I'm considering doing a science fair project on natural selection in humans, but from the quick research that I've done in the past few days, it seems that there isn't a good way to study such a subtle thing without gene analysis and such. I don't really have the time or resources for something like that. Does anyone here have any ideas?

Natural selection in humans isn't documented often. As a species, we've evolved to the point where we are dependent of natural selection and Darwinian evolution. For your science fair, however, you could write something on why the human population doesn't follow the "natural selection trend."

it can b performed--especially for genetically affected populations

i've heard my teacher talking abt it..........

[quote="rituraj007"]it can b performed--especially for genetically affected populations[/quote]

It's not possible; to study an isolated population, with no genetic flow, is quite difficult. To do this, you must find an isolated population from the world, with no technological benefits. This most likely points to Africa. However, you must also find a large population, within the ten to hundreds of thousands, to eliminate genetic drift. I doubt that you can find a hundred thousand people in a tribe that doesn't migrate in Africa.

Well, not entirely possible. But some isolated countries up north (maybe Iceland) have a fairly limited/controlled population with minimum genetic variation.
Some people have already researched there, but records are hard to find.

I have this other question (sorry if it's obvious): Is evolution constant? like does it happen at a constant "rate" or does it depend on the need for change?

i wud say it depends on need for change.....

but the thing is a change is required :D

if we see evolution takes plac only when need 4 change is req,that means it wants to become stable
 and if the system becomes stable,well u can think what can happen if body metabolism comes in equilibrium

i hope u understand wat i mean  :)

[quote="oasisxyz"]I have this other question (sorry if it's obvious): Is evolution constant? like does it happen at a constant "rate" or does it depend on the need for change?[/quote]

There is no "need" but evolution is not constant. Evolution is the result of [i]random[/i] factors that influence natural selection. If a drought doesn't hit a population of plants in a long time, that's a result of randomness, [i]not[/i] because the Earth itself deems those plants "perfect."

hmm...
i don't know much about evolution (but i do know the names of the bones and muscles :D (thank you science teacher!))

anyway, from what i know, evolution is a result of the organism adapting to its environment
whatever happens in the environment, the organism has to adapt to the change in order to survive
however, evolution can only happen through the generations, which means although one organism can slightly adapt, it doesn't mean it's offspring will do well in the change
with that said, i would think as the environment changes, the organism does too, since no organism is completely identical to its parent
although it's very slowly, i think evolution is a constant process

however, as i said before, i do not know much about evolution
my science teacher taught us a little, but not enough for me to give you an accurate answer (yes, it's the same teacher that taught us the names of the muscles and bones)
anyway, i might be completely wrong, but that is what i think

One known example of evolutionary pressures effecting a human genetic trait - and it's a fairly cruel example - is the sickle cell trait. This trait is a variant of the gene for hemoglobin; I think it differs from "normal" hemoglobin by a single amino acid. Sickle-cell hemoglobin molecules are much more likely to clump together than normal hemoglobin.

A person who is heterozygous for the sickle cell trait has a mixture of normal and sickle cell hemoglobin. Under nearly all normal circumstances, the clumping never happens, and the person lives a normal lifespan in normal health. The one exception happens when the person is infected with malaria. Under the stress of the malaria parasite, the hemoglobin sticks together, disrupting and destroying red blood cells - preferentially infected red blood cells. Compared to a person with only normal hemoglobin, this person is less likely to die of malaria and more likely to be less debilitated and to recover sooner. In an environment in which malaria is endemic, being heterozygous for the sickle cell trait is an evolutionary advantage.

But here comes the cruel part. A person who is homozygous for the sickle cell trait has only the variant hemoglobin. That person's hemoglobin tends to clump together under mild and commonplace stresses, and to cause severe health crises. This person has the disease called sickle cell anemia. In the absence of modern medicine, this person will probably not live to adulthood and will probably not have children.

In the absence of malaria, the early deaths of those homozygous for the trait would cause the sickle cell gene to gradually disappear from the gene pool (although it would take many generations.) But in the presence of endemic malaria, the advantage shared by those who are heterozygous for the trait sets up an equilibrium favoring the continuation of the trait at some level within the population.

In the last 500 years or so, human populations have been pretty thoroughly mixed across the globe, but there was a time when humans were less mobile. The sickle cell trait is found in people who trace their ancestry to regions of the world in which malaria has long been endemic - most commonly lowland tropical Africa. (There are some other, similar, hemoglobin variants called thallasemias that occur in people with ancestries in other endemic malaria locales.)

[quote="Kent Merryfield"]One known example of evolutionary pressures effecting a human genetic trait - and it's a fairly cruel example - is the sickle cell trait. This trait is a variant of the gene for hemoglobin; I think it differs from "normal" hemoglobin by a single amino acid. Sickle-cell hemoglobin molecules are much more likely to clump together than normal hemoglobin.[/quote]

Yes, this is the most common textbook example of [b]heterozygous advantage[/b].

You can research this by doing an experimenting of your own, similar to the AP Biology Population Genetics lab. With proper variables and limits, you could simulate a large population with the sickle-cell anemia (for your science fair). However, sickle-cell anemia was only recognized for it's heteroyzgous advantage, and little real population studies have been done without interference (medicine, technology, etc.)

Thanks for the ideas   :)
It's a very open question...still thinking...


linear algebra
matrix
linear algebra unsolved

Let $M$ be a set of matrix $A_{3,3}$ that has elements from ${ { 1, 2,3,4,5,6,7,8,9} }$. Prove the fallowing:
1) If A is From M, prove that $-a\leq det(A)\leq a$, a from R (and find "a")
3) Show that, if B is from M then $B^{-1}$ is $not$ from M
3) Show that, if D is from M then ranck(A) belongs to ${2,3}$
4) What is the number of elements of M
5) Show that M has at least 18 matrix with det(x)=0
6) If A is from M, what can you say about the elements of $A^{2006}$ ? (Do they belong to {1,2,3,4,5,6,7,8,9}? )
7) If A is from M, and a=det(A), is "a" divisible with a certain number ?
8) If A is from M, is det(A) from $Z$  ?


calculus
function
geometry
similar triangles
calculus computations

I don't really know where this question belongs because all of the other questions here appear much more complicated so I guess it should be moved wherever necessary

A swimming pool is 15ft wide, 40 ft long and 3 feet deep atone end, and 10ft deep at the other. Wateris being added to it at a rate of 25ft^3/min
a)How fast is the water level rising when there is 2000ft^3 of water in the pool?
b)Same as above but when there is 3000ft^3 of water

so dx/dt=25ft^3/min and x=2000 and 3000
Now I have no clue what to do, I got a hint stating to use similar triangles so i drew a 40 by 7 triangle but i really dont understand what this does for me
Can somebody guide me through this?

This sort of question goes in the "Calculus Computations and Tutorials" subforum. [Edit- it was moved while I was composing this post]

The difficulty here comes from the fact that we are given volume rather than depth- you'll need to find the depth in terms of the volume.
This is a piecewise defined function- we have one formula for less than $ 2100$ cubic feet and a bottom that isn't fully covered, and another for more than $ 2100$ cubic feet and a bottom that is covered.


Let $a\ge b>0$ such that $b\in [\frac{1}{a^{2}},\frac{1}{\sqrt a}]$. Prove that $a^{2}+b^{2}-1\ge ab(a+b-1)$

If $a \geq 1$, then $b \leq max\{ \frac{1}{\sqrt{a}},\frac{1}{a^{2}}\}= \frac{1}{\sqrt{a}}\leq 1$.
If $a < 1$, then $b \geq min \{ \frac{1}{\sqrt{a}},\frac{1}{a^{2}}\}= \frac{1}{\sqrt{a}}\geq 1$.
For both cases, we obtain $(a-1)(b-1) \geq 0$.
Since $a+b+1 > 0$, then $(a-1)(b-1)(a+b+1) \geq 0 \Rightarrow a^{2}+b^{2}-1 \geq ab(a+b-1)$ as desired.  :)


Incidentally you get the handy little formula out of that:

$\displaystyle \tan^{-1}a + \tan^{-1}b=\tan^{-1}{\frac{a+b}{1-ab}}$

thanks alot for that! really appreciated!


Positive real numbers $a,b,c$ satisfy the relation $abc=1$. Prove the inequality:
$\frac{a+3}{(a+1)^{2}}+\frac{b+3}{(b+1)^{2}}+\frac{c+3}{(c+1)^{2}}\geq3$.

Nice, I just checked the 2 identities you wrote. They are useful, although I didn't use them. Thx. Actually in the second identity $\geq$ should be replaced by $=$. :)

$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq\frac{1}{1+ab}$

what's the quickest way you know of proving that? ...not using the "dumb-***" (multiplying everything out) method...

they all still use the "dummy" method i was talking about... multiplying stuff out and completing squares... i was hoping for a smooth, cauchy-schwartz type proof, hehe.

thanks though!

Another method. Use:
 $\sum \frac 1 {1+a}+\sum \frac 1 {(1+a)^2}\ge \sum \frac 1 {(1+a)}+\frac 1 2 \sum \frac 1 {1+a^2}$
And renoramlizing ($a=\frac x y...$)we arrive to a well known ineq : $\sum \frac x {x+y}\ge \frac 3 2$. Which can be solved with Cauchy, twice.

[quote="yookon88"][quote="silouan"]
$\sum\frac{1}{a+1}+\sum\frac{1}{1+ab}\geq 3$  which is an identity using the fact that $abc=1$

I think it is nice ;[/quote]

I don't understand this part[/quote]
Actually, equality holds.
$\sum \frac{1}{1+a}+\sum\frac{1}{1+ab}=\sum\frac{1}{1+a}+\sum\frac{1}{1+\frac{1}{c}}=$
$\sum(\frac{1}{1+a}+\frac{a}{a+1})=3$

[quote="yookon88"][quote="silouan"]
$\sum\frac{1}{a+1}+\sum\frac{1}{1+ab}\geq 3$  which is an identity using the fact that $abc=1$

I think it is nice ;[/quote]

I don't understand this part[/quote]

Very easy ! they must be equal find this and the problem is finised 1/1+ab = c/1+c hence abc=1.
Then summing up the terms we are done ...

Davron

The initial problem is old and was discussed in a more general form at http://www.mathlinks.ro/Forum/viewtopic.php?t=39549 .

For the inequality

$\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}\geq\frac{1}{1+xy}$

for nonnegative reals x and y, in fact I have always been dumb455ing this one, but now, being asked, I see a quite neat proof:

Cauchy-Schwarz yields

$\left(y+x\right)\left(1+xy\right)\geq\left(\sqrt{y\cdot 1}+\sqrt{x\cdot xy}\right)^2=\left(\sqrt{y}+x\sqrt{y}\right)^2=y\left(1+x\right)^2$,

so that $\frac{1}{\left(1+x\right)^2}\geq\frac{y}{y+x}\cdot\frac{1}{1+xy}$, and similarly $\frac{1}{\left(1+y\right)^2}\geq\frac{x}{x+y}\cdot\frac{1}{1+xy}$. Adding these two inequalities, we get

$\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}\geq\frac{y}{y+x}\cdot\frac{1}{1+xy}+\frac{x}{x+y}\cdot\frac{1}{1+xy}$
$=\left(\frac{y}{y+x}+\frac{x}{x+y}\right)\cdot\frac{1}{1+xy}=\frac{1}{1+xy}$,

qed..

@Xirti: Nope, $\sum \frac x {x+y}\ge \frac 3 2$ is not a correct inequality...

  Darij

[quote="xirti"]Of course :D  I nedeed $\le$ instead of $\ge$. Now is in the corect form.[/quote]

$\frac{4}{4+2}+\frac{2}{2+1}+\frac{1}{1+4}>\frac32$.

Next attempt?

  darij

i knew cauchy schwartz could do it! very nice!!

I nedeed $\le$ somewhere else. This is the final attempt. Sorry for mistakes.

[quote="silouan"]Quite well-known .
 The best solution i think is the following
$\frac{a+3}{(a+1)^2}+\frac{b+3}{(b+1)^2}+\frac{c+3}{(c+1)^2} \ge 3$

<=> $\sum\frac{1}{a+1}+2\sum\frac{1}{(1+a)^2}\geq 3$.

Now use the well-known ineq $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}\geq\frac{1}{1+ab}$  . So it suffices to prove that

$\sum\frac{1}{a+1}+\sum\frac{1}{1+ab}\geq 3$  which is an identity using the fact that $abc=1$

I think it is nice ;[/quote]
hoho, $L.H.S. \ge \sum\frac{1}{a+1} + \sum\frac{1}{1+ab} = \sum\frac{1}{1+a} + \sum\frac{a}{1+a} = 3$

[quote=Dgurschi]Is it possible to solve this inequality with Jensen?[/quote]
Probably not. Or at least if it exists then it's not a straightforward Jensen.

Let $a=x/y$ cyc. Then the LHS becomes
$$\sum_{cyc}\frac{\left(x+3y\right)y}{\left(x+y\right)^2}=\sum_{cyc}\frac{y}{x+y}+\frac{2y^2}{(x+y)^2}$$
It's easy to prove $\sum_{cyc}\frac{y}{x+y}\ge 3/2$. Now $$\sum_{cyc}\frac{2y^2}{(x+y)^2}\ge 2\frac{(x+y+z)^2}{\sum_{cyc}(x+y)^2} =\frac{(x+y+z)^2}{\sum_{cyc}x^2+xy}$$
$$\ge \frac{(x+y+z)^2}{\sum_{cyc}x^2}\ge \frac{3}{2}$$
Adding them up we get the desired inequality.


The altitude drawn to the base of an isosceles triangle is $ 8$, and the perimeter $ 32$. The area of the triangle is:

$ \textbf{(A)}\ 56\qquad
\textbf{(B)}\ 48\qquad
\textbf{(C)}\ 40\qquad
\textbf{(D)}\ 32\qquad
\textbf{(E)}\ 24$

Let $ a$ be the length of one of two congruent sides and $ b$ be the length of the base, such that $ P\equal{}2a\plus{}b\equal{}32\implies \dfrac{b}{2}\equal{}16\minus{}a$.

