conics
parabola
integration
function
calculus
calculus computations

Dear all,

I am having a hard time in solving this problem. Would you please help?

"Consider the region R bounded by y = 4x - x^2 and the x-axis. Find the equation of the line that intersects the origin and divides R into two regions of equal area."

Thanks a lot! Enjoy the weekend.

Let the parabola be $ f$ and let the line be $ y \equal{} ax$. The intersection of $ y$ and $ f$ is found as follows:

$ ax \equal{} \minus{} x^2 \plus{} 4x$
$ a \equal{} \minus{} x \plus{} 4$
$ x \equal{} 4 \minus{} a$

If you draw a graph, you'll see that $ y$ divides $ R$ into an upper and a lower region. The upper region goes from $ x \equal{} 0$ to $ x \equal{} 4 \minus{} a$ and the lower goes from $ x \equal{} 0$ to $ x \equal{} 4$. The resulting equation is as follows:

$ \int_0 ^{4 \minus{} a} ( \minus{} x^2 \plus{} 4x \minus{} ax)\,dx \equal{} \int_0 ^{4 \minus{} a}ax \, dx \plus{} \int_{4 \minus{} a}^4( \minus{} x^2 \plus{} 4x)\,dx$

The right-hand side uses the bounds it does because of which function bounds the lower region on certain intervals. This is easier to see if you draw a graph. To find $ a$, we evaluate these integrals:

$ \minus{} \frac {1}{3}x^3 \plus{} 2x^2 \minus{} \frac {1}{2}ax^2 \Big]^{4 \minus{} a}_0 \equal{} \frac {1}{2}ax^2 \Big]^{4 \minus{} a}_0 \plus{} \Big[ \minus{} \frac {1}{3}x^3 \plus{} 2x^2 \Big]^4_{4 \minus{} a}$

$ \minus{} \frac {1}{3}(4 \minus{} a)^3 \plus{} \left(2 \minus{} \frac {1}{2}a\right)(4 \minus{} a)^2 \equal{} \frac {1}{2}a(4 \minus{} a)^2 \minus{} \frac {64}{3} \plus{} 32 \minus{} \left[ \minus{} \frac {1}{3}(4 \minus{} a)^3 \plus{} 2(4 \minus{} a)^2\right]$

Note that the terms on the left are also on the right (except they are negative on the right). So we combine them:

$ \minus{} \frac {2}{3}(4 \minus{} a)^3 \plus{} 2\left(2 \minus{} \frac {1}{2}a\right)(4 \minus{} a)^2 \equal{} \frac {32}{3}$
$ \minus{} \frac {2}{3}(4 \minus{} a)^3 \plus{} (4 \minus{} a)(4 \minus{} a)^2 \equal{} \frac {32}{3}$
$ \minus{} \frac {2}{3}(4 \minus{} a)^3 \plus{} (4 \minus{} a)^3 \equal{} \frac {32}{3}$
$ \frac {1}{3}(4 \minus{} a)^3 \equal{} \frac {32}{3}$
$ (4 \minus{} a)^3 \equal{} 32$
$ 4 \minus{} a \equal{} \sqrt [3]{32}$
$ a \equal{} 4 \minus{} \sqrt [3]{32}$

So, the desired line is $ \boxed{y \equal{} (4 \minus{} \sqrt [3]{32})x}$.


function
calculus
integration
symmetry
real analysis
real analysis unsolved

let f be reimann integrable function from [a,b] to R and g be function from[a,b] to R which agrees with f on all but finitely many points of[a,b].show g is reimann integrable and integral of f from a to b=integral of g from a to b.does the same result hold when finitely many is replaced by countably many?

Every upper and lower step function for $f$ is also an upper and lower step function for $g$, except that you may possibly have to modify it in the finitely many points. However, that does not change the values of the integrals of the step functions. By symmetry you can replace $f$ and $g$ in the above statement. So, since the riemann integral definition involves supremum and infimums of such integrals, the statement immediately follows.

The statement is not true for countably many points. Let $f(x) = 0$ for $x\in [0,1]$ and let $g(x) = 1$ if $x$ is rational, and let $g$ agree with $f$ otherwise. Since the rationals are dense in the real numbers it is clear that any lower step function must be $\le 0$ everywhere except for on finitely many points, while any upper step function must be $\ge 1$ everywhere except on finitely many points. So the inf of upper functions is not equal to sup of lower functions and hence the function is not riemann integrable.


Given any $n+1$ integers between $1$ and $2n$, show that two of them are relatively prime. Is this result best possible, i.e., is the conclusion still true for $n$ integers between $1$ and $2n$?

[hide]
We just need to show there must be two consecutive numbers, which we know are relatively prime.  The most numbers we can get that aren't consecutive would be:
1, 3, 5, ..., 2n-3, 2n-1
Which is a total of n numbers.  Since there are n+1 numbers, one of the numbers in between must be filled, thus there will be a pair of consecutive integers.
[/hide]

And the same conclusion is not true for $n$ (as we could choose $n$ even numbers).


Can anyone find the longlists of the IMO 2003, 2004 ? If yes caould you give some details about this? Thank you .

Some people, e.g. people on the IMO Problem Selection Committee, still may have them though the policy is to return the non-ISL problems to the respective proposer countries and not to collect or to keep track of them. Well that means [b]NO[/b] to your question. The rule is partially based on the lack of good competition problems world-wide and to some extent to the lower quality of some of the proposed problems, lots of "well-known", wrong, standard, boring, too easy, too hard problems, too long problem descriptions etc.

[quote="silouan"]Thank you Orl and something else. What is the latest IMO in which we have Longlist ?

1988 ?[/quote]

Actually 1989 should have been the latest IMO which made the ILL publically available. But due to a lot of chaos in Moscow's IMO 1992 it was handed out again. But on the other hand a main member (from New Zealand) of the IMO 1992 Problem Selection Committee denied the fact that there was an ILL. Lots of problem were sent in to late back then and were not considered for the contest so the ILL might have been produced after the actual IMO. But anyway I got the ILL 1992 from the extremely helpful American IMO team leader back then. Well I got the ISL and ILL 1992 from different people and it seems not every ISL problem can be matched with an ILL problem, e.g. the ISL is not a subset of the ILL. BTW did you know that the German Democratic Republic (GDR) actually was supposed to administrate the IMO in 1992 but it was too late. At this time GDR's reunification already took place and the event moved to Russia and they obviously were not prepared to handle all the issues which may have to be addressed at an IMO.  :)

Will you post the ILL from 1989 and 1992 to the forum ?

BTW thank you for all the informations.

[quote="silouan"]Will you post the ILL from 1989 and 1992 to the forum? [/quote]

ILL 1989 is partially already on the forum. Search for it! If you are looking for the ILL 1989 and ILL 1992 as distributed to the IMO team leaders check right this forum for the .pdf-files. BTW ILL 1992 is acknowledged to be an ILL of high quality. Eventually I will have posted all available IMO, ISL and ILL problems on ML/AoPS. Check the Resources section frequently! :)

[quote="silouan"]BTW thank you for all the information.[/quote]

You are welcome! :)

[quote="silouan"]Will you post the ILL from 1989 and 1992 to the forum ?
[/quote]

See [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=18&year=1989]here[/url] and [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=18&year=1992]here[/url].


Hap, the crossing guard, is very popular with the students at Springdale Elementary School. Along with his usual morning smile, Hap has a daily riddle. One of the favorites was this one: How do 5 and 9 more make 2?

umm...9 hours after 5 o'clock is 2 o'clock?

im not very sure this is a math problem.

uh.... sounds more like a riddle than a math problem like it says in the description but...... otherwise, I think the time one is right. I can't think of any other way.


linear algebra
matrix
linear algebra unsolved

Is this homework?  :maybe:  :o

What do you mean the solutions?  Have same sign?  The question is far from clear.


limit
inequalities
probability and stats

Suppose that $ P_{n}$ and $ P$ are prob. measures on $ \mathbb{N}_{0}\equal{}\{0,1,2,\,. \,. \,.\}$ such that $ \liminf_{n\to \infty}P_{n}(j)\ge P(j)$ for every $ j$. How can we conclude from this that $ \lim_{n\to \infty}P_{n}(j)\equal{}P(j)$ for every $ j$?

Just note that $ P(j)\equal{}1\minus{}\sum_{i: i\ne j}P(i)$ and similarly for $ P_n$. Now fix $ \varepsilon>0$ and choose a finite set $ S\subset\mathbb N\setminus\{i\}$ such that $ 1\minus{}\sum_{i\in S}P(i)\le P(j)\plus{}\varepsilon$. Then $ \limsup_{n\to\infty}P_n(j)\le 1\minus{}\sum_{i\in S}\liminf_{n\to\infty}P_n(i)\le 1\minus{}\sum_{i\in S}P(i)\le P(j)\plus{}\varepsilon$. The rest should be clear.

This argument can be shortened a bit by using the Fatou lemma, of course.

Yes I got it. Thanks fedja. I tried to use Fatou's lemma but I couldn't do it. If possible, can you give some hints to do by Fatou's lemma?

Fatou's lemma says that
$ 1 \equal{} \sum_j P(j) \leq \sum_j \liminf_n P_n(j) \leq \liminf_n \sum_j P_n(j) \equal{} 1$

Hence the equality must hold. However, as far as I can see, fedja is doing exactly this (invoking Fatou's lemma in the leftmost inequality), so the only difference is that he didn't dare do it for an infinite sum.

All you have proved now is that $ \liminf_{n\to\infty}P_{n}(j)\equal{}P(j)$ for every $ j$. But how to prove that $ \liminf_{n\to\infty}P_{n}(j)\equal{}\limsup_{n\to\infty}P_{n}(j)$

Whoops, wrote my response too hastily. I was just trying to say that fedja's solution already uses Fatou's lemma. The calculation he did is
$ \varlimsup_n P_n(i) \equal{} 1 \minus{} \varliminf_n \sum_{j \neq i} P_n(j) \leq 1 \minus{} \sum_{j \neq i} \varliminf_n P_n(j) \leq 1 \minus{} \sum_{j \neq i} P(j) \equal{} P(i)$


calculus
integration
trigonometry
induction
abstract algebra
calculus computations

Evaluate

\[{\int_0^\frac{\pi}{2}} \frac{\sin 2005x}{\sin x}dx\]

[quote="Jimmy"]Again, I noticed a pattern:

For odd integers $n$, $\int_0^{\frac{\pi}{2}} \frac{\sin(nx)}{\sin x} \ dx = \frac{\pi}{2}$.

So the answer is $\frac{\pi}{2}$.[/quote]

Let's prove that by induction:

To begin, $\sin(nx+2x)=\sin nx\cos 2x+\sin 2x\cos nx$

Since $\frac {\sin 2x\cos nx}{\sin x}=2\cos x\cos nx$, and since $\frac {\cos 2x}{\sin x}=\csc x-2\sin x$, we are left with

$\sin nx\csc x+2(\cos nx\cos x-\sin nx\sin x)=\sin nx\csc x+2\cos((n+1)x)$

The latter half of this is zero when integrated, so this establishes the induction.

This is the Dirichlet kernel from Fourier analysis.

Claim:

$\sum_{k=-n}^ne^{2ikx}=1+2\sum_{k=1}^n\cos(2kx)= \frac{\sin((2n+1)x)}{\sin x}.$

This can be proved either by appropriately manipulating geometric sums or by other trigonometric identites.

Then

$\int_0^{\frac{\pi}2}\frac{\sin((2n+1)x)}{\sin x}\,dx = \int_0^{\frac{\pi}2} 1+2\sum_{k=1}^n\cos(2kx)\,dx$

The constant term integrates to $\frac{\pi}2$ and each of the other terms integrates to zero.

[quote="Jimmy"]Again, I noticed a pattern:

For odd integers $n$, $\int_0^{\frac{\pi}{2}} \frac{\sin(nx)}{\sin x} \ dx = \frac{\pi}{2}$.

So the answer is $\frac{\pi}{2}$.[/quote]

Yes, you are right.

Let $I_n=\int_0^{\frac{\pi}{2}} \frac{\sin (2n+1)x}{\sin x}dx\ (n=0,1,2\cdots)$, $I_n$ satisfies the following equality...

Probably this integral is classical. It is said that this integral was made an examination at Tokyo University in 1950.


While following a treasure map, you start at an old oak tree. You first walk 825 meters  directly south, then turn and walk 1.25km  at 30.0  west of north, and finally walk 1.00 km  at 40.0  north of east, where you find the treasure: a trophy!!!

To return to the old oak tree, in what direction should you head ?

To return to the old oak tree, how far will you walk?


function
combinatorics proposed
combinatorics

Find a function $f:N\rightarrow N$ such that for all $k\in N$ every graph of average degree at leas $f(k)$ has a bipartite subgraph of minimum degree at least $k$

Must we find the least possible $f(k)$?

If not, it is clear that some such function exists: at first, we may choose a bipartite subgraph of our graph $G$ without loss more then a half of all edges. Indeed, divided it into two groups  such that the number of edges, which ljoin vertices of different groups is maximal. Then each vertex $v$ of degree $d$ is joined with at most $d/2$ vertices of the same group (else move it to other group).

After that, we get a bipartite subgraph with $n$ vertices and at least $nf(k)/4$ edges.

Delete vertices of degree at most $k-1$ step by step. After some step, we either get a graph, in which all degrees are at least $k$, or an empty graph. The first case is desired, assume that the second case holds. After each step, we deleted at most $k-1$ edges, so the number of edges in initial garph does not exceed $(k-1)n$, which contradicts to an upper estimate above for $f(k)>4(k-1)$.


hmm... can't help much but here is a link to the mathworld's site [url]http://mathworld.wolfram.com/topics/VectorAlgebra.html[/url]

Also, vectors are not getting started but I'll leave that to the moderators to deal with.

An informational question like this can usually stay in whatever forum it starts out in. The threads that I "promote" to other forums are challenge problems that are clearly at too high a level for participants who are just getting started. A set of problems at the end of an expository article will also usually be left alone.


