In a small town of 50,000 inhabitants , it has been counted that 42 percent of the males and 28 percent of the females married people from their own town. Assuming these numbers have remained fairly constant over the years , how many males are there in the town ?

[hide] Let $ x$ be the number of males, and $ y$ the number of females. Since only those males and females who lived in the town could have married each other, we have

$ x\frac{42}{100}\equal{}y\frac{28}{100}$

$ \implies 42x\equal{}28y \implies y\equal{}\frac32x$.

Since $ x\plus{}y\equal{}50000$, we have

$ \frac32x\plus{}x\equal{}50000 \implies \frac52x\equal{}50000 \implies x\equal{}\boxed{20000}$.[/hide]


[quote="sks119"]A train moving with a constant velocity of 27.2 m/s travels 146 m north in 12.5 s and an undetermined distance to the west. Calculate the direction of the train's motion counterclockwise relative to north? How far west has the train traveled in this time?
can someone please give me a hint on how to go about doing this question?[/quote]
Apply the Pythagorean theorem.

$(27.2)^{2}= v_{n}^{2}+v_{w}^{2}$

$v_{n}= \frac{146}{12.5}= 11.68$

$\cos\theta = \frac{11.68}{27.2}$

$\vec{w}=v_{w}\cdot (12.5)\hat{w}$


[quote="Jimmy"]Given triangle $ABC$, let $O_{A}$ be the excircle of $ABC$ opposite side $A$ with center $O$, prove that $OA$ bisects angle $BAC$.[/quote]
[hide]Draw perpendiculars from O to the extensions of AB and AC. Call the points where the radii of the excircle meet the extensions B' and C', respectively. Then because AB' and AC' are tangents, they are equal. Two angles are 90, and radii are equal (B'O and C'O). By SAS AB'O is congruent to AC'O. Then the angles are all equal so then angle A is bisected.[/hide]

[hide="Hmmm?  :?: "]Isn't the excenter exactly [i]defined[/i] as the intersection point of the bisectors of one internal and remaining two external angles?[/hide]

You get a choice: either there is the excircle (the circle tangent to 1 side and 2 extensions of sides) and then you can ask about its center, or there is the excenter (the intersection of an internal and two external angle bisectors) and you can ask about circles centered at it.  So the problem is not definitional as given.

[quote="Farenhajt"]But opposite any given vertex the excircle is unique (hence there are three of them for any given triangle), therefore it's center necessarily must be the excenter, which lies on the vertex's internal angle bisector.[/quote]

Which is a theorem, not a definition.

As much as "prove that internal bisectors are concurrent" is a theorem - I mean, sure it IS a theorem, but it goes into triangle geometry everyday basics.

[hide]when we draw $OB , OC , OA$we see that $OA$ is prependicular to side $BC$ and $\angle OBC=\angle OCB= \alpha$becuase $\triangle OBC$ is equilateral triangle then $\angle BAO=\angle CAO=2\alpha-90$AND its prove but maybe i didn't understand it correctly but if i its true i think its very easy!!!!!!!!!!!!!!![/hide]

Yes, you're right this was extremely easy.  I was messing around with Geometer's Sketch Pad, and thought this looked interesting before I tried to prove it. Sorry about that.

no problem!!!because maybe its not so easy for someone!!!


Need help solving the following problem:

In how many distinct ways can you arrange the letters VICISSITUDES?

This is a classic counting problem that will appear many times in math contests.

[hide="Solution"]The method we will adopt is Correcting for Overcounting.

First, lets count the total number of ways to rearrange the letters, without distinction. For the first letter, there are 12 letters to choose from...for the second letter, since you've chose one already, there's only 11 to choose from...and so on.... Now we multiply the numbers together...getting $ 12*11*.....*2*1 \equal{} 12!$ if you knew already.

But, there are some letters, which you noticed, I hope, that are repeated. Let's look at the S's for example, and call them S(1), S(2) and S(3). VICIS(1)S(2)ITUDES(3) is different from VICIS(2)S(1)ITUDES(3), but if the S's had no distinction, the two spellings would be the same.

It turns out, that the $ n$ number of times a letter repeats, we have to divide by $ n!$ to correct for overcounting.

So...in our word here, the letter I has repeated 3 times, and the letter S has also repeated 3 times. Therefore, we have to divide our total, $ 12!$ by $ 3!$ TWICE.

Our equation looks like $ \frac {12!}{3!*3!} \equal{} \boxed{13,305,600}$[/hide]

[quote="Suou"][quote="SweetMath"]How may four letter words can you make from VICISSITUDES? Is there any formula?[/quote]
It is $ \dbinom {12}{4}$

I am with a question, how many words without I's consecutives are there?[/quote]

I don't think that is right because there are multiple I's and S's.

Yes, it is wrong. I can not find a formula, i believe that you need separate in cases.

As LincolnMS said, you have to divide n times for every time a letter gets repeated.  So I guess the formula would be something like
$ \frac{n!}{x_1!*x_2!*x_3!*...}$
where n is the amount of letters in the word and $ x_1$ is the amount of repetitions of the first repeating letter, $ x_2$ is the amount of repetitions of the second repeating letter, etc.

I'm not sure if this is correct, though.  :|

I try for the case $ AABC$ choose $ 3$ letters, are $ 12$ cases $ 3.2.1 \plus{} AAB \equal{} 3 \plus{} AAC \equal{} 3$, then $ 6 \plus{}3\plus{}3\equal{}12$ , if i put letter $ D$, $ AABCD$ there are $ 33$ cases $ \equal{} 12\plus{}6\plus{}4\plus{}4\plus{}7\equal{}4.3.2 \plus{} 3\plus{}3\plus{}3$, then for VICISSITUDES are $ 8.7.6.5 \plus{} 8.4\equal{}1712$
I think that is right   :maybe:

I do believe you'll have to go case by case, snowpaw, because not EVERY SINGLE 4-letter word contains the repeated letter!  :P

First case: All letters are different. $ 8*7*6*5\equal{}1680$

Second case: 2 letters are the same....For each pair, either I's or S's, there are $ \binom{7}{2}\equal{}21$ possible choices. Of those choices, each one has $ 6*2\equal{}12$ possible arrangements. Plus, there are 2 different pairs, either 2 I's or 2 S's. So, it's $ 21*6*2\equal{}252$

Third Case: Two pairs are both in the word. $ \binom{4}{2}\equal{}6$

$ 6\plus{}1680\plus{}252\equal{}\boxed{1938}$ I'm pretty sure.

[quote="LincolnMS nerd herd"]First case: All letters are different.
Second case: 2 letters are the same....
Third Case: Two pairs are both in the word...
$ 6 \plus{} 1680 \plus{} 252 \equal{} \boxed{1938}$ I'm pretty sure.[/quote]

You didn't count for three Is and three Ss, which would increase the word count by 2*4*7 = 56

Ahh, good point, Sweetmath. It's good to have someone always there to check your answers. then it'll be $ 1938\plus{}56\equal{}1994$....I think.


I need laughs! I need you to post jokes+riddles for me and others to laugh and work their brains.

This definitely belongs to Games&Fun Factory.

Answers: Do not look unless u really have to.
[hide] im serious [/hide]
[hide] dont look unless u have solved all the riddles. [/hide]
[hide] ok here goes [/hide]
[hide] 1. All the men were married so not a "single" man got wet--hahahaha
2.she had put sugar in her tea previously, she tasted the new cup and found the tea was sweet.
3.coal
4.a river
5.counterfeit money
6.NOON
7.only once, the 2nd time u r subtracting from 20
8.he was completely bald
9.nothing
10. a match [/hide]
 :roll:  :P  :rotfl:

math jokes: an engineer, a physician, and a mathematician are betting on horses. the physician's horse comes in last. "I don't understand it," he exclaims. "I have determined the health and strength of each horse!" The engineer's horse does a little better, but still fails miserably. "What! I determined the strength of the horses' hooves in relation to the properties of the track."
They both look at the mathematician whose horse came in 1st. He says, "Well, first I assumed all horses were identical and spherical..."

[quote]Three turtles are racing on a non-circular track. Once the race is over, one turtle said that when he finished, he saw 0 turtles ahead of him. Another turtle said that when he finished, he saw 1 turtle ahead of him. However, the other turtle said he also saw 1 turtle ahead of him when he finished. How could this be? [/quote]

The turtle that saw 0 turtles ahead of him got first and the two turtles that saw 1 turtle ahead of them tied?

or....

one of the turtles that said they saw one turtle lied ;]

mz94 is right.. i heard that one a long time ago.

[quote="erabel"]@ pacman: Ark. And Noah had the Ark. Not Moses.[/quote]

Thanks! :lol:

Hey, does anybody here have the "proof" of 1=2? I'd like to see it.

Look [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=1569&cat_id=384]here[/url] towards the bottom.

Anyone know any other faulty proofs, interesting paradoxes, or interesting subsets of math riddles?

[quote="CyberGenius"]I am going to start a sort of riddle quiz.
1.  Four men were in a boat on the lake. The boat turns over, and all four men sink to the bottom of the lake, yet not a single man got wet! Why?
2. A woman was horrified to find a fly in her tea. The waiter took her cup and went into the kitchen and returned with a fresh cup of tea. She shouted, "You brought me the same tea!" How did she know?
3. Black I am and much admired, men seek me until they're tired. When they find me, they break my head, and take from me my resting bed. What am I?
4. What can run but never walks, has a mouth but never talks, has a head but never weeps, has a bed but never sleeps?
5. Whoever makes it, tells it not. Whoever takes it, knows it not. And whoever knows it wants it not.
6. What word can be written forward, backward or upside down, and can still be read from left to right?
7. How many times can you subtract the number 5 from 25?
8.  Samuel was out for a walk when it started to rain. He did not have an umbrella and he wasn't wearing a hat. His clothes were soaked, yet not a single hair on his head got wet. How could this happen?
9. I can be found where anything cannot; dead men eat me all the time, but if a living man eats me, he'll die. What am I?
10. Take one out and scratch my head, I am now black but once was red. What am I ?

That should be enough for a while. This is fun![/quote]
[hide="my answers"]1. The four men were inside the boat, and the boat was upside down, so water would have to go up to reach the men, which is scientifically impossible.
2. She put sugar/honey into it, and she noticed the difference.
3. Iouo
4. Iouo
5. Iouo
6. Iouo
7. You can subtract 5 from 25 as many times as you'll want, but you'll always get 20 out.
8. BALDY.
9. Nothing.
10. Iouo[/hide]

[quote="CyberGenius"]you can post math riddles on here too. I'll start you guys off with a riddle. No math required.
Three turtles are racing on a non-circular track. Once the race is over, one turtle said that when he finished, he saw 0 turtles ahead of him. Another turtle said that when he finished, he saw 1 turtle ahead of him. However, the other turtle said he also saw 1 turtle ahead of him when he finished. How could this be?[/quote] You could also have the turtles in a line, so the back turtle can only see the one directly in front and the middle turtle is blocking the back turtle's view of the front turtle.


analytic geometry
algebra
function
domain
geometry proposed
geometry

[quote="huymaick54"]A convex pentagon in a plane coordinate system with both five vertexs of its have integer coordinates. Show that there is at least one point which has integer coordinates in its domain, not on its egdes?[/quote]
Can anyone help me?


Putnam
LaTeX
calculus
calculus computations

Hi-every body
Please post the solution of this problems  there .
                                                                      Best wishes :) .

The first is (I think) an old Putnam problem and is definitely on the forum somewhere.  I personally am usually to lazy to go look at documents people attach to their posts, so you might consider typing the text of the problems here.  (The forum is LaTeX-equipped, so it's possible to make nice, pretty equations.)

Conviniently please link one solution for me .

Is there exist any one to solve this i am very unhapy because no one didnot  post any solution for this and it is very bad.
Please send the solution for me.

navid: I make it a habit of not clicking on the links when people post problems as attached files rather than by typing them in (preferably in $\text{\LaTeX},$ but I'll read it in straight text.) So I still don't know what your problem is, and I don't plan to find out. I'm probably not the only one here who does that. You'll get better responses if you do your own typing.


How are you using the symbol $\Sigma$? It isn't in the normal summation fashion since $a>0$ is a single number so $\sum\frac{1}{a}=\frac{1}{a}\neq 3$ unless $a=\frac{1}{3}$, which is besides the point. Mainly, why would you sum a single number when it is the same thing as the number itself?

And then $c$ isn't even in the problem after it is defined to be positive.

From what I understand,

Let $a,b,c>0$ and $\frac1{a}+\frac1{b}+\frac1{c}=3$

Prove that $\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}\ge1$

[quote="tranthanhnam"]let $a,b,c >0$ and $\sum \frac{1}{a}=3$
Prove that :$\sum \frac{a}{b^{2}+1}\geq 1$[/quote]
I think, if $a,$ $b$ and $c$ are positive numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ then
$\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}\geq\frac{9}{7}.$


function
topology
real analysis
real analysis unsolved

Find an example of a contracting function f: X --> X on a metric space X where f has no fixed points on X.

(A contracting function f: X --> X is defined as a function that satisfies

$ d(x, y) > d(f(x), f(y))$

where d is the metric and x and y are any elements of the metric space X.)

