ummagumma wanted a hard problem, so here it is: :)

We have a hollow sphere of inner radius $ R_1$ and outer radius $ R_2$.Find the deformation of the hollow sphere if the pressure inside it is $ p_1$ and the pressure outside is  $ p_2$

i know nothing about deformation...I'll need some (some?? :lol: ) days to learn it

I'm supposing the presure insite the small sphere is $ p_1$ the pressure out-side is $ p_1$ and the presure in the interface of the 2 spheres $ p_2$

you only give 2 pressures, the correct number would be 3. so for simplicity i assumed that.

In real life this kind of deformation would not be symmetrical and uniform. One point of the sphere would be more fragile and would deformate more etc etc.

here we dont have that. the pressure is equal in the surface of the sphere. so it will deformate (theorically) symmetrically and uniformly.

Let's make the process very slow by increscing $ p_2$ from $ p_1$ to $ p_3$, and the difference would not be much high.

Let $ V$ be the volume of the gas that have between the interface of the spheres.

and $ \Delta V$ be the modulus of the difference of the volume of the gas.

So we can use bulk modulus to solve for the variation of the volume.

For a more complex case we can use the same thought.

in 2 dimensions.

Suppose you have a circle of volume $ V$ and internal pressure $ p$.

Suppose you put it $ h$ meters inside the water. This circle is very elastic, can be tought as a ballon, just like the problem above. so it will deforme untill the pressure outside and inside be the same. (remember we have a gradient of pressure that is never non-zero, so the pressure that is equal is in the interface)


Suppose $ A$ and $ B$ are digits for which the product of the two-digit numbers $ \underline{2A}$ and $ \underline{AB}$ is $ 1,971$. Find the sum of $ A$ and $ B$.

You could go algebraic here, using $ 20\plus{}A$ and $ 10A\plus{}B$, but in the time limits of FTW, it's better to go for a lucky guess (with brute force and guess and check): $ 27\times73\equal{}1971$, and $ 7\plus{}3\equal{}\fbox{10}$.

There are 2 possibilities for A, B since they should multiply to give 1 or 21. No other multiplication results in 1 on the unit. We immediately see 21 * 11 is too low. So it's either 27 * 73 or 23 * 37.


group theory
abstract algebra
function
inequalities
superior algebra
superior algebra unsolved

Let $G$ be an infinite group. Is it true that the cardinality of the set of subgroups of $G$ is always precisely $\text{card} \, G$?

[Edit: As pointed out by jmerry, $(\mathbb{R},+)$ is an obvious counterexample to the statement that the set of subgroups has cardinality at most $\text{card} \, G$. Prove the converse: it has cardinality [i]at least[/i] $\text{card} \, G$.  ]

Certainly not. The additive group of $\mathbb{R}$ has more subgroups than that.

For an easier example, consider the group of functions from $\mathbb{R}$ to $\mathbb{Z}$ which are zero at all but finitely many points. The group has continuum cardinality, but we can define a subgroup for every subset of $\mathbb{R}$ by taking the functions which are zero outside this set. The cardinality of this set of subgroups clearly matches the power set of $\mathbb{R}$.

We can also show that your first example has $2^c$ subgroups ($c$ is the continuum cardinality). We choose a basis of $\mathbb R$ over $\mathbb Q$. Such a basis has $c$ elements, and for each subset $A$ of the basis, we can find a subgroup of $\mathbb (R,+)$ formed precisely by those elements of $\mathbb R$ which are only rational linear combinations of elements of $A$ (i.e. don't have elements of the basis outside $A$ in their decomposition).

In fact, this looks a lot like your second construction :).

That's right, $(\mathbb{R}, +)$ is the obvious example to the statement that $G$ may have more subgroups than $\text{card} \, G$  :blush: . On the other hand, the cardinality of the set of subgroups of $G$ is always at least $\text{card} \, G$ (and, of course, at most $2^{\text{card} \, G}$  :) ). Disregard the other (false) part of the problem.

Problem. Let $G$ be an infinite group and let $S(G)$ be the set of all subgroups of $G$. Then:
\[
\text{card} \, G \leq \text{card} \, S(G) \leq 2^{\text{card} \, G}.
\]

So we should prove the first inequality, right? :)

For each element $a$ of $G$, consider the group generated by $a$. We get a multiset of subgroups having cardinaltity $|G|$. Now, each subgroup obtained like this appears at most $\aleph_0$ times, so if $t$ is the cardinality of the set of cyclic subgroups, then $|G|\le \aleph_0\cdot t=t$ (this last equality is true for all infinite cardinals $t$).

Is it ok?

Nice! It's o.k.  ;)  Again simpler than what I had done!  :(  :lol: .


[quote="Resources Section"]India has been participating in the  International Mathematical Olympiad since 1989. :)
The selection of the Indian team is a three-tier process: a state level exam in November - 300 from all over India sit for the next phase - Indian National Mathematical Olympiad (INMO). 30 selected in INMO attend IMO Training Camp to select the Indian team and train them. For the whole year from the training camp upto the next training camp, the students are given postal coaching, and a few of these are selected to attend the camp again the following year.[/quote]
Where do you live?  :oops:

in tamil nadu we are eligible to write rmo only after clearing  the amti finals.but this is not so in other states.

Potla[quote]Where do you live? [/quote]

I am living in the USA, but I am moving back to India pretty soon.

[quote="Nlakkamraju"]How exactly are people choosen for the imo? is there an Indian equivalent for the AMC 10,12, AIME and the USAMO?[/quote]

As far as I know...The Indian equivalant for the AMC 10,12 and the AIME is the [url=http://en.wikipedia.org/wiki/Regional_Mathematical_Olympiad]Regional MAthematical Olympiad(RMO) ......[/url]

Unlike the AMC or the AIME , the RMO does not ask us to find values to a problem.....it requires detailed and rigorous proofs to questions of moderate difficulty......

Next comes the [url=http://en.wikipedia.org/wiki/Indian_National_Mathematical_Olympiad]INMO -- The Indian National MAthematical Olympiad[/url].....equivalent to the USAMO......but has questions far below the level of the USAMO......

Total 30 students are selected as the winners of these two exams .....who attend the [url=http://en.wikipedia.org/wiki/International_Mathematical_Olympiad_Training_Camp]IMOTC[/url]---the training camp for the IMO....From this 6 Students are selected from the 30 and are sent to the IMO to represent INDIA......

PM me if you want to ask anything else......

BTW in which class/grade are you. That matters since the ones in 12th have a disadvantage.
Only 6 students from 12th in the RMO out of 30. Also in the INMO 6 students are selected of 12th.
Also when are you coming to India....Registration is not going to last long......

When does regristration start? I willl be starting 9thin India next school year. What are the essential courses needed fot the RMO and INMO?

[quote="nuclear_alchemist"]It ends on 20th oct here. You need attested marksheets from principal etc. So HURRY!!![/quote]
Sorry, but from my information the delhi region's last date is 30th......

[quote="Nlakkamraju"]I will be moving th AP, India. I heard that RMO varies from region to region. What about AP? And anyone know if i could take the RMO in Hydrabad? What is the exact format of the test? Does anyone have past tests and solutions?[/quote]
Yes, it varies from region to region...read the wikipedia topic for RMO carefully......Also to get the format you can take a look in the resources section......


Hi,

Anyone has this [url=http://www.amazon.com/Art-Craft-Problem-Solving/dp/0471789011/ref=reader_req_dp/102-1019131-0052133]book[/url] here ? Can you describe it, please ?  :lol:

Thanks you,

Best regards.

I had it, but then ragnarok23 stole it.
PM him, or IM him, or email him.

I have it, but havent read it yet. So I dont know.

psh. some help you guys are. :)
anyway, the book is what the title advertises. it really gets into the essence of problem solving, and proposes a wide variety of problems (however many have no solution...)

I teach a college class entitled "Honors Seminar in Problem Solving" that aims at the Putnam exam. There's not exactly a textbook for the course - just problem sets - but the book I recommend to my students is Zeitz.

it is somewhat efficient, but isnt the name kindof copyrighted?

Most people say that it should be read by people who have done well enough on the AIME, like over 6 or so, and/or have qualified for USAMO.  buzzer11 stated that many problems in the book don't have solutions, but to clarify, the book just doesn't give the solutions; hints are on a website.  There's some overlap between the book and AoPS books.


#8

To win a game with a fair coin, you must flip (n+1) heads in a row, where (n) is the total number of tails flipped so far.  So if your first flip is heads, you win and the game is over; if your first flip is tailed but youur next two are heads, you win and the game is over; and so on.  What is the probability that you win by flipping exactly four heads in a row?  Express your answer as a common fraction.

H=Heads
T=Tails

TTTHHHH
THTTHHHH
TTHHTHHHH
TTHTHHHH
THTHTHHHH
THTHHTHHHH

Add up the probability of each to get $ \frac{21}{1024}$

[hide="solution"]

You know that the first flip must be tails and the last five flips are THHHH (you can't have five heads in a row, since that would've already finished the game)

Now, you must fit in another tail in between to get the required n=3.

We can just list out now, or use casework (the x is either a head or a tail, we're letting it take up a spot).
[hide="Case 1, T x T H H H H"] Clearly this can only be TTTHHHH.[/hide]
[hide="Case 2, T x x T H H H H"] You can have either T H or HT.[/hide]
[hide="Case 3, T x x x  T H H H H"] H T H or T H H.[/hide]
[hide="Case 4, T x x x x  T H H H H"] H T H H.[/hide]

Our total probability is $ \frac{1}{128}\plus{}\frac{2}{256}\plus{}\frac{2}{512}\plus{}\frac{1}{1024}\equal{}\frac{8\plus{}8\plus{}4\plus{}1}{1024}\equal{}\frac{21}{1024}$.[/hide]

UH, dude... this was already on my topic Mathcounts Battlefield-The Reality..... and it is 21/1024

Sorry.  I don't check on the Battlefield. But, thanks.


geometry
geometric transformation
reflection
MATHCOUNTS
Pythagorean Theorem

A variation of this was on our school's math team tryout.

In Flatland, all rivers are straight. The Axis River runs in a east-west direction.

A man is 8 miles due north of the Axis River. From this point he is 4 miles south and 5 miles east of his house. He must go to the Axis River to collect water and then return to his house. What is the minimum possible distance he can walk in order to do this?

[hide="hint"]The answer is not 21 miles.[/hide]

[quote="mathgeniuse^ln(x)"][hide]Notice that the house can be reflected across the river.

Then draw a line connecting the man and the house.

Then reflect the part of the line on the other side of the river to this side and get that as your path.

So the answer is just the distance from the reflected point to the man and that is 13.[/hide][/quote]
The idea is correct but the answer is not 13.

I doubt this is right...
[hide]
Like mathgeniuse^ln(x) said, we can reflect the house across the river. Now we can draw the straight line paths to create a right triangle whose legs have lengths 20 and 5. By the pythagorean theorem, the answer is$\sqrt{425}$ which can be simplified to $5\sqrt{17}$[/hide]

[quote="Ubemaya"]That's what I got.

How did you come up with the reflection trick?[/quote]
I personally have seen this problem before on mathcounts, so I know about the reflection trick.
Maybe the same for math genius.

Sorry, I thought it was 4 below not above.

So the answer is $\sqrt{20^{2}+5^{2}}=\sqrt{425}=5\sqrt{17}$


A topological space is termed metrizable if there exists a metric such that the topology arises as the topology of the metric.
Prove that any metrizable space is Hausdorff, i.e. for every two distinct points $ x$ and $ y$, there exist disjoint neighbourhoods $ U$ and $ V$ of $ x$ and $ y$ respectively.

This is trivial.  Let the neighborhoods be open balls of radius $ \frac {d(x, y)}{3}$.


geometry
rectangle
quadratics
algebra
quadratic formula

$ABCD$ is a rectangle with area 180 and $E$ is the midpoint of side $\overline{CD}$. If $AF = 4$, with $F$ on side $\overline{AB}$, $BC = 2x-1$ and $CE = 4x+1$, for some $x$, find the numerical area of $\triangle AEF$.

[hide]Since CE = 4x+1, CD = 8x+2. That means $(8x+2)(2x-1)=180$
$16x^2-4x-2=180$, combining like terms and dividing by 2, $8x^2-2x-91=0$
$(4x+13)(2x-7)=0$, $x=\frac{7}{2},-\frac{13}{4}$

Since lengths are positive, $x=\frac{7}{2}$.
We know the height of $\triangle{AEF}$ is 4, now we know its base is $\frac{7}{2}\cdot2-1=6$.

$A=\frac{1}{2}bh=12$. [/hide]

[hide]$K_{ABCD} = 180 = (8x+2)(2x-1) \Rightarrow x=3.5, \therefore K_{\triangle AFE}=4/2(2x-1)=2(6)=12$. (Go google calculator! :P )[/hide]

[hide]Okay.  Start with the rectangle.  Then you put the lengths on it.  Then form the equation (8x+2)(2x-1)=180.  Solve it using the quadratic formula and take the positive value so x=3.5.  Plug the value in and you get the height of the triangle(6)  Do 6*4=24/2 = 12
12 is the answer.[/hide]


Two boys and four girls are officers of the Math Club. When the photographer takes a picture for the school yearbook, she asks the club's six officers and the faculty sponsor to sit in a row with the faculty sponsor in the middle and the two boys not next to one another. How many different seating arrangements are possible?

Deal with the sponsor, then you have $ 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$ ways for them to be arranged, minus the 4 (BBGSGGG, GBBSGGG, GGGSBBG, and GGGSGBB) ways that the boys are sitting next to each other gives you $ \boxed{716}$

The answer is 528. the girls can be arranged $ 4 \cdot 3 \cdot 2 \cdot 1 \equal{} 24$ ways. The boys can be arranged 22 ways with the sponsor in the middle in positions $ 1 2 3 S 4 5 6$:
3,4; 3,5; 3,6; and the other way ( 4,3; 4,2; 4,1 ) with a total of 6.
2,4; 2,5; 2,6; and the other way, with a total of 6.
1,4; 1,5; 1,6; and the other way, with a total of 6.
1,3; 3,2; 4,6; 6,4; with a total of 4.

