.)

Ok, I'll do the calculations again, if I reach a correct solution I'll change my post.

[quote="Djole"]
The velocity of the first quark in the center-of-mass system is the initial speed of the oscillations, and the energy of these oscillations is the kinetic energy of that quark in the C-system:
\[U = \frac{m_{2}(v_{q}-v_{c})^{2}}{2}= \frac{4m_{1}m_{2}m_{3}^{2}E}{(m_{1}+m_{2})^{2}(m_{2}+m_{3})^{2}}. \]
The total kinetic energy of the meson is
\[T = E_{q}-U = \frac{m_{2}(v_{q}-v_{c})^{2}}{2}= \frac{4m_{1}m_{2}^{2}(m_{2}+2m_{3}) E}{(m_{1}+m_{2})^{2}(m_{2}+m_{3})^{2}}. \]
[/quote]
I think the kinetic energy of the meson is the same as the kinetic energy of the center-of-mass of the two quarks. Hence the kinetic energy of the meson is
\[E_{k}=\frac12(m_{2}+m_{3})v_{c}^{2}=\boxed{\frac{4m_{1}m_{2}^{2}}{(m_{1}+m_{2})^{2}(m_{2}+m_{3})}E}\]
Moreover, the kinetic energy of the first quark, i.e. the energy of the meson, is the sum of the kinetic energy and the internal energy of the meson. Therefore, the internal energy of the meson is
\[E_{i}=E_{q}-E_{k}=\boxed{\frac{4m_{1}m_{2}m_{3}}{(m_{1}+m_{2})^{2}(m_{2}+m_{3})}E}\]

That sounds ok... I can't figure out where I went wrong in my solution, it was ok until the final few equations  :huh: But your solution looks better than mine (good work!) so I must've made a mistake I can't find now.


[quote="KawaiiZia"]Okay, how much is 49.5 Ounces of water in glasses of water?
x_X;[/quote]

well first you have to define what how much a "glass" is.

[hide="solution"]

$\frac{49.5}{x}$ where $x$ how much a glass is in ounces

[/hide]

49.5 Ounces of water in glasses of water?

to explain better, heres an example

Lets say that one glass is $5.5$ ounces.
then how many glasses is $49.5$?

it would basically be $\frac{49.5}{5.5}$ which equals to $9$

[hide]given a normal glass that holds 8 ounces, we take $\frac{49.5}{x}= \frac{49.5}{8}= 6.18$ but because we are looking at whole glasses, we have to round up and we get $7$ glasses of water[/hide]


Paul Dreyer sent out an email to coaches with a form on which he solicited feedback. By all means, mail that form to him - but I thought I'd start a topic here for people to talk about it.

I left the form itself on my other computer (someone who has it could post it here), but to paraphrase a few of the questions:

On a round by round basis: was this round too hard, too easy, or about right?

Do you like the relays of mixed sizes or would you rather have all relay teams one standard size?

Was early April a good time of year for it?

Would you do something like this in the fall?

Would you do it again next year? Would you do it again next year if there were an entry fee (maybe $ \$30 \minus{}\$50$ per school, with maybe a discount for hosting multiple schools)?

May I ask what this is? I have never heard of it before these past few threads.

[quote="Kent Merryfield"]On a round by round basis: was this round too hard, too easy, or about right?[/quote]
Team: Okay.
Theme: Too easy.
Individual: Okay.
Relay: Seemed harder than standard ARML's relays, though I haven't looked at many of them. Our team really screwed this part up, but apparently a lot do in general. In conclusion I don't think I can judge the difficulty well...

[quote="Kent Merryfield"]Do you like the relays of mixed sizes or would you rather have all relay teams one standard size?[/quote]
I would prefer it not to be so complicated for teams to figure out how to divide people and order them. I guess one standard size is better then.

[quote="Kent Merryfield"]Was early April a good time of year for it?[/quote]
I'm not sure.

[quote="Kent Merryfield"]Would you do something like this in the fall? Would you do it again next year? Would you do it again next year if there were an entry fee (maybe $ \$ 30 \minus{}\$ 50$ per school, with maybe a discount for hosting multiple schools)?[/quote]
First two: Probably, if I could still do high school contests next year. :P I'm not the one that can answer the last one. And way to get lucky with the accidental LaTeX tags.

To the above poster: ARML Local is a new preliminary competition they started this year. Basically an easier version of ARML with 6 person teams as a warmup for the real ARML. For example, the Theme Round (Power Round equivalent) was answer based instead of proof based.

In general, I enjoyed the contest; thank you to the ARML people for running it.

I think that perhaps the theme round had material that was too well-known to students who prepare for olympiads.  Euler's theorem is a part of everybody's olympiad preparation.  To many of us, it was fairly easy application of standard knowledge.  But very few students learn anything close to this material in school, and I think perhaps that teams consisting of such students were probably hurt disproportionately.  Of course, for any topic, somebody will have background knowledge.  But I think that Euler's theorem was too universally studied by experienced students and too universally unknown to the others.  I did enjoy the fact, however, that students who tried to apply routine substitution code-breaking procedures might meet with some success, but slowly, and not as much as students who found the encoding equation.

Other than that, I had no complaints.  Perhaps one more challenging individual question would have been in order, but that's all.  I liked the different-sized relay teams; they added variety and I think kept people more on their toes in general.  I think April was a good time of year; I would participate in the fall if I could, but that is a much busier time of year for my school and me, and we would have more trouble getting a team together.  An entry fee would pose no obstacle to my school.

I also enjoyed it. The difficulty was just right. However, the 6-person relay was definitely larger than we've had before, and although it wasn't hard, it's frustrating that a single mistake can be worth a significant six points, especially if I made that single mistake and my team was three points away from second place.

Here's the actual survey form:

[b][size=150]2008 ARML Local Competition [/size][/b]

[b]Thanks and Post-Contest Survey[/b]

To all coaches:

Thank you so much for your time and effort in helping put on this inaugural ARML Local contest.  I hope that your students found the questions challenging but not impossible, and that they enjoyed competing.  One final request: please fill out the survey below and mail it, along with a copy of all of the team score sheets, to [Paul Dreyer]. Thanks again.

[b]Question 1:[/b]  For each round, please rate the questions from 1 (too easy) to 5 (too hard).  Feel free to make any additional comments about the questions below:

Team: _____  Theme: _____ Individual: _____ Relays: _____
[b]Comments: [/b]

[b]Question 2:[/b] For the relay rounds, would you prefer to keep the mixed team size format or use the traditional ARML format of three per team?
[b]Mixed / Constant[/b]
[b]Comments:[/b]

[b]Question 3:[/b] Concerning the timing and frequency of the event, is early April a good time for the contest (as an ARML warm-up, for example)?  If there were a second contest in the fall, would you compete in that as well?
[b]Comments: [/b]

[b]Question 4:[/b] Were the proctoring, relay, and scoring instructions clear?
[b]Yes /  No
Comments: [/b]

[b]Question 5:[/b] Our Bernoulli Division was very small this year (3 teams).  There are several options, either to have a single division, expand eligibility for the Bernoulli Division (currently, non-selective schools with enrollments of fewer than 250/year), or have teams self-select (with a promotion system for winners of the lower division).  How do you think this should be handled?
[b]Comments: [/b]

[b]Question 6:[/b] Next year, we plan to hold ARML Local again, but will charge a fee of $ \$30\minus{}\$50$/team (possibly with a discount for teams who host multiple schools).  Do you think your school will compete?  [b]Yes / No[/b]
[b]Comments: [/b]

[b]Overall Comments:[/b]

Regarding to #3: The early April timing did not work for us - it fell into Spring Break (I was out of town), and that Saturday also overlaps with State Science Olympiad (near Spokane) at which I did see ARML students.  So we did not take ARML Local.
The new relay variation seems to be fun, but might be too much work to just configure team alignment for each round.


geometry
rectangle
perimeter
ceiling function
floor function
combinatorics proposed
combinatorics

Let $ F_n$ be a set of all possible connected figures, that consist of $ n$ unit cells. For each element $ f_n$ of this set, let $ S(f_n)$ be the area of that minimal rectangle that covers $ f_n$ and each side of the rectangle is parallel to the corresponding side of the cell. Find $ max(S(f_n))$,where $ f_n\in F_n$?

Remark: Two cells are called connected if they have a common edge.

It is easy to see that the perimeter of the minimal rectangle is $ P\le 2n\plus{}2$. Hence, the area is $ A\le\left(\frac{P}4\right)^2\le\frac{(n\plus{}1)^2}4$, the maximum of $ A$ occurs when the dimension is $ \frac{n\plus{}1}2\times\frac{n\plus{}1}2$. If $ n$ is even, the maximum occurs when the dimension is $ \frac{n}2\times\frac{n\plus{}2}2$. Hence $ \max(S(f_n))\equal{}\left\lceil\frac{n\plus{}1}2\right\rceil\plus{}\left\lfloor\frac{n\plus{}1}2\right\rfloor$.

Without the proof of this fact, $ P\leq 2n\plus{}2$, only 2 points were awarded.

[quote="Erken"]Without the proof of this fact, $ P\leq 2n \plus{} 2$, only 2 points were awarded.[/quote]I thought it's trivial, it turned out that this fact is not so obvious. :P  How's this proof:

Call a figure "almost connected" iff each pair of cells have a vertex in common. It's obvious that the maximum perimeter of the minimal covering the figure is $ 4n$. To make the figure connected, we need to shift the $ n\minus{}1$ cells, then shift the $ n\minus{}2$ cells, and so on. Each time we shift them, the perimeter reduces by at least $ 2$. We shift them $ n\minus{}1$ times, hence the perimeter is reduces by at least $ 2n\minus{}2$, and hence the perimeter of the connected figure is at most $ 4n\minus{}(2n\minus{}2)\equal{}2n\plus{}2$.

Besides that, you need to prove that any figure can be obtained by "shifting".


find  :D

$ \left\{ \left(\frac{2}{1}\right)^\pi \; \frac{1}{1\plus{}\pi} \right\} \cdot \left\{ \left(\frac{3}{2}\right)^\pi \; \frac{2}{2\plus{}\pi} \right\} \cdot \left\{ \left(\frac{4}{3}\right)^\pi \; \frac{3}{3\plus{}\pi} \right\} \cdot \left\{ \left(\frac{5}{4}\right)^\pi \; \frac{4}{4\plus{}\pi} \right\} \cdot \left\{ \left(\frac{6}{5}\right)^\pi \; \frac{5}{5\plus{}\pi} \right\} \cdot \quad \ldots\quad \ldots$

hello, i have no prove now but i think
$ \prod_{n\equal{}1}^\infty\left(\frac{n\plus{}1}{n}\right)^\pi\left(\frac{n}{n\plus{}\pi}\right)\equal{}(\pi\minus{}3)(\pi\minus{}2)(\pi\minus{}1)\pi\cdot\Gamma\left(\pi\minus{}3\right)$.
Sonnhard.

Direct application of the famous formula

$ \Gamma(z) \equal{} \lim_{n\to\infty} \frac{n^z}{z (1 \plus{} z) \cdots (1 \plus{} \frac{z}{n})}$

will immediately yield the answer $ \Gamma(\pi \plus{} 1)$.


Since Astronomy is most related to Physics, I consider this might be the best place for this post.
How many of you are interested in [b]Stargazing[/b]? How serious are you? How many [b]hours do you spend [/b]on an average per week? What resources/instruments do you use? Any interesting observations ? Do you maintain a notebook? Do you take pictures??!! What else do you do? What does one have to do / have to be a [b]serious astronomer[/b]? Any software/hardware you'd recommend??

Let's discuss everything related to the wonderful activity you can [b]spend hours over and feel you just passed a minute[/b]!!!

My 'Stats' : -

Average 2 hours per week (some weeks no stargazing!!!  :o )
Serious since 2 years
Orion AltAzimuth Dobsonian 4.5 '' Reflector with Solar Filter, lenses from 4mm to 32mm , filters, etc.
Don't maintain a notebook yet
Don't have picture taking equipment
Want to be [b]Really Serious[/b] - dunno how?? Help!!!  :help:

Star gazing?

I get to like it only last year.