Btw f has to be continuous, right? Simple epsilon delta proof taking delta = epsilon.

$ f(x) \equal{} \sqrt{1 \plus{} x^2}$ on $ \mathbb{R}$.

That's what we get by weakening one hypothesis in the contraction principle: that $ d(f(x),f(y))\le K\cdot d(x,y)$ with $ K<1$.
How about the other hypothesis? Can you find an example which is a contraction in the stronger sense, but on a space that isn't complete?

$ f: \mathbb{R}_{*} \to \mathbb{R}_*$ where $ f(x) \equal{} \frac{x}{2}$ shows that if the space is not complete, that does not work, even if $ f$ is  a $ \frac{1}{2}$-contraction.

$ \mathbb{R}_* \equal{} \mathbb{R} \smallsetminus \{0\}$?  (Insert general complaint about ambiguous notations here.)


Whenever I try to log into the AoPSWiki, it says

[quote]Login error:
You have not specified a valid user name.[/quote]

Is there anyway to fix this?

Are you using the same username/password as you use on AoPS?

I think it doesn't like the underscores in your username.  I'll see if I can fix it.

That was a tough one, but I think I fixed it.  That said, I may well have broken something else, so let me know either way.

Well, now there's another problem.  Whenever I try to log on, it says

[quote]A database query syntax error has occurred. This may indicate a bug in the software. The last attempted database query was:

    (SQL query hidden)

from within function "User::addToDatabase". MySQL returned error "1062: Duplicate entry 'H_s_potter2002' for key 2 (localhost)".[/quote]


function
integration
calculus
derivative
real analysis
real analysis unsolved

Prove or find a counter example: Let F be any measurable function that is differentiable almost everywhere on [a,b]. Then $ F(x) \equal{} \int_a^x F' \plus{} F(a)$.

My answer (But I don't know if I am right):

[hide]Counter example
F(x)= x(x-c)/(x-c)  F(c) = 0

a<c<b

F'(x) = 1

but $ F(c) \neq \int_a^x F' \plus{} F(a)$[/hide]

That's a little cheap, but you can do that with anything with a jump discontinuity. Let's try asking the question that could have been asked. Let $ F$ be a [i]continuous[/i] function that is differentiable almost everywhere. Is $ F(x) \equal{} \int_a^xF' \plus{} F(a)?$

I'll even throw in another condition: make $ F$ monotone.

The Cantor function? It has a derivative like  constant function but is not constant.

Yes, you need to impose absolute continuity.

OK. so to make the prop true all we need is abs. cont.? Yes-- in fact, that is a theorem in Royden.

It's a standard theorem of measure theory and differentiation.  Usually called something like the Fundamental Theorem of Calculus for Lebesgue Integrals.

I totally missed the point of that theorem until now.


The salary of a salesperson was a $ 4\%$ commission on all sales. Her commission rate is then raised to $ 4.5\%$. What will be the percent of increase in her salary, assuming she sells the same amount of merchandise?


I am trying to do some research into panic attacks and anxiety dissorder, as I suffer from both along with many other people. Anyway, as long as I am having these panic attacks and such, I thought I might as well take advantage of it and study my behaviors (The scientist in me). So I am kind of stumped by a few things,

 + What causes chest pain typically experience when having a panic attack?
 + Usually a lump in the throat developes, what actually [i]is[/i] this lump?
 + What could cause the numbness in the fingers and toes?

If anybody knows anything about this or has opinions/hypotheses about these ideas, please reply!

Thanks,

This problem has a lot to do with psychology and a lot to do with biology. Since I don't want to be too verbose, I am just going to give you some tips on where to find this stuff.

Go to the nearest biology book and check out the sympathetic nervous system and especially its effects on vasoconstriction, release of adrenaline and noradrenaline, and how it diverts blood more to motor muscles instead of places that the blood is not immediately needed. Go to the nearest psychology book and then check out how the "fight-or-flight" response can alter behavior in people who become anxious. This should help.

Anxiety is a defense mechanism. For some people, it's a lot more powerful than others.


[hide]just powers of 2 and or 5
I get 457[/hide]

Yeah its easy i just wrote really hard so people would read it :P

I agree with mathmom.

Indeed, every rational number can be written as a repeating decimal.

Those rationals that can be written as terminating decimals can also be written as repeating decimals in either of two ways (with an infinite string of zeros at the end or a infinte string of 9s at the end.)

So, e.g., 1/5 can be written either as 0.2000000...... with an infiite string of zeros OR as 0.19999999.... with an infinite string of 9s.

Those rationals that do not have a terminating decimal  representation have only a single repeating decimal representation.

Biffanddoc probably means can be written as a terminating decimal. In this case, instead of .999999... we write 1 and instead of .1999999... we write .2. Of course you are both right, and Biffanddoc worded that part strangely, but I think the problem is pretty clear.

[quote="biffanddoc"]What is the sum of the integers x 1-100 inclusive that cannot be written 1/x as a repeating decimal?[/quote]

[hide="Solution"]
Only the ones that end in zeros, which would be any combination of 2s and 5s.

1
2
4
5
8
10
16
20
25
32
40
50
64
75
80

Too lazy to add 'em all up.
[/hide]

[quote="mathmom"][quote="biffanddoc"]cannot be written 1/x as a repeating decimal?[/quote]

What does that mean?  Doesn't the definition of "repeating decimal" allow for the repeated digit(s)  to be zero?  (that is, 1/5 = .2000....)  If so, then any rational number can be expressed as a repeating decimal, and the answer to this problem would be 0.[/quote]
If $\frac{1}{5}$ is .2000..., then it would not be a "repeating" decimal.  Here's why.  If .2000 was counted as a repeating decimal, then it would be more like this: .2000..2.  The reason is that a repeating decimal is more like: .191919191....--It eventually has to repeat every number.  0.135713571357....  would also be a repeating decimal.

[quote="mathgeek2006"]... a repeating decimal ... eventually has to repeat every number ...[/quote]
No.  That is not correct.  $\frac{1}{6}=0.1\overline{6}\approx0.166666$

[quote="mathgeek2006"]Well, there are some exceptions... :blush: ...but I don't think that .2000 would be counted as a repeating decimal($.2\overline{0}$).[/quote]
Perhaps mathmom raised the question so that people would be more precise in the way they asked a question.  According to [url=http://mathworld.wolfram.com/RepeatingDecimal.html]MathWorld[/url]:

[quote]Numbers such as 0.5 are sometimes regarded as repeating decimals since $0.5=0.5\overline{0}=0.4\overline{9}$[/quote]
Definitional questions arise quite frequently in math contests.  Both the contest taker and the contest writer must be very careful.

Yeah, they should clarify those things, I guess.  They(and me) Probably think of it as common sense about repeating decimals.... :blush:


Ok, then, I will present my solution. I thought it's a nice problem, but it seems I was wrong:
   First of all, it's clear that x_i/(x_i+1+2(n-1)x_i+x_i+2)>x_i/(2(n-1)(x_1+...+x_n)), so the expression is strictly greater than 1/2(n-1). Now comes the hard part.
   We have x_i/(x_i+1+2(n-1)x_i+x_i+2)<=1/2*x_i/(sqrt(x_i+1*x_i+2)+(n-1)x_i)). So, if we take a_i=sqrt(x_i+1*x_i+2/x^2_i), then prod a_i=1 and the expression is at most 1/2*sum 1/(n-1+a_i). It's easy to prove that sum 1/(n-1+a_i)<=1, so the expression is at most 1/2. Let x>0 and take x_i=x^i. The expression becomes a function f(x)=(n-2)/(x^2+x+2n-2)+x^n-2/(x^n-1+(2n-2)x^n-2+1)+x^n-1/(x+1+2(n-1)x^n-1). This function is 1/2 in x=1, is continuous and has limit 1/2(n-1) when x tends to zero (we took n>2). So, f takes all the values in (1/2(n-1),1/2] and this is the set of values for our expression.

Harazi,this is a beautiful problem like its beautiful solution but it is very hard.

Congratulation for your solution.


Is this true? Let $a=a_{0}+a_{1}\cdot 10^{1}+\cdots+a_{n}\cdot 10^{n}$ be a base 10 representation of $a$. Prove that there exists a positive integer $k$ such that if $b_{0}+b_{1}\cdot 10^{1}+\cdots+b_{m}\cdot 10^{m}$ is the base 10 representation of $ka$ then either $b_{0}\leq b_{1}\leq \cdots \leq b_{m}$ or $b_{0}\geq b_{1}\geq \cdots \geq b_{m}$.

LEMMA 1) for any $(a,10)=1$ there exist a $ka$ composed of only 1's digits

A) if $a\nmid 9$:

we have that $a|1+10+10^{2}+...+10^{\phi(a)-1}$ because $1+10+10^{2}+...+10^{\phi(a)-1}=\frac{10^{\phi(a)}-1}{9}$ and $a|10^{\phi(a)}-1$ but $a\nmid 9$

B) if $a|9$ then $a|9|111,111,111$

Now let's prove the problem:

Let $a=d*2^{b}*5^{c}$ where$(d,10)=1$

By LEMMA 1) there is a $kd$ composed of only 1's digits. $a|kd*10^{b+c}$ and the first digits of $kd*10^{b+c}$ are 1's, the others are 0's, so they are in a decreasing order.


function
trigonometry
algebra open
algebra

Let $f,g: I\to R$ $f$ is convex and $g$ is concave.It's true that the equiation $f(x)=g(x)$ has at most 2 real roots ?

$f(x) = 1+x^{2}$
$g(x) =-1-x^{2}$

$f = g$ does not have real roots

your problem is "has at most 2 complex or 2 real roots"?

if is complex we need another argument

yes. but i know that there are such equiations without real root..my question is   that if there are real root there are maximum 2 root ?

$g( \frac{a+b}{2}) > f( \frac{a+b}{2})$

if this condiction is true. so, there is at least 2 roots on  $f=g$

can we re-formulate the problem like that?!

$a,b$ are some numbers that the function is continuous

I didn't quite understand what you wrote there.Read my quiestion again :D

read again what i said

i just put a correction, because can be complex roots

with my correction will 2 or more real roots

with this the intersection will every have real roots

but,  if  $g( \frac{a+b}{2}) = f( \frac{a+b}{2})$

we will have only one real roots

see  $f(x) = x^{2}$  $g(x) =-x^{2}$

Yes i forgot to say that the equiation $f(x)=g(x)$ has at most 2 REAL roots..so this is my question .this equiation has more than 2 real root ?

this is bug?!

you say that $f=g$  has more than 2 roots and after you ask "$f=g$ has more than 3 roots?

if you that this have, so have

because this is not true for a lot of $f,g$s

$f$ and $-g$ are convex, so their sum $f+(-g)$ is convex.

Now, if at least one of $f$ and $-g$ is strictly convex, then $f-g$ is strictly convex.

It's well-known that a strictly convex function reaches a value (like $0$) at most two times, so we're done.

f=g has max 2 solutions only if f is STRICT convex and g STRICT concave

Let`s have one more question:

f:I -> R and g:J -> R  f,g are STRCIT convex( we can chose as weel concave). Is it true that f=g has a maximum of 3 solutions? think to the graphic

[quote="Zamfirmihai"]f:I -> R and g:J -> R  f,g are STRCIT convex( we can chose as weel concave). Is it true that f=g has a maximum of 3 solutions? think to the graphic[/quote]
No, $f(x)=x^{2}+\sin x$, $g(x)=x^{2}$. Then $f''(x),g''(x)>0$ for all $x\in\mathbb{R}$, but $f(k\pi)=g(k\pi)$ for all $k\in\mathbb{Z}$.

sin x is not convex for all x.....for example it is convex on [pi; 2 pi].

i don`t now about f``(x) or g``(x), I have to learn more math yet

$x^{2}+\sin x$ really is convex on $\mathbb R$, although $\sin x$ isn't.

[quote="Zamfirmihai"]i don`t now about f``(x) or g``(x), I have to learn more math yet[/quote]
$f(x)$ is strictly convex on $I\Leftrightarrow f'(x)$ is strictly increasing on $I\Leftrightarrow f''(x)>0$ on $I$.
In case $f(x)=x^{2}+\sin x$ we have $f''(x)=2-\sin x>0$ for all $x\in\mathbb{R}$.


limit
trigonometry
logarithms
integration
calculus
calculus computations

lim as x->0 for sin3x/xe^x

i dont know what to do with the e^x!

Let's not worry about the exponential, because when x --> 0, e^x --> 1. Now I ask you this: what is the value of the following limit,

$ \lim_{x\to 0}\frac{\sin x}{x}$?

it is?! why

the limit of sinx/x is just 1

[quote="A_marshall_13"]it is?! why[/quote]

$ e^{0}\equal{} 1$.

