How many pairs of positive integral values are there for $ x$ and $ y$ satisfying the following conditions?

$ 4x \minus{} 17y \equal{} 1$
$ x \le 1000$

hello, we have $ 4\cdot13\minus{}3\cdot17\equal{}1$, so all solutions of your equation have the form $ x\equal{}13\plus{}17k,y\equal{}3\plus{}4k$ with $ k \in \mathbb{Z}$, and your condition gives $ 13\plus{}17k \le 1000 \Leftrightarrow 0<k \le 58$.
Sonnhard.

I agree with Sonhard's formulas.[quote="Dr Sonnhard Graubner"]all solutions of your equation have the form $ x \equal{} 13 \plus{} 17k,y \equal{} 3 \plus{} 4k$ with $ k \in \mathbb{Z}$[/quote]After all, if $ 4x\minus{}1\equal{}17y$ with $ x$ and $ y$ integers, then
$ 4x\minus{}1\equal{}17y$ must be a multiple of $ 17$ that's $ 1$ short from a multiple of $ 4$, so it must be $ 4x\minus{}1\equal{}17y\equal{}(4\cdot17)k\plus{}(3\cdot17) \equal{}68k\plus{}51$ for some integer $ k$
$ (68k\plus{}17$ and $ 68k\plus{}34$ are multiples of $ 17$, but are not $ 1$ short from a multiple of $ 4)$.
Solving $ 4x\minus{}1\equal{}17y\equal{}68k\plus{}51$ for $ x$ and $ y$ we get
$ x \equal{} 13 \plus{} 17k$, and $ y \equal{} 3 \plus{} 4k$.
However,
$ 0 < 13 \plus{} 17k \le 1000 \Leftrightarrow 0 \le k \le 58$
I count $ 59$ solutions.


[quote="zhaobin"][quote="Nameless"][quote="duongqua"]FOr all $a,b,c,x,y,z$ positive real numbers such that $xy+yz+zx=3$ we have
    $\frac{a}{b+c}(y+z)+\frac{b}{c+a}(z+x)+\frac{c}{a+b}(x+y)\geq 3$[/quote]
Assume that $x+y+z=1$
$\sum\frac{a}{b+c}x+\sqrt{3(xy+yz+zx)}\le\sqrt{\sum(\frac{a}{b+c})^2}\sqrt{\sum x^2}+\sqrt{\frac{3(xy+yz+czx)}{4}}+\sqrt{\frac{3(xy+yz+zx)}{4}}\le \sqrt{\sum(\frac{a}{b+c})^2+\frac{3}{2}}\sqrt{\sum x^2+2\sum yz}=\sqrt{\sum(\frac{a}{b+c})^2+\frac{3}{2}}$
But we have: $\sqrt{\sum(\frac{a}{b+c})^2+\frac{3}{2}}\le \sum\frac{a}{b+c}\Leftrightarrow\sum\frac{bc}{(a+b)(a+c)}\ge\frac{3}{4}$[/quote]
I can't see what had you proved...[/quote]
$\sum\frac{a}{b+c}x+\sqrt{3(xy+yz+zx)}\le\sum\frac{a}{b+c}\Rightarrow\sqrt{3(xy+yz+zx)}\le\sum\frac{a}{b+c}(1-x)=\sum\frac{a}{b+c}(y+z)$

[quote="Nameless"][/quote][quote="zhaobin"][quote="Nameless"][quote="duongqua"]FOr all $a,b,c,x,y,z$ positive real numbers such that $xy+yz+zx=3$ we have
    $\frac{a}{b+c}(y+z)+\frac{b}{c+a}(z+x)+\frac{c}{a+b}(x+y)\geq 3$[/quote]
Assume that $x+y+z=1$
$\sum\frac{a}{b+c}x+\sqrt{3(xy+yz+zx)}\le\sqrt{\sum(\frac{a}{b+c})^2}\sqrt{\sum x^2}+\sqrt{\frac{3(xy+yz+czx)}{4}}+\sqrt{\frac{3(xy+yz+zx)}{4}}\le \sqrt{\sum(\frac{a}{b+c})^2+\frac{3}{2}}\sqrt{\sum x^2+2\sum yz}=\sqrt{\sum(\frac{a}{b+c})^2+\frac{3}{2}}$
But we have: $\sqrt{\sum(\frac{a}{b+c})^2+\frac{3}{2}}\le \sum\frac{a}{b+c}\Leftrightarrow\sum\frac{bc}{(a+b)(a+c)}\ge\frac{3}{4}$[/quote]
I can't see what had you proved...[/quote][quote="Nameless"]
$\sum\frac{a}{b+c}x+\sqrt{3(xy+yz+zx)}\le\sum\frac{a}{b+c}\Rightarrow\sqrt{3(xy+yz+zx)}\le\sum\frac{a}{b+c}(1-x)=\sum\frac{a}{b+c}(y+z)$[/quote]
why you can assume $x+y+z=1$ :?:

[quote="zhaobin"]sorry,I am wrong.
maybe you rewrite the inequality as:
$\frac{a}{b+c}(y+z)+\frac{b}{c+a}(z+x)+\frac{c}{a+b}(x+y)\geq \sqrt{3(xy+yz+zx}$
sorry again.[/quote]
Yes. :)


geometry
complex numbers
number theory proposed
number theory

Do there exist any right triangles with integer sides and angles of rational degree? Its easy to show using geometry that if one such triangle exists infinitely many do.

This is equivalent to having a $n$-th root of unity such that it's real and imaginary part are rational.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=61379 , then you get that even one of them rational is rare.


geometry
parallelogram
analytic geometry
graphing lines
slope
rectangle
inequalities

WITH FULL SOLUTIONS PLEASE

Fractions a/b and c/d are called neighbor fractions if their difference (ad - bc)/(bd) has a numerator of positive or negative 1, that is, ad - bc = positive or negative one. Prove that

1. In this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);

2. If a/b and c/d are neighbor fractions, then (a + b)/(c + d) is between them and is a neighbor fraction for both a/b and c/d; moreover,

3. no fraction e/f with positive integer "e" and "f" such that f is less than b + d is between a/b and c/d. [/img]

[hide="Hint"] Consider the quadrilateral $ (0, 0), (b, a), (d, c), (b \plus{} d, a \plus{} c)$ and [url=http://en.wikipedia.org/wiki/Pick%27s_theorem]Pick's theorem[/url]. [/hide]

[quote="theax"]WITH FULL SOLUTIONS PLEASE

Fractions a/b and c/d are called neighbor fractions if their difference (ad - bc)/(bd) has a numerator of positive or negative 1, that is, ad - bc = positive or negative one. Prove that

1. In this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);[/quote]

[hide="Hint"]Note that saying a/b can be simplified is the same as saying a and b have a common divisor greater than 1. Prove that assuming such a common divisor leads to a contradiction.
[hide="Answer"]Lets call the common divisor k.Now you claim, for your "neighbor fractions", |ad - bc| = 1. Since (k divides a and hence ad) and (k divides b and hence bc), this means k divides |ad - bc|=1. But this [b]cannot[/b] happen because we already assumed k > 1.[/hide][/hide]

[hide="Solution"] The area of the above parallelogram is $ ad \minus{} bc$ (can you prove it?).  We are given that this area is $ 1$.  By Pick's Theorem, and since the quadrilateral has at least $ 4$ boundary lattice points, it has exactly $ 4$ boundary lattice points and no interior lattice points.

The fact that it has no boundary lattice points implies #1.  Otherwise, if $ \frac {a}{b} \equal{} \frac {a'}{b'}$ is a simplification then $ (b', a')$ is on the side between $ (0, 0)$ and $ (b, a)$, and the same for the other side.

The fact that it has no interior lattice points implies #3.

The fact that the parallelogram is convex and that there are no other interior and boundary lattice points implies #2.  The line from the origin to $ (b \plus{} a, d \plus{} c)$ is between the other two lines, so its slope is also strictly between the slopes of the other two lines.  And of course the parallelogram generated by $ (b \plus{} a, d \plus{} c)$ and either of the other two points is still a parallelogram of area $ 1$, again by Pick's Theorem.[/hide]
See [url=http://en.wikipedia.org/wiki/Farey_sequence]Farey sequence[/url].

Thanks a lot...
I wonder if there is another way of solving this

You can do everything with just algebra, but it's not quite as nice.  The geometric interpretation is a much more interesting framework to view this problem.

[quote="theax"]You said: The area of the above parallelogram is  (can you prove it?). Why?
Could you elaborate more on your explanations, please?[/quote]
If you surf the site enough, you should see a proof without words for it in that flash applet on the left side of the screen. :D

Start by proving that a triangle with coordinates (0,0),(a,b),(c,d) has area
$ \frac{|ad \minus{} bc|}{2}$

You can prove it geometrically by drawing the rectangle with sides $ a \plus{} b, c \plus{} d$ around the parallelogram and subtracting the appropriate areas, but I find that proof unsatisfactory.  The fun proof is with [url=http://en.wikipedia.org/wiki/Cross_product]cross products[/url].

Thanks...

Is there an easier way to solve these series of problems using just algebra?

[hide="an algebraic method"]
Say $ \frac {a}{b} > \frac {c}{d}$. As they are neighbor fractions, $ ad \minus{} bc \equal{} 1$. Now if $ \frac {e}{f}$ is greater than $ \frac {c}{d}$ but less than $ \frac {a}{b}$, then $ \frac {e}{f} > \frac {c}{d}\Rightarrow ed \minus{} cf > 0$.

But also $ ed \minus{} cf$ is an integer, so $ ed \minus{} cf\geq 1$. Similarly, from $ \frac {a}{b} > \frac {e}{f}$, we have $ af \minus{} be\geq 1$.

We want to bound $ f$, so eliminate $ e$ from the above two inequalities: $ b(ed \minus{} cf) \plus{} d(af \minus{} be)\geq b \plus{} d$. The terms with $ e$ cancel and what remains on the left hand side is $ f(ad \minus{} bc)$, which is just $ f$. So $ f\geq b \plus{} d$, which is what we wanted.
[/hide]


linear algebra
matrix
geometry
geometric transformation

i need help in finding  determinant D  (nxn  n>1)
determinant D in Attachments

The matrix is:

$ \left [\begin{array}{ccccccc}1&2&3&\cdots&n-2&n-1&n\\\noalign{\medskip}2&3&4&\cdots&n-1&n&n\\\noalign{\medskip}3&4&5&\cdots&n&n&n\\\noalign{\medskip}\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\noalign{\medskip}n-1&n&n&\cdots&n&n&n\\\noalign{\medskip}n&n&n&\cdots&n&n&n\end{array}\right ]$

Now just use elementary column operations to simplify the matrix but preserve the value of the determinant.

(i)  Take $ -1$ times the $ n-1$st column and add it to the $ n$th column to get $ D_{1,n}=1$ and $ D_{1,i}=0$ for $ i>1$.

(ii) Expand about the last column, but since you have all zeros except for the first element, you have $ |D|=(-1)^{(n+1)}|D_{1,n}|$.

(iii) Now take $ -1$ times $ n-2$nd column and add it to the $ n-1$st column to get all zeros in the final column of the $ |D_{1,n}|$ matrix with the exception of $ 1$ in the upper right position.

...

Continuing in like fashion, I believe you get $ |D|=\left( (-1)^{\frac{(n+1)(n+2)}{2}-1}\right) n$.

Npare wrote:

[quote]thank you very much

but>>>>

what do you mean this:
n-1 st ,n th,n-2 nd
any column ????[/quote]

This is basically just an English construct.  So for example:

$ n$ th  $ \rightarrow$  column $ n$
$ (n-1)$ st $ \rightarrow$ column $ n-1$

and so forth.

I remember bewildering some Francophone students with the phrase $ \text{i-i}\grave{e}\text{me}$ as a translation of $ i$ th.  Sorry.

To add some more detail,

Call the matrix with last row first column n $ A_{n}$. Notice that the last row has the same entries in each column. If $ a_{1}$ is the first column, add $ \minus{}a_{1}$ to columns 2 through n. This does not change the determinant so you may expand along the last row. The determinant $ \left|{A_{n}}\right| \equal{} (\minus{}1)^{n\plus{}1}n|B|$ where B is some matrix whose determinant is to be taken. Let us denote the last row of B by $ B_{(n\minus{}1)}$. Note that the ith row of $ A_{n}\minus{}1$, $ 1\le i < n\minus{}1$, is given by $ (i\minus{}1)*B_{(n\minus{}1)}$ + $ B_{i}$ and  that $ B_{(n\minus{}1)}$is n*$ A_{(n\minus{}1)}$. In other words, $ |B| \equal{}\frac{1}{{n\minus{}1}}|A_{n\minus{}1}|$. Put this back in the other equation and telescope to see that

$ \left|{A_{1}}\right|\equal{}1$ and $ \left|{A_{m}}\right| \equal{}\minus{}m(\minus{}1)^{\frac{{m(m\plus{}1)}}{2}\minus{}1}$ for m > 1.


algebra
polynomial
vector
superior algebra
superior algebra unsolved

Let W equal the set of all polynomials in F[x] with degree less than n such that the limit of the sum of the coefficients of the terms x_i is 0.  Show that W is a subspace of V_n and find a basis of W over F.