Using the pythagorean theorem we see that: $ a^2\equal{}8^2\plus{}\left(\dfrac{b}{2}\right)^2\implies a^2 \equal{} 64 \plus{} \left(16\minus{}a\right)^2 \implies 32a\equal{}64\plus{}256\implies a\equal{}10$.

Now that we know $ a$ the value of $ b$ is $ 32\minus{}2a \equal{} 32\minus{}20 \equal{} 12$.

Hence the area of the triangle is $ \dfrac{8\cdot 12}{2} \equal{} \boxed{48} \implies \boxed{\text{ Ans. } \mathbf{(B)}}$


function
limit
integration
topology
real analysis
real analysis unsolved

$ f$ is a continuous function defined on $ [0,\plus{}\infty)$.
For $ \forall \,\alpha >0$, $ \lim_{n\to \plus{}\infty} f(n\alpha)\equal{}0$.
Prove: $ \lim_{x\to \plus{}\infty} f(x)\equal{}0$.

Thanks

Naive reply (which means I am probably missing something): take $ \alpha\equal{}1$?

Moreover, I'm able to construct a function $ f\in C([0,\infty))$ s.t. $ f(x)\not\to 0$, $ x\to\infty$, but $ f(n\alpha)\to 0$, $ n\to\infty$ for almost all $ \alpha$.

Maybe it need that six equivalent propositions of continuity of real numbers.
The six equivalent propositons are so abstract. How do you master it ?

Thanks

So how I construct an example

First, note that it is that the statement is equivalent to the one for $ [1,\infty)$. Next, by considering $ f(1/x)$, we are to show that for $ f\in C((0,1])$ one has $ f(x)\to 0$ whenever $ f(x/n)\to 0, n\to\infty$ for all $ n$.

Now take an open set $ A\subset (0,1]$ with $ \lambda(A\cap (0,1/n)) \equal{} O(1/n^3)$, when $ n\to \infty$. (Something like $ \bigcup(1/k,1/k \plus{} 1/k^4)$.

Define $ A_n \equal{} nA\cap [0,1] \equal{} \{x: x/n\in A\}$. We have $ \lambda(A_n) \equal{} O(1/n^2)$, so by Borel-Cantelli lemma almost all $ x$ belong to only finite number of $ A_n$. That means that a continuos function, which equals $ 1$ in a midpoint of each interval $ A$ consists of and $ 0$ outside $ A$, is a good example of a function I spoke of.

Though I don't know how to construct a counterexample to the initial problem (or how to prove it). It might seem that the proof should use Baire categories if it exists, but then why it should not work for $ e^n$? That's why I think there is no proof, though I'm not sure at all. Extremely difficult problem indeed.

Probably, it is not so hard and probably it can be proved by considering the expression
\[ \int_0^1 \sum_{k\equal{}1}^\infty \frac1{nx}1_{f(nx)\ge a}dx\]

[quote="zhoraster"]Yes, [b]BlueVelvet[/b] you are missing something. Here $ n$ means a natural number. The problem seems to me very tough. Say, if $ f(e^n x)\to 0, n\to\infty$ for all $ x$, then we cannot conclude that $ f(x)\to 0$, $ x\to\infty$.[/quote]

$ f(e^n x)$ is not satisfied with assumption.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70016]Link to a previous discussion.[/url]

A proof can also be found in [url=http://books.google.nl/books?id=hGYficzfWyQC&pg=PT187&lpg=PT187&dq=Crofts+Lemma&source=bl&ots=TtWK1Ifn8w&sig=T7fGvMYnoAY7NKPB7OdPHc85phg&hl=nl&ei=a7nZSpLkDZHH-Qbf4Oi7DQ&sa=X&oi=book_result&ct=result&resnum=10&ved=0CEcQ6AEwCQ#v=onepage&q=&f=false]this book[/url].

I have no experience with it, but maybe [url=http://www.codeplex.com/GoogleBookDownloader]this[/url] will work. But I hope you can read the page I linked to?


A variant of another inequality that I had posted.

For a natural number $ n$ show that $ \frac{1}{n}\plus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}....\plus{}\frac{1}{3n\plus{}1}>1$.

[quote="manjil"]A variant of another inequality that I had posted.

For a natural number $ n$ show that $ \frac {1}{n} \plus{} \frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 2} \plus{} .... \plus{} \frac {1}{3n \plus{} 1} > 1$.[/quote]
The following inequality is true too.
For all $ n\in\mathbb N$ prove that:
\[ \frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 3} \plus{} ... \plus{} \frac {1}{3n \minus{} 1}\geq\frac {1}{2}
\]
The denominators constitute an arithmetic progression.

[quote="arqady"]The denominators constitute an arithmetic progression.[/quote]
Is that the general clue to construct such ineq.s, arqady?
 :)


A quadrangle $ABCD$ is inscribed in the circle with the radius $\frac{65}{8}.$ If the perimeter of the quadrangle is $44,$ and the side length of $BC$ and $CD$ is $13,$ then find the lengths of $AB$ and $DA.$

[hide="A start"]Drawing the radii and altitudes to sides BC and CD gives 3-4-5 triangles, and the distance from the center to each of those sides is $\frac{39}{8}$.  Maybe trig would be useful?[/hide]

this has always puzzle me, what is a quadrangle and what are its differences from a quadrilateral??? I only know that the duel of a quadrilateral is a quadrangle... If possible, can someone also provide a picture??

[hide]
Using Brahmagupta's theorem:
\[ \frac{65}{8}\cdot 22 = 9\sqrt{(22-AB)(22-DA)}  \]
since we have that $AB + BC +CD +DA = 44$ we know that $AB + DA = 44 - 26 = 18$
now we can substitue that information back to the first equation:
\[ \frac{65\cdot11}{4} = 9 \sqrt{22^2 - 22(AB + DA) + (AB)(DA)} = 9\sqrt{22^2-22(18) + (AB)(DA)  }\]
and then we make the substitution that $AB = 18 - DA$ and solve...which I am too tired to do. So I'll either solve this tomorrow or some else can finish up :P[/hide]


number theory
prime numbers
number theory unsolved

Let $ p_{i}$ be consecutive prime numbers for $ i\equal{}1,2,...$

Prove that:
$ \sum_{i\equal{}1}^{\infty}\frac{1}{p_{i}}\equal{}\infty$

Hmm, I was hoping that there was a "prettier" solution for this problem, but I must admit it's quite beautiful. Anyway, thank you  :wink:


symmetry
combinatorics proposed
combinatorics

There are much water and much pure alcohol. There are also two buckets: one of capacity 3 litres and one of capacity 5 litres. How can we obtain 1 litre of vodka (vodka contains 40% pure alcohol and 60% water)?

This problem is not very difficult, but it's cool, in my opinion.

This sort of problem appears everywhere! I've even seen it used in movies, like "Die hard" :).

Yes, but this problem little more difficult than usual problems of this sort.

And I still don't know whether we can obtain 1 litre of 50% solution.

[quote="cm"]r we allowed to do it infinitely?[/quote]
Of course, no! We have to obtain it by finite number of steps.

This is how you can get 1 liter of 50% solution:

[hide="click here to show solution"]
Fill the 5 L bucket with water.
Put the empty 3 L bucket gently into the 5 L one so that 3 L spill over the top.
Remove the 3 L bucket and pour the remaining 2 L of water into it from the 5 L bucket.
Fill the 5 L bucket with alcohol.
Put the 3 L bucket into the 5 L bucket gently so that 3 L of alcohol spill over the top of the 5 L bucket, be careful not to spill anything into the 3 L bucket!
Remove the 3 L bucket. Pour the 2 L of water from it into the large bucket.
The large bucket now contains 4 L of 50% solution.
Pour 3 L of it into the small bucket. The 5 L bucket now contains 1 L of 50 % solution.
[/hide]

[quote="JanSiwanowicz"]This is how you can get 1 liter of 50% solution:

Fill the 5 L bucket with water.
Put the empty 3 L bucket gently into the 5 L one so that 3 L spill over the top.
Remove the 3 L bucket and pour the remaining 2 L of water into it from the 5 L bucket.
Fill the 5 L bucket with alcohol.
Put the 3 L bucket into the 5 L bucket gently so that 3 L of alcohol spill over the top of the 5 L bucket, be careful not to spill anything into the 3 L bucket!
Remove the 3 L bucket. Pour the 2 L of water from it into the large bucket.
The large bucket now contains 4 L of 50% solution.
Pour 3 L of it into the small bucket. The 5 L bucket now contains 1 L of 50 % solution.[/quote]

You have only one 3 L bucket! If you have 2 L of water in it, you can't delete 3 L of pure alcohol from 5 L bucket preserving 2 L of water!

The problem with 50% solution is hard! And I even don't know the answer!

Hehe..yes you can, but you have to use the outside of the bucket not the inside :)

If you lower a small bucket with 2 L of water in it INTO the big bucket, the inside of the small bucket will not be affected. As you are lowering the small bucket INTO the large bucket, the large bucket's contents will be displaced and will spill over the side of the big bucket and onto the floor, but not into the small bucket because the rim of the small bucket is still above the rim of the large bucket.

Read my solution again, carefully :)

Think outside the bucket...I mean box...

[quote="JanSiwanowicz"]If you lower a small bucket with 2 L of water in it INTO the big bucket, the inside of the small bucket will not be affected. As you are lowering the small bucket INTO the large bucket, the large bucket's contents will be displaced and will spill over the side of the big bucket and onto the floor, but not into the small bucket because the rim of the small bucket is still above the rim of the large bucket.[/quote]

That is really cool! :D I do wonder whether it is a "legal" move, however. ;)

JanSiwanowicz, I think you also forgot that buckets also have a volume...small, but still....So by placing a bucket of 2L inside one of 3L, you will displace a little more that 2L ;)
I also remember a similar question about buckets and so on, and the oficial solution said that you just take a bucket of 2L, which was a perfect cilinder, and spill it, so that it is exactly filled by half.... I really don't know if this is possible.... :oops:   :?:

No, I did not forget, I was just ignoring it. :)

I was using mathematical buckets that have no volume :)

As for the tipping over, I know what problem you are referring to...that had barrels, not buckets. Barrels have point symmetry about their center, so you can tip them over so that the liquid level just reaches the top of the bottom flat surface of the barrel - then you have half the contents of the barrel remaining. However, buckets lack such symmetry so I did not use those types of moves.

Not sure if one CAN get to 1 L of 50% without shinanigans.

Does the source of the problem also have a solution he/she is willing to share?

Here's a solution to the original problem :) :

1) Fill the small bucket with ethanol.
2) Pour the ethanol in the small bucket into the large bucket. The small bucket is now empty, and there is 3L of ethanol in the large bucket.
3) Fill the small bucket with ethanol.
4) Fill the large bucket with ethanol from the small bucket. There is 1L of ethanol left in the small bucket.
5) Empty the large bucket.
6) Pour the ethanol in the small bucket into the large bucket. The small bucket is now empty, and there is 1L of ethanol in the large bucket.
7) Fill the small bucket with water.
8) Pour the water in the small bucket into the large bucket. The small bucket is now empty, and there is now 3L of water and 1L of ethanol in the large bucket.
9) Fill the small bucket with ethanol.
10) Fill the large bucket with ethanol from the small bucket. There is now 3L of water and 2L of ethanol in the large bucket, while there is 2L of ethanol left in the small bucket.
11) Fill the small bucket with vodka from the large bucket. There is 4L of vodka left in the large bucket.
12) Empty the small bucket.
13) Fill the small bucket with vodka from the large bucket. There is 1L of vodka left in the large bucket.


prove:$ x\minus{}y|x^{2n\plus{}1}\minus{}y^{2n\plus{}1} \forall n,x,y \in \mathbb{N}$

Not really Pre-Olympiad material.  I think the statement you want is $ x \plus{} y | x^{2n\plus{}1} \plus{} y^{2n\plus{}1}$, since $ x \minus{} y | x^k \minus{} y^k$ is true for all $ k$, not just odd $ k$.

I am in my school's library with nothing to do other than solving problems and exercises on Mathlinks, so I might as well clarify some things.  As t0rajir0u stated, you are probably asking for a proof that $ x\plus{}y|x^{2k\plus{}1}\plus{}y^{2k\plus{}1}$.  The other, easier, exercise is completed similarly.

[u]Proof.[/u] Consider the expression $ x^{2k\plus{}1}\plus{}y^{2k\plus{}1}$ modulo $ x\plus{}y$.  Since $ \minus{}x\equiv y\pmod{x\plus{}y}$, $ x^{2k\plus{}1}\plus{}y^{2k\plus{}1}\equiv x^{2k\plus{}1}\plus{}(\minus{}x)^{2k\plus{}1}\equiv 0\pmod {x\plus{}y}$, as required. $ \Box$

Additional Proof:

Consider $ f(x,y)\equal{}x^{2n\plus{}1}\plus{}y^{2m\plus{}1}$ and observe that $ f(x,\minus{}x)\equal{}0\forall x$ and thus $ f(x,y)\equal{}(x\plus{}y)g(x)$, where $ g(x)$ is a polynomial with integer coefficients.


Point M is taken on the diameter of a circle of radius R in such a way that the distance between point M and the center of the cirlce is [i]a[/i]. Find the sum of squares of the distances from point M to the endpoints of any chord parallel to this diameter.

let $ AB$ be a chord parallel to the given diameter, with $ A,B$ on the circle. let $ C$ be the midpoint of $ AB$ and let $ H$ be the foot of the perpendicular from $ M$ to $ AB$. note that $ AC^2 \equal{} R^2 \minus{} h^2$, where $ h$ is the length of $ MH$. then $ MA^2\plus{}MB^2 \equal{} h^2\plus{}AH^2\plus{}h^2\plus{}HB^2 \equal{} 2h^2\plus{}(AC\plus{}a)^2\plus{}(AC\minus{}a)^2 \equal{} 2h^2\plus{}2AC^2\plus{}2a^2 \equal{} 2h^2\plus{}2R^2\minus{}2h^2\plus{}2a^2 \equal{} 2(R^2\plus{}a^2)$.