Too bad my house is in a city where light pollution is so heavy that you can't even see Libras here

Anyway, i just like it


calculus
integration
function
trigonometry
complex analysis
complex analysis unsolved

Hi guys --- does anyone know how to represent the elliptic integral \[ K(x) \equal{} \int_0^{\frac \pi 2} \frac {\,d\varphi} {\sqrt{1 \minus{} x^2\sin^2\varphi}}\] as a contour integral of an analytic function?  (Here $ x$ is only a parameter.)

We could just substitute $ z\equal{}e^{i\varphi}, d\varphi\equal{}\frac{\minus{}i}{z}\,dz$:
$ K(x)\equal{}\int_C \frac{\minus{}i}{z\sqrt{1\minus{}x^2\left(\frac1{2i}(z\minus{}\frac1z)\right)^2}}\,dz \equal{}\minus{}i\int_C \frac1{z\sqrt{1\plus{}\frac{x^2}4(z^2\minus{}2\plus{}\frac1{z^2})}}\,dz$
$ \equal{}\minus{}i\int_C \frac2{xz\sqrt{z^2\minus{}2\plus{}\frac4{x^2}\plus{}\frac1{z^2}}}\,dz\equal{} \frac{\minus{}2i}{x}\int_C \frac1{\sqrt{z^4\plus{}\left(\frac4{x^2}\minus{}2\right)z^2\plus{}1}}\,dz$, where $ C$ is a quarter-circle and we choose the branch of the square root which is a positive multiple of $ z$ on the circle.

If you multiply by four and make it an integral around the whole circle, there are branch cuts both inside and outside, so no version of the residue theorem can apply.


geometry
geometric transformation
reflection

Has anyone else noticed that in English class you rarely learn about grammar and word choice? All I have ever done in English is read and analyze. We have never opened a grammar book to learn about the proper time to use "who" or "whom", "they" or ", which", etc.

Now I think that reading and analyzing is important, but I think grammar and word choice should really be taught more in English classes. Deprived of proper grammar instruction, I find it difficult to do well on some of the SAT English questions (improving sentences, etc.).

Who else thinks so?

Actually....[soapbox] the so-called 'grammar' rules you cite, such as the use of who/whom, singular 'they,' and which/that, bear no relation to the way the English language has ever been used, past or present.  They only perpetuate themselves through prescriptive grammar books, overzealous copy-editors, and standardized tests.  Great authors don't observe them, and neither does the English language as people actually use it.  (See, for instance, any of the many Language Log posts on the topic, [i]Merriam-Webster's Concise Dictionary of English Usage[/i], [i]A Student's Introduction to English Grammar[/i], the OED, or [i]The Cambridge Grammar of the English Language[/i]) [/soapbox]

I wish that grammar would be taught better, but in a way that reflects real English usage and that is more directed towards improving the style and clarity of writing.

It depends on your school and your teacher, I guess. I got enough grammar in high school. If you're so worried about the SAT's, though, you could always pick up a grammar book and look out for instances of grammar rules and errors in real life and you'd do fine. I would suggest [u]A Writer's Reference[/u] by Diana Hacker.

[quote="12markkram34"]Improving sentences skills are necessary for doing well on SAT.[/quote]If the only reason you're taught something is to do well on the SAT, then frankly, the test is flawed. The test should not be on completely pointless stuff that [b]nobody[/b] ever uses

Well, unfortunately, whether or not we take the test is not our decision. Flawed or not flawed, students are forced to study the grammar of standardized tests if they wish to "succeed".

You can peer-review essays with a friend for practice on improving sentences. It will also give you ideas to improve your own writing.

There is nothing to learn in theory about sentence completion. Ask the following four questions (in this order), stopping to fix any problems you might run into and starting over:

1. Is the sentence saying what it was meant to say?
2. Is that message clear?
3. Is the sentence appropriately placed/used in the larger context of the passage?
4. Is the sentence grammatically correct?

[quote="12markkram34"]I never said that. Those skills [b]are[/b] useful in real life. However, the most urgent use is usually for success on the SAT. I say that they aren't teaching it enough in school. It's not pointless and I didn't say that.[/quote]Take the first 3 lines from the previous poster:
1. Is the sentence saying what it was meant to say?
2. Is that message clear?
3. Is the sentence appropriately placed/used in the larger context of the passage?

Now instead switch the 4th to
4. Is the sentence grammatically correct to [b]an overwhelming majority[/b] of English speaking people?
If that's correct, then there is very little reason to cover it in our schools.

[quote="PKLove"]The purpose of grammar is to make the thoughts you're saying clear. Being overzealous about grammar rules can be just as harmful as not using grammar rules at all.[/quote]QFT

The message contains only quote and code tags. Please add some text of your own.


suppose $P(x)$with  integer coefficients such that the polynomial.suchthat $P(a)P(b)=-(a-b)^2$ for some distinct a,b.prove that $P(a)+P(b)=0$ :lol:

Let P(a)=c (mod a-b). Then P(b)=c (mod a-b), because (a-b)|(P(a)-P(b)). Let p prime divisor of (a-b) and $ord_p(a-b)=k>0$. If ord_p(c)<k, then ord_p(P(a)P(b))<2k it give conradition. Therefore $c=0, P(a)=\pm (a-b),P(b)=\mp (a-b).$ It give P(a)+P(b)=0.


Triangle ABC is isosceles with vertex angle $ B$ and legs $ \overline{BC}$ and $ \overline{AB}$, as shown. What is the perimeter of triangle $ ABC$?

[asy]draw((-10,0)--(10,0)--(0,7)--cycle);
label("$A$",(-10,0),SW);
label("$C$",(10,0),SE);
label("$B$",(0,7),N);
label("8cm",(0,0),S);
label("5cm",(5,3.5),NE);[/asy]

Since the triangle is isoceles and leg BC=AB, leg AB is 5.
5+5+8=18 cm.

Since $ AB$ and $ BC$ are congruent, you have a $ 5$, $ 5$, $ 8$ triangle, with perimeter $ \boxed{18}$.


function
real analysis
real analysis unsolved

Let $ f,g : [\minus{}1,1]\minus{}>[0,2]$ continuous functions,$ g$ is surjective(the image of $ g$ is $ [0,2]$).
Prove that the equation $ f(x^2\minus{}1)\equal{}g(x)$ has at least one real root in $ [\minus{}1,1]$.

This is simple( maybe I'm wrong )
$ t(x)\equal{}f(x^2 \minus{} 1) \minus{} g(x)$
$ g$ surjective ,so $ g(a)\equal{}2$ and $ g(b)\equal{}0$
$ t(a)\equal{}f(a^2 \minus{} 1) \minus{} g(a)\equal{}f(a^2 \minus{} 1) \minus{} 2 \le 0$
$ t(b)\equal{}f(b^2 \minus{} 1) \minus{} g(b)\equal{}f(a^2 \minus{} 1) \minus{} 0 \geq 0$


3 times 5 is 15, and $ \frac {4}{6} \equal{} \frac {2}{3}$. We add that to 15 to get $ \frac {47}{3}$, which, when added to $ \frac {8}{3}$, yields $ \frac {55}{3}$.

EDIT: Reading phail. Thanks guys.

I don't think thats right.. Correct me If I am wrong but Izzy did $ {15\plus{}4/6\plus{}8/3}$ while the question is $ {15\minus{}4/6\plus{}8/3}$
THe answer is 17.

[quote="GameBot"]Simplify: $ 3(5) \minus{} \frac46 \plus{} \frac83$.[/quote]
$ \frac46\equal{}\frac23$ so

$ 15\minus{}\frac23\plus{}\frac83$

$ 15\plus{}\frac63$

$ 15\plus{}2\equal{}17$

So 17 is correct, KingJamesMvP. :)


Which terms must be removed from the sum
\[ \frac12 \plus{} \frac14 \plus{} \frac16 \plus{} \frac18 \plus{} \frac1{10} \plus{} \frac1{12}
\]if the sum of the remaining terms is equal to $ 1$?

$ \textbf{(A)}\ \frac14\text{ and }\frac18 \qquad \textbf{(B)}\ \frac14\text{ and }\frac1{12} \qquad \textbf{(C)}\ \frac18\text{ and }\frac1{12} \qquad \textbf{(D)}\ \frac16\text{ and }\frac1{10} \qquad \textbf{(E)}\ \frac18\text{ and }\frac1{10}$

$ \frac12 \plus{} \frac14 \plus{} \frac16 \plus{} \frac18 \plus{} \frac1{10} \plus{} \frac1{12}\equal{}\frac{60\plus{}30\plus{}20\plus{}15\plus{}12\plus{}10}{120}$. Clearly, the $ 15$ and the $ 12$ must be removed from the numerator for the numberator to sum to $ 120$. So the answer is $ \boxed{(E)}$


Find the locus of the orthocentres of all triangles inscribed in a fixed circle

The distance $OH$ between the circumcenter $O$ and the orthocenter $H$ is given by

$OH^2 = 9R^2 - (a^2 + b^2 + c^2)$

$OH = 3R \sqrt{1 - \frac{a^2 + b^2 + c^2}{9R^2}}$

where $R$ is the circumradius and $a, b, c$ the sides of the triangle. Since the sides $a, b, c$ can become arbitrarily small as the vertices $A, B, C$ move on the given circle, the locus of all possible orthocenters is a disk with radius $3R$ and center $O$.


superior algebra
superior algebra solved

can anyone help with:

how do you prove an n-cycle has order n?


What is the smallest two-digit positive integer such that the product of its two digits is one-half the integer?

All of these numbers $ \overline{ab}\equal{}10a\plus{}b$ satisfy $ 2ab\equal{}10a\plus{}b\implies(2a\minus{}1)(b\minus{}5)\equal{}5$. Since $ b\leq9\implies b\minus{}5\leq4$, we are sure that $ b\minus{}5\equal{}1\implies b\equal{}6$ and $ 2a\minus{}1\equal{}5\implies2a\equal{}6\implies a\equal{}3$, so our answer is $ \boxed{36}$.


Initially a  particle lies at a point P on  horizontal line.
Each time a balanced coin is tossed  and a head appears the particle moves  a distance a units from its present position in a direction to the right of P while if a tail appears it is moved from its presnt position a distance  b units to the left .

If L is the distance of the particle from P atfter n tosses of the coin, find
           E(L), the expected value of L.

would it not just be:

bn/2 - an/2 along the x? or vice versa or watever...


I want to learn more than the AoPS vol. 2 chapter on Graph Theory, especially colorings and how to prove those types of questions with out brute forcing. Are there any good websites or resources that you would suggest? Any help would be appreciated.

[u]Pearls in Graph Theory[/u], by Hartsfield and Ringel, is a very good book for graph theory.


A best-of-five series ends when one team wins three games. The probability of team A defeating team B in any game is  1/6 . What is the probability that team A will win the series? Express your answer as a fraction in lowest terms.

[hide]
Ways for A to win
The letter denotes the winner
AAA
AABA
ABAA
BAAA
BBAAA
BABAA
BAABA
ABBAA
ABABA
AABBA
For the first one, there is a $\frac 16 ^3=\frac{1}{216}$ chance.
For each of the next three, there is a $\frac 56 \cdot \frac 16^3=\frac{5}{1296}$
For each of the last ones, there is a
$\frac 16^3 \cdot \frac 56 ^3=\frac{25}{1776}$
Sum them all to get $\frac{23}{648}$[/hide]


analytic geometry
graphing lines
slope
number theory
prime factorization
AMC

For some reason I can't figure out what I did wrong on these two problems, can anyone explain how to do them? I know they shouldn't be too hard but I think I've looked at them too many times to see what I'm missing. thanks!

For number 14, you can factor 45.  Once you do that, you need to find five different factors which simplify to those factors.