[quote="A_marshall_13"]the limit of sinx/x is just 1[/quote]

Ok, so you are now able to compute your limit.

okay thanks!
i have one more question though, but i won't write down the question cause i gotta figure out some of this by myself.. what do you do when theres natural logs in the limit question? haha sorry..i really suck at calc

[quote="Carcul"]Well, that depends on the particular limit. One possibility is using the fact that

$ \lim_{x\to 1}\frac{\ln[f(x)]}{f(x)\minus{}1}\equal{} 1$,

if f(1) = 1.[/quote]

Not exactly : "If $ \lim_{x\to 1}f(x)\equal{}1$" instead of "If $ f(1)\equal{}1$"
Or "If $ f(1)\equal{}1$ and $ f(x)$ is continuous at $ 1$"

You can use L'Hospital
$ (xe^{x})'\equal{}e^{x}(x\plus{}1)$
$ (sin 3x)'\equal{}3 cos3x$
So $ \lim_{x\to 0}\equal{}\frac{1}{3}$

woww okay..confusing
well heres the question haha
lim as x->0 ln(x-5)/ln(x-5)^3

and i havent learned l'hopitals rule yet so there has to be a way to do it that i know.. haha but thanks!

Does it  mean ,I can see you problem.
If
$ \frac{ln(x\minus{}5)}{ln(x\minus{}5)^{3}}$

I was assuming f(x) continous, and so $ \lim_{x\to 1}f(x) \equal{} f(1)$. Now:

$ \lim_{x\to 0}\frac{\sin 3x}{xe^{x}}\equal{}\lim_{x\to 0}3\cdot\frac{\sin 3x}{3x}\cdot\frac{1}{e^{x}}\equal{} 3\cdot 1\cdot 1 \equal{} 3$.

A_marshall_13, your last limit doesn't seem to make sense. Are you sure of that?

http://www.mathlinks.ro/Forum/latexrender/pictures/b/f/f/bfff8bff8e1a658a723383a7226d26d9065873ed.gif

do you mean theres a problem with this one?
its as x->6 btw

The problem is that $ \ln (x\minus{}5)$ isn't defined for $ x$ sufficiently close to 0. Of course, you could fix this by replacing with $ \ln (5\minus{}x)$. There's also the question of whether the $ 5\minus{}x$ or the natural log is being cubed. We have
\[ \lim_{x\rightarrow 0}\frac{\ln (5\minus{}x)}{\ln ((5\minus{}x)^{3})}\equal{}\lim_{x\rightarrow 0}\frac{\ln (5\minus{}x)}{3\ln (5\minus{}x)}\equal{}\frac{1}{3}\]
\[ \lim_{x\rightarrow 0}\frac{\ln (5\minus{}x)}{(\ln (5\minus{}x))^{3}}\equal{}\lim_{x\rightarrow 0}\frac{1}{(\ln (5\minus{}x))^{2}}\equal{}\frac{1}{(\ln 5)^{2}}\]
You probably wanted the first one, since the second one doesn't use anything about the natural logarithm other than continuity for positive reals. Even then, there isn't anything "mysterious" (indeterminate forms like $ \frac{0}{0}$ or $ \frac{\infty}{\infty}$) going on here. A few nice ones are
\[ \lim_{x\rightarrow 0}\frac{\ln (a\plus{}x)\minus{}\ln a}{x}\equal{}\frac{1}{a}\ (a>0)\]
\[ \lim_{x\rightarrow\infty}\frac{\ln x}{x^{\epsilon}}\equal{}0\ (\epsilon >0)\]
\[ \lim_{x\rightarrow 0^{\plus{}}}x^{\epsilon}\ln x\equal{}0\ (\epsilon >0)\]
The first is the definition of the derivative. The second can be done using the bound
\[ 0\leq\ln x\equal{}\int_{1}^{x}\frac{1}{x}\ dx\leq\int_{1}^{x}\frac{1}{x^{1\minus{}\epsilon /2}}\ dx\equal{}\frac{2}{\epsilon }(x^{\epsilon/2}\minus{}1)\]
for $ x\geq 1$, or L'Hopital's Rule. The third is equivalent to the second (substitute $ y\equal{}1/x$).

[quote="A_marshall_13"]do you mean theres a problem with this one? its as x->6 btw[/quote]

In that case: $ \lim_{x\to 6}\frac{\ln(x\minus{}5)}{\ln(x\minus{}5)^{3}}\equal{}\lim_{x\to 6}\frac{\ln(x\minus{}5)}{3\ln(x\minus{}5)}\equal{}\frac{1}{3}$.


Roslyn has ten boxes. Five of the boxes contain pencils, four of
the boxes contain pens, and two of the boxes contain both pens
and pencils. How many boxes contain neither pens nor pencils?

PIE leads to $ 5\plus{}4\minus{}2$ or $ 7$ boxes that have either pencils and/or pens in them.  Then, we subtract that from $ 10$, for an answer of $ \boxed{3}$.


search percent geometry ARML college HMMT \/closed

I happened to look at **********'s profile and even though his # of posts is 3, when I click on "find all posts by **********" I get 5142 posts. Most of the posts in the list isn't even **********'s. How is this possible?

Heh.  That's pretty funny.  '*' is a wildcard character in most database search programs.  Thus, the username '************' would match everybody when the program did a match by username.  Pretty amusing.  This shouldn't happen with anyone else, but you're going to have a tough time finding *********'s work here.

i just checked mr. r's profile and it said that you had 225 posts, joined may 28, but only had 1.63 posts per day. i did a quick estimate, and it should be just under 3 posts per day. not that any of this really matters. but i was just wondering if everybody's was off. or somat like that. im also too lazy to check the percent of total posts. but if you can answer my questions. twould help the nagging in my head. thanks you

Check my profile again - I think it says 'Mar 28', not May 28...

March 28th to August 13th would be 138 days, and Mr. Rusczyk had 226 posts, so he had 1.6376811594202898550724637681159 posts per day.
I'd say it's correct.

ah. well i guess i overlooked that. hehe. i forgot that you were involved wit the site and all. my bad.

[quote="ComplexZeta"]There's something funny about that though. In **********'s post list, it says there are 5218 posts, but the forum has 5537 posts. I was trying to figure out yesterday why these numbers weren't the same, but I couldn't figure it out.[/quote]
People with their name longer than 10 letters wouldn't be included (there is only 10 wild card, *, in his/her name), so that might make the difference.

[quote="rrusczyk"]There are forums that aren't visible to everybody (specifically, there are boards just for the online classes that are in session, and there is a board for our San Diego Math Circle that's only visible to members of the circle).  This is why there's a disparity in the numbers...[/quote]
What is "San Diego Math Circle"?

[quote="Tare"]People with their name longer than 10 letters wouldn't be included (there is only 10 wild card, *, in his/her name), so that might make the difference.[/quote]

That's definitely not true because my name is longer than 10 letters, and I have posted more than the disparity.

[quote="Tare"]What is "San Diego Math Circle"?[/quote]

A local group of middle and high school math students that Richard and I work with.

[quote="Syntax Error"]sigh. i wish i could live in san diego. instead of lowly oklahoma. sigh. back to aops![/quote]
Me too.
At least the weather will be nicer than here...

[quote="ComplexZeta"]Hm. We'll have to arrange something like that for the Bay Area people. It sounds like a good idea.[/quote]
Let's also arrange it for people in East Coast :)

Don't you already have like 3 of them in the Bay Area?

We don't have anything like what it seems you have. In the spring we have practices for ARML, and the rest of the year we have Berkeley Math Circles, which most of the top SFBA math people don't attend with any sort of regularity, if at all. It would be great to get some semi-formal math problem solving training going on in the Bay Area on a weekly basis or something. Some people around here seem to be interested in starting something like that, and we might have a chance to beat TJ at ARML if we can get that going, so that would be a nice goal to have.

Why don't many of the top SFBA students go to the Berkeley Math Circles - looks like you have a who's who as far as instructors are concerned?

I think the problem is that they don't usually discuss math competition topics, and many of the top students in this area are more interested in competition math than in topics that don't show up on competitions. I went to BMC for about half a year, but I had a lot of trouble physically getting there since I don't drive, and my parents didn't like the commute -- an hour each way. We had a carpool arranged, but there were only two of us who wanted to go on a regular basis, so after a while the plans fell through, and I stopped going.

[quote="Fierytycoon"][quote]Let's also arrange it for people in East Coast[/quote]

Oh yeah!!! High five Tare!

We don't have math circles around here in Massachusetts :([/quote]
We do have that ****y Tournament. It's not worth my time though. I mean, I was the top scorer for 6th grade (in an 6-8 grade cluster) even though I quit halfway since it was so dull.

eh...used to live near San Diego...if u count San Francisco as near...I liked it there...none of this Texas heaat...

hmm, aren't really many math circles here in NY, either

hafta ask Joel bout it


modular arithmetic
Support
AoPSwiki
number theory

Help requested on two of them.

1)
How many of the first 1000 natural numbers have a remainder of 1 when divided by 5, a remainder of 2 when divided by 6, and a remainder of 5 when divided by 7?

2)
If 14 times a natural number has a units digit of 2 when expressed in base 12, what is the units digit of the natural number when expressed in base 6?

The first one has something to do with some type of Remainder Theorem?

thanks.

1.  http://www.cut-the-knot.org/blue/chinese.shtml

A useful hint:  $n+4$ is divisible by $30$.  :)

2.  Rephrase in modular arithmetic, or better yet, algebraically.

[quote="20000th User"]Help requested on two of them.

1)
How many of the first 1000 natural numbers have a remainder of 1 when divided by 5, a remainder of 2 when divided by 6, and a remainder of 5 when divided by 7?
[/quote]

[hide="Solution"]
We can use modular arithmetic to solve the first one.  Set up the equations described by the problems:

$x\equiv 1 \pmod{5}$
$x\equiv 2 \pmod{6}$
$x\equiv 5 \pmod{7}$

Now we solve it as a system of modular equivalences.

Conver the first equation into the form:
$x=5a+1$

Plug that into the second equivalence:

$5a+1\equiv 2 \pmod{6}$
$5a\equiv 1 \pmod{6}$
$a\equiv 5 \pmod{6}$

So therefore $a=6b+5$.  Now,  plug that back into the first equation:
$x=5(6b+5)+1$
$x=30b+26$

Now, we plug this into our last congruence.

$x\equiv 5 \pmod{7}$
$30b+26\equiv 5 \pmod{7}$
$30b+26\equiv 5 \pmod{7}$
$30b\equiv-21 \pmod{7}$
$30b\equiv 0 \pmod{7}$
So the smallest possible value of b is $0 \pmod{7}$.  This implies that the possible values for b are simply the multiples of 7.

Scrolling up, we see we earlier arrived at the equation: $x=30b+26$.  Plug in the first few possible values of b:

$x=30(0)+26=26$
$x=30(7)+26=236$
$x=30(14)+26=446$
$x=30(21)+26=656$
$x=30(28)+26=866$
$x=30(35)+26=1076$ IS TOO LARGE

We count up the number of values of b that keep x under 1000, and see that only $\boxed{5}$ work.

[/hide]

I think that was a bit unclear, so post any questions/comments.  For follow-up problems of a similar nature, visit [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80124]this thread[/url].

[quote="mysmartmouth"][quote="20000th User"]Help requested on two of them.

1)
How many of the first 1000 natural numbers have a remainder of 1 when divided by 5, a remainder of 2 when divided by 6, and a remainder of 5 when divided by 7?
[/quote]

[hide="Solution"]
We can use modular arithmetic to solve the first one.  Set up the equations described by the problems:

$x\equiv 1 \pmod{5}$
$x\equiv 2 \pmod{6}$
$x\equiv 5 \pmod{7}$

Now we solve it as a system of modular equivalences.

Conver the first equation into the form:
$x=5a+1$

Plug that into the second equivalence:

$5a+1\equiv 2 \pmod{6}$
$5a\equiv 1 \pmod{6}$
$a\equiv 5 \pmod{6}$

So therefore $a=6b+5$.  Now,  plug that back into the first equation:
$x=5(6b+5)+1$
$x=30b+26$

Now, we plug this into our last congruence.

$x\equiv 5 \pmod{7}$
$30b+26\equiv 5 \pmod{7}$
$30b+26\equiv 5 \pmod{7}$
$30b\equiv-21 \pmod{7}$
$30b\equiv 0 \pmod{7}$
So the smallest possible value of b is $0 \pmod{7}$.  This implies that the possible values for b are simply the multiples of 7.

Scrolling up, we see we earlier arrived at the equation: $x=30b+26$.  PLug in the first few possible values of b:

$x=30(0)+26=26$
$x=30(7)+26=236$
$x=30(14)+26=446$

Now we see a pattern.  We count up the number of values of b that keep x under 100, and see that only $\boxed{5}$ work.

[/hide]

I think that was a bit unclear, so post any questions/comments.  For follow-up problems of a similar nature, visit [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80124]this thread[/url].[/quote]
Actually, to make the solution a bit simpler, you can use the fact that $x\equiv 1\equiv-4\pmod{5}$ and $x\equiv 2\equiv-4\pmod{6}$ means that $x\equiv-4\pmod{30}$, as t0rajir0u said.

[quote="10000th User"]OK, who is this funny guy trying to imitate me by putting a negative color on my Gohan avatar? You got too much time on your hands to be doing this... Also, just to let you know if you really want to imitate me, I never type "thanks" without a capital T. ;)[/quote]
This is starting to get weird. There's a new user named "rruszyk" with the same avatar as the real rrusczyk, and there's another new user named "Iversonfan2007" impersonating Iversonfan2005. Is this all the same person?  :huh:  If it is, please do not try to use someone else's username as your own. It's not funny, just confusing, and it eventually gets really irritating.  :wink:

I am not here to "imitate" you. Please don't spam (:D)!