So $ 6\plus{}6\plus{}6\plus{}4\equal{}22$ and $ 22 \cdot 24\equal{}528$ ways. :)

Ebob's way will also work.  One just has to minus those four cases multiplied by how many ways one can arrange the boys and the girls.

We can solve this problem with complementary counting. There are a total of $6!=6*5*4*3*2*1=720$ arrangements for the boys and girls. Then, we figure out how many arrangements are there with the two boys together. The seating row already looks like this. $***S***$, where "S" is the sponsor. You see that the two boys can be placed in the asterisks in $4$ different ways, $BB*S***$, $*BBS***$, $***SBB*$, $***S*BB$. There are a total of $4!=4*3*2*1=24$ ways to seat the four remaining girls. Now, there are two ways to seat the distinguishable boys. The arrangement $B_1B_2*S***$ is different from $B_2B_1*S***$. $B_1$ is boy number one and $B_2$ is boy number two. Multiplying all the criteria, we get $24*4*2=192$. Subtracting this from $720$ reveals that the answer is $720-192=\boxed{528\text{ total possible arrangements.}}$


I was doing my homework and one of the problems said:
Simplify this fraction:$ \cfrac{\cfrac{1}{b(abc+a+c)}-\cfrac{1}{a+\cfrac{1}{b+\cfrac{1}{c}}}}{\cfrac{1}{a+\cfrac{1}{b}}}$
This is what i did and i was wondering if it is possible to simplify it even more.
Thanks
[hide="what i got"]
\begin{eqnarray*}
\displaystyle
\cfrac{\cfrac{1}{b(abc+a+c)}-\cfrac{1}{a+\cfrac{1}{b+\cfrac{1}{c}}}}{\cfrac{1}{a+\cfrac{1}{b}}}&=&\left(\cfrac{1}{b(abc+a+c)}-\cfrac{1}{a+\cfrac{1}{b+\cfrac{1}{c}}}\right)\cdot{a+\cfrac{1}{b}}\\
&&\\
&&\\
&=&\left(\cfrac{1}{b(abc+a+c)}-\cfrac{1}{a+\cfrac{1}{\cfrac{1+bc}{c}}}\right)\cdot{a+\cfrac{1}{b}}\\
&&\\
&&\\
&=&\left(\cfrac{1}{b(abc+a+c)}-\cfrac{1}{\cfrac{abc+a+c}{1+bc}}\right)\cdot{a+\cfrac{1}{b}}\\
&&\\
&&\\
&=&\left(\cfrac{1}{b(abc+a+c)}-\cfrac{1+bc}{abc+a+c}\right)\cdot{a+\cfrac{1}{b}}\\
&&\\
&&\\
&=&\cfrac{-b^2c-b+1}{b(abc+a+c)}\cdot\cfrac{1+ab}{b}\\
&&\\
&&\\
&=&\boxed{\cfrac{-ab^3c-b^2c-ab^2+ab-b+1}{b^2(abc+a+c)}}\\
\end{eqnarray*}
[/hide]

Hmm...is there anything similar to the first three terms of the numerator and the expansion of the denominator?

If you expand the numerator:
$ \frac{b^2(abc\plus{}a\plus{}c) \plus{} ab\minus{}b \plus{}1}{b^2(abc\plus{}a\plus{}c)}$
But you can't cancell... can you?

Well, if this is called simplifying, you can turn it into $ \frac{ab\minus{}b\plus{}1}{b^2(abc\plus{}a\plus{}c)}\minus{}1$.


[hide="Incomplete attempt"]

If all 3 are perfect powers of 12, then it is possible to take the roots perfectly.
So x y and z can be expressed as m^12, n^12 and p^12.

m^4 - n^4 - p^4 = 16
m^3 - n^3 - p^3 = 8
m^2 - n^2 - p^2 = 4

solve for m, n and p then take their 12th powers.

[/hide]

[hide="Guess"]

Is x=2^12, y=z=0 the only solution?
Probably not, i'll have to do some more work.

[/hide]

This problem is in AOPS Volume 2, initially it was an MOP problem.


quadratics
algebra
real analysis
real analysis unsolved

(Appearently I posted in the wrong forum. If any mod reads this, u could delete the other one if you want to.)

Hi, I have two things here that I really can't solve.

1) Prove that if the equation
ax^4 + bx^3 + cx^2 - bx + a = 0 (a != 0)
has a solution x = r, then x = -1/r is also a solution to the equation.

2) Solve the equation
6x^4 - 35x^3 + 62x^2 - 35x + 6 = 0
Start by dividing everything by x^2.
Bring together all terms with the same coefficient and assume that
x + 1/x = y.

What I don't get in the second one is the how to do the second step.

(1) Since $ar^4+br^3+cr^2-br+a=0$,

we have $a\left(-\frac{1}{r}\right)^4+b\left(-\frac{1}{r}\right)^3+c\left(-\frac{1}{r}\right)^2-b\left(-\frac{1}{r}\right)+a=\frac{1}{r^4}(a-br+cr^2+br^3+ar^4)=0$. Q.E.D.

(2) Since $x=0$ is not the root of the equation, dividing this by $x^2$ both of sides, we have

$6\left(x^2+\frac{1}{x^2}\right)-35\left(x+\frac{1}{x}\right)+62=0$   ....

Following up on #2

$6(x^2 + \frac{1}{x^2})-35(x + \frac{1}{x})+62=0$given, $y=x+\frac{1}{x}$

based on the original definition for y:

$y^2=x^2+2+\frac{1}{x^2}$

If add two and subtract two on $6(x^2 + \frac{1}{x^2})$ you get $6(x^2 + \frac{1}{x^2} + 2) - 12$.   Now the equation looks like: $6(x^2 + \frac{1}{x^2} + 2)-35(x + \frac{1}{x})+50=0$ and you can substitute $y$ and $y^2$ into the equation to get: $6y^2 -35y + 50 = 0$.  Now it is a simple quadratic equation to solve for y.  Once you find y, plug your solutions into the equation $y=x+\frac{1}{x}$. This equation can be then solved by multiplying the entire thing by x, then using the quadratic formula.


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that if H is a subgroup of S(n) that contains the transposition (1 2) and the n-cycle a = (1 2 ... n), then H = S(n).

Hi,

(1,...,n)(i,i+1)(1,...,n)^{-1} = (i+1,i+2)

That means since you have (1,2) in your subgroup you'll also have (a,a+1) for all a =1..n-1, and since (a,c) = (a,a+1)(a+1,c)(a,a+1) (a<>c), any transposition will be in your subgroup. Since the set of all transpositions generates $S_{n}$, you are done.


The set of integers is painted by $k$ colours in arbitrary way. Prove that for some of this colours there exists integers $c_1, c_2, \ldots, c_k$ such that it is possible to represent every integer in the form $a-b+c,$ where $a, b$ are some integers of selected colour and $c$ is one of the numbers $c_1, c_2, \ldots, c_k.$

It can get easy if you use a lemma that appeared in an old shortlist:

One of the colors, has arbitrary long sequences of the form $a_1,a_2...,a_k$ such that $0<a_{j+1}-a_j<k+1$.

However I think proving it is not a hard task.

Does someone has a proof for that lemma?

Did I remember it properly?

good, rela lemma is with the differences bounded by some constant not related with $k$. And in this case the lemma doesnt help.


calculus
integration
function
limit
real analysis
logarithms
calculus computations

I'm having a bit of difficulty proving that the upper sum and the lower sum of the integral of 1/x^k (where k is a non-negative real number) are equal on the interval [0,1] which is split by a partition of equal subsections. I'm having problems mainly because I'm not sure how to evaluate the upper sum using Riemann sums when x=0, as the function is not defined.

Is it possible to define a new interval [a, 1] and get take the limit of a to approach 0 from the right?

And in general, how would you integrate such functions that approach an infinite value at a specific point?

Thanks

First: an unbounded function is not Riemann integrable. Period. The upper sum will be infinite; the Riemann sums will be unbounded.

What you then deal with is not the Riemann integral itself but rather an add-on to the theory called improper integrals.

So: $\int_{0}^{1}\frac1{\sqrt{x}}\,dx$ does not exist as a Riemann integral.

For each $a>0,\ \int_{a}^{1}\frac1{\sqrt{x}}\,dx$ does exist as a Riemann integral.

So, by convention (and we'll have to agree on the terms of this convention), we define the improper integral as

$\int_{0}^{1}\frac1{\sqrt{x}}\,dx=\lim_{a\to 0^{+}}\int_{a}^{1}\frac1{\sqrt{x}}\,dx$

In this case, that integral exists and we call this a convergent improper integral; as with anything defined by a limit, we have to allow for the possibility that the limit may not exists.

The definitions are quite different for the Lebesgue integral, but you're not there yet.


Additive combinatorics
convex polygon
algebra
incircle
polygon
IMO Shortlist

[quote="orl"]Show that in the plane there exists a convex polygon of 1992 sides satisfying the following conditions:

[i](i)[/i] its side lengths are $ 1, 2, 3, \ldots, 1992$ in some order;
[i](ii)[/i] the polygon is circumscribable about a circle.

[i]Alternative formulation:[/i] Does there exist a 1992-gon with side lengths $ 1, 2, 3, \ldots, 1992$ circumscribed about a circle? Answer the same question for a 1990-gon.[/quote]

Nice problem !

[hide="My solution"]
Let $ \{a_i\}_{i\equal{}1}^n$ the $ n$ ordered sides of such a polygon ($ \{a_i\}$ is a permutation of $ 1,2,...,n$).
Let R the radius of the circle about which the polygon is circumscribed and $ O$ its center.

1) [b][u]This situation is impossible if $ n\equal{}2\pmod 4$ (and so for example for $ n\equal{}1990$)[/u][/b] :
Let $ A_1,...,A_n$ the n vertices ($ A_{n\plus{}1}\equal{}A_1$) with $ A_kA_{k\plus{}1}\equal{}a_k$.
Let $ \Delta_i$ the distance between $ A_i$ and the contact (tangent) between the circle and the side $ A_iA_{i\plus{}1}$.

We have $ \Delta_{k\plus{}1}\equal{}a_k\minus{}\Delta_k$ and so $ \Delta_p\equal{}\sum_{k\equal{}1}^{p\minus{}1}(a_k(\minus{}1)^{k\minus{}1})\plus{}(\minus{}1)^{p\minus{}1}\Delta_1$

So $ \Delta_1\equal{}\Delta_{n\plus{}1}\equal{}\sum_{k\equal{}1}^{n}(a_k(\minus{}1)^{k\minus{}1})\plus{}(\minus{}1)^{n}\Delta_1$

So, if $ n\equal{}2\pmod 4$, we have $ n$ even and so $ 0\equal{}\sum_{k\equal{}1}^n(a_k(\minus{}1)^{k\minus{}1})$ which implies $ \sum_{\text{odd }k\in[1,n]}a_k$ $ \equal{}\sum_{\text{even }k\in[1,n]}a_k$

So $ \sum_{k\in[1,n]}a_k$ even, so $ \frac{n(n\plus{}1)}{2}$ even, which is wrong when $ n\equal{}2\pmod 4$
Q.E.D.

[b][u]2) This situation is always possible if $ n\geq 4$ and $ n\equal{}0\pmod 4$ (and so for example for $ n\equal{}1992$)[/u][/b] :
Choose $ \Delta_1\equal{}\frac 12$ and $ a_k\equal{}k$ which gives :

$ \Delta_{2p}\equal{}p\minus{}\frac 12$
$ \Delta_{2p\plus{}1}\equal{}p\plus{}\frac 12$

It remains to choose the radius as root of the equation $ \sum_{k\equal{}1}^n\arctan(\frac{k}{x})\equal{}\pi$ which clearly has root as soon as $ n\geq 3$
Q.E.D.

[/hide]


integration
calculus
calculus computations

Yes, I have:

[hide=" Click to reveal hidden content"]

For the limits $ 0$ to $ 2$ and $ a^{2} \equal{} 5$,

$ nI_{n} \equal{} 3\cdot 2^{n\minus{}1} \minus{} 5(n\minus{}1)I_{n\minus{}2}$

Therefore, $ I_{5} \equal{} \frac{48}{5} \minus{} 4I_{3}$

$ \equal{}\frac{48}{5} \minus{} 4(4\minus{}\frac{10}{3}I_{1})$

Now $ I_{1} \equal{} \int_{0}^{2} \frac{x}{\sqrt{5\plus{}x^{2}}}\ dx$

$ \equal{} \sqrt{5\plus{}x^{2}}$ between $ 0$ and $ 2$

$ \equal{}3\minus{}\sqrt{5}$

Therefore, $ I_{5} \equal{} \frac{168}{5} \minus{} \frac{40\sqrt{5}}{3}$.

[/hide]


geometry
inequalities
email
function
trigonometry
angle bisector
inequalities proposed

[quote="2424"][quote="quykhtn-qa1"]Let $ l_a,l_b,l_c$ be three lengths of internal angle bisector of triangle with area $ S$,then we have:
$ (l_al_b \plus{} l_bl_c \plus{} l_cl_a)^2 \ge \frac {64p^3S^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
And ,we have an inequality than stronger this inequality.I will post later.[/quote]
I think this is Tran Quang Hung's ineq,not of you! :lol:[/quote]
Of course.But i have an inequality than stronger and i speak with my teacher Tran Quang Hung :wink: .I will post my inequality :)

[quote="2424"]
ok,can you post full your proof and your ineq?Thank you. :)[/quote]
I hope you like this inequality:
$ (l_al_b\plus{}l_bl_c\plus{}l_cl_a)^2 \ge \frac{16p^4l_al_bl_c}{l_a(b\plus{}c)^2\plus{}l_b(c\plus{}a)^2\plus{}l_c(a\plus{}b)^2} \ge \frac{64p^3S^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
Regards.