Suppse that $ABCDEF$ be the convex hexagon. Let $P=(AB\cap EF), Q=(EF\cap CD), R=(CD\cap AB)$ and $S=(BC\cap DE), T=(DE\cap FA), U=(FA\cap BC).$ Prove that if \[ \frac{AB}{PR}=\frac{CD}{RQ}=\frac{EF}{QP}  \] then \[ \frac{BC}{US}=\frac{DE}{ST}=\frac{FA}{TU}.  \]

Posted at

http://www.mathlinks.ro/Forum/viewtopic.php?t=4822
http://www.mathlinks.ro/Forum/viewtopic.php?t=55028

  darij


Let $a,b>1$. Solve in $\mathbb{R}$
$a^{bx}+b^\frac{a}{x}=a^b+b^a$.

Let $f(x)$ be the function $f(x) = a^{cx} + b^\frac{d}{x} $ where $a,b>1$ and $ c,d > 0 $ . It's trivial that  $f(x)=Q > 1$ has only positive solutions.

For $x\in  ]0, +\infty[$, we have $f''(x)>0$, so $f'(x)$ increasing, so $f'(x)=0$ has only one root, so  $f(x)=Q$ has at most 2 solutions.

It's easy to verify that $f(u)=f( v)$ , where $v= \frac{d* ln(b)}{u* c* ln(a)}$. In our case $c=a,d=b$ and  $u=1$  is a solution, so the oher one is $v=  \frac{a*ln(b)}{b* ln(a)}$ .
 :cool:   I have already seen/solved this kind of problem  :cool:


How many ways can change be made for a quarter using standard U.S. coins?

Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1396323#1396323]this[/url]; the answer is $ \boxed{13}$. Unless you can't make change for a quarter with a quarter, in which case it would be $ \boxed{12}$.

I don't think that a quarter can be changed for a quarter because it wouldn't "change" ;). Also, I did this in a sprint round and it said $\boxed{12}$ was correct.

The number of ways to arrange pennies with nickles to from a quarter is: $25/5 = 5$
The number of ways for pennies with dimes is: $25/10 = 2$
The # of ways to arrange pennies to form a quarter + # of nickles: $2$
The number of ways for dimes and nickles: $2$
The number of ways for all of them is $1$
Therefore $1+2+2+2+5 = \boxed {12}$


Assuming that $ D_n\in\text{M}_n{(p)}$ such that
$ D_n\equal{}\left| \begin{array}{ccccc}
a & b \\
c & a & b \\
   & c & a & \ddots\\
   &    & \ddots &\ddots& b\\
   &    &            &c         &a
\end{array}
\right|$
Then calculate $ D_n$.

We have : $ D_{n\plus{}1}\equal{}aD_{n}\minus{}bcD_{n\minus{}1},n \geq 2$,$ D_1\equal{}a,D_2\equal{}a^2\minus{}bc$, easy calculate $ D_n$ !


algebra
polynomial
search
Galois Theory
superior algebra
superior algebra unsolved

Is there somebody who knows how to write a code in kash to find the galois group of a given polynomail?

A search directly tells my that you want "Galois(f);"  :wink:

No it doesnot work, it gives an error that f must have a value.

Do you know any computer program which can help me to find Galois group of an irreducible polynomail of degree 12.

It's quite obvious that you need to set $ f$ to you specific polynomial  :wink:
Try Galois(X^3-2);
Or in short: RTFM!

[quote="peteryellow"]
Do you know any computer program which can help me to find Galois group of an irreducible polynomail of degree 12.[/quote]format C:

ZetaX, i am not taht stupid definitely I have tried with a polynomail insted of f, but that is not working, and they have a pdf-file which tells that how can I use command but the problem is that there is nothing about Galois group.

But otherwise thanks for your help.

I [b]tried[/b] that command [b]myself[/b].
Maybe you still use kash version 2, then it should be GaloisGroup(x^3-2); or something like that. And still I'm pretty sure it's stated in one of the manuals.


Okay. We have this number that is in form of:

$3^{1001} 7^{1002} 13^{1003}$

And what is the unit digit of this huge number? :D

 :P I changed the wording to make it less boring.

[hide]

First, note the patterns:

\[3 \rightarrow 3, 9, 7, 1\]
\[7 \rightarrow 7, 9, 3, 1\]
\[13 \rightarrow 3, 9, 7, 1\]

We notice that each of the four integers repeat their units digit every 4 terms.

\[1001 \equiv 1 (\displaystyle{mod} 4)\]
\[1002 \equiv 2 (\displaystyle{mod} 4)\]
\[1003 \equiv 3 (\displaystyle{mod} 4)\]

Thus, $3^{1001}$ ends in $3$, $7^{1002}$ ends in $9$, and $13^{1003}$ ends in $7$. Hence, the units digit of their product is the units digit of

\[3 \cdot 7 \cdot 9=189\]

Therefore the units digit of $3^{1001} \cdot 7^{1002} \cdot 13^{1003}$ is $9$.

[/hide]

Let's change up the question. What is the leftmost digit of $3^{1001} \cdot 7^{1002} \cdot 13^{1003}$? :P  :P  :P

Hint:
[hide]
$3<x<5$ :P
[/hide]

interesting hint...\
i think we may be in for some logs

You don't have to use mod there. It's just remainder. ;)

Same result in this case but if you can use the easy one, why not use the easy one?

That sounds insane o_o ...

How about the second to last digit?

[hide]3^x can end in 3,9,7,1 and in this case, it'll end in 3 because 1001/4 = 250R1

7^x can end in 7,9,3,1 and 7^1002'll end in 9 b/c 1002/4 = 250R2

13 ends in 3 so it's the same as 3^x so it's 3,9,7,1 and 1003/4 = 250R3 and so it ends in 7

so it's 3*9*7 = 189 so the answer is 9[/hide]

[hide]1001/4 for 3 R1 in cycle of 3,9,7,1 so 3 is unit digit
1002/4 for 7 R2 in cycle of 7,9,3,1 so 9 is unit digit
1003/4 for 13 is the same for 3 so R3 gives us 7 as unit digit.
We multiply together to get 3*9*7 which has a unit digit of 9.[/hide]


[quote="AstroPhys"]What does "=)" mean?[/quote]
(+-) means plus or minus
(the plus sign (+) above the minus sign (-))

[hide="interpreting ) as 0"]Simply expanding $(x+2)(x-3-2\imath)(x-3+2\imath)$, we have $(x+2)(x^{2}-6x+5)=x^{3}-4x^{2}-7x+10$.

Because the coefficient of $x^{3}$ in the given equation is 2, we must multiply this equation by 2 also, giving:

$2x^{3}-8x^{2}-14x+20$, so $a=-8$, but then, because $a,b,c$ are constants, there cannot be a solution (no $x$ term in the given expression).[/hide]

Not the plus or minus the [quote]=)[/quote].
If it is zero then b-flat's solution works.

[quote="linzk3"]An equation of the form 2x^3 + a*x^2 +ab+c=) has roots of -2 and 3(+-)2i.  Determin (a,b,c).

(+-) means plus or minus[/quote]

Please use $\LaTeX$. Do you mean this:

An equation of the form $2x^{3}+ax^{2}+bx+c = 0$ has roots $-2$ and $3\pm 2i$. Determine the ordered triple $(a,b,c)$
[hide="If so, then the solution is:"]We must try to find the required cubic. Since the leading coefficient is $2$, we use the cubic:
\[2(x+2)(x-3-2i)(x-3+2i) \]

\[= 2(x+2)((x-3)^{2}+4) \]

\[= 2(x+2)(x^{2}-6x+13) \]

\[= 2(x^{3}-4x^{2}+x+26) \]

\[= 2x^{3}-8x^{2}+2x+52 \]
So $(a,b,c) = (-8, 2 , 52)$
[/hide]

@ b-flat: $(x-3-2i)(x-3+2i) = x^{2}-6x+13$

[quote="vishalarul"]
@ b-flat: $(x-3-2i)(x-3+2i) = x^{2}-6x+13$[/quote]

It doesn't really matter; there still is no solution based on the original problem, though thanks for pointing that out.

Also, you should probably have $x^{3}$ in your last equation  :wink:


oke!!!
i have seen this inequality
i have solved it, but my solution is long
I need a short solution!! who can help me!!
 :P  :)
with x,y,z>0 prove

1/(x²+yx+y²) + 1/(x²+xz+z²)+1/(y²+yz+z²) ≥ 1/(x+y+z)²

Actually there is a stronger inequality, for ex : $ \displaystyle \sum _{cyc} \frac 1{x^2 \plus{} xy \plus{}y^2} > \displaystyle \sum _{cyc} \frac 1{x^2 \plus{} 4xy \plus{}y^2} \geq  \frac 9{2(x\plus{}y\plus{}z)^2}$ :D
Because of $ \frac 1t$ is convexe you can write  $ \displaystyle \sum _{cyc} \frac 1{x^2 \plus{} 4xy \plus{}y^2} \geq 9/( \displaystyle 2\sum_{cyc} x^2 \plus{}4\sum_{cyc} xy) \equal{} \frac 9{2(x\plus{}y\plus{}z)^2}$
And more general for $ n$ positive numbers :  $ \displaystyle \sum _{1\leq i<j\leq n} \frac 1{x_i^2 \plus{} 2(n\minus{}1)x_ix_j \plus{}x_j^2} \geq  \frac {n^2(n\minus{}1)}{4(x_1\plus{}...\plus{}x_n)^2}$.. Do you follow me  ?
 :cool:

I hate to point this out, but any one of those three terms on the left hand side of 1/(x²+yx+y²) + 1/(x²+xz+z²)+1/(y²+yz+z²) ≥ 1/(x+y+z)² is ≥ 1/(x+y+z)² since it rearranges trivially...  :wink:

Other ineq:
Given $ a;b;c \geq 0$.Prove that:
$ \sum_{cyc} \frac{1}{a^2\plus{}ab\plus{}b^2} \geq \frac{9}{(a\plus{}b\plus{}c)^2}$

oke!! thanks for yours!!
I wrote carefully  sorry!!!
it's is \frac{9}{(x+y+z)²}
and I have a inequality like that
[mat]\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \ge \frac{9}{(MA+MB+MC)^2}[/mat]

with a,b,c are 3 side of triangle ABC , M is tolicelli point!!!

thanks for yours!!!

[quote="chien than"]Other ineq:
Given $ a;b;c \geq 0$.Prove that:
$ \sum_{cyc} \frac {1}{a^2 \plus{} ab \plus{} b^2} \geq \frac {9}{(a \plus{} b \plus{} c)^2}$[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=6364

  darij


"Find the smallest prime that is the fifth term of an increasing arithmetic sequence, all 4 preceding terms also being prime."

Why is it that if the common difference is not a multiple of 3, then either the middle prime would be a multiple of 3 or two of the primes would be multiples of 3 (impossible)?

If the common difference is not a multiple of three, then when you take the terms (mod3), at least one of the terms is a multiple of 3.


inequalities
number theory
Diophantine equation
number theory open

Solve this diophantine equation in $\mathbb{N}$ (or more generally in $\mathbb{Z}$):

$\sum_{k=1}^{n}{a_{k}}=\prod_{k=1}^{n}{a_{k}}$

A trivial solution : $a_1 = a_2 = ... = a_n = 0$ :)

First of all, the am-gm isn't a very good approximation in the set of natural numbers and also it still need tons of checking and it does not give general solutions. Also, if this problem wasn't very hard be sure i wouldn't post it here. ;)

There are an infinite number of solutions $\in \mathbb{N}$.

Denote $S = a_1 + a_2 + ... + a_n$.  Denote $P = a_1a_2a_3...a_n$.  Then we want the solutions to $P = S$.

Suppose $P = S$ is some prime.  Then the only possible values for $a_n$ are $a_1 = P, a_2 = a_3 = ... = a_n = 1$.  This makes $S = (n-1) + P$, which only has solutions for $n = 1$.

However, suppose $P = S$ is not prime.  We can arrive at an arbitrary solution by choosing a set of integers $b_1, b_2, ... b_k$ where $b_i \neq 1$ such that $\prod_{i=1}^{k} b_i = P$ and $\sum_{i=1}^{k} b_i < P$ (I suspect this is always true as long as $b_i \neq 1, k > 2$).  Then the solution can be generated as follows:

Set $a_1 = b_1, a_2 = b_2, ... a_k = b_k$.  At this point, without any other terms set ($k = n$), we have $S < P$.  Since we are working $\in \mathbb{N}$, we know that $P - S$ must be an integer, $d$.  The solution can be generated by setting $a_{k+1} = a_{k+2} = ... = a_{k+d} = 1$.  This increases $S$ by $d$ while keeping $P$ the same, so we have $S = P$ and a solution.  Q.E.D.

Demonstration:  Suppose we want $S = P = 12$.  We arbitrarily choose $b_1 = 2, b_2 = 2, b_3 = 3$, which gives us $2 + 2 + 3 = 7 < 2 \times 2 \times 3 = 12$.  We have a difference of $5$, so our arbitrary solution is $1 + 1 + 1 + 1 + 1 + 2 + 2 + 3 = 1 \times 1 \times 1 \times 1 \times 1 \times 2 \times 2 \times 3$, which gives us a solution.

 :)

You didn't prove anything significant. You could have just taken $1,1,1,...,2,n$.

Just a question: what exactly do you want to hear¿ A list of all solutions¿ Then good luck...

Some more trivia: The number of $k$ such that $a_k=1$ is at last $n-\mbox{ln_2} (n)-1$.

I could just describe the solutions; for $\mathbb{N}$, the solutions for $n > 1$ consist of all sequences $a_1, a_2, ... a_n$ such that there are at least two terms $\neq 1$, with a suitable number of $1$s thereafter.

(I realize the proof was not particularly significant, but that's because the problem is not particularly significant.  My method will, however, generate all nontrivial solutions.  :))

For $\mathbb{Z}$, the solutions can be generated like so:

As before, take any wanted product, $P$, and choose a series of integers $b_1, b_2, ... b_k$ such that $b_i \neq 1$ for any $1 \le i \le k$, and such that $| \prod_{i=1}^{k} b_i | = |P|$ and $| \sum_{i=1}^{k} b_i | \le \sum_{i=1}^{k} |b_i| < |P|$.  As before, we set $a_1 = b_1, ... a_k = b_k$.