[hide]The prime factorization is 3,3,5.  However, you need two extra factors.  Try making one of the above factors negative and adding 1 and -1 to the list, and you'll end up with 45.  Since they are distinct, there is only one option for which factor out of 3,3, and 5 you make negative.[/hide]

[quote="Art of Owna"]For the mouse one, when the mouse is running up the line, he is getting closer, closer, and closer, until he reaches his closest point, then after that he will get farther and farther away from that point, and the line when the mouse is the closest is the line perpendicular to the given line and also goes through where the cheese is.  You can use point slope to find the equation of that line.  Now you have two equations for lines for two variables.. then solve[/quote]

[hide]Yup, take the negative reciprocal of the slope (perpendicular), solve for b in y=mx+b by substituting (12, 10) for x and y, and where the two lines intersect, you'll find your answer.[/hide]

oh woah thanks....wow i somehow forgot that a negative and a positive doesnt give a positive for #13, but I guess thats what happens when you take a test with a fever and then trust your work when later trying to figure out what you did wrong, and #14 i missed making a second line, i did something weird with more irrational numbers, not sure how i got there, anyways thanks again, hope i'm more awake for the test this wednesday


algebra
polynomial
number theory
greatest common divisor
modular arithmetic
vector
linear algebra

Let $a_{i}$ and $b_{i}$ ($i=1,2, \cdots, n$) be rational numbers such that for any real number $x$ there is:
\[x^{2}+x+4=\sum_{i=1}^{n}(a_{i}x+b)^{2}\]
Find the least possible value of $n$.

I think it is 5:

For $n=5$ we have $(a_1,a_2,a_3,a_4,a_5)=(1/2,1/2,1/2,1/2,0)$ and $(b_1,b_2,b_3,b_4,b_5)=(1,-1,1,0,1)$, this works.

Assume it were possble with 4 (or less), then I will prove this implies $a_1=a_2=a_3=a_4=0$ or $b_1=b_2=b_3=b_4=0$, which leads to a contradiction.

Thus, assume there were rationals for which $(2a_1)^2+(2a_2)^2+(2a_3)^2+(2a_4)^2=4 =b_1^2+b_2^2+b_3^2 +b_4^2$, and $\sum_{cyc}(2a_1)(b_1)=1$.

The existence of such rationals implies the existence of integer solutions (other than the trivial solution) to $\{\begin{array}{c} w^2+x^2+y^2+z^2=v^2 \\ a^2+b^2+c^2+d^2=e^2 \\ 4(ax+by+cz+wd) = ev\end{array}$.

Now we have $(x^2+y^2+z^2+w^2)(a^2+b^2+c^2+d^2)=e^2v^2 =16(ax+by+cz+wd)^2$. [list][*]For $n<4$: As we're dealing with integers, at least 1 factor in the LHS must be a multiple of 4, implying that either $(a,b,c)$ or $(x,y,z)$ are all even, wlog $(a,b,c)$.
[*]For $n=4$: assume both factors are not divisible by 8, then a,b,c,d,x,y,z,w are all odd, since each factor then is 4 mod 8. But from $(e/4)^2(v/4)^2 =(ax+by+cz+wd)^2$ it follows that at least one of $\frac e4,\frac v4$ is even, since $ax+by+cz+wd$ is even (the sum of 4 odd integers). Thus also here one of the factors has all its components even, wlog $(a,b,c,d)$ are 4 even integers.[/list]

Now, if a solution existed, we would have that $(x^2+y^2+z^2+w^2)(a^2/4+b^2/4+c^2/4+d^2/4)= 16(ax/2+by/2+cz/2+dw/2)^2$, thus $(a/2,b/2,c/2,d/2,x,y,z,w)$ is a smaller solution. Thus, by descent, either $a=b=c=d=0$ or $x=y=z=w=0$, which is impossible.

The minimum is thus $n=5$.

By suitable transformation (letting $x=\frac{y-1}{2}$ and then multiplying through by noted square number 4), we are really looking for rational polynomials whose squares add up to $y^2+15$.

Obviously $n=5$ works (one of many examples is in Peter's post).

Lemma: 3 rational squares (one or two of which may be 0) cannot sum to $15$.

Proof: Multiplying through by their GCD, we have relatively prime integer solutions to $x^2+y^2+z^2=15d^2$. If $d^2=1 \pmod 8$, then the RHS is $7 \pmod 8$ and can't be expressed as a sum of three squares. If $d^2=0,4 \pmod 8$, then $x,y,z,d$ must all be even, contradicting that they are relatively prime. Thus the lemma is proven.

This obviously rules out $n=1,2,3$.

Assume there exists a solution with $n = 4$.

Then we have $a_1^2+a_2^2+a_3^2+a_4^2=1$, $a_1b_1+a_2b_2+a_3b_3+a_4b_4=0$ and $b_1^2+b_2^2+b_3^2+b_4^2=15$

Consider the vectors $a=(a_1,a_2,a_3,a_4)$ and $b=(b_1,b_2,b_3,b_4)$, with $a \cdot b=0$, $|a|=1$, $|b|=15$.

There exists a matrix transformation which takes $a$ to $(1,0,0,0)$ that will have rational coefficients (EDIT: see below), and will necessarily preserve orthoganality and magnitude, so it will take $b$ to $(0,x,y,z)$, where $x^2+y^2+z^2=15$ and $x,y,z$ are rational, which we know to be impossible by our lemma.

Thus, the minimal value is $n=5$.

EDIT: mssmath has pointed out to me that my original argument for why this existed didn't work, but
$\left[\begin{array}{cccc}
a_1 & -a_2 & -a_3 & -a_4       \\
a_2 & a_1 & a_4 & a_3       \\
a_3 & -a_4 & a_1 & a_2       \\
a_4 & a_3 & -a_2 & a_1
\end{array}\right] ^{-1}$ does

Can somebody kindly explain how one gets the explicit matrix transform described by Mewto55555?


[hide]
3x+1 must equal either 1 or 0, since those are the only numbers whose square root is equal to that number.

so
3x+1=1
x=0
3x+1=0
x=-1/3
So the solution is[b] 0, -1/3[/b][/hide]

Yeah, I almost missed out on x = -1/3...

I guess if you square both sides and divide by (3x + 1), you lose a value of x...

[hide]

sqrt(3x + 1) = 1

3x + 1 = 1

3x = 0

x = 0

but you can't divide by zero, which comes in the form of sqrt(3x + 1), where x = -1/3, so...

[/hide]

Here's how I did it.

[hide]$\sqrt{3x+1}^2 = {(3x+1)}^2$
${3x+1} = {9x^2+6x+1}$
${9x^2+3x} = 0$
${3x(3x+1)} = 0$
$3x = 0$ or $3x+1=0$
$x=0$ or $x = \frac{-1}{3}$[/hide]

Are the participants in MathCounts expected to know how to factor prolynomials?

Can anybody think of a nice way to use substitution that helps make the problem look simpler?

Well, you could let $y=3x+1$, so that $\sqrt{y}=y$, or $y^2-y=0$, which obviously has solutions 1 and 0, so $3x+1=1$, $x=0$, or $3x+1=0$, so $x=\frac{-1}{3}$.


The number $ 25^{64}$ times $ 64^{25}$ is the square of a positive integer N. In decimal representation, the sum of the digits of N is what?

I think we're looking for the square root of your answer.

$ 2^{11} * 10^{64}$

Which is 2048 and all the zeroes.

So 4+ 8+ 2 = 14


The game consists in flipping a coin, all the way to reaching 2 next tails, in addition the maximum number of throws amounts 8. What probability of making 8 throws is?

Alright, here's my solution:
[hide]
Let's first decide on the terminology. Let "heads winning" be the result of the game when after exactly $ 8$ throws there are less than $ 2$ tails. Also, let "tails winning" be the result of the game when after $ 8$ or less throws there are exactly $ 2$ tails. After that is achieved, the game stops.
[b]Lemma 1.[/b]
Claim: For $ 2\leq n\leq 8$, the chances of tails winning in exactly $ n$ throws are $ \frac{n\minus{}1}{2^n}$.
Proof: For tails to win in exactly $ n$ throws, the $ n$-th throw must be a tail. Other than that, exactly $ 1$ throw among $ n\minus{}1$ previous throws must be a tail. So, we can choose one of these throws in exactly $ n\minus{}1$ ways. Chances for achieving each throw just the way we want it to be is $ \frac{1}{2}$. There are $ n$ throws, so the probability of tails winning in exactly $ n$ throws is $ \frac{n\minus{}1}{2^n}$
[b]The conclusion[/b]
Tails can win in $ 2$, $ 3$, $ ...$, $ 8$ throws, while heads can win only in $ 8$ throws. So, the probability of heads winning is $ \frac{1}{2^8}\plus{}\frac{8}{2^8}$. Why is that?
Heads can win in $ 8$ throws when all throws are heads - as noted in the lemma, the probability for that is $ \frac{1}{2^8}$. Also, heads can win in $ 8$ throws when exactly $ 7$ throws are heads - we can choose the throw which ended with tails in $ 8$ ways. So, the probability of that happening is $ \frac{8}{2^8}$.
Altogether, tails win in 8 throws + heads win in 8 throws give us the result of $ \frac {7\plus{}1\plus{}8}{2^8}\equal{}\frac{1}{16}$. All possible outcomes give us the result of $ \frac{1}{4}\plus{}\frac{2}{8}\plus{}\frac{3}{16}\plus{}...\plus{}\frac{7}{256}\plus{}\frac{8}{256}\plus{}\frac{1}{256}\equal{}1$, which is pretty self obvious (just checking, really). So, the probability of making 8 throws is $ \frac{1}{16}$.
[/hide]

I think that's it. I might have made a mistake somewhere, but it seems OK to me right now.


geometric inequality
geometry
circumcircle
trigonometry
function
incenter
inequalities

Denote the triangle sides $a = BC$, $b = CA$, $c = AB$, the angles $\alpha = \measuredangle BAC$, $\beta = \measuredangle ABC$, $\gamma = \measuredangle BCA$ and the altitude feet from the vertices $A, B, C$ as $D, E, F$. Let $S , S_A, S_B, S_C$ be the areas, $O, O_A, O_B, O_C$ the circumcenters and $R, R_A, R_B, R_C$ the circumradii of the triangles $\triangle ABC, \triangle BPC, \triangle CPA, \triangle APB$. The area of the triangle $\triangle ABC$ is

$S = \frac{abc}{4R}$

and similarly of the other triangles. As long as the point P is not on any side of the all acute angle triangle $\triangle ABC$, the inequality can be written as

$S_A R_A + S_B R_B + S_C R_C \ge S R$

If P is the orthocenter of the $\triangle ABC$, then $\measuredangle BPC = \pi - \alpha = \beta + \gamma$, $\measuredangle CPA = \pi - \beta = \gamma + \alpha$, $\measuredangle APB = \pi - \gamma = \alpha + \beta$. Since the angle $\alpha$ spans the longer arc $BC$ of the circle $(O)$ and the angle $\pi - \alpha$ spans the shorter arc BC of the circel $(O_A)$, their radii are equal, $R_A = R$. Similarly, $R_B = R_C = R$ and the inequality turns into a trivial equality

$S_A + S_B + S_C = S$

If the point P is anywhere in the triangle interior, then $R_A = \frac{a}{2 \sin{\measuredangle BPC}}$, $R_B = \frac{b}{2 \sin{\measuredangle CPA}}$, $R_C = \frac{c}{2 \sin {\measuredangle APB}}$ and ${\measuredangle BPC + \measuredangle CPA + \measuredangle APB = 2 \pi}$. Consequently,

$\arcsin {\left( \frac{a}{2 R_A} \right)} + \arcsin{\left( \frac{b}{2 R_B} \right)} + \arcsin{\left( \frac{c}{2 R_C} \right)} =$
$= \pi - \measuredangle BPC + \pi - \measuredangle BPA + \pi - \measuredangle APB = \pi$

Using Lagrange multipliers $\lambda_1, \lambda_2$, we can find maxima or minima of the function

$f(S_A, S_B, S_C, R_A, R_B, R_C) = S_A R_A + S_B R_B + S_C R_C$

subject to the constraints

$g_1(S_A, S_B, S_C) = S_A + S_B + S_C - S = 0$
$g_2(R_A, R_B, R_C) = \arcsin{\left( \frac{a}{2 R_A}\right)} + \arcsin{\left( \frac{b}{2 R_B}\right)} + \arcsin{\left( \frac{c}{2 R_C} \right)} - \pi = 0$