[quote="quykhtn-qa1"]Let $ l_a,l_b,l_c$ be three lengths of internal angle bisector of triangle with area $ S$,then we have:
$ (l_al_b \plus{} l_bl_c \plus{} l_cl_a)^2 \ge \frac {64p^3S^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
And ,we have an inequality than stronger this inequality.I will post later.[/quote]
Since this inequality,we have an nice inequality of Walther Janous
$ l_al_b\plus{}l_bl_c\plus{}l_cl_a \ge 3\sqrt{3}S$

 :)

[quote="quykhtn-qa1"][quote="quykhtn-qa1"]Let $ l_a,l_b,l_c$ be three lengths of internal angle bisector of triangle with area $ S$,then we have:
$ (l_al_b \plus{} l_bl_c \plus{} l_cl_a)^2 \ge \frac {64p^3S^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
And ,we have an inequality than stronger this inequality.I will post later.[/quote]
Since this inequality,we have an nice inequality of Walther Janous
$ l_al_b \plus{} l_bl_c \plus{} l_cl_a \ge 3\sqrt {3}S$

 :)[/quote]
@Quytuan:[hide] Anh có thể cho em nick thầy Quang Hùng ko ạ, em cũng học thày ý nhưng ko biết nick chat ạ :D Và lại email thầy em lại để  trong sách cho bạn mượn rồi ạ :D[/hide]

[quote="can_hang2007"][quote="quykhtn-qa1"]Let $ l_a,l_b,l_c$ be three lengths of internal angle bisector of triangle with area $ S$,then we have:
$ (l_al_b \plus{} l_bl_c \plus{} l_cl_a)^2 \ge \frac {64p^3S^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
And ,we have an inequality than stronger this inequality.I will post later.[/quote]
Dear friends, I think this is a consequence of the Holder's Inequality. What do you think? Anyway, I think you guys should not spam anymore. It makes the topic become long and hard to read. If you guys want to chat each other, I think you can do it through the private message function![/quote]
With this inequality my Solution use Schur inequality
could you please post your solutions:
thank you very much

[quote="quykhtn-qa1"]Let $ l_a,l_b,l_c$ be three lengths of internal angle bisector of triangle with area $ S$,then we have:
$ (l_al_b\plus{}l_bl_c\plus{}l_cl_a)^2 \ge \frac{64p^3S^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
[/quote]
Yes, it's true! I have proved this inequality by Holder and $uvw$.

[quote="quykhtn-qa1"]Let $ l_a,l_b,l_c$ be three lengths of internal angle bisector of triangle with area $ S$,then we have:
$ (l_al_b\plus{}l_bl_c\plus{}l_cl_a)^2 \ge \frac{64p^3S^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
And ,we have an inequality than stronger this inequality.I will post later.[/quote]

nice and stonger!

BQ

[quote="quykhtn-qa1"]Let $ l_a,l_b,l_c$ be three lengths of internal angle bisector of triangle with area $ S$,then we have:
$ (l_al_b\plus{}l_bl_c\plus{}l_cl_a)^2 \ge \frac{64p^3S^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
And ,we have an inequality than stronger this inequality.I will post later.[/quote]

The proposed inequality is equivalent to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=327619][b]this one[/b][/url] ... :)

[quote="quykhtn-qa1"]Let $ l_a,l_b,l_c$ be three lengths of internal angle bisector of triangle with area $ S$,then we have:
$ (l_al_b\plus{}l_bl_c\plus{}l_cl_a)^2 \ge \frac{64p^3S^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
And ,we have an inequality than stronger this inequality.I will post later.[/quote]

and you do:

$l_bl_c+l_cl_a+l_al_b \geq \frac{4}{3}(\frac{al_a(s-a)}{s}+\frac{bl_b(s-b)}{s}+\frac{cl_c(s-c)}{s})+(3\sqrt{3}-\frac{8}{3})S.$

BQ

[quote="xzlbq"]and you do:

$l_bl_c+l_cl_a+l_al_b\geq \frac{5}{3}S(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2})+\frac{4\sqrt{3}}{3}S.$

BQ[/quote]

and for max(m):

$l_bl_c+l_cl_a+l_al_b\geq mS(\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2})+\sqrt{3}(3-m)S.$

and you do:

$l_bl_c+l_cl_a+l_al_b \geq \frac{5}{3}(\frac{l_bl_c}{l_a}+\frac{l_cl_a}{l_b}+\frac{l_al_b}{l_c})r+\frac{4}{3}\sqrt{3}S.$

BQ

Note:

$\frac{l_bl_c}{l_a}+\frac{l_cl_a}{l_b}+\frac{l_al_b}{l_c}\geq \sqrt{3}p,p=\frac{1}{2}(a+b+c).$

and for max(m):

lb*lc+lc*la+la*lb >= m*(lb*lc/la+lc*la/lb+la*lb/lc)*r+(3-m)*sqrt(3)*S.
BQ


From the University of Maryland Math Competition, Part 1, 2001

Suppose that 100 more than the sum of n consecutive integers is equal to the sum of the next n consecutive integers. Then n equals:
a. 10   b. 31   c. 50   d.100   e. 200

I wish my high school registered for this competition this year. :(

[hide]
\[ 100+\frac{n(2a_1+n-1)}{2}=\frac{n(a_1+n+a_1+2n)}{2}  \]

Since this is a multiple choice test, we can simply plug in the answers to check if they produce reasonable $a_1$. We find that $n=10$ is the only one that has a solution. :D :D If this was AIME or something, then you could actually work it out. BTW, this is the hard way...
[/hide]

Yep, I dont why you guys decided to evaluate the sum of the arithmetic series first, Algebraic manipulation is sufficient to solve the problem:

Suppose the sum of the n consecutive integer is: $N$

  To find the sum of the next n consecutive integers we must observe the difference of the first terms of each series remains constant, that is $n$.  Hence, the second series is simply the sum of the first series $N$, plus their difference, $n \cdot n=n^2$.

  Thus,

$100 + N = N +n^2$

$100=n^2$

$n=10$

  And consequently, there are 10 consecutive integers.


[quote="Whatthedeuce"]In the checkout section when buying a book or paying for a class, paying with an Art of Problem Solving Gift Certificate is an option. I was just wondering, where/how can I get one and how much money is it normally worth?[/quote]
Some competitions (USAMTS and HMMT, probably others) give certificates to students who perform well.

You can also buy one from us (only via phone right now) if you want to give it to someone as a gift.


calculus
integration
quadratics
inequalities
linear algebra
matrix
Support

Which graphing calculator do you guys think is the most useful? I think I am going to get a new one and I want to make sure that it will be useful for a while... Thanks :)

TI-83 or TI-84 most definitely. They're the most common and widely accepted for all kinds of tests (SAT etc.) and I don't know whether or not the 89's are, but 89's are tons more complicated.
However, if you take the time to learn a ton of stuff about the 89, it can be a powerful tool.

[quote="Hokkage"]I don't know whether or not the 89's are, but 89's are tons more complicated.
However, if you take the time to learn a ton of stuff about the 89, it can be a powerful tool.[/quote]Ti-89's are accepted on SATs, AP tests, AMCs, so yeah. Get that.  :)

It's loads more powerful, and will be really efficient for you when you're doing calculus.

Yeah, 89s are the best.  They are a little harder to use, but I already have the basics down, and thats from 5 minutes messing with a friend's, since I only have an 83+, and am still just wishing for an 89...

i would say go for the 89 just remember that it does have its limitations.  ;)
however, the 84/83+ is also a good be. if you get one of these you can write/download programs that will enable this calculator to do many of the same things as the 89, however it will take longer to do more complicated computations and the answer will not be in exact form.

Go for the 89 if you're doing calculus or something like that. If you won't be doing that stuff, go for an 83 or 84.

I recommend the ti-89, I think it's the best of the ti calculators and it is allowed on most major tests (like the SAT).

TI-89 Titanium is a must it takes alittle bit to get used to it but after that it leves the 83's 84's in the dust.

TI 83 or 84 SILVER lol Silver makes it so much faster :D

Its not like you need an 89 for the SAT anyways, im sure we can all do the SAT math problems in our heads.

But 89 will last you longer than an 83+, but do make sure that your teacher allows an 89, some dont.

I suggest getting the new Microsoft Origami. Its not really a caculator but its a computer almost as big as a TI and im sure you can buy some great graphing caculator programs. But you cant use it on the SAT but still its going to be sweet.

Even if you aren't allowed to use an 89 on a test, it is worthwhile when the time comes to check your homework and such.  I once had an 83+, but would never consider anything less than an 89 after using one...

Although the TI-89 is a good calculator, I might reccomend something in the realm of the TI-83+ just because the TI-89 does integrals and sometimes (at least from what I've seen) people rely too much on their calculators to do things for them, when it comes time for the no-calculator sections, they tend to bomb those.  But the 89 is still a great calculator.

89 is faster, more powerful, does more (so much more that it seems cheating :) )  And you can program assembly on that w/e :lol:
Oh, and not allowed on Act.

So definitely get an 83 and do math.  :lol:

Lets see I have used a:
TI-81 (NO... they don't make these for a reason),
82 (Haha, no.),
83+ SE (Good calc... loads better in 84 in design... but if you want its equivalent, go for the new TI ugliness),
86 (Most people never even think about this calculator, and while unintuitive at first... much, much, much, better than 83+/84 family, I mean poly and simult have saved my life countless times),
89 (Frankily don't get a Titanium... not worth the extra cash, and its ugly.  Classic edition is wonderful, but I only use it on mathematical competitions and probably in calculus.  Its kind of banned in algebra classes for var manipulation :roll: )
92+ (uhm... this is completely banned on standardized tests and I just used it in class a couple times for our teachers teaching us some quadratic manipulation... plus its not for most peoples budget... especially not the modern V200 equivalent)
And now for Casio calcs:
The incredibly well CFX-9850GB PLUS... Now lots of people say, "Not TI, must be crap..." Its actually good because it has a much nicer price (get the modern equivalent tho...), and mind you it is as good as 86 (Thats more powerful than the 83+ for all you 83+ standardized non creative dull people...) and actually has some advantages over it (Like color graphing! And intutive inequality graphing!).  And now for the warning: Lack of programs, lower speed (mainly in graphing and programs), a weaker poly than 86 (better than 83+ which has none), weaker matrix manipulation, and lower hardware specs (32 KB of mem just limits me... plus no archive, maybe newer model is better tho), and really unintutive.

Final Selection:
If standardized testing is concern: TI-86 is the way to go.  Very powerful, just one level under the 89.  Its one of the strongest calculators allowed by ACT, and while at first confusing, it is actually much faster than the dumb 83 menus.
If you don't have to worry about calc restrictions: TI-89 (Not titanium but you can choose that). Man you will love it.  I mean you will become lazy in doing some algebra equations, but it is truly raw power.

If you don't want to spend much and like to do something crazy: Get whatever modern equivalent of the CFX-9850 GB PLUS.  Its crazy, but good.

The hardcore ones get an hp-50g, cuz it supports RPN.

I recommend 89 titanium.  I have a 83+ and and 89 titanium and I'd say the 89 is better.  It has more functions and allows you to do more than the 83+.

Titanium if you don't care about spending extra cash

83+ if you need something worth its price

[quote="kstan013"]84s are the most user-friendly[/quote]

I have an 84. It was nice, but kind of old, so whenever i needed some special function for statistics i couldn't find it on my TI-84 (just because the navigation was different from the TI-83)

What's a TI-92 like anyways? I've always wanted to see what it can do.

[quote]also, the 83/84s are ugly[/quote]
i'd have to say the 86's are uglier.  :D

[quote="SplashD"]Why do people care if the calculator is ugly or not, if it can do the work then get it[/quote]
yes, so get an 89. most normal ppl should be able to figure out how to use it, and when they do, it just pwns.(plus, its prettier)

89. They do calculus for you, which is really useful in AP physics.

[quote="drunner2007"]What's a TI-92 like anyways? I've always wanted to see what it can do.[/quote]

TI-92 has been replaced with the Voyage 200.  I have a TI-92 from MathCounts.  It's a little on the slow side and bulky, so I don't use it.  It does make for an interesting conversation piece, though.

[quote="junggi"][quote="kstan013"]84s are the most user-friendly[/quote]
uh no. I'd say theyre equally user-friendly(actually I think 89s are more convenient to use than 83/84s. accessing functions, programs, apps, etc aer easier).
people say 89s are harder to use, but most of them are just too stupid or lazy to learn how to use it. just skim thru first few pages of manual, or just play around with it and you'll get how to use it in 10 minutes or less.
and yes, 89s are much much better than 83(more functions, faster, etc). also, the 83/84s are ugly[/quote]

more bang for your buck with 84+

89s are banned from a lot of things including use for math courses in my school and all local competitions. Also, they're too expensive and the only real upgrade is the algebra solver which is for lazy people.

If you don't want to program or anything then I'd recommend a TI 83 Plus, if you wanted to program or just want a new calculator, get the TI 84 Plus Silver Edition

Get a TI-84 Plus Silver edition.
Bid carefully and you'll get it for $\$60$ on ebay. :wink:

[quote="SplashD"][quote="b-flat"][quote="kimmystar94"]ti 89 like rox..
it graphs 3d and it's cool...
it pwns everything else under it...
:10:[/quote]

You can get an application for the 83/84s that graphs 3D faster than the 89s.

Also, haoye, please do not revive old topics, thanks.[/quote]

You mean $\text{Graph}^{3}$? All the applications take up a lot of room and they're built into the TI-89s[/quote]

Yes, but not really that much (16k), even on a regular 83+.

[quote="b-flat"][quote="kimmystar94"]ti 89 like rox..
it graphs 3d and it's cool...
it pwns everything else under it...
:10:[/quote]

You can get an application for the 83/84s that graphs 3D faster than the 89s.

Also, haoye, please do not revive old topics, thanks.[/quote]

The topic isn't that old, just three days. :? Unlike the topic that I inadvertently revived last year, which was all the way from 2003 :oops_sign:

[quote="Ignite168"]If you don't want to program or anything then I'd recommend a TI 83 Plus, if you wanted to program or just want a new calculator, get the TI 84 Plus Silver Edition[/quote]

How so? The 83+ and 84+ are practically identical when it comes to programming... They're practically identical in all other aspects too, escept maybe the 84 is faster and has a cool clock. :cool_speak:

The 84+ has more memory for 1 (3 times as much), and the 84+SE has 9 times as much...