(V_n is the set of all polynomials in F[x] with degree less than n)

I know that to show W is a subspace of V, I must show that under the operations of V, W is a vector space over F[x].  But I don't know how to prove that in W, distribution exists.

Also, how do I find the basis?

I have no idea what you mean by "limit" here.

It still doesn't make sense, but at least there's enough there to guess. Do you mean "such that the sum of the coefficients of the $ x^i$ terms is zero"?

When trying to prove that a subset is a subspace, some of the axioms are automatic. You only have to worry about whether operations stay in the subspace, since the other properties are inherited from those properties in the full space.

[quote="rawkerxxx"]But I don't know how to prove that in W, distribution exists. [/quote]Huh?

[quote="jmerry"]It still doesn't make sense, but at least there's enough there to guess. Do you mean "such that the sum of the coefficients of the $ x^i$ terms is zero"?[/quote]

Yes.

When trying to prove that a subset is a subspace, some of the axioms are automatic. You only have to worry about whether operations stay in the subspace, since the other properties are inherited from those properties in the full space.

[quote="rawkerxxx"]But I don't know how to prove that in W, distribution exists. [/quote]Huh?[/quote]

Well, I need to show that W is itself a vector space over F[x], so I need to show that the axioms are satisfied.  So is it just automatic that since F[x] is a field then I don't need to show distribution?[/url]


function
limit
integration
real analysis
real analysis unsolved

Let $ f: [0; \infty) \to \mathbb{R}$ a derivable function. Prove that if $ \lim_{x \to \infty} f(x)$ exists and $ \lim_{x \to \infty } \int_0^x f(t) dt \equal{}0$ then $ \lim_{x \to \infty} f(x)\equal{}0$.

If $ \lim_{x \to \infty} f(x) \equal{} L \ne 0$ then $ \forall \varepsilon \exists x_0$ s.t. $ L \minus{} \varepsilon < f(x) < L \plus{} \varepsilon$ for $ x>x_0$ and then $ \int_{x_0}^{\infty} (L \minus{} \varepsilon)dt < \int_{x_0}^{\infty} f(t)dt < \int_{x_0}^{\infty} (L \plus{} \varepsilon)dt$. If $ \varepsilon$ is chosen small enough, we get a contradiction.

[quote="CommandoGuard"]If $ \lim_{x \to \infty} f(x) \equal{} L \ne 0$ then $ \forall \varepsilon \exists x_0$ s.t. $ L \minus{} \varepsilon < f(x) < L \plus{} \varepsilon$ for $ x > x_0$ and then $ \int_{x_0}^{\infty} (L \minus{} \varepsilon)dt < \int_{x_0}^{\infty} f(t)dt < \int_{x_0}^{\infty} (L \plus{} \varepsilon)dt$. If $ \varepsilon$ is chosen small enough, we get a contradiction.[/quote]
I'm not convinced by CommandoGuard's proof  : $ x_0$ is depending on $ \epsilon$ :P
$ f$ is continuous so $ \forall_n,\exists_{n < x_n < n \plus{} 1},\int_n^{n \plus{} 1}f(t)dt \equal{} f(x_n)$. We can write  $ \int_0^{\infty} f(t)dt \equal{} \sum_{n \equal{} 0}^{ \plus{} \infty}\int_n^{n \plus{} 1}f(t)dt \equal{} \sum_{n \equal{} 0}^{ \plus{} \infty}f(x_n)$
$ \sum_{n \equal{} 0}^{ \plus{} \infty}f(x_n)$ is convergent because $ \int_0^{\infty} f(t)dt$ is convergent,   so $ \lim_{n\rightarrow \plus{} \infty}f(x_n) \equal{} 0$, where $ x_n\rightarrow \plus{} \infty$ . But $ \lim_{x\rightarrow \plus{} \infty}f(x)$ exists, so $ \lim_{x\rightarrow \plus{} \infty}f(x) \equal{} 0$
 :cool:

I don't think there is any problem with the first solution. We are not taking anything like $ \\epsilon \rightarrow 0$; just fix $ \epsilon$ so that the two bounds are of the same sign and diverge to the infinity then that's okay.

Ok, I agree  :D  . Finally the proof of CommandoGuard is better, because  only the convergences of $ f(\plus{}\infty)$ and $ \int_0^{\plus{}\infty}f(t)dt$ are necessary.
 :cool:


While traveling home from school, Alyssa fell asleep halfway through the journey. When she awoke, she still needed to travel one-fourth of the distance that she had traveled while sleeping. For what part of the journey was she awake? Express your answer as a common fraction.

Let $ x$ be the distance she traveled while asleep.

Let $ y$ be the total distance.

$ y \equal{} x \plus{} \frac{1}{4}x \plus{} \frac{1}{2}y \implies y \equal{} \frac{5}{4}x \plus{} \frac{1}{2}y$

$ \frac{1}{2}y \equal{} x \plus{} \frac{1}{4}x \implies \frac{1}{2}y \equal{} \frac{5}{4}x$

What we want to find is the simplified value of $ \frac{1}{2}y \plus{} \frac{1}{4}x$.

We can find $ x$ in terms of $ y$.

$ \frac{1}{2}y \equal{} \frac{5}{4}x$

$ \frac{2}{5}y \equal{} x$

We want to find $ \frac{1}{2}y \plus{} \frac{1}{4}x$, which we can  now simplify to:

$ \frac{1}{2}y \plus{} (\frac{1}{4} \cdot \frac{2}{5}y)$

or

$ \frac{1}{2}y \plus{} \frac{1}{10}y$

Thus, our answer is $ \frac{1}{2} \plus{} \frac{1}{10}$ or $ \boxed{\frac{3}{5}}$.

You can also imagine the trip itself. For the first 1/2 the way, Alyssa was awake. Then, she travels some distance, and when she wakes up, she needs to travel 1/4 of the the way that asleep time. Thus, 5/4 of the asleep time is equal to one half the total time, so the total time is 10/4 times the asleep time. The first half is 5/4 of the asleep time, and the woke up time is 1/4 as stated earlier, so she was awake for 6/4 of the asleep time. The fraction of time she was awake is $ \frac{\frac{6}{4}}{\frac{10}{4}}\equal{}\frac{6}{10}\equal{}\frac{3}{5}$ of the total time.


Question

A rectangular area can be completely tiled with 200 square tiles.
If the side length of each tile was increased by 1cm, it would take only 128 tiles to tile the area.
Find the side length of each tile.

If we let $ a$ and $ b$ be the lengths of the sides of the rectangle in centimeters and let $ x$ be the length od each side of the square tile in centimeters,
The number of tiles needed to cover the rectangle is $ \frac{ab}{x^2}\equal{}200$.
If the length of each side is increased by 1 cm then the side length is $ x\plus{}1$.
The number of tiles that will be required now will be $ \frac{ab}{(x\plus{}1)^2}\equal{}128$.
Simplifying we get:
                           $ ab\equal{}200(x^2)$
                           $ ab\equal{}128(x\plus{}1)^2$

By the Transitive property, we get:
                           $ 200x^2\equal{}128x^2\plus{}256x\plus{}128$
                           $ 72x^2\minus{}256x\minus{}128\equal{}0$
                            $ 9x^2\minus{}32x\minus{}32\equal{}0$
                            $ (9x\plus{}4)(x\minus{}4)\equal{}0$
                            $ x\equal{}\minus{}\frac{9}{4}\text{ or }4$

Therefore, the length of each side of the square tile is 4 cm.


Show that for y'''' = (a^4)y, y(0) = 0, y'(0) = 0, y(L) = 0, y'(L) = 0, to have a non-trivial solution cos(aL)cosh(aL) = 1. Assume a does not equal 0.

I have the following hint: Show that a basis for the null space of L = x'''' - a^4 is (cosh(ax), sinh(ax), cos(ax), sin(ax)).

How would I show this?

Thanks for any help in advance.

I do it and I get that cos(aL) = cosh(aL), but that must be wrong.

I let y(x) = Acosh(ax) + Bsinh(ax) + Ccos(ax) + Dsin(ax) for some constants A,B,C,D.
For y(0) = 0, I get that C = -A and y'(0) = 0, it seems that D = -B.
I'm not sure what to do after this.


Square ABCD is located inside of a circle with center [i]O [/i]and radius R in such a way that verticies B and C belong to the circle and side AD contains point [i]O[/i]. Find [i]r[/i][size=59]BOC[/size]

well you only need to note that the inradi of a triangle with sides $ a,b,c$ and semiperimter $ s$ is given by:
$ r\equal{}\sqrt{\frac{(s\minus{}a)(s\minus{}b)(s\minus{}c)}{s}}$ and use Pithagoras in triangle $ ODC$.

 :)


limit
trigonometry
real analysis
real analysis unsolved

Can anyone help me with these two problems?

lim x^2cos10x as x->0

and the second:

lim (1/t^2)sin^2(t/2) as t->0

$ \lim_{x\to 0}x^{2}\cos(10x) \equal{} 0\cdot 1 \equal{} 0$.

$ \lim_{t\to 0}\frac{1}{t^{2}}\sin^{2}\left(\frac{t}{2}\right) \equal{}\frac{1}{4}\lim_{t\to 0}\left[\frac{\sin\left(\frac{t}{2}\right)}{\frac{t}{2}}\right]^{2}\equal{}\frac{1}{4}\cdot 1^{2}\equal{}\frac{1}{4}$.

In the second one I used the fact that $ \lim_{x\to 0}\frac{\sin x}{x}\equal{} 1$.


Let w be a polynomial of degree two with integer coeff. Suppose that for each integer x the value w(x) is the square of an integer. Prove that w is a square of a polynomial.

This one appeared many times here.
The last and better result on that problem is Peter Scholze's proof of Harazi conjecture, namely :
If $P$ has integer coefficients and $P(2^n)$ is a square for all non-negative integers $n$, then $P$ is the square of a polynomial with integer coefficients.

Look here :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=polynomial+square&t=15102

Pierre.


algebra
polynomial
linear algebra
linear algebra unsolved

Suppose that $ P_n$ be the space of all polynomials whose degrees $ \leq n$ with coefficients $ \in \mathbb{R}$. We define the inner product for given $ f(x)\equal{}\sum_{i\equal{}0}^n a_ix^i ,  g(x)\equal{}\sum_{i\equal{}0}^n b_ix^i \in P_n$ as following : $ \left\langle f,g\right\rangle\equal{}\sum_{i\equal{}0}^n a_ib_i$. Suppose that $ L \subseteq P_n$ be the set of all polynomials like $ f$ s.t. $ f(1)\equal{}0$ (obviously $ L$ is a subspace). For $ g \in P_n$ we define the distance between $ g$ and $ L$ as $ \rho (g,L)$. Prove that $ \forall g \in P_n , \rho (g,L)\equal{}\frac{g(1)}{\sqrt{n\plus{}1}}$.