$h = f + \lambda_1 g_1 + \lambda_2 g_2$

$\frac{\partial h}{\partial S_A} = R_A + \lambda_1 = 0$

$\frac{\partial h}{\partial S_B} = R_B + \lambda_1 = 0$

$\frac{\partial h}{\partial S_C} = R_C + \lambda_1 = 0$

$\frac{\partial h}{\partial R_A} = S_A - \lambda_2 \frac{a}{2 R_A \sqrt{R_A^2 - \left( \frac a 2\right) ^2}} = S_A - \lambda_2 \frac{\tan{(pi - \measuredangle BPC)}}{R_A} = 0$

$\frac{\partial h}{\partial R_B} = S_B - \lambda_2 \frac{b}{2 R_B \sqrt{R_B^2 - \left( \frac b 2\right) ^2}} = S_B- \lambda_2 \frac{\tan{(\pi - \measuredangle CPA)}}{R_B} = 0$

$\frac{\partial h}{\partial R_C} = S_C - \lambda_2 \frac{c}{2 R_C \sqrt{R_C^2 - \left( \frac c 2\right) ^2}} = S_C - \lambda_2 \frac{\tan{(\pi - \measuredangle ABP)}}{R_C} = 0$

From the first 3 equations, it immediately follows that $-\lambda_1 = R_A = R_B = R_C$, which is possible only for the orthocenter of a triangle and then $R_A = R_B = R_C = R$. Substituting this to the last 3 equations and using the fact that for the orthocenter, $\alpha = \pi - \measuredangle BPC$, $\beta = \pi - \measuredangle CPA$, $\gamma = \pi - \measuredangle APB$

$\lambda_2 = \frac{S_A R}{\tan \alpha} = \frac{S_B R}{\tan \beta} = \frac{S_B R}{\tan \gamma}$

$\frac{S_A}{a} \cos \alpha = \frac{S_B}{b} \cos \beta = \frac{S_B}{c} \cos \gamma$

$PD\ \cos \alpha = PE\ \cos \beta = PF\ \cos \gamma$

which is true for the orthocenter ($PA = \frac{PE}{\cos \gamma} = \frac{PF}{\cos \beta}$, etc.). To see if the orthocenter is the minimum or maximum of the function f, we calculate its value at some other point of the triangle interior, for example, at the circumcenter O of the all acute angle $\triangle ABC$. In this case, $OA = OB = OC = R$ and

$f = R^2 a + R^2 b + R^2 c = \frac{abc}{4} \left(\frac{1}{\sin \beta \ \sin \gamma} + \frac{1}{\sin \gamma\ \sin \alpha} + \frac{1}{\sin \alpha\ \sin \beta}\right)$

The orthocenter extrem is a minimum, if we can show that

$\frac 1 4 \left( \frac{1}{\sin \beta\ \sin \gamma} + \frac{1}{\sin \gamma\ \sin \alpha} + \frac{1}{\sin \alpha\ sin \beta} \right) \ge 1$

${\sin \alpha + \sin \beta + sin \gamma \ge 4\ \sin \alpha\ \sin \beta\ \sin \gamma}$

Using $\gamma = \pi - (\alpha + \beta)$, $\sin \gamma = \sin{(\alpha + \beta)}$ and going to half-angles, this can be written as

$F(\alpha, \beta,\gamma) = 8\ \sin{\frac{\alpha}{2}}\ \sin{\frac{\beta}{2}}\ \cos{\frac{\alpha + \beta}{2}} = 8\ \sin{\frac{\alpha}{2}\ \sin{\frac{\beta}{2}}\ \sin{\frac{\gamma}{2}} \le 1}$

We can again use Lagrange multiplier $\lambda$ to find a maximum of this function subject to the constraint $G(\alpha, \beta,\gamma) = \alpha + \beta + \gamma - \pi = 0$:

$H = F + \lambda G$

$\frac{\partial H}{\partial \alpha} = 4\ \cos{\frac{\alpha}{2}}\ \sin{\frac{\beta}{2}}\ \sin{\frac{\gamma}{2}} + \lambda = 0$

$\frac{\partial H}{\partial \beta} = 4\ \sin{\frac{\alpha}{2}}\ \cos{\frac{\beta}{2}}\ \sin{\frac{\gamma}{2}} + \lambda = 0$

$\frac{\partial H}{\partial \gamma} = 4\ \sin{\frac{\alpha}{2}}\ \sin{\frac{\beta}{2}}\ \cos{\frac{\gamma}{2}} + \lambda = 0$

It follows that

$-\frac{\lambda}{4} = \cos{\frac{\alpha}{2}}\ \sin{\frac{\beta}{2}}\ \sin{\frac{\gamma}{2}} = \sin{\frac{\alpha}{2}}\ \cos{\frac{\beta}{2}}\ \sin{\frac{\gamma}{2}} = \sin{\frac{\alpha}{2}}\ \sin{\frac{\beta}{2}}\ \cos{\frac{\gamma}{2}}$

and if $\sin{\frac{\gamma}{2}} \neq 0$, $\sin{\frac{\alpha}{2}} \neq 0$, $\sin{\frac{\beta}{2}} \neq 0$,

$\cos{\frac{\alpha}{2}}\ \sin{\frac{\beta}{2}} = \sin{\frac{\alpha}{2}}\ \cos{\frac{\beta}{2}}$
$\cos{\frac{\beta}{2}}\ \sin{\frac{\gamma}{2}} = \sin{\frac{\beta}{2}}\ \cos{\frac{\gamma}{2}}$
$\cos{\frac{\gamma}{2}}\ \sin{\frac{\alpha}{2}} = \sin{\frac{\gamma}{2}}\ \cos{\frac{\alpha}{2}}$

$\sin{\frac{\alpha - \beta}{2}} = \sin{\frac{\beta - \gamma}{2}} = \sin{\frac{\gamma - \alpha}{2}} = 0$

Hence, either one angle is zero, the other 2 add up to $\pi$ and $F = 0$, or all 3 angles are equal to each other and to $\frac{\pi}{3}$ and $F = 1$. The first case represents a minimum for a "triangle" with 2 sides parallel and the corresponding vertex at $\infty$, the other to a maximum for an equilateral triangle. But in an equilateral triangle, the orthocenter and circumcenter are identical. As a result, in a non-equilateral all acute angle triangle, the value of the function f at the circumcenter is greater than that at the orthocenter, i.e., the extrem at the orthocenter is a minimum. It is still necessary to find a point in an equilateral triangle, where the function f has a value (or a limit on one of the triangle sides) greater than that at the orthocenter. We can choose the foot of one altitude, say D. But then

$AB = BC = CA = a$, $DB = DC = \frac a 2$, $DA = \frac{\sqrt 3}{2} a$

$DA \cdot DB \cdot a + DB \cdot DC \cdot a + DC \cdot DA \cdot a = a^3 \left( \frac 1 4 + \frac{\sqrt 3}{2} \right) > a^3$

so that even in an equilateral triangle, the value of the function f at the orthocenter is a minimum.

I'll just make a small $\LaTeX$ notice: use \sin, \tan or \arcsin for functions and \left( .... \right) to adjust the height of the () paranthesis: $\sin { \widehat{BOC} } = \tan { \left( \frac A2 \right) }$ looks better than  $sin { \widehat{BOC} } = tan { ( \frac A2 ) }$.

And this doesn't look like a pre-olympiad, but rather an olympiad problem (it's also from China! :) ) therefore it is moved in the advanced forums.

[quote="Michael Niland"]P is any point inside an acute-angled triangle ABC

Prove that \[ PA \cdot PB \cdot AB + PB \cdot PC \cdot BC + PC \cdot PA \cdot CA \geq AB \cdot BC \cdot CA \]

Prove there is equality if and only P is the orthocentre.[/quote]

The inequality $PA \cdot PB \cdot AB + PB \cdot PC \cdot BC + PC \cdot PA \cdot CA \geq AB \cdot BC \cdot CA$ holds for any triangle ABC (not necessarily acute), and for any point P in its plane. It can be rewritten in the equivalent form

$\frac{PA\cdot PB}{BC\cdot CA}+\frac{PB\cdot PC}{CA\cdot AB}+\frac{PC\cdot PA}{AB\cdot BC} \geq 1$,

and it was proven on http://www.mathlinks.ro/Forum/viewtopic.php?t=6037 (as the inequality $P+Q+R\geq 1$ in posts #6-#9), on http://www.mathlinks.ro/Forum/viewtopic.php?t=5889 (in post #3) and http://www.mathlinks.ro/Forum/viewtopic.php?t=21016 (as Lemma 2 in post #14). The only problem is that most of the proofs of this inequality don't make it easy to identify when equality holds.

  darij

Thanks a lot, Myth. It took me some time to check it out. This is the best inversion problem I have ever seen, but there is no chance I would have figured out your solution.

yetti

1913,  [b]T. Hayashi [/b] Prove that

 \[\frac{PB}{CA}\frac{PC}{AB}+\frac{PC}{AB}\frac{PA}{BC}+\frac{PA}{BC} \frac{PB}{CA}\geq1.\] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&p=1966226#p1966226]here[/url]

[size=150]1993,   [b]I [/b] Prove that [/size]
 \[\frac{PA}{BC}( \frac{{P'}{B'}}{{C'}{A'}}
+\frac{{P'}{C'}}{{A'}{B'}}  )
+\frac{PB}{CA}(\frac{{P'}{C'}}{{A'}{B'}}
+\frac{{P'}{A'}}{{B'}{C'}}  ) +\frac{PC}{AB}(\frac{{P'}{A'}}{{B'}{C'}}+\frac{{P'}{B'}}{{C'}{A'}}  )\geq2.\]
Here，  $ P $  is any point of the plane with a triangle $ ABC $ ， $P'$   is any point of the plane with  a  triangle  $ A'B'C '$ .

$\frac{PB}{CA}\frac{PC}{AB}+\frac{PC}{AB}\frac{PA}{BC}+\frac{PA}{BC} \frac{PB}{CA}\geq1\Rightarrow \frac{PA}{BC}+\frac{PB}{CA}+\frac{PC}{AB} \geq \sqrt{3}$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&p=2589173#p2589173
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=500544&p=2842999#p2842999
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=225313&hilit=inequality
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=49&t=6073&hilit=Brazil+inequality

 http://www.artofproblemsolving.com/community/c6h1192524p5823133
Let $P$ is a point in the plain $ABC$,and $BC=c,CA=b,AB=c.$Prove that$$\frac{PA}{2a+b}+\frac{PB}{2b+c}+\frac{PC}{2c+a}\geq \frac{\sqrt{3}}{3}.$$

[quote=Michael Niland]P is any point inside an acute-angled triangle ABC

Prove that \[ PA \cdot PB \cdot AB + PB \cdot PC \cdot BC + PC \cdot PA \cdot CA \geq AB \cdot BC \cdot CA \]

Prove there is equality if and only P is the orthocentre.[/quote]

This is Hayashi's inequality. We can prove it using complex or vector.

[quote=Myth]You are a hero, Yetti! Moreover, you are a great Darij's follower. :lol:

I have a geometric proof.

Consider inversion with center $P$ then the inequality will transrom to $PA^2\cdot BC+PB^2\cdot AC+PC^2\cdot AB\geq AB\cdot BC\cdot CA$. But $PA^2\cdot BC+PB^2\cdot AC+PC^2$ is a moment of inertia of $P$ wrt to mass-points $(BC)A$, $(CA)B$ and $(AB)C$. Due to Steiner's theorem $I_P=I_I+(AB+BC+CA)\cdot IP^2$, there $I$ is an incenter of $ABC$ (center of gravity of $(BC)A$, $(CA)B$ and $(AB)C$). Thus "minimal" point $P$ is an incenter $I$ !!!
Performing inversion again we see that $I$ transforms to orthocenter! And the equality is evident for orthocenter, since $R_{ABC}=R_{PBC}=R_{PAC}=R_{PAB}$, so applying formula $S_{ABC}=AB\cdot BC\cdot CA/R_{ABC}$ we immediatelly obtain the result.[/quote]
http://www.artofproblemsolving.com/community/c6h1331039p7176818

[quote=sqing]Let $P$ be an interior point of a triangle $ABC$ with side lengths $ a, b, c $ . Prove that$$  PA+PB+PC \geq \sqrt[3]{a^2PA+b^2PB+c^2PC}$$(Marcel Chiriţă)[/quote]

Actually [url=https://artofproblemsolving.com/community/c6h1491070]this one[/url] written as
\[\begin{array}{c}(PA + PB + PC)(PA\cdot PB + PB\cdot PC + PC\cdot PA)\ge a^2 PA + b^2PB + c^2PC\end{array}\]
 is slightly stronger and holds for [b]any[/b] point $P$ in the plane. ;). But this is just a particular case of the [b]polar moment of inertia inequality[/b], i.e.
\[\boxed{(x + y + z)\left(xPA^2 + yPB^2+zPC^2\right) \ge yza^2 + zxb^2 + xyc^2}\]
for $x = PB\cdot PC$, $y = PC\cdot PA$ and $z = PC\cdot PA$.