Also, they have better archiving routines so you don't have to GarbageCollect as much & they have a nicer display.

And, the SE calcs have 128kb of RAM, even though only 32k is available to the user (unless you know z80 assembly).

Umm, this topic went from April to October...  Isn't that a little more than 3 days?

[quote="chronoflare"]
Final Selection:
If standardized testing is concern: TI-86 is the way to go.
[/quote]

The 86 has issues when it comes to memory addressing that make it slower than the 85 (the calculator it is an "upgrade" to) in some areas.  That alone should knock it out of the running.

Are ti-92's any good? Or is it just better to get a ti-89?

The 83+ and 84+ are both great calculators, and easy to figure out.  If you get the 89, its kinda hard to work with until you really get the hang of it.  [hide="secret"]the 83/84+ have more games made for them if you care about that at all lol[/hide]

[quote="anirudh"]The TI-89. It is allowed in MC. How is the Voyage 200 in MC?[/quote]
As a general rule, anything w/a typewriter style keyboard is not allowed in mathcounts, I think

[quote="DanK"]The only major thing in my area that 89s are banned from is the ACT, but if you're getting a new calculator, then you should still have that for taking the other test anyway, plus the 89 easily beats any other TI models in what it can do...  Definitely go with that if you can afford it.  Otherwise, 83+ and 84+ aren't bad.[/quote]
I am almost positive the 89 is allowed. Then again, maybe not...

i use a TI-89 titanium, but the TI-84s have little masks for them, so you can get like a pink TI-84.. i wish they did that 4 all calculators..
but anyway..
i don't use many advanced fuctions on the TI-89 except 4 the solve and factor,
and unless you download the program, you can't graph non-functional graphs (i think so)
but the TI-84 has a built-in app for the program. .it's called inequalz.. it's good..

i recommend the TI-84 or TI-89 titanium...

Please do not revive old topics!

Thanks and locked.

[color=red]b-flat[/color]


Show that the perpendiculars let fall from any point on the straight line $ 2x\plus{}11y\minus{}5\equal{}0$ upon the two straight lines $ 24x\plus{}7y\minus{}20\equal{}0$ and  $ 4x\minus{}3y\minus{}2\equal{}0$ are equal to each other. Thanks a lot for your help.

I'm not exactly sure what you mean. What does the word upon mean?

("...2x+11y-5=0 upon the two straight lines...")


Thnks xcenario. I had the following:

$ x^{2} \plus{} y^{2} \equal{} 17$, by adding the equations and dividing by $ 2$.

Therefore $ xy \equal{} 4$.

$ (x\plus{}y)^{2} \equal{} x^{2} \plus{} y{2} \plus{} 2xy \equal{} 25$

and $ (x\minus{}y)^{2} \equal{} 9$

$ x\plus{}y \equal{} \pm5$

$ x\minus{}y\equal{} \pm 3$

$ (x, y) \equal{} (\minus{}4, \minus{}1), (\minus{}1, \minus{}4), (1, 4)$ or $ (4, 1)$

$ (x^2\plus{}xy\plus{}y^2)\plus{}(x^2\minus{}xy\plus{}y^2) \equal{} 2(x^2\plus{}y^2)\equal{}21\plus{}13\equal{}34$  i.e.  $ x^2\plus{}y^2\equal{}17$
$ (x^2\plus{}xy\plus{}y^2)\minus{}(x^2\minus{}xy\plus{}y^2) \equal{} 2xy\equal{}21\minus{}13\equal{}8$  i.e.  $ xy\equal{}4$

So, $ \left\{\begin{array}{l}x\plus{}y\equal{}\pm5 \\ xy\equal{}4 \end{array}\right.$

Therefore
$ \left\{\begin{array}{l}x\plus{}y\equal{}\minus{}5 \\ xy\equal{}4 \end{array}\right. \iff t^2\plus{}5t\plus{}4\equal{}0 \iff t\equal{}\minus{}1, \minus{}4$
$ \left\{\begin{array}{l}x\plus{}y\equal{}5 \\ xy\equal{}4 \end{array}\right. \iff t^2\minus{}5t\plus{}4\equal{}0 \iff t\equal{}1, 4$

Thus, $ (x, y)\equal{}(\minus{}4, \minus{}1), (\minus{}1, \minus{}4), (1, 4), (4, 1)$.

[quote="AndrewTom"]Solve, for real numbers $ x$ and $ y$,

$ x^{2} \plus{} xy \plus{} y^{2} \equal{} 21$
$ x^{2} \minus{} xy \plus{} y^{2} \equal{} 13$.[/quote]

Add the equations to get;

$ 2x^2 \plus{} 2y^2 \equal{} 34$
$ x^2 \plus{} y^2 \equal{} 17$ (Equation 1)

Plugging this into the first equation we have;

$ 17 \plus{} xy \equal{} 21$
$ xy \equal{} 4$
$ x \equal{} \frac{4}{y}$

Plugging this into Equation 1, we get;

$ (\frac{4}{y})^2 \plus{} y^2 \equal{} 17$
$ (\frac{16}{y^2} \plus{} y^2 \equal{} 17$

Let $ y^2 \equal{} w$

Then;

$ \frac{16}{w} \plus{} w \equal{} 17$
$ 16 \plus{} w^2 \equal{} 17w$
$ w^2 \minus{} 17w \plus{} 16 \equal{} 0$
$ (w \minus{} 16)(w \minus{} 1) \equal{} 0$

So either;

$ w \minus{} 16 \equal{} 0$ or $ w \minus{} 1 \equal{} 0$

so either;

$ w \equal{} 16$ or $ w \equal{} 1$

so either;

$ y^2 \equal{} 16$ or $ y^2 \equal{} 1$

so according to Equation 1, either;

$ x^2 \equal{} 1$ or $ x^2 \equal{} 16$

But xy = 4, so the possible solutions are - (-1, -4), (1, 4), (-4, -1), and (4, 1).


Peter selected a few stamps from his collection; one of each of the face values: 1,2,...,37 forints. His plan is to arrange them in groups such that the sum of the face values is the same in each group. How many different arrangements can be made?

Enjoy!

Here's my solution, I think it's right.

[hide]Sum up the values of all the stamps, using $\frac{n(n+1)}{2}$, and you get $37 \cdot 19$. Since we want the face values of each group to sum up to the same number, that number and the number of groups multiply to $37 \cdot 19$, so there can be one group worth 703 or 19 groups woth 37 ({37},{1,36},{2,35},...,{18,19}). You can't have 37 groups of 19 because there would be one stamp in each group, all obviously of different values. Same with 703 groups, since you don't have that many stamps.
Therefore, there are two arrangements.[/hide]


number theory
greatest common divisor
algorithm
algebra proposed
algebra

[quote="pureblack2003"]f,g from Q[x], $f=x^{3}-x^{2}-ax+4$ , $g=x^{3}-5x^{2}+6x-4$. if $gcd(f,g)$ is a polynom of degree 2, then :

a)a=-2 ; b)a=4 ; c)a=3; d)a=0; e)a=1

note: if it doesn't work out, than maybe instead of 6, the coefficient in g, there may be b, a real number. i can't make it out from the paper this exercice is written on ...[/quote]

a=4 gcd is $x^{2}-3x+2$ and $b=8$

Any typo?
No such $a$ exist.
By the 1st step of Euclid algorithm we have:
$f(x)=g(x)+d(x).$
where $d(x)=4x^{2}-(a+6)x+8$ must be gcd because it has degree 2.
But there is no $a$ which allows for $d(x)$ be divisor of $g(x)$.

P.S. Missing at 1st reading:
If we read 6 as b then after 1st step of algoritm we have:
$d(x)=4x^{2}-(a+b)x+8$
By the 2nd one:
$4x^{3}-20x^{2}+4bx-16=x\cdot d(x)+d_{1}(x)$
$d_{1}(x)=(a+b-20)x^{2}+(4b-8)x-16$
So $d(x)$ and $d_{1}(x)$ must be collinear:
$\frac{a+b-20}{4}=\frac{8-4b}{a+b}=\frac{-16}{8}$
This gives: $a=4, b=8$
So b) $a=4$ is right.

P.P.S. Sorry [b]cckek[/b] I've realised that there  was your solution  at the 3d reading only. Before it was surprising for me because the roots 1 and 2 of your gcd (the same of mine of course at the end of ends - I'm bad in arithmetic) does not satisfy g(x)=0 if b=6.
And now it is a pity to delete ...  :lol:


I want to take the MIT Delve AP Calculus class in 9th grade. Bad idea?

if you've had the prerequisite material and you have ample time its fine  :)
don't get too stressed though


combinatorics unsolved
combinatorics
RMO

DARN IT! :mad: I did a poor mistake in this so simple question. dont know what i was doing. maybe even this is wrong.

[hide="in short:"] initial sum of 4 digits =6, so, possibilities for 5th and 6th digits are (0,4), (1,3), (2,2).
combinations for i)$ \frac{4*5!}{2!} \equal{} 240.$  ii)$ \frac{5*5!}{2*2} \equal{} 120$ iii) $ \frac{5*5!}{3!} \equal{} 60$. summing up , we get 400.[/hide]

The different combinations are 0,1,2,3,4,0 ; 0,1,2,3,3,1 ;0,1,2,3,2,2
In the first case numbers is 4 * (5!/2!) =240 in second case 5*(5!/2!2!)=150
and similarly in third case 100. Thus total no of numbers is 490.

Yes second and third case aren't right analytic as sa1 mentioned. Thank God I'll get full score on this one I hope, if they don't subtract for explanation-quality.

yeah, my brain has gone mad today, too silly calc. mistakes, thats wut i did there.

i completely messed up in calculations

i don't know how,but my answer to case 1 was $ 60$

probably i missed one $ 4$ during multiplication. :(

[hide="@Arvind:"][quote="Aravind Srinivas L"]the answer is 490[/quote]
Pls check the posts before you before posting. [/hide]

guys i just messed with the calculations
i just did upto the sequence of 0,1,2,3 and (3,1) (4,0),(2,2)
and did a mathematical error in my calculations
will i get any marks

romil u r expecting too much if u proceeded in the wrong way..
But if u were a skand, u would be getting 11 atleast out of 14.

i know that is y my chances are very low

gosh are my calculations so weak,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,????

there are three possible sets of numerals satisfying the given condition
 they are :
 1___(0,1,2,3,0,4)
 2___(0,1,2,3,3,1)
 3___(0,1,2,3,2,2)
in the first case ,
             _ _ _ _ _ _ the first underline can be filled in 4 ways only. and the restof them can be filled in 5!/2! ways therefore,total of 4x 120/2 = 240
in second case ,
             _ _ _ _ _ _ let us calculate total number of possible numbers i e 6!/2!2! now we will subtract from this the numbers startimg with zero i.e. 5!/2!2!  :idea:
 = 720-120/4 = 150
in thitrd case,
             _ _ _ _ _ _ here the total number of possible numbers are 6!/3! . we shall subtract the numbers starting with zero i.e. 5!/3! :10:
 = 720/6  -  20  = 100
there fore total number of possible solutions are 240+150+100=490 :D
 combinatorics is easy if u read ncert 11th text

[quote="Akashnil"]Find the number of all $ 6$-digit natural numbers such that the sum of their digits is $ 10$ and each of the digits $ 0,1,2,3$ occurs at least once in them.

[14 points out of 100 for the 6 problems][/quote]

There are 3 possibilities
#1: (0,1,2,3,2,2), #2: (0,1,2,3,1,3), #3: (0,1,2,3,0,4).
For #1, There are 5*5*4*3*2*1 ways to make 6 digit numbers. But there are three repetitions of 2 so you divide by 3!. Eaqualling: 100
For #2, There are again 5*5*4*3*2*1 ways to make 6 digit numbers. But 1 and 3 appear twice so we divide by (2! *2! ). Yielding: 150
For #3, There are again 4*5*4*3*2*1 ways to make 6 digit numbers. Here 0 appears twice so divide by 2!. giving 240
Sum: 490

Why isn't anybody trying multinomial?
(Don't downvote)
Ohkay sorry...
I saw the loophole in my solution...
Using multinomial yeilds a lot of cases and thus should be omitted.
[b]Thus for those who try this- Multinomial(stars and bars technique) is not useful here. [/b]


What is the number of units in
the area of the circle
with center at $ P$ and passing
through $ Q$? Express your answer
in terms of $ \pi$.

[asy]size(150); pair P = (-3,4), Q=(9,-3); string stringpair(pair p){return "$("+string(p.x)+", "+string(p.y)+"$)";}
draw((-15,0)--(15,0),Arrows(4)); draw((0,-15)--(0,15),Arrows(4));
dot("$Q$"+stringpair(Q),Q,SE,linewidth(3)); dot("$P$"+stringpair(P),P,NW,linewidth(3));[/asy]

The distance between P and Q is our radius, which we can use to find area.

$ \sqrt{(4 \minus{}\minus{} 3)^2 \plus{} (\minus{}3 \minus{} 9)^2} \equal{} \sqrt{49 \plus{} 144} \equal{} \sqrt{193}$

$ \pi r^2 \equal{} 193 \pi$

Notice that the distance between the two points would be the radius of the circle, since the radius connects the center to a point on the circle. According to the distance formula, the distance $ PQ$ is $ \sqrt {(9 \minus{} ( \minus{} 3))^2 \plus{} ( \minus{} 3 \minus{} 4)^2} \equal{} \sqrt {12^2 \plus{} 7^2} \equal{} \sqrt {144 \plus{} 49} \equal{} \sqrt {193}$, so the radius of the circle measures $ \sqrt {193}$ units long. Because the area of any circle is $ \pi$ times the square of the length of its radius, the circle referred to in the problem has an area of $ \pi(\sqrt {193})^2 \equal{} \boxed{193\pi}$, when expressed in terms of $ \pi$.


Let $ ABC$ be an acute-angled triangle, and let $ O$ be its circumcentre. The circle through $ A, O$ and $ B$ is called $ S$. The lines $ CA$ and $ CB$ meet the circle $ S$ again at $ P$ and $ Q$ respectively. Prove that the lines $ CO$ and $ PQ$ are perpendicular.