The codimension of $ L$ is 1. Therefore, $ \dim L^{\perp} \equal{} 1$. It is easy to see that $ h(t) \equal{} 1 \plus{} t \plus{}\dots \plus{}t^n\in L^{\perp}$ since $ \langle f(t), h(t)\rangle \equal{} f(1)$. Therefore, $ L^{\perp} \equal{} \{\lambda h\}$.
Now the distance from $ g$ to $ L$ equals the length of the projection of $ g$ onto $ L^{\perp}$. We have,
\[ \rho(g, L) \equal{} \frac{|\langle g, h\rangle|}{\|h\|} \equal{} \frac{|g(1)|}{\sqrt{n\plus{}1}}.\]


[hide]$ {\cos 37^\circ\plus{}\sin 37^\circ\over\cos 37^\circ\minus{}\sin 37^\circ}\equal{}\frac{\sqrt{2}\sin(45^\circ\plus{}37^\circ)}{\sqrt{2}\sin(45^\circ\minus{}37^\circ)}\equal{}{\sin 82^\circ\over\sin 8^\circ}\equal{}\cot 8^\circ$[/hide]


[quote="aznmonkey1992"]little jacob had a pie and gave 1/4 of it to his dear little mommy
then, he gave 1/3 of what was left to his dear little daddy  :rotfl:
lastly, he ate 3/16 of the portion left
whats left? (fraction)[/quote]

[hide]
$\frac{1}{4}+\frac{3}{4}\cdot\frac{1}{3} = \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
$\frac{1}{2}\cdot\frac{3}{16}=\frac{3}{32}$
$\frac{16}{32}+\frac{3}{32}=\frac{19}{32}$
$1-\frac{19}{32}=\boxed{\frac{13}{32}}$
[/hide]

[quote="aznmonkey1992"]little jacob had a pie and gave 1/4 of it to his dear little mommy
then, he gave 1/3 of what was left to his dear little daddy  :rotfl:
lastly, he ate 3/16 of the portion left
whats left? (fraction)[/quote]
$1-(1/4+1/4+3/32)=1-(16/32+3/32)=13/32$

[quote="aznmonkey1992"]little jacob had a pie and gave 1/4 of it to his dear little mommy
then, he gave 1/3 of what was left to his dear little daddy  :rotfl:
lastly, he ate 3/16 of the portion left
whats left? (fraction)[/quote]
[hide]

1-1/4=3/4
2/3x3/4 = 6/12 = 1/2
13/16 x 1/2 = 13/32
[/hide]

[quote="aznmonkey1992"]little jacob had a pie and gave 1/4 of it to his mommy
then, he gave 1/3 of what was left to his daddy
lastly, he ate 3/16 of the portion left
whats left? (fraction)[/quote]

he started with 1. then after his mommy, he still had 3/4. He gave 1/3 of 3/4 to his daddy, or 1/4. So he had 1/2 left. Then he ate 3/16 of 1/2 or 3/32 of what was left. Therefore, there was 13/32 of the pie left.


inequalities
number theory solved
number theory

Find all $x,y,z,\in Z^+$ satisfying the equation:
$\sqrt {xyz} - \sqrt x - \sqrt y - \sqrt z =2$

There's a standard way to tackle this, by using inequalities after setting $x\le y\le z$. However, we can do it algebraically. The number on the left is a member of the field $\mathbb Q[\sqrt x,\sqrt y,\sqrt z]$, and it can be shown that such numbers are rational iff the coefficients of each irrational among $\sqrt x,\sqrt y,\sqrt z$ and the combinations formed by the multiplications of two or three of these ($\sqrt {yz},\sqrt{zx},\sqrt{xy},\sqrt{xyz}$) is $0$. If $\sqrt x$ is irrational (for example), then $\sqrt {yz}=1$, and this doesn't lead to a solution. This means that $x,y,z$ are, in fact, squares, so we can rewrite the equation in positive integers: $mnp-m-n-p=2$.

It's now much easier to employ the usual methods: $\frac 2{mnp}+\frac 1{mn}+\frac 1{np}+\frac 1{pm}=1$, and we may set $m\le n\le p$. From here we get $m\le 2$, and if $m=2$, then $n=p=2$, so this is a solution: $(m,n,p)=(2,2,2)\Rightarrow (x,y,z)=(4,4,4)$. If $m=1$, then $\left(\frac 3n+1\right)\left(\frac 3p+1\right)=4$, so $n\le 3$. $n=3\Rightarrow p=3$. Here's another solution: $(x,y,z)=(1,9,9)$. $n=2\Rightarrow p=5$, so another solution is $(x,y,z)=(1,4,25)$. Finally, if $n=1$, then there's no solution.

The only solutions are thus $(x,y,z)=(4,4,4),(1,9,9)$ or $(1,4,25)$, and their permutations.

I hope it's Ok.


AMC USA(J)MO USAMO inequalities algebra polynomial function

Hello, Please comment which country conducts the hardest team selection test  and national olympiad.
                                          THANKS

china, obvious. Basically CMO is harder than IMO

[quote="galois"]yeah basically the countries are china,romania ,usa.but i've never seen the vietnamese selction test nor the russian ones.duz ne1 have ne idea where they cud be found.and ,siuhochung,u have china 2002-2003 official solution? :?
best regards[/quote]

i dun hv, but you may ask members from china, or charlie, siutsz, they have joined cmo 2003

For China MO official solutions they were given as a talk for leaders only, during the 2nd day of contest. I could possibly get the solution for 2002 but not 2003 because our leader in 2003 didn't attend the talk at all :( I will try to find the Macau team leader for the solution

In france, we have two olympiads :
-the academic math olympiad, really easy (designed for common high school students ot the 2nd year)
-the general competition, really difficult, for high school last year students, but not like the imo : there's a big problem with many many sub-questions to be solved.

for the vietnamese selction test of 1994 : (2*4h)

1/ ABCD is a parallelogram. E is a point of BC and F a point of CD so that ABE and BDF have the same area. BD crosses AE at the point M and AF at N. Prove that BM, MN, ND follows the triangular inequality, and that E and F moves so that MN goes smaller, then the radius of the circle meeting the 3 summits (sorry, don't know how to say in English...) of a triangle which have for lengths BM, MN, ND also goes smaller.

2/ Let the equation x+y+z+t-Nxyzt-N=0 (*) (N integer >0).
Prove that for an infinity of values of N, (*) have a solution in positive integers, and that for N=(4^k)(8m+7) (k, m nonnegative integers), (*) has no solutions in positive integers.

3/ P is a polynomial function of degree 4, with 4 real roots.
Prove that the equation (1-4x)P(x)/x+(1-(1-4x)/x)P'(x)-P''(x)=0 has also 4 real roots.

4/ ABC is a triangle with AB=BC=CA (dunno what's this in English...), and M is a point. D is the midpoint of AD. DM and AP crosses at point N. The images by the symetry of center M of the points A, B, C are A', B', C'. Prove that there is an unique point P so that AA=PB', PB=PC', PC=PA', and that when M moves while being different of D, the circle meeting M, N, P keeps meeting a point which is not moving.

5/ Find all the functions f:R->R so that :
f(x sqrt(2))+f(x (4+3sqrt(2))=2f(x (2+sqrt(2)).

6/ Compute sum(1/(n1!n2!...n1994!(n2+2n3+...+1993n1994)!)), with the sum on all the 1994-uplets (n1,...,n1994) of nonnegative integers so that n1+2n2+...+1994n1994=1994.

Difficult, isn't it ?

I've got a selection of past Vietnamese nat'l Tests. If anyone cares, I can post them here. General impression is that they are hard ;)  ( and kind of requiring university-level knowledge, if you ask me )

it depends what you define university level by. for example in romania we do in 11th and 12th grades what people in us do in 1st and 2nd year of college. (you can see the problems in the college playground).

hi guys
i just wanna say that there are  selection tests are very very easy
also the preparation for the imo is very short not efficace
Morrocan team is selected after just 3tests (asy tests)
i hope that this problem will be solved as i was a victime of the easy tests :?
any way !
i just wanna ask for the mathematical program in college or high schools in romania or Russia
i think our program is  :blush:
thank for all guys by!

[quote="Anonymous"][quote="galois"]yeah basically the countries are china,romania ,usa.but i've never seen the vietnamese selction test nor the russian ones.duz ne1 have ne idea where they cud be found.and ,siuhochung,u have china 2002-2003 official solution? :?
best regards[/quote]

the USAMO 2003 was fairly easy (at least for four questions).  if i can do them, then it is only fair to classify as fairly easy.   :D  :D  :D
I have looked at ChMO, and they are very difficult.

The CMO (canadian) from 2003 was a joke...  the trival questions, two standard (aka easy) questions, and one question that is a little bit more difficult but is still easy.[/quote]

Yeah, I live in Canada and I can see students in my class using a calculator for 2-3 (-1). It's sad. No offence! There are also lots of intelligent people, but hard to find.

[quote="boxedexe"]Yeah, I live in Canada and I can see students in my class using a calculator for 2-3 (-1). It's sad. No offence! There are also lots of intelligent people, but hard to find.[/quote]

Of course, the relationship between CMO qualifiers and the students in your class using a calculator to subtract numbers is nonexistent.  So your post is pointless.

But, more importantly, don't resurrect year old threads in the future.

What about Argentinian TST for IMO??? (spanish version)

http://www.oma.org.ar/enunciados/index.htm#imo

[quote="blahblahblah"]Do you even read threads all the way through?  Don't bump old threads![/quote]

what do u mean? :!:

if u a means every one, use Quote.

Romania have very good problems, but are and easy(some of problems) :) ! Iran and China have very big problems :( .

i dont khow a lot about other countries.but i suppose IRAN have one of most hard selections.
 :(

[quote="kelvinkklee"]Britain has many selection tests, but the first few ones are really easy, which is pointless to select students for the IMO team members. The first selection test is even ridiculous, which is composed of 25 IQ-like muliple choice questions.[/quote]

The first few are meant to be easy. The SMC is designed to be accessible to as many people as possible.

BMO 1 is also suppose to be easy so that it can interest people who've never heard of the Olympiads before and also to test a few of the better people (though not much).

BMO 2 is the first test when it starts getting hard and this year was no exception. With 55 people getting 2 or less out of 180 entrants, I think its fair to say that its hardly "too easy".

(Although compared to USAMO etc it is easy). However, there are many more TSTs once the Trinity lot have had their training and after Oundle.

[b]Edit: Whoa, I didn't see how old this thread was![/b]

[quote="SimonM"][b]Edit: Whoa, I didn't see how old this thread was![/b][/quote]Almost 5 years :)

Except someone else had responded to it more recently (like yesterday) which confused me.

It's still a valid topic nevertheless. Even in our modern times  :lol:

What are people's opinions now? Is China still Numero Uno?

I don't think India's selection tests are the hardest of the lot.. But I believe others (Soumyashant, Chappli, etc) would be better suited to comment on this than myself.

[quote="Amazigh"][quote="Johan Gunardi"][quote="galoisj"]
... all guys of national training team have the ability to get IMO medals ...[/quote] :o  Indonesia has been taking part in IMO for 20 years. But we have yet to get gold. There are only two students who got silver, in 2001 and 2007. What do you think the differences between China and Indonesia?[/quote]

1) Math olympiad history and the structure of the olympiad in China (math camps etc)
2) Number of students that participate[/quote]

Yeah,you are right. :)

[quote="galoisj"]I am from China
Yes,I think CMO is little harder than IMO,but I should clarify that CMO is not the IMO team selection test.

CMO is just used to form the national training team which has about 35 students(seems have expanded to 45 in the last two years)

And in national training team,there are generally 6 small tests and 2 big ones.
Because all guys of national training team have the ability to get IMO medals.The problems of national training team tests are harder than those of CMO.
Some problems are even much much harder which almost no one from China can solve out.[/quote]

[url=http://www.mathlinks.ro/resources.php?c=37&cid=47]China TST[/url] has been regarded as very challenging though not always the most beautiful problems. And the [url=http://www.mathlinks.ro/index.php?f=10]Mathlinks contest[/url] which only took place in rare cases in the most recent years. As you could see we have a few years of the China TST in the resources section. Can you tell us more about the "national training team tests". Do you know the original sources of those problems. Possibly you can post those problems here.
Romania TST 2008 looks odd. Most problems are based on AMM problems. What is going on? Are they running out of ideas or good people who propose interesting problems. Maybe they think people are training with ISLs anyway, and why not motivate those students to read AMM more in detail?!

[quote="Valentin Vornicu"]Well I was kind of upset on the Romanian selection tests this year. I sent in a few problems of my own, which didn't get selected (I used most of them for MLC anyway in the end.)[/quote]

I can't seem to locate all problems from the recent MLCs. Where are they? I saw one of your problems (the 8th problem in the list which is a geometry one) as proposal for RMC 2008. For the different contests and levels 71 problems were proposed in total. I mean they have some alternatives so I don't understand they use so many known problems from AMM.


For $ A \equal{} \left( \begin{array}{cc} 1 & 2 \\
3 & 4 \end{array} \right),\ B \equal{} \left( \begin{array}{cc} 0 & \minus{} 1 \\
\minus{} 2 & 3 \end{array} \right),$ find $ (A \plus{} B)^2, A^2 \plus{} 2AB \plus{} B^2.$

Question from kunny:

What should we learn from the problem?

[quote="kunny"]Question from kunny:

What should I learn from the problem?[/quote]
Once again, commutativity - or rather the lack of commutativity - is the issue.

Needless to say, that's right.

What would become $ (A\plus{}B)^2$ for matrices $ A,\ B$ generally?

I had asked a question like this before. I do know the answer.