If $S$ and $P$ have opposite sign, add the terms $a_{k+1} = 1, a_{k+2} = -1$, which reverses the sign of $P$ (this step is optional; see next step).

From there (regardless of whether $S$ and $P$ originally had opposite sign or not), if $S < P$, then add the term $a_{k+3} = 1$ until $S = P$.  If $S > P$, then add the terms $a_{k+3} = -1, a_{k+4} = -1, a_{k+5} = 1$ until $S = P$.

This will generate all solutions, so I think it counts as a valid answer to the problem, even if it does seem trivial.   :)

[quote="ZetaX"]For the solutions in $\mathbb{N}$: apply AM-GM inequality.
And for the general case: there are a lot of 'trivial' solutions...[/quote]...................apply AM-GM inequality?


Who do you think will win it all???

Most think the Gators will win, but im going with the Buckeyes.

Let's go OSU!!! :first: :thumbup:

for Flordia :winner_second:

35-24

What now, all you doubters, what now?

Huh. Guess my awesomeness is proven once again.

[quote="footballrocks41237"]NOOOOOOOO!  19.8 seconds left and OSU is down by 10!  IT CAN'T BE!  IT CAN'T BE!

EDIT:  WTF?  I HATE THE FLORIDA GATORS!  They beat [b]THE OHIO STATE UNIVERSITY[/b] in both Football and Basketball.  WTF.  WTF.[/quote]
OMG NOOOOOO
ITS ALL A CONSPIRACY TO OVERTHROW THE GOVERNMENT

lol, i was going for the gators, but my dad was going for osu
i told him that osu was hopeless by halftime, but he wouldn't listen to me =)

my bracket owned it had a 1510/1920 score

jorian


Ok it says:

In the video game duck shoot, the duck moves in a straight line from point A to B at a rate of 7cm/sec. Bullets fired from point O travel at 25 cm / sec. If a player shoots as soon as a duck appears at A, at which angle [i]a[/i] should the gun be aimed in order to score a direct hit?

This is my attempt at drawing the thing how it looks in the exercise (pretty bad drawing, but is for u to understand where is point O and all those things):

http://img171.imageshack.us/img171/7639/duckshootme3.png

Erm, correct me if I'm wrong, but isn't the distance that the gun is away from the duck important?

(unrelated, silly, note: if it's a video game, why are there actual bullets?)

It doesnt show the distance from the gun to the thing... and i dont know if there are real bullets are all :P well whatever how would I know.

Wait, darn it, never mind, ignore the above post. I was treating the speeds as lengths for some odd reason.

[hide="solution"]
Let the distance the duck travels be $ t$, the distance the bullet travels be $ p$.
We want $ a$ such that $ \sin a\equal{}\frac{t}{p}$
Now, $ \frac{p}{25}\equal{}\frac{t}{7}$ due to the d=rt thing. Thus, $ \frac{t}{p}\equal{}\frac{7}{25}$, so $ a\equal{}\arcsin \frac{7}{25}\equal{}\text{I'm too lazy to get out a calculator right now}$[/hide]


geometry
parallelogram
ratio
geometric transformation
reflection
3D geometry
pyramid

This is for the hardcore geometry fans out there. Arranged roughly in order of difficulty.

1. . Let P be an interior point of the parallelogram ABCD. Prove that
angleAPB + angleCPD = 180◦ if and only if anglePDC = anglePBC.

2.  Points K and L are chosen on the sides AB and AC of an equilateral
triangle ABC such that BK = AL. Segments BL and CK intersect at
P. Determine the ratio AK : KB for which the segments AP and CK
are perpendicular.

3. The diagonals of a convex quadrilateral ABCD are orthogonal and inter-
sect at point E. Prove that the reflections of E in the sides of quadrilateral
ABCD lie on a circle.

4. The base of a regular quadrilateral pyramid π is a square with side length
2a and its lateral edge has length a(17^.5). Let M be a point inside the
pyramid. Consider the five pyramids which are similar to π, whose top
vertex is at M and whose bases lie in the planes of the faces of π. Show
that the sum of the surface areas of these five pyramids is greater or equal
to one fifth the surface of π, and find for which M equality holds.

(I don't know how to do #3 or #4, so if someone could post solutions for those I would very much appreciate them.)

[quote="Arrange your tan"][quote="soltixnj"]17^1/2[/quote]You must put grouping symbols around the
fractional exponent, because what you have is equivalent to 17/2.[/quote]

I'd recommend just using latex. Saves trouble.

[quote="soltixnj"]This is for the hardcore geometry fans out there. Arranged roughly in order of difficulty.

1. . Let P be an interior point of the parallelogram ABCD. Prove that
$ \angle{APB} + \angle{CPD} = 180 ^{\circ}$ if and only if $ \angle{PDC} = \angle{PBC}$.

2.  Points K and L are chosen on the sides $ \overline{AB}$ and $ \overline{AC}$ of an equilateral
triangle $ \triangle{ABC}$ such that $ \overline{BK} = \overline{AL}.$ Segments $ \overline{BL}$ and $ \overline{CK}$ intersect at
P. Determine the ratio $ \frac{\overline{AK}}{\overline{KB}}$ for which the segments $ \overline{AP}$ and $ \overline{CK}$
are perpendicular.

3. The diagonals of a convex quadrilateral ABCD are orthogonal and inter-
sect at point E. Prove that the reflections of E in the sides of quadrilateral
ABCD lie on a circle.

4. The base of a regular quadrilateral pyramid $ \pi$ is a square with side length
$ 2a$ and its lateral edge has length $ a\sqrt{17}$. Let M be a point inside the
pyramid. Consider the five pyramids which are similar to $ \pi$, whose top
vertex is at M and whose bases lie in the planes of the faces of $ \pi$. Show
that the sum of the surface areas of these five pyramids is greater or equal
to $ \frac{1}{5}$ the surface of $ \pi$, and find for which M equality holds.

(I don't know how to do #3 or #4, so if someone could post solutions for those I would very much appreciate them.)

[color=red][size=59]~LaTeXed[/size][/color] [/quote]


[hide]let $A=\sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \cdots }}}}$
Therefore $A^2=\frac 94+ \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \cdots }}}}$[/hide]
[hide="Click here to see rest of solution"]so then $A^2-\frac 94=A$

so $A=\frac{1+\sqrt 10}{2}$[/hide]

[quote="Silverfalcon"]As usual, if you know how to solve this type of problem, leave this to beginners so they can learn from it.

The number represented by the infinite sequence $\sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \cdots}}}}$ is equal to what rational number?

Hint: The answer won't be an integer.[/quote]

[hide]
we can set it equal to S.
$S=\sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \sqrt{\frac{9}{4} + \cdots\\}}}}$
We can replace it with itself in this statement:
$S=\sqrt{\frac{9}{4}+S\\}$
Squaring...
$\displaystyle S^2=\frac{9}{4}+S\\
S^2-S-\frac{9}{4}=0\\
S=\frac{1 \pm \sqrt{1+4*1*\frac{9}{4}}}{2}=\frac{1\pm\sqrt{10}}{2}$
The negative solution is illegal, so the answer is $\frac{1+\sqrt{10}}{2}$
[/hide]

I know most people on the forum don't like to have their terminology or notation corrected (I have no idea why), but this is actually a nested (or infinite) radical. A repeated infinite fraction is generally written in the form\[a_0+\frac{1}{a_1+\frac{1}{a_2+\ldots}}\]Also, it's not really an infinite sequence, which is something that looks like $\{a_n\}^\infty$. For instance, $\{2n\}_{n=1}^\infty=\{2,4,6,\ldots\}$. However, it can be interpreted as the limiting value of an infinite sequence.

Let $s_1=\sqrt\frac94$ and $s_{n+1}=\sqrt{\frac94+s_n}$ for $s\ge1$.

Then the number we want to find is $\lim{s}_n$.


Let $ a_{1},a_{2},\ldots,a_{100}$ and $ b_{1},b_{2},\ldots,b_{100}$ be permutations of the positive integers from 1 to 100. Show that for some positive integers $ 1\le i,j\le100$, with $ i\neq j$, we have $ a_{i}b_{i}\equiv a_{j}b_{j}\pmod{100}$.

Is the following correct?

[hide]We have 100 products and 100 slots to put them in, so we need exactly one product per slot.  In particular, this means that exactly 50 of our products must be even.  For any even $ a_{i}$, the product $ a_{i}\cdot b_{j}$ will be even regardless of the parity of $ b_{j}$, and likewise if $ b_{j}$ is even.  Thus, the only way we can have 50 even products is if each even $ a_{i}$ is paired up with an even $ b_{j}$.  However, we then have that each of the even products is divisble by 4 (and this is true even mod 100 because 100 is divisible by 4), so in particular we won't achive a product equal to 2, contradiction.[/hide]

That proves (with obvious modifications) the following stronger result: if $ p^{2}$ divides $ n$ for some prime $ p$, then for any permutation $ \sigma$ of the set $ \{0, 1, \ldots, n-1\}$, the set $ \{0\cdot \sigma(0), 1\cdot \sigma(1), \ldots, (n-1)\cdot \sigma(n-1)\}$ contains two elements which are congruent $ \pmod n$.

Now, if $ n = p$ is prime, the result still holds, from an entirely different argument using Wilson's Theorem (and possibly other arguments as well).

This leaves open the question of what happens in the case of squarefree non-primes.  Anyone else want to give it a go?


[url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=13162]this person's[/url] avatar is irritating me. moderators or admin, please do something with this user's avatar.

Well there's nothing there anymore so I would assume it was removed.

it seems he might mean his signature or profile
it seems like it could...be... you know

His interests and signature are both quite racist.


number theory
Diophantine equation
combinatorics unsolved
combinatorics

@ cckek : Are you sure that is right ?

I did 3 cases a1=1 or a2=1 or a3=1 (Clearly a4,a5,....an cannot be equal to one for then we have more than two inversions)

We get a recurrence relation that gives n(n-1)/2 -1 after solving.

Note: (3,1,2), (2,3,1) give two inversions for n=3
(1,4,2,3), (1,3,4,2), (2,1,4,3), (3,1,2,4), (2,3,1,4) are the 5 inversions for n=4

Correct me if I am wrong.
I'll post the complete solution in a couple of hours.

[quote="cckek"][quote="Sathej"]Let A = (a1,a2,...,an) be a permutation of (1,2,...,n). A pair (ai,aj) is said to correspond to an inversion of A, if i<j but ai > aj. How many permutations of (1,2,...,n), n>=3 have exactly two inversions?[/quote]

A little more general. If in the $\sigma$ permutation in lower row before $1$ stand $t_{1}$ elements grater than $1$, $0\le t_{1}\le n-1$, before $2$ stand $t_{2}$ elements greater than $2$, $0\le t_{2}\le n-2$,..., before $k$ stand $t_{k}$ elements greater than $k$, $0\le t_{k}\le n-k$,..., before $n-1$ stand $t_{n-1}$ elements grater than $n-1$, $0\le t_{n-1}\le 1$, then the number of inversions is in the permutation $\sigma$  is given by the sum $t_{1}+t_{2}+\cdots+t_{k}+\cdots+t_{n-1}$.  So the number of permutation wich has $p$ inversion is given by the number of solutions of $t_{1}+t_{2}+\cdots+t_{k}+\cdots+t_{n-1}=p$ diophantine equation where $0\le t_{i}\le n-i, \forall i=%Error. "upperline" is a bad command.
{1,n}.$ Its very easy to observe that if we note with $S_{p}^{n}$ the number of permutations of rage $n$ wich has $p$ inversions than $S_{p}^{n+1}=\sum_{i=0}^{p}S_{i}^{n}$.
So the answer for ur problem is: $S_{2}^{n+1}=S_{0}^{n}+S_{1}^{n}+S_{2}^{n}=1+n+S_{2}^{n}\Rightarrow S_{2}^{n+1}-S_{2}^{n}=n+1.$ Calcultaing term by term and summed we got $S_{2}^{n+1}-S_{2}^{3}=n-2+\frac{n(n+1)}{2}-1-2=\frac{(n-2)(n+5)}{2}$, so $S_{2}^{n}=\frac{(n+4)(n-3)}{2}+2.$[/quote]

a little mistake:)

$S_{1}^{n}=n-1$ so we got $S_{2}^{n+1}-S_{2}^{n}=n.$ therefore
$S_{2}^{n+1}-S_{2}^{3}=\frac{n(n+1)}{2}-1-2$ so  $S_{2}^{n}=\frac{(n-1)n}{2}-1.$

This is what I did though it basically amounts to just what you did

Let $I_{n}$ denote number of permutations of $(1,2,3,...n)$ with two inversions.

We see that $a_{i}\ne 1 \forall i \ge 4$

Case 1] $a_{1}= 1$
Then we must have two inversions in $(a_{2},a_{3},...a_{n})$
Thus it can be done in $I_{n-1}$ ways.

Case 2] $a_{2}= 1$
Again $a_{1}\ne i \forall i \ge 4$ else we will have atleast 3 inversions

Thus a) $a_{1}= 2$
Then we want one inversion in $(a_{3},a_{4},...,a_{n})$
Thus n-3 ways to do this.

     b) $a_{1}= 3$
Then we are forced to put $(a_{1},a_{2},...a_{n}) = (3,1,2,4,5,...,n)$
Thus one way to do this

Case 3] $a_{3}= 1$
Then we are forced to put $(a_{1},a_{2},...a_{n}) = (2,3,1,4,5,...,n)$
Thus one way to do this

Thus $I_{n}= I_{n-1}+(n-3)+1+1$
Thus $I_{n}= I_{n-1}+n-1$

We see that $I_{3}= 2$

Thus we have $I_{n}= \frac{(n-1)n}{2}-1.$

Hello Vihang,
    How many problems did you get correct this year?
Sathej


We have a sequence $ x_{n\plus{}3} \equal{} x_{n} \plus{} x_{n\plus{}1}x_{n\plus{}2}$ for positive integers $ x_{j}$ .Given
$ x_{1} \equal{} x_{2} \equal{} x_{3}\equal{}1$...Is it possible to convert this into a linear recursive relation ...?