Let $P$ is any point inside an acute-angled triangle $ABC$ . Prove that \[ \frac{PA}{asin2A}+\frac{PB}{bsin2B}+\frac{PC}{csin2C} \geq 2. \]

[quote=sqing]Let $ABC$ be a triangle and let $P$ be an arbitrary point. Prove that
$$PA+PB+PC\geq \frac{1}{2}(m_a+m_b+m_c+3r)$$
$$\left(\frac{PB \cdot PC}{m_ah_a}\right)^2+\left(\frac{PC \cdot PA}{m_bh_b}\right)^2+\left(\frac{PA \cdot PB}{m_ch_c}\right)^2\geq \frac{16}{27}$$[/quote]
Let $ABC$ be a triangle and let $P$ be an arbitrary point in space. If $x$, $y$ and $z$ are non-zero real numbers, [url=https://artofproblemsolving.com/community/c6h1609583p10048777]prove that[/url]
\[\boxed{ \left(\frac {PB\cdot PC}x\right)^2 + \left(\frac {PC\cdot PA}y\right)^2 + \left(\frac {PA\cdot PB}z\right)^2 \ge \frac {\left(abc\right)^2}{\left(ax\right)^2 + \left(by\right)^2 + \left(cz\right)^2}}\]

1913, T. Hayashi Prove that\[\frac{PB}{CA}\frac{PC}{AB}+\frac{PC}{AB}\frac{PA}{BC}+\frac{PA}{BC} \frac{PB}{CA}\geq1.\]
Let $P\equiv G$- the centroid of triangle $ABC$ [url=https://artofproblemsolving.com/community/c6h1748539p11396887]we get[/url]
\[\frac{m_b\cdot m_c}{bc}+\frac{m_c\cdot m_a}{ca}+\frac{m_a\cdot m_b}{ab}\ge\frac{9}{4}.\]Now we apply this for the triangle whose sides $m_a,m_b,m_c$ to obtain
\[\frac{bc}{m_b\cdot m_c}+\frac{ca}{m_c\cdot m_a}+\frac{ab}{m_a\cdot m_b}\ge4\]
https://artofproblemsolving.com/community/c6h1748539p11397720
$$\frac{m_am_b}{a^2+b^2}+\frac{m_bm_c}{b^2+c^2}+\frac{m_cm_a}{c^2+a^2}\geq\frac{9}{8}$$
 [url=https://artofproblemsolving.com/community/c6h407568p2276600]here[/url]

If you apply the following link on the quadruple $(A,B,C,P)$ instead of $(A,B,C,D)$ one gets $AB\cdot PC^2+BC\cdot PA^2+CA\cdot PB^2\ge AB\cdot BC\cdot CA$
https://artofproblemsolving.com/community/u346763h1770343p11614441


superior algebra
superior algebra unsolved

Let G be some group with four conjugacy classes C1,C2,C3,C4, with C1={1}, C2 having two elements and both C3 and C4 having three elements.  The character table for G:

[img]http://www.geocities.com/jacobin2001/Charactertable2.jpg[/img]

where p is a primitive root of unity.

Is this character table possible?!

source unknown.

Hi.

The first two columns are not orthogonal, so this can't be an character table. If there would be a -1 instead of -2, you'd have the character table of A_4.

$ G$ does not exist as it is decribed and I do not refer to the character table for this claim. According to the description of conjugacy classes $ G$ contains $ 9$ elements, but has a conjugacy class of length $ 2$, a contradicition


Triangle $ABC$ has an incenter $I$. Let points $X$, $Y$ be located on the line segments $AB$, $AC$ respectively, so that $BX \cdot AB = IB^2$ and $CY \cdot AC = IC^2$. Given that the points $X, I, Y$ lie on a straight line, find the possible values of the measure of angle $A$.

From $BX \cdot AB = IB^2$, it follows that $\dfrac{BX}{IB} = \dfrac{IB}{AB}$. Since in addition, the angle $\angle IBA \equiv \angle IBX$ is common, the triangles $\triangle IBX \sim \triangle ABI$ are similar by SAS and the angle $\angle BIX = \angle BAI = \dfrac{\angle A}{2}$. Likewise, from $CY \cdot AC = IC^2$, it follows that the triangles $\triangle ICY \sim \triangle ACI$ are also similar and the angle $\angle CIY = \angle CAI = \dfrac{\angle A}{2}$. From the triangle $\triangle BCI$, the angle $\angle BIC = 180^\circ - \left( \dfrac{\angle B}{2} + \dfrac{\angle C}{2}\right) = 90^\circ + \dfrac{\angle A}{2}$. Since the points X, I, Y are collinear, the angle $\angle XIY = 180^\circ$ is straight. But it is equal to the sum of the following angles:

$\angle XIY = \angle BIX + \angle BIC + \angle CIY = \dfrac{\angle A}{2} + 90^\circ + \dfrac{\angle A}{2} + \dfrac{\angle A}{2} = 180^\circ$

Solving for the angle $\angle A: \ \ \frac 3 2\ \angle A = 90^\circ,\ \ \angle A = 60^\circ$.


[quote="myc"]What positive integers N have the property that $N^{2}$ has remainder N when divided by 100?

I have a feeling this has to do with modulos.  I know N has to end in either 1, 3, or 5.

Thanks[/quote]

[hide]It has to be a two digit number because it obviously cant be a 3 digit number and if it was a one digit number, the square wuld not exceed 100.

The possible numbers can be written as $10a+1, 10a+3, 10a+5$.

$100a^{2}+20a+1 \equiv 20a+1\bmod 100$.

$\Rightarrow 20a+1\equiv 10a+1\bmod 100$

$\Rightarrow 20a\equiv 10a\bmod 10$

$\Rightarrow 2a\equiv a\bmod 10$

i dont think there is a solution for $10a+1$.

do it for the other ones too

one number that i know works is 25.[/hide]

Both of you made the same mistake: it doesn't work for 10a+3, notice that $(10a+3)^{2}=100a^{2}+60a+9$, which ends with a nine, not a 3.  However, 10a+6 works.  I will complete jli's solution here:
[hide]jli wasn't right about the fact that 10a+1 doesn't work.   If a=0, then it works, thus, a=$1$ works.  Now, for 10a+5, we have that
\[(10a+5)^{2}\equiv 10a+5\bmod 100\to 100a^{2}+100a+25\equiv 10a+5\bmod 100\Rightarrow 90a+20\equiv 0\bmod 100\Rightarrow 9a+2\equiv 0\bmod 10\Rightarrow 9a\equiv 8\bmod 10\Rightarrow 9(9a)\equiv 9(8)\bmod 10\Rightarrow a\equiv 2\bmod 10 \]
a cannot be greater than 10, so a=2.  Thus, $25$ works.  Now, let's look at 10a+6:
\[(10a+6)^{2}\equiv 10a+6\bmod 100\Rightarrow 100a^{2}+120a+36\equiv 10a+6\bmod 100\Rightarrow 110a+30\equiv 0\bmod 100\Rightarrow 11a+3\equiv 0\bmod 10\Rightarrow 11a\equiv 7\bmod10\Rightarrow a\equiv 7\bmod 10 \]
Since a cannot be greater than 10, a=7, so $76$ works.  Thus our answers are $1, 25, 76$.[/hide]

One way to think about this is to notice two items:

1. Prime factors of 100 are $2^{2}\cdot 5^{2}$
(Think of 100 as  $4 \cdot 25)$
2. Solutions for $N^{2}= N$  are $N=0$ or $N=1$
(Because $N^{2}= N \implies N(N-1)=0 \implies N=(0,1)$

Rest is simple

[hide]You want $N \equiv 0$ or $N \equiv 1$ (Both modulo 4 and 25) and CRT gives the answer as $0,1,25,76$
Basically 0 mode 25 are (0,25,50,75) out of which only 0 and 25 works (First is 0 mod 4 and second, that is 25 is, 1 mod 4)

Similarly, for 1 mod 25,  after 1, only other solution (out of 26,51,76) is 76 which is 0 mod 4)

If you want "positive" numbers only then disregard 0 and the rest of the numbers are: 1,25,76.

[/hide]

$Why is$ \[10a+5\Rightarrow 90a+20\equiv 0\bmod 100 ? \]

$90(15)+20$ does not equal  $0 (mod 100)$.

Thanks

Oh, sorry for making that unclear.  I meant that since
\[(10a+5)^{2}\equiv 10a+5\bmod 100\Rightarrow 100a^{2}+100a+25\equiv 10a+5\bmod 100 \]
we can subtract 10a+5 from both sides to get that
\[100a^{2}+90a+20\equiv 0\bmod 100 \]
Now, since $100a^{2}$ is divisible by 100, we can disregard it (since it is equal to 0mod100).  From here, you can read my solution above and try to figure out the rest (if my solution even makes sense; I am terrible at explaining).


Each integer 1-9 is squared. What is the sum of the resulting units digits?

We need to find the units digit of $ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} ... \plus{} 8^2 \plus{} 9^2$
We can use the sum of squares formula, namely $ 1^2 \plus{} 2^2 \plus{} ... \plus{} n^2 \equal{} \frac{(n)(n\plus{}1)(2n\plus{}1)}{6}$
So $ 1^2 \plus{} 2^2 \plus{} 3^2 \plus{} ... \plus{} 8^2 \plus{} 9^2 \equal{} \frac{9*10*19)}{6} \equal{} 285$, so the sum of the resulting units digit is $ \boxed{5}$

Or we can do

$ 1\plus{}4\plus{}9\plus{}6\plus{}5\plus{}6\plus{}9\plus{}4\plus{}1$
$ 45$
ends in 5.


MATHCOUNTS FTW Mafia AMC USA(J)MO USAMO Stanford

Rate the person above you about how much you have heard their name on AOPS from 1-10.

1=New player
10=Like rrusczyk

Rate me!

hmm... 7.5?

I think this belongs to FF... just my opinion...

5. Ok if a mod moves it. Im seriously a 4?

5. Dunno that much about you, besides what I remember from the Mathcounts course.

:diablo:
---------= 6.66= My rate on xpmath
100

EDIT: LOL, my math fails...

4 for AIME15

Hmm, perhaps we should post why we're giving the rating that we are? It would make this a bit more interesting.

5. I don't see you around much, but I think you were in my mc class. What happened to your avatar?

8.  I see you in ernie's sigg and spamming topics.  :D  and on this thread

I'm in ernie's sig?

9 because:
1) Numb theorey class
2) SJMC and Vanteus
3) FTW
4) Common contrib
5) 2000 rating

I meant cool FTW guy, srry
or he is a mod?

5. Same reasons

So youve seen me a lot! This is not how much you like the person, but how much youve seen them!

4

Actually, for me, it's not whether I like you more than someone else or if I've seen you on more than someone else, etc. I'm rating based on how you post and act.

Moved to G&FF.

Ah, and for the sake of not ending the game: 4.

8

we formed an alliance in "Guess who will post next!"

9

We did not form an alliance in Guess who posts next, but I helped you beat budi

9

chenhsi gave me a 9, so I gave chenhsi a 9 right back

1 - because I didn't..., I actually was the one that started putting the spam


If $S(n)$ means the sum of digits in base $10$ of $n$, prove that
$\frac{S(n)}{S(n^2)}$ is unbounded

Indeed, very well known, but very very nice. I knew it for base 2 (saw it in an old shortlist), but the construction is practically the same.

Here is the number for base 2: consider $f(n)=2^n$ and the number $ f(f(k)-1)-f(f(k)-f(1))-f(f(k)-f(2))-...-f(f(k)-f(k-1))-1$.