I proved that O is the orthocenter of triangle PQC so PO is perpendicular QC

Hi
APOQB is cyclic
so : OPQ=OBQ=90-BAC
       PQC=BAC
then : OPQ+PQC=90
         it means that OP and QC are perpendicular
QO and PC are perpendicular in the same way, so O is the orthocentere of QPC


If one writes a solution that would be given a 5 and goes on a tangent or into extra detail at the end in order to recieve a commendation, and the additional information is incorrect will the person be penalized?

If the extra work is crucial to your solution to the question that is asked, it will be a problem.  However, if it is just an extra exploration beyond the question that is asked, and your correct answer to the question does not in any way depend on this work, then the grader should not deduct points.


ARGENTINA:
ARG1 PABLO BLANC 4 P 0 7 1 0
ARG2 NICOLÁS RODRÍGUEZ CASTRO 0 P 0 6 0 0
ARG3 PABLO ZIMMERMANN 7 P 0 1 0 0
ARG4 MAURO SCHILMAN 3 P 0 7 0 0
ARG5 ROBERTO MORALES 2 P 0 7 7 0
ARG6 FERNANDO VIDAL 5 P 3 7 0 0

BRASIL:
BRA1 Guilherme Phillipe Figueiredo 3 1 P 7 1 0
BRA2 Ramon Moreira Nunes 4 7 P 7 1 0
BRA3 Régis Prado Barbosa 7 7 P 7 1 0
BRA4 Henrique Pondé de Oliveira Pinto 6 1 P 7 6 1
BRA5 Rafael Sampaio de Rezende 3 6 P 7 1 0
BRA6 Rafael  Tupynambá Dutra 3 2 P 7 1 1

COLOMBIA:
COL1 Diego Cifuentes 3 1 P P P 0
COL2 Santiago Cuellar 6 2 P P P 0
COL3 Jose Alejandro Samper 7 1 P P P 0
COL4 Miguel Acosta 3 1 P P P 0
COL5 Jorge Alberto Olarte 7 0 P P P 0
COL6 Federico Castillo 7 1 P P P 0

MEXICO:
MEX1 Isaac  Buenrostro Morales P 7 0 7 P 1
MEX2 Aldo Pacchiano Camacho P 2 0 7 P 0
MEX3 Marco Antonio Avila Ponce de Léon P 2 0 7 P 0
MEX4 Cristian Manuel Oliva Avilés P 7 0 6 P 0
MEX5 Fernando Campos García P 7 0 7 P 0
MEX6 Manuel  Jesús Novelo Puc P 1 0 7 P 0

PERU:
PER1 DANIEL SONCCO 7 7 0 6 5 P
PER2 FERNANDO MANRIQUE 3 0 0 7 7 P
PER3 JOSSY ALVA  3 1 0 7 0 P
PER4 CESAR  CUENCA 3 1 0 7 0 P
PER5 AMILCAR  VELEZ 0 1 0 7 1 P
PER6 RICARDO RAMOS 3 2 0 7 6 P

Felicidades por sus resultados, en especial perú que por lo que veo no se viene nadie con las manos vacias.
A mi país le falta mucho por mejorar, partiendo de la administración

solo tengo una duda ¿Uds, los solvers de la imo cuanto entrenan?


For question 1 ,try a=0.001,b=0.001,c=1  :blush:

1. I think your problem is wrong..
It's should be :
$ 1 \plus{} \frac {3}{ab \plus{} bc \plus{} ca} \ge \frac {6}{a \plus{} b \plus{} c}$
Which can be done by simple AM-GM..
2. $ \sum 2\frac {a}{b} \plus{} \frac{b}{c} \ge \sum 3 \frac {a^2}{bc} \ge \sum 3a$

The first is true if $ abc\equal{}1$
My solution is no hard
$ a\equal{}1/x$, $ b\equal{}1/y$, $ c\equal{}1/z$ and $ xyz\equal{}1$
$ 1\plus{}\frac{3}{xy\plus{}yz\plus{}zx}\ge\frac{6}{x\plus{}y\plus{}z}$
We know, that $ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca ... (a\plus{}b\plus{}c)^2\ge3(ab\plus{}bc\plus{}ca)$
$ 1\plus{}\frac{3}{xy\plus{}yz\plus{}zx}\ge 1\plus{}\frac{9}{(x\plus{}y\plus{}z)^2}$
By AM-GM we have
$ 1\plus{}\frac{3}{xy\plus{}yz\plus{}zx}\ge 1\plus{}\frac{9}{(x\plus{}y\plus{}z)^2}\ge2\sqrt{1*\frac{9}{(x\plus{}y\plus{}z)^2}}\ge\frac{6}{x\plus{}y\plus{}z}$
QED

It's easy
$ 1\plus{}\frac{3}{xy\plus{}yz\plus{}zx}\equal{}1\plus{}\frac{9}{3(xy\plus{}yz\plus{}zx)}\ge 2\sqrt{1*\frac{9}{3(xy\plus{}yz\plus{}zx)}}\equal{}$
$ \frac{6}{\sqrt{3(xy\plus{}yz\plus{}zx)}}$
We know, that $ (a\plus{}b\plus{}c)^2\ge3(ab\plus{}bc\plus{}ca)$
$ \frac{6}{\sqrt{3(xy\plus{}yz\plus{}zx)}}\ge\frac{6}{\sqrt{(x\plus{}y\plus{}z)^2}}\equal{}\frac{6}{x\plus{}y\plus{}z}$

[quote=sqing][quote=Potla]Can anyone please help me solve the following?(NB: I think they are easy but I cannot solve them :( )
 Prove that for $ a, b, c\in\mathbb{R}^\plus{}$,
$ \boxed{1}$ $ 1\plus{}\frac{3}{a\plus{}b\plus{}c}\geq\frac{6}{ab\plus{}bc\plus{}ca}$
$ \boxed{2}$ $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq a\plus{}b\plus{}c$ given that $ abc\leq 1$

Please reply ASAP :)[/quote]
Mathematics bulletin (China Beijing) NO.8(2015) problem2256:
If $a,b,c \in \mathbb{R}^{+}$ such that $abc=1$ prove that $$\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq a\plus{}b\plus{}c$$
[url=https://artofproblemsolving.com/community/c6h1410747p7924226]here[/url]
Old:
If $a,b,c \in \mathbb{R}^{+}$ such that $abc=1$ prove that $$\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq a\plus{}b\plus{}c\geq \frac{a+1}{b+1}\plus{}\frac{b+1}{c+1}\plus{}\frac{c+1}{a+1}$$[/quote]

$a=\frac{y}{z}$ same with $b,c$
$a+b+c\ge\sum \frac{a+1}{b+1}\implies \sum \frac{x}{y}\ge \sum\frac{xz+yz}{y^2+yz}$
$\Leftrightarrow \sum \left(\frac{x}{y}-\frac{xz+yz}{y^2+yz}\right)\ge 0$
$\Leftrightarrow \sum \frac{xy+xz-xz-yz}{y^2+yz}= \sum \frac{xy-yz}{y(y+z)}=\sum\frac{x-z}{y+z}\ge 0$
$\Leftrightarrow \sum \frac{x+y}{y+z}\ge 3$ which is true by AM-GM

[quote=Potla]Can anyone please help me solve the following?(NB: I think they are easy but I cannot solve them :( )
 Prove that for $ a, b, c\in\mathbb{R}^\plus{}$,
$ \boxed{1}$ $ 1\plus{}\frac{3}{a\plus{}b\plus{}c}\geq\frac{6}{ab\plus{}bc\plus{}ca}$
$ \boxed{2}$ $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq a\plus{}b\plus{}c$ given that $ abc\leq 1$

Please reply ASAP :)[/quote]

For the second problem,
Just consider $$\frac {a}{b} + \frac{a}{b} + \frac {b}{c} \geq 3  \sqrt[3]{\frac {a^2}{bc}} = 3a$$

Adding the inequalities we get the desired.
Note: I have taken $abc=1$ which is stronger


Let $ A=(a_{ij})\in M_{n}(\mathbb{R})$ such that $ a_{ii}=2007\forall i=\overline{1,n}$ and $ a_{ij}\in\{4,20\}\forall i\not =j;i,j=\overline{1,n}$. Solve system $ Ax=0$.

$ A$ is nonsingular, so the only solution can be $ x=0$ where $ 0$ is the null vector.

To see this, it is necessary to realize that if $ \det A=0$, then $ \det A \equiv 0 \hspace{1 mm}\mod 2$.  Rewriting $ A$ in $ \mod 2$ gives the identity matrix which is clearly nonsingular.

Number theory... :(

Note, if you further knew $ n\leq 100$ you could just use a diagonal dominance argument which is more generally useful to know.

We have $ |A|=\text{even}+2007^{n}=\text{odd}$ therefore $ |A|\not = 0$. Ok?  :D


Αφού κανείς δεν ενδιαφέρθηκε να η λύση .
Θέτουμε $a=\frac{x}{y+z}$   :)

Τότε  $2\sum ab=\sum\frac{2xy}{(x+z)(y+z)}\geq\frac{3}{2}$ αφού η τελευταία είναι ισοδύναμη με $x(y-z)^2+y(z-x)^2+z(x-y)^2\geq 0$  που προφανώς ισχύει

Επίσης $\sum\sqrt{ab}=\sum\sqrt{\frac{xy}{(x+z)(y+z)}}\leq\frac{3}{2}$

αφού προκύπτει από την ΑΜ-ΓΜ αφού για παράδειγμα

$\sqrt{\frac{xy}{(x+z)(y+z)}}\leq\frac{\frac{x}{x+z}+\frac{y}{y+z}}{2}$  ;)

Άρα $2\sum ab\geq\frac{3}{2}\geq\sum\sqrt{ab}$  :lol:

Θανάση βάζω και μία που έφτιαξα σήμερα στο σχολίο στο ίδιο στυλ αλλά με 6 μεταβλητές .

Λοιπόν έστω  $p,q,r,s,t,w >0$  τέτοιοι ώστε  $\frac{1}{1+p}+ \frac{1}{1+q}+ \frac{1}{1+r}=2$  και  $\frac{1}{1+s}+ \frac{1}{1+t}+ \frac{1}{1+w}=2$

Να δείξετε ότι  $p^2s^2+q^2t^2+r^2w^2\geq\frac{3}{16}$

 :lol:  :o

σιλουανε μπραβο.Είσαι ο πρωτος που τη λυνει..... :)

μολις λυθει η δικια σου θα βαλλω και εγω μια παρομοια, φρεσκια κατασκευη..... :lol:

Σιλουανέ
 η ανισότητα σου δεν ισχύει για
        (p ,q , r)=(1 , 1/2 , 1/5) και (s , t , w) = (1/5 , 1/2 , 1)  !!
 Δες την ξανά .

Κάνεις λάθος στις πράξεις !

[quote="nikos"]Κάνεις λάθος στις πράξεις ![/quote]

Δείξε μου που.......Γράψε σε παρακαλώ τις πράξεις και γιατί αλλάζει η φορά .

Έχεις δίκιο . Πρέπει μάλλον να είμαι αρκετά χαζός .Τελος πάντων συγγνώμη για τη λάθος άσκηση και το χρόνο που σας έφαγε . :blush:
Θανάση βάλε τη φρεσκια τη δικιά σου τώρα

Σιλουανέ

 μην απογοητεύεσαι , διορθώνεται το πρόβλημα με τον περιορισμό p ≥  q ≥  r  και  s ≥ t ≥ w .

 Νίκος

Ευχαριστώ Νίκο .Τώρα έχεις και τη λύση ?

Aναφερομαστε στο

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=75568[/url]

 :)


The question is not asking for $f(2)$, it's asking for $f'(x)$ which is calculus and doesn't belong here.

And inom calls this [i]easy[/i] :(  :mad:

anirudh, don't get the wrong idea about calculus. Not all of it is extremely difficult (though much of it is).

Let's just pretend inom meant $f(2)$ and forget it. :D

[hide="BTW"] Given $f'(x)=5\cdot15x^{4}+4x^{3}-3\cdot2x^{2}+1$, you could find $f'(2)$. :D [/hide]

[quote="anirudh"][quote="inom"]$f(x)=15x^{5}+x^{4}-2x^{3}+x-29$
Find
$f'(2)$[/quote]

how would you do that :maybe:[/quote]
I don't even know what that means. :rotfl:

ok anirudh this might clear things up for you.

$f'(x)=\lim_{h\to{0}}\frac{f(x+h)-f(x)}{h}$

this is the "long way" to get the equation for $f'(x)$ but once u have it you can easily find $f'(2)$


Hi, I'm new to these forms so I'm not sure whether this is the right place to post this question or not. Basically, it was on a maths SAC at my school (SAC = School Assesed Coursework), and pretty much no one could do it. I've spent about an hour fiddling around with it trying to work it out, but without any success. Here it is:

a) Show that if z^5 - 32 = (z - 2)(z^2 - pz + 4)(z^2 - qz + 4), then: p^2 + 2p - 4 = 0

b) Express the solutions of the equation z^5 = 32 in the form z = rcisx, where -pi < x < pi

c) Use part (b) to write z^5 - 32 in the form (z -2)(z^2 - 4cos(m)z + 4)(z^2 - 4cos(n)z + 4) where you should state the values of 'm' and 'n'

d) Hence, write down the exact values of: cos (2/5 pi) and: cos (4/5 pi)

Well, good luck to all who try!

well, starting with the roots would help
$2e^{i\frac{2}{5}\pi*k}$ with k = -2,-1,0,1,2 (since we are limitied on the values of pi, we need any 5 values of k and these are the only 5 consecutive values that keep it in the restraints)
then for the breakdown of this, for a quadratic with the lead coefficient of 1, the product of the roots is the constant
so we want 2 pairs of roots that multiply to 4 (since we know that it breaks down into 2 quadratic factors with the constant being 4)
i will not tell you the answer, but just think about it, i hope i have given you enough to start, it is better if you solve the rest on your own, that way you could solve another problem similiar to this

Ahhh that makes sense :) Ok I shall attempt that now. Finding the 5 roots of z^5 = 32 is the only part I can do without outside assitance, so I will make a start from there. Now that you mention it, it makes perfect sense about the pairs of roots multiplying to give the constant of 4 (hits head DOH!). I think the reason the question is so confusing is the first part...would you suggest its better to start off with part b) and work backwards to prove part a)?

you can use synthetic division to factor out the $(z-2)$ and then you have to factor it again, i suppose that is manageable just expand $(z^2 - pz + 4)(z^2 - qz + 4)$ and compare the coefficients with the division that you get from taking out the $(z-2)$

Ok thank you SO much for your little hints, they helped set me on the right path to solving this devilish (well I think it's devilish) problem. I also went to my teacher for some help (my teacher is the guy who invented the problem...he is a Doctor of Maths:O) and he basically told me what you said about multiplying out conjugate roots.