If you have a problem, please take it to the "Support" section of the forum or the "Games and Fun Factory". Better yet, employ the fully functional Private Message system.

10000th User... if you dislike my use of this avatar or if you feel it violates some type of law (copyright infringement, etc.), please alert me and I'll change it immediately.

Now, this thread has really gone off topic.

For the first one, I have:

$14x \equiv 2 \mod 12$
$x \equiv y \mod 6$

I guess I'll have to brush up on my modular arithmetic. (Yay for [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic]AoPSWiki[/url]!)

Note: I am/was actually the 20000th User, but the new server changed the "memberlist" functionality.

Also, I am in no way affiliated to "Iversonfan2007" and to the account with the misspelling of Mr. Rusczyk's name.

[quote="mysmartmouth"]Iversonfan2005 is Iversonfan2007.  I don't know why he switched, but see iversonfan2005's blog.[/quote]

err..cuz it's past 2005..duh :wink:

[quote="20000th User"]
$14x \equiv 2 \mod 12$
$x \equiv y \mod 6$[/quote]

I think you mean the second one, and I suggest you rewrite it this way.

$14x = 2+12k$

:)

[quote="20000th User"]
2)
If 14 times a natural number has a units digit of 2 when expressed in base 12, what is the units digit of the natural number when expressed in base 6?
[/quote]

let $n$ be the natural number

then clearly

$14n\equiv 2n\equiv 2\mod 12$
$n\equiv 1\mod 6$
so the units digit is 1

(i just used basic mod properties, they are easy to prove it you take the number into the form

$2n=2+12q$ for $q\in\mathbb{Z}$, divide, and look in mod 6

[quote="10000th User"]OK, who is this funny guy trying to imitate me by putting a negative color on my Gohan avatar? You got too much time on your hands to be doing this... Also, just to let you know if you really want to imitate me, I never type "thanks" without a capital T. ;)[/quote]

Lol I thought it was you that made the thread at first. Then i was like... "10000th user is not at intermediate nt level? I thought he was better at math".


Any ideas on who the 9th grader and 10th grader who got 144 on the 10A?

How about the <=6th grader who got 126 on the 10B?

Or the 10th grader with a 132 on the 12B?

Sorry if I've just missed it, but I don't remember us coming up with a definitive answer on those. Congrats to Santhosh, Eric, and Sitan, who I know are winners. (insert drivel about everyone being a winner here :P)

The 9th grader and 10th grader with the 144s were Tianqi Wu and Boru Wang respectively, I believe.

The 6th grader is Edward Park from Dickerson. And yes, he's coming to Walton!  :D (At least he better, otherwise I'll have to flunk his brother! Who's pretty good himself.)

Whos the second tenth grader with the 12b? i took the 12b too...

If I had to guess, I would say Robert Grosse from Chamblee...


geometry
incenter
inradius
geometric transformation
reflection
geometry theorems

Let $ ABC$ be a triangle, and let $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$ be the midpoints of its sides $ BC$, $ CA$, $ AB$. Let $ I_{a}$, $ I_{b}$, $ I_{c}$ be the $ A$-excenter, the $ B$-excenter, the $ C$-excenter of triangle $ ABC$, respectively.

Let $ X_{b}$, $ X_{c}$, $ Y_{c}$, $ Y_{a}$, $ Z_{a}$, $ Z_{b}$ be the orthogonal projections of the points $ B$, $ C$, $ C$, $ A$, $ A$, $ B$ on the lines $ I_{b}I_{c}$, $ I_{b}I_{c}$, $ I_{c}I_{a}$, $ I_{c}I_{a}$, $ I_{a}I_{b}$, $ I_{a}I_{b}$, respectively. (So, altogether, each point is projected onto two lines, and on each line we project two points!)

Prove the following properties of this configuration:

[b](a)[/b] The points $ Y_{c}$ and $ Z_{b}$ lie on the line joining the points where the $ A$-excircle of triangle $ ABC$ touches its sidelines $ CA$ and $ AB$.

[b](b)[/b] The points $ Y_{a}$ and $ Z_{a}$ lie on the line $ B^{\prime}C^{\prime}$.

[b](c)[/b] Let $ H_{a}$ be the orthocenter of triangle $ I_{a}BC$. Then, the line $ C^{\prime}A^{\prime}$ touches the circle with diameter $ H_{a}I_{a}$ at the point $ Z_{b}$.

[b](d)[/b] Let $ S$ be the incenter of triangle $ A^{\prime}B^{\prime}C^{\prime}$, let $ r$ be the inradius of triangle $ ABC$, and let $ s=\frac{a+b+c}{2}$ be the semiperimeter of triangle $ ABC$. Then, the points $ X_{b}$, $ X_{c}$, $ Y_{c}$, $ Y_{a}$, $ Z_{a}$, $ Z_{b}$ lie on the circle with center $ S$ and radius $ \frac{1}{2}\sqrt{r^{2}+s^{2}}$.

Note that I call this circle the [i]second Spieker circle[/i] of triangle $ ABC$. The point $ S$ is called the [i]Spieker point[/i] of triangle $ ABC$.

[b](e)[/b] The circle with center $ S$ and radius $ \frac{1}{2}\sqrt{r^{2}+s^{2}}$ is the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4327]Taylor circle[/url] of triangle $ I_{a}I_{b}I_{c}$.

[b](f)[/b] The point $ H_{a}$ is the reflection of the point $ A$ in the point $ S$.

I posted parts of this at http://www.mathlinks.ro/Forum/viewtopic.php?t=5958 , but the reaction was not particularly interested. I proved both parts [b](b)[/b] and [b](d)[/b] in http://www.mathlinks.ro/Forum/viewtopic.php?t=5958 , and now I will post a quick proof of [b](c)[/b] (using [b](b)[/b]) because I need it in a different topic:

[hide="Proof of c"][i]Proof of [b](c)[/b].[/i] From part [b](b)[/b], we know that the point $ Z_{a}$ lies on the line $ B^{\prime}C^{\prime}$. Similarly, the point $ Z_{b}$ lies on the line $ C^{\prime}A^{\prime}$. Hence, the line $ A^{\prime}Z_{b}$ coincides with the line $ C^{\prime}A^{\prime}$.

Now we consider the triangle $ I_{a}BC$. The midpoint of its side $ BC$ is $ A^{\prime}$. The foot of the $ B$-altitude of this triangle is $ Z_{b}$. The orthocenter of this triangle is $ H_{a}$. Thus, from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=157491]a problem I just posted elsewhere[/url], we see that the circle with diameter $ H_{a}I_{a}$ touches the line $ A^{\prime}Z_{b}$ at the point $ Z_{b}$. Since the line $ A^{\prime}Z_{b}$ coincides with the line $ C^{\prime}A^{\prime}$, this yields that the circle with diameter $ H_{a}I_{a}$ touches the line $ C^{\prime}A^{\prime}$ at the point $ Z_{b}$. This proves property [b](c)[/b].[/hide]


algebra
polynomial
quadratics
Vieta
articles
quadratic formula

$ 1)$ The equation $ x^{4}+Ax^{3}+Bx^{2}+Cx+D=0$ with integers $ A, B, C, D$ has as two of its roots $ -2i+5$ and $ 3+i$. What is the value of $ A$?

$ 7)$ The equation $ x^{3}\#+4x^{2}+4x+4=0$ has one real root. This root lies between what two consecutive integers?

[hide="my progress on 1 so far"]form the factors with the given roots $ (x\minus{}(\minus{}2i\plus{}5))\cdot(x\minus{}(3\plus{}i))\Longrightarrow (x\plus{}2i\minus{}5)\cdot(x\minus{}3\minus{}i)\Longrightarrow(x^{2}\minus{}8x\plus{}ix\plus{}9\minus{}i)$ now all u have to do is find the complementary factor, which im finding hard to do[/hide]

[quote="Bicameral"]$ 1)$ The equation $ x^{4}\plus{}Ax^{3}\plus{}Bx^{2}\plus{}Cx\plus{}D \equal{} 0$ with integers $ A, B, C, D$ has as two of its roots $ \minus{}2i\plus{}5$ and $ 3\plus{}i$. What is the value of $ A$?[/quote]

[hide]
From my experience if a polynomial with integer - or even real for that matter - coefficients has complex factor $ a\plus{}bi$, then $ a\minus{}bi$ is also a solution (the positive and negative roots of the radical in the quadratic formula, respectively), so if $ 5\minus{}2i$ and $ 3\plus{}i$ are roots, than so are $ 5\plus{}2i$ and $ 3\minus{}i$. Since our polynomial is of the fourth degree, from Vieta's we know that $ \minus{}A$ is sum of all of its roots, and thus $ A \equal{}\minus{}[(5\plus{}2i)\plus{}(5\minus{}2i)\plus{}(3\plus{}i)\plus{}(3\minus{}i)] \equal{}\minus{}16$. [/hide]

[quote="mihail911"][quote="Anaxerzia"][quote="Bicameral"]$ 1)$ The equation $ x^{4}\plus{}Ax^{3}\plus{}Bx^{2}\plus{}Cx\plus{}D \equal{} 0$ with integers $ A, B, C, D$ has as two of its roots $ \minus{}2i\plus{}5$ and $ 3\plus{}i$. What is the value of $ A$?[/quote]

[hide]
From my experience if a polynomial with integer - or even real for that matter - coefficients has complex factor $ a\plus{}bi$, then $ a\minus{}bi$ is also a solution (the positive and negative roots of the radical in the quadratic formula, respectively), so if $ 5\minus{}2i$ and $ 3\plus{}i$ are roots, than so are $ 5\plus{}2i$ and $ 3\minus{}i$. Since our polynomial is of the fourth degree, from Vieta's we know that $ \minus{}A$ is sum of all of its roots, and thus $ A \equal{}\minus{}[(5\plus{}2i)\plus{}(5\minus{}2i)\plus{}(3\plus{}i)\plus{}(3\minus{}i)] \equal{}\minus{}16$. I might be wrong in my original assumption though...
[/hide][/quote]

hmm..i never thought of that. i don't know, it's plausible. what does someone else think?[/quote]

yea thats a perfectly fine solution...dont see anything wrong with it

Yes, it's called the Conjugate Root Theorem and here's a sketch of the proof:

$ P(x) \equal{} a_{0}\plus{}a_{1}x\plus{}......\plus{}a_{n}x^{n}$ such that $ a_{1},a_{2},........,a_{n}\in\mathbb{R}$. Let $ z$ be a complex root of $ P(x)$ hence:

$ a_{0}\plus{}a_{1}z\plus{}.......\plus{}a_{n}z^{n}\equal{} 0$ and take the conjugate of both sides and see what happens.

[quote="mihail911"]what im asking is, does that assumption/conjecture that anaxerzia made really hold true for all such cases?[/quote]

[b]Conjugate Root Theorem:[/b]  If a polynomial $ P(x)$ with [i]real[/i] coefficients has a root $ a\plus{}bi$, where $ a, b$ are real, then $ a\minus{}bi$ is also a root of $ P(x)$.

[hide="Proof 1"] [b]Lemma:[/b]  $ \overline{z\plus{}w}\equal{}\bar{z}\plus{}\bar{w}$

Proof:  Let $ z \equal{} a\plus{}bi, w \equal{} c\plus{}di$.  Then $ \overline{z\plus{}w}\equal{} (a\plus{}c)\minus{}(b\plus{}d) i \equal{} (a\minus{}bi)\plus{}(c\minus{}di) \equal{}\bar{z}\plus{}\bar{w}$.

[b]Lemma:[/b]  $ \overline{zw}\equal{}\bar{z}\cdot\bar{w}$

Proof:  $ \overline{zw}\equal{}\overline{ (ac\minus{}bd)\plus{}(ad\plus{}bc) i}\equal{} (ac\minus{}bd)\minus{}(ad\plus{}bc) i \equal{} (a\minus{}bi)(c\minus{}di) \equal{}\bar{z}\cdot\bar{w}$, as desired.

[b]Lemma:[/b]  Let $ P(x)$ be a polynomial with real coefficients.  Then $ \overline{P(z)}\equal{} P(\bar{z})$.

Proof:  Let $ P(x) \equal{} a_{0}\plus{}a_{1}x\plus{}...\plus{}a_{n}\plus{}x^{n}$.  Then

$ \overline{P(z)}\equal{}\overline{a_{0}\plus{}a_{1}z\plus{}... }\equal{}\overline{a_{0}}\plus{}\overline{a_{1}z}\plus{}... \equal{} a_{0}\plus{}a_{1}\bar{z}\plus{}a_{2}\bar{z}^{2}\plus{}...$ by the first two lemmas.