Because $ x_n$ supexponential, increase as $ c^{2^n}$, all lineal recurent sequense $ O(c^n)$.

:( sorry can you explain it detailed and express $ x_{n}$ in the form $ a_{1}x_{k} \plus{}......$ or some similar linear form .... :?:

any replies...or explanation for Rust's statement ...?..Is it that dumb that people dont answer it..?If that is the case..I am sorry :blush:

What Rust said you is that all linear recurrences $ x_{n\plus{}k}\equal{}\sum_{i\equal{}1}^ka_ix_{n\plus{}k\minus{}i}$ imply solutions $ x_n\equal{}\sum_{i\equal{}1}^k\alpha_iz_i^n$ whose behaviour when $ n\to\plus{}\infty$ is $ \approx aA^n$

And obviously your sequence does not have this behaviour when $ n\to\plus{}\infty$

So, we'll surely be unable to find a formula $ x_{n\plus{}k}\equal{}\sum_{i\equal{}1}^ka_ix_{n\plus{}k\minus{}i}$ for your sequence.

But maybe this is not what you call a "linear recurrence" ? Maybe you accept non fixed summations ? Try to be more clear about the problem.
In what test / contest / class did you find it ?

oh..ok....it was not asked in any contest ...you can say it is a kind of my own problem.I was trying some another problem where I wanted to get a solution by finding the $ n'th$ term of this sequence. What do you mean by non fixed summations ? Anything  from which I will be able to find the $ n'th$ term is fine ? I wanted something like a linear recurrence (homogeneous or inhomogeneous) as we know the general way to solve such recurrences .So just find a general expression for $ x_{n}$


geometry
geometric transformation
rotation
reflection
trigonometry
parallelogram
vector

Due to earth's rotation, a plumb bob may not hang exactly along the direction of the Earth's gravitational force on the plumb bob but may deviate slightly from that direction. (a) show that the deflection  $\theta$is given by $\theta = (\frac{2\pi^2R}{gT^2})sin(2L)$, where R is the radius of the Earth and T is the period of the Earth's rotation.

This problem is very hard. I try to get a = $\frac{2\pi(RcosL)^2}{T^2RcosL}$ and then must get F so I try to separate the thingie into components byut then Fx and Fy get into a mess and I'm confused. Help!

Thanks

Look at the attached figure. The circle with the center E and radius R is the Earth. The bob is at the point B with the lattitude angle $\angle L$. The gravitational force on B is $\overrightarrow  F_{BG} = \overrightarrow{BG}$ with the magnitude $F_{BG} = mg$. The radius of rotation at the lattitude L is the perpendicular distance $r = R \cos \widehat L$ of B from the axis of rotation EN (N is the north pole). The centrifugal force on B is $\overrightarrow F_{BC} = \overrightarrow{BC}$ with the magnitude

$F_{BC} = m \omega^2 r = \frac{4 \pi^2 mr}{T^2} = \frac{4 \pi^2 mR \cos \widehat L}{T^2}$

where T is the rotation period of the bob around the axis EN and $\omega = \frac{2\pi}{T}$ the angular velocity. You can get the magnitude of the sum of these these 2 forces as the diagonal BR of the parallelogram BGRC or as the 3rd side BR of the triangle $\triangle BGR$ by placing the vector $\overrightarrow F_{BC} = \overrightarrow {BC}$ tail to the head of the vector $\overrightarrow F_{BG} = \overrightarrow{BG}$, i.e., into $\overrightarrow F_{BC} = \overrightarrow{GR}$. In either case, you have the triangle $\triangle BGR$ with 2 known sides BG, GR = BC and the known angle between them $\angle BGR = \widehat L$. Using the cosine theorem, you can calculate the side BR, and using the sine theorem next, you can calculate sine of the angle $\angle GBR = \theta$ (using formulas, not numbers). You can simplify the task by expressing the centrifugal force $\overrightarrow F_{BC} = \overrightarrow{BC}$ as the sum of its horizontal component $\overrightarrow F_{BH} = \overrightarrow{BH}$ and vertical component $\overrightarrow F_{BV} = \overrightarrow{BV}$ with the magnitudes $F_{BH} = F_{BC} \sin \widehat L$ and $F_{BV} = F_{BC} \cos \widehat L$. The vectors $\overrightarrow F_{BG}, \overrightarrow F_{BV}$ are (anti)parallel and the magnitude of their resultant $\overrightarrow F_{BD} = \overrightarrow {BD} = \overrightarrow F_{BC} + \overrightarrow F_{BV}$ can be found algebraically by subtraction:

$F_{BD} = F_{BG} - F_{BV} = F_{BG} - F_{BC} \cos \widehat L = mg - \frac{4 \pi^2 mR \cos^2 \widehat L}{T^2}$

The resultant $\overrightarrow F_{BR} = \overrightarrow{BR}$ is now the hypotenuse of the right angle triangle $\triangle BDR$ and it can be calculated by the Pythagorean theorem. But it is not necessary. The tangent of the angle $\angle DBR = \theta$ is the ratio of the known legs BD, BH of the right angle triangle $\triangle BDR$:

$\tan \theta = \frac{DR}{BD} = \frac{BH}{BC - BV} = \frac{BC \sin \widehat L}{BG - BC \cos \widehat L} =$

$= \frac{\dfrac{4\pi^2 mR \cos \widehat L \sin \widehat L}{T^2}}{mg - \dfrac{4\pi^2 mR \cos^2\widehat L}{T^2}} = \frac{2\pi^2 R \sin 2\widehat L}{gT^2} \cdot \frac{1}{1 - \dfrac{4\pi^2 R \cos^2\widehat L}{gT^2}}$

Now comes the 1st approximation. In the 2nd factor $\frac{1}{1 - q}$ in the expression for $\tan \theta$

${{q = \frac{4\pi^2 R \cos^2\widehat L}{gT^2}} \le \frac{4\pi^2 R}{gT^2}} \doteq \frac{40 \cdot 6.4 \times 10^6\ \text m}{10\ \text{m/sec}^2 \cdot 86400^2\ \text {sec}^2} \doteq \frac{25.6}{86.4^2} \doteq 0.0034$

If we are content to calculate the small angle $\theta$ with accuracy 0.5 %, the 2nd factor is so close to 1 that it can be neglected. Hence,

$\tan \theta \doteq \frac{2\pi^2 R \sin 2\widehat L}{gT^2}$

The 2nd approximation is that for small angles, $\tan \theta \doteq \theta$, where $\theta$ is in radians. This is easily seen if you draw a vertical tangent to a unit circle at its intersection U with the positive x-axis and a ray from the origin at an angle $\theta$ with the positive x-axis. This ray intersects the unit circle at a point W and the tangent at a point T. The segment UT on the tangent is $\tan \theta$ and the arc UW of the unit circle is the angle $\theta$ in radians. For small angles, the segment UT and the arc UW are about the same, but I would have to use calculus to calculate how close they are. In our case, $\tan \theta$ is really small (about half of the above q):

$\theta \doteq \tan \theta \doteq \frac{2\pi^2 R \sin 2\widehat L}{gT^2} \le \frac{2\pi^2 R}{gT^2} \doteq 0.0017\ \text{rad} \doteq 0.1^\circ = 6'$

hey, thank you so much for the detailed reply, yetti. But where is the diagram?

The diagram is at the bottom of my reply, in from of my eyes. The mathlinks web site does not display attachments to messages, if you just browse as a guest, you have to log in using your member name and password. But you certainly have to log in to post messages, how come you do not see the attached drawing ?

[quote="yetti"]But you certainly have to log in to post messages, how come you do not see the attached drawing ?[/quote]I think what happened was that he was browsing as a guest, didn't see the diagram, logged in to reply and say that he does not see the diagram, without noticing that once logged in he is able to see it :)


An infinite road has traffic lights at intervals of 1500m. The lights are all synchronised and are alternately green for $\frac 32$ minutes and red for 1 minute. For which $v$ can a car travel at a constant speed of $v$m/s without ever going through a red light?

you are wrong. Let t(sek) time needed for one interval, then
$\{\frac{tm}{150}\}<\frac 35$ for all natural m. It give $t=75k,v=v_0/k,v_0=20(m/sek).$

I think v is any of the set 1, 2, 5, 10, 20.
Evidently v needs to be less or equal to 25. Secondly it must be an integer divisor of 10 or be a multiple of 10, but less than 25.

The answer depends on the starting point of the car. For example, if the car starts at point $x_0=0$, the velocity should be of the form $v=\dfrac{20}{2n+1}$ m/s for some non-negative integer $n$. So, answers like $20$ m/s, $20/3$ m/s, $4$ m/s would do.

I write my answer here: https://nivotko.wordpress.com/2022/12/21/avoiding-traffic-lights/


Let $ x,y,z$ be real numbers, find the greatest value of $ a$ such that
\[ x^6\plus{}y^6\plus{}z^6\plus{}a(x^5y\plus{}y^5z\plus{}z^5x)\geq0.\]

Have you a solution to it, pvthuan ? I think it is not easy.

[quote="Akashnil"][quote="pvthuan"]Let $ x,y,z$ be real numbers, find the greatest value of $ a$ such that
\[ x^6 \plus{} y^6 \plus{} z^6 \plus{} a(x^5y \plus{} y^5z \plus{} z^5x)\geq0.
\]
[/quote]
Are you sure there isnt a typo,like the third "$ \plus{}$" sign or there is some condition on $ x,y,z,a$[/quote]
why i think the quetsion is right see it readsreal numbers and not positive real numbers if that is ur point


function
calculus
derivative
real analysis
real analysis unsolved

If a continuous function on [0,1] has zero derivative almost everywhere on the interval (note that we do not require it to be differentiable on [0,1]), what can we say about it in general (existence, for example)?

Aside from being nowhere strictly monotonic. :) In fact, you can add the property that this function is nowhere monotonic.

That certainly works. :)

I was hoping there was a non-constructive proof that there isn't such a function which is also nowhere monotonic (assuming that is what 'no you can't' means). I want to exercise my Baire Category muscle.

I was providing a counterexample to your implicit statement "such a function can't be strictly monotonic". We apparently misunderstood each other.

Now, if you want to ask whether there are any such functions which are nowhere monotonic, it shouldn't be too hard to modify one of those constructions to create one. Even subtracting two different strictly monotonic solutions is likely to work.

I see now. I wasn't clear before. What I meant originally in the first post was that the function is differentiable almost everywhere on [0,1] but the derivative is always zero at any point of differentiability (it is never positive or negative). Moreover, the function can be assumed to be nowhere monotonic. And the question is does such a function exist?


This was discussed in another topic:
[quote="K81o7"]$z=\frac{1}{0}$

The flaw comes in saying this.  Equality is something we only can assign to certain mathematical objects, ie numbers, vectors, matrices, etc.  This mathematical object does not fall under any of thoe categories, so we can't define equailty for it.[/quote]

[quote="jclarkemathL314159"](1/0)*0=?
Is it 0, 1, or infinity?[/quote]
I believe it's indeterminate, meaning we don't know what it is.

[quote="jclarkemathL314159"](1/0)*0=?
Is it 0, 1, or infinity?[/quote]

The answer is....

UNDEFINED!!!!

I am pretty sure it is
[hide]$UNDEFINED!$[/hide] :D


The decimal expression of the natural number $a$ consists of $n$ digits, while that of $a^3$ consists of $m$ digits. Can $n+m$ be equal to $2001$?


A circle of diameter 10 is inscribed in a trapezoid ABCD where angles A and B are right angles and AD=8. Find the length of longest side of this trapezoid. Express your answer as an improper fraction reduced to lowest terms.

[hide="hint"]
Use the fact that AB+CD=AD+BC[/hide]

[hide="Solution"]Call the P point of tangency to side CD.

Let CP=x. CD can be expressed as $ \sqrt {10^2 \plus{} (x \minus{} 3)^2}$

$ 10 \plus{} \sqrt {10^2 \plus{} (x \minus{} 3)^2} \equal{} 8 \plus{} 5 \plus{} x$
$ \sqrt {10^2 \plus{} (x \minus{} 3)^2} \equal{} x \plus{} 3$
$ 100 \plus{} x^2 \minus{} 6x \plus{} 9 \equal{} x^2 \plus{} 6x \plus{} 9$
$ 12x \equal{} 100$
$ x \equal{} \frac {25}{3}$

Thus, the answer is $ \frac {25}{3}\plus{}5\equal{}\frac{40}{3}$[/hide]

[hide]
[asy]pathpen = black + linewidth(0.62); pointpen = black;
pair A=(0,0),B=(0,8),D=(10,0),C=(10,40/3),O=(5,5);
D(MP("A",A)--MP("B",B,NW)--MP("C",C,NE)--MP("D",D)--cycle);
path p = Circle(O,5); D(p); pair F= IP(p,B--C);
D(B--MP("B'",(C.x,B.y),E)--C--cycle, blue+linewidth(0.8));
D(MP("E",D(F),NW)--D(O));
D(D(IP(p,C--D))--D(IP(p,A--B)));
MP("10",(A+D)/2); MP("5",(A+IP(p,A--B))/2,W);  MP("3",(B+IP(p,A--B))/2,W); MP("3",(B+F)/2,NW); MP("x",(C+F)/2,NW); MP("x-3",(C+(D.x,5))/2,NE); MP("10",(B+(10,B.y))/2,NE);[/asy]
Let $ E$ be the point of tangency between the inscribed circle and $ \overline{BC}$, and $ B'$ be the foot of the altitude from $ B$ to $ \overline{CD}$. By two-tangent, we have $ BE \equal{} 8 \minus{} 5 \equal{} 3$. Let $ EC \equal{} x$, by two-tangent again $ B'C \equal{} x\minus{}3$. The Pythagorean Theorem on $ \triangle BB'C$ yields
\[ (x\minus{}3)^2 \plus{} 10^2 \equal{} (x\plus{}3)^2 \Longrightarrow x \equal{} \frac{25}{3}\]Hence the longest side is $ CD \equal{} 5 \plus{} x \equal{} \boxed{\frac{40}{3}}$.
[/hide]


A janitor has to wash two round floors. If the time to wash the floor is proportional to its size, and the time required to wash the first floor is 4 hours, the time required to watch the second floor is 1 hour, what is the diamater of the first floor if the diamater of the second floor is 20 m.