[quote="kunny"]

What would become $ (A \plus{} B)^2$ for matrices $ A,\ B$ generally?[/quote]

[quote="rob22"]$ A^2 \plus{} AB \plus{} BA \plus{} B^2.$ For some $ A$ and $ B$, and in this case,$ AB \neq BA,$ so you cannot simplify this using commutivity of multiplication, since it is generally NOT true given two randomly selected matrices.[/quote]

Perfect! :)


inequalities
Euler
inequalities proposed

Prove that

$ 8R^2\minus{}5r^2\leq r_a^2\plus{}r_b^2\plus{}r_c^2\leq16R^2\minus{}24Rr\plus{}11r^2$

[quote="Ligouras"]Prove that

$ 8R^2 \minus{} 5r^2\leq r_a^2 \plus{} r_b^2 \plus{} r_c^2\leq16R^2 \minus{} 24Rr \plus{} 11r^2$[/quote]

we have

\[ \sum {r_a^2 }  \equal{} (4R \plus{} r)^2  \minus{} 2s^2\]
and use Gerretsen Inequality $ 4R^2 \plus{} 4Rr \plus{} 3r^2 \ge s^2 \ge 16Rr \minus{} 5r^2$ and Euler Inequality $ R \ge 2r$ can prove it.


trigonometry
inequalities
inequalities proposed

Let $a,b,c$ be the angles of a triangle. Prove that

$ 8\sin{\frac{a}{2}}\sin{\frac{b}{2}}\sin{\frac{c}{2}} \leq \sqrt{\cos{\frac{a-b}{2}}\cos{\frac{b-c}{2}}\cos{\frac{c-a}{2}}}$

[quote="manlio"]Let $a,b,c$ be the angles of a triangle. Prove that

$ 8\sin{\frac{a}{2}}\sin{\frac{b}{2}}\sin{\frac{c}{2}} \geq \sqrt{\cos{\frac{a-b}{2}}\cos{\frac{b-c}{2}}\cos{\frac{c-a}{2}}}$[/quote]

Dear manlio,
Your inequality is very nice but $ \geq\ $ should be changed to $ \leq\ $

To prove it use these indentities :
$ \sin{\frac{a}{2}}\sin{\frac{b}{2}}\sin{\frac{c}{2}} = \frac{r}{4R} $
${ cos{\frac{a-b}{2}}cos{\frac{b-c}{2}}\cos{\frac{c-a}{2}}} = \frac{p^2+2Rr+r^2}{8R^2} $

Our inequality becomes : $ p^2+2Rr \geq\ 31r^2 $ which is very easy to prove.


LaTeX
integration
calculus
calculus computations

In my haste to learn latex I made a horrible mistake.  :blush: I apologize for it. I meant to ask you all to solve
$\int\frac{x^{5}}{(x^{3}+1)^{2}}dx$

I do thank you for your efforts to solve

$\int\frac{5x}{(x^{3}+1)^{2}}dx$

Actually I think the top integral is easier to solve than the bottom one.

It's far easier. Substitute $u=x^{3}+1$.

i think this is simple integration
substitute $x^{3}= t$ then $3x^{2}= dt$
now my integration becomes $\int{\frac{tdt}{3(t+1)^{2}}}$
which can be written as $\frac{1}{6}({\int{\frac{(2t+2)dt}{t^{2}+2t+1}}-2\int{\frac{dt}{(t+1)^{2}}}})$
on integrating this we get
$\frac{1}{3}(ln|t+1|+\frac{1}{t+1})+c$
on substituting the value of t we get
$\frac{1}{3}(ln|x^{3}+1|+\frac{1}{x^{3}+1})+c$

my approach is to factor $\frac{x^{5}}{(x^{3}+1)^{2}}$
as

$\frac{x^{3}}{3}\cdot\frac{3x^{2}}{(x^{3}+1)^{2}}$

Then integrate by parts

I don't have the solution in front of me at present and my latex skills are not that good right now so maybe someone could show this integration by parts. BTW there is a substituition after the integration by parts but  it is trivial

Back to the latex  to try to figure out how to write out integration by parts.

It is interesting how these integrals can have more than one solution. Is there a word for that?


How do we calculate $100!$, of course without using any tools?

More specifically, what is the "lest most significant digit" for $100!$?

[quote="xxxyyyy"][quote="Kurt Gödel"]I guess you mean "right most significant digit"  :wink:

Similar problem posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=ravi&t=102715]here[/url].[/quote]

The right most digit is $0$, isnt it?

I meant the "left most digit" only... thats why I asked for a method to evaluate this monster...[/quote]
right most [b]significant[/b] (>0) digit

ok well 100! end in 24 zeros, so to get the rightmost non zero number we must divide by 10^24. Then, look at the leftover numbers mod 10 and simplify.  It's not a hard process, but I won't actually do it as 100 is kindof large for this, I think when i saw this it was 24! instead.

and to find the leading digit you would take the logarithm of $100!$, then take the fractional part of that number, and raise $10$ to that power. the leading digit of the final number is the leading digit of $100!$.


logarithms
geometric series
real analysis
real analysis unsolved

[quote="ffrogg"]Denote by $m(n)$ the largest prime factor of $n$,where $n$ is an integer.
Prove that $\sum_{n=2}^{\infty}\frac{1}{n \cdot m(n)}$ converges.[/quote]

$\sum_{n=2}^{\infty}\frac{1}{n \cdot m(n)}\le \sum_{n=2}^{\infty}\frac{1}{p_{n}}$ where $p_{n}$ is the n_th prime.

But $\sum_{n=2}^{\infty}\frac{1}{p_{n}}$ seems to be divergent.

All terms are positive, so I can rearrange at will.  Pull out all those values for which $m(n) = p$ for a fixed $p$.  The sum of these terms is $\frac{1}{p^{2}}\sum_{n \in S_{p}}\frac{1}{n}$ where $S_{p}$ is the set of all integers whose prime factors are all less than or equal to $p$.  This is $\prod_{q \leq p}\frac{q}{q-1}$ where the product is over prime $q$.  Thus our total sum is $1+\sum_{p}\frac{1}{p^{2}}\prod_{q \leq p}\frac{q}{q-1}$.  I'm not immediately sure how to show that this converges; comparison with a geometric series fails, and I don't remember any of the cleverer convergence tests.

Incidentally, the limit appears to be substantially less than 3, but harmonic-y things tend to require a lot of terms to get big, so it could be substantially larger.

A simple estimate: All primes greater than 2 are odd, so $\prod_{q\le p}\frac{q}{q-1}\le \frac21\cdot \frac32\cdot \frac54\cdot \frac76\cdot \frac98\cdots \frac{p}{p-1}\le \sqrt{(\frac21)^{2}\cdot(\frac32)^{2}\cdot\frac43\cdot\frac54\cdot \frac65\cdot \frac76\cdots \frac{p}{p-1}}=\sqrt{3p}$

The $n$th prime is greater than $n$, so the sum is smaller than $\sum_{p}\frac1{p^{2}}\sqrt{3p}=\sum_{p}\sqrt{3}p^{\frac32}< \sqrt{3}\sum_{n}n^{\frac32}$. This converges.

With stronger estimates, we have something like $\frac{1}{n^{2}\log n}$.


Let be given a circle with diameter AB and point X on AB. Let P be a arbitrary point of this circle, different from A and B.
Prove that

 [tan(APX)]/[tan(PAX)] does not depend on choice of P.

at reference frame

[tan(APX)]/[tan(PAX)]=AX/XB  does not depend on choice of P.
[/u][/quote][/b][/i]


induction
complex numbers
algebra unsolved
algebra

Q1 Show that $ (a_1 ^2\plus{}b_1 ^2)(a_2 ^2 \plus{}b_2 ^2 )...(a_n ^2 \plus{}b_n ^2)$ can be written as sum of two squares

Q2 Let z1 and z2 be two complex numbers such that $ \left|z_1 \right| < 1 < \left|z_2 \right|$ Show that $ \left| \frac {1 \minus{} \bar{z_1}z_2}{z_1 \minus{} z_2} \right| < 1$

[quote="jackie-chan01"]Q1 Show that $ (a_1 ^2 \plus{} b_1 ^2)(a_2 ^2 \plus{} b_2 ^2 )...(a_n ^2 \plus{} b_n ^2)$ can be written as sum of two squares[/quote]

$ (a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ac\plus{}bd)^2\plus{}(ad\minus{}bc)^2$ and an immediate induction gives the required result.

[quote="jackie-chan01"]Q2 Let z1 and z2 be two complex numbers such that $ \left|z_1 \right| < 1 < \left|z_2 \right|$ Show that $ \left| \frac {1 \minus{} \bar{z_1}z_2}{z_1 \minus{} z_2} \right| < 1$[/quote]

We get $ (1\minus{}|z_1|^2)(1\minus{}|z_2|^2)<0$

$ \implies$ $ 1\plus{}z_1\overline{z_1}z_2\overline{z_2}$ $ <z_1\overline{z_1}\plus{}z_2\overline{z_2}$

$ \implies$ $ (1\minus{}\overline{z_1}z_2)(1\minus{}z_1\overline{z_2})$ $ <(z_1\minus{}z_2)(\overline{z_1}\minus{}\overline{z_2})$

$ \implies$ $ |1\minus{}\overline{z_1}z_2|^2$ $ <|z_1\minus{}z_2|^2$

Hence the result


[quote="shortlist"][size=150]Suppose that $ a_0,a_1,...a_n$ are real numbers such that for all $ \minus{} 1 \equal{} < x < \equal{} 1$ , $ \minus{} 1 \equal{} < a_0 \plus{} a_1x \plus{} ... \plus{} a_nx^n < \equal{} 1$
Show tht for all  $ \minus{} 1 \equal{} < x < \equal{} 1$   then  $ \minus{} 2^{n \minus{} 1} \equal{} < a_n \plus{} a_{n \minus{} 1}x \plus{} ...... \plus{} a_0x^n < \equal{} 2^{n \minus{} 1}$[/size][/quote]

That's wrong : choose as counter-example $ n\equal{}0$ and $ a_0\equal{}1$


algebra
polynomial
real analysis
real analysis unsolved

Find a polynomial $f ( X, Y ) \in \mathbb{R} [ X, Y ]$ such that $\mbox{Im } f = \mathbb{R^{+}}-\{0\}$ for all $x, y \in \mathbb{R}$.
For which degrees (on $X$ and $Y$) such a polynomial exist ? (I answered only in the case of $\mbox{deg}_X f(X,Y) = 2$)

p(x,y)=x^2+y^2+1
Could you check your statement ?

I'm sorry. I was kind of "distracted" while writting the statement ;)


Find all polynomials of degree $n$ with real coefficients $a_0$,...,$a_n$ and $n$ real zeroes $x_1$,...,$x_n$ so that $\left\{a_i\right\}_{0\leq i\leq n}$ and $\left\{x_i\right\}_{1\leq i\leq n}$ are two arithmetic sequences.

Misha

If you know a solution then you can post it in http://www.mathlinks.ro/Forum/viewtopic.php?t=24565 ;)

Misha, do you know [url=http://www.mathematik-olympiaden.de/IMOs/imo04pic.htm]Michail Shkolnikov ?[/url]   :)


hello, rewriting your first equation in the form
$ (x\plus{}y)(x^2\minus{}xy\minus{}y^2)\equal{}35$
and simplifying we get
$ x^2\minus{}xy\minus{}y^2\equal{}7$
plugging
$ y\equal{}5\minus{}x$ in the equation above we get
$ x^2\minus{}x(5\minus{}x)\minus{}(5\minus{}x)^2\equal{}7$
solving this equaton and deriving $ y$ we get
$ (3,2)$ or $ (2,3)$ as the unique solution of our system.
Sonnhard.

You're right, of course, Sonnhard.  :lol:


function
real analysis
real analysis unsolved

Find all functions $ f: [0,1]\rightarrow \mathbb{R},f\in C([0,1]),~f$ derivable on $ \langle 0,1 \rangle$ such that
$ f(0)\equal{}1,f(1)\equal{}2,~~f'(x)\plus{}2f(x)\geq 2x\plus{}3, ~\forall x\in \langle 0,1 \rangle$.

Put $ f(x)\equal{} x\plus{}1\plus{}K(x)e^{\minus{}2x} \Longrightarrow K'(x)\geq 0$ so $ K(0)\leq K(x) \leq K(1)$
$ (f(1)\equal{}1 \Longrightarrow  K(0)\equal{}0)$ and $ ( f(2)\equal{}2 \Longrightarrow  K(1)\equal{}0 )$
Finally $ K(x)\equal{}0$, so $ f(x)\equal{}x\plus{}1$
:cool:


Find all solutions to the system

x = 3y - 2
y = 3z - 2
z = 3x - 2.

Anyone get this one?

Hmm...no trig substitutions are involved, right?  :roll:

I believe

[hide](-2,-2,-2) and (1,1,1) are the only solutions.[/hide]

[color=#00FF00]Can you support that belief?[/color]

would anyone actualy CARE TO POST SOLUTIONS??????
answers are WORTHLESS!!!!!!!

Solution:

[hide]

Summing the three equations, we obtain

x^3 + y^3 + z^3 - 3x - 3y -3z + 6 = 0

(x^3 - 3x + 2) + (y^3 - 3y + 2) + (z^3 - 3z + 2) = 0

(x - 1)^2 * (x + 2) + (y- 1)^2 * (y + 2) + (z - 1)^2 * (z + 2) = 0.

We can obtain the solutions (1, 1, 1) and (-2, -2, -2) from this equation immediately.

Note that at least one of x, y and z has to be less than -2. Suppose x :le: -2. Then by x^3 = 3y - 2, y :le: -2 as well. Similarly z :le: -2. Therefore when variable is less than -2, all three are, and LHS is negative. So there are no other solutions.[/hide]

My mind is too small for both math and grammar.

OK ... there's a shorter proof anyway :) Less calculations !
Suppose x, y, z are not all equal. Then you"ll get a contradiction. :D
Try it !