[quote="stephen38"]$ \sum_{cyclic}{\frac{x^{2}\minus{}yz}{x^{2}\plus{}2yz}}\geq\sum_{cyclic}{\frac{x^{2}\minus{}yz}{x^{2}\plus{}y^{2}\plus{}z^{2}}}\geq0$ :D .[/quote]

how can you assume that all $ x^{2}\minus{}yz$'s are positive?

[quote="apollo"]$ \sum_{cyclic}{\frac{x^{2}\minus{}yz}{x^{2}\plus{}2yz}}$ $ \ge$ 0[/quote]
$ \sum_{cyclic}{\frac{x^{2}\minus{}yz}{x^{2}\plus{}2yz}}\geq0\Leftrightarrow\sum_{cyc}(x^{3}y^{3}\minus{}x^{2}y^{2}z^{2})\geq0.$ :D

[quote="Sung-yoon Kim"]

how can you assume that all $ x^{2}\minus{}yz$'s are positive?[/quote]

$ \frac{x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}xy\minus{}xz\minus{}yz}{x^{2}\plus{}y^{2}\plus{}z^{2}}\geq 0$
Sorry for my incomplete solution. :blush:  :D

[quote="stephen38"][quote="The soul of rock"][quote="stephen38"]$ \sum_{cyclic}{\frac{x^{2}\minus{}yz}{x^{2}\plus{}2yz}}\geq\sum_{cyclic}{\frac{x^{2}\minus{}yz}{x^{2}\plus{}y^{2}\plus{}z^{2}}}\geq0$ :D .[/quote]
This isn't true or you didn't finish it.[/quote]
Is it true now,soul of rock?????? :maybe:[/quote]

what i meant is that you can't deduce without proof that the first inequality holds. if you thought that $ \frac{x^{2}\minus{}yz}{x^{2}\plus{}2yz}\geq\frac{x^{2}\minus{}yz}{x^{2}\plus{}y^{2}\plus{}z^{2}}$, then you gotta notice that is's false when $ x^{2}\minus{}yz$ is negative.

[quote="Sung-yoon Kim"][/quote][quote="stephen38"][quote="The soul of rock"][quote="stephen38"]$ \sum_{cyclic}{\frac{x^{2}-yz}{x^{2}+2yz}}\geq\sum_{cyclic}{\frac{x^{2}-yz}{x^{2}+y^{2}+z^{2}}}\geq0$ :D .[/quote]
This isn't true or you didn't finish it.[/quote]
Is it true now,soul of rock?????? :maybe:[/quote][quote="Sung-yoon Kim"]

what i meant is that you can't deduce without proof that the first inequality holds. if you thought that $ \frac{x^{2}-yz}{x^{2}+2yz}\geq\frac{x^{2}-yz}{x^{2}+y^{2}+z^{2}}$, then you gotta notice that is's false when $ x^{2}-yz$ is negative.[/quote]

You're right :blush: .Anyway,thanks for your remark.


Prove or disprove:n>2
For any n positive reals we have:
$(2n-2)(\sum_{i=1}^n x_{i})(\sum_{i=1}^n x_{i}^{n-1})$ $\leq$ $n(n-1)(\sum_{i=1}^n x_{i}^{n})+(\sum_{i=1}^n x_{i}^{n-2})(\sum_{i=1}^n\sum_{j=1}^n x_{i}.x_{j})$
Sorry for my Latex technique! ;)

The last sum has [i]n[/i]^2 or [i]n[/i]([i]n[/i]-1) terms ?
In the last case, it must [i]j[/i]<>[i]i[/i], isn't it ?

Yes, I mean i<>j. For n=2,it is an equality;n=3 it is Schur inequality. But with n=4,this ineq have the opposite sign!... :maybe:
But now I can put another question:If we change the expression $\sum x{i}x{j}$ by a relevant $(\sum x{i})^2$ then is it correct?Because,the ineq is correct if we change $\sum x{i}x{j}$ to $\sum x{i}^2$ on the RHS.(I am sure!).


Phaloganes lived $ \frac{1}{5}$ of his life as a child, $ \frac{1}{4}$ as a boy and $ \frac{1}{3}$ as a working man. He then spent his final 13 years in retirement. When he died he was $ x$ years old. Find the sum of the digits of his age at death.

First, we add $ \frac{1}{5}$, $ \frac{1}{4}$, and $ \frac{1}{3}$ together.  That yields $ \frac{47}{60}$.  Therefore, $ \frac{13}{60}$ of his life must have been in retirement--hence $ 13$ years.  So Phaloganes lived $ \frac{60}{13} \cdot 13$ years--thus $ 60$ years.  The sum of his age's digits is $ 6 \plus{} 0 \equal{} \boxed{6}$.

$ \frac {1}{5} \plus{} \frac {1}{4} \plus{} \frac {1}{3} \equal{} \frac {47}{60}$

$ \frac {13}{60}x \equal{} 13 \Rightarrow x \equal{} 60$

Sum of digits is $ 6 \plus{} 0 \equal{} \boxed{6}$.

EDIT: Beaten, darn. :(


This is the same as:

$ (5-\sqrt{(25-9})^2 = (5 - 4)^1 = 1$

So the answer is 1.

[color=blue]Fixed $ \text{\LaTeX}$.

~Izzy

[/color]


see title. because of me we came in 6th instead of 2nd. then i came in 16th instead of 9th.

that team round was mean
MEAN

Wow, you people are all so much better than me and you still get 6th place team.
Next year, if I make the team, Mass will be like 57th.
Good job though, Carl, 16th is really really good.

wah wah wah. I'm carl. i win everything. i'm in the top 20 in the country and i'm stupid. wah wah wah. i wish i had proved fermats last theorem. i'm terrible at set(which he actually is). i complain hat i'm better than pretty much everyone, but not quite everyone. wah wah wah.

Yeah, I agree. Massachusetts really needs to get its act together.

Hopefully some people will come out of nowhere next year, b/c i don't see that much awesomeness right now. Were there any ther 7th graders in CD besides kevin? I can't remember...

There probably was one more 7th grader.  I know that I was 12th and was 3rd highest non-eighth grader.  So if not, that 7th grader was 11th.

I was 8th this year. Got a 35, messed up on the trapezoid problem on the target.
Oh, sorry, I thought 13375P34K43V312 was Carl.  :blush: :oops:
Can't see myself doing very well in Nats though.

Edit: Wait... how is that possible? Carl got 16th. Do you have alt accounts?

1337 is Carl, I was mocking his complaints. I was 7th this year, but second last year. I was the one who actually fell for their elementary trick and said 8 for the geometric mean thing. I think I may have faced u in CD-I can't remember if u won ur first round.

No I didn't. I hate CD. I didn't answer a single question, bet everyone was laughing at me.
Thankfully CD doesn't count.

That opinion would be different if you'd been 7th before CD, and 4th afterward...

Hi guys! What's up!? I just wanna know what's up!

(/annoying)

the humidity. not really. actually sorta. u and ur illinois...


A prime $ p$ is the sum of two integer squares if and only if $ p\equal{}2$ or $ p\equal{}4k\plus{}1$.

To prove this, I started with the right direction by assuming that $ a^{2}\plus{}b^{2}\equal{}p\equal{}4k\plus{}3$.  So then in $ \mathbb{Z}/4\mathbb{Z}$, this would mean that $ a^{2}\plus{}b^{2}\equal{}3$.  However, the squares in $ \mathbb{Z}/4\mathbb{Z}$ are $ 0^{2}\equal{}2^{2}\equal{}0$ and $ 1^{2}\equal{}3^{2}\equal{}1$ so this is impossible.  Therefore $ p\equal{}4k\plus{}1$.

However, if we assume that $ p\equal{}4k\plus{}1$, how do we show that $ p\equal{}a^{2}\plus{}b^{2}$ for some $ a,b\in\mathbb{Z}$?


Can someone show me how to generalize the sequence $1^2+2^2+3^2+4^2+5^2+...+n^2$ using finite differences? (I tried searching it, but I couldnt find anything)

[quote="surge"]$1^2=1$
$1^2+2^2=5$
$1^2+2^2+3^2=14$
$1^2+2^2+3^2+4^2=30$
$1^2+2^2+3^2+4^2+5^2=55$
$1^2+2^2...+6^2=91$

Constant is 2 @ Cubic.

Subsitute values into 5 equations, solve for coefficients.[/quote]

Don't you only need 4 equations for 4 variables?

I don't know what you mean by "finite differences" but there's a picture proof thing on the left which says the answer is $\frac{n(n+1)(2n+1)}{6}$.

The extra equation is there for moral support. Duh!


Find all functions $f: \mathbb N \to \mathbb N$ such that for every positive integer $n$,
\[ f(f(n)) + f(n) = 2n+2001 \quad \text{or} \quad 2n + 2002.\]

[quote="MJ GEO"]$ f$ from $ N$ to $ n$ and $ f(f(n)) \plus{} f(n) \equal{} 2n \plus{} 2001 or 2n \plus{} 2002$[/quote]

Suppose we have $ ax \plus{} b\le f(x)\le cx \plus{} d$ $ \forall x\in\mathbb N$. Then :

$ af(x) \plus{} b\le f(f(x))\le cf(x) \plus{} d$

$ (a \plus{} 1)f(x) \plus{} b\le f(f(x)) \plus{} f(x)\le (c \plus{} 1)f(x) \plus{} d$

Then, $ f(f(x)) \plus{} f(x)\le 2x \plus{} 2002$ $ \implies$ $ (a \plus{} 1)f(x) \plus{} b\le 2x \plus{} 2002$ $ \implies$ $ f(x)\le \frac {2}{a \plus{} 1}x \plus{} \frac {2002 \minus{} b}{a \plus{} 1}$

Same, $ f(f(x)) \plus{} f(x)\ge 2x \plus{} 2001$ $ \implies$ $ (c \plus{} 1)f(x) \plus{} d\ge 2x \plus{} 2001$ $ \implies$ $ f(x)\ge \frac {2}{c \plus{} 1}x \plus{} \frac {2001 \minus{} d}{c \plus{} 1}$

So $ ax \plus{} b\le f(x)\le cx \plus{} d$ $ \forall x$ $ \implies$ $ \frac {2}{c \plus{} 1}x \plus{} \frac {2001 \minus{} d}{c \plus{} 1}$ $ \le f(x)\le$ $ \frac {2}{a \plus{} 1}x \plus{} \frac {2002 \minus{} b}{a \plus{} 1}$ $ \forall x$

And since $ 1\le f(x)\le 2x \plus{} 2001$, we can build four sequences $ a_n,b_n,c_n$ and $ d_n$ :
$ a_1 \equal{} 0$, $ b_1 \equal{} 1$, $ c_1 \equal{} 2$ and $ d_1 \equal{} 2001$

$ a_{n \plus{} 1} \equal{} \frac {2}{c_n \plus{} 1}$

$ b_{n \plus{} 1} \equal{} \frac {2001 \minus{} d_n}{c_n \plus{} 1}$

$ c_{n \plus{} 1} \equal{} \frac {2}{a_n \plus{} 1}$

$ d_{n \plus{} 1} \equal{} \frac {2002 \minus{} b_n}{a_n \plus{} 1}$

And $ a_nx \plus{} b_n\le f(x)\le c_nx \plus{} d_n$ $ \forall x\in\mathbb N$

It's easy to show that $ \lim_{n\to \plus{} \infty}a_n \equal{} 1$, $ \lim_{n\to \plus{} \infty}b_n \equal{} 666 \plus{} \frac 23$, $ \lim_{n\to \plus{} \infty}c_n \equal{} 1$ and $ \lim_{n\to \plus{} \infty}d_n \equal{} 667 \plus{} \frac 23$

So $ x \plus{} 666 \plus{} \frac 23\le f(x)\le x \plus{} 667 \plus{} \frac 23$ $ \forall x\in\mathbb N$

And so the unique possibility, which indeed is a solution : $ \boxed{f(n) \equal{} n \plus{} 667}$ $ \forall n\in\mathbb N$


geometry
projective geometry
trigonometry
symmetry
invariant
circumcircle
geometric transformation

As this is a fact with a number of applications to geometry problems, here is a short note about it.

[color=blue][b]Theorem 1, the Steiner theorem.[/b] If ABC is a triangle, and D and E are two points on its sideline BC such that the lines AD and AE are symmetric to each other with respect to the angle bisector of the angle CAB, then we have

$\frac{BD}{DC}\cdot\frac{BE}{EC}=\frac{AB^2}{CA^2}$

(with directed segments).[/color]

This theorem is widely ascribed to Steiner, but I was told it appeared in Pappus, Collection, Bk VI, proposition 12 (haven't checked).

Note that two lines passing through the vertex A of triangle ABC and being symmetric to each other with respect to the angle bisector of the angle CAB (like our lines AD and AE) are called [i]isogonal cevians[/i].

Let me present two proofs of Theorem 1.

[hide="First proof of Theorem 1"][i]First proof of Theorem 1.[/i] The most straightforward way to prove Theorem 1 is to use the Sine Law:

At first, we know that the lines AD and AE are symmetric to each other with respect to the angle bisector of the angle CAB. But the lines CA and AB are also symmetric to each other with respect to the angle bisector of the angle CAB. Since a symmetry leaves (non-directed) angles invariant, we thus conclude that the angle between the lines AE and AB equals to the angle between the lines AD and CA. In other words, < BAE = < DAC. Similarly, < BAD = < EAC.