The ratio of the areas is the square of the ratio of the diameters, and thus the ratio of the times is the square of the ratio of the diameters as well, so \[ \frac{4}{1}\equal{}\left(\frac{x}{20}\right)^2 \Rightarrow x\equal{}\boxed{40}\]

[quote="5849206328x"]The ratio of the areas is the square of the ratio of the diameters, and thus the ratio of the times is the square of the ratio of the diameters as well, so
\[ \frac {4}{1} \equal{} \left(\frac {x}{20}\right)^2 \Rightarrow x \equal{} \boxed{40}
\]
[/quote]

How do you know its the areas and not the circumference that its talking about?

By the way, shouldnt it be x/20= (1/4)^2 and not the other way around? I'm really confused :(

Hi helpmeplz,

I think 5849206328x did the problem incorrectly. In fact, since the time required to wash is proportional to its SIZE, that implies circumference. Therefore, if we denote the diameter of the first circle with x, then:

(Time to wash the first floor)/(Time to wash the second floor) = (x)/(20, diameter of second circle)

4/1 = x/20 which implies that x = 80. Therefore the diameter of the first circle is 80. Notice how there is nothing raised

to a power. This is only true when if constructing proportions with area and side length (similar polygons).

Hope it makes sense,

rts2007

If size meant circumference then the circumference is what the janitor washes.  That usually is not very helpful :wink:

[quote="rts2007"]Hi helpmeplz,

I think 5849206328x did the problem incorrectly. In fact, since the time required to wash is proportional to its SIZE, that implies circumference. Therefore, if we denote the diameter of the first circle with x, then:

(Time to wash the first floor)/(Time to wash the second floor) = (x)/(20, diameter of second circle)

4/1 = x/20 which implies that x = 80. Therefore the diameter of the first circle is 80. Notice how there is nothing raised

to a power. This is only true when if constructing proportions with area and side length (similar polygons).

[b]Hope it makes sense,[/b]

rts2007[/quote]

Uh... no.

If someone asks you your size, would you measure your circumference and give it to them?

Well then, how big is that book over there?

Would you answer the perimeter?

So BOGTRO and 5849206328x,

How would you interpret the problem? And, by the way, when someone asks me my size I don't say "blah blah" square inches. You can't interpret size as being area. And also, that book is huge! Would you say that the book's size is, for example, 45 square inches?

Thanks,

rts2007

Yes, you would.
You wouldn't say like $ 15$ inches, referring to its perimeter.
Something's size is a measure of how much space it takes up,
a measure that in this case is two-dimensional and is therefore area.

You wouldn't say 45 square inches nor would you say 15 inches, referring to its perimeter. This is a weird problem.


geometry
trigonometry
articles
LaTeX
trig identities
Pythagorean Theorem
Law of Cosines

1) The perpendicular bisectors of two of the sides of triangle $ ABC$ intersect the third side at the same point.  Prove that the triangle is right-angled.

2) Segments $ AD$ and $ BE$ are medians of right triangle $ ABC$, and $ AB$ is its hypotenuse.  If a right triangle is constructed with legs $ AD$ and $ BE$, what will be the length of its hypotenuse in terms of $ AB$?

3) If $ \tan{x} \equal{} \frac {2ab}{a^{2} \minus{} b^{2}}$, where $ a > b > 0$ and $ 0^\circ < x < 90^\circ$, then find $ \sin{x}$ in terms of $ a$ and $ b$.

1
[hide]Let the penpendicular bisectors bisect AB and AC. That implies thatt their intersection point O is equidistant from B, C, and A, so triangles ABO and AOC are isoceles, so A = B+C. since A+B+C = 180, A=90.[/hide]

[quote="elcric"]1) The perpendicular bisectors of two of the sides of triangle $ ABC$ intersect the third side at the same point.  Prove that the triangle is right-angled.

2) Segments $ AD$ and $ BE$ are medians of right triangle $ ABC$, and $ AB$ is its hypotenuse.  If a right triangle is constructed with legs $ AD$ and $ BE$, what will be the length of its hypotenuse in terms of $ AB$?

2) If $ \tan{x} \equal{} \frac {2ab}{a^{2} \minus{} b^{2}}$, where $ a > b > 0$ and $ 0^\circ < x < 90^\circ$, then find $ \sin{x}$ in terms of $ a$ and $ b$.[/quote]

Hint
[hide](1) Answered above.

(2) Let $ AC \equal{} 2a$ and $ BC \equal{} 2b$. Use Pythagoras to find $ AD$ and $ BE$.

(3) Construct a right triangle with sides opposite and adjacent to angle $ x$ be $ 2ab$ and $ a^2 \minus{} b^2$ respectively. [/hide]

[quote="elcric"]I still couldn't get numbers 2 and 3.  Could someone please help?  Thanks a lot![/quote]
for 3, after you construct a right triangle with sides opposite and adjacent to angle $ x$ be $ 2ab$ and $ a^2\minus{}b^2$ respectively.
we get $ tanx\equal{}2ab/{a^2\minus{}b^2}$ and $ sinx$ is easy to determine.

for the three the hint is to use the known: $ \sin^2x \equal{} \frac{\tan^2x}{1 \plus{} \tan^2x}$
the solution is here though:
[hide]
$ \tan^2x\equal{} \frac{(2ab)^2}{a^4 \plus{} b^4 \minus{}2a^2b^2}$
$ < \equal{} >$
$ \frac{\tan^2x}{1\plus{}\tan^2x} \equal{} (\frac{2ab}{a^2 \plus{} b^2})^2$
$ < \equal{} >$
$ \sin x\equal{}\frac{2ab}{a^2 \plus{} b^2}$ because sinx>0[/hide]

[quote="Athinaios"]for the three the hint is to use the known: $ \sin^2x \equal{} \frac {\tan^2x}{1 \plus{} \tan^2x}$
the solution is here though:
[hide]
$ \tan^2x \equal{} \frac {(2ab)^2}{a^4 \plus{} b^4 \minus{} 2a^2b^2}$
$ < \equal{} >$
$ \frac {\tan^2x}{1 \plus{} \tan^2x} \equal{} (\frac {2ab}{a^2 \plus{} b^2})^2$
$ < \equal{} >$
$ \sin x \equal{} \frac {2ab}{a^2 \plus{} b^2}$ because sinx>0[/hide][/quote]

How did you get $ \sin^2x \equal{} \frac{\tan^2x}{1 \plus{} \tan^2x}$?  Thanks!

[quote]How did you get $ \sin^2x \equal{} \frac{\tan^2x}{1 \plus{} \tan^2x}$? Thanks![/quote]

One way is just to use trig identities

$ \sin^{2}x \plus{} \cos^{2}x \equal{} 1$; divide by $ \sin^{2}$ to get

$ 1 \plus{} \frac{1}{\tan^{2}x} \equal{} \frac{1}{\sin^{2}x}$, so

$ \sin^2x \equal{} \frac{\tan^2x}{1 \plus{} \tan^2x}$

[quote="Athinaios"]for the three the hint is to use the known: $ \sin^2x \equal{} \frac {\tan^2x}{1 \plus{} \tan^2x}$
the solution is here though:
[hide]
$ \tan^2x \equal{} \frac {(2ab)^2}{a^4 \plus{} b^4 \minus{} 2a^2b^2}$
$ < \equal{} >$
$ \frac {\tan^2x}{1 \plus{} \tan^2x} \equal{} (\frac {2ab}{a^2 \plus{} b^2})^2$
$ < \equal{} >$
$ \sin x \equal{} \frac {2ab}{a^2 \plus{} b^2}$ because sinx>0[/hide][/quote]

I don't get how $ \frac {\tan^2x}{1 \plus{} \tan^2x} \equal{} (\frac {2ab}{a^2 \plus{} b^2})^2$ was gotten.

Thanks!

there is a known id that says: if $ \frac{a}{b} \equal{}\frac{c}{d}$ then it is also true that
$ \frac{a}{b \plus{} a} \equal{}\frac{c}{d \plus{}c}$(easy to prove) so using this you know
$ \frac{tan^2x}{1} \equal{}\frac{(2ab)^2}{a^2 \plus{} b^2 \minus{}2ab}$
$ < \equal{} >$
$ \frac{tan^2x}{1 \plus{} tan^2x} \equal{}\frac{(2ab)^2}{a^4 \plus{} b^4 \minus{}2a^2b^2 \plus{} 4a^2b^2}$
$ \equal{} \frac{(2ab)^2}{a^4 \plus{} b^4 \plus{} 2a^2b^2}\equal{}$
$ \equal{}(\frac{2ab}{a^2 \plus{} b^2})^2$ i think now it is clear

[quote="Athinaios"]there is a known id that says: if $ \frac {a}{b} \equal{} \frac {c}{d}$ then it is also true that
$ \frac {a}{b \plus{} a} \equal{} \frac {c}{d \plus{} c}$(easy to prove) so using this you know
$ \frac {tan^2x}{1} \equal{} \frac {(2ab)^2}{a^2 \plus{} b^2 \minus{} 2ab}$
$ < \equal{} >$
$ \frac {tan^2x}{1 \plus{} tan^2x} \equal{} \frac {(2ab)^2}{a^4 \plus{} b^4 \minus{} 2a^2b^2 \plus{} 4a^2b^2}$
$ \equal{} \frac {(2ab)^2}{a^4 \plus{} b^4 \plus{} 2a^2b^2} \equal{}$
$ \equal{} (\frac {2ab}{a^2 \plus{} b^2})^2$ i think now it is clear[/quote]

Thanks, I got number 3 now.  Does anyone have a solution to problem 2?

For #2,

Use Stewart's Theorem
$ (dad\plus{}man\equal{}bmb\plus{}cnc)$

To obtain

$ AD^{2}\equal{}\frac{1}{2}AC^{2}\plus{}\frac{1}{2}AB^{2}\minus{}\frac{1}{4}BC^{2}$,
$ BE^{2}\equal{}\frac{1}{2}BC^{2}\plus{}\frac{1}{2}AB^{2}\minus{}\frac{1}{4}AC^{2}$,
So
$ AD^{2}\plus{}BE^{2}\equal{}HYPOTENUSE^{2}\equal{}\frac{1}{4}AB^{2}\plus{}AB^{2}\equal{}\frac{5}{4}AB^{2}$
So
$ HYPOTENUSE \equal{} \frac{\sqrt{5}}{2}AB$

[quote="iniquitus"]For #2,

Use Stewart's Theorem
$ (dad \plus{} man \equal{} bmb \plus{} cnc)$

To obtain

$ AD^{2} \equal{} \frac {1}{2}AC^{2} \plus{} \frac {1}{2}AB^{2} \minus{} \frac {1}{4}BC^{2}$,
$ BE^{2} \equal{} \frac {1}{2}BC^{2} \plus{} \frac {1}{2}AB^{2} \minus{} \frac {1}{4}AC^{2}$,
So
$ AD^{2} \plus{} BE^{2} \equal{} HYPOTENUSE^{2} \equal{} \frac {1}{4}AB^{2} \plus{} AB^{2} \equal{} \frac {5}{4}AB^{2}$
So
$ HYPOTENUSE \equal{} \frac {\sqrt {5}}{2}AB$[/quote]

What's Stewart's Theorem?  How does it work and what does it do?  Thanks a lot!

Wikipedia article
http://en.wikipedia.org/wiki/Stewart's_theorem
(URL tags dont work :S) copy and paste into browser

umm....i dont really understand anything above.... >.< BUT I HAVE A SOLUTION for 3 except its sort of differentish

and i just learned that trig stuff so i might have taken some long ways when there was a shorter way thats obvious

[hide="click here for a solution without using that crazy formula!! :P "]
make a right triangle and since that tangent stuff is opposite over adjacent (yay for sohcahtoa!! :D ) you label the leg opposite the angle with a measure of x 2ab and the adjacent leg a^2-b^2 (sorry i cant use latex so the rest will be hard to read.....)

Using the Pythagorean Theorem we can say that the length of the hypotenuse is
sqrt((a^2-b^2)^2+(2ab)^2)
=sqrt(a^4-2(a^2)(b^2)+b^4+4(a^2)(b^2)
=sqrt(a^4+2(a^2)(b^2)+b^4)
=sqrt((a^2+b^2)^2)
=a^2+b^2

sin(x)=opposite/hypotenuse ( :D )
=(2ab)/(a^2+b^2)

yay!! this was more algebra than geometry :lol: [/hide]

[quote="elcric"]

What's Stewart's Theorem?  How does it work and what does it do?  Thanks a lot![/quote]

Here you have [b]Stewart's theorem [/b]and also a proof:
[quote][color=green]Let the triangle $ ABC$ and $ M$ a point on $ BC$.Prove that
\[ AB^2\cdot MC \plus{} AC^2\cdot BM \equal{} AM^2\cdot BC \plus{} BM\cdot MC\cdot BC
\]
[/color][/quote]
[hide="Proof"]Use the law of cosines:
$ AM^2 \equal{} AB^2 \plus{} BM^2 \minus{} 2AB\cdot BM \cos B$
$ AC^2 \equal{} AB^2 \plus{} BC^2 \minus{} 2AB\cdot BC \cos B$
Then multiply with $ BC$ the first relation and with $ \minus{} BM$ the second relation,and by summing will obtain the proof[/hide].


If $ \minus{}2x\plus{}1<3$, what is the least integer in the solution set?

$ \minus{}2x\plus{}1<3\implies\minus{}2x<2\implies x>\minus{}1\implies x\ge\boxed0$.


geometry
geometric transformation
reflection
parallelogram
rectangle
rotation
circumcircle

Let $ABC$ be a right-angled triangle ($AB\perp AC$) and the mobile points $M\in AB,\ N\in AC$ such that $BN=CM$. Prove that one from the bisectors of the angles formed between the lines $BN,\ CM$ passes through a fixed point.