Hence if $ P(z) \equal{} 0$ then $ \overline{ P(z) }\equal{} P (\overline{z}) \equal{} 0$, which completes the proof. [/hide]

[hide="Proof 2"] Consider the quotient of $ P(x)$ by $ (x\minus{}(a\plus{}bi))(x\minus{}(a\minus{}bi))$.  We have

$ P(x) \equal{} (x^{2}\minus{}2ax\plus{}a^{2}\plus{}b^{2}) Q(x)\plus{}(cx\plus{}d)$

Where $ P(a\plus{}bi) \equal{} c(a\plus{}bi)\plus{}d \equal{} 0$.  Then $ bc \equal{} 0\implies c \equal{} 0\implies d \equal{} 0$, so we have $ (x^{2}\minus{}2ax\plus{}a^{2}\plus{}b^{2}) | P(x)$ as desired. [/hide]

[quote="Bicameral"]$ 1)$ The equation $ x^{4}+Ax^{3}+Bx^{2}+Cx+D = 0$ with integers $ A, B, C, D$ has as two of its roots $ -2i+5$ and $ 3+i$. What is the value of $ A$?

$ 7)$ The equation $ x^{3}\#+4x^{2}+4x+4 = 0$ has one real root. This root lies between what two consecutive integers?[/quote]

The key to number 1 is that each coeffiecient and D is an integer. By Vieta's we know that we can get the value of A, B, C, and D by adding the products of the roots. Now if we allow the other two conjugates to be our factors the imaginary parts will cancel out. I believe there was a similar problem to this one on 2007's AMC 12 A.

[quote="sonny"]The key to number 1 is that each coeffiecient and D is an integer. [/quote]

Actually, the key to number 1 is that each coefficient is real, so that the conjugate root theorem can be applied.  The fact that they are integer can be readily deduced from the given roots, and is in fact extraneous information given that they are real.

Can someone explain Vieta's rules for me? I don't understand the wiki article about it...

EDIT:...um...never mind. I worked it out myself, makes perfect sense.


I would like the older students to leave this problem for those who are entering the 6th or 7th grade this next schoolyear.

In a certain game played with marbles, the game begins when all the marbles are arranged in piles of 12 and the excess marbles are removed.

When the game is played with Aaron's marbles, 7 marbles are removed after the piles of 12 are made.

When the game is played with Pei-Hsin's marbles, 11 are removed after the piles of 12 are made.

When the game is played with Zack's marbles, 8 are removed after the piles of 12 are made.

If Aaron, Pei-Hsin, and Zack combine their marbles into one stack to play, how many marbles are removed after the piles of 12 are made?

It occurred to me that most people here on these forums (and I mean people who read and post regularly) are either rising 8th graders (Martin, Neal) or high schoolers ("alumni" as mathcounts.org called us) or beyond that.  I haven't seen hide or hair (wait...or is that I have seen neither hide nor hair :?: ) of any 6th or 7th graders.  But maybe I can get my brother to answer this one (he's 5th grade)

Yeah, I had noticed the lack of response to this one.  I left it partially to encourage new younger students.  If your brother is interested in math, by all means, see if he enjoys the challenge problems and the site!

I seriously doubt there are any 6th or 7th graders here, I tried posting tons of easy problems like you suggested but so far almost no response to any...*sigh*

Perhaps we could get freerangegraphics to do an ad for us :)

I gave this problem to my daughter Meena, who is in 4th grade.  I'll let her type in her solution, starting with the next paragraph:

First I read the problem over.  Than I treated it like a division problem.  12 was the divisor for all the kids.  I wrote down the remainders next.  They were Aaron has 7 as a remainder,  Pei-Hsin had 11 as a remainder and Zach had 8 as a remainder.  I added up the remainders to get a total of 26 marbels.  12 goes into to 26 2 times with a remainder of  2.  My answer is there are two marbels removed after there marbels are combined.

[quote="Tare"]I seriously doubt there are any 6th or 7th graders here, I tried posting tons of easy problems like you suggested but so far almost no response to any.[/quote]

There is definitely at least one sixth-grader on the board, but he is busy. He lurks more than he posts here. (Hmmm, maybe he is taking to heart the information in the thread on the College Forum about the importance of getting homework done.  :wink:  )Federal law in the United States makes Web sign-up for persons under 13 years of age a little bit laborious, but there is an established procedure for that on this site, and I doubt that the sixth grader of my acquaintance is the only person that young on this site.

Personally, I would look for a beginner's problem in the Getting Started Forum (where I lurk regularly, but hardly ever post) rather than on the Mathcounts Forum. That might be one reason why the question has not received many replies, besides the fact that the board membership has increased a great deal since the problem was first posted.

I'm in 7th grade, and can find lots of problems in the mathcounts series to work, but most of the 6th-8th graders from my team that visit the site have a pretty hard time with finding almost any questions to answer.  They visit the site at school, but get frustrated with the difficulty.

My parents coach my team and like to try to find problems on this site for them, so any easier level problems would be great to get started.

Getting started forum really is plenty hard for kids truly "getting started".

Than again, it's nice to have a site made that really challenges.

actually, there are a lot of 7th graders here, but most don't bother looking at the mathcounts or getting started forums, only the miller mathcounts. at least thats what i've noticed.

shall i start posting easier problems then. im trying to push it a little bit because the people who frequent the aops forum are pretty good so if i want anyone to come to mc the problems have to be harder than regular mc problems. but ill do easier problems if that means your math teams will start coming and solving them

haha forgot to solve the problem myself (i'm in seventh grade and trying out for the CHAMPION Miller Middle School Team, currently made it to the second round of tryouts):

just assign each to a number of marbles, for example 19, 23, and 20, respectively (i forgot the names). add the number together, you get 62, divide by 12, and you get 2 as the remainder... so 2 is the answer. props to the fourth grader.

I'm in 6th grade but I just got back from a vacation and notfunny already solved it the way I would and it's 2.

There's another way to do this problem also, using modular arithmetic.  Then again I didn't know how to do that in seventh grade really...

Well.. its 99% identical to the solution already given.. I mean saying that 19+23+20 :equiv: 2 (mod 12) or 19+23+20 = 62 which is 5 12s + 2, not much difference really.


I draw a card from a standard 52-card deck. If I draw an Ace, I win 1 dollar. If I draw a 2 through 10, I win a number of dollars equal to the value of the card. If I draw a face card (Jack, Queen, or King), I win 20 dollars. If I draw a $ \clubsuit$, my winnings are doubled, and if I draw a $ \spadesuit$, my winnings are tripled. (For example, if I draw the $ 8\clubsuit$, then I win 16 dollars.) What would be a fair price to pay to play the game? Express your answer as a dollar value rounded to the nearest cent.

Okay, we just add up the sum of the values, and divide by $ 52$.

[i]Case 1: The card is a diamond or a heart.[/i]

$ 2([1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 8 \plus{} 9 \plus{} 10] \plus{} [20 \plus{} 20 \plus{} 20])$

[i]Case 2: The card is a club.[/i]

$ 2([1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 8 \plus{} 9 \plus{} 10] \plus{} [20 \plus{} 20 \plus{} 20])$

[i]Case 3: The card is a spade.[/i]

$ 3([1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 8 \plus{} 9 \plus{} 10] \plus{} [20 \plus{} 20 \plus{} 20])$

_________________

So our answer is $ \frac {(2 \plus{} 2 \plus{} 3)([1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 8 \plus{} 9 \plus{} 10] \plus{} [20 \plus{} 20 \plus{} 20])}{52}$.

Let's break this big fraction up into little pieces, particularly the numerator.

$ (2 \plus{} 2 \plus{} 3) \equal{} 7$

$ [1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} 8 \plus{} 9 \plus{} 10] \equal{} \frac {10 \cdot 11}{2} \equal{} 55$

$ [20 \plus{} 20 \plus{} 20] \equal{} 20 \cdot 3 \equal{} 60$

So we have $ \frac {7 \cdot (55 \plus{} 60)}{52} \implies \frac {7 \cdot 115}{52} \implies \frac {805}{52} \implies \boxed{\approx 15.48}$


How many different three-digit numbers can be made by choosing three distinct digits from the set $ \{2,3,4,5\}$?

There are 4 choices for the first digit, 3 for the second, and 2 for the third. This totals to $ 4 \times 3 \times 2 \equal{} \boxed{24}$ distinct numbers.


prove that there are infinity prime in the from $ 8k\plus{}5$

without using Dirichlet's theorem,how to prove it?
I have one solution,but,I don't sure it's true.
[hide="My solution"]Let $ p_1, p_2, ... , p_n$ be any finite collection of primes of the form
$ 8n\plus{}5$.  Consider the number $ (2*p_1*p_2*...*p_n)^2 \plus{} 1$.  It cannot be
divisible:by $ 2$, because it's odd;by any $ 4n\plus{}3$ prime, because $ \minus{}1$ is a quadratic residue;solely by $ 8n\plus{}1$ primes, because it's of the form $ 8n\plus{}5$.Therefore it has a prime factor of the form $ 8n\plus{}5$.$ \therefore$ contradiction[/hide]

You could extend $ 8k+5$ to $ 1 +2^i(mod 2^(^i^+^1^))$, using a similar argument.

There is a similar argument for $ a \mod n$ iff $ a^2 \equiv 1 \mod n$.

Could you post that proof here? I don't even know the proof of the case -1(mod q) where q is prime.

That proof (the general one) is really long (about 50 pages I guess if you want to do it completely elementary). The one for $ \minus{}1\mod q$ is about 10-15 pages (I once wrote it down a year ago).


Mort walks each day for exercise and likes to vary his path. He starts each day at position 1 on the circle shown, then walks in a straight line to another numbered point on the circle, and continues, visiting each numbered point exactly once, before returning to his starting point. How many different possible paths can Mort take?
[asy]
import olympiad; size(100); import geometry; import graph; defaultpen(linewidth(0.8));
draw(Circle(origin,1));
dot(Label("$1$",align=N),dir(80));
dot(Label("$2$",align=NE),dir(45));
dot(Label("$3$",align=SE),dir(-30));
dot(Label("$4$",align=SE),dir(280));
dot(Label("$5$",align=SW),dir(200));
dot(Label("$6$",align=NW),dir(150));[/asy]

I think, but am not sure, that it is 120.
Since he visits each of the five sites once,
I believe it would be 5! or 120.

The answer is 120. Here's the reasoning:
Mort can go to any of the positions except for the one he is standing on. Which is 6 - 1 = 5.
From that position, he can go to any of the positions except for the one he just went to, or the one he is currently on. 6-2=4.
Continuing this process, we find that he can go to 3 more positions, then 2, then 1 because he has to get to his starting point.
5*4*3*2*1=120.
Good job blibberinghum

After he has visited all the points, he can go right or left to go back to point 1, giving us 5!*2=240.

[quote=yellowswampy2008]whats asy?[/quote]

The asymptote. It's red in the gamebot post

ok. What do you think the answer to the question is?

[quote=raagavbala]I think there's something wrong with it, but i don't know what's wrong[/quote]

It's the geometry_dev package; there's no such thing. Maybe there's been an asymptote update that made this code bad. MinnesotaMath reported and gave the fixed version.

[quote=GameBot]Mort walks each day for exercise and likes to vary his path. He starts each day at position 1 on the circle shown, then walks in a straight line to another numbered point on the circle, and continues, visiting each numbered point exactly once, before returning to his starting point. How many different possible paths can Mort take?
[asy]import olympiad; size(100);  import graph; defaultpen(linewidth(0.8));
draw(Circle(origin,1));
dot(Label("$1$",align=N),dir(80));
dot(Label("$2$",align=NE),dir(45));
dot(Label("$3$",align=SE),dir(-30));
dot(Label("$4$",align=SE),dir(280));
dot(Label("$5$",align=SW),dir(200));
dot(Label("$6$",align=NW),dir(150));[/asy][/quote]

[quote=GameBot]... then walks in a straight line...[/quote]

I don't think the walk back to point 1 is on the circle but rather in a straight line. So 120. But it is 240 if Mort walks on the circle instead.


What do you think about MathLinks Contest?

It's cool. (most of the time, when the questions are not too well-known)

I often give them a try but apparently I always end up forgetting to hand them in in time. :blush:

Gonna have to hurry though cuz at the age of 20 it's done.. :?

[quote="Valentin Vornicu"][quote="Myth"]These are provocative answers, beause they don't allow to criticize ML Contest  :P  :D  :D[/quote]I allow critics of course ... :) ... the fact that I don't really take them into consideration is a secondary thing :D :P :D[/quote]
I really hope you are kidding...

[quote="Myth"]I really hope you are kidding...[/quote]:D have a look at the post's subject :P

I have no idea of what you are talking about ...

I want to know that when will the next Edition? I also want to participate

Yeah, I agree it'd be nice to see another edition of the contest. :) Maybe appoint a committee that cares about the contest if you have too much work, Valentin?

[quote="moogra"]what is mathlinks anyways[/quote]This site.

Even if I tell you it's meant to be above IMO level? ;)

[quote="Peter VDD"]Even if I tell you it's meant to be above IMO level? [/quote]
I can't say that it is harder than IMO,but I like the problems more than the once of IMO :lol: ,especially Geometry problems :P.
It is really a very cool contest and I am looking forward for the next one,though it would be better if the contest start after January because till then I will be busy with my applications :blush: .
To speak about the week points of the contest,I would mention that the results of the contests being anounced too late :| .