Now, the sine law in triangles ADB and ADC yields $\frac{BD}{AB}=\frac{\sin\measuredangle BAD}{\sin\measuredangle ADB}$ and $\frac{CA}{DC}=\frac{\sin\measuredangle CDA}{\sin\measuredangle DAC}$. Since < ADB + < CDA = 180°, we have $\sin\measuredangle ADB=\sin\measuredangle CDA$; thus,

$\frac{BD}{AB}\cdot\frac{CA}{DC}=\frac{\sin\measuredangle BAD}{\sin\measuredangle ADB}\cdot\frac{\sin\measuredangle CDA}{\sin\measuredangle DAC}=\frac{\sin\measuredangle BAD}{\sin\measuredangle CDA}\cdot\frac{\sin\measuredangle CDA}{\sin\measuredangle DAC}=\frac{\sin\measuredangle BAD}{\sin\measuredangle DAC}$.

Similarly,

$\frac{BE}{AB}\cdot\frac{CA}{EC}=\frac{\sin\measuredangle BAE}{\sin \measuredangle EAC}$.

Multiplying these two equations, we get

$\left(\frac{BD}{AB}\cdot\frac{CA}{DC}\right)\cdot\left(\frac{BE}{AB}\cdot\frac{CA}{EC}\right)=\frac{\sin\measuredangle BAD}{\sin\measuredangle DAC}\cdot\frac{\sin\measuredangle BAE}{\sin\measuredangle EAC}$.

But since < BAE = < DAC and < BAD = < EAC, we have $\frac{\sin\measuredangle BAD}{\sin\measuredangle DAC}\cdot\frac{\sin\measuredangle BAE}{\sin\measuredangle EAC}=1$, and thus

$\left( \frac{BD}{AB}\cdot\frac{CA}{DC}\right) \cdot\left( \frac{BE}{AB}\cdot\frac{CA}{EC}\right)=1$.

In other words,

$\left( \frac{BD}{DC}\cdot\frac{BE}{EC}\right) \cdot\frac{CA^2}{AB^2}=1$,

so that

$\frac{BD}{DC}\cdot\frac{BE}{EC}=\frac{AB^2}{CA^2}$,

and Theorem 1 is proven.[/hide]

[hide="Second proof of Theorem 1 (by Sprmnt21)"][i]Second proof of Theorem 1 (by Sprmnt21).[/i] As in the first proof, we show that < BAD = < EAC. Now, let the circumcircle of triangle ADE meet the lines AB and CA at the points D' and E' (apart from A). Then, the equation < BAD = < EAC becomes < D'AD = < EAE'. But since the points A, D, E, D', E' lie on one circle, < D'AD = < D'E'D and < EAE' = < EDE'; thus, < D'E'D = < EDE'. Hence, D'E' || DE; in other words, D'E' || BC. Hence, by Thales, $\frac{BD^{\prime}}{CE^{\prime}}=\frac{BA}{CA}$. On the other hand, by the intersecting chords theorem, $BD\cdot BE=BD^{\prime}\cdot BA$ and $CD\cdot CE=CE^{\prime}\cdot CA$. Hence,

$\frac{BD\cdot BE}{CD\cdot CE}=\frac{BD^{\prime}\cdot BA}{CE^{\prime}\cdot CA}=\frac{BD^{\prime}}{CE^{\prime}}\cdot\frac{BA}{CA}=\frac{BA}{CA}\cdot\frac{BA}{CA}=\frac{BA^2}{CA^2}=\frac{AB^2}{CA^2}$,

so that

$\frac{BD}{DC}\cdot\frac{BE}{EC}=\frac{BD\cdot BE}{DC\cdot EC}=\frac{BD\cdot BE}{CD\cdot CE}=\frac{AB^2}{CA^2}$.

This proves Theorem 1 again.[/hide]

Here is another result often cited on MathLinks:

[color=blue][b]Theorem 2, the isogonal conjugate theorem.[/b] Let P be a point in the plane of a triangle ABC. Then, the reflections of the lines AP, BP, CP with respect to the angle bisectors of the angles CAB, ABC, BCA concur at one point.[/color]

(This point can be a usual Euclidean point, but it also can be an infinite point. Anyway, this point is called the [i]isogonal conjugate[/i] of the point P with respect to triangle ABC. More about isogonal conjugates can be found in http://www.mathlinks.ro/Forum/viewtopic.php?t=18472 .)

Theorem 2 is often derived from the Ceva theorem in the trigonometric form, but it can also be easily shown using the Ceva theorem in the standard form, once we know Theorem 1:

[hide="Proof of Theorem 2"][i]Proof of Theorem 2.[/i] Let the lines AP, BP, CP meet the lines BC, CA, AB at the points A', B', C', and let the reflections of the lines AP, BP, CP with respect to the angle bisectors of the angles CAB, ABC, BCA meet the lines BC, CA, AB at the points A", B", C".

The line AA" is the reflection of the line AA' with respect to the angle bisector of the angle CAB. In other words, the lines AA' and AA" are symmetric to each other with respect to the angle bisector of the angle CAB. Hence, Theorem 1 yields

$\frac{BA^{\prime}}{A^{\prime}C}\cdot\frac{BA^{\prime\prime}}{A^{\prime\prime}C} =\frac{AB^2}{CA^2}$.

Similarly,

$\frac{CB^{\prime}}{B^{\prime}A}\cdot\frac{CB^{\prime\prime}}{B^{\prime\prime}A} =\frac{BC^2}{AB^2}$;
$\frac{AC^{\prime}}{C^{\prime}B}\cdot\frac{AC^{\prime\prime}}{C^{\prime\prime}B} =\frac{CA^2}{BC^2}$.

Hence,

$\left( \frac{BA^{\prime}}{A^{\prime}C}\cdot\frac{CB^{\prime}}{B^{\prime}A}\cdot\frac{AC^{\prime}}{C^{\prime}B}\right) \cdot\left( \frac{BA^{\prime\prime}}{A^{\prime\prime}C}\cdot\frac{CB^{\prime\prime}}{B^{\prime\prime}A}\cdot\frac{AC^{\prime\prime}}{C^{\prime\prime}B}\right)$
$=\left( \frac{BA^{\prime}}{A^{\prime}C}\cdot\frac{BA^{\prime\prime}}{A^{\prime\prime}C}\right) \cdot\left( \frac{CB^{\prime}}{B^{\prime}A}\cdot\frac{CB^{\prime\prime}}{B^{\prime\prime}A}\right) \cdot\left(\frac{AC^{\prime}}{C^{\prime}B}\cdot\frac{AC^{\prime\prime}}{C^{\prime\prime}B}\right)$
$=\frac{AB^2}{CA^2}\cdot\frac{BC^2}{AB^2}\cdot\frac{CA^2}{BC^2}=1$.

But since the lines AA', BB', CC' concur at one point (namely, at the point P), the Ceva theorem yields

$\frac{BA^{\prime}}{A^{\prime}C}\cdot\frac{CB^{\prime}}{B^{\prime}A}\cdot\frac{AC^{\prime}}{C^{\prime}B}=1$;

thus, we must also have

$\frac{BA^{\prime\prime}}{A^{\prime\prime}C}\cdot\frac{CB^{\prime\prime}}{B^{\prime\prime}A}\cdot\frac{AC^{\prime\prime}}{C^{\prime\prime}B} =1$,

so that, by the Ceva theorem, the lines AA", BB", CC" must also concur. But these lines AA", BB", CC" are simply the reflections of the lines AP, BP, CP with respect to the angle bisectors of the angles CAB, ABC, BCA. Hence, the reflections of the lines AP, BP, CP with respect to the angle bisectors of the angles CAB, ABC, BCA concur, and Theorem 2 is proven.[/hide]

Another trivial consequence of Theorem 1 is the following well-known fact:

[color=blue][b]Theorem 3.[/b] Let ABC be a triangle. The angle bisector of its angle CAB intersects the line BC at a point W. Then, $\frac{BW}{WC}=\frac{AB}{CA}$.

In words: The angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides.[/color]

[hide="Proof of Theorem 3"][i]Proof of Theorem 3.[/i] Each line is symmetric to itself with respect to itself. The line AW is the angle bisector of the angle CAB; hence, we can say that the line AW is symmetric to itself with respect to the angle bisector of the angle CAB. Thus, applying Theorem 1 with D = W and E = W, we get $\frac{BW}{WC}\cdot\frac{BW}{WC}=\frac{AB^2}{CA^2}$, i. e. $\left(\frac{BW}{WC}\right) ^2=\left( \frac{AB}{CA}\right) ^2$. Since the point W lies inside the segment BC, the signed ratio $\frac{BW}{WC}$ is positive. Furthermore, the ratio $\frac{AB}{CA}$ is also positive, since it is a ratio of non-directed segments. Hence, the equation $\left(\frac{BW}{WC}\right) ^2=\left( \frac{AB}{CA}\right) ^2$ leads to $\frac{BW}{WC}=\frac{AB}{CA}$, and Theorem 3 is proven.[/hide]

As a counterpart to Theorem 3, we can show:

[color=blue][b]Theorem 4.[/b] Let ABC be a triangle. The external angle bisector of its angle CAB intersects the line BC at a point W'. Then, $\frac{BW^{\prime}}{W^{\prime}C}=-\frac{AB}{CA}$.

In words: The external angle bisector of an angle of a triangle divides the opposite side externally in the ratio of the adjacent sides.[/color]

[hide="Proof of Theorem 4"][i]Proof of Theorem 4.[/i] As we know, the internal and the external angle bisectors of an angle are perpendicular to each other. Since the line AW' is the external angle bisector of the angle CAB, it thus must be perpendicular to the internal angle bisector of the angle CAB, and therefore it remains fixed under the reflection with respect to this internal angle bisector. In other words, the line AW' is symmetric to itself with respect to the angle bisector of the angle CAB. Hence, we can apply Theorem 1 to D = W' and E = W', and obtain $\frac{BW^{\prime}}{W^{\prime}C}\cdot\frac{BW^{\prime}}{W^{\prime}C}=\frac{AB^2}{CA^2}$, i. e. $\left( \frac{BW^{\prime}}{W^{\prime}C}\right) ^2=\left( \frac{AB}{CA}\right) ^2$.

Since the point W' always lies on the extension of the segment BC, the signed ratio $\frac{BW^{\prime}}{W^{\prime}C}$ is negative. But the ratio $\frac{AB}{CA}$ is positive, being a ratio of non-directed lengths. Hence, the equation $\left( \frac{BW^{\prime}}{W^{\prime}C}\right) ^2=\left( \frac{AB}{CA}\right) ^2$ yields $\frac{BW^{\prime}}{W^{\prime}C}=-\frac{AB}{CA}$, and Theorem 4 is proven.[/hide]

Finally, the most important application of Theorem 1 is a result about symmedians:

Let ABC be a triangle. The reflection of the median of triangle ABC from the vertex A with respect to the angle bisector of the angle CAB is called the [i]symmedian[/i] of triangle ABC from the vertex A. Now, we will prove:

[color=blue][b]Theorem 5.[/b] Let ABC be a triangle, and let its symmedian from the vertex A intersect the line BC at a point R. Then, $\frac{BR}{RC}=\frac{AB^2}{CA^2}$.

In words: The symmedian from a vertex of a triangle divides the opposite side in the ratio of the squares of the adjacent sides.[/color]

[hide="Proof of Theorem 5"][i]Proof of Theorem 5.[/i] Let M be the midpoint of the segment BC; then, the line AM is the median of triangle ABC from the vertex A. On the other hand, the line AR is the symmedian of triangle ABC from the vertex A, i. e. it is the reflection of the median AM with respect to the angle bisector of the angle CAB. Hence, the lines AM and AR are symmetric to each other with respect to the angle bisector of the angle CAB. Hence, by Theorem 1, applied to the points D = M and E = R, we conclude that $\frac{BM}{MC}\cdot\frac{BR}{RC}=\frac{AB^2}{CA^2}$. But since M is the midpoint of the segment BC, we have $\frac{BM}{MC}=1$, and thus we are left with $\frac{BR}{RC}=\frac{AB^2}{CA^2}$. Hence, Theorem 5 is proven.[/hide]

  Darij

thanks Darij! I really have benefited from that.
 But how do you get the time to write such elaborate and nice posts?
    Ashwath

I have another nice proof of the Steiner theorem.
Draw the circumcircle of $ \triangle ABC$, let AD and AE meet it at F and G, respectively. Since $ \angle BAF \equal{} \angle GAC$, $ BF \equal{} CG$, so BFGC is an isoceles trapezoid. By similar triangles, we have the ratios
$ \frac {BD}{BF} \equal{} \frac {AD}{AC}$

$ \frac {CE}{CG} \equal{} \frac {AE}{AB}$

$ \frac {BE}{AB} \equal{} \frac {EG}{CG}$

$ \frac {CD}{AC} \equal{} \frac {DF}{FB}$
Combining all of these, and using that $ BF \equal{} CG$, we have
$ \frac {BD}{DC} \cdot \frac {BE}{EC} \equal{} \frac {AB^2}{AC^2} \cdot \frac {AD}{DF} \cdot \frac {EG}{AE}$
But since $ DE \parallel FG$, $ \triangle ADE \sim \triangle AFG$. Thus, $ \frac {AD}{DF} \equal{} \frac {AE}{EG}$, so
$ \frac {BD}{DC} \cdot \frac {BE}{EC} \equal{} \frac {AB^2}{AC^2}$


AMC AIME geometry probability USA(J)MO USAMO ratio

when will the MAA post solutions??

and meanwhile, could someone post the solution for 1. (license plates problem) and also verify that these are correct? Nobody else I know took it so I have no idea.

2. 200
3. 598 [edit: 578]
4. 450
5. 888
8. 284
12. 091

for number three (easiest problem on the test, in my opinion), extend lines FC and EB out and call their intersection X.  Triangle BCX is congruent to triangles CDF and ABE, by ASA: Angle XCB is complementary to FCD because they're on a straight line with a right angle, and since FCD is also complementary to FDC, XCB=FDC.  you can do the same thing with angles CBX and ABE and EAB to get the other angle.  the hypotenuse is also equal, since it's a side 13 of the square.  Therefore, EF is the hypotenuse of the isosceles right triangle EFX, with sides 17-17-17root2, so EF square is (17root2)^2 = 598 [edit: 578].

i thought the aime II was a lot easier than the aime I.

and yeah, can someone do 6 and 7?