Let $ x,y,z \in R$ Prove that $ (x \plus{} y \plus{} z)^2$ be a factor of $ 6(x^5 \plus{} y^5 \plus{} z^5) \minus{} 5(x^3 \plus{} y^3 \plus{} z^3)(x^2 \plus{} y^2 \plus{} z^2)$

[quote="apollo"]$ 6(x^5 \plus{} y^5 \plus{} z^2)$[/quote]

I think you meant $ 6(x^5\plus{}y^5\plus{}z^5)$, am I wrong?

In that case
$ 6(x^5 \plus{} y^5 \plus{} z^5)\minus{}5(x^3 \plus{} y^3 \plus{} z^3)(x^2 \plus{} y^2 \plus{} z^2)$
$ \equal{} (x\plus{}y\plus{}z)^2 (x^3\plus{}y^3\plus{}z^3\minus{}2(x^2y\plus{}xy^2\plus{}x^2z\plus{}xz^2\plus{}yz^2\plus{}yz^2) \plus{} 6xyz)$.

Pierre.

[quote="pbornsztein"][quote="apollo"]$ 6(x^5 \plus{} y^5 \plus{} z^2)$[/quote]

I think you meant $ 6(x^5 \plus{} y^5 \plus{} z^5)$, am I wrong?

In that case
$ 6(x^5 \plus{} y^5 \plus{} z^5) \minus{} 5(x^3 \plus{} y^3 \plus{} z^3)(x^2 \plus{} y^2 \plus{} z^2)$
$ \equal{} (x \plus{} y \plus{} z)^2 (x^3 \plus{} y^3 \plus{} z^3 \minus{} 2(x^2y \plus{} xy^2 \plus{} x^2z \plus{} xz^2 \plus{} yz^2 \plus{} yz^2) \plus{} 6xyz)$.

Pierre.[/quote]
yes,I've edited it.


[quote="mhuarancca"]hallar todos los numeros primos $p$ , $q$ tal que:

$\frac{x^{3pq}-x}{3pq}$

es un numero entero , espero soluciones[/quote],
que es $x$, ??

Primero, enunciare una afirmación:
AFIRMACIÓN: Sean $p$ primo y $a$ un entero positivo. Entonces $p-1|a$ si y solamente si $x^{a}\cong 1mod(p)$ para todo entero $x$ no divisible por $p$.

Es fácil ver que $p, q> 2$, ya que en caso contrario tendriamos (para $x=2$) $3pq$ par y (para $x=2$) $2^{3pq}-2\cong 2 mod(3)$ (no divisible por $3pq$).
Si $p$ no divide a $x$, entonces $x^{3pq-1}\cong 1mod(p)$ y, por la AFIRMACIÓN, $p-1|3pq-1$$\implies$ $p-1|$

Creo que a jonfv le falto un pedazo de la solucion. Que sigue?

Bueno, ahora si va el post completo.
Primero, enunciare una afirmación:
AFIRMACIÓN: Sean $p$ primo y $a$ un entero positivo. Entonces,
 $p-1|a$ si y solamente si $x^{a}\cong 1mod(p)$ para todo entero $x$ no divisible por $p$.

Es fácil ver que $p, q> 2$, ya que en caso contrario tendriamos $3pq$ par y (para $x=2$) $2^{3pq}-2\cong 2 mod(3)$ (no divisible por $3pq$).
Si $p$ no divide a $x$, entonces $x^{3pq-1}\cong 1mod(p)$ y, por la AFIRMACIÓN,
$p-1|3pq-1\implies p-1|p(3q-1)\implies p-1|3q-1$
Analogamente, $q-1|3p-1$.
Por la simetría, podemos suponer $p\geq q$.
Sea $3q-1=k(p-1)$ y veamos los posibles valores de $k$:
Si $k\geq 5$; $3q-1=k(p-1)\geq 5(p-1)\geq 3q-1+2p-4>3q-1$, absurdo.
Si $k=4$; $3q-1=4(p-1)\geq 3q-1+p-3$ y entonces $p=q=3$.
Si $k=3$; $3|3(p-1)=3q-1$, absurdo.
Si $k=2$; $p-2|2(p-2)=3(q-1)|3(3p-1)\implies p-2|15$. Así, los posibles valores de $(p,q)$ son $(17,11)$ y $(5,3)$.

Es fácil verificar que todos los posibles valores dados para $(p,q)$ satisfacen las condiciones del problema.

Sobre la AFIRMACIÓN, basta demostrar que si $a$ es un entero positivo que satisface $x^{a}\cong 1 mod(p)$ para todo entreo $x$ no divisible por $p$, entonces $a\geq p-1$.
Por ahora, sólo se me ocurre una prueba algébrica para ello.
Sean $P,Q\in R[x]$ polinomios no nulos de una variable con coeficentes en $R$ (numeros reales) de grados $m$, $n$. Sabemos que:
$I)$ Existen $S,T\in R[x]$ tales que $P=S.Q+T$ y, $n> t$ o $T=0$, donde $t$ es el grado de $T$.
$II)$ Si $\alpha\in R$ es raíz de P, entonces existe $V\in R[x]$ tal que $P=(x-\alpha).V$.
$III)$ $P$ posee a lo sumo $m$ raíces reales distintas.
Es fácil probar $I)$ y $I)\to II)\to III)$.
El caso es que $R$ puede ser substituido por cualquier campo (conjunto con suma, producto y donde todo elemento posee inverso multiplicativo), por ejemplo, $C$ (complejos), $Q$ (racionales) o $\frac{Z}{pZ}$ (restos módulo $p$: $\overline{0},\dots,\overline{p-1}$).
En nuestro problema, el polinomio $\overline{1}x^{a}-\overline{1}$ posee $p-1$ raíces distintas $\overline{1},\dots,\overline{p-1}$ y entonces $a\geq p-1$


geometry
geometric transformation
homothety
ratio
circumcircle
reflection
Euler

Dunno if it's famous.
For a concyclic quadrilateral, draw 4 simson lines (the simson line with respect to D and triangle ABC, that respect to C and triangle ABD and so on)
Prove that these 4 lines concur at the anticenter of the quadrilteral.

Let $M,N$ be the midpoints of $DB,DC$ respectively, and let $\ell_M,\ell_N$ be the perpendiculars from $M,N$ to $AC,AB$ respectively. We have to show that $\ell_M,\ell_N$ concur on the Simson line of $D$ wrt $ABC$. When we apply a homothety of center $D$ and ratio $2$, this comes down to showing that the image of the Simson line of $D$ wrt $ABC$ through this homothety passes through the orthocenter of $ABC$, which is a very well known property of the Simson line.

how to prove this well-known property of the simson line?

Let $ABC$ be a triangle, $P$ a point on its circumcircle, $H$ its orthocenter, $P_b,P_c$ the reflections of $P$ in $CA,AB$ respectively. We have to show that $H,P_b,P_c$ are collinear.

Let $B',C'$ be the intersections of the altitudes from $B,C$ respectively with the circumcircle of $ABC$ (the intersections different from $B,C$). Now $HP_b$ is the reflection of $PB'$ in $CA$, and $HP_c$ is the reflection of $PC'$ in $AB$. The rest should be simple angle chasing.

[quote="siuhochung = Soarer"]For a concyclic quadrilateral, draw 4 simson lines (the simson line with respect to D and triangle ABC, that respect to C and triangle ABD and so on)
Prove that these 4 lines concur at the anticenter of the quadrilteral.[/quote]

I just wanted to remark that this problem is part of a somewhat broader theory. In the note "The Euler point of a cyclic quadrilateral" on [url=http://de.geocities.com/darij_grinberg]my website[/url], I showed some facts from this theory. In particular, Theorem 2 says that the Euler point of the cyclic quadrilateral ABCD lies on the Simson lines of the vertices A, B, C, D with respect to the triangles BCD, CDA, DAB, ABC, and Theorem 7 shows that this Euler point coincides with the anticenter; hence, your problem is solved. Also, the Euler point aka anticenter is proved to lie on the nine-point circles of triangles BCD, CDA, DAB, ABC, and a few more properties are shown (yet this is not all one can say).

  darij


A rising number, such as $ 34689$, is a positive integer each digit of which is larger than each of the digits to its left. There are $ \dbinom{9}{5} \equal{} 126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, the $ 97$th number in the list does not contain the digit

$ \textbf{(A)}\ 4\qquad
\textbf{(B)}\ 5\qquad
\textbf{(C)}\ 6\qquad
\textbf{(D)}\ 7\qquad
\textbf{(E)}\ 8$

[hide]The smallest numbers in the list are the ones that begin with $ 1$. For five-digit rising numbers beginning with $ 1$, the remaining $ 4$ digits are taken from $ (2,3,4,5,6,7,8,9)$.

For every $ 4$-number combination, there exists exactly $ 1$ permutation such that the numbers are in increasing order. Thus there are $ \binom{8}{4} = 70$ five digit rising numbers that begin with $ 1$.

Since $ 70<97$, we know the $ 97$th number does not begin with $ 1$. We move on to five-digit rising numbers starting with $ 2$. Using the same reasoning as above, there are $ \binom{7}{4} = 35$ five digit rising numbers that begin with $ 2$.

Since $ 70+35 = 105 > 97$, we know the $ 97$th number begins with $ 2$.

Now we focus on the second digit. There are $ \binom{6}{3}=20$ numbers beginning with $ 23$. $ 70+20 = 90 < 97$ so the $ 97$th number does not begin with $ 23$.

There are $ \binom{5}{3} = 10$ numbers that begin with $ 24$. Since $ 70+20+10 = 100 > 97$, we know the $ 97$th number begins with $ 24$.

We have narrowed our choices enough so now we can just list them out to find the $ 97$th number. The first number to begin with $ 24$ is $ 24567$, and it is the $ 91$st number. It follows that:

\begin{align*}
91 &\leftrightarrow 24567 \\
92 &\leftrightarrow 24568 \\
93 &\leftrightarrow 24569 \\
94 &\leftrightarrow 24578 \\
95 &\leftrightarrow 24579 \\
96 &\leftrightarrow 24589 \\
97 &\leftrightarrow 24678
\end{align*}

The $ 97$th number is $ 24678$, which does not contain the digit $ 5$. Answer: $ \boxed{\textbf{(B)}}$[/hide]

Can somebody explain the logic behind $\binom84=70$ rising numbers starting with $1$?
Same with $\binom74$ starting with 2...?
TIA

Once we choose the 1, we can choose 8 digits larger than 1. Now, we can pick them in any order, since we will order them eventually, by our restriction, so we just choose 4 out of thee 8, hence $\binom{8}{4}$.

Does anyone know the source for this? It seems like an AMC 10 problem.

Seriously, #24? Lol, that's easy for #24, even if there are 30 problems.


MATHCOUNTS
geometry
rectangle
ratio
perimeter
geometric transformation
rotation

there are a bunch of problems that i don't know how to do and i want to learn how to do all of these before mc state. so if you guys could post solutions it would be super helpful and appreciated! these are just a few....

1. in rectangle ABC, H is the midpoint of BC, E lies on AD, F lies on AB. in rectangle CEFG, H lies on FG and HG=3 inch. angle DEC=45. what is the number of square inches in the positive diff between the areas of the two rectangles?

2. the minute hand ona 12-hour clock is 12 cm long and the hour hand is 6cm long. what is the ratio of the greatest distance traveled by any point on the hour hand in 12 hours to the greatest distance traveled by any point on the minute hand in 12 hours?

3. a circle is tangent to two sides of a square and its diagonal. given that the length of a side of the square is four units, how many units are in the radius of the circle? express your answer as a common fraction in simplest radical form.

4. 2 factors have a product of 3*9^2x - 6*27^x. one of the factors is 3^x - 2. what is the other factor?

thanks!

Number 4:
edit: it seems my help is not wanted here.

Honestly, MMM, you are solving problems in the Mathcounts forum now? Even if eqn1190 needs help, it would only take a few more minutes for some less qualified people to solve them. :)

Uh, as far as I know it, when people say they need help, everybody is entitled to answer them.. I mean, if people want to try them as challenges, thats fine, but people want to know how to do them as fast as possible..

Number 1:[hide]

Using the fact that the sum of the angles in a straight line is 180 :^o:, we find that AEF=180-FEC-DEC=180-90-45=45.  Using the fact that the sum of the angles in a triangle is 180 :^o:, we find that AFE=180-FAE-AEF-180-90-45=45.  Using these rules, we find that AEF=AFE=BFH=(by vertical angles)CHG=GCH=45 :^o:.