No, my own solution does NOT involve trig substitution :mrgreen:

Hey guys, i take the time to post solutions only after i find out they are correct.

my solution:

The proof by contradiction was the basis that I used. I did not however prove it myself, I will let someone else take that. Knowing that the variables all must be equal, we have:

x=y=z
or

x^3=3x-2

or

(x-1)^2(x+2)=0

x=1 and -2, and they are the only solutions.... that was how I obtained my solution.
k, bye.

[quote="Arne"]So here goes my solution :

Suppose two of x, y, z are not equal. Then WLOG x < y.
Hence x < y => 3y - 2 < 3z - 2 => y < z.
So y < z => 3z - 2 < 3x - 2 => z < x.
So x < y < z < x, which is impossible.
Therefore x, y z are equal.
They must be equal to one of the roots of the equation x - 3x + 2 = 0.
So the only solutions are (1, 1, 1) and (-2, -2, -2). :)[/quote]
yay that's what i did (i couldn't post before b/c my phone line broke AGAIN :() anways, note that if any two of the variables are equal, all three are automatically equal and you get the two solutions already found.

there was something like this on an AIME i think. it had a lot of 17s in the problem, that's all I remember...

Oh, and there are some typos in your solution Osiris ... Doesn't matter in fact.

Edited, sorry... ^^" I guess I really shouldn't do math at 1 AM.


number theory
greatest common divisor
superior algebra
superior algebra unsolved

If gcd (a,b)=1, show that gcd (a+2b, 2a+b) = 1 or 3

I've been working on this problem for so long but I can't really get anywhere.

any hints?

$ \rm{HINT}\; (a\plus{}2b,2a\plus{}b) \supset (3a,3b) \equal{} 3(a,b) \equal{} 3$

Wouldn't it be the opposite containment? If a is larger than b, then gcd(a+2b, 2a+b) is less than gcd(a+2b, 2a+b)...no?

My hint uses ideals. If they're not familiar then read $ (x,y)$ as $ {\rm gcd}(x,y)$ and "contains" as "divides". Then it amounts to saying that $ d|a\plus{}2b,2a\plus{}b \implies d|3a,3b \implies d|{\rm gcd}(3a,3b) \equal{} 3\; {\rm gcd}(a,b) \equal{} 3$.


Let $n\in\mathbb{N}^*$. Solve the equation $\sum_{k=0}^n C_n^k\cos2kx=\cos nx$ in $\mathbb{R}$.

note that

$\sum_{k=0}^n C_n^k\cos2kx= \mbox{Re}(\sum_{k=0}^n C_n^ke^{2kxi}) = \mbox{Re}[(e^{2xi}+1)^n]$.

now we wish to write $e^{2xi}+1 = 1+\cos 2x+i\sin 2x$ in trigonometric form...

its norm is $\sqrt{2+2\cos 2x}$, and its argument, $\gamma$, is such that

$\cos\gamma = \frac{1+\cos 2x}{\sqrt{2+2\cos 2x}} = \cos\frac{ 2x}{2} = \pm \cos x$. so if $x$ ends in $[-\pi/2, \pi/2]$, we take $\gamma = x$ and if $x$ ends in $(\pi/2, 3\pi/2)$ we take $\gamma = x+\pi$. (because in all cases $\sqrt{2+2\cos 2x}\cos\gamma = 1+\cos 2x$ must be $\geq 0$.)

now we have

$\mbox{Re}[(e^{2xi}+1)^n] = \sqrt{2+2\cos\2x}^n\cos n\gamma$.

if $n$ is even, then the original equation becomes $\sqrt{2+2\cos\2x}^n\cos nx = \cos nx$.

then either $\cos nx = 0$, in which case $x = \frac{(2k+1)\pi}{2n}, \ k\in \mathbb Z$, or $2+2\cos\2x = 1$, in which case $\cos 2x = -\frac12$, i.e. $x = k\pi \pm \frac{\pi}{3}, \ k\in \mathbb Z$.

if $n$ is odd and $x$ ends in $(\pi/2, 3\pi/2)$, the original equation becomes $\sqrt{2+2\cos\2x}^n\cos nx = -\cos nx$. if $\cos nx \neq 0$ then we get $\sqrt{2+2\cos\2x}^n = -1$ which is impossible. otherwise, $\cos nx = 0$, in which case $x = \frac{(2k+1)\pi}{2n}, \ k \in \mathbb Z$. (all these are solutions, whether they end in the presumed interval or not...)

if $n$ is odd and $x$ ends in $[-\pi/2, \pi/2]$, the original equation becomes $\sqrt{2+2\cos\2x}^n\cos nx = \cos nx$, in which case either $\cos nx = 0$, in which case we have the solutions mentioned above or $\cos 2x = -\frac12$, i.e. $x = 2k\pi \pm \frac{\pi}{3}$ (because we assumed $x$ ends in $[-\pi/2, \pi/2]$)


integration
trigonometry
calculus
calculus computations

:(  i asked this in exam yesterday and only 3 students got it... what   is going on ?  :huh:

show

$ \sum_{n\equal{}1}^\infty\;\dfrac{n^2}{2^n\cdot \dbinom{2n}{n} }\;\;\equal{}\;\;\boxed{\dfrac{8}{343}\left(16 \plus{} \dfrac{37}{\sqrt 7}\;\cot^{\minus{}1} \sqrt 7\right)}$

Using

$ \bold{(i)} \quad \binom{2n}{n}^{ - 1} = \frac {n}{4^n} \beta (n, \tfrac{1}{2} ),$
$ \bold{(ii)} \quad \sum_{n = 1}^{\infty} n^3 x^n = \frac {x(x^2 + 4x + 1)}{(1 - x)^4},$

we have

\begin{eqnarray*} & & \sum_{n = 1}^{\infty} \binom{2n}{n}^{ - 1} \frac {n^2}{2^n} \\
& = & \sum_{n = 1}^{\infty} \frac {n^3}{8^n} \beta (n, \tfrac{1}{2} ) \\
& = & \sum_{n = 1}^{\infty} \frac {2n^3}{8^n} \int_{0}^{\frac {\pi}{2}} \sin^{2n - 1} \theta \, d\theta \\
& = & \int_{0}^{\frac {\pi}{2}} \frac {2}{\sin \theta} \sum_{n = 1}^{\infty} n^3 \left( \frac {\sin^2 \theta}{8}\right)^n \, d\theta \\
& = & 16 \int_{0}^{\frac {\pi}{2}} \frac {\cos^4 \theta - 34 \cos^2 \theta + 97) \sin \theta}{(\cos^2 \thehta + 7)^4} \, d\theta \\
& = & 16 \int_{0}^{1} \left( \frac {384}{(7 + t^2)^4} - \frac {48}{(7 + t^2)^3} + \frac {1}{(7 + t^2)^2} \right) \, dt \quad ( t = \cos \theta). \end{eqnarray*}

Well, we can evaluate the last integral using $ t = \sqrt {7} \tan \varphi$ or something else, but I think that is not an important step of this calculation. (Frankly, my head was messed up with bunch of calculation before finishing it... [img]http://myhome.shinbiro.com/~sosQED/images/m_0503.gif[/img])


[quote="bos1234"]$\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3}}...}}}}}$ =  $3$

Find x[/quote]

[hide]\[\sqrt{x+3 \cdot 3}= 3 \implies \boxed{x=0}. \][/hide]

[quote="bos1234"]$\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3}}...}}}}}$ =  $3$

Find x[/quote]

[hide]

$\left({\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3}}...\right)^{2}}}}}}}$ =  $(3)^{2}$

$x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3}}...}}}}$ =  $9$

$3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3}}...}}}}$ =  $9-x$

Notice that dividing the sum by 3 will bring you back to the original sum since this is goes on 4ever

 $\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3}}...}}}}$ =  $\frac{9-x}{3}$

$\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3\sqrt{x+3}}...}}}}}$ =  $\frac{9-x}{3}$

Notice that the LHS is equal to 3 as given in the question

$3$ =  $\frac{9-x}{3}$
solve for that
$x=0$[/hide]


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

Hi,I am a beginner on groups.Here are two questions,thanks for help.

Problem 1:Let $|G|=p^{n}$($p$ is a prime number) and let $H$ be a subgroup of $G$ such that $|H|=p^{m}$,where $m\le n$.Show that if $m\le k \le n$,then there exists a subgroup $K$ of $G$ such that $|K|=p^{k}$ and $H \subseteq K$.
Hint in my book
[hide]Use the fact that,by the class equation,$Z(G)$ has a subgroup of order $p$.[/hide]

Problem 2:Suppose  $|G|=p^{n}$($p$ is a prime number),$H \lhd G$,and $|H|=p$.Prove that $H \subseteq Z(G)$,where $Z(G)$ is the center of $G$.

[quote="ma_go"]as regarding the second question..
$H$ is generated by a non-unity element of its, say $h$.
now, take $g\in G\setminus H$ of order $p$ (there exists one): $ghg^{-1}\in H$, so $ghg^{-1}= h^{k}$ for some integer $0<k<p$. you have $h=g^{p}hg^{-p}=h^{k^{p}}=h^{k}$, by induction+fermat theorem.[/quote]
Thanks,Nice.
But I think we should to prove that for every $g \in G\setminus H$
we have:$gh=hg$
but Never mind,Just notice that $p|o(g)$ then we let $o(g)=mp$
thus as what you did,we can proved that $h=h^{k^{pm}}=h$.
 :P
Thank you again,and what about problem 1?

[quote="ma_go"][quote="zhaobin"]notice that $p|o(g)$ then we let $o(g)=mp$
thus as what you did,we can proved that $h=h^{k^{pm}}=h$.[/quote]

uhm.. i think you meant $\text{o}(g) | |G| = p^{n}$, so that $h = h^{k^{p^{m}}}= h^{k}$, right? :)

i'm thinking about problem 1, but i'm too blind to see how hint can be used... and, actually, i'm not even that convinced of the truth of the statement :D
(that doesn't mean at all it's false, of course :oops: )[/quote]
Oh,Yes...I also made a stupid mistake. :oops:

[quote="zhaobin"]No one :(,thank you[/quote]
Ehm, what about patience ;).

ad Problem 1: By induction, we may assume $m=k+1$. It is enough to find for every proper subgroup $H$ of $G$ a subgroup $K$ s.t. $H \subseteq K$ and $[K: H] = p$. Now, by [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=104015[/url], $N_{H}/ H$ is a nontrivial $p$-group. Hence it has a subgroup $U$ of order $p$. Then there's a subgroup $K$ s.t. $H \subseteq K \subseteq N_{H}$ and $U = K / H$, and we're done.

[quote="-oo-"][quote="zhaobin"]No one :(,thank you[/quote]
Ehm, what about patience ;).

ad Problem 1: By induction, we may assume $m=k+1$. It is enough to find for every proper subgroup $H$ of a subgroup $K$ s.t. $H \subseteq K$ and $[K: H] = p$. Now, by [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=104015[/url], $N_{H}/ H$ is a nontrivial $p$-group. Hence it has a subgroup $U$ of order $p$. Then there's a subgroup $K$ s.t. $H \subseteq K \subseteq N_{H}$ and $U = K / H$, and we're done.[/quote]
Sorry,I can understand  you,Can you explian more?
What do you mean by $m=k+1$ or it should be $k=m+1$?
And I aslo still don't know what did you said left...
Sorry,AND thank you.

Yeah I meant $k=m+1$ of course. When you can extend every subgroup with index $p$, it works for every power of $p$, therefore we can assume $k=m+1$ (you can also prove this by induction on $k$) What is else unclear?

I can't understand anything left...
Have you mistaken $H$ for $K$....?
And what is the final subgroup you find such that $H \subseteq K$,and $|K|=p^{m+1}$
Thanks for responsing.

I have not mistaken $H$ for $K$, and I've writen how to choose $K$ ...

Thank you very much.I understand now. :)


geometry
incenter
parallelogram
geometry unsolved

Let $ A_1,B_1,C_1$ be the foot of the altitudes from $ A,B,C$ respectively in acute triangle $ ABC$. Let $ O_A$ be the incenter of triangle $ AB_1C_1$, and $ T_A$ be point where the incircle of triangle $ ABC$ touches $ BC$. Define $ O_B,O_C,T_B,T_C$ analogously. Show that the hexagon $ T_AO_CT_BO_AT_CO_B$ have all sides equal.

IOb*IB=IB^2*(1-COS∠B)= 2*IB^2*(sin^2(∠B/2))=2*r^2
Analogously, IOc*IC=2*r^2
IOb*IB= IOc*IC  => ObOcCB are concyclic  => ∠IObOc=∠C/2  ∠IOcOb=∠B/2
Analogously ∠IOcOa=∠A/2  ∠IOaOc=∠C/2
IOb⊥OaOc  IOa⊥ObOc   IOc⊥OaOb  =>  I is the orthocenter of △OaObOc
We have TcTb ⊥ IA , so  TcTb//ObOc

ObOc=BC*(IB/IC)* 2*(sin^2(∠B/2))=2*BC* sin(∠B/2)* sin(∠C/2)
TcTb=2*r*cos(∠A/2)

Notice in any triangle , r=4R*sin(∠B/2)* sin(∠C/2)* sin(∠A/2) BC=2R* sin∠A
So ObOc= TcTb

=> TcObOcTb is a parallelogram  TcOb=TbOc
Analogously, the six sides are all equality


Prove that for $ n>1$ the $ n$-th harmonic number $ H_{n} \equal{} 1\plus{} \frac{1}{2} \plus{} \frac{1}{3} \plus{} \cdots \plus{}\frac{1}{n}$ is not an integer.