[quote="gimmemath"]when will the MAA post solutions??

and meanwhile, could someone post the solution for 1. (license plates problem) and also verify that these are correct? Nobody else I know took it so I have no idea.

2. 200
3. 598
4. 450
5. 888
8. 284
12. 091[/quote]
I have the same answers as you for the following ones:

2. 200
4. 450
5. 888
12. 091

My confirmed list  so far is the following:

1. 372
2. 200
3. 578
4. 450
5. 888
12. 091

[quote="gimmemath"]great, did you get different answers for the others are just not get them?

did anyone by any chance get 372 for #1? ahh[/quote]
I got 372 at the contest for #1, but I came home and revalidated it and got a different answer (342 ??). I felt so bad because this is such an easy give away. If someone can confirm it either way, that will be great.

I have different answers for the others on your list. I am not sure if I am right, because I have not yet revalidated them again.

Yeah I got 372 too.

Does anyone know the answer for the set probability problem?

[quote="shubidubab"]hm btw (17*root2) is not 598 it's 578.... hope it was just ur typo.[/quote]
Yes, I got 578 too.

i answered 9...

1.   372
2.   200
3.   578
4.   450
5.   888
8.   284 <- this i got wrong... ZOMG 1 extra calculation would've
9.   259     saved me from choosing the wrong answer.... :-(
11. 179
12. 091

btw the way i got #1 was considering two cases:
1. when there is 1 or 0 zero, then it's 7*6*5*4*3
2. when there are 2 zeros, then it's 10 (for choosing 2 zero's placement) times 6*5*4 for the rest 3 spots...
so 7*6*5*4*3+10*6*5*4 = 3720.
I hope i got it right.

btw i scored 118.5 in amc 12. If i get this all 9 correct, then
is there a chance for me to advance to USAMO?

OK my answers:
1. 372
2. 200
3. 578
4. 450
5. 888
6. 256
7. 553
9. 259
12. 091

10. 711 but I am not sure.

The rest I didn't check too well or didn't solve so my answers are probably wrong.

Btw will an index of 116 be enough?

[quote="PenguinIntegral"]I need some solutions. Specifically:

A solution without cyclic quadrilaterals (too hard for a 3?) and bruteforce/coords for 3.

A nice, elegant solutions for 5 and 7. (Which aren't hard to do by just some sample cases and bashing your head against the table until ideas pop into your head, but something simple?)

I kind of case-bashed 6, is there anything better?

A [i]nice Euclidian geometry[/i] solution to 9 and 11.

Solutions to 14 and 15.[/quote]

5 was the number of unit squares below the line right? I think you can draw a bijection between the number of such unit squares and the number of lattice points in the interior of the triangle bounded by the given line and the two axes. Then you can calculate that with Pick's theorem.

For 6 (this was the even odd digit things right?), I think you can create a recursion: let $a_{n}$ be the numbert of n-digit numbers that satisfy the given property. To add another digit in front, it has to be less than the first digit and odd, or greater and even. There are always 4 such digits.

For 7, you know the numbers are $(k^{2})k, (k^{2}+1)k, ...(k^{2}+69)k$... etc. Set the last term less than $(k+1)^{3}$.

The only tricky part to 11 is when the large circle rolls over the small one...you can find the arclength required to roll over and the distance between the centers traveled.

Phelpedo, do you know the answers for 7 and 10?

Most agree that zero is a multiple of every number.

I think that the answer of Problem 13 must be 640.

oh come on... 0 is a multiple of 3...that is perfectly clear

[quote="8parks11"]try posting problems / solutions at
http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_II

too[/quote]

[url=http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_II]Up[/url] for the most part, I think a couple of solutions are missing.

For the ones I thought I did correctly, I agree with the answers at http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_II.  So that's 1-10 and 13.  For the others, I either made mistakes of which I was aware right after the test or didn't know how to approach the problem.

As far as the multiples of 3 go, zero is definitely a multiple of 3 and every real  number.

It seems like this sounds a lot like the mistake a lot of people made on #7 on the AIME I; not counting $1$ as a power of $2$.
Just an observation.

$0$ is not a power of $2$, though. Maybe you meant $1$.

Yeesh. That's what I meant. *edits post*

[quote="DPopov"]I would say 8 or 9 is AIME II cutoff. [/quote]

I really hope the floor isn't a 9...I got an 8. Do you think it could be a 9 with the expansion and everything? I think that seems kind of high after a floor of 9 two years ago before any expansion.

[quote="Max300"]I really hope the floor isn't a 9...I got an 8. Do you think it could be a 9 with the expansion and everything? I think that seems kind of high after a floor of 9 two years ago before any expansion.[/quote]

No, it should not be a 9.  Last year, with the smaller expansion, the floor was 8 -- 432 people were invited to write the USAMO.  This year, the expansion is even greater, to 500 people, so the floor should be 8 or below, which will reflect the larger expansion.

(Thanks AoPS!  :)  )

Could you please tell me how to solve the problem 14?

[quote="syntax_ERROR"]How did anyone do Number 13?[/quote]

You could use the binomial theorem.(From the last row to the first row.)

[quote="jli"]does anyone know where the problems are online?[/quote]

http://www.artofproblemsolving.com/Wiki/index.php/2007_AIME_II


Hello All,

I'm new to this Forum and I do consider myself of intermediate level about LaTeX. I've typed a few documents (I am a computer scientist - research engineer) with it (and I also used the Lout formatter).

First, I don't understand what are the rules about mixing environments (apart from the obvious requirement of balanced begin&end). What is the exact requirements for mixing environments.

More specifically, I'm busy coding some slides with \documentclass{slides} (I want to explain about HTTP protocol) and I have the following [code]
\begin{tabular}[t]{p{4cm} c {p{4cm}}
\hline
%
\fbox{\begin{minipage}{4cm}
\begin{boxedverbatim}
GET /path/file.html HTTP/1.0
From: someuser@jmarshall.com
User-Agent: HTTPTool/1.0

\end{boxedverbatim}
\end{minipage}} %
& gives & %
\fbox{\begin{minipage}{4cm}
\begin{boxedverbatim}
HTTP/1.0 200 OK
Date: Fri, 31 Dec 1999 23:59:59 GMT
Content-Type: text/html
Content-Length: 1354

Happy New Millennium!

\end{boxedverbatim}
\end{minipage}}\\
%
\textbf{request} & & \textbf{reply}
\hline
\end{tabular}
[/code]

However, LaTeX complains at the blank line before the first \end{boxedverbatim}
saying [list]
Runaway argument?
{p{4cm} c {p{4cm}} \hline \fbox {\begin {minipage}{4cm} \begin {boxed\ETC.
! Paragraph ended before \@array was complete.
<to be read again>
                   \par
l.180

[/list]

Any clues?

Thanks

one of my errors has been the tabular argument. [list]
\begin{tabular}[t]{p{7.5cm} c p{7.5cm}}[/list]

In my original code, the last p had extra braces.

Sorry for my mistake.

Regards.


geometry
trigonometry
circumcircle
perimeter
trig identities
Law of Sines
area of a triangle

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths $3$, $4$, and $5$. What is the area of the triangle?
$\textbf{(A)}\ 6 \qquad
\textbf{(B)}\ \frac{18}{\pi^2} \qquad
\textbf{(C)}\ \frac{9}{\pi^2}\left(\sqrt{3}-1\right) \qquad
\textbf{(D)}\ \frac{9}{\pi^2}\left(\sqrt{3}+1\right) \qquad
\textbf{(E)}\ \frac{9}{\pi^2}\left(\sqrt{3}+3\right)$

Can you change the post so that the year and problem number appear on the source field. Thanks.

Oh, and I forgot to answer the question.

[hide="View"]

First, we find the radius of this circle. Since the three arcs have sum of 12, this is the circumference. So, we get $r = \frac{6}{\pi}$. Now, the triangle with arc of 4 has central angle of 120 degrees (you can find this by setting (4/12)(360) = 120). So, we know the radius, angles, so all we have to do is use Sine Area Formula and combine the all three. The result, when simplified, is $\frac{9}{\pi} (\sqrt{3} + 3)$ or E. [/hide]

This is not a very satisfying way of solving this, but I'll show you it.

[hide="not very cute solution"]
Ok so first off, the radius of the circle is the circumradius of the triangle.

The perimeter of the circle is $3+4+5 = 12 = 2 \pi r \rightarrow r = \frac{6}{\pi}$.

For each arc there is a corresponding angle that subtends that arc (I hope I'm using proper terminology):
$3 \rightarrow 90^\circ$
$4 \rightarrow 120^\circ$
$5 \rightarrow 150^\circ$

For each arc, there is a corresponding side (LoC):
Arc of 3: $2r^2(1 - \cos(90^\circ)) = 2r^2 = a^2 \rightarrow a = \sqrt{2} r$
Arc of 4: $2r^2(1 - \cos(120^\circ)) = 3r^2 = b^2 \rightarrow b = \sqrt{3} r$
Arc of 5: $2r^2(1 - \cos(150^\circ)) = (2 + \sqrt{3})r^2 = c^2 \rightarrow c = \frac{\sqrt{2}+\sqrt{6}}{2} r$

Therefore the area of the triangle is $\frac{abc}{4R} = \frac{\sqrt{6} (\sqrt{2}+\sqrt{6}) r^2}{8} = \frac{(6+2\sqrt{3}) 6^2}{8 \pi^2} = \frac{9}{\pi^2} (\sqrt{3}+3) = \boxed{E}$
[/hide]

It was a quick solution so oh well for the niceties.

[quote="Elemennop"]Nice method, but you could've ended it after you got the circumradius by noting that the area of the triangle is $2R^2\sin A\sin B\sin C$ (this is gotten by substituting Law of Sines into the formula $\frac{1}{2}ab\sin C$). Plugging it in would've ended the problem much earlier, though your method is fine enough.[/quote]

Would you mind going through the proof? I'm stuck. :blush:

Sure.

We start by noting that the area of $\triangle ABC=[ABC]=\frac{1}{2}ab\sin C$.

By the Extended Law of Sines, we know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$.

We wish to substitute the values of $a$ and $b$ out of our equation for the area of a triangle, so we substitute in $\frac{a}{\sin A}=2R\Longrightarrow a=2R\sin A$, and $b=2R\sin B$.

We now have $[ABC]=\displaystyle\left(\frac{1}{2}\right)(2R\sin A)(2R\sin B)\sin C=2R^2\sin A\sin B\sin C$, and we are done.

[quote="1234567890"][quote="Elemennop"]
By the Extended Law of Sines, we know that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$.
[/quote]

Wow, I feel even dumber. I thought it was 4R. Sorry to bother you.

Agolsme, you can change it in the options.[/quote]

It's still not appearing for me in this forum; although, it does appear in the preolympiad forum.  :huh:

[quote="agolsme"]where is this source field?

I remember seeing it a month ago or so when I posted topics but now it doesn't seem to be there. (It used to be right under the subject field for me, but now it's not)[/quote]

I think that in some forums still, only the moderators can see the Source field.  But that should be fixed soon...Valentin said he is going to make it such that people can have the option whether or not they want to see the source field.

Anyway, I already edited the source on this problem ;)

[quote="AlanLiangTheAsian"]ok so...how did you get the law of sines equal to the diamter (2R)?[/quote]

that is the extended law of sines.

For proof, please see here:

[url]http://mathworld.wolfram.com/LawofSines.html[/url]

Alternatively my solution is posted on the problem page.


Try to solve this problem :
  Find all positive integers n for which the equation a + b + c + d = n√(abcd) has a solution in positive integers.

The equation is equal to:
[tex] (a+b+c+d)^2=n^2abcd [/tex](1)
[tex] \Leftrightarrow a^2-a(n^2bcd-2b-2c-2d)+(b+c+d)^2=0 [/tex].
Assume that [tex] (a,b,c,d) [/tex] is a solution of the equation where $a \geq b \geq c \geq d $and $a+b+c+d$ is the least sum among the sum of another solutions.
We can see that if $ a $ is a solution then $ n^2bcd-2b-2c-2d-a $ is a solution too.
However [tex] n^2bcd-2b-2c-2d-a \geq a [/tex] then $ n^2bcd \geq 2(a+b+c+d) $.
Hence,from (1),we have:
$ (a+b+c+d)^2 \geq 2a(a+b+c+d) $ or $ b+c+d \geq a $.
So,we have:
[tex] 36b^2\geq (a+b+c+d)^2 \geq n^2b^2cd \geq n^2b^2 [/tex].So $ n \leq 6 $.Now,we must check $ n\in\lbrace{1,2,3,4,5,6}\rbrace $.I haven't done it yet,but I think it is not too difficult,since we just guess the solutions.


I received this in the mail today:

FROM ALLY M. MALINGA
TEL. NUMBER : +34 654438665
EMAIL ADDRESS : allymamoud2004@yahoo.com

You may be surprised to receive this letter from me since you do not know me personally. I am ALLY M. MALINGA, the first son of TAWANDA MALINGA,the most popular black farmer in Zimbabwe who was murdered in the land dispute in my country. I got your contact through network online hence decided to write you.

Before the death of my father, he had taken me to Johannesburg to deposit the sum of U.S$12.6 million (Twelve Million, Six Hundred thousand United States dollars), in one of the private security company, as he foresaw the looming danger in Zimbabwe. This money was deposited in a box as gem stones to avoid much demurrage from security company. This amount was meant for the purchase of new machines and chemicals for the Farms and establishment of new farms in Swaziland. This land problem came when Zimbabwean President Mr. Robert Mugabe introduced a new Land Act Reform wholly affecting the rich white farmers and some black farmers, and this resulted to the killing and mob action by Zimbabwean war veterans and some lunatics in the society. In fact a lot of people were killed because of this Land reform Act for which my father was one of the victims.