[u]Remark.[/u] I shall be pleased to see a syntetical solution.
This problem, with analytical geometry, has a easy solution.

[u]Indication.[/u] The fixed point $F$ is the reflection of the point $A$ w.r.t. the middlepoint of the side $[BC]$ and the distance of the point $F$ to the lines $KB,KC$ is equal to $\frac{AB\cdot AC}{l}$ where $K\in BN\cap CM,\ BN=CM=l$.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=57850

[color=blue][b]Lemma.[/b][/color] Let PQRS be a parallelogram inscribed in another parallelogram ABCD, such that $P \in AB, Q \in BC, R \in CD, S \in DA$. Let the lines AQ, CP meet at a point M and the lines AR, CS at a point N. Then the lines $BN \parallel DM$ are parallel to the lines $SP \parallel QR$.

Using parallel projection, project the parallelogram PQRS inscribed in the parallelogram ABCD into a rectangle P'Q'R'S' inscribed in a rectangle A'B'C'D' and skip the prime notation. The right angle triangles $\triangle PQB \cong \triangle RSD$ are congruent and the right angle triangles $\triangle QRC \cong \triangle SPA$ are also congruent. In addition, the the 2 triangle pairs are similar. Let $QU, SV$ be lines parallel to the rectangle sides $AB \parallel CD$ and $RW$ a line parallel to the rectangle sides $BC \parallel DA$. Let the points $K \equiv QU \cap RW$ and $L \equiv SV \cap RW$. The rectangles $BVLW \cong QCRK$ are congruent and they have parallel diagonals $BL \parallel QR$. Let the line AR, BL, CS meet the lines BC, CA, AB at points E, F, G, respectively. Put a cross-hair through the point F, intersecting the rectangle sides AB, BC, DA at points W', V', U'. Denote $AB = CD = h_0,\ BC = DA = w_0$ and $BW = QK = VL = CR = h_1$, $BV = WL = AS = QC = KR = UD = w_1$ the heights and widths of the appropriate rectangles. We have

$\frac{GA}{GB} = \frac{AS}{BC} = \frac{w_1}{w_0}$

$\frac{EB}{EC} = \frac{AB}{CR} = \frac{h_0}{h_1}$

The rectangles $BV'FW' \sim BVLW \cong QCRK$ are similar, hence,

$\frac{FV'}{FW'} = \frac{LV}{LW} = \frac{h_1}{w_1}$.

Since the point F is on the line CA, we also have

$\frac{FW'}{FU'} = \frac{BC}{AB} = \frac{w_0}{h_0}$

Consequently,

$\frac{FC}{FA} = \frac{FV'}{FU'} = \frac{FV'}{FW'} \cdot \frac{FW'}{FU'} = \frac{h_1 w_0}{w_1 h_0}$

$\frac{GA}{GB} \cdot \frac{EB}{EC} \cdot \frac{FC}{FA} = \frac{w_1}{w_0} \cdot \frac{h_0}{h_1} \cdot \frac{h_1 w_0}{w_1 h_0} = 1$

Thus the lines $AE \equiv AR,\ CS \equiv CG,\ BL \equiv BF$ are concurrent. But since $N \equiv AR \cap CS$, this is the concurency point and as a result, $QR \parallel BL \equiv BN$.

Let $\triangle ABC$ be a right angle triangle with the angle $\angle A = 90^\circ$. Let BN, CM be arbitrary cevians of this triangle intersecting at a point G in its interior, $M \in AB, N \in AC$. Let O be the midpoint of the hypotenuse BC. Reflect the figure in the point O (i.e., rotate it by $180^\circ$ around the point O). The result is a rectangle $ABA'C$ with the inscribed parallelogram MNM'N', where the lines BN, CM meet at the point G and the lines CM', BN' meet at a point G'. According to the above lemma, the lines $AG' \parallel A'G \parallel MN' \parallel NM'$ are parallel. From a number of similar triangles formed by the lines BC, BN, CM, MN cut by the parallels $AG' \parallel A'G \parallel MN' \parallel NM'$, it quickly follows these lines are cut in the same ratios. For example, let the parallel AG' cut the lines BC, BN, CM, MN at points P, U, V, X and let the parallel A'G cut the lines BC, MN at a point Q, Y. Then form BP = CQ, BQ = CP and MX = NY, MY = NX, we quickly find that

$\frac{BU}{BN} = \frac{CG}{CM},\ \ \frac{UG}{BN} = \frac{GV}{CM},\ \ \frac{GN}{BN} = \frac{VM}{CM}$

and it is clear that UG = VG iff BN = CM. In other words, the triangle $\triangle GUV$ is isosceles iff BN = CM. This triangle is isosceles iff its base line $AG' \equiv UV$ is perpendicular to the bisector of the angle $\angle UGV$ and parallel to the bisector of the complementary angle $\angle BGC$, which is then identical with the line A'G, where A' is a reflection of the vertex A in the midpoint O of the hypotenuse BC, i.e., the diametrally opposite point of the vertex A on the triangle circumcircle (O). Even simpler is to consider one of the congruent triangles $\triangle BNM' \cong \triangle CN'M$ with the sides BN, BM' = CM and CM, CN' = BN. The 1st triangle is isosceles with BN = BM' = CM iff its base line NM' is perpendicular to the bisector of the angle $\angle NBM' = \angle NGC$ and parallel to the bisector of the complementary angle $\angle BGB = 180^\circ - \angle NGC$, i.e., iff the lines A'G and the bisector of the angle $\angle BGC$ are identical.

It is useless to suppose that ABC is right-angled.
If P is the reflection of A wrt the midpoint of BC, clearly, the triangles PBN, PCM and ABC have the same area.
So, the distance from P to the line BN is 2.Area(ABC)/BN and the distance from P to the line CM is 2.Area(ABC)/CM
Hence, if BN = CM, P lies on a bisector of (BN,CM)

The problem was also solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=38363 .

  darij

Jpe, you try to solve the my "dear" problem mentioned at
http://www.mathlinks.ro/Forum/viewtopic.php?t=57850

Ph-An introduced the right angle here in an attempt to camouflage the fact that

   BN and CM are common internal tangents of 2 circles. Just move one of them

   parallel to the other towards the midpoint. Angle bisectors direction will not change.

   Now it is not hard to see fixed points.

   Thank you.

    M.T.


list three times in which a properly working clock's minute and hour hand makes an angle of 180 degrees.

[hide]Minute hand moves six degrees per minute. Hour hand moves 1/2 of a degree per minute. Minute hand catches up 5.5 degrees every minute.

In the second hour, the minute hand needs to catch up 60 degrees then move ahead 180 degrees for a total of 240 degrees. The number of minutes needed to do this is

$\frac {240}{5.5} = 43 \frac 7{11}$

So at $2:43$ and $\frac 7{11}$ minutes, the hour hand and the minute hand will form a 180 degree angle.

Six o'clock is another time when they form a 180 degree angle.

In the first hour, the hour hand needs to catch up 30 and move ahead 180 for a total of 210.

$\frac {210}{5.5} = 38 \frac 2{11}$

Another time, then, is one o'clock and thirty-eight and two-elevenths minutes.[/hide]


Hi all,

Please click on "Watch this topic link" on the right upper (or lower) side of the topic so that you will be informed when updates about this contest appear.

For now problems of Round 1 have been posted.

The forum for discussions (asking questions about notations, or whatever, is [url=http://www.mathlinks.ro/Forum/index.php?f=382]MathLinks Contest 6th Edition, Fall 2005[/url].

[quote="perfect_radio"]in problem 2, $a_k = x_k$, right?[/quote]Yeah, that's correct  :blush: The file was modified.

[quote="perfect_radio"]in problem 2, $a_k = x_k$, right?[/quote]
[quote="Valentin Vornicu"]Yeah, that's correct  The file was modified[/quote]

I didn't see what was modified. What about the case when $k=n$. We have to determine $a_n$ . I'm sorry but i have a problem with english, can me explane the mean of $consecutive$ :blush:  Does it mean for example $a_1=p,a_2=p+1$ etc. ?

Should we give the possible values of $a_1$ in terms of $n$, or the possible values $a_1$ can take over all values of $n$?

[quote="Genrih"]I didn't see what was modified. What about the case when $k=n$. We have to determine $a_n$ . I'm sorry but i have a problem with english, can me explane the mean of $consecutive$ :blush:  Does it mean for example $a_1=p,a_2=p+1$ etc. ?[/quote]Yes, that's what consecutive means. If you determine $a_1$ of course you have determinated all the others :D

@Saul: all possible values of $a_1$. I did not say anothing similar with "in terms of $n$" ;)

[quote="amir2"][quote="silouan"]Where can I find the problems of the second round ?Please reply. Thanks[/quote]
U can find it here:[/quote]

I cannot find it. Could you please post the adreese of the topic. Please mr Valentin.

The link is http://www.mathlinks.ro/Forum/contest.php

[quote="Valentin Vornicu"]Yes, that's what consecutive means. If you determine $a_1$ of course you have determinated all the others :D

[/quote]
Thank you

Ok, a few more typos in problem 2. Nothing big, as I assume everyone has understood the problem as they should have. Namely the new statement is

[quote]Let $a_1,a_2,\ldots,a_{n-1}$ be $n-1$ consecutive positive integers  [b]in increasing order[/b] such that $\displaystyle k { n \choose k } \equiv 0 \pmod {a_k}$, for all $k\in\{1,2,\ldots,n-1\}$. Find the possible values of $a_1$.[/quote]

Does K have 5 elements, or do the subsets in F have 5 elements?

Round 5 has been posted. You can discuss about Round 4 now.

A small typo (which I am sure many of you noticed anyway) which was signaled by Jin: in problem 2, $x_1,x_2,\ldots, x_n$ are integers, not real numbers. The PDF is corrected now.

The statement of problem 2 was updated as follows:

[quote="New statement"]A $n\times n$ matrix is filled with non-negative real numbers such that on each line and column the sum of the elements is 1. Prove that one can choose $n$ positive entries from the matrix, such that each of them lies on a different line and different column.[/quote].

And also problem 3, a small typo (thanks perfect_radio)

[quote="Problem 3"]Let $\mathcal{C}_1$, $\mathcal{C}_2$ and $\mathcal{C}_3$ be three circles, of radii 2, 4 and 6 respectively. It is known that each of them are tangent exteriorly with the other two circles. Let $\Omega_1$ and $\Omega_2$ be two more circles, each of them tangent to all of the 3 circles above, of radius $\omega_1$ and $\omega_2$ respectively.

Prove that $\omega_1 + \omega_2 = 2\omega_1 \omega_2$.[/quote]

A new edition will start soon. Check the main portal page for details!

[url=https://artofproblemsolving.com/community/c1997017_mathlinks_contest_6th]here [/url] is the related post collection


How many positive integer factors of $ 2^2 \times 3^2 \times 5$ are multiples of 12?

12 = 2^2*3.

There are 4 factors of 15. So, 15 is our answer.

sorry to revive this old thread but can someone tell me what is the right answer  ?

Hello,

$ 12=2^{2}3$. Therefore we get as possible solutions:
$2^{2}3=12$
$2^{2}*3*5=60$
$2^{2}3^{2}=36$ and
$2^{2}3^{2}*5=180$.

Thus we have $4$ positive integer factors, which are multiplies of $12$.

Best wishes from Leipzig/Germany.

Besides counting, actually, there is a faster solution. 12 requires a 2, another 2, and a 3. Thus there is a 3 and 5 left. Using the method to find the factor pairs we get $(1+1)(1+1) = \boxed {4}.$


solve

Find the remainder when 2^{1990} is divided by 1990.

[quote="AIME15"]Google is your friend.  This question has been asked at least $ 5$ times:

[url=http://www.mathhelpforum.com/math-help/number-theory/15484-find-remainder.html]Here[/url], [url=http://answers.yahoo.com/question/index?qid=20081110084811AA640xQ]here[/url], [url=http://answers.yahoo.com/question/index?qid=20081211041845AAZCdHD]here[/url], [url=http://www.goiit.com/posts/list/algebra-find-the-remainder-94670.htm]here[/url], and [url=http://www.pagalguy.com/forum/quantitative-questions-and-answers/852-number-system-402.html]here[/url].[/quote]

ummmmmmmmmmmmmmmmmmm actually I got it from contests on forum.

doesn't matter, it's already been posted then if it was from a contest on the forum :wink:

is this a statement or question :huh: :huh:

[quote]saszs I have a B- in english
 [/quote]
Isn't this spam??
English has nothing to do with math, espicially if ur talking about ur own grade.

[quote="saszs"]anyway I didn't see the solution.

is the answer 1024.[/quote]

Maybe clicking the links may help?

Try the first link...if you're willing to read more than two sentences?

If you're asking for the answer and you already know it, then what's the use? You don't need to share these stuff. AoPS is mainly for posted problems. And there's an emphasis on problems. Cause you don't have to post if you don't have a problem. Unless you're sharing something that is productive and helps us learn.

[quote="jonathanchou711"]AoPS is mainly for posted problems. [/quote]

Er. No.


The radius of semicircle $ AFB$ is $ \frac {AB}{2} \equal{} 2$, implying that its area is $ 2\pi$. Since $ ABCE$ is a square, its area is $ 4^2 \equal{} 16$. Lastly, the area of triangle $ CDE$ is $ \frac {4\times{4}}{2} \equal{} 8$. Adding these up, we have $ 16 \plus{} 8 \plus{} 2\pi \equal{} \boxed{24 \plus{} 2\pi}$.


hmm set it up this way

[hide="hint"]
ABCD

where BC must be greater than 5
and the number ABCD must be greater than 2999.

Sorry for not using LaTeX I feel really sleepy now  :oops:
[/hide]

[hide]100 possibilities for middle digits of numbers.
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
21
22
31
41
51
20
30
40
50
60
70
80
90
are the only possibilities that the product would be equal to or less, and this, i believe, are 29 possibilities,
so 100-29=71, and since 9999 is the last 4-digit number, there are 71 per 1000 starting from 3000.
3
4
5
6
7
8
9 so 71*7, 497, so 497 possibilities, but there are 10 possibilities for each ones place, so 4997*10=4970.
$\boxed{4970}$ [/hide]

edited: is it right now?