[hide="Another possible approach"]We use induction.

Base cases: $ H_2\equal{}\frac{3}{2}$, not an integer.

Inductive step:
Assume $ H_p$ is not an integer for a prime $ p$.  Let $ k$ be a number such that $ p<k<2p$.

We can say that $ H_k\minus{}\frac{1}{p}\equal{}\frac{p}{p}\left(H_k\minus{}\frac{1}{p}\right)\equal{}\frac{p}{p}\left(1\plus{}\frac{1}{2}\plus{}\cdots\plus{}\frac{1}{p}\plus{}\cdots\plus{}\frac{1}{k}\minus{}\frac{1}{p}\right)\equal{}\frac{px}{py}$ for some integer $ x$ (it doesn't matter) and $ y\equal{}\frac{k!}{p}$.  It is obvious that only one number from $ 1$ to $ k$ inclusive is a multiple of $ p$, hence $ p\nmid y$.  Therefore $ H_k\equal{}\frac{px}{py}\plus{}\frac{y}{py}\equal{}\frac{px\plus{}y}{py}$.  However, it is obvious that the numerator is not a multiple of $ p$ (because of that $ y$ term), hence $ H_k$ cannot be an integer.

Finally, by Bertrand's Postulate, at least one possible value of $ k$ is prime, and the induction is complete. $ \blacksquare$
[/hide]


Source: Individual Test 1997 Fall Classic - Mu Alpha Theta

#32
In triangle ABC, angle ABC = 120 degrees, AB = 3, and BC = 4. If perpendiculars constructed to side AB at A and to side BC at C meet at D, then CD = ?

No calculators allowed. (So no using law of cosines/law of sines a million times)

(The solution was simple, but I suspect it would've taken me forever to think of it)

That was surprisingly easy.
[hide]Draw a diagram.
Make a perpendicular from BC at B.
Label the point of intersection of the perpendicular and AD as F.
Make a perpendicular from BF at F.
Label the point of intersection of the 2nd perpendicular and CD as G.
By subtraction (120-90=30), <ABF is 30 degrees. This makes ABF a 30-60-90 triangle.
If AB=3, then BF=2sqrt(3).
Since opposite sides of a rectangle are congruent, FG=4 and CG=2sqrt(3).
Also buy subtraction (180-90-60=30), <DFG is 30 degrees. This makes ABF a 30-60-90 triangle.
Since FG=4, DG=4sqrt(3)/3.
Lastly, since CD=CG+GD, CD=2sqrt(3)+4sqrt(3)/3

So $CD=\frac{10\sqrt{3}}{3}$ [b]ANS.[/b][/hide]

Or extend sides AD and BC so they meet at a point E.  And then EAB and ECD are similar.  From there it's easy.


Find the smallest real number $ C$ so that $ 1\plus{}(x\plus{}y)^2 \le C(1\plus{}x^2)(1\plus{}y^2)$ for all real numbers $ x$ and $ y$.

Can some mod please either lock this or edit the title and subject description in the way of http://www.mathlinks.ro/viewtopic.php?t=206052 (and then delete that one)¿


[quote="warut_suk"]Do there exist $a, b, c, d\in N$ such that $ab=cd$ and $a+b+c+d$ is a prime number?[/quote]

[hide]
It cannot be done .

Setting $ab=cd=k$ for some integer positive $k$ then we have

$a+b+c+d = a+c+k\left(\frac{1}{a}+\frac{1}{c}\right)=\frac{(a+c)(k+ac)}{ac}$

If the expression has to be a prime , it is nessecary for $a+c=ac$ , (*since $k+ac=ac$ will only result in $k=0$ which contradict with the assumption) .

Here , we see that $c|a$ and $a|c$ which comes to conclusion that $a=c=2$

However , this gives us $b=d$ and plugging inside the equation gives us

$a+b+c+d=4+2b=2(b+2)$ which is not a prime .

Another case is when $ac=1$ or $a=c=1$ the result also the same .

Hence no prime number can be formed .

 :) [/hide]

[hide="Generalization"]Let a, b, c, d and n be positive integers. If $ab=cd$, then $a^n+b^n+c^n+d^n$ is not prime.
[u]Proof[/u]
Let $\frac{a}{c}=\frac{d}{b}=\frac{x}{y}$ where $(x, y)=1$ and $x, y\in\mathbb{N}$
and $a=ux$, $c=uy$, $d=vx$, $b=vy$, $u, v\in\mathbb{N}$
then we have $a^n+b^n+c^n+d^n=u^nx^n+v^ny^n+u^ny^n+v^nx^n=(u^n+v^n)(x^n+y^n)$
Since $u^n+v^n>1$, $x^n+y^n>1$, the conclusion follows.[/hide]


ratio
geometry
circumcircle
trigonometry
angle bisector
perpendicular bisector

Let the line PN meet the (extended) triangle side CA at R. The cross-ratio of 4 points is preserved in a central projection of the line BC to the line PR, hence,

$\frac{NB}{NC} \cdot \frac{MC}{MB} = \frac{NP}{NR} \cdot \frac{QR}{QP}$

M is the midpoint of BC, N is the midpoint of PR, hence,

$\frac{NB}{NC} = \frac{QR}{QP}$

The angle bisector AN meets the circumcircle of the triangle $\triangle ABC$ at the midpoint K of the arc BC opposite to the vertex A. Since $PR \perp AN, PO \perp AB$, it follows that $\angle OPR = \angle KAB = \angle KAC = \angle KBC$ and the isosceles triangles $\triangle ROP \sim \triangle BKC$ together with the points $Q, N \in RP$ resp. $N, M \in BC$ on their bases are similar. Hence, the angles $\angle NOQ = \angle NKM$ are equal, which means that the lines $OQ \parallel KM$ are parallel and $KM \perp BC$ is the perpendicular bisector of BC.

[quote="mr.danh"][hide="Another solution"]Let $ L$ be the second intersection of $ AN$ and the circumcircle of triangle $ ABC$. Let $ D$ on $ AB$ and $ E$ on $ AC$ such that $ LD\perp AB$, $ LE\perp AC$. Easy to see that the Simson line $ (DMN)$ is perpendicular to $ AN$. So $ DM\parallel PQ$. Apply Thales's theorem to pairs of parallel lines $ DM\parallel PQ$, $ LD\parallel OP$, we have that $ \frac {AO}{AL} \equal{} \frac {AP}{AD} \equal{} \frac {AQ}{AM}$.
Hence $ OQ\parallel LM$. And $ OQ\perp BC$ (since $ LM\perp BC$).[/hide][/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=297174]Posted[/url]

Let $PN$ hit $AC$ at $R$. Note that there exists a circle $\omega$ tangent to $AP$ and $AR$. Let $Q'$ be the intersection of $PR$ and the line through $O$ perpendicular to $BC$. $Q'$ lies on the polar of $A$, so $A$ lies on the polar $\ell$ of $Q'$. Let $E$ be the intersection of $PR$ and $\ell$, and let $M'$ be the intersection of $AQ$ and $BC$. Because $E$ lies on the polar of $Q'$, $(P, R; Q', E) = -1$. $\ell$ and $BC$ are both perpendicular to $OQ'$, so they are parallel. Hence, $(P, R; Q', E) = A(P, R; Q', E) = A(B, C; M', \infty) = -1$, i.e., $M'$ and $\infty$ are harmonic conjugates, whence $M' = M$ and $Q' = Q$.

Another solution:

Let $AN$ and $AM$ cut the circumcirle of $ABC$ with centre $H$ at $F$ and $G$, respectively.
It is obvious that $H, M, F$ are collinear and $HF\perp BC$. Let $HF$ cut the circumcircle again at $D, DG$ cut $BC$ at $K$.
Easy to see that $APOE\sim DBFC$, and since $\angle OAQ =\angle FDK$, line $AQ$ corresponds to $AK, OQ$ corresponds to $FK$.
Thus $\angle NOQ = \angle MFK$, but $\angle MFK = \angle MGK$ (since $MKGF$ is cyclic)$= \angle AFD$.
$\angle NOQ = \angle AFD$, this means that $OQ\parallel FD$, thus $OQ\perp BC$.

[hide="Solution"]
We do not use isothermal coordinates.

Let $PN \cap AC=R$, let the perpendicular from $O$ to $BC$ meet $PR$ at $Q'$, and let $AQ'$ meet $BC$ at $M'$.  It is sufficient to prove that $M'=M$.

Clearly $\triangle APO \cong \triangle ARO$.  Then by the [url=http://www.artofproblemsolving.com/blog/81708]Ratio Lemma[/url], we have $\dfrac{PQ}{QR}=\dfrac{PO}{OR}\cdot\dfrac{\sin \angle POQ}{\sin \angle ROQ}=\dfrac{\sin \angle ABC}{\sin \angle ACB}=\dfrac{AC}{AB}$.  Then, since $AP=AR$, by the Ratio lemma we have $\dfrac{\sin \angle BAQ}{\sin \angle CAQ}=\dfrac{AC}{AB}$.  Finally, the Ratio Lemma gives us that $\dfrac{BM'}{M'C}=\dfrac{AB}{AC}\cdot \dfrac{\sin \angle BAQ}{\sin \angle CAQ}=1$, as desired.

QED  [/hide]

Here's a solution for those that aren't scared of diving into a bit of algebra.  It's not an efficient solution at all (NAA's above beats it with a very similar idea), but hey, it works.

[hide="Intense Trig Bash"]
Let $R=AC\cap PQ$ and $X=QO\cap BC$.  I first propose a lemma.

[b]LEMMA:[/b] Suppose $a,b,c,d$ are angles such that $\dfrac{\sin a}{\sin b}=\dfrac{\sin c}{\sin d}$ and $a+b=c+d$.  Then $a=c$ and $b=d$.

[i]Proof.[/i] Let $k=a+b=c+d$.  Then $b=k-a$ and $d=k-c$.  Cross-multiplying and expanding gives \begin{align*}\sin a\sin(k-c)&=\sin c\sin(k-a)\\\sin a(\sin k\cos c-\sin c\cos k)&=\sin c(\sin k\cos a-\sin a\cos k)\\\sin a\sin k\cos c-\sin a\sin c\cos k&=\cos a\sin c\sin k-\sin a\sin c\cos k\\\sin a\cos c&=\sin c\cos a\\\sin(a-c)&=0\\a&=c,\end{align*} as desired.  $\blacksquare$

Note that it is easy to see that $\triangle PAR$ is isosceles, which implies that $\angle RAN=\angle PAN=\angle NPO\implies AROP$ is cyclic.  Therefore $\angle PAO=\angle PRO$ and $\triangle POR$ is also isosceles.

Applying the Ratio Lemma to $\triangle ABC$ gives \[\dfrac{AB\sin\angle BAM}{AC\sin\angle MAC}=\dfrac{BM}{CM}=1\implies \dfrac{AB}{AC}=\dfrac{\sin\angle QAR}{\sin\angle PAR}.\]  Next, applying the Ratio Lemma to $\triangle PAR$ gives $\tfrac{PQ}{QR}=\tfrac{AP\sin\angle PAQ}{AR\sin\angle QAR}=\tfrac{\sin\angle PAQ}{\sin\angle QAR}$ and applying it to $\triangle POR$ gives $\tfrac{PQ}{QR}=\tfrac{PO\sin\angle POQ}{OR\sin\angle QOR}=\tfrac{\sin\angle POQ}{\sin\angle QOR}$.  Therefore \[\dfrac{\sin\angle POQ}{\sin\angle QOR}=\dfrac{PQ}{QR}=\dfrac{\sin\angle PAR}{\sin\angle QAR}=\dfrac{AC}{AB}.\] In addition, by LoS we have $\tfrac{AB}{\sin\angle ACB}=\tfrac{AC}{\sin\angle ABC}$, so all in all we have $\tfrac{\sin\angle POQ}{\sin\angle QOR}=\tfrac{\sin\angle ABC}{\sin\angle ACB}$.  But note that since $ARPO$ is cyclic we have $\angle POQ+\angle QOR=180^\circ-\angle BAC=\angle ABC+\angle ACB$.  Therefore by our Lemma, $\angle ABC=\angle QOP$.  It follows that $BXOP$ is cyclic, so $\angle BXO=180^\circ-\angle BPO=90^\circ$ and $BC\perp QO$, as desired.  $\blacksquare$[/hide]

This is similar to some of the previous solutions, but uses homothety to simplify one step.