I think it is harder than IMO.( Of course ,The 6th problem this year is too hard,but the other problem is(maybe much) easier than ML contest).
But ML contest is not harder than ( or as hard as) China TST and USAMO.
though  some of the problem in ML contest areso hard ,and there are not any solutions :( )


linear algebra
matrix
trigonometry
LaTeX

Evaluate:

$ \begin{vmatrix}\tan\frac {\pi}{4} & 2\sin\frac {\pi}{6} \\
\minus{} 4 \cos \frac {\pi}{3} & \minus{} \cot\frac {\pi}{4}\end{vmatrix}$

EDIT: sorry, I copied the Latex and forgot to delete the $ 8$

wait is that a determinant?
if so, how can you evaluate it, if uv already solved it? 8?

By multiplying. Why are we doing his homework for him?

[quote="xpmath"]Why are we doing his homework for him?[/quote]
Because we're not?

[quote="jeez123"]Well, I got a contest on all of these next Saturday, and I am stuck on a lot of problems. Thankfully, I doesn't have to wait for every weekly math team practice to get help.  :D[/quote]


A 100mm cube has a thin and rigid wire inserted into 20mm from each of two adjacent edges ,as shown
The cube contains a sphere that is not cut by a wire. What is the largest possible radius to the nearest mm, of hte sphere that just touches the wire?

Hi joshisunny16. Thanks for your message. I don't know how to post a diagram but here is my solution.

Looking at the horizontal cross-section, halfway up the cube, which the wire cuts it at the point $ W \equal{} (80\sqrt {2}, 80\sqrt {2})$, we can see that the centre of the required sphere lies on the diagonal of this square cross-section through $ W$. Let $ r$ be the radius of the sphere. Then perpendiculars dropped to the $ x \minus{}$ and $ y \minus{}$axes where the origin is the bottom left-hand corner (the corner furthest from $ W$) are of length $ r$. Therefore the length of this diagonal is

$ r\sqrt {2} \plus{} r \plus{} 20\sqrt {2} \equal{} 100\sqrt {2}$

$ \implies r \equal{} \frac {80\sqrt {2}}{1 \plus{} \sqrt {2}}$

This is approximately $ 46.86291501$ or $ 47$ to the nearest mm.

Hope this is clear enough for you to follow.


explain yourself  :)  I think it's obvious cause the function is convex, and when $x=0$ it's equal $2$..?

Using AM-GM this is greater than or equal to $2\sqrt[4]{4x^{4}+1}$, clearly greater than or equal to 2.

[quote="arqady"][quote="ashrafmod"]. Prove that for any real number x,
$\sqrt{x^{2}+( x-1 )^{2}}+\sqrt{x^{2}+( x+1 )^{2}}\geq 2$[/quote]
Inequality of a triangle!  :D[/quote]

Consider the triangle with vertices $(0, 1)$, $(0,-1)$, and $(x, x)$.

EDIT: Whoops, arqady is right.

[quote="paladin8"]
Consider the triangle with vertices $(0, 1)$, $(0,-1)$, and $(x, 0)$.[/quote]
Consider the triangle with vertices's $(0, 1)$, $(0,-1)$ and $(x, x)$ :wink:


modular arithmetic
IMO Shortlist
sums
Additive Number Theory
number theory

Is it possible to find  $100$ positive integers not exceeding $25,000$, such that all pairwise sums of them are different?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

I have just constructed (with aid of computer) a set of 100 numbers with required property, but last number in this set is 26780. All my numeric experiments lead to result that the largest number is about 27000.
So we have two possibilities:
1) Estimate 25000 is quite exact and there are no such set;
2) my very simple "greedy" algorithm is nearly to optimal.

Approximately behaviour of my sequences is $\frac{n^2\sqrt{n}}{4}$.

This b)  ;)

Look at WOP to India # ? and France # 6.

Pierre.

Have a look at this [url=http://www.artofproblemsolving.com/Forum/files/imo_2001_shortlist.pdf]IMO 2001 ShortList document, page 70.[/url]

I believe that trick with prime numbers is far from evidence. Hence France 6 is much more easier than this one.

The solution given in WOP for France 6 (and as I see in the ISL01) is due to Erdos and Turan.

Pierre.

[quote="Zhero"][hide="Solution by math154 and me"]Let $p = 101$, let $[n]_p$ denote the remainder when $n$ is divided by $p$, and let $f(n) = 2pn + [n^2]_p$. We claim that the set $\{f(1), f(2), \ldots, f(100)\}$ satisfies the conditions of the problem.

For all $i \leq 100$, $f(i) = 2pi + [i^2]_p < 20200 + 101 < 25000$. In addition, for $1 \leq i,j \leq 100$, $f(i) = f(j)$ implies $|2p(i-j)| = |[i^2]_p - [j^2]_p| < p$, whence $i=j$. It follows that all of the numbers $f(1), f(2), \ldots, f(100)$ are distinct and less than 100.

We will now show that if $f(a) + f(b) = f(c) + f(d)$ for $1 \leq a, b, c, d \leq 100$, then $\{a,b\} = \{c,d\}$. If the equation holds, then $2p((a+b)-(c+d)) = [c^2]_p + [d^2]_p - ([a^2]_p + [b^2]_p)$. The right-hand side is divisible by $2p$ and has absolute value less than $2p$, so both sides are 0. Hence, $a+b=c+d$ and $a^2 + b^2 \equiv c^2 + d^2 \pmod{p}$. But $a^2 + b^2 + 2ab = c^2 + d^2 + 2cd$, so $ab \equiv cd \pmod{p}$. Since $a+b \equiv c+d \pmod{p}$ as well, $c$ and $d$ must be the two roots of the quadratic $t^2 - (a+b)t + ab$ mod $p$, whence $\{a, b\} \equiv \{c, d\} \pmod{p}$.

Without loss of generality, suppose that $a \equiv c \pmod{p}$. Since $|a-c| < 100$, we must have $a=c$, and therefore that $b=d$. The result follows immediately. [/hide][/quote]

Do you not only look to one possible set?

[quote="Zhero"]What do you mean?[/quote]

They constructed a set with such functions, and show it isn't a good one.
But they didn't prove it is so with all ones.

[quote=SCP]
Do you not only look to one possible set?[/quote]

They found a sequence which complies with the requirement. Which means that it is possible to find a set of 100 numbers with that property

PUTNAM 1994 B6 kills this problem instantly.

[hide=Solution (Always look for general cases)]We will prove that for any $p$,we can find $p-1$ numbers $a_1,.....,a_{p-1}$ less than $2p^2$ with their pairwise sums different (then we just plug in $p=101$ to get the result)
If $a$ is an integer then define $a'$ to be the remainder obtained when $a$ is divided by $p$
Now choose $a_n=2np+n^{2'}$
Clearly,$\max(a_i)=a_{p-1}=2p(p-1)+(p-1)^{2'}<2p(p-1)+p<2p^2$ and it remains to check that the pairwise sums are all distinct.
Now for the sake of contradiction assume that $a_n+a_m=a_k+a_l$ for some $n \neq m$ and $k \neq l$ between $1$ and $p-1$.
Note that this is equivalent to $2p(n+m-k-l)=k^{2'}+l^{2'}-m^{2'}-n^{2'}-------(1)$
The right hand side is between $2-2(p-1)$ and $2(p-1)-2$,so $n+m=k+l$
Now $n^2+m^2 \equiv k^2+l^2 \pmod p$(from $(1)$) and combining this gives $nm \equiv lk \pmod p$
Thus to conclude by the general method for such cases consider a polynomial $P(X)=(X-m)(X-n)-(X-l)(X-k)$ are multiples of $p$,hence choose $m=l$ and $n=k$ or $m=k$ and $n=l$ for stewing up a contradiction,hence we have exhausted all possibilities and the result is clear by substituting $p=101$
[/hide]


inequalities
analytic geometry
graphing lines
slope
geometry
incenter
circumcircle

please respond in spoiler. extra points for generalizing a problem. the deadline for replying is Sunday, September 21, 2003 at 6:00 PM EST.

1. The students in a class of 60 students were given a test. 37 of the students were male and 23 were female. The average score was 80. The average score for those who passed was 84 and the average score for those who failed was 60. How many students failed?

2. Which of the following triplets could not be the sides of a triangle?
(7,7,8); (3,4,6); (2,7,11); (17,18,25); (1023,2168,3040). Of the remaining 4 triples, which triples form the sides of an obtuse triangle?

3. If the graphs of 2x+y+3=0 and 3y+ax+2=0 are to meet at right angles, then what is a?

4. In a circle of radius r, a central angle of 4.5 radians subtends an arc of 108 feet. What is r?

5. Let ABC be an equilateral triangle with sides x. Let P be the point where the angle bisectors of ABC meet. What is AP, in terms of x?

[hide]

1. Note that the average of a test scoreis the total sum of all the scores divided by the number of people who took the test. Having that said, let A = the sum of the passing scores B = the sum of the failing scores x = the number of people who passed y = the number of people who failed. The average of the whole class is

(A+B)/(x+y) = 80  because everyone either passed or failed.

A/x = 84

B/y = 60

A = 84x

B = 60y

Substituting these in, we get

(84x + 60y)/ (X+y) = 80

84x+60y = 80x+80y

4x = 20y

x=5y

We also know that there are a total of 60 people, where a number of them failed, but the others did not. The number of girls and boys is irrelevant.

x+y = 60

6y = 10

y= 10 people who failed.

2. From the triangle inequality we have a+b>c, where a b and c are sides of a triangle. If we set a and b as the shortest sides, we we can see if such a triangle exists. Applying this to each one, we know that a triangle with sides (2;7;11)  can not exist.

For an obtuse triangle, a:^2:+b:^2: < c:^2: where a  and b are the shortest sides. Applying this, we see that (7;7;8), (17;18;25), and (1023; 2168; 3040) are obtuse triangles.

3. Two lines are perpendicular iff the slope of one line is the negative reciprocal of the other. Putting the equations in slope intercept form, we get

y = -2x-3

y = -ax/3 - 2/3

-a/3 = 1/2

a = -3/2

4. 4.5/2:pi: * 2:pi:r = 108

4.5r = 108

r = 24

5. We can plainly see that P is the incenter of triangle ABC, also it's circumcenter, centroid, and orthocenter because it is an equilateral triangle. Using these facts, drop a perpendicular from P to AB, and call it D. It's clear that AD = DB, because Triangles ACD and BCD are congruent, so AD = x/2. We know angle PAD = 30 degrees, so we have a 30-60-90 triangle. PD = :rt3:x/2, so AP = :rt3:x

[/hide]

[hide]1. 60*80 = 84(60 - f) + 60f

4800 = 5040 - 84f + 60f

24f = 240

f = 10

gerneralize:

n = # of people in class

f = failed people

a = average score of whole class

s = average score for failing

p = average score for passing...

an = p(n-f) + sf

an = pn - pf + sf

an - pn = f(s-p)

f = (an-pn)/(s-p)  or n(a-p)/(s-p)

2. a. (2,7,11) wouldn't be because 2+7 < 11

    b. (7,7,8) because 7 :^2: +7 :^2: > 8 :^2:

generalize: wouldn't it just be...

in a triangle with sides a, b,c

the sum of the lengths of the two sides have to be greater than the length of the remaining side, and the difference between the lengths of two sides have to be less than the length of the remaining side

If a triangle with sides a,b,c and a<b<c

and a :^2:  + b :^2:  > c :^2:

then it is obtuse.

3. In slope intercept form:

y = -2x - 3

y = -(a/3)x - 2/3

-a/3 = 1/2

-2a = 3

a = -2/3

generalize: if 2 lines intersect in right angles then:

m_1 = -1/m_2

4. 4.5 radians is 810 degrees. So then...

(810/360) * C = 108

2.25 :pi: 2r = 108

r  :approx: 7.64

generalize:

l = length the arc the central angle subtends.

r = radius

a = angle in radian

the angle in degrees is 180a = d

(180d/360) * C = l

(d/2) * 2r :pi: = l

r = l/(d :pi: )

5. The angle bisector of a eq. triangal is also the altitude, perp. bisector and median... and the meeting point of the medians seperates for example AD into AP = 2/3 AD and PD = 1/3 AD

And

AD = ( :rt3: /2)x

So AP = ( :rt3: /2)x * (2/3)

= ( :rt3: /3)x

and wouldn't that be generalizing it??[/hide]

ill post solutions for this tomorrow, i have to scan the two towers quickly for my book report. though ive already read it before, but pfft, im screwin the skoo system

this round is over tomorrow, after that is when i'll give the answer and feedback

20 minutes left all!

edited: this contest is now over. you may read the solutions given, comment, etc. the solutions will be out shortly.

Chinaboy, I think you made a couple really dumb mistakes.

I agree, although you made some too, so don't use "dumb" anymore. Call it "careless" or something.

Some? I'm pretty sure I only made one. Do you know which one that is?

*with a British accent*"Are you mocking me?" :)
 It's number 5, and I used "some" since you could've made some mistake if you generalized the problems. I apologize for that assumption.
I'm just trying to advise people to use nicer words... :cry:

Is the official solution going to be posted soon? today?


Suppose that the number of people entering a department store in a given day is a random variable with mean 50. Suppose further that the amount of money spent by these customers are random variables with common mean 8. Assume also that the amount of money spent by a customer is independent of the total number of customers that have entered the store. What is the expected amount of money spent in the store on a given day.

Well, what do can you say about $ \mathbb{E}XY$ for independent $ X$ and $ Y$?