One final question which I still don't understand however...in the final part of the problem, where you have to find exact values of cos 2pi/5 and cos 4pi/5, you solve a quadratic and get -1/4(1 +/- root 5). I know that these are the exact values, but how do you know which exact value belongs to which cos angle? (besides using trial and error, and knowing that the negative result will belong to cos 4pi/5 ). Is there any other mathematical reasoning?

well, think about the graph of the cosine function $0>x>\pi$, for the larger the angle, what happens to the value of the cosine? (i.e. as you go right, what happens to the value of the cosine)


geometry
modular arithmetic
combinatorial geometry
coordinate geometry
IMO
IMO 1987

Let $n\ge3$ be an integer. Prove that there is a set of $n$ points in the plane such that the distance between any two points is irrational and each set of three points determines a non-degenerate triangle with rational area.

We will construct our set $\{A_n=(x_n,y_n)\}$ as a set of lattice points. the area of every triangle will automatically be rational (half an integer, actually), so all we need to do is ensure the fact that no three points are colinear and the distance between any two of them is irrational.

The base step $n=3$ is easy: take $A_1=(0,0),\ A_2=(1,1),\ A_3=(3,2)$. Now suppose $n\ge 4$ and we have constructed the lattice points $A_1,A_2,\ldots,A_{n-1}$. For $i=\overline{1,n-1}$ choose primes $p_i\equiv 1\pmod 4$, and choose $a_i,b_i$ s.t. $p_i\not|a_i,b_i,\ p_i|a_i^2+b_i^2$. After that, we can find $k_i$ s.t. $p_i\|(a_i+k_ip_i)^2+b_i^2$ (i.e. $p_i$ divides $(a_i+k_ip_i)^2+b_i^2$, but $p_i^2$ does not). Finally, by the Chinese Remainder Theorem, we can find $x_n,y_n$ s.t. $x_n-x_i\equiv a_i+k_ip_i\pmod{p_i^2},\ y_n-y_i\equiv b_i\pmod{p_i^2},\ \forall i\in\overline{1,n-1}$. The numbers $(x_n-x_i)^2+(y_n-y_i)^2$ will be divisible by $p_i$ but not $p_i^2$, meaning that all the distances $d(A_n,A_i),\ i=\overline{1,n-1}$ are irrational.

This process may yield colliniarities $A_n,A_i,A_j,\ i\ne j\in\overline{1,n-1}$, but this can easily be repaired. Take another prime $p$ which divides none of the differences $x_i-x_j,y_i-y_j,\ i\ne j\in\overline{1,n-1}$, different from $p_i,\ \forall i$ and we also throw in the conditions $p|x_n-x_i,\ p\not|y_n-y_i,\ \forall i\in\overline{1,n-1}$. This will ensure that $\frac{y_n-y_i}{x_n-x_i}\ne\frac{y_i-y_j}{x_i-x_j},\ \forall i\ne j\in\overline{1,n-1}$.

take $C$ an integer very superior to $n$.
Then the points $A_1=(C+1,C^{C+1})$, $A_2=(C+2,C^{C+2})$,..,$A_n=(C+n,C^{C+n})$.

Of course the triangles will not be degenarated. And if $C$ is large enough,
$d(A_i,A_j)^2=(i-j)^2+(C^{C+i}-C^{C+j})^2$ is an integer but not a perfect square.

Take an equilateral triangle with area 1 and thus side length $ 2\sqrt {\frac {1}{\sqrt {3}}}$, and take any set of points with rational areal co-ordinates with respect to this triangle and no three collinear. I claim this set satisfies the desired properties.

Proof:

Rational areal co-ordinates, and the rationality of the area of our triangle of reference, imply that every triangle comprised of three points of such a set, being the determinant of a rational matrix multiplied by a rational scalar, is rational in area.

The distance d between any two rational points (a,b,c),(p,q,r) (with rational displacement vector (a-p,b-q,c-r)) is given by:
\[ d^2 \equal{} \frac {4}{\sqrt 3} ( \minus{} (a \minus{} p)(b \minus{} q) \minus{} (b \minus{} q)(c \minus{} r) \minus{} (c \minus{} r)(a \minus{} p)) \not \in \mathbb{Q}
\]

I have another solution: just take the integer points of a parabola $ y \equal{} x^2$.
The area is clearly rational (in fact, the area of lattice polygons is always rational), but the distances between them are not:

$ (a,a^2),(b,b^2); (a \minus{} b)^2 \plus{} (a^2 \minus{} b^2)^2 \equal{} (a \minus{} b)^2(1 \plus{} (a \plus{} b)^2)$ is not a perfect square.

I think I can do it for other polynomials, but it wasn't test :)

[hide = Solution]Let $A$ be the set of all points of the form $\left( a , \frac{a^2+a}{2} \right)$, for $a\in \mathbb{Z}^+$.

Claim 1: Any 3 elements of $A$ have rational area.
[hide = Proof]Applying Shoelace formula to the given triangle gives a rational expression, since the points are lattice points.
[/hide]

Claim 2: The distance between any two elements of $A$ is irrational.

[hide = Proof]We take two points of $A$, say $\left( a , \frac{a^2+a}{2} \right)$ and $\left( b , \frac{b^2+b}{2} \right)$. Applying the distance formula gives:

\begin{align*}
d&=\sqrt{(a-b)^2+\frac{(a^2-b^2+a-b)^2}{4}}\\
&=\sqrt{(a-b)^2+\frac{\left( (a-b)(a+b)+(a-b)\right)^2}{4}}\\
&=\sqrt{(a-b)^2+\frac{(a-b)^2(a+b+a)^2}{4}}\\
&=(a-b)\sqrt{1+\frac{(a+b+1)^2}{4}}\\
\end{align*}

Since we wish to prove $d$ is irrational, and we know $(a-b)$ is rational, then we can ignore it. So we wish to prove the following expression is irrational:

\[\sqrt{1+\frac{(a+b+1)^2}{4}}.\]

Let $a+b+1=c\geq 1$, and assume the expression is rational, with $p$ and $q$ relatively prime:

\begin{align*}
&\frac{p}{q}=\sqrt{\frac{c^2}{4}+1}\\
\implies & \frac{p^2}{q^2}=1+\frac{c^2}{4}\\
\implies & 4p^2=q^2(4+c^2).
\end{align*}

Since $p$ and $q$ are relatively prime, $q\mid 2$. So $q=2$ or $q=1$.

Case 1, $q=1$:

The equation becomes
\begin{align*}
 & (2p)^2=4+c^2\\
\implies & (2p)^2-c^2=4,
\end{align*}
which is impossible.

Case 2, $q=2$:

The equation becomes

\begin{align*}
 & 4p^2=16+4c^2 \\
\implies & p^2-c^2=4,
\end{align*}

which is also impossible.

Therefore $\frac{p}{q}=\sqrt{\frac{c^2}{4}+1}$ is impossible, and the distance between the two points is irrational, as desired.[/hide]

We now take $n$ distinct elements from $A$. By Claims 1 and 2, these $n$ points all have irrational distances between them, and any three points create a triangle with rational area, which means we are done. :coolspeak:[/hide]

By shoelace, any triangle which has three lattice points as vertices must be rational. Thus, it is sufficient to find a set of $n$ lattice points which have pairwise irrational distance. For any lattice point $(x,y)$, we let $S_{(x,y)}$ denote the set of lattice points which have an irrational distance to the point $(x,y)$. Notice that $S_{(x,y)}$ is just $S_{(0,0)}$ shifted by $x$ and $y$.

[b]Claim:[/b] For any positive integer $N$, there exists an $N \times N$ lattice grid which is contained entirely in $S_{(0,0)}$

Clearly true since points $(a,b) \in \mathbb{Z}^2$ where $\sqrt{a^2+b^2}$ is an integer has density $0$. $\square$

Thus, we proceed with induction. Take a set of $k$ points $\{P_1,P_2, \ldots, P_k \}$ which can be enclosed by a square of sidelength $M$. Then, we can find the same squares in $S_{P_1}, S_{P_2}, \ldots S_{P_k}$ of sidelength $M+1$. All of these squares have at least common lattice, and thus has irrational distance to all points chosen in the set. $\blacksquare$

Solution from [i]Twitch Solves ISL[/i]:

Note that any triangle with vertices which are lattice points has rational area. So we'd be done if we could prove:
[b][color=red]Claim:[/color][/b] Given a finite set $S$ of lattice points, we can add one lattice point $P$ such that $P$ is irrational distance from every point in $S$.
[i]Proof.[/i] Suppose $S$ lies inside $[1,n] \times [1,n]$. Let $a^2+b^2=c^2$ by a primitive Pythagorean triple with the additional property that $a/b \neq x/y$ for any $(x,y) \in S$; this is possible there are infinitely many primitive Pythagorean triples.
Consider large $R$; I claim that $P = \left( aR, bR \right)$ is okay if $R$ is sufficiently large. The squared distance from $P$ to a point $(u,v)$ is \[ D = \left( aR-u \right)^2 + \left( bR-v \right)^2 = (cR)^2 - 2(au+bv)R + (u^2+v^2) \] So for this to be square, it must be inside the set \[ D \in \{(5R-(a+b)n)^2, (5R-(a+b)n+1)^2, \dots, (5R+2n)^2 \}. \] In other words, if $R$ is large enough, then $D$ is not a square unless $D$ is actually identically the square of a polynomial in $R$. For that to be the case, we would need \[ D = \left( cR - \sqrt{u^2+v^2} \right)^2 \] which would only happen if \[ c\sqrt{u^2+v^2} = au+bv \implies 0 = c^2(u^2+v^2) - (au+bv)^2 = (bu-av)^2 \] which doesn't happen because we assured $a/b \neq u/v$.
Hence for each $(u,v) \in S$, there exists a constant $C_{u,v}$ for which $R > C_{u,v}$ guarantees $P$ is irrational distance from $(u,v)$. Take the max of these finitely many constants to finish. $\blacksquare$

Let $L_t=n^{n^{n^{n^{n^t}}}}.$ Then, suppose the points are $(1,L_1),(2,L_2)\cdots (n,L_n)$. The distance between the points with x-coordinates a and b is $$\sqrt{(a-b)^2+(L_a-L_b)^2}.$$ Note that $$(a-b)^2<<<<<<<2(L_a-L_b)+1,$$ so the thing inside the root is not a perfect square, hence the distance is irrational. By shoelace, all areas are multiples of 1/2, hence done.

(alternatively bary bash with respect to an equilateral triangle trivializes the problem :coolspeak:)

Pick $n$ arbitrary non-collinear lattice points (Shoelace implies rational area). If $D$ is the largest squared distance among these, if $P>D$ is prime, scaling everything by $\sqrt{P}$ makes all lengths irrational but areas rational.

When your solution is shorter than the problem statement...

Place on coordinate plane; induct: $n \longrightarrow n+1$. Take $A_{n+1}=(10^{10^{ \mid \max \{x_i\} \mid +10}}+\max \{x_i\},\max \{y_i\}+1)$ over $1 \leq i \leq n$. Trivially works by Shoelace and Pythag.

Consider the space $\mathbb{R}^{2n}$ of of $n$-tuples of points in the plane. Now, for each nonnegative rational $a$, we are excluding all tuples with points $P_1$, $P_2$, and $P_3$ forming a triangle of area $a$; for fixed points and fixed $a$ observe that that is a set with one less dimension so it has measure $0$; unioning over all triples of points and all nonnegative rationals $a$, noting this is a finite union, observe that we are excluding a set of measure $0$ and thus there must be an element outside of the set we are excluding, and this is the $n$-tuple of points we want (we take its underlying set).

There are lots of ways to do this, but here's one that uses $\nu_2$.

We will use lattice points only, so the area condition is automatically satisfied. Now, let $(x_1, y_1), \dots, (x_n, y_n)$ be the $n$ points, and let $d_i = x_{i+1} - x_i$ and $e_i = y_{i+1} - y_i$. Pick $(d_1, d_2, \dots, d_{n-1}) = (1, 1, \dots, 1)$ and $$(e_1, e_2, e_3, \dots, e_{n-1}) = (1, 2^n + 1, 2 \cdot 2^n+1, \dots, (n-2) \cdot 2^n + 1).$$ Notice that for two points $(x_i, y_i)$ and $(x_j, y_j)$, it suffices to show that $$(d_{i+1}+d_{i+2}+\cdots+d_j)^2 + (e_{i+1}+e_{i+2} + \cdots + e_j)^2$$ is never a perfect square. This is clear because both have $\nu_2$ equal to $\nu_2(j - i)$ as all the $d_i$ and $e_i$ are $1$ mod $2^n$. Hence we have mod $4$ issues upon factoring out the $2^r$ term, done.

VerySus

Consider the points $(x,x^2)$ for $x = 1,2,\dots,n.$ If $1 \le r,s \le n,$ then $$\sqrt{(r-s)^2 + (r^2-s^2)^2} = (r-s) \sqrt{(r+s)^2 + 1}.$$ The expression in the square root is never a perfect square except when $r + s = 0,$ which obviously can't happen. The area of a lattice triangle is always rational, so this set of points satisfies the problem condition.