Let $D$ be the other intersection point of $AN$ and the circumcircle of $ABC$. Then $DM\perp BC$ by the South Pole Theorem. Let $E$ and $F$ be the points on $AC$ and $BC$ such that $DE\perp AC$ and $DF\perp BC$, respectively. Note that there is a homothety centred at $A$ sending $O$ to $D$, $P$ to $F$ and (since clearly $OR\perp AC$) $R$ to $E$. Since $FME$ is the Simson line of point $D$, it sends the entire segment $PR$ to $FE$, and thus $Q$ to $M$. Now it is clear that $QO\parallel MD$, so $QO\perp BC$.

I will be using homothety. Extend PN to meet AC at J.Let $O*$ be the second intersection of AO with circumcircle of $ABC$. Consider the homothety centred at $A$ that maps $O$ to $O*$. Drop $OX\perp AB$ and $OY\perp AC$.$P$ maps to $X$ and $J$ maps to $Y$.So $PJ$ maps to $XY$. So, $AM\cap  PJ$ maps to $AM\cap  XY$.Invoking Simpson, easy to see that $X,M,Y$ collinear.So, $Q$ maps to $M$. $OQ\perp BC$ iff $O*M\perp BC$ . But this is trivial.                                                                                Q.E.D.  Please check my solution whether it is correct.

[quote=jayme]Dear Mathlinkers,
let X be the second point of intersection of AD with the circumcircle of ABC.
The problem can be solved without calculation by considering the circle passing through M, N and X.
Sincerely
Jean-Louis[/quote]

 Can you please elaborate on your method.It seems very nice.

You can also use Cartesian coordinates with, say, $N=(0,0)$, $A=(-1,0)$, $P=(0,k)$, and $O=(k^2,0)$. Since $AN$ is the angle bisector, it's actually easy to find the equation of $AC$.

Dynamic points:
We will vary the point $B$ on $AP$ ($A,P,O,N$ will all remain fixed). Let $X=BN\cap QO$ and $X'=BN\cap (ON)$. We have $B\mapsto C\mapsto M \mapsto Q$ is a projective transformation, and so lines $BN$ and $OQ$ each have degree 1, therefore their intersection $X$ has degree 2. Since $(ON)$ is fixed, $X'$ has degree 1. Now to prove $X=X'$ we only must check 4 points for $B$. These four points will be $AB_{\infty}, P, A, K$ where $K\in AB$ with $KN\parallel AC$. For each of these points the result is trivial, thus $X=X'$ and this implies $XO\bot XQ$.

[quote=shobber]Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively. And $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced.

Prove that $QO$ is perpendicular to $BC$.[/quote]
Proof-
Let $R$ denote intersection of $PN$ with $AC$. Let $\omega$ denote circumcircle of triangle $APR$.
Note by symmetry about $AN$ that $O\in\omega$ and is the $A-antipode$. Let $OQ\cap\omega=S,AQ\cap\omega=T$.
Now $H(A,O,P,R)\xRightarrow{\text{Q}}H(T,S,R,P)\xRightarrow{\text{A}}H(AT,AS, AP,AR)\implies H(AS,AM, AB,AC)$ and hence $AS\parallel BC$ and we know $AS\perp OS$
So $OQ\perp BC$
Hence proved.

Extend $AO$ to meet $(ABC)$ at $L$. Let $D$ be the altitude from $L$ to $AB$, $E$ be the altitude from $L$ to $AC$, and $T$ be the altitude from $O$ to $AC$. Note that by Simson line wrt $\triangle{ABC}$, we have $D, M, E$ collinear, because $LM\perp BC$. Also, it is easy to see $APOT$ is cyclic, so by Simson line wrt $\triangle{APT}$, we have $P, N, T$ collinear; hence, $Q\in PT$. Finally, let $h$ be the homothety about $A$ mapping $L$ to $O$. Because $PO||LD, OT||LE$, we have that $h(D)=P$ and $h(E)=T$; hence, $h(M)$ lies on both $AM$ and $PT$, so $h(M)=Q$. Thus, by similar triangles we see $QO||ML$, meaning that $QO\perp BC$, as desired.

Here is my solution:

Let the perpendicular from $O$ to $BC$ meet $PR$ at $Q'$. We wish to show $Q'=Q$.
Let the parallel through $Q'$ to $BC$ meet $AB$ and $AC$ at $S$ and $T$, respectively.
Then since $OQ' \perp TS$, and $\angle OPA=\angle ORA=90$, both $PSQ'O$ and $RTOQ'$ are cyclic.
Then $\angle Q'OS= \angle Q'PS = \angle Q'PA = \angle Q'RA = \angle Q'OT$.
Thus, since $Q'O \perp ST, Q'S=Q'T$, in which case $Q'$ is on the median from $A$ and $Q'=Q$.

Redefine $Q$ as $T$, let $Q = PN \cap AC,$ note that $OQ \perp AC$. now let $X$ be arc midpoint of $BC$, and let $E$ and $F$ be the foot of altitude from $X$ to $AB$ and $AC$. Now by simson line we know that $EMF$ is collinear and $EMF \parallel PQ,$ so there's a homothety now at A sending $P \to E, Q \to F, T \to M, O \to X$ implying the desired result.

Oops it looks like I missed the homothety, and kinda rederived homothety?

Let $D$ be the arc midpoint of $BC$. Let the foot from $D$ to $AB$ be $X$, and the foot to $AC$ be $Y$. Let $R=XY\cap AN$.

Then, by Simson's theorem $XMY$ are collinear, call this line $L_1$. We will attempt to show that $L_1\parallel PNQ$. It suffices to note that
\[\angle ARX = 180 - \angle XAR - \angle AXR = 180-\angle DAC-\angle BXM = 180-\angle DBM - \angle BDM = 90\]
Thus, $AR\perp XY$, but since we also have $AR\perp PNQ$, then $XY\parallel PNQ$.

Thus, we may now create a series of ratios since $PQ\parallel XM$
\[\frac{AQ}{AM}=\frac{AP}{AX}=\frac{AO}{AD}\]
Thus, by SAS-ratio similarity, we have $\triangle AQO\sim \triangle AMD$. Thus, $QO\parallel MD$ and since $MD\perp BC$, we clearly have $QO\perp BC$ and we are done.

[hide=Solution]
[asy]
unitsize(0.8cm);
pair A, B, C, B1, C1, M, N, P, Q, O;
A=(2,3sqrt(5));
B=(0,0);
C=(8,0);
M=(4,0);
N=(3.5,0);
P=(29/4,3sqrt(5)/8);
Q=(127/32,3sqrt(5)/64);
O=(127/32,-15sqrt(5)/16);
B1=(-31/16,-93sqrt(5)/32);
C1=(221/16,-93sqrt(5)/32);
draw(A--B1--C1--A--(-1/4,-3sqrt(5)/8)--P--A--B--C--P--O--A--M--Q--N--P--N);
draw(circle(O,63sqrt(5)/32));
label("$A$", A, dir(90));
label("$B$", B, W);
label("$C$", C, NE);
label("$B'$", B1, SW);
label("$C'$", C1, SE);
label("$M$", M, S);
label("$N$", N, SSW);
label("$O$", O, S);
label("$P$", P, NE);
label("$Q$", Q, NNE);
[/asy]

Consider the circle centered at $O$ passing through $P$. Then, this circle is tangent to $AB$ and $AC$. Let $B'C'$ be the line such that $BC\parallel B'C'$ and $B'C'$ is tangent to the circle. Then, since $PN\perp AO$, this means that the reflection of $P$ over $N$ lies on $AB$, $PN$, and the circle. Since $AM$ is a median of $AB'C'$, this means that $PN$, $AM$, and the line through $O$ perpendicular to $B'C'$ are concurrent, so this means that $OQ\perp B'C'$, which means $QO\perp BC$.

Let $\ell \equiv \overline{PQ}$ and $R = \ell \cap \overline{AC}$. Loot at the isosceles $\triangle APR$ (with $AP = AR$). By symmetry we have $\overline{OR} \perp \overline{AC}$. Let $O' = \overline{AO} \cap \odot(ABC)$ and $\ell'$ be the Simson line of $O'$ wrt $\triangle ABC$ ; $\ell'$ intersect $\overline{AB},\overline{AC}$ at $P',R'$, respectively. As $\overline{O'M} \perp \overline{BC}$ so $M \in \ell$.
[asy]
size(200);
pair A=dir(145),B=dir(-150),C=dir(-30),M=1/2*(B+C),Op=dir(-90),Pp=foot(Op,A,B),Rp=foot(Op,A,C),N=extension(A,Op,B,C),P=extension(A,B,N,N+M-Rp),R=extension(N,P,A,C),Q=extension(N,P,A,M),O=extension(Q,foot(Q,B,C),A,N);
draw(circumcircle(A,B,C),red);
dot("$A$",A,dir(A));
dot("$B$",B,dir(B));
dot("$C$",C,dir(C));
dot("$M$",M,dir(-60));
dot("$O'$",Op,dir(Op));
dot("$P'$",Pp,dir(Pp));
dot("$R'$",Rp,dir(-90));
dot("$N$",N,dir(N));
dot("$R$",R,dir(90));
dot("$P$",P,dir(-170));
dot("$Q$",Q,dir(90));
dot("$O$",O,dir(-140));
draw(Pp--A--C--B^^Op--A--M,red);
draw(P--O--R^^Q--O,blue);
draw(Pp--Op--Rp^^Op--M,blue);
draw(P--R^^Pp--Rp,green);
[/asy]
Consider the homothety $\mathbb H$ at $A$ sending $O \to O'$. Note $\mathbb H(P) = P'$ and $\mathbb H (Q) = Q'$ as $\overline{OP} \parallel \overline{O'P'}$ and $\overline{OR} \parallel \overline{O'R'}$. Thus $\mathbb H(\ell) = \ell'$. It follows $\mathbb H(Q) = M$ as $$\mathbb H (Q) = \mathbb H ( \ell \cap \overline{AM}) = \mathbb H(\ell) \cap \mathbb H(\overline{AM}) = \ell' \cap \overline{AM} = M$$
Now since $\overline{O'M} \perp \overline{BC}$ and $\overline{OQ} \parallel \overline{O'M}$. Hence $\overline{OQ} \perp \overline{BC}$, as desired. $\blacksquare$

[hide = for storage]Let the circumcircle of $\triangle{ABC}$ be $\Omega$, let AO meet $\Omega$ at D.
Let E,G be the foot of the perpendiculars from D to AB and AC respectively. Now note that E,M,G are collinear by Simson's Line .Let F be the intersection of PN with AC. Note that since AN is the angle bisector of $\angle{BAC}$, quad APOF is a cyclic kite.
Now notice that triangles POF and EDG are similar and thus we have,
\\ $\frac{PO}{DE} = \frac{PF}{EG} = \frac{OF}{DG}$
\\Also, since we have $OP \parallel DE$ and $OF \parallel DG$,
\\ we have $\frac{AF}{AG} = \frac{OF}{DG}$ \\and
\\ $\frac{AP}{AE} = \frac{OP}{DE}$
\\Therefore, \\$\frac{AP}{AE} = \frac{AF}{AG} = \frac{PF}{DG}$
$\Rightarrow PF \parallel DG \Rightarrow PQ \parallel EM $.
\\Now using Thales theorem on the triangles AEM and AMD we will get $QO \parallel LM$.
[/hide]

[hide=Highly recommended if you are in such a desperate situation] Very efficient coordinate bash here: https://www.math.hkust.edu.hk/excalibur/v5_n3.pdf [/hide]

one of the finest use of phantom points!

consider $PN \cap AC=R$ and perpendicular to $BC$ from $O$ to hit $BC$ at $X$ such that $OX \cap PR=Q'$ and $AQ' \cap BC=M'$\\
Join $OR$ and $OC$ for later use.\\

now beforehand solving this problem we propose a lemma:\\

[color=#073763][b]Lemma:-[/b][/color] In a triangle $ABC$ denote $AX$ to be a cevian, s.t. $\angle{BAX}=\alpha$ and $\angle{XAC}=\beta$ then $\frac{BX}{XC}=\frac{AB}{AC}\cdot \frac{\sin{\alpha}}{\sin{\beta}}$\\

[color=#134F5C][b]Proof:-[/b][/color] denote $\angle{AXB}=\theta$ then from sine rule we get $\frac{BX}{\sin{\alpha}}=\frac{AB}{\sin{\theta}}$ and $\frac{CX}{\sin{\beta}}=\frac{AC}{\sin{\theta}}$ , dividing these we get $\frac{BX}{CX}=\frac{AB}{AC}\cdot \frac{\sin{\alpha}}{\sin{\beta}}$ $\square$\\

mark $\angle{BAQ'}=x$ and $\angle{Q'AR}=y$\\

now we notice that since $AN$ is angle bisector and $AN\perp PR$ we get $\triangle{APN}\cong \triangle{ARN}$ hence we get :\\