[quote="turboturtle"][hide]100 possibilities for middle digits of numbers.
00
01
02
03
04
05
06
07
08
09
10
11
12
13
14
15
21
22
31
41
51
are the only possibilities that the product would be equal to or less, and this, i believe, are 21 possibilities,
so 100-21=79, and since 9999 is the last 4-digit number, there are 79 per 1000 starting from 3000.
3
4
5
6
7
8
9 so 79*7, 553, so 553 possibilities[/hide][/quote]
[hide="correction"]
You should rethink your answer a bit.  If 79/100 possibilities are valid, then how come only 553 of the 7000 numbers are valid?
For a four-digit number ABCD, A must be in [3, 9], while D must be in [0, 9].  So there are 7 possible values for A, 10 possible values for D, and 79 possible values for BC.
7 * 10 * 79 = 5530.
[/hide]

nope, they didn't miss that many(that don't work)

how is the answer not 497?
-jorian[/hide]

The total number of numbers greater than 2999 is 7*10*10*10=7000. The ones that don't work are the numbers with the middle digits whose product is five or less. They are:
00-09: 10
10-15: 6
20-22: 3
30-31: 2
40-41: 2
50-51: 2
60-90: 4
There are 29 for each of the 7 thousands. 29*7=203. 7000-203= [hide]6797 numbers total.[/hide]

[quote="easyas3.14159..."][quote="davidli"]What did I do wrong?[/quote]
Try adding 10+6+3+2+2+2+4 up again. I don't think you'll get 29. :wink:[/quote]

Yeah...before making a post like that check YOUR math first. :wink:

no, he said [hide]497[/hide]was wrong...i asked if[hide]4970[/hide]was right
-jorian

[quote="13375P34K43V312"]davidli, you are VERY, VERY close. You forgot about the last digit, for your penultimate step.[/quote]
BUt there are only 7000 four digit numbers greater than 2999...
I got [hide]4970[/hide] but it has already been said that that is wrong... I used [hide]complementary counting[/hide]...

[quote="13375P34K43V312"]How many four-digit numbers greater than 2999 are there such that the product of the two middle digits exceeds 5?[/quote]
[hide="Solution"]
products > 5:
1: 4
2: 7
3: 8
4: 8
5: 8
6: 9
7: 9
8: 9
9: 9
71 possible numbers for the 2 digits
71*10=710 for every thousand
710*7=$4970$ Numbers
[/hide]

[quote="jhredsox"]it has to be about [hide]4970[/hide] because [hide]29%[/hide] of the time, u get a number with middle digits having a product equal to or less than 5

lets check....

[hide]$7000*.29=2030$

$7000-2030=4970$[/hide]
-jorian[/quote]

How did you get 29%? Also, you have to factor in the first digit.

[hide]First we have to find the number of middle digits with product less than or equal to 5.

We have

00, 01, 02, 03, 04, 05, 06 , 07, 08 ,09
10, 11, 12, 13, 14, 15
20, 21, 22
30, 31
40, 41
50, 51
60
70
80
90

So, we have 29 possibilities that aren't allowed so there are 71 possibilities that are allowed.

Also, since it must be greater than 2999, the thousands digit is greater than or equal to 3, and the units digit can be anything.

So the answer is 7*71*10=4970

$4970$[/hide]


how many zeros at the end of 1000!  ?
exlanation+answer please.
i got 249, but my teacher says its wrong.

I don't see why it should be wrong...
But you can look here: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=1000%21&t=17157

thanks my teacher is probbly doing more druigs


linear algebra
matrix
induction
superior algebra
superior algebra solved

You can also do this using Hamilton-Cayley's relation! :D :D
Or if you want to waste time and try guessing, you can try induction! :D
cheers!

Or like this:
A=2I_2+B
where B=
-1 -1
1   1

Then observe that B^2=O_2 so B^k=O_2
now apply Newton's formula and get that A^n=2^n*sum(k=0,n) (n,k)B^k=2^n(I_2+nB) and get what Moubinool found! :D :D

(n,k)=n!/[k!(n-k)!]

cheers! :D :D


Define an infinite sequence $ x_n$ by setting $ x_1 \equal{} 1$ and $ x_{n\plus{}1}\equal{}\frac{x_n}{1\plus{} \sqrt{x_n}}$ for all $ n\geqslant 1$. Show that $ x_n$ converges, and determine its limit.

[quote="Pythagorean12"]Define an infinite sequence $ x_n$ by setting $ x_1 \equal{} 1$ and $ x_{n \plus{} 1} \equal{} \frac {x_n}{1 \plus{} \sqrt {x_n}}$ for all $ n\geqslant 1$. Show that $ x_n$ converges, and determine its limit.[/quote]

$ x_1 > 0 \Rightarrow x_2 > 0 \Rightarrow ... \Rightarrow x_n > 0 \forall n \in N$
$ x_{n} \minus{} x_{n \plus{} 1} \equal{} x_{n \plus{} 1} \sqrt {x_n} > 0$ , so $ x_n$ is decreasing and bounded below $ \Rightarrow$ $ x_n$ converges , the limit exists and $ \ge 0$ .

So let the limit $ \equal{} L^2$ with $ L \ge 0$,  we have $ L^2 \equal{} \frac {L^2}{1 \plus{} L} \Rightarrow L^2(1 \minus{} \frac {1}{1 \plus{} L}) \equal{} 0 \Rightarrow L \equal{} 0$

So the limit $ \equal{} 0$


this problem involves the unit circle i think. does anyone know how to solve this???

√3tan3x= -1

only the 3 is square rooted

Not the correct forum.  You want

http://www.artofproblemsolving.com/Forum/index.php?f=149

Anyway,

$\tan 3x =-\frac{1}{ \sqrt{3}}$
$3x = 150^{\circ}, 330^{\circ}$ (or the corresponding radian measures)
$x = 50, 110, 170, 230, 290, 350$ (or the corresponding radian measures; I've repeatedly added $120^{\circ}= \frac{360^{\circ}}{3}$ to get all the solutions in the circle.)


[quote="ambros"]Ισως κάποιοι να με κατηγορήσουν ότι καταστρέφω το φόρουμ με άσχετα με τα μαθηματικά τόπικ αλλά η περίσταση ειναι πολύ ειδική κι ας με συγχωρήσουν  όλοι

ας γράψουμε όλοι τα συναισθήματα κι την συμπαράσταση μας για τους 65 νεκρούς κι τα 3,5 εκατομμύρια στρέματα καμένης γής

ειμαστε όλοι μαζί τους[/quote]

 Αν υπάρχουν στην καμένη περιοχή παιδιά της Γ΄λυκείου ή που ασχολούνται με ολυμπιάδες και παρακολουθούν το φόρουμ , ας μας στείλλουν PM για να τους χαρίσουμε βιβλία για τη μελέτη τους ή τις εξετάσεις. Αλλά και όποια βοήθεια περνάει από το χέρι μας , θα τους την προσφέρουμε με χαρά.

 [b]Σημείωση[/b]

 Αμβρόσιε , τέτοιες παρεμβάσεις όχι μόνο δε χαλάνε το φόρουμ αλλά αντίθετα μας δίνουν δύναμη να ασχολούμαστε με περισσότερο πάθος. Διότι τα μαθηματικά είναι πάνω από όλα ανθρωπισμός  !!!


geometry
3D geometry
number theory unsolved
number theory

find the smallest number expressible as the sum of two cubes in two different ways

1. Bad statement: can a cube be negative.
2. _not_ hard, just trying in worst case.
3. The Ramanujan-Hardy-story is well known, and so is the taxi number.
4. I lock this thead for the enormous lack of a sensefull title (and I will lock others if it happens again).
5. I need a prime number of points.


function
search
induction
algebra unsolved
algebra

Find all functions $f: \mathbb{N} \longrightarrow \mathbb{N}$ such that for all $m,n \in \mathbb{N}$,

$(2^m+1)f(n)f(2^mn)=2^m f(n)^2 +f(2^mn)^2+(2^m-1)^2n$

[hide]
It is easy to prove that for all primes $p$ we have $f(p)=p+1$, also $f(2^m)=2^m+1$ and $f(1)=2$, so intuition tells me ( :D ) that function $f(n)=n+1$ is the only one, though I don't know how to "make the final blow" having informations above  :?
[/hide]

I believe the answer is:
\[f(2^mn)=p+2^mq,\] where $m$ is any nonnegative integer, $n$ is any positive odd integer, and $p,q$ are any positive integers that satisfy $pq=n$.

Thus the function is not unique; i.e., it can vary according to $p$ and $q$.

I know that [tex]2f(n)>f(2n)[/tex] implies [tex]f(n)=n+1[/tex]. Homever I was unable to prove that [tex]2f(n)>f(2n)[/tex] ...

I believe that $f(2n)<2f(n)$ does hold. However, I do not see how it implies $f(n)=n+1$, as my solution above also satisfies the relation.

Pity the fool ! My proof is wrong. I'll try to fix it.
(Don't botter with the useless things I said before)

[b]k2c901_1[/b]: do you have full proof?

Yes, I do. I post it a little bit later...

After manipulating the original equation we see that \[(2^mf(n)-f(2^mn))(f(2^mn)-f(n))=(2^m-1)^2n.\] Note that the two factors in the LHS have a sum that is a positive integer, and hence must both be positive. Setting $n=pq$ for any odd integer $n$ and positive integers $p,q$, setting $m=1$ above yields \[(2f(n)-f(2n))(f(2n)-f(n))=pq.\] We see that it is possible for $f(n)=p+q$ for any $p,q$ by simply setting $(2f(n)-f(2n))=p$ and $(f(2n)-f(n))=q$. Now replacing $f(n)=p+q$ in the first equation, and setting $a=(2^mf(n)-f(2^mn))$ and $b=(f(2^mn)-f(n))$, it is easy to see that either $a=(2^m-1)p$ or $a=(2^m-1)q$ for all $m$. Substitution then gives $f(2^mn)=p+2^mq, q+2^mp$. However, replacing $n$ by $2n$ in the original equation, and setting $f(2n)=p+2q$ (since $p$ and $q$ are symmetric), we see that only  $f(2^mn)=p+2^mq$ is possible. It is easy to check that this solution is consistent with the initial condition.

It has been posted before. You can search it in the in the forum.

[quote="k2c901_1"] it is easy to see that either $ a \equal{} (2^m \minus{} 1)p$ or $ a \equal{} (2^m \minus{} 1)q$ for all $ m$. [/quote]

How is it easy? I don't see...

I told you that a solution is posted [url=http://www.mathlinks.ro/viewtopic.php?search_id=2146709638&t=6268]here[/url]:

Notice that:
\[ (2^mf(n) \minus{} f(2^mn))(f(2^mn) \minus{} f(n)) \equal{} n(2^m \minus{} 1)^2
\]
So the value of $ f(2^mn)$ can be computed by the value of $ f(n)$. Now by setting $ m \equal{} 1$ and $ n$ to be an odd number we see that we can set
\[ f(n) \equal{} g(n) \plus{} \frac n{g(n)}
\]
where $ g(n)$ is an arbitrary divisor of $ n$. So we get by induction on $ m$ that
\[ f(2^nm) \equal{} 2^mg(n) \plus{} \frac n{g(n)}
\]
and the problem is solved because every natural number can be written as product $ 2^mn$ where $ n$ is an odd number.


The$ 2^n$ rows of a$ 2^n . n$ table are filled with all the different n-tuples of the numbers+1,-1. Then some numbers were replaced by 0-s.Prove that one can choose a nonemptyset of rows so that the sum of all the numbers written in them is 0

I think this was proven by dzeta a while back.


which $ \ce{H}$ of Slicylic Acid is most acidic knowing that this reaction does take place
Salicylic acid + acetyl chloride -> Aspirin

I think you would probably look at the resonance stabilized structures when each hydrogen is deprotonated.

yes that gives H attached to the OH group i smore acidic sinve the EWG attached to benzene ring stabilizes it but the case is not the same for -COOH but
Me-COOH is more acidic than Phenol because of greater stabilization of Carboxylate ion by Resonance
so does that play a role here

carboxylic acids are more acidic than phenols as the negative charge is delocalised on the oxygens.
Benzoic acid is more acidic than phenol.
So, the hydrogen of -COOH must be more acidic.

obviously the h of cooh is more acidic.
regarding the reaction
[quote]
Salicylic acid + acetyl chloride -> Aspirin
[/quote]
there is no way you can predict the acidities of the hydrogens using this reaction as this is not an acid base reaction

yes this is not a acid base reaction.
but is -COOH is more acidic than -OH then why does the oxygen of -COOH attack the C of acetyl chloride?

well , the solvent over here(water usually) acts as a base and so cooh mainly exists as coo-( i know it is not very acidic, but it is more acidic than oh) which makes it better nucleophile than oh (which exists mainly as oh). so cooh reacts with c of acetyl chloride. also it is less sterically hindered than oh

sorry this is wrong.
 oye pardesi you confused me in fact oh group reacts with acetyl chloride( also esterification is usually, not always tho, acid catalyzed so the above explanation is wrong).

Wait, I believe aspirin formation is an esterification process so shouldn't it be the oxygen atom of the hydroxyl group rather than the carboxy group?

P.S. This seems to support the view that the $ \minus{}COOH$ proton is more acidic:

http://www.chemheritage.org/EducationalServices/pharm/asp/acid03.htm

neways wat does this have to do with acidity. it is based on the nucleophilicity of oh or cooh

The bottom of the link I posted indicates an acid base reaction where $ \minus{}\ce{COOH}$ is deprotonated.

[quote="Pardesi"]but is -COOH is more acidic than -OH then why does the oxygen of -COOH attack the C of acetyl chloride?[/quote]

That's precisely the reason! If $ \minus{}CO_2H$ is more acidic than $ \minus{}OH$, that means $ \minus{}CO_2^\minus{}$ is a weaker base than $ \minus{}O^\minus{}$, and this also means that $ \minus{}O^\minus{}$ is a better nucleophile than $ \minus{}CO_2^\minus{}$.

yes that is ...
i was looking more from a probablistic point of view

By the way an intramolecular hydrogen bond forms in Salicylic Acid so that might explain why the $ \ce{\minus{} OH}$ is harder to deprotonate.