It is against this background that, I and my family fled Zimbabwe for fear of our lives and are currently staying in madrid ( spain) where we are seeking political asylum and moreso have decided to transfer my father?s money to a more reliable
foreign account since the law of spain, prohibits a refugee (asylum seeker) to open any bank account or to be involved in any financial transaction throughout the territorial zone of spain.

As the eldest son of my father, I am saddled with the responsibility of seeking a genuine foreign account where this money could be transferred without the knowledge of my government who are bent on taking everything we have got. The South African government seems to be playing along with them.

I am faced with the dilemma of moving this amount of money out of South Africa for fear of going through the same experience in future, both countries have
similar political history. As a businessman, I am seeking for a partner who I have to entrust my future and that of my family in his hands, I must let you know that this transaction is risk free. If you accept to assist me and my family, all I want you to do for me, is to make an arrangements with the security company to clear the consignment(funds) from their afiliate office here in madrid as I have already given directives for the consignment to be brought to madrid from South Africa. But before then all modalities will have to be put in place like change of ownership to the consignment and more importantly I intend to use this money for investmentment after being received.

I have two options for you. Firstly you can choose to have certain percentage of the money for nominating your account for this transaction. Or you can go into partnership with me for the proper profitable investment of the money in your country. Whichever the option you want, feel free to notify me. I have also mapped out 5% of this money for all kinds of expenses incurred in the process of this transaction. If you do not prefer a partnership I am willing to give you 20% of the money while the remaining 75% will be for my investment in your country. Contact me with the above telephone number or my E-mail address while I implore you to maintain the absolute secrecy required in this transaction.

Thanks, GOD BLESS YOU
Yours Faithfully,
ALLY M. MALINGA

Thin I should take 20% of 12.6 million USD? :D

[quote="grobber"]I received this in the mail today:

FROM ALLY M. MALINGA
TEL. NUMBER : +34 654438665
EMAIL ADDRESS : allymamoud2004@yahoo.com

You may be surprised to receive this letter from me since you do not know me personally. I am ALLY M. MALINGA, the first son of TAWANDA MALINGA,the most popular black farmer in Zimbabwe who was murdered in the land dispute in my country. I got your contact through network online hence decided to write you.

If you do not prefer a partnership I am willing to give you 20% of the money while the remaining 75% will be for my investment in your country. Contact me with the above telephone number or my E-mail address while I implore you to maintain the absolute secrecy required in this transaction.

Thanks, GOD BLESS YOU
Yours Faithfully,
ALLY M. MALINGA

Thin I should take 20% of 12.6 million USD? :D[/quote]
WOW. If this mail is for real  ;)  then you shouldn't have post it because he says his in "danger". So it would be best to delete this post  :maybe:  :lol: . However from my point of view you should contact him, maybe just maybe (1 chance out of 10000000... :D  :maybe: ) you'll get 20% out of 12 mil. dol. = 2.4 mil dol it's your choice. I can't help thinking why did he sent this mail to you and not to anyone else.  :maybe:  It might be just a bad joke.  :lol: . This is very strange indeed ... :? BY THE WAY HE IMPLORES YOU TO MAINTAIN ABSOLUTE SECRECY... so... :P  :lol:  ;)

I receive different variants of this e-mail at least 5 times a day ... so just delete it :)

i receive such spam for more than a year... but this one is strange cause he writes in more or less correct english...

  dg

You want to know my opinion about such messages:
I think that this is just a way to find a person to wash dirty money, from drugs and others.....
So never believe this e-mails.
I read on Darij Grinberg's (sorry for the spelling)site how he writes he's e-mail. Check it out. It is really ingineous. :D:D :) :lol:
http://de.geocities.com/darij_grinberg/

[quote="DusT"]I read on Darij Grinberg's (sorry for the spelling)site how he writes he's e-mail. Check it out. It is really ingineous. :D:D :) :lol:
http://de.geocities.com/darij_grinberg/[/quote]

Thanks for the compliments... but try not to use exactly the same method as me: if too many people use the same kind of protection, then spammers will react by "improving" their search machines... use =, &, or just insert some signs in the mail address and say these signs must be deleted, etc.. or you can also save the address in a GIF file rather than in ASCII.

  Darij

14 year bump
The amount of 14 year bumps I have now is crazy


A woman is chosen at random among all women that have two children. She is asked do you have at least one boy, and she answers 'yes.' What is the probability her other child is a boy? Assume every pregnancy has a 50/50 chance to be a boy or a girl.

sorry thats incorrect, nice try though =)

possibilities:
BG GB BB
probability: 1/3

It's a very basic example of a Monte Hall problem. The answer is 1/3.

yeah... i agree... why wouldn't it be one-half?

There are 4 equally likely possibilities:

BB
BG
GB
GG

There are three possibilities with at least one boy and only one with two boys. $\frac{1}{3}$

NoSoupForYou is correct, the answer is unfortunately 1/3. ( joking)  :P

No because either one of the boys could be the boy she was talking about.

It is $\frac 12$ ..

Conditional probability.

I believe it would be 2/3. It's just like that problem on the Sprint round that was 6/7.

No, the answer is 1/3.

bb, bg, gb

out of these 1 choice works, so the answer is [b] 1/3[/b]

Read 19.4 of AoPS, Vol. 2. It explains this problem in detail.

Yeah, misread the problem. I meant 1/3, and I probably confused myself because the answer to that Sprint round question was (7 - 1)/7.


The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $ 20$, the father is $ 48$ years old, and the average age of the mother and children is $ 16$. How many children are in the family?

$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

[quote]The Dunbar family ...[/quote]
I remember late-90's AHSME's in which "Walter" was a recurring character. The AMC director (or whatever his title was) at the time was Walter Mientka. I guess traditions will be served.

[quote="mysmartmouth"]Let there be x children.  So there are x+2 family members.  The dad is 48, and the x+1 other family members are an average age of 16.  So:

$\frac{48+16(x+1)}{x+2}= 20$
$48+16x+16= 20x+40$
$4x = 24$
$x=6$
Answer: E[/quote]
That is a much better way than how I did it, but they are quite similar:
[hide] Say that the mother's age was m, there were n children, and the children's average age was x.  We have that
\[\frac{48+m+nx}{n+2}=20\Rightarrow 48+m+nx=20n+40\Rightarrow m+nx=20n-8 \]
We also have that
\[\frac{m+nx}{n+1}=16\Rightarrow m+nx=16n+16 \]
Notice that both right hand sides of the equations are equal to $m+nx$, so
\[16n+16=20n-8\Rightarrow 4n=24\Rightarrow n=6 \]
Thus, the answer is $E$. [/hide]

i did the exact same thing as mysmartmouth except i divided wrong and somehow got 4..... :wallbash_red:

[quote="mysmartmouth"]Do the Dunbars really have 6 children?[/quote]
pm mr. Dunbar?

I agree.

On an old vestavia geometry test, I saw a question that gave the answer that their teacher/coach was 98 years old, even though she's not.

This was a rather simple problem to be the 10th problem in the AMC10

[quote="randomdragoon"]Somone at my school (jokingly) remarked that this problem [i]might[/i] give people with knowledge of Mr. Dunbar's family an unfair advantage.

Though, I doubt that a question from the AMC would have any grains of truth in it.[/quote]

Sorry for reviving this thread, but I would like to make a point here.

In AoPS Volume 1, there is a question regarding the ages of past US presidents like Washington in a certain year. Well, AoPS is a US book and mostly US students would read it. How do you expect US students not to know about these famous historical figures? I don't know because I'm from Singapore, but honestly...

ehh your question doesn't really warrant reviving this post.. a new topic would've sufficed..

The person doesn't matter.. it could've been a random president of some random place (like Mr. Dunbar, though he's not exactly random, and not from a random place at all :D), and it's not like we memorize the birthyears of all the presidents anyway. At any rate, you'd have to be some kind of fan (why in the world..?) to bother with the birthyear of a president.. :huh:

The problem was found in a book, not on a contest. If you're going to claim to have solved a problem in a book that's supposed to be helping you because you know the birthyears of the presidents, you're not really learning any math. With contests, they have to do everything to make sure everyone is on a level playing field, people's priority is generally to get as high of a score as possible, not to learn as much as possible during the test timeframe.

I had a cool but not as efficient solution.
let the Sum of the Kids ages = S the Sum of the Moms age = M Let the number of kids = n

then we have S + M / n + 1 = 16
and S + M + 48 / n + 2 = 20

Multiplying the second equation out we get 20n + 40 = S + M +48

now dividing by n+1 we get : 20n + 40 / n+1 = 16 + 48/n+1

so 4n = 24 and n=6 i thought it was cool how i could solve 2 equations with three variables


There are $ k$ kids involved. The total number of grapes eaten can be represented by the equation:
$ n + (n - 2) + (n - 4) + ... + [n - 2(k - 1)] = 2009\Rightarrow kn - 2[\frac {k(k - 1)}{2}] = 2009$, which simplifies to $ kn - k^2 + k = 2009\Rightarrow k(n - k + 1) = 2009$.

Now we want to minimize $ n$. We notice that $ 2009 = 7\times7\times41$. Let's try some algebra bashing/brute force substitution to see our possibilities:

If $ k = 1$, then we have $ n = 2009$.

If $ k = 7$, we have $ 7(n - 6) = 2009\Rightarrow n - 6 = 287\Rightarrow n = 293$.

If $ k = 41$, we have $ 41(n - 40) = 2009$, so $ n = 89$.

If $ k = 49$, we have $ 49(n - 48) = 2009$, so $ n = 89$.

If $ k = 287$, we have $ n - 286 = 7$. This can't be the smallest $ n$.

If $ k = 2009$, we have $ n - 2008 = 1$. Ignore that, it can't be the smallest.

Therefore the smallest possible value of $ n$ is $ \boxed{089}$.

I'm so dumb. I didn't realize that the $ k$ meant every contestant. I thought it referred to only one arbitrary contestant....

so did I....

which makes the answer so trivial.

It's not really good wording though. Very confusing.

[quote="un_o"]If you know a little bit of Calculus, this question becomes so much easier.

Just take the derivative and let it = 0 and the rest is just simple arithmetic. :)[/quote]

But it must be an integer, and it's a single equation with two variables.

dont you just do if a=average grapes eaten (middle term) and b=number of contestants ab=2009 and a>=b so the minimum value of a is 49 and the winner eats 49+40=89 grapes

I got this problem wrong because I forgot the sum of the first odd numbers  :P Oh darn.

Anyways, after looking at the problem for a few minutes, we can see the terms of the sequence we have will be n, n-2, n-4...5, 3, 1, but we can always cross out a few of the end terms later if need be. We will see why soon.

Okay, so how many odd numbers do we need to reach 2009? We see that $ 45^2 \equal{} 2025$ and that will be the smallest perfect square greater than 2009.

Therefore, we will need the sum of the first 45 odd numbers, starting from 1. Computing, we see the last term will be 89. Our sequence is:

$ 1, 3, 5, 7, ... 85, 87, 89$. This will yield a sum of 2025. To account for the difference between 2025 and 2009, we can simply remove the first 4 terms, 1, 3, 5, and 7.

Nevertheless, our answer is [b]89[/b] because that is the final term of the sequence.

Once you get $ k(n\minus{}k\plus{}1)\equal{}2009$, you can just use AM-GM to get the minimal $ n$:
\[ \frac{n\plus{}1}{2}\ge\sqrt{2009}\]

You don't even need AM-GM, you can just do this:

[hide="Solution"]
Suppose there are $ m$ kids.

$ \sum_{k\equal{}1}^{m}n\plus{}2\minus{}2k\equal{}mn\plus{}2m\minus{}m(m\plus{}1)\equal{}2009$

$ m^2\minus{}m(1\plus{}n)\plus{}2009\equal{}0$

The discriminant is greater than or equal to zero:

$ (n\plus{}1)^2\geq 4\cdot 2009\iff n\geq 2\sqrt{2009}\minus{}1$.

It's not hard to find that $ 44<\sqrt{2009}<45$, so that $ 2\sqrt{2009}\minus{}1<\boxed{089}$.
[/hide]

That's equivalent... you just essentially use a proof of 2-variable AM-GM.

I hate when AIME problems tell you that they got a solution by proofs.

lol wait... so the kth student is the last student? Wow... they could've just said "last student"

Well no, the kth student refers to any student (1st, 3rd, 29th, 56th, etc). If they told you that the kth student was the last, it would have made the problem very trivial.

Oh, thanks I understand the question now... the wording is so confusing

[quote="worthawholebean"]A group of children held a grape-eating contest. When the contest was over, the winner had eaten $ n$ grapes, and the child in $ k$th place had eaten $ n\plus{}2\minus{}2k$ grapes. The total number of grapes eaten in the contest was $ 2009$. Find the smallest possible value of $ n$.[/quote]

[hide="Solution"]
If there were $x$ kids, the total number of grapes eaten were $\sum_{i=1}^{x} [(n+2)- 2x] = x(n+2) - 2\left(\sum_{i=1}^{x} x\right) = x(n+2) - x(x+1) = x(n-x+1) = 2009$.

The possible values of $x$ are the positive factors of $2009$, namely $1$, $7$, $41$, $49$, $287$, and $2009$.

If $x = 1$, $n-1+1 = 2009 \implies n = 2009$
If $x = 7$, $n-7+1 = 287 \implies n= 293$
If $x = 41$, $n-41+1 = 49 \implies n = 89$
If $x = 49$, $n-49+1 = 41 \implies n= 89$
If $x = 287$, $n - 287 + 1 = 7 \implies n = 293$
If $x = 2009$, $n - 2009 + 1 = 1 \implies n = 2009$.