Consider the points of the form $(x,x^2)$ for $x=1,2,\ldots n$, as they are lattice points the area of any triangles formed by this points it´s rational,  the distance between any of this points it´s equal to
\[\sqrt{(x-y)^2+(x^2-y^2)^2}=(x-y)\sqrt{(x+y)^2+1}\]
which is never rational

The answer is yes. Define $y_x = 1434^{1434^{1434^{1434^x}}}$. Then, consider the points $(i, y_i)$ for $1 \leq i \leq 1987$. Because $y_i$ is an integer for all integers $i$, then every point is a lattice point and all areas are rational by shoelace (in fact, the denominator must be 2 if present).

However, the distance between any two points, $\sqrt{(i-j)^2 + (y_i - y_j)^2}$ cannot be an integer since $0 < (i-j)^2 \ll 2(y_i-y_j)+1$, so we are done.


calculus
integration
geometry
number theory proposed
number theory

Does there exist a triangle with sides of integral lengths such that its area is equal to the square of the length of one of its sides?

I don't have a solution to this problem but I have some idea.

Let the sides be $x+y$, $y+z$ and $z+x$. The area of the triangle is $\sqrt{xyz(x+y+z)}$. Suppose $\sqrt{xyz(x+y+z)}= (x+y)^{2}$. That is $xyz(x+y+z)= (x+y)^{4}$. Set $\lambda = \frac{z}{x+y}$. This gives us $\lambda^{2}+\lambda = \frac{(x+y)^{2}}{xy}$. But $\frac{(x+y)^{2}}{xy}\ge 4$. Hence $\lambda^{2}+\lambda \ge 4$. As $\lambda$ is positive we should have $\lambda \ge \frac{\sqrt{17}-1}{2}$ (I hope I am right). Now we may use a computer search.  :(


symmetry
inequalities
function
logarithms
calculus
inequalities proposed

(At least I think it's new.)

Let $n \ge 2$ be an integer. Prove that for all reals $a_1, a_2, \dots, a_n > 0$ and reals $p, k \ge 1$,

\[
\left(\frac{a_1+a_2+\cdots + a_n}{a_1^p+a_2^p+\cdots +a_n^p}\right)^k \ge \frac{a_1^k+a_2^k+\cdots +a_n^k}{a_1^{pk}+a_2^{pk}+\cdots +a_n^{pk}}
\]

With equality iff $p=1$ or $k=1$ or $a_1=a_2=\cdots = a_n$. Moreover, the inequality flips if $p < 1$, and flips (possibly again) if $k < 1$. ($p$ and $k$ still positive.)

I guess it [i]is[/i] new, and maybe even a respectable match for the adroit inequality specialists who roam these forums... :starwars:

A hint:
[hide]1. The idea involves majorization.

Another hint:
[hide]2. See if you can apply Karamata's inequality.

One last hint:
[hide]3. Use some algebra to introduce sums over a concave function.[/hide][/hide][/hide]

Vasc, can you post a solution for this one you proved? ;)

[quote="Vasc"]Mildorf, long ago I proved that if $(x_1,x_2,...,x_n) \succ (y_1,y_2,...,y_n)$, then

$(a_1^{x_1}+...+a_n^{x_1})(a_1^{x_2}+...+a_n^{x_2})...(a_1^{x_n}+...+a_n^{x_n}) \ge  (a_1^{y_1}+...+a_n^{y_1})(a_1^{y_2}+...+a_n^{y_2})...(a_1^{y_n}+...+a_n^{y_n}).$

Your inequality is a nice consequence of this result for $k,p$ positive integers.[/quote]

An astonishingly general result! I think I see what you are saying - just a bit of continuity work will be needed to get $p, k \in \mathbb{R}^{+}$. My proof is as follows:

Taking the $k$th root of both sides, we see that the inequality is equivalent to
\[
\sum_{i=1}^n \sqrt[k]{\frac{a_i^k}{a_1^k + a_2^k + \cdots + a_n^k}} \ge \sum_{i=1}^n \sqrt[k]{\frac{a_i^{pk}}{a_1^{pk} + a_2^{pk} + \cdots  a_n^{pk}}}
\]
WLOG, suppose that $a_1 \ge a_2 \ge \cdots \ge a_n$. We prove a lemma.  Let $S_i = \frac{a_i^p}{a_1^p + \cdots + a_n^p}$ and $T_i = \frac{a_i^q}{a_1^q + \cdots + a_n^q}$ for $i = 1, 2, \dots, n$ where $0 < q < p$. Then the sequence $S_1, S_2, \dots, S_n$ majorizes the sequence $T_1, T_2, \dots, T_n$.
\par\indent To prove the claim, we note that $S_1 \ge \cdots \ge S_n$ and $T_1 \ge \cdots \ge T_n$ and have, for $m \le n$,

\[
\sum_{i=1}^{m} S_i \ge \sum_{i=1}^{m} T_i \\
\iff \left(a_1^p + \cdots + a_m^p\right)\left(a_1^q + \cdots + a_n^q\right) \ge \left(a_1^q + \cdots + a_m^q\right)\left(a_1^p + \cdots + a_n^p\right) \\
\iff \left(a_1^p + \cdots + a_m^p\right)\left(a_{m+1}^q + \cdots + a_n^q\right) \ge \left(a_1^q + \cdots + a_m^q\right)\left(a_{m+1}^p + \cdots + a_n^p\right) \\
\iff \sum_{(i,j) |\, \{1 \le i \le m < j \le n\}} a_i^pa_j^q - a_i^qa_j^p \ge 0
\]
Which is obvious. In particular, $m = n$ is an equality case, and the claim is established. But now the desired is a direct consequence of the Majorization inequality applied to the sequences in question and the function $f(x) = \sqrt[k]{x}$.

The "new idea" I refer to is this symmetric fraction majorization, which implies a considerably more general result than what I have posted, namely:

\[ f\left(\frac{a_1^p}{a_1^p+\cdots + a_n^p}\right) + \cdots + f\left(\frac{a_n^p}{a_1^p + \cdots + a_n^p}\right) \le f\left(\frac{a_1^q}{a_1^q + \cdots + a_n^q}\right) + \cdots + f\left(\frac{a_n^q}{a_1^q + \cdots + a_n^q}\right) \]
 iff $\mbox{XOR}\left\{\, p > q;\, f\, \mbox{is convex}\,\right\}$.

Vasc, may we see a complete solution to your gem?

[quote="Mildorf"]The "new idea" I refer to is this symmetric fraction majorization, which implies a considerably more general result than what I have posted, namely:

\[ f\left(\frac{a_1^p}{a_1^p+\cdots + a_n^p}\right) + \cdots + f\left(\frac{a_n^p}{a_1^p + \cdots + a_n^p}\right) \le f\left(\frac{a_1^q}{a_1^q + \cdots + a_n^q}\right) + \cdots + f\left(\frac{a_n^q}{a_1^q + \cdots + a_n^q}\right) \]
 iff $\mbox{XOR}\left\{\, p > q;\, f\, \mbox{is convex}\,\right\}$.[/quote]

Mildorf, I am very sorry to say this, but I remember your method being used by Iura in http://www.mathlinks.ro/Forum/viewtopic.php?t=23752 post #3. However, you have generalized this method in quite a nice way.

By the way, thanks a lot for including my solution of problem 25!

[quote="Vasc"]A complete solution to my "gem" is the following:
Apply Karamata's Inequality to the convex function
$f(x)=\ln(a_1^x+a_2^x+...+a_n^x)$.
That's all.[/quote]

This is a pretty nice proof (and the same as mine). Just wanted to note that the convexity of the function $f\left(x\right)=\ln\left(a_1^x+a_2^x+...+a_n^x\right)$ is a simple application of the Hölder inequality: We have to prove that, for positive reals $a_1$, $a_2$, ..., $a_n$, nonnegative x and y, and nonnegative u and v such that u + v = 1, we have

$u\ln\left(a_1^x+a_2^x+...+a_n^x\right)+v\ln\left(a_1^y+a_2^y+...+a_n^y\right)\geq\ln\left(a_1^{ux+vy}+a_2^{ux+vy}+...+a_n^{ux+vy}\right)$.

By taking the exponent, we transform this into

$\left(a_1^x+a_2^x+...+a_n^x\right)^u\cdot\left(a_1^y+a_2^y+...+a_n^y\right)^v\geq a_1^{ux+vy}+a_2^{ux+vy}+...+a_n^{ux+vy}$.

But this is a trivial consequence of Hölder's inequality (since u + v = 1). I am posting this proof only to show a further example on how often convexity of certain functions can be shown without any use of calculus.

  Darij

[quote="darij grinberg"][quote="Mildorf"]The "new idea" I refer to is this symmetric fraction majorization, which implies a considerably more general result than what I have posted, namely:

\[ f\left(\frac{a_1^p}{a_1^p+\cdots + a_n^p}\right) + \cdots + f\left(\frac{a_n^p}{a_1^p + \cdots + a_n^p}\right) \le f\left(\frac{a_1^q}{a_1^q + \cdots + a_n^q}\right) + \cdots + f\left(\frac{a_n^q}{a_1^q + \cdots + a_n^q}\right) \]
 iff $\mbox{XOR}\left\{\, p > q;\, f\, \mbox{is convex}\,\right\}$.[/quote]

Mildorf, I am very sorry to say this, but I remember your method being used by Iura in http://www.mathlinks.ro/Forum/viewtopic.php?t=23752 post #3. However, you have generalized this method in quite a nice way.
[/quote]

Good thing I kept it in quotes! :blush: My apologies.

I am surprised nobody has yet answered http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48247 now that this idea has once again surfaced.  ;)

[quote="siuhochung"]Well, I think the well known idea of compression reduces the inequality to two-variable case. I mean, we only need to prove if $(x_1, x_2) \ge (y_1, y_2)$, the inequality is true, then it's true. But this is simple because we can expand both sides and muirhead/amgm.[/quote]

Can anyone mind checking my idea? I haven't really used the idea of compression before so I don't really know. By compression I am refering to an idea posted by darij grinberg on the forum before where he proved muirhead revisited or rearrangement revisited (I don't really remember that)

[quote="darij grinberg"]But this is a trivial consequence of Hölder's inequality (since u + v = 1). I am posting this proof only to show a further example on how often convexity of certain functions can be shown without any use of calculus.[/quote]

I think that, in order to prove Holder, we must use calculus one way or the other. (i'm refering to limits) If you know a proof that doesn't make use of limits, please share. I remember that I tried to prove it some time ago, but I failed...


Triangle ABC has sides with lengths 10, 23 and 27 units. The perimeter of a particular square is 60% of triangle ABC’s perimeter. What is the area of the square?

Triangle ABC has perimeter 60, so the square has a perimeter of 36. Thus each side is 9, and the area of the square is $ \boxed{81}$.


geometry trapezoid trigonometry ratio parallelogram AMC AIME

[hide="Hint"]
The lengths of the bases suggest a certain relationship.
[/hide]

[hide="Solution"]
It shouldn't be hard to use trigonometry to bash this, but in fact there is a much easier way. Extend $ \overline{AD}$ and $ \overline{BC}$ to meet at point $ E$:

[asy]size(250);
defaultpen(0.8);
pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13);
draw(A--B--C--D--cycle);
draw(D--F--C,dashed);
label("\(A\)",A,S);
label("\(B\)",B,S);
label("\(C\)",C,NE);
label("\(D\)",D,W);
label("\(E\)",F,N);
label("39",(C+D)/2,N);
label("52",(A+B)/2,S);
label("5",(A+D)/2,E);
label("12",(B+C)/2,WSW);[/asy]

Since $ \overline{AB} \parallel{} \overline{CD}$ we have $ \triangle AEB \sim \triangle DEC$, with the ratio of proportionality being $ \frac {39}{52} = \frac {3}{4}$. Thus
\begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\
\frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*}
So the sides of $ \triangle CDE$ are $ 15,36,39$, which we recognize to be a $ 5 - 12 - 13$ right triangle. Therefore (we could simplify using side squared - area ratio),
\[ [ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{210} \Rightarrow \mathrm{(C)}.
\]
[/hide]

[hide]Drop heights from D and C, to points X and Y respectively. Let both be h. Now, let AX=x, and BY=13-x. We have
$ h^2\plus{}x^2\equal{}25$ and $ h^2\plus{}(13\minus{}x)^2\equal{}144$. Subtracting the first from the second, we have 169-26x=119 so 26x=50, and x=25/13. Plug in to find h=60/13 and the area is $ \frac{30}{13}(52\plus{}39)\equal{}210$. So C.[/hide]

I hate to revive an old thread, but I found a nice easy way to do the problem with minimal algebra:[hide]make the trapezoid into a parallelogram using points D, C and B and a triangle with sides 5, 12, and 52-39=13. But we know a 5, 12, 13 triangle is right, so we find the area to be 30, and subsequently find the height to be 60/13. Then we find the area of the parallelogram to be 39 x 60/13=180, and 180+30=210.[/hide]

How would you do this problem in the case it [b]wasn't [/b] a right triangle? (whats the general strategy after extending the sides)?

Probably using Law of Cosines to find the angle and then using the sine area formula.

What happens if you can't explicitly find the angle? Like on the AIME I test the problems with finding the area of the triangle which have vertices on the faces used a technique of finding cos theta and sin theta then squaring, and adding to eliminate theta).  Would that work here?

Or, you can just use Heron's formula to find the area, and then use the dumb formula $bh/2$ to find the height.

Sorry, but how would you use heron's? I mean, you could split the trapezoid with a diagonal into two triangles, but then the diagonal is still undetermined...

Oh, maybe you are misunderstanding the trick the he is describing. He is not constructing a diagonal. He is [i]constructing a parallelogram[/i]. Here's another example trapezoid: $AB || CD$, $AB = 20, BC = 13, CD = 34, AD = 15$. Construct the point $P$ so that $ABPD$ is a parallelogram. Now look at the triangle $BPC$. It has sides of length $13, 15, 34-20 = 14$. Then, its area is 84, so you can find that the height to $BC$ in this triangle is $84 \cdot 2 /14 = 12$. So the area of the trapezoid would be $(20+34) \cdot 12 / 2 = 324$.