Why an intramolecular hydrogen bond implies that deprotonation is harder?


algebra
polynomial
geometry
3D geometry
search
algebra open

For arbitrary $ x$ being integer polynomial $ P(x)\equal{}ax^2 \plus{} bx \plus{}c$ is a perfect square.

Prove that there exist positive integers $ d$ and $ e$ for which $ P(x)\equal{}(dx \plus{} e)^2$

1) How can this problem appear in the open question section???? :wink:
Please look of what "open question" means!

2) There are many problems in this forum where it is proved that if $ P(x)$ is a square, a cube, etc... for all integers, or for some judiciously chosen integers, etc... then $ P(x)$ is itself a square, a cube, etc...
Try a search!

Pierre.


One can holds $ 12$ ounces of soda.  What is the minimum number of cans to provide a gallon ($ 128$ ounces) of soda?

$ \textbf{(A)}\ 7 \qquad
\textbf{(B)}\ 8 \qquad
\textbf{(C)}\ 9 \qquad
\textbf{(D)}\ 10 \qquad
\textbf{(E)}\ 11$


A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=609[/img]

$ \textbf{(A) } 15 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } \frac {35}{2} \qquad \textbf{(D) } 18 \qquad \textbf{(E) } \frac {55}{3}$

[hide]

*HINT* Use ratio of triangles, knowing that the triangles have the same height. Therefore, the ratio of the area of the triangles is equal to the ratio of the area. I'll elaborate more when I have time to redraw their diagram.

[/hide]

This was a USAMTS problem a few years back.  It surprises me it is on the 10.

Anyway,
[hide]
$(D)$
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=623[/img]

Rearrange the image as shown.  The ratio of the areas of two triangles who share an altitude is equal to the ratio of their bases.  Thus,
\[ \frac{x}{y} = \frac{7}{7} = 1 \implies x = y  \]
Now, since $\triangle{DFE}$ and $\triangle{FEC}$ share an altitude, the ratio of their areas is also equal to 1 so $[DEF] = 3$.
Using the same concept, $\triangle{DEB}$ shares an altitude with $\triangle{DCE}$ and also $\triangle{AEB}$ shares an altitude with $\triangle{ACE}$. Thus,
\[ \frac{[BED]}{[EDC]} = \frac{BE}{EC} = \frac{[AEB]}{[ACE]}  \]
Substituting,
\[ \frac{10+[BED]}{10} = \frac{[BED]}{6}  \]
\[ 60 = 4[BED]  \]
\[ 15 = [BED]  \]

The area of the quadrilateral is then $15 + 3 = \boxed{18}$.[/hide]

Could someone verify this?  I don't have the answer, so I just want to be sure.

[quote="SnowStorm"]Could someone verify this?  I don't have the answer, so I just want to be sure.[/quote]
[hide="I get the same answer"]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=632[/img]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=630[/img]
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=631[/img]
\begin{eqnarray*} u+3 &=& v \\ 7u &=& 3v+21 \end{eqnarray*}
Solving these two simultaneous equations:
\begin{eqnarray*} u &=& 15/2 \\ v &=& 21/2 \\ u+v &=& 18 \end{eqnarray*}
[/hide]

During the competition I solved it this way:

[hide]Using SnowStorm's diagram, connect $BF$. Let the area of $\triangle {BEF}$ be $X$ square units and let the area of $\triangle {BFD}$ be $Y$ units.

Now $\triangle {DFA}$ and $\triangle {FCA}$ both share the same altitude, so the ratio of their bases is the same as the ratio of their bases.

Thus $DF = FC = x$ for some positive integer $x$. Similarly we have $EF = 3y$ and $FA = 7y$ for some positive integer $y$.

$\triangle {BCF}$ and $\triangle {BFD}$ share the same altitude so the ratio of their areas is the ratio of their bases. Therefore $\frac{X + 3}{Y} = \frac{x}{x} = 1 \Rightarrow Y = X + 3$.

$\triangle {BFA}$ and $\triangle {BEF}$ also share the same altitude, so the ratio of their areas is also the ratio of their bases. So $\frac{X}{Y + 7} = \frac{3y}{7y} = \frac{3}{7} \Rightarrow 7X = 3Y + 21$.

Substituting $Y = X + 3$ in the last equation, $7X = 3(X + 3) + 21$, $4X = 30$, $X = 7.5$.

So the area of quadrilateral $BEFD = X + Y = X + (X + 3) = 2X + 3 = 2(7.5) + 3 = 15 + 3 = 18$,

hence $(D)$.[/hide]

Lol it's exactly the same is rcv's solution... I didn't see his.  :blush:

[quote="aidan"]During the competition I solved it this way:

[hide]
Now $\triangle {DFA}$ and $\triangle {FCA}$ both share the same altitude, so the ratio of their bases is the same as the ratio of their bases.
[/hide]
[/quote]

Typo? lol  :P

[quote="SnowStorm"]This was a USAMTS problem a few years back.[/quote]

Also see [url=http://www.kalva.demon.co.uk/aime/aime85.html]AIME 1985, Problem 6[/url], which I believe was the inspiration for that USAMTS problem. ;)

a really easy solution lies in mass points chess64  :)

[hide]
Let the biggest triangle be $\triangle ABC$, going around clockwise with $A$ in the lower-left hand corner. Let the important points on $\overline{AB}$ and $\overline{BC}$ be $D$ and $E$ respectively, and let the intersection of $\overline{AE}$ and $\overline{CD}$ be $X$. Last, let $AB=c$, $AC=b$, $BC=a$, $AD=d$, $CE=e$, and let the area of polygon $PQR$ be $[PQR]$ (for any number of points $P,Q,R$).

Consider $\triangle ACX$ and $\triangle XCE$. With respect to $\overline{AE}$, they have the same height and area, so we have $AX=XE$. Drawing in $\overline{DE}$, we also have $[DEX]=[ADX]=3$. Now, using $\triangle AEC$ and $\triangle ABC$ we have $\frac ae=\frac{[ABC]}{[AEC]}=\frac{[ABC]}{14}$. From $\triangle ADC$ and $\triangle ADE$, we also have $\frac{[ADE]}{[ADC]}=\frac{6}{10}=\frac{a-e}a=1-\frac ea=1-\frac{14}{[ABC]}$, so $\frac{14}{[ABC]}=1-\frac6{10}=\frac25$, which means that $[ABC]=14\frac52=35$. Finally, $[DBEX]=[ABC]-[ADX]-[AXC]-[XEC]=35-3-7-7=\fbox{18}$.[/hide]

Spreading the joy of mass points  :D

[hide]Draw triangle $ABC$ with cevians $CD$ and $BE$. Let the intersection point of this cevian be $P$. Assign the following mass points:

$B_m = 1$

Since $[BCP] = [ECP]$, $EB$ is bisected by cevian $CD$. So,

$E_m = 1$
$P_m = B_m + E_m = 2$

We try to use the ratio of $DP$ to $PC$ by comparing the areas of $BDP$ and $CPB$. Let the mass point of $D$ be $7x$ and the mass point of $C$ be $3x$. Since $7x + 3x = 2$ because of point $P$, we have $x = \frac{1}{5}$. So, the mass points of:

$C_m = \frac{3}{5}$
$D_m = \frac{2}{5}$

We will now draw cevian $AF$ which goes through point $P$ so as to split the quadrilateral into two triangles. The mass point of $F$ is $B_m + C_m = \frac{8}{5}$. As a result, the mass point of $A$ is $F_m - P_m = \frac{2}{5}$. The area of $ADP$ is solved by:

$1 * 3 = \frac{2}{5} * [ADP]$

and $AEP$ by:

$\frac{3}{5} * 7 = \frac{2}{5} * [AEP]$

Solving for them, we get the areas to be $\frac{15}{2}$ and $\frac{21}{2}$, respectively. By adding, we get:

$\boxed{18 \implies D}$[/hide]

can someone explain how rcv made the relationships he did about the areas of the various triangles? there's probably some theorem I don't know.

It's all of the same-base, same-altitude stuff from AoPS. Basically, just give variables to a few of the lengths that matter (as bases) and the altitudes, then write the equations for areas and you'll see the relationship.

[quote="crptone"]Can someone post a solution to this problem? It was the only problem that I missed :([/quote]

Me too.

I did it using rcv's method at home in about 3 mins. If only I had seen it earlier.

Mod Edit: Please keep this kind of stuff to PM.

It's when a moderator edits another person's post. The previous poster quoted nearly the entire thread to say something that should've been in  a PM, therefore his post got deleted.


Each double desk is for 2 students. There are 30 students and 15 desks in a class. Given that half of the girls sit with the boys. Is it possible to rearrange the sitting plan so that half of the boys sit with the girls?

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Since [i]half[/i] the girls are sitting with boys, there must be an even number of girls. There must be either $ 4n$ or $ 4n \plus{} 2$ girls. Let there be $ 4n \plus{} 2$ girls. Then $ 2n \plus{} 1$ girls sit with the boys, and $ 2n \plus{} 1$ girls sit together. However, it's impossible for $ 2n \plus{} 1$ girls to sit in pairs. Then there are $ 4n$ girls. Similarly, there must be $ 4m$ boys for it to be possible for the rearrangement to be possible. However, there are $ 30 \minus{} 4n$ boys, which is congruent to $ 2\mod 4$. Therefore, it is impossible to rearrange it so that half the boys are sitting with girls.
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There are $ n$ people and there are $ 6$ comissions. Each comission has at least $ n/4$ members. Prove that there are two comissions and a group at least $ n/30$ members belonging to this two comissions.

It has been post before and the solution is:
Suppose that any two comissions has less than $ \frac{n}{30}$ common members.By using the principle of inclusion and exclusion we have ; $ 6 \frac n4\minus{}\binom{6}{2} \frac {n}{30}\plus{}...>n$ and it's contradiction.


$f: [0,1]\rightarrow\mathbb{R}$ is continous.and:
1)$\forall{x}\in[0,1], f(x)\geq0$
2)$f(1)=1$
3)if $x,y,x+y\in[0,1]$ then $f(x)+f(y)\leq{f(x+y)}$
find the minimum c such that if $x\in[0,1]$ then $f(x)\leq{cx}$

c=2,Prove the following lemma by indution:
$\forall n,\forall x\in[\frac{1}{2^{n}},\frac{1}{2^{m-1}}],we have f(x)\leq 2x$( for n=1,it is triavial)
for c<2,it not true ,just let $f(x)-\to \frac12,x\geq \frac12,f(x)=0,x<\frac12$

[quote="Hawk Tiger"]for c<2,it not true ,just let $f(x)-\to \frac12,x\geq \frac12,f(x)=0,x<\frac12$[/quote]
I dont understand this step.

[quote="weiquan"][quote="ali666"][quote="Hawk Tiger"]Sorry,I am wrong.
It should be $f(x)=1,x>\frac12,f(x)=0,x\leq \frac12$
I have a bad memory...[/quote]
f shoud be continous.[/quote]

I think he means $f(x)=0 \,\forall\, x\leq \frac{1}{2}-\epsilon$, $f(x)=1 \,\forall\, x\geq \frac{1}{2}+\epsilon$ and $f(x)=\frac{1}{2}+\frac{1}{2\epsilon}\left(x-\frac{1}{2}\right) \,\forall\, \frac{1}{2}-\epsilon\leq x\leq \frac{1}{2}+\epsilon$ for some arbitrarily small $\epsilon$[/quote]
I don't mean that(what I mean is wrong :D ).But your example is vailid.


geometry
circumcircle
modular arithmetic
geometric transformation
homothety
geometry proposed

Dear everyone.

prove the following problem.

if from any point O on the circumcircle of triangle ABC, straight lines OA', OB', OC' be inflected to meet
BC, CA, AB in A', B', C' respectively, and to make with them equal angles,
then the three points A', B', C' are collinear.

I think it is easy but not bad.
thanks.

[quote="77ant"]
[b]and to make with them equal angles[/b]
[/quote]

I cant understand what you mean...?
Can you tell it more clearly???

Thank you
Dimitris

The normal projections of O on the sides determine the [b] Simson Line[/b]; if, instead of perpendicular we draw lines making the same $ \angle \alpha$ with the sides, the property is still valid. The proof is quite similar as for Simson line.

Best regards,
sunken rock

We present two proofs for the problem.

[b]First solution.[/b]
Using directed angles, we have:
$ (A'B',A'C')\equiv (A'B',A'O) \plus{} (A'O,A'C')$
$ \equiv (CA,CO) \plus{} (BO,BA)$
$ \equiv (CA,CO) \plus{} (OC,OB) \plus{} (BO,BA) \plus{} (OB,OC)$
$ \equiv (AC,AB) \plus{} (OB,OC)$
$ \equiv (AC,AB) \plus{} (AB,AC)$
$ \equiv 0 \pmod{\pi}$
Therefore $ A',B',C'$ are collinear, QED.

[b]Second solution.[/b]
Using the fact about Simson's line. Let $ A_0,B_0,C_0$ be the orthogonal projection of $ O$ on $ BC,CA,AB$ then $ A_0,B_0,C_0$ are collinear (Simson's line). Since $ (OA',BC) \equiv (OB',CA) \equiv (OC',AB) \pmod{\pi}$, the three triangles $ \triangle OA_0A',\triangle OB_0B',\triangle OC_0C'$ are similar in the same direction. Thus, consider the similar $ \mathcal{S}$ takes $ A_0\mapsto A',B_0\mapsto B',C_0\mapsto C'$, we have the desired result.

With the second solution, we can still point out the line which is the image of line $ \overline{A'B'C'}$ through the homothety $ \mathcal{H}(O,k \equal{} 2)$ passes through $ H\equiv \mathcal{S}(H)$ where $ H$ is the orthocenter of $ \triangle ABC$. Moreover, $ H'$ is the point satisfies that $ (AH',BC)\equiv (BH',CA)\equiv (CH',AB) \pmod{\pi}$. This is an extension of Steiner's line.