Therefore, the smallest possible value of $n$ is $\boxed{089}$.
[/hide]

[hide]We see that: 1st place has n grapes, 2nd place has n-2 grapes, ..... kth place (let k be last place) has n+2-2k grapes.

The sum of these grapes is (n-k+1)(k)=2009=$7^2*41$

We see that for n to be at a minimum,k must equal either 49 or 41, which both yield $\boxed{089}$[/hide]

[hide="Another fast Solution"]Just follow the above mentioned steps to arrive at:

$k(n-k-1) = 2009 \Longrightarrow -k^2+k(n-1)-2009= 0 \Longrightarrow$

$(n-1)^2+4(2009) \ge 0 \Longrightarrow n^2-2n \le 8037$

Notice that $90^2= 8100$

So we just try $n=90$ first and see that $k(89-k)=2009$ is not an integer so we go one below and try $n=\boxed{089}$ which gives an integer $k$ and we are done.

[/hide]

p15: Say there were $p$ children. Then the last place child ate $n-(2p-2)$ grapes, so the total was $pn - p(p-1) = p(n-p+1) = 2009 = 7^2 \cdot 41$. We minimize $n$ by selecting $p = 41$, giving $n = \boxed{89}$.


conics
ellipse
geometry
geometry unsolved

Points $ B$ and $ D$ lie on an ellipse with foci $ K,L$ such that $ KB, LD$ meet outside the ellipse at $ A$ and $ KD, LB$ meet inside the ellipse at $ C$.
Prove that $ ABCD$ is a quadrilateral in which a circle can be inscribed.

[hide]Let $ (I)$ be, the incircle of the triangle $ \bigtriangleup ABL.$

Through the point $ K,$ we draw the second tangent line of $ (I),$ which intersects the line segments $ BL,\ AL,$ at ponts $ C',\ D',$ respectively.

In the configuration now, of the circumscribed quadrilateral $ ABC'D',$ we have that $ KB \plus{} BL \equal{} KD' \plus{} D'L$ $ ,(1)$ as well ( and easy to prove ).

From $ (1)$ we conclude that the point $ D'$ lies on the ellipse $ (O)$ with foci the points $ K,\ L$ and the point $ B$ lying on it.

Hence, we conclude that $ D'\equiv D\equiv (O)\cap AL$ $ \Longrightarrow$ $ C'\equiv C\equiv BL\cap KD$ and the proof is completed.[/hide]
Kostas Vittas


Jared has nine coins in his pocket. They all look alike, but one coin is counterfeit and weighs less than the others. What is the least number of weighings on a balance scale needed to guarantee that the counterfeit coin is found?

$ \boxed2$ weighings.

Divide the $ 9$ coins into three groups. Take two groups and weigh them: if they are the same weight, take two coins from the unweighed group and compare them (if one of them is lighter, that's the counterfeit coin and if they are the same, then the untouched coin is the counterfeit coin), and if they are of different weights, take the lighter group and compare two of them (if one of them is lighter, that's the counterfeit coin and if they are the same, then the untouched coin is the counterfeit coin).

It is easy to see that one weighing is not sufficient.

For more information on this, refer to [url=http://en.wikipedia.org/wiki/Definitive_Solution_to_the_Counterfeit_Coin_Problem#The_counterfeit_coin_problem]this[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1317183#1317183]this[/url].


\[ a,b,c \geq0 \]

then prove

\[ \sum_{cyc}\frac{\sqrt{a^2 + 2b^2 + c^2}}{a^2 + 2bc} \geq \frac{6}{a+b+c} \]
 :lol:


inequalities
algebra
polynomial
inequalities unsolved

Let $ a,,b,c$ be positive numbers such that:
       $ a\plus{}b\plus{}c\equal{}$9
  prove that: $ abc$+54 $ \geq$ 3 $ (ab\plus{}bc\plus{}ca)$

It is equivalent to $ f(a,b,c) \equal{} abc \plus{} 54.\frac {(a \plus{} b \plus{} c)^3}{9^3} \minus{} 3(ab \plus{} bc \plus{} ca)\frac {a \plus{} b \plus{} c}{9}\geq 0$. Now need only check that $ f(1,1,1),f(1,1,0),f(1,0,0)$ are $ \geq 0$.

[quote="hoangclub"]Let $ a,,b,c$ be positive numbers such that:
       $ a \plus{} b \plus{} c \equal{}$9
  prove that: $ abc$+54 $ \geq$ 3 $ (ab \plus{} bc \plus{} ca)$[/quote]

Wrong.Try $ 4,4,1$

[quote="N.T.TUAN"]Now need only check that $ f(1,1,1),f(1,1,0),f(1,0,0)$ are $ \geq 0$.[/quote]

I wonder why  :maybe:

[quote="FOURRIER"]wich page or wich chapter?

thanks :)[/quote]
3.4 Polynomial Inequalities with Degree 3
pages 33 and 34...


Is it possible for each of the 25 students in a class to shake hands with eactly 7 classmates?

[hide="hint"]  think of how handshakes can be distributed, or "counted" for that matter [/hide]

It was answered in the marathon ...

Basically he said it's impossible to have $25 \cdot 7/2$ handshakes because it would result in a fractional number of handshakes.


inequalities
number theory proposed
number theory

Trivial with Hensel's Lemma!

The number of factors $2$ of $3^{x}-1$ is $f(x)+1$, where $f(x)$ is the number of factors $2$ of $x$.
So, $x \leq f( x )+1$, that implies x = 0 or 1

My solution: We easy get $x=0,y=0,x=1,y=1,x=2,y=2,x=4,y-5$, and for $x=3, x=6$ we do not have solutions. Now suppose$x\ge 7$. We have of course that $x$ is even(easily get mod8), so $x=2k$, we have $(3^k-1)(3^k+1)=2^{2k}y$, but $gcm((3^k-1),(3^k+1)=2)$ So, let's see what happens when k is even. We get $k=2^ut,gcm(t,2)=1.$$3^k+1=2(mod4)$, so denote $v_2(w)$ the exponent of $2$ in $w$, we have $v_2(3^k-1)\ge 2k-1$, but $3^{2^ut}-1=(3^t-1)(3^t+1)(3^{2t}+1)...(3^{2^{u-1}t}+1)$, but we easy get $3^{2x}+1=2(mod4)$ so the inequality can be written ${u+v_2(3^t+1)\ge 2^{u+1}t-1}$, obvios false, because $3^t+1\le 4^t$, and $v_2(4^t)=2t$. So, we are done when $k$ is even. Suppose now that $k$ is odd, $3^k-1=2(mod4)$, so we have${v_2(3^k+1)\ge 2k-1}$, which is false(analogous with what i've done before)

See here (from 2005 Romanian TST) : http://www.artofproblemsolving.com/Forum/viewtopic.php?p=198299&sid=8e00e8b252767dc670e5c9bd7a604079#p198299

[quote="Agr_94_Math"]Could some one explain Hensel's formula? What does he mean by number of factors 2?[/quote]
This is actually well-known as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=721&t=401494]Lifting The Exponent[/url] Lemma

Can you also solve it by dividing by 2^x and taking away 2^-x and show x has to be small for there to be no fractions in the binomial expansion?

The equation $3^x=2^xy+1$ implies that $3^x\equiv 1$(mod $2^x$) and by Euler-Fermat $x\equiv 0$ (mod$2^{x-1}$) which means that $2^{x-1}\le x$, so $x\le2$.

Substituting, we get that two only possible solutions are $(1,1)$ and $(2,2)$.

[quote=N.T.TUAN]Find all natural solutions of  $3^{x}= 2^{x}y+1.$[/quote]
We claim that the only solutions are $\boxed{(1,1)},\boxed{(2,2)},\boxed{(4,5)}$.We now prove that these are the only.
If $x=1$ then we get $(x,y)=(1,1)$ is a solution otherwise for $x\geq 2$ we get $x$ is even by looking modulo $4$.Now $$\nu_2(3^x-1)=\nu_2(3+1)+\nu_2(3-1)+\nu_2(x)-1=\nu_2(x)+2$$.Hence $\nu_2(x)+2\geq x \Longleftrightarrow x\leq 2+\lceil \log_2x \rceil \implies \lceil \log_2x \rceil\geq x-3 \implies x\geq 2^{x-3}$ which clearly fails for $x\geq 6$.Thus $x<6$.Now checking the permissible range we get the claimed solutions.$\blacksquare$

We have trivial solutions $(x,y)=(0,0);(1,1);(2,2)$. We need to find another solution. See that $3^x \equiv 1 \pmod {16}$ if $3^x \equiv 0 \pmod {4}$, so we can write $x=4k$, for some $k$.
Our equation becomes $3^{4k}-1=2^{4k}y$ or $(3^{2k}-1)(3^{2k}+1)=2^{4k}y$. It is easy to see that $y$ is odd or even. So we have two cases
$I$ case is $y \equiv 0 \pmod {2}$: Let's define $z$ and $d$ such $zd=y,$ so $3^{2k-1}=4^{2k}z$ and $3^{2k}+1=d$. This has obviously no solution.
$II$ case is $y \equiv 1 \pmod {2}$: Again, $3^{2k}=4^{2k-1}2z$ and $3^{2k}+1=2d.$ This has solutions when $k<3$ and that is for $k=1 \implies z=1, d=5$ and finally $x=4, y=5$.
So, all solutions are $(x,y)=(0,0);(1,1);(2,2);(4,5)$ :)


calculus
derivative
calculus computations

I just finish a test, but I don't understand a question.
T(x) is the number of seconds it takes an average athlete to swim 100 meters freestyle at age x years.

 Write a sentence to interpret T?22)= -0.5.

What is the ? next to the T means if the question has no problem?

[quote="JRav"]I have never seen such notation.  Are you sure you are writing it correctly?  Could you please explain it a little more clearly?[/quote]
 My teacher's answer to this question is "At age 22 the swimmer抯 time for swimming 100 m is decreasing ?sec for each year抯 increase in age." But I don't know how come it be?

Also, I think of this symbol "抯" should be a 's or something, because sometime my computer would get mix up with different languages.

[quote="jmerry"]From the answer, that should be T'(22). The notation indicates a derivative, and derivatives are rates of change.[/quote]
then is -0.5 means a velocity or acceleration?

Since the units of T are neither distance nor velocity, I wouldn't call it either. It is a rate of change, and it's a first derivative rather than a second.

One more question: why in her official answer, she doesn't used the number 0.5 or -0.5? Where should this number go?

[quote="jmerry"]We often indicate sign by talking about direction; increase or decrease in this case.[/quote]
I know negative sign means decrease and positive sign means  increase. my question is I don't see .5 this number use in the interpret sentence.

[quote="linj"] "At age 22 the swimmer抯 time for swimming 100 m is decreasing ?sec for each year抯 increase in age." But I don't know how come it be?[/quote]

Sorry to butt in, but I'd guess the "decreasing ?sec" is supposed to be "decreasing 0.5sec"

I would interpret it to mean that at age 22, the change drops by .5.


conics
parabola
symmetry
geometry solved
geometry

Two parabolas with perpendicular axis of symmetry intersect at 4 points. Prove that this points lie on one circle.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=16944 .

  darij


I'm not quite sure if I understand how to use generating functions. What are some cases where they are used (example problems, etc.)?

(Also, I think the AopsWiki article on it could do with some additions.  :wink: )

For a more detailed explanation of generating functions you should consider getting Intermediate Probability and Counting..

Generating functions can help in some problems where heavy casework might be required.

More specifically, you can find an explicit formula for a recursion, like the Fibonacci numbers.

Taken almost word-for-word from Intermediate Counting and Probability:
[quote]
Suppose that $ a_k$ is a sequence of numbers. The corresponding generating function for this sequence is
\[ f(x)\equal{}a_0\plus{}a_1k\plus{}a_2x^2\ldots\]
with each $ a_k$ matched with $ x^k$. These can be used in combinatorial problems using the concepts from binomial theorem.
[/quote]
Here's example 14.1 from the book, again almost word-for-word: [quote]

There are three urns on a table. One has 2 red balls, one has 2 green, one has 3 blue. All balls of the same color are indistinguishable. In how many was can I choose four balls from the urns?
This problem can be solved using casework obviously, but generating functions are fun. From the red urn, we can choose $ 0,1,$ or $ 2$ balls. We have $ 1$ to choose each of these possibilities, so the generating function for the red one is $ 1x^0\plus{}1x^1\plus{}1x^2\equal{}1\plus{}x\plus{}x^2$. The coefficient of the $ x$ with exponent $ k$ is the number of ways to choose $ k$ objects. We do similar things for the other two urns, multiply them, then find the coefficient of $ x^4$. (Each product corresponds to a way to choose the four balls, which is why this works)[/quote]


$ xy\plus{}yz\plus{}xz\minus{}2xyz\equal{}\frac{1}{4}(1\minus{}2x)(1\minus{}2y)(1\minus{}2z)\plus{}\frac{1}{4}$.

How do you get from the LHS to the RHS, assuming you don't know what the RHS is at the start, how do you manipulate the LHS into the RHS?

Thanks!

is x+y+z=1? i dont think it is correct otherwise..

well then,
divide the whole by 4 ,
$ \sum\frac{4xy}{4} \plus{} \frac {1}{2} \plus{} \frac { \minus{} 8xyz}{4} \minus{} \frac {2x \plus{} 2y \plus{} 2z}{4} \equal{} \frac {1}{4}\prod(1 \minus{} 2x) \plus{} \frac {1}{4}$

Sorry but what do you mean divide the whole thing by 4?