[b]
I don't quite follow you.
Do you mean Geometry prove or Induction or Inequality?[/b]

Generally, for geometry, you need to draw a figure, lable what is given and what you need to find. I recomend using a 2-column proof style i.e:

Statement | Reason
--------------|-----------
statement1|given
statement2|reason

etc.
You end up with a nice proof.

[quote="shinwoo"]what do we have to do when we try to prove something?[/quote]Regardless of whether it involves geometry or not, to prove a mathematical proposition you need to leave no doubt in a mathematician's mind that your proposition is true. That is, show that your proposition is true using precise mathematical statements that are either well-known, obvious, or verified by you. That's how I think of it, at least.


Someone seemed to have fun with valuations on number fields :D (see http://www.problem-solving.be/pen/viewtopic.php?t=216 ).

Note that
\[4a^{3}+2b^{3}+c^{3}-6abc = (\sqrt[3]{4}a+\sqrt[3]{2}b+c)(2 \sqrt[3]{2}a^{2}+\sqrt[3]{4}b^{2}+c^{2}-2ab-\sqrt[3]{2}bc-\sqrt[3]{4}ac ).\]
Lets write simply $d$ for the second term on the right hand side.

We see that $4a^{3}+2b^{3}+c^{3}-6abc =0$ has only the trivial solution $a=b=c=0$ by infinite descent $\mod 2$.
Thus $|4a^{3}+2b^{3}+c^{3}-6abc | \geq 1 = \frac{|d|}{|d|}$ and division by $|d|$ gives us that $|\sqrt[3]{4}a+\sqrt[3]{2}b+c| \geq \frac 1{|d|}$.

We are thus left to show that $|d| \leq 4a^{2}+3b^{2}+2c^{2}$, and I'm not in the mood to do this ($d=\| \sqrt[3]{4}\zeta_{3}^{2}a+\sqrt[3]{2}\zeta_{3}b+c \|^{2}>0$ is not hard to show).

In general for any nonzero algebraic integer $z$, the norm $N(z)$ is a nonzero integer, thus $|N(z)| \geq 1$.

[quote="Peter"]Let $ a, b, c$ be integers, not all equal to $ 0$. Show that
\[ \frac{1}{4a^{2}\plus{}3b^{2}\plus{}2c^{2}}\le\vert\sqrt[3]{4}a\plus{}\sqrt[3]{2}b\plus{}c\vert.\]
[/quote]

The reference for this problem "Belarus 2001" is not correct.
The correct reference is [b]"CRUX(No. 7, Volume 25, 1999), Problem A240, Mohammed Aassila"[/b]

[quote="ideahitme"]...
The reference for this problem "Belarus 2001" is not correct.
The correct reference is [b]"CRUX(No. 7, Volume 25, 1999), Problem A240, Mohammed Aassila"[/b][/quote]
Both references are incorrect.
The problem appeared in the [b]KöMaL magazine in December, 1996 (problem N.122.).[/b] See [url=http://db.komal.hu/scan/1996/12/99612548.tif.png]here[/url]. I was the author, but neither I can be sure that nobody ever constructed a similar exercise earlier...


During Juanita's last basketball season she scored a total of 12 points in each of two games, 8 points in each of three games, and 11 points in one game. Express her mean number of points per game played as a decimal to the nearest tenth.

$ Mean \equal{} \frac{2*12 \plus{} 3*8 \plus{} 11}{6} \equal{} \frac{59}{6} \equal{} 9.8\overline{3}$

So that to the nearest tenth is $ \boxed{9.8}$


Real numbers $ x,y$ are such that $ x^p\plus{}y^q$ is rational for any different odd primes $ p,q$. Prove that $ x,y$ are rational.

For $ p,q \in \{3,5,7 \}$ we have $ y^p \minus{} y^q \equal{} (x^{11} \plus{} y^p) \minus{} (x^{11} \plus{} y^q)$ is rational.
Thus, $ y^5(y^2 \minus{} 1)$ and $ y^3(y^2 \minus{} 1)$ are rational.
Assume that $ y \notin \{ \minus{} 1,0,1 \}$ otherwise we are done. Thus $ y^2 \equal{} \frac {y^5(y^2 \minus{} 1)}{y^3(y^2 \minus{} 1)}$ is rational.
It follows that $ y^2 \minus{} 1$ is rational, so $ y \equal{} \frac {y^3(y^2 \minus{} 1)}{y^2(y^2 \minus{} 1)}$ is rational.

In the same manner, we prove that $ x$ is rational.

Pierre.


analytic geometry
graphing lines
slope
geometry
perpendicular bisector

(1) How many points are there in a plane that are 4 cm. from a given line and also 5 cm. from a given point on that line?

(2) How many points are there that are equidistant from two given points A and B and also 2 inches from the line passing through A and B?

(3) Given <ABC, what is the number of points found in the interior of the angle, equidistant form the sides of the angle, and also equidistant from points A and B?

(4) The number of points equidistant from two given parallel lines and at the same time equidistant from two given points on oneof these line is

(A) 1 (C) 3
(B) 2 (D) 4

(5) Line AB and CD are paralllel and are 6 inches apart. Point P is on line AB. The number of points equidistant from these two lines and also 5 inches from point P is

(A)1 (C)0
(B)2 (D)4

(6) The triangle whose vertices are A(1,2), B(7,2), and C(5,6).

(PART A) From the locus points equidistant from A and B, write an equation of the locus.
(PART B) Write an equation of the line passing through A and C.
(PART C) Find the coordinates of point P if P is a point on line AC that is equidistant from points A and B.

For number 1, There are either 2, 1, or 0 because if the points are far apart or very close apart, as illustrated, the lines would not connect. But if they are the right distance they would connect twice. They could also touch at one point. [geogebra]97c327210aaef21f00c76df01ac94dd892d8503b[/geogebra]
Hope thats clear.
(2) needs to be clarified. When it says "and also 2 inches from the line passing through A and B" , that doesnt make sense. What part?
(3)Still doesnt make Sense. There are infinite. Its a line
(4) one

For #1, there are 4. They lie where the circle of radius 5 and the center at the given point intersect the two lines parallel to the first one, either 4 inches above or below.

For #2, there are 2 points. They both lie on the perpendicular bisector, and there is one two inches away on either side of AB.

For #3, there is only one. It lies where the perpendicular bisector of AB intersects the angle bisector.

For #4, the answer should be (A) 1. The point lies where the perpendicular bisector of the segment connecting the two points intersects the line parallel to and halfway between the other two lines.

For #5, the answer is (B) 2. The points lie where the circle with center P and radius 5 intersects the line parallel to and halfway between AB and CD.

For #6A, the locus is the perpendicular bisector of AB. The solution is easy because the y-coordinates are the same. $ x\equal{}4$
For #6B, let the slope be $ m$.

$ m\equal{}\frac{6\minus{}2}{5\minus{}1}\equal{}\frac{4}{4}\equal{}1$

Using point-slope form:

$ y\minus{}2\equal{}1(x\minus{}1)$
$ y\minus{}2\equal{}x\minus{}1$
$ y\equal{}x\plus{}1$

For #6C, P is located at the intersection of the perpendicular bisector of AB and the line AC. We have the two equations from parts A and B:

$ x\equal{}4$
$ y\equal{}x\plus{}1$

Therefore,

$ y\equal{}4\plus{}1\equal{}5$

So the solution is $ P(4,5)$


If $p$ is a perfect square, then the next perfect square greater than $p$ is:

A) $p^2+1$
B) $p^2+p$
C) $2p+1$
D) $p+2 \sqrt p + 1$
E) None of these

You may challenge yourself or look at my solution:

[hide="I think this is answer"]

Name n as $k^2$, then we can rewrite all the terms in terms of k.

The fourth one is equal to this:

$p + 2 \sqrt p + 1\\
k^2 + 2k + 1\\
(k+1)^2$

So, this is a next perfect square.

[color=darkred]No need for underscores.[/color][/hide]
If there wasn't an E choice, I could use process of elimination (disproving all of the other choices), but I guess not...

[hide="easy ans"]d[/hide][hide="explanation"]
We assign n to  :sqrt: p, and we are tasked with finding (n+1)2, which just happens to be n 2 +2n+1. We substitute  :sqrt: p back in to get p+2 :sqrt: 5+1, which is [d].[/hide][hide="tip"]It is a nifty trick that if you have n and n2 and need to find (n+1)2 you add n, n+1, and n2 together to get it.[/hide]

[quote="Silverfalcon"]If $p$ is a perfect square, then the next perfect square greater than $p$ is:

A) $p^2+1$
B) $p^2+p$
C) $2p+1$
D) $p+2 \sqrt p + 1$
E) None of these[/quote]

[hide="my solution"]plug is one.
d is the only one that works
let p=x^2
x^2+2x+1=(x+1)^2
if p is a perfect square, x is an integer.
therefor (x+1)^2 will always be the next perfect square [/hide]


calculus
integration
calculus computations

Suppose g'(x) < 0 for all x>= 0 and F(x)= the integral of tg'(t)dt from x to 0 for all x>=0. Which of the following is FALSE?

a) F takes on negative vaules.
b) F is continuous for all x>0
c) F(x) = xg(x) - integral of tg(t)dt from x to 0.
d) F'(x) exists for all x>0
e) F is an increasing function.

Do you really mean $ F(x)\equal{}\int_x^0 tg'(t)\,dt$ or should that be $ F(x)\equal{}\int_0^x tg'(t)\,dt$?

It is a little ambiguous

For a) F doesn't take negative value at 0 so I'm not sure if that's supposed to be a trick question or not.

mlok's question pertains to e)

counterexample shows c to be false


induction
pigeonhole principle
combinatorics unsolved
combinatorics

On a $ 50 \times 50$ board, the centers of several unit squares are colored black. Find the maximum number of centers that can be colored black in such a way that no three black points form a right-angled triangle.

[hide="is it correct?"]Induction on $ m\plus{}n$ for an $ m \times n$ board with $ m,n \ge 2$ with maximum for $ f(m \times n)\equal{}m\plus{}n\minus{}2$

Base case: $ m\plus{}n\equal{}4$, $ m\equal{}n\equal{}2$
Trivial with an example painting the two diagonal squares then $ f(2 \times 2 )\equal{} 2\equal{}m\plus{}n\minus{}2$

Assume that it is true for every $ m \times n$ square with $ 4 \le m\plus{}n < k$ and $ m,n>2$

For the case of $ m\plus{}n\equal{}k$, one example with $ f(m \times n)\equal{}m\plus{}n\minus{}2$ is painting the first row and the first column without its intersection.

Suppose that there is a painting with $ f(m \times n)>m\plus{}n\minus{}2$ then by pigeonhole principle there is a column with two painted squares in row $ i$ and $ j$.

Case 1) $ i\equal{}1$, $ j\equal{}m$ then $ f(m \times n)\le 2\plus{}f((m\minus{}2) \times n)$ no more painted squares on the $ i$ or $ j$ row.
If $ m>3$ then $ m\minus{}2\ge 2$ then by induction hypothesis $ f(m \times n)\le 2\plus{}f((m\minus{}2) \times n)\equal{}2\plus{}m\minus{}2\plus{}n\minus{}2\equal{}m\plus{}n\minus{}2$ Contradiction.
If $ m\equal{}3$ then $ f(m \times n)\le 2\plus{}f(1 \times n)$
if it is painted the square of that column on the remaining row then there is no other square painted on that row, $ f(m \times n)\le 2\plus{}f(1 \times n)\equal{}2\plus{}1\equal{}3$ Contradiction.
If that square it is not painted then $ f(1 \times n) \le n\minus{}1$ then $ f(m \times n)\equal{}f(3 \times n) \le 2\plus{}n\minus{}1\equal{}1\plus{}n\minus{}2$ Contradiction.
If $ m\equal{}2$ then $ f(m \times n)\equal{}2$ no more painted squares on the $ i$ or $ j$ row. Contradiction.

Similarly for the other cases with two or three bands.
Then $ f(50 \times 50)\equal{}98$[/hide]


On average, how often do the minute and hour hands meet on an ordinary clock?

Do you mean how often? In that case, it should be[hide]Every 720/11 Minutes[/hide]

Would not it just simply be every [hide="EDIT Wrong"] $65$ minutes?[/hide]

Well, unless I'm over thinking this, I think it should be

[hide]Well, it wouldn't be every 65 minutes because the hour hand goes farther away.  At 12 o'clock they would be together, but they wouldn't meet up again until 1:05 and 1:06 because the hour hand would be at the 1 at 1:00 sharp, but it would be 1/12 of the way between the 1 and the 2 when it meets the minute hand.  So 1/12 of 5 is 5/12, so the answer is on average, $65\frac{5}{12}$[/hide]
The best part of doing math is making it seem like you're the one that's right, even when you have no clue as to what the answer is.

Oh okay now the other answer makes sense.

Magixter's answer is correct.

[hide="Solution"]Every 12 hours, the hands meet 11 times. Therefore, they meet every $\frac{12}{11}$ hours, or $\frac{720}{11}$ minutes, on average.[/hide]


A volume of 0.50 L of water at 16 degrees Celsius is put into an aluminum ice-cube tray of mass 0.250 kg a the same temperature. How much energy must be removed from this system by the refrigerator to turn the water into ice at -8.0 degrees Celsius?