Since, in a 45-45-90 triangle, the legs are congruent and the hypotenuse is  :rt2: times the length of the legs, we find that:

CG=3

CH=3 :rt2:

BC=2(CH)=6 :rt2:

EA=3/ :rt2: =3 :rt2: /2

FA=EA=3 :rt2: /2

AB=FA+FB=FA+CH=9 :rt2: /2

FG=3+FH=3+(FB)* :rt2:=9

Since we have AB,BC,CG, and FG, we find the areas of the rectangles and subtract:

(AB)(BC)-(CG)(FG)

54-27

27[/hide]

Number 3:

[hide]

This circle is inscribed in a right,45-45-90 triangle with leg 4 and hypotenuse 4 :rt2: . As proven in AoPS, volume I, the area of a triangle can be given by A=bh/2, where b and h are the measures of a triangle's base and height.  AoPS I also proves that the area of a triangle can be given by A=rs, where r is the radius of the triangle's incircle and s is half the perimeter, or the semiperimeter.  So we want to find r.

We know that b and h both equal 4, and that s= (4+4+4 :rt2: ), or 8+2 :rt2: .  So we have:

4*4/2=r(8+2 :rt2: ).

8=r(8+2 :rt2: )

r=8/(8+2 :rt2: )

r=(8-2 :rt2: )/7 [/hide]

Number 2:[hide]

The tip of the hour hand travels a distance of one full rotation, or the circumference of a circle with radius 6. It therefore travels a distance of 2 :pi: r, or 12 :pi: .

The tip of the minute hand travels a distance of 12 full rotations, or 12 times the circumference of a circle with radius 12. It therefore travles a distance of 12(2 :pi: r), or 288 :pi: .

Finally, we take 12 :pi: /288 :pi: and get 1/24[/hide]

By the way, where are these problems from?

these are national and state sprints i believe.

yeah, i just did problem three a couple days ago, it was a state sprint from, like 95 or something

[quote]Since, in a 45-45-90 triangle, the legs are congruent and the hypotenuse is  times the length of the legs, we find that:

CG=3
CH=3
BC=2(CH)=6
EA=3/  =3  /2
FA=EA=3  /2
AB=FA+FB=FA+CH=9  /2
FG=3+FH=3+(FB)* =9
[/quote]

how do you know what cg is...they only tell you that fg=3. also, how do you know if hg is perpendicular to bc? lastly, you said that hcg = bfg by way of vertical angles. how are they vertical angles?


calculus
integration
real analysis
calculus computations

I' m not sure this is a suitable place to talk about this. :oops:
Does anyone know what the name of this below mathematical formula is ?If possible,please give me full information about it,who is the author and any applications of it .
[img]http://img396.imageshack.us/img396/8657/49152651zq6.jpg[/img].
Thanks in advance!

Well if you take the integral from negative to positive infinity, you get the convolution of w and k


geometry
3D geometry
tetrahedron
vector
geometry unsolved

A set $ E$ of points in the 3D space let $ L(E)$ denote the set of all those points which lie on lines composed of two distinct points of $ E.$  Let $ T$ denote the set of all vertices of a regular tetrahedron. Which points are in the set $ L(L(T))?$

Let the tetrahedron be ABCD.
Consider [b]x=AB, y=BC, z=CA.[/b]
clearly any point P in space with [b]AP=p[/b] can be given by
[b]p[/b]=a[b]x[/b]+b[b]y[/b]+c[b]z.[/b]

Now,$ L(T)$ consists of:
all points with position vector  $ \alpha x$
all points with position vector  $ \beta y$
all points with position vector  $ \gamma z$
all points with position vector  $ \alpha_1 x \plus{} (1\minus{}\alpha_1)y$
all points with position vector  $ \beta_1 z \plus{} (1\minus{}\beta_1)y$
all points with position vector  $ \gamma_1 x \plus{} (1\minus{}\gamma_1)z$
So $ L(L(T))$ consists of points of the form
$ \alpha l x\plus{}(1\minus{}l)(\beta_1 z \plus{} (1\minus{}\beta_1)y) \\ \beta ym\plus{}(1\minus{}m)(\gamma_1 x \plus{} (1\minus{}\gamma_1)z) \\ \gamma z n\plus{}(1\minus{}n)(\alpha_1 x \plus{} (1\minus{}\alpha_1)y)$

Now let us prove that at least one of the above three equation gives a[b]x[/b]+b[b]y[/b]+c[b]z.[/b]
Now out of a,b,c at least one is nonzero.[Obviously pts. A,B,C,D are not considered as they already belong to given set]. let c be nonzero.
Also let b+c be not equsl to 0 or 1.
Then
$ \alpha l x\plus{}(1\minus{}l)(\beta_1 z \plus{} (1\minus{}\beta_1)y)\equal{}ax\plus{}by\plus{}cz.$ gives
$ \alpha\equal{}\frac{a}{1\minus{}(b\plus{}c)} \\ \beta_1\equal{} \frac{b}{b\plus{}c} \\ l\equal{}1\minus{}(b\plus{}c)$ and we are done. Easy to check that if c+b is 0 or 1, they are in L(L(T)).


Can someone please tell me how to get the three dots logical symbols, because and therefore, in LaTex? Thanks.

$\therefore$

Closely related is $\because$


modular arithmetic
quadratics
number theory proposed
number theory

Friends, here is a problem which i created long time when there is a topic about a 1998 Bulgary problem. Now let me remind it:

[b]Problem[/b]: Let $m,n$ are positive integers such that $3m|(m+3)^n+1$. Prove or disprove that $m\equiv2\pmod{12}$.

[b]Note[/b]: It's well-known that $\frac{(m+3)^n+1}{3m}$ is an odd number!

I hope to see some nice detail solutions! :D

This seems boring problem, friends ?  :(

[quote="Lilian"]We obtain that n is odd and that m=M12+2[/quote]

really it's the hardest solution which i know, can u explain more details man ? :D

Here is my approach:
Since $3|3m|(m+3)^n+1\to n$ is odd because 3 is a prime which has form 4k+3..
According Newton's expanding we deduce that:$3m|m^n+3^n+1$
  This yields that:$m|3^n+1$.Then if prime $p|m\to p|3^n+1\to 3^n=-1(\mod p)\to 3^{n+1}=-3(\mod p)\to -3$ is quadratic residue mod p which impies that:$p$ has form $6k+1$
  Hence m has also this form.If k is odd->m has form 12t+7,unless m has one 12t+1
 So my result is  wrong,i.e $m\neq 2 (\mod 12)$
To Prulycomplex:Could u post your solution for my problem?I haven't yet solved it... ;)
http://www.mathlinks.ro/Forum/viewtopic.php?t=57096&highlight=

[quote="keira_khtn"]Here is my approach:
 So my result is  wrong,i.e $m\neq 2 (\mod 12)$
[/quote]

Yes your's wrong!
Here's what i though:

1) firstly we will show that $m$ is even. Indeed, if not as works of [b]keira-khtn[/b] we get $m\equiv1\pmod{6}$. But then $(m+3)^n+1\equiv 2\pmod 3$ so $(m+3)^n+1$ is not divisible by $3$. That's why $m$ must be odd.

2) Now we write $m=2k$. If $k$ is even then $(m+3)^n+1\equiv 3^n+1\equiv 2\pmod{4}$ so $(m+3)^n+1$ is not divisible by $m$. So $k$ must be odd. Then cause $k\mid 3^n+1$ and use the same above we have $k\equiv1\pmod{6}$. So $m\equiv2\pmod{12}$.
We're done! :D ;)

I think there something unclear in pluricomplex'solution. Here is my approach:

[b] lemma [/b]. If $p=3k-1$ is a prime and $p|3^2+y^2$ for some $(x,y)$ then $p|x,p|y$.

Proof: If $p$ does not divide $xy$,  $-3x^2\equiv\ y^2\leq >\ -3\equiv\frac{x^2}{y^2}$ then $-3$ is quadratic residue mod $p=3k-1$ (absurd). So the lemma is proven.

Let get back to our problem. We must have $3m| m^n+3^n+1$.  It easy to see that $m\equiv\ -1 [3]$ and n is odd (because $m^n+1\vdots\ 3$).let $n=2j+1$.
we have 2 cases:
[b] case 1 [/b]: $m=2$ :trivial
[b] case 2 [/b]: $m>2$: then let $q$ be a atribary odd prime divisor of m. So if $q$ in the form $3l-2$, $p|3^n+1=3*3^{2j}+1$ which leads to $p| 1$ (absurd).
So any odd prime divisor of $m$ are in the form $3k+1$. But $m\equiv\ -1 [3]$. So we must have $m=2^{2i+1}*t$ where $t$ is the product of primes in the form $3k+1$. But we have $3^n+1=3*3^{2j}+1\equiv\ 4 [8]$ so $i=1$, which completes our proof.

[this is my second post for this one, i wonder where my first one  was???!!!]


Let $ (x_n) (n \equal{} 1,2,...)$ be a sequence,determined by the following recursive formula:
$ x_1 \equal{} \frac {1}{2}$
$ x_{n \plus{} 1} \equal{} x_{n} \minus{} x_{n}^2 \plus{} x_{n}^3 \minus{} x_{n}^4 \plus{} ... \plus{} x_{n}^{2007} \minus{} x_{n}^{2008} (n\in N*)$
Find the limit $ lim_{n\rightarrow \infty} nx_{n}$

$ x_{n\plus{}1}(1\plus{}x_n)\equal{}x_n\minus{}x_n^{2009}$.
Because $ 0<x_1<1$ we get $ 0<x_{n\plus{}1}<x_n$.
Let $ y_n\equal{}nx_n$, then $ y_{n\plus{}1}\equal{}\frac{(n\plus{}1)(x_n\minus{}x_n^{2009})}{1\plus{}x_n}<\frac{(n\plus{}1)y_n}{n\plus{}y_n}$.
Therefore from $ y_n<1$ we get $ y_{n\plus{}1}<1$. It give $ y_n<1 \ \forall n\in N$, because $ y_1<1$.
Let $ z_n\equal{}1\minus{}y_n$, then $ z_{n\plus{}1}\equal{}\frac{z_n\plus{}(n\plus{}1)x_n^{2009}}{1\plus{}x_n}>\frac{nz_n}{n\plus{}1}$.
Therefore $ z_n>\frac{z_1}{n}\equal{}\frac{1}{2n}$.
It give $ 1\minus{}\frac{1}{2n}<y_n<1$, therefore $ \lim_{n\to\infty}y_n\equal{}1$.

already posted : http://www.mathlinks.ro/viewtopic.php?t=191955


I got this from my trig book and I did get answers but my book doesn't have answer key for these and I'm not sure whether my answer's right. I guessed them but yeah..  :?

1. Our architect, Joe Miklinon, is designing a new parking garage for the city of Kuinkin. The floors on the garage are to be 10 feet apart. The exit ramps between each pair of floors are to be 75 feet long. What is the measurement of the angle of elevation of each ramp?

[hide="My Answer"]
If I let angle of elevation to $\theta$, then:

$\cos \theta = \frac {10}{75}\\
\theta = 82^\circ$[/hide]

2. A car is traveling up a long hill on a road that makes me an angle of 8 degrees with the horizontal. How far will the car have to travel to reach the top of the hill if the horizontal distance traveled is 715 feet?

[hide=" My answer"]

$\cos 8^\circ = \frac {715}{x}\\
x = 722 \text {feet}$[/hide]

For number 1, I think you would use $\sin$ instead of $\cos$.

Your answer looks correct on number 2 though.