Where did the yz+zx+xy go? =S


modular arithmetic
number theory proposed
number theory

Find all solutions for positive integers $ x$ and $ n$ and primes $ p$

$ x^5 \plus{} x^4 \plus{} 1 \equal{} p^n$

$ x^{3}\equiv 1\pmod{x^{2} \plus{} x \plus{} 1}$  $ \implies$  $ x^{2} \plus{} x \plus{} 1|x^{5} \plus{} x^{4} \plus{} 1$
$ \implies$  $ (x^{5} \plus{} x^{4} \plus{} 1) \equal{} (x^{2} \plus{} x \plus{} 1)(x^{3} \minus{} x \plus{} 1) \equal{} p^{n}$
$ \implies$  $ x^{2} \plus{} x \plus{} 1 \equal{} p^{u},x^{3} \minus{} x \plus{} 1 \equal{} p^{v}$
$ p|x^{2} \plus{} x \plus{} 1$  $ \implies$  $ p|x^{3} \minus{} 1$  $ \implies$  $ x\equiv 2\pmod{p}$
$ \implies$  $ p|7$  for $ u,v\geq 1$
$ p|7$  $ \implies$  $ p \equal{} 7$
$ x^{2} \plus{} x \plus{} 1 \equal{} 7^{u},x^{3} \minus{} x \plus{} 1 \equal{} 7^{v}$
if $ v > u$  $ \implies$  $ x^{2} \plus{} x \plus{} 1|x \minus{} 2$  $ \implies$  $ x \minus{} 2 > x^{2} \plus{} x \plus{} 1$  Contradiction!
$ \implies$  $ v\leq u$  $ \implies$  $ x^{2} \plus{} x \plus{} 1\geq x^{3} \minus{} x \plus{} 1$  $ \implies$
$ x^{3} \minus{} x^{2} \minus{} 2x\leq 0$ $ \implies$  $ x(x^{2} \minus{} x \minus{} 2)\leq 0$
$ x > 0$  $ \implies$  $ x^{2} \minus{} x \minus{} 2\leq 0$  $ \implies$  $ (x \minus{} 2)(x \plus{} 1) < 0$  $ \implies$  $ 1\leq x \leq 2$
if $ x \equal{} 1$  $ \implies$   $ 7^{u} \equal{} 3$  Contradiction!
if $ x \equal{} 2$  $ \implies$  $ u \equal{} v \equal{} 1$  $ \implies$ $ n \equal{} 2$
if $ u \equal{} 0$  $ \implies$  $ x^{2} \plus{} x \plus{} 1 \equal{} 1$  contradiction!
if $ v \equal{} 0$  $ \implies$  $ x^{3} \equal{} x$  $ \implies$  $ x \equal{} 1$ but $ 1^{2} \plus{} 1 \plus{} 1 \equal{} 7^{u}$  Contradiction! :wink:

we obviously have $ x^5 \plus{} x^4 \plus{} 1 \equal{} (x^2 \plus{} x \plus{} 1)(x^3 \minus{} x \plus{} 1) \equal{} p^n$.
i won't consider trivial cases.
$ x^2 \plus{} x \plus{} 1 \equal{} p^k,x^3 \minus{} x \plus{} 1 \equal{} p^l,l \plus{} k \equal{} n \implies x(x \plus{} 1) \equiv \minus{} 1 \pmod {p},x(x \minus{} 1)(x \plus{} 1) \equiv \minus{} 1 \pmod {p} \implies x \minus{} 1 \equiv 1 \pmod {p} \implies p \mid 2^2 \plus{} 2 \plus{} 1 \equal{} 7 \implies p \equal{} 7$.
the rest is trivial. :)
too late :oops:

how about a generalization :
$ x^{3k\plus{}2}\plus{}x^{3k'\plus{}1}\plus{}1\equal{}p^n$.
have u ever thought about this ?
 it seems interesting...  :huh: :?:

Fast work!
And I wont hold you to task for missing out one simple solution  :)


[b]Let $ABC$ be a triangle. Prove that[/b] $\{\begin{array}{cccc}1\blacktriangleright & & \boxed{ \cos A\le \frac{R(p-a)}{2ar}}& \\\\ 2\blacktriangleright & (a-b)(a-c)\le 0 & \implies & \frac{p-a}{a}\le\cos A\le \frac{R}{2r}\cdot\frac{p-a}{a}\\\\ 3\blacktriangleright & b+c=2a & \implies & \frac{1}{2}\le \cos A\le\frac{R}{4r}\\\\ 4\blacktriangleright & & R\ge 2r & \end{array}$

first note that $a^{2}=b^{2}+c^{2}-2bccosA$ so we would have $cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc}$ so we have:
$cosA=\frac{b^{2}+c^{2}-a^{2}+2bc}{2bc}-1$ so $cosA=\frac{(b+c+a)(b+c-a)}{2bc}-1=\frac{p(b+c-a)}{bc}-1$
--------------------------------
now in $2$,from $(a-b)(a-c)\leq 0$ we get that $a^{2}+bc\leq ab+ac$
we want to prove that $\frac{p-a}{a}\leq cosA$ so we have to prove that $\frac{p-a}{a}\leq \frac{p(b++c-a)}{bc}-1$
so we have to rpove that $\frac{p}{a}\leq \frac{p(b+c-a}{bc}$ so we have to show that $bc\leq a(b+c-a)$
so we have to show that $a^{2}+bc\leq ab+ac$ which is proven in the begining of the proof...
--------------------------------
now in $3$,from $b+c=2a$ we get that $b+c-a=\frac{b+c}{2}$ and that $a+b+c=\frac{3(b+c)}{2}$ so $p=\frac{3(b+c)}{4}$
so we have $cosA=\frac{3(b+c)^{2}}{8bc}-1$
ok now we want to show that $\frac{1}{2}\leq cosA$ so we have to show that $1\leq \frac{(b+c)^{2}}{4bc}$
which is trivial because of $AM-GM$...
-------------------------------
a stronger version of $4$ is the folowing inequality:
$\frac{R}{r}\geq \frac{a}{b}+\frac{b}{a}$ where $a,b$ are the lengths of two arbitary sides of the triangle...

For the first:
$\frac{R(p-a)}{2ar}=\frac{\frac{abc}{4S}(p-a)}{2a\frac{S}{p}}=\frac{bcp(p-a)}{8S^{2}}=\frac{\cos^{2}\frac{A}{2}b^{2}c^{2}}{8S^{2}}=\frac{\cos^{2}\frac{A}{2}}{2\sin^{2}A}=\frac{1}{8\sin^{2}\frac{A}{2}}\le\cos A\Leftrightarrow 8\sin^{2}\frac{A}{2}\cos A-1\le 0\Leftrightarrow 4(1-\cos A)\cos A-1\le 0 \Leftrightarrow (2\cos A-1)^{2}\ge 0$ which is true.

[quote="Virgil Nicula"][b]Let $ABC$ be a triangle. Prove that:[/b] $\boxed{ \cos A\le \frac{R(p-a)}{2ar}}$[/quote]
[color=darkblue][b]Proof. (A little different than gemath's solution.)[/b]  :D
We have: $\frac{R(p-a)}{2ar}=\frac{R}{4R\sin A}\cdot\frac{p-a}{r}=\frac{\cot\frac{A}{2}}{4\sin A}$
Hence $\cos A\le\frac{R(p-a)}{2ar}$ $\iff\cos A\le \frac{\cot\frac{A}{2}}{4\sin A}\iff 8\sin^{2}\frac{A}{2}\cos A-1\le 0\iff \left(2\cos A-1\right)^{2}\ge 0$[/color]

[color=darkred][u][b]My proof :[/b][/u] [/color]$\frac{r}{p-a}=\tan\frac{A}{2}=\frac{1-\cos A}{\sin A}=\frac{2R(1-\cos A)}{a}$ $\implies$ $\frac{R(p-a)}{2ar}=\frac{1}{4(1-\cos A)}\ge \cos A$.


A car traveled 75% of the way from town A to town B at an average speed of 50mph. The car travels at an average speed of S mph for the remaining part of the trip. The average speed for the entire trip was 40mph. What is S?

What is wrong with the following setup:

(50x + 50x/3) / (x + x/3*50/S) = 40 ?

Don't know how to put it in lex :(

[quote="dear_xavier"]A car traveled $ 75\%$ of the way from town A to town B at an average speed of $ 50$mph. The car travels at an average speed of $ S$ mph for the remaining part of the trip. The average speed for the entire trip was $ 40$mph. What is $ S$?

What is wrong with the following setup:

$ \frac {50x \plus{} \frac{50}3x}{x \plus{} \frac x3\cdot\frac{50}S} \equal{} 40$ ?

Don't know how to put it in lex :([/quote]
Let the distance from A to B be $ d$.
The first $ 75\%$ of the way: distance of $ \frac34d$, time taken $ \frac {\frac34d}{50} \equal{} \frac3{200}d$.
The last $ 25\%$: distance of $ \frac14d$, time taken $ \frac {\frac14d}{40} \equal{} \frac1{160}d$.
Putting these together, we have $ S \equal{} \frac {\frac34d \plus{} \frac14d}{\frac3{200}d \plus{} \frac1{160}d} \equal{} \boxed{\frac {800}{17}}$.

*Note: we could have just assumed the $ d \equal{} 1$, since it is not important.

Alternatively, we could note that this is the harmonic mean of $ 50,50,50,40$, or $ \frac4{\frac1{50} \plus{} \frac1{50} \plus{} \frac1{50} \plus{} \frac1{40}} \equal{} \frac {800}{17}$. There are $ 3$   $ 50$'s because the distance traveled at $ 50$mph is $ 3$ times the distance traveled at $ 40$mph.

I think the problem states that the average is 40 and the remaining quarter's average is S, not the other way around.
So, just say it's 40 miles away. The average is 40mph, so it is one hour.
The first 30 miles are 50 mph, so it would take 36 minutes.
Then, 10 miles in 24 minutes is needed, or [b]25 mph.[/b]

math154, average speed of the entire trip is already given - 40 mph.

S = 25 is the official answer.

Ttocs45, I did it the same way as you  - plugging in, but I still dont know what is wrong with my algebric approach?

Sorry for misreading the question! I was in a hurry.

Try this:

$ \frac4{\frac1{50}\plus{}\frac1{50}\plus{}\frac1{50}\plus{}\frac1x}\equal{}40\implies x\equal{}\boxed{25}$.

Yes. Both solutions should work well.
I'm not sure how your calculating it, though.

$ \frac4{\frac1{50} \plus{} \frac1{50} \plus{} \frac1{50} \plus{} \frac1x} \equal{} 40$

$ \implies\frac1{\frac3{50}\plus{}\frac1x} \equal{} 10$

$ \implies1 \equal{} \frac35 \plus{} \frac {10}x$

$ \implies\frac25 \equal{} \frac {10}x$

$ \implies\frac52 \equal{} \frac x{10}$

$ \implies\boxed{25} \equal{} x$

Thanks, math154.
I don't understand how you tried to do it, dear_xavier. Did you organize your equations incorrectly?

[quote="dear_xavier"]What is wrong with the following setup:

(50x + 50x/3) / (x + x/3*50/S) = 40 ?[/quote]

I believe dear_xavier is trying to divide the distance by time to get the speed and equating, where $ x$ is the time it took to complete the first stretch of $ 75\%$ from A to B.

That would mean the distance should be $ 50x\plus{}\frac x{\frac3{50}}\equal{}\frac{200}3x$ instead.

The denominator should be the time this takes, which is $ x\plus{}\frac x{3\cdot\frac S{50}}\equal{}x\plus{}\frac{50}{3S}x$.

Dividing the first by the second, we get $ \frac{\frac{200}3x}{x\plus{}\frac{50}{3S}x}\equal{}\frac{\frac{200}3}{1\plus{}\frac{50}{3S}}\equal{}\frac{200}{3\plus{}\frac{50}S}\equal{}40$.

Multiplying through by $ 3\plus{}\frac{50}S$, we get

$ 200\equal{}120\plus{}\frac{2000}S$, which means that

$ 80\equal{}\frac{2000}S$, which gives $ S\equal{}\boxed{25}$.

Yes, I agree that this is not the easiest solution to think out.

You can use brute force for this, but if you want to use a variable for distance then you can.
3d/(50*4) + d/4s = d/40
Multiply the whole equation by 200/d to get 3 +50/s = 5 and then solve for s to get s = 25

In case anybody doesn't know, I was only explaining what dear_xavier did wrong and corrected his setup. I have posted a solution above...

And jhangil's post in $ \text{\LaTeX}$:

[quote="jhangil"]You can use brute force for this, but if you want to use a variable for distance then you can.
$ \frac {3d}{50\cdot4} + \frac d{4s} = \frac d{40}$
Multiply the whole equation by $ \frac {200}d$ to get $ 3 + \frac {50}s = 5$ and then solve for $ s$ to get $ s = 25$.[/quote]

Thanks yall,

My setup is actually correct, it was some arithmetic error (40*50 = 200) that gave me wrong result, so I thought it's all wrong.

"Yes, I agree that this is not the easiest solution to think out."

Most of the time, i find complicated ugly solutions :(  :)


Define a selfish set to be a set that has its own cardinality as an element. Find, with proof, the number of subsets of $ \{1,2, \cdots, n\}$ that are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.

Can anyone run me through how to attempt/solve this question? I'm quite stuck ;(

Refine the conditions for a subset to be a minimal selfish subset. not only must its cardinality be an element, it must be its smallest element...

Let $f_n$ denote the number of minimal selfish subsets of $\{1, \ldots n\}$. We have $f_1=1$ and $f_2=2$. For $n>2$, the number of minimal selfish subsets of $\{1, \ldots n\}$ not containing $n$ is equal to $f_{n-1}$. But, for any minimal selfish set containing $n$, by removing $n$ from the set and subtracting $1$ from each remaining element, we obtain a minimal selfish subset of $\{1, \ldots n-1\}$. On the other hand any minimal selfish subset of $\{1, \ldots , n-1\}$ gives a minimal selfish subset of $\{1,\ldots n\}$ containing $n$ by the inverse procedure. Hence the number of minimal selfish subsets of $\{1,\ldots, n\}$ containing $n$ is $f_{n-2}$. Finally, we see that \[f_n=f_{n-2}+f_{n-2}\] which together with the initial values $f_1=f_2=1$ implies that $f_n=F_n \Box$