There must be other data in order to solve this problem.

Well, is this all the other info you need?

Specific heat of Aluminum 920 J / (kg C).
Specific heat of Water 4186 J / (kg C).
Specific heat of Ice 2100 J / (kg C). (do I need this one?)
Heat of Fusion for Water $ 3.33 \times 10^{5}$ J / kg


[quote="Magic Man"]im still struggling with logs of base e i cant seem to figure out how to write them out! for example:
how do u solve the following

1)e^1-x = 17.5
2)3e^2x-1 = 0

any help is greatly appriciated [size=150]thank you very much![/size]
Magic Man  :ninja:[/quote]

The notation for the log of base e is $ \ln$. If $ e^a\equal{}b$, $ a\equal{}\ln b$.

I'm assuming you meant
$ 1) e^{1\minus{}x}\equal{}17.5$
$ 2) 3e^{2x}\minus{}1\equal{}0$
(otherwise, the first one becomes trivial)

[hide]For the first one, try to take the natural log of both sides. The second one is a little harder but it's basically the same thing. Now try to solve it yourself.[/hide]

that dosent really explain much could some one explain it in more detail please
thank you

For the first one:

$ e^{1 \minus{} x} \equal{} 17.5$

If you take the natural log of both sides, you get

$ \ln e^{1 \minus{} x} \equal{} \ln 17.5$

Because $ \ln e$ cancels out, you are left with

$ 1 \minus{} x \equal{} \ln 17.5$

Just solve for $ x$.


function
parameterization
parametric equation

Two girls start moving at the same time from the same point A and go in opposite directions on a track whose graphical representations is $ \frac{(x\minus{}7)^2}{3} \plus{} \frac{(y\plus{}19)^2}{5} \equal{} 1$. Their respective speeds are $ 7 \frac{1}{3}$ feet per second and $ 9 \frac{1}{5}$ feet per second. When the girls meet at point A again for the first time, find the number of times they passed each other. (Note: their leaving A together for the first time and meeting at point A the last time are not passes.)

I read the solution and they mentioned parametric equations(?) in the first line. I need help understanding what it is. Thanks!

It's when variables are expressed as functions of some other parameter.
Ex: Suppose we had the graph y=2x. That's the same as if we have y(t) = 2t and x(t) = t, and allow t to vary over the reals.

[hide="Expressing a graph parametrically has several advantages"]
- It shows how the graph is drawn. For example, saying that a truck takes the path y=2x gives much less information than saying that a truck takes the path given by x=t^2, y=2t^2 for nonnegative t - the second might relate the position of the truck to the time, and would give you important information, such as acceleration. It also tells which direction the graph is being drawn in, if there is any retracing, and stuff like that
- It shows causality - it may make sense to say that y and x vary with time rather than saying y is caused by x
- It often makes it easier to write equations for graphs, especially if writing them either explicitly or implicitly would require the use of awkward inverse functions[/hide]

In this case, parametric form would allow the expression of each girl's position in terms of time, because in this case, we are looking at position in the context of time.

_A_ parametric equation for the unit circle is y(t) = sin(t), x(t) = cos(t), with 0<=t<2pi. See if you can use this to parametrize your ellipse.


From BMO (1993):

[quote]
Find, showing your method, a six-digit integer n with the
following properties: (i) n is a perfect square, (ii) the
number formed by the last three digits of n is exactly one
greater than the number formed by the ﬁrst three digits
of n. (Thus n might look like 123124, although this is not
a square.)
[/quote]

Call $ n$ the number formed by the first three digits ($ 100\le n \le 998$)

We have
$ 1001n + 1 = g^2$

$ n = \frac {g^2 - 1}{1001}$

So, $ 1001 | g^2 - 1$, which is

$ \begin{cases} g^2 \equiv 1 (7) \\
g^2 \equiv 1 (11) \\
g^2 \equiv 1 (13)\end{cases}$

So, we have:

$ g \equiv 1  or  - 1 (7)$
$ g \equiv 1  or  - 1 (11)$
$ g \equiv 1  or  - 1 (13)$

But I really don't know how to go on... I suppose I have to use the Chinese Remainder Theorem, but I have not completely understood how to find the value for it!
Or is there a better way to solve the problem?

Thanks in advance!  :)

[quote="Gabuntu94"]
So, we have:

$ g \equiv 1  or  - 1 (7)$
$ g \equiv 1  or  - 1 (11)$
$ g \equiv 1  or  - 1 (13)$

But I really don't know how to go on... I suppose I have to use the Chinese Remainder Theorem, but I have not completely understood how to find the value for it!
[/quote]
yes, you have to use Chinese Remainder theorem only from here...though I didn't think of your method when I first saw this problem... :P
Ok now that you have obtained these three congruences....Find the quadratic Residues mod 7,11,and 13....
The quadratic residues :-
1) mod 7 - 0,1,4,2
2) mod 11 - 0,1,4,9,5,3
3) mod 13 - 0,1,4,9,3,12,10

Therefore we have Two cases --- $ g \equiv 1 or -1 (mod \ 13)$
Take these cases, solve them and get the answer....(I haven't done that yet.... :) )

[quote="Rijul saini"]
yes, you have to use Chinese Remainder theorem only from here...though I didn't think of your method when I first saw this problem... :P
Ok now that you have obtained these three congruences....Find the quadratic Residues mod 7,11,and 13....
The quadratic residues :-
1) mod 7 - 0,1,4,2
2) mod 11 - 0,1,4,9,5,3
3) mod 13 - 0,1,4,9,3,12,10

Therefore we have Two cases --- $ g \equiv 1 or \minus{} 1 (mod \ 13)$
Take these cases, solve them and get the answer....(I haven't done that yet.... :) )[/quote]

Sorry... How do you obtain that?  :blush:
Which cases should I take in consideration?

 :)

No, just do what you were doing.  You already did all of what Rijul saini said, so don't worry about it.

Also, all the chinese remainder theorem says is that there is one and only one solution, not how to find it.  You use it to show that there is a unique solution, not on a problem like this.  To finish the problem and find all $ g$'s, we consider the 3 pairs of numbers $ (715, 286), (364, 637), (924, 77)$.  The pairs have the property that the numbers in each pair are $ 0\mod$ 2 of either 7, 11, or 13 and $ \pm 1\mod$ the third one.  For example, $ 715\equiv 0\mod 13\equiv0\mod11\equiv1\mod7$.

So if we add one number from each pair, we will get a number that is $ \pm 1\mod 7, 11, 13$.  We take the numbers mod 1001 so they are 3 digit numbers.  The numbers we get are listed as follows:  $ 1, 155, 428, 274, 846, 1000, 727, 573$ whose squares are $ 1, 24025, 183184, 75076, 715716, 1000000, 528529, 328329$.  The parenthesis in the following are just to denote the 3-digit numbers, not multiplication.  I don't know if $ (024)(025)$ or $ (075)(076)$ are allowed, but since $ (000)(001)$ and $ (999)(1000)$ aren't allowed, I suspect they aren't either.  So the list that works is $ 183184, 715716, 528529, 328329.$

Wow! Thank you very much tenniskidperson3!! :10:
I have understood!! :w00t:

But... how did you find the three pairs of numbers? (I mean $ (715, 286), (364, 637), (924, 77)$)

Again, thank you!!   :)


Prove that
$ \dfrac{a}{\sqrt{1\plus{}a^2}}\plus{}\dfrac{b}{\sqrt{1\plus{}b^2}}\plus{}\dfrac{c}{\sqrt{1\plus{}c^2}}\le\dfrac{3\sqrt{3}}{2}$  for $ a,b,c\in R^\plus{}$ such that $ a\plus{}b\plus{}c\equal{}abc$

Try subsituting a, b and c by tan(x), tan(z) and tan(y), there x, y and z are the angles of a triangle. The expression reduces to the sum sin(x)+sin(y)+sin(z). Since sin(x) is concave on the interval 0<x< pi, the maximum can be achieved using jensens inequality!


[quote="gpapargi"]Το παρακάτω πρόβλημα (βλέπε συνημμένο) έπεσε στο πολυτεχνείο Κρήτης και  μου το έστειλε ένας συνάδελφος μου πληροφορικός. Το να λυθεί με κυκλική λίστα (όπως περιγράφει η εκφώνηση) είναι μια τετριμμένη άσκηση για το μάθημα «δομές δεδομένων». Το ζητούμενο (και ο λόγος που το θέτω στο forum) είναι να βρεθεί [b]αλγεβρική λύση[/b].

[b]Δηλαδή ζητείται να βρεθεί η συνάρτηση που της δίνεις το πλήθος των κρατούμενων και σου επιστρέφει τον αριθμό του επιζώντα.[/b]

Το έψαξα λίγο και βρήκα τον τύπο που δείχνει να ισχύει. Θέλει όμως και απόδειξη. Μέχρι στιγμής τον έχω αποδείξει για κάποιες μεμονωμένες περιπτώσεις αλλά όχι γενικά. Είπα λοιπόν να το θέσω και στους άλλους και βλέπουμε. Δεν ξέρω αν είναι δύσκολο ή εύκολο πρόβλημα για κάποιον που ασχολείται με διακριτά μαθηματικά και θεωρία αριθμών.

ΥΓ
Αν δεν είναι για αυτόν τον πίνακα, please Sil μετέφερε  στους φοιτητές.
[/quote]

Αν έχω καταλάβει σωστά το πρόβλημα, τότε: $ f(n) \equal{} \begin{cases}2k \minus{} 1, & n \equal{} 2k \minus{} 1 \\
2k \minus{} 1, & n \equal{} 2k\end{cases}$

Είμαι σίγουρος ότι θα υπάρχει και πιο κομψός τρόπος να το δει κανείς, αλλά αρκεί να παρατηρήσεις ότι στον πρώτο γύρο εκτελέσεων πεθαίνουν οι άρτιοι ενώ στο δεύτερο οι περιττοί.

Cheerio,

Durandal 1707

Edit: πολύ αιματηρό πρόβλημα ρε George!

Πράγματι όσοι το είδαν αυτό σχολίασαν, και μάλιστα κυκλοφόρησε γύρω στην 28 Οκτωβρίου.

Δεν είμαι σίγουρος ότι κατάλαβα τη σκέψη σου Παναγιώτη. Υπάρχουν πολλοί γύροι εκτελέσεων ανάλογα με το πλήθος n. Στον πρώτο γύρο σκοτώνονται άρτιοι και σε όλους στους επόμενους περιττοί. Πως προκύπτει η τιμή 3 (αρχικός αριθμός επιζώντα) για n=9;

Durandal, εγώ διαφορετική λύση βρήκα. Σίγουρα η λύση σου δεν είναι ορθή αφού ο gpapargi μας λέει πως $ f(10) \equal{} 3$.  :lol:

Πάντως μια χαρά ολυμπιακό πρόβλημα. Νομίζω καλύτερα να μείνει εδώ

Παιδιά, άκυρο! Κατάλαβα ΕΝΤΕΛΩΣ διαφορετικό πρόβλημα, apologies!!!

Cheerio,

Durandal 1707

Edit: Demetres, σε παρακαλώ πες μου πως έκανες typo και ότι ο George μας λέει ότι $ f(9) \equal{}3$ αντί για $ f(10)$. Τρίτη εσφαλμένη ανάγνωση δε θα την αντέξω :lol:

[quote="Demetres"]ο gpapargi μας λέει πως $ f(10) \equal{} 3$.  :lol:
[/quote]

Προσοχή Δημήτρη: $ f(9) \equal{} 3$  :)

Γιώργος

Ok, πείτε μου αν τουλάχιστον αυτήν τη φορά κατάλαβα σωστά το πρόβλημα (μα μιλάμε για κρίση ηλιθιότητας! :lol:) και αν βρήκατε το ίδιο αποτέλεσμα (spoiler warning):
[hide]Δε θέλω να γράψω την απόδειξη για να μην τη χαλάσω σε όσους δεν έχουν ασχοληθεί ακόμα με την άσκηση. Ωστόσο, το αποτέλεσμα μου (και το hint ταυτόχρονα) είναι ότι αν $ n \equal{} 2^k \plus{} m$, όπου $ 0\leq m\leq 2^k \minus{} 1$, τότε η διαδικασία ολοκληρώνεται σε $ k \plus{} 1$ loops και $ f(n) \equal{} 2m \plus{} 1$.[/hide]
Cheerio,

Durandal 1707

Ναι, typo έκανα. $ f(9) \equal{} 3$, $ f(10) \equal{} 5$

Συμφωνώ με την απάντηση του Durandal. Δεν έκατσα να υπολογίσω σε πόσα loops τελειώνει, αλλά $ \pm 1$ είσαι σίγουρα μέσα.  :)

Τόσο το βρήκα και εγώ απλά εκφρασμένο στη μορφή $ f(n)\equal{}...$ για να μπει σε υπολογιστή αμέσως. Με ενδιαφέρει όμως η γενική  απόδειξη καθώς έχω βρει μόνο για μερικές περιπτώσεις. Ας τη στείλει κάποιος εδώ ή στο mail μου (φαίνεται στο profile μου).