1)[hide]I get a different answer for #1. I think you're misunderstanding the question a bit. If you draw it out, you get a right triangle with a height of 10 and a hypotenuse of 75. Since the angle of elevation $\theta$ is [i]opposite[/i] the height, we instead have:

$\theta=\sin^{-1}(\frac{10}{75})\approx7.66^{\circ}$
[/hide]

2) Looks correct to me :D


inequalities
function
integration
calculus
derivative
inequalities unsolved

Let $p$, $q$ be two integers, $q\geq p\geq 0$. Let $n \geq 2$ be an integer and $a_0=0, a_1 \geq 0, a_2, \ldots, a_{n-1},a_n = 1$ be real numbers such that \[ a_{k} \leq \frac{ a_{k-1} + a_{k+1} } 2 , \ \forall \ k=1,2,\ldots, n-1 . \] Prove that \[ (p+1) \sum_{k=1}^{n-1} a_k^p \geq (q+1) \sum_{k=1}^{n-1} a_k^q . \]

[quote="xirti"]Remark: It is equivalent with :

Given a convex function $f: [0,1]\to \mathbb{R}$ with $f(0)=0, f(1)=1$, prove that $(p+1)\int_0^1f^p(x)dx\le (q+1)\int_0^1f^q(x)dx$.

Can someone prove the ineq for $a_k=\frac k n$ and $q=p+1$? In contest I reduced the problem to this case but I didn't have enough time to finish it. If the answer is positive I can write my (piece of ) solution.[/quote]
Nice idea.
what do you mean by "If the answer is positive "?
means $f(x)>0$
PS:can you post you solution?
 :)

[quote="xirti"] Can someone prove the ineq for $a_k=\frac k n$ and $q=p+1$? [/quote]

If the answer of this question is positive... (positive means affirmative; yes)

Anybody interests?I read the official solution of this problem,I think it was amazing.And this is why I bounced the topic to the top to see your nice solutions :)

[quote="gb2124"]Where do you find the official solutions,Can you give a link?[/quote]
It's for you :)

It's actually quite simple in the integral version mentioned above.
The inequality we have to prove is \[ (p\plus{}1)\int_0^1 f^p(x)dx\geq (q\plus{}1)\int_0^1f^q(x)dx\]
Where $ f$is convex and increasing. I will assume that $ f$ is differentiable even though it can be avoided  :) The inequality is equivalent to \[ \int_0^1x^qdx\int_0^1f^p(x)dx\geq \int_0^1x^pdx\int_0^1f^q(x)dx\]
and in turn equivalent to
\[ \int_0^1\int_0^1 f^q(x)f^p(y)f'(x)dxdy\geq \int_0^1\int_0^1 f^q(x)f^p(y)f'(y)dxdy\]
which can be rewritten as
\[ \int_0^1\int_0^1 f^p(x)f^p(y)(f^{q\minus{}p}(x)\minus{}f^{q\minus{}p}(y))(f'(x)\minus{}f'(y))dxdy\geq 0\]
Because $ f$ is convex $ f'$ is increasing and we are done.

Dear Albanian eagle:
I don't quite understand the solution(since i'm not good at integral).
Since ak is non continuable,how can you use that?
And furthermore i don't understand the last sentence you wrote.
Why is that inequality is true since f is convex and f' is increases.
Can you explain more to me?
Thanks in advance.
And to hxy09
Thanks for the material.

dear gb2124, I proved the inequality claimed in the second post by xirti. He claimed an equivalence to the original problem.
From my solution you can get the original problem if you convert each step to it's discrete analog. Keep in mind that integrals are just "continuous summations". Derivatives will correspond to finite differences etc.
If you do that you'll see a motivation as to why one uses the Abel summation in the official solution.
As to the last line if you look carefully the function being integrated is non-negative and therefore the integral is non-negative.

Dear albanian eagle
Oh yes!
I forgot the condition that f is also increasing too.
I think you are right.


abstract algebra
invariant
vector
group theory
superior algebra
superior algebra unsolved

Let $ \mathcal{K}_1, \cdots, \mathcal{K}_r$ be the conjugacy classes of the finite group $ G$ and for each $ \mathcal{K}_i$ let $ K_i$ be the element of $ RG$ that is the sum of the members of $ \mathcal{K}_i$. Prove that an element $ \alpha \in RG$ is in the center of $ RG$ (i.e. commutes multiplicatively with every element of $ RG$) if and only if $ \alpha \equal{} a_1K_1 \plus{} \cdots \plus{} a_rK_r$ for some $ a_1, \cdots, a_r \in R$.

The implication $ \Leftarrow$ is fairly easy but I have no clue how to approach $ \Rightarrow$.

$ \alpha$ is in the center if and only if it is fixed under conjugation, and every element fixed under conjugation is of the form $ \frac{1}{|G|} \sum_{g \in G} g \beta g^{\minus{}1}$.  Now split the summation up into conjugacy classes.

What result are you using there? I'm not sure how that expression is even in $ RG$, since $ |G|$ is a number. Also, where does $ \beta$ live?

Sorry, I was thinking of the group algebra.  In any case, this is a standard trick used in representation theory, although I wanted a slightly different argument: it is true more generally that given a group action of a group $ G$ on a set $ X$ linearized to a representation on the free $ R$-module over $ X$, the fixed points of the action of $ G$ are linear combinations of (sums of every element in) orbits of this action.  (Here $ X \equal{} G$, the action is conjugation, and the orbits are conjugacy classes.)

This is straightforward.  Given an orbit $ O$, linear combinations of elements of $ O$ form an invariant subspace.  $ G$ acts transitively on $ O$, so any vector that isn't a scalar multiple of $ \sum_{s \in O} s$ can't be fixed by the action of $ G$.

I almost understand what you're saying, but you're using a few words I'm unfamiliar with. I'm working through an algebra text sequentially and at this point only the most basic ring theory has been introduced (no modules, representations, etc.). I have done work with group actions, so I'm closest to following you on that stuff. If you could explain on a more elementary level or restate the above in less advanced terminology it would help a lot.

Take a specific example like G the symmetric group on 3 points and R=Z the integers.  Let s=(1,2,3) and t=(1,3,2) be the two distinct elements of G of order 3, and let a=1, b=2 be two distinct elements of R.

Then a*s + b*t = 1*(1,2,3) + 2*(1,3,2) is in RG.  Is it in the center of RG?

Well if it was, then for any g in G, we should find that g*(a*s + b*t) = (a*s+b*t)*g.  Since g is invertible in RG, we can move the g's to one side only:  we should find that a*s + b*t = g^-1 * (a*s+b*t) * g.

Ok, well what happens when we try this with g=(1,3)?

Well, g^-1*(a*s + b*t)*g = a*( g^-1*s*g ) + b*( g^-1*t*g ), so the g does not change the "coefficients", it just changes who they are coefficients of.

Specifically, (1,3) * (  1*(1,2,3) + 2*(1,3,2) ) * (1,3) = 1*(1,3,2) + 2*(1,2,3).

Uh oh, using g=(1,3) switched (1,2,3) and (1,3,2) but left their coefficients in place.  Since the coefficients were different, the result was different and a*s + b*t was not in the center.  If we had chosen a=b, then we would not be able to tell the coefficients had moved around, and the element would have been in the center.

Let $ \alpha \equal{} \sum_{h\in G}\alpha_h\cdot h$. Then, we have the equivalence of assertions:

($ \alpha$ lies in the center of $ RG$)
$ \Longleftrightarrow$ ($ \alpha u \equal{} u\alpha$ for every $ u\in RG$)
$ \Longleftrightarrow$ ($ \alpha g \equal{} g\alpha$ for every $ g\in G$) (because every $ u\in RG$ is a formal linear combination of elements of $ G$)
$ \Longleftrightarrow$ ($ \sum_{h\in G}\alpha_h\cdot h\cdot g \equal{} g\cdot\sum_{h\in G}\alpha_h\cdot h$ for every $ g\in G$)
$ \Longleftrightarrow$ ($ \sum_{h\in G}\alpha_h\cdot hg \equal{} \sum_{h\in G}\alpha_h\cdot gh$ for every $ g\in G$)
$ \Longleftrightarrow$ ($ \sum_{i\in G}\alpha_{ig^{ \minus{} 1}}\cdot i \equal{} \sum_{h\in G}\alpha_{g^{ \minus{} 1}i}\cdot i$ for every $ g\in G$)
(here we substituted $ i \equal{} hg$ in the left hand sum and $ i \equal{} gh$ in the right hand sum)
$ \Longleftrightarrow$ ($ \alpha_{ig^{ \minus{} 1}} \equal{} \alpha_{g^{ \minus{} 1}i}$ for every $ g\in G$ and every $ i\in G$)
$ \Longleftrightarrow$ ($ \alpha_{u} \equal{} \alpha_{v}$ for any two elements $ u$ and $ v$ of $ G$ that lie in the same conjugacy class)
(because any two such elements $ u$ and $ v$ have the form $ ig^{ \minus{} 1}$ and $ g^{ \minus{} 1}i$ for suitably chosen $ i$ and $ g$, while conversely, $ ig^{ \minus{} 1}$ and $ g^{ \minus{} 1}i$ are conjugated for any $ g\in G$ and $ i\in G$)
$ \Longleftrightarrow$ ($ \alpha \equal{} a_1K_1 \plus{} a_2K_2 \plus{} ... \plus{} a_rK_r$ for some elements $ a_1$, $ a_2$, ..., $ a_r$ of $ R$).

While t0rajir0u's approach is quicker, it requires $ \left|G\right|$ to be invertible (or at least a non-zero-divisor) in $ R$, so it is not as general as it could be.

  darij


function
number theory unsolved
number theory

Find all pairs of positive integers $m,n$ such that $\phi(n)|m$ and $\phi(m)|n$, where $\phi$ denotes Euler's function.

Rust you are writing like you have seen this problem
before and you only have to check that the solution is
this,but how are the other solutions not good ?
i mean how did you get to your answer ?

It i not hard. Let $k=ord_{2}(n)$, then $ord_{2}(\phi (n))=k-1+\sum_{p|n}ord_{2}(p-1)\le r=ord_{2}(m)$ and $r-1+\sum_{p|m}ord_{2}(p-1)\le k$.
Therefore $|k-r|\le 1$ and n,m had no more one odd prime divisor. It is easy to check, if p is odd prime divisor nm, then p=3.

[quote="Rust"]$ord_{2}(\phi (n))=k-1+\sum_{p|n}ord_{2}(p-1)$[/quote]

How come do you have this property?


inequalities
floor function
number theory proposed
number theory

We know that
$ a_{1}\equal{}2, a_{2}\equal{}3, a_{3}\equal{}5, a_{4}\equal{}6, a_{5}\equal{}7, a_{6}\equal{}8, a_{7}\equal{}10...$ Natural numbers without squares.
Find $ a_{n}$

[quote="inny"]We know that
$ a_{1} \equal{} 2, a_{2} \equal{} 3, a_{3} \equal{} 5, a_{4} \equal{} 6, a_{5} \equal{} 7, a_{6} \equal{} 8, a_{7} \equal{} 10...$ Natural numbers without squares.
Find $ a_{n}$[/quote]

 Let $ \sqrt{a_n}\equal{}p_n\plus{}y_n$ with $ p_n\equal{}[\sqrt{a_n}]\in\mathbb N$ and $ y\in(0,1)$

We get $ n\equal{}a_n\minus{}[\sqrt{a_n}]\equal{}(p_n\plus{}y_n)^2\minus{}p_n$ and so $ p_n^2\minus{}p_n<n<(p_n\plus{}1)^2\minus{}p_n$

For a given $ n$, this last inequality has always a unique solution $ p_n$ (suppose $ p$ and $ p\plus{}1$ are solutions and it would imply $ (p\plus{}1)^2\minus{}(p\plus{}1)<n<(p\plus{}1)^2\minus{}p$, which is impossible).

But $ p_n^2\minus{}p_n<n<(p_n\plus{}1)^2\minus{}p_n$ $ \iff$ $ \frac{\sqrt{4n\minus{}3}\minus{}1}2<p_n<\frac{\sqrt{4n\plus{}1}\plus{}1}2$

and so $ p_n\equal{}1\plus{}[\frac{\sqrt{4n\minus{}3}\minus{}1}2]$

Hence the answer : $ \boxed{a_n\equal{}n\plus{}1\plus{}\left\lfloor\frac{\sqrt{4n\minus{}3}\minus{}1}2\right\rfloor}$


limit
logarithms
trigonometry
calculus
calculus computations

Ascertain $ \lim_{x\to\frac {\pi}{2}}\frac {\ln [\sin (4m + 1)x]}{\ln [\sin (4n + 1)x]}$, where $ \{m,n\}\subset \mathcal Z\ .$