[quote="irantst"]Let x,y,z\in\mathbb R^{+} and x+y+z=3. Prove that:
\frac{x^3}{y^3+8}+\frac{y^3}{z^3+8}+\frac{z^3}{x^3+8}\geq\frac19+\frac2{27}(xy+xz+yz)[/quote]
http://www.mathlinks.ro/viewtopic.php?p=1241795#1241795


Do we need to mail in an entry form for Round 2?  If not, do we need to mail in any other forms/cover sheets?

You only need to send in the Entry form once.  If you have already sent it, you don't need to send it (or any other form) again.

hey
i can't open up the solution form pdf
it says something like
"does not start with %pdf"
is it just me?

What page are you trying to access the solution form from?


geometry
area of a triangle
Heron\u0027s formula

1. (1991 #3) What is the first time between 4:00 and 5:00 that the hour hand and the minute hand are exactly 10 degrees apart?

2. (1991 #10) A videotape can record 2 hours on short play, 4 hours on long play, or 6 hours on extra-long play. after recording 32 minutes on short play and 44 minutes on long play, how many minutes can it record on extra-long play?

3. (1992 #1) How many times does the digit "4" appear in the integer part of the quotient 10^100 / 7 ?

4. (1992 #5) One side of a triangle is divided into segments of lengths 6 cm and 8 cm by the point of tangency of the inscribed circle. If the radius of the circle is 4cm, what is the lenght, in cm, of the shortest side of the triangle?

5. (1993 #1) Find the sum of the digits in the decimal representation of 789000003450000^2 .

Answers:

[hide]1. 4:20
2. 198
3.17
4. 13
5. 54[/hide]

[hide]
1. What is the first time between 4:00 and 5:00 that the hour hand and the minute hand are exactly 10 degrees apart?
1. The angle between the hands of a clock at h:m is 30h-11m/2. So, since h must be 4, 120 - 11m/2 = 10, 11m/2 = 110 and m = 20. 4:20.

2. (1991 #10) A videotape can record 2 hours on short play, 4 hours on long play, or 6 hours on extra-long play. after recording 32 minutes on short play and 44 minutes on long play, how many minutes can it record on extra-long play?
2. 1 minute short play = 1/120
1 minute long play = 1/240
1 minute X-long play = 1/360
1 - (32/120 + 44/240) = 66/120. (66/120)/(1/360) = 198 minutes.

3. (1992 #1) How many times does the digit "4" appear in the integer part of the quotient 10^100 / 7 ?
There are 99 digits in this, as one can find out. Every six digits, there will be one 4. So, there are 16 4's and 3 digits left. Looking at 1/7, 4 is the second digit so in the last three digits there is another 4, making a total of 17 fours.

4. (1992 #5) One side of a triangle is divided into segments of lengths 6 cm and 8 cm by the point of tangency of the inscribed circle. If the radius of the circle is 4cm, what is the lenght, in cm, of the shortest side of the triangle?
Ah, I'm too lazy to write this one out. Let the other segment be x. Then use a = rs and Heron's formula to get two different equations and solve for x.

5. (1993 #1) Find the sum of the digits in the decimal representation of 789000003450000^2 .
This comes out to 789*789|zeroes|2*789*345|zeroes|345*345|zeroes.
Finding each of these, we get a digit-sum of 54.[/hide]

[hide="1"] so the hour hand and the minute stand start 4/12=x/360 degrees away from each other, or 120. The hour hand moves forward 360 degrees in 12 hours, or 30 in an hour, or .5 in a minute. The minute hand travels 360 degrees in an hour, or 6 in a minute. So the minute hand gains 5.5 degrees on the hour hand a minute. Since it start 120 degrees a part, it needs to make 110 degrees, and 110/5.5=20 minutes, so 4+20=4:20

[/hide]

[hide="2"]

so, since it recorded 32 minutes on short play out of 120, it records 16/60 of the space. Since it recorded, 44 minutes on long play out of 240 minutes, it records 11/60 of the space. so there is 1-27/60 or 33/60 space left. 33/60 of 6 hours or 360 minutes is 198 minutes.
[/hide]

[hide="3"] So find a pattern. You find that 1/7=.142857 repeating, but no integer digits. Since there are 6 digits repeating, you can see that every 6 digits, a new four comes. You just have to see where the first four comes.
so 10^1/7=1.42857 repeating, still no four. 10^2/7=14.2857, a integer 4!

so there is 98 more tens left, so 98/6=16 remainder 2. So 16+ the original 4 is 17 fours.  [/hide]

[hide="4"]

I am too dumb to do this one

[/hide]

[hide="5"]
so basically it is 789^2+345^2+789*345+345*789. So just do that, and the digits add up to 54[/hide]


modular arithmetic
number theory unsolved
number theory

$ p $ is a prime number , $ n, a \in Z $ are not divisible by p . Prove that :
If the congruence equation $ x^n \equiv a $ [mod p] has a solution ,then the congruence equation $ x^n \equiv a $ [mod $ p^r $] has a solution for all $ r \in N $

By induction.
Suppose $x^n\equiv a\pmod{p^k}$. Consider $y=x+p^kt$. Then $y^n\equiv x^n+nx^{n-1}p^kt\pmod{p^{k+1}}$, where $x^n-a=p^k\cdot r$. So $y^n-a=p^k(r+nx^{n-1}t)$. Threfore we just need to find $t$ s.t. $r+nx^{n-1}t\equiv 0\pmod{p}$. We can do it since $nx^{n-1}\not\equiv 0\pmod{p}$.


Challenge:

Sequences [tex]\{a_n\}[/tex] and [tex]\{b_n\}[/tex] satisfy

[tex]a_0=1[/tex]

[tex]\displaystyle a_{n}=\frac{2a_{n-1}}{1+2a_{n-1}^2}[/tex]

[tex]\displaystyle b_{n}=\frac{1}{1-2a_n^2}[/tex]

Prove that [tex]b_n[/tex] is the square of an integer for all [tex]n>0[/tex].

(EDIT) Furthermore, find a closed expression for [tex]a_n[/tex] and [tex]b_n[/tex] in terms of n.

If we take [tex]x_n=\sqrt{b_n}[/tex] then we get [tex]x_n=2x_{n-1}^2-1[/tex] and since [tex]x_1=3[/tex], we just continue by induction.

The solution is correct, but not enough work is shown.

Hm. I guess I'll have to make it a bit harder.

[quote="grobber"]I plotted a few terms with Mathematica (I couldn't have done it by hand, as the terms increase very quickly) and I spotted the recurrence relation :).[/quote]

Well that's no fun!

Here's an explicit solution:

[tex]\displaystyle {2a_n^2 = \frac{8a_{n-1}^2}{4a_{n-1}^4 + 4a_{n-1}^2 + 1}}[/tex]

[tex]\displaystyle {1-2a_n^2 = \frac{4a_{n-1}^4 - 4a_{n-1}^2 + 1}{4a_{n-1}^4 + 4a_{n-1}^2 + 1}}[/tex] [tex]\displaystyle {= \frac{(2a_{n-1}^2 - 1)^2}{(2a_{n-1}^2 + 1)^2}}[/tex]

Therefore [tex]\displaystyle {b_n = \Big[ \frac{2a_{n-1}^2 + 1}{2a_{n-1}^2 - 1} \Big]^2}[/tex] [tex]\displaystyle {= \Big[ 1 + \frac{2}{2a_{n-1}^2 - 1} \Big]^2}[/tex] [tex]\displaystyle = (1 - 2b_{n-1})^2[/tex].

So by induction, all of the [tex]b_n[/tex] are integers.

(Okay, that turned out not to be so fun after all. lol)


$ \frac {dy}{dx} \equal{} \sqrt {y}$

$ \frac {dy}{\sqrt {y}} \equal{} dx$

$ 2\sqrt {y} \equal{} x \plus{} 2C$

$ y \equal{} \left(\frac {x}{2} \plus{} C\right)^2$

$ C \equal{} \sqrt {y_0}$

$ y \equal{} \left(\frac {x}{2} \plus{} \sqrt {y_0}\right)^2$

It's not unique, because there is one solution just demonstrated by Yuriy, and another is $ y \equal{} 0$.

Yuriy, beware, the technique of separating variables does not in itself guarantee uniqueness of the solutions obtained. More intricate arguments are required for that.

In fact, Yuriy's function is never a solution over all of $ \bf R$ due to sign problems.

Here should be $ y_0> 0$ and $ y$ never should be $ 0$ and solution exist if $ x > \minus{} 2\sqrt {y_0}$

oh sorry. the condition is y0>0.

And that doesn't show that the answer is unique.


calculus
conics
geometry
function
analytic geometry
graphing lines
slope

there are two tangent lines to the curve y= 4x-x^2 that go through (2,5). Find the equations of both of them.

this is a textbook style CALCULUS PROBLEM (unless it is some type of conic geometry)
plz do not ask me about some of my steps b/c they are very uber complex (at least the way i do it)

the derivatite of the function is $ 4\minus{}2x$.
so tangent lines will pass through $ (a,4a\minus{}a^2)$ with slope $ 4\minus{}2a$
the point (2,5) should also be on this line
the equation of the line passing through $ (a,4a\minus{}a^2)$ with slope $ 4\minus{}2a$ is $ y\equal{}(4\minus{}2a)x\plus{}a^2$
plugging in $ (2,5)$ for $ (x,y)$ gives $ 5\equal{}8\minus{}4a\plus{}a^2$ which yields solutions a=1 and 3
with equations
$ y\equal{}2x\plus{}1$ and $ y\equal{}\minus{}2x\plus{}9$


Hi... I'm new to the latex thing and I can't find the way to put in a congruence sign... thanks!

I'm not sure which congruence sign you want, but you can choose between $\cong$ and $\equiv$, mainly depending on whether this is a geometry problem or a number theory problem.


ARML
geometry
3D geometry
modular arithmetic
algebra
polynomial
logarithms

Might as well post this somewhere - the problem is on one side of the ARML mug and the answer on the other.

Compute the largest odd prime factor of $ 2009^3 \minus{} 6025.$

EDIT: Yes, I misread the fine print on the mug. The problem is the largest odd prime factor of $ 2009^2\minus{}6025.$

Although the problem with the cube has some entertainment value, except for the part about deciding whether $ 2011$ is prime or not.

[quote="Kent Merryfield"]... and the answer on the other. [/quote]
That's actually the wrong answer.

Whoops- wrong problem. It's actually $ 2009^2\minus{}6025$ that we wish to find the largest prime factor of.

[hide]
Note that $ 2009^2 \equiv 1 \pmod{2008}$ and $ 6025 \equiv 1\pmod{2008}$.

Thus, $ 2008$ is a factor of $ 2009^2 \minus{} 6025$. So, let $ x \equal{} 2008$.

Then, we have: $ 2009^2 \minus{} 6025 \equal{} (x \plus{} 1)^2 \minus{} (3x \plus{} 1) \equal{} x(x\minus{}1) \equal{} 2008\cdot 2007 \equal{} (2^3\cdot251) \cdot (3^2 \cdot 223)$.

Since $ 251$ is prime, the answer is $ \boxed{251}$. [/hide]

That was much less interesting.

And as jmerry mentioned, the answer printed on the mug is wrong - they have the factorization given by Xantos but then report the answer as 223.

Awesome -- I think that they've had typos or other errors more often than not on those mugs, at least from the handful of years I've seen.  (The one I had tea out of this morning asked me to computer the value of $ x \minus{} y$.)

Just wondering, are there any efficient ways in determining whether some large numbers are prime (of course when they don't satisfy some well-known divisibility tests first)? First example, it would probably take me a while to determine if 2011 is prime (as I would try to divide it by 7, 13, 17, 19, 23 etc until about the largest prime less than the square root of 2011, or maybe I would just give up half-way and guess it's a prime with doubt)...

Google "primality testing."  It was relatively recently proven that there are efficient (i.e., taking time polynomial in $ \log p$) deterministic algorithms for primality testing; the original example is based on derandomizing an efficient randomized algorithm.  I don't know if they're efficient enough to use in practice for truly large primes.

But this is a small-ish number. Forget the fancier tests, can we do the sieve?

$ 47^2 \equal{} 1600 \plus{} 560 \plus{} 49 > 2011,$ so we only need to check $ 2,3,5,7,11,13,17,19,23,29,31,37,41,43.$

$ 2,3,$ and $ 5$ you know already. And $ 7,11,13$ can be done by the $ 1001$ test: $ 11\minus{}2\equal{}9$ is not divisible by any of $ 7,11,$ or $ 13.$ That leaves 8 other primes to check.

And when checking those, you don't have to actually determine the quotient, just whether it's divisible. For instance, $ 37:$ Since $ 37\cdot3 \equal{} 111,$ subtract $ 111$ from $ 2011$ to get $ 1900.$ Since $ 1900 \equal{} 19\cdot 10^2,$ it's obviously not divisible by $ 37.$

(Actually, we could have done $ 37$ by checking mod $ 999.$ We have $ 2011\equiv 2\plus{}11\equal{}13\mod{999}$ so it's not divisible by $ 37.$)

And so on. Yes, it's annoying, but it's well within the range of what you can do by hand.


Simplify: $ \sum_{k=10}^{2006}{k\choose{10}}$
(Answer should contain no summations but may still contain binomial coefficients/combinations).

[hide="Solution"]Add $ \binom{10}{11} \equal{} 0$ to the beginning, and then repeatedly use $ \binom{n}{m} \plus{} \binom{n}{m \plus{} 1} \equal{} \binom{n \plus{} 1}{m \plus{} 1}$, and the whole sum collapses to $ \binom{2007}{11}$

[b]Note:[/b] [i]The positions of these binomials on Pascal's Triangle have earned the generalization of this technique the name "Hockey Stick Theorem"[/i][/hide]


See http://www.mathlinks.ro/Forum/index.php?f=57&ppage=50&sort=lastpost&order=DESC , at the moment there is a new topic by dna to be seen.But when clicking on it , it says it doesn't exist!

(FireFox 1.5.0.4, Windoof XP)

Edit: It works now... what is going on¿

It's a bug allright, but not yet clear how it arrises  :?


Find the sum of the positive odd whole numbers less than 30.