$\angle{APN}=\angle{ARN}$ and hence $\angle{OPN}=90-\angle{APN}$ and as $N$ is mid-point of $PR$ we get $\triangle{OPN} \cong \triangle{ORN}$, giving us $\angle{ORN}=90-\angle{APN}$\\

hence we get $OR \perp AC$ and therefore points $C,R,X,O$ are concyclic which gives $\angle{XOR}=C$\\

also points  $B,P,O,X$ are concyclic which gives us $\angle{XOP}=B$\\

from [color=#073763][b]Lemma[/b][/color] we get $\frac{PQ'}{Q'R}=\frac{AP}{AR}\cdot \frac{\sin{x}}{\sin{y}}$ and since $AP=AR$ we get $\frac{PQ'}{Q'R}=\frac{\sin{x}}{\sin{y}}$\\

also similarly we have in $\triangle{POR}$ applying [color=#073763][b]Lemma[/b][/color] we get $\frac{PQ'}{Q'R}=\frac{OP}{OR}\cdot \frac{\sin{B}}{\sin{C}}$ and since $OP=OR$ we get $\frac{\sin{x}}{\sin{y}}=\frac{\sin{B}}{\sin{C}}$\\

and now we again apply [color=#073763][b]Lemma[/b][/color] in $\triangle{ABC}$ we get$ \frac{AM'}{M'C}=\frac{AB}{AC}\cdot \frac{\sin{x}}{\sin{y}}=\frac{AB}{AC}\frac{\sin{B}}{\sin{C}}=\frac{2R}{2R}=1$ hence we get $AM'=M'C$ and hence $M'$ is midpoint of $BC$ which gives $AM'$ to be the median of $\triangle{ABC}$ and hence the problem statement follows as $M=M'$ and $Q=Q'$ $\blacksquare$

Solved with a hint to construct DE//BC

The circle centered at O through P is tangent to AB,AC, since it lies on the angle bisector. Let DE be the unique line //BC with DE tangent to this circle, with R the foot from O onto AD; since PR perp. NO (look at kite APOR, PR diagonal) along with the well known fact that AM (median in ADE), the line perp. to DE through O are concurrent, this finishes because Q is indeed the concurrence point. $\blacksquare$

wait what

Let $L$ be the midpoint of arc $BAC$ and let $K$ be the midpoint of arc $BC$.

Notice that $ANML$ is cyclic so we get $APQO\sim LCNK$.

This means
\[\measuredangle(OQ,KN)=\measuredangle(PQ,CN)\implies \measuredangle QOK=\measuredangle CNP.\]

Subtract $\measuredangle BNA$:
\[\measuredangle(OQ,BC)=\measuredangle(OQ,KN)-\measuredangle(KN,BC)=\measuredangle QOK-\measuredangle BNA\]
\[\measuredangle QOK-\measuredangle BNA=\measuredangle CNP-\measuredangle BNA=\measuredangle BNP-\measuredangle BNA=90^{\circ}\]
done!

we can also solve as follows:

Let $R$ be on $(APQ)$ where $AR\parallel BC$. Let $T=PN\cap AC$.
\[-1=(\infty_{BC},M;B,C)\stackrel{A}{=}(R,AQ\cap (APO);P,T)\]
so $\frac{RP}{RT}=\frac{QP}{QT}$ and $\overline{RQO}$ bisects $\angle PRT$.

Let $AN \cap (ABC) = E$. By Simson, the line connecting the projections from $E$ to $AB$ and $AC$ passes through $M$. Hence the homothety at $A$ sending this line to $PN$ and $E$ to $O$ also sends $M$ to $N$. Thus
\[QO \parallel ME \perp BC. \quad \blacksquare\]


A 20-gallon container is filled halfway with a mixture that is $ 90\%$ vinegar and $ 10\%$ water. How many gallons of water must be added for the mixture to become $ 60\%$ vinegar and $ 40\%$ water?

Since only half the 20 gallon container is filled, 10% water is equal to 1 gallon and 90% vinegar is 9 gallons. In the wanted result, the 9 gallons is 60%, so the 40% water would be 6 gallons. Thus, you must 5 gallons of water to the container to get 60% vinegar and 40% water.


Θέλω να πω και κάτι που ίσως φαίνεται πολύ απλοϊκό, αλλά μου έκανε τεράστια εντύπωση. Είναι μια ανισότητα που μοιάζει με την B. C. S., αλλά είναι ποιο ισχυρή για αριθμούς μικρότερους της μονάδας:
$ ( x1^{2} \plus{} y1^{2} )^{2}( x2^{2} \plus{} y2^{2} )^{2} > \equal{} ( x1x2 \plus{} y1y2 )^{4}$.
Η απόδειξη που είδα είναι η εξής:
έστω διανύσματα $ a \equal{} (x1, y1), b \equal{} (x2,y2)$, τότε η ανισότητα γραφεται
$ |a|^{4}|b|^{4} > \equal{} (ab)^{4} < \equal{} > |a|^{4}|b|^{4} > \equal{} |a|^{4}|b|^{4}|cos(a, b)|^{4}$
$ < \equal{} > 1 > \equal{} |cos(a, b)|$, που ισχύει.

Πρέπει να πω ότι η ανισότητα B. C. S.  είναι η εξής (στην πιο απλή μορφή της):
$ ( x1^{1} \plus{} y1^{1} )^{2}( x2^{1} \plus{} y2^{1} )^{2} > \equal{} ( x1x2 \plus{} y1y2 )^{2}$.
Θα γράψω και την καινούρια ανισότητα για να γίνει πιο εμφανής η διαφορά:
$ ( x1^{2} \plus{} y1^{2} )^{2}( x2^{2} \plus{} y2^{2} )^{2} > \equal{} ( x1x2 \plus{} y1y2 )^{4} < \equal{} >$
$ ( x1^{2} \plus{} y1^{2} )^{1}( x2^{2} \plus{} y2^{2} )^{1} > \equal{} ( x1x2 \plus{} y1y2 )^{2}$.

Στην καινούρια, όταν είναι $ x1, x2, y1, y2 > 0$, προκύπτει η ποιο ισχυρή ανισότητα,
γιατί είναι $ ( x1^{2} \plus{} y1^{2} )^{1}( x2^{2} \plus{} y2^{2} )^{1} < \equal{} ( x1^{1} \plus{} y1^{1} )^{2}( x2^{1} \plus{} y2^{1} )^{2}$

Επίσης, μιας και η απόδειξη έγινε διανυσματικά, μπορεί να γενικευτεί για μεγαλύτερο αριθμό μεταβλητών χρησιμοποιώντας διανύσματα με περισσότερες διαστάσεις. Για μεγαλύτερο αριθμό παρενθέσεων, αρκεί μια απλή επαγωγή.

[quote="maniopas"]Θέλω να πω και κάτι που ίσως φαίνεται πολύ απλοϊκό, αλλά μου έκανε τεράστια εντύπωση. Είναι μια ανισότητα που μοιάζει με την B. C. S., αλλά είναι ποιο ισχυρή για αριθμούς μικρότερους της μονάδας:
$ ( x_1^{2} \plus{} y_1^{2} )^{2}( x_2^{2} \plus{} y_2^{2} )^{2} > \equal{} ( x_1x_2 \plus{} y_1y_2 )^{4}$.
[/quote]

Αυτή η ανισότητα δεν είναι σαν την B. C. S. ...είναι η B. C. S. απλά υψώνοντας τα 2 μέρη της ανισότητας στο τετράγωνο :!: Θα μπορούσαμε εξίσου να υψώσουμε σε οποιαδήποτε δύναμη αλλά αυτό δεν σημαίνει ότι αλλάζει η ανισότητα!

Δικό μου λάθος: μπέρδεψα τον τύπο της B. C. S. Αυτό που στην πραγματικότητα έκανα είναι να αποδείξω τη ίδια την B. C. S. Η απόδειξη όμως παραμένει εντυπωσιακή. Η σύγκριση έχει γίνει με μια άλλη ανισότητα.


Hola a todos.

Como seguramente saben por estos días es la XXIV Olimpíada Iberoamericana de Matemáticas en Santiago de Querétaro (México).

En este momento se está desarrollando la prueba del primer día.

Estaría muy agradecido si alguien puede subir los enunciados cuando ésta termine, lo mismo mañana, y después los puntajes o demás novedades que vayan teniendo, así los que no estamos allá nos mantenemos informados  :D

Gracias!

alguien ya puede subir los temas, solo se que fueron combinatoria, teoría de números y geometría

[quote="msecco"]Problema 1-Seja n um natural maior que 2.Suponhamos que n ilhas estejam localizadas ao redor de um círculo e que entre cada duas ilhas vizinhas haja duas pontes.Partindo da ilha x_1, de quantas maneiras se podem percorrer as 2n pontes passando por cada ponte exatamente uma vez?
Problema 2 - Para cada inteiro positivo n, definimos a_n=n+m, em que m é o maior inteiro tal que 2^(2^m)<=n*2^n.Determinar quais inteiros positivos não aparecem na sequência a_n.
Problema 3- Sejam C1 e C2 duas circunferencias de centros O1 e O2, com o mesmo raio, que se intersectam em A e B.Seja P um ponto sobre o arco AB de C2 que está dentro de C1.A reta AP intersecta C1 em C, a reta CB intersecta C2 em D e a bissetriz de <CAD intersecta C1 em E e C2 em L.Seja F o simétrico do ponto D em relação ao ponto médio de PE.Demonstrar que existe um ponto X que satisfaz <XFL=<XDC=30º e CX=O1O2.[/quote]

Obrigado!  :)

[quote="msecco"]Problema 1-Seja n um natural maior que 2.Suponhamos que n ilhas estejam localizadas ao redor de um círculo e que entre cada duas ilhas vizinhas haja duas pontes.Partindo da ilha x_1, de quantas maneiras se podem percorrer as 2n pontes passando por cada ponte exatamente uma vez?
Problema 2 - Para cada inteiro positivo n, definimos a_n=n+m, em que m é o maior inteiro tal que 2^(2^m)<=n*2^n.Determinar quais inteiros positivos não aparecem na sequência a_n.
Problema 3- Sejam C1 e C2 duas circunferencias de centros O1 e O2, com o mesmo raio, que se intersectam em A e B.Seja P um ponto sobre o arco AB de C2 que está dentro de C1.A reta AP intersecta C1 em C, a reta CB intersecta C2 em D e a bissetriz de <CAD intersecta C1 em E e C2 em L.Seja F o simétrico do ponto D em relação ao ponto médio de PE.Demonstrar que existe um ponto X que satisfaz <XFL=<XDC=30º e CX=O1O2.[/quote]

Como te fue Secco?

[quote="Joao Pedro Santos"]¿Alguien puede colocar los problemas del segundo día?[/quote]

El primer día ya está en la sección de problemas: http://www.mathlinks.ro/Forum/resources.php?c=1&cid=29&year=2009

El segundo día parece que no lo han subido aún, pero igual aquí están los links por si los quieren revisar.

http://www.mathlinks.ro/Forum/viewtopic.php?t=302565
http://www.mathlinks.ro/Forum/viewtopic.php?t=302566
http://www.mathlinks.ro/Forum/viewtopic.php?t=302567

Gracias a todos los que subieron los enunciados!

Supongo que por estas horas ya se saben los resultados, al menos el programa decía que anoche ya era la reunión de cortes, así que los que tengan la información si pueden compartirla estaremos muy agradecidos!  :)  Especialmente quisiera saber cómo les fue a los países de Argentina, Brasil y Perú.

Otra cosa: ¿se sabe qué países propusieron los problemas? Vi que el 3 era de El Salvador, pero me gustaría saber los demás también.

Muchas gracias!

Peru:

PER1(julian): 34
PER2(amilcar): 36
PER3(jesus): 38
PER4(percy): 38

Brasil:

BRA1(renan): 41
BRAX: 38
BRAY: 33
BRAZ: 33
donde
$ \{X,Y,Z\} \equal{} \{2,3,4\}$. Eso es todo lo que se. No se como fueron los cortes. Supongo que bra1, per3, per4 y brax son oros. Bueno, otra buena noticia: si suman puntajes, notaran que PERU consiguio la progresion aritmetica:
2006: 4lugar
2007: 3lugar
2008: 2lugar
2009: 1lugar

JAJAJA :rotfl:
de los otros participantes no se nada, si tienen algo, escriban. :P

Creo que los puntajes de Ecuador fueron algo asi:

ECU1 (Christian) 19
ECU2 (Gabriel) 24
ECU3 (Jenny) 14
ECU4 (Miguel) 16..

Eso me contaron, y que los cortes no estaban hasta ayer...

Colombia
jorge olarte 39
nicolas del catillo 35
david arcila 24
carlos diaz 23

Estos son los resultados de El Salvador:

               P1  P2  P3  P4  P5    P6  total
Alberti      7     7    0   7    4     0    25     Bronce
Julio         6     6    0   7    7     0    26     Bronce
Tony        6     7    2    7    1    0    23     Mención
Nahomy   3     1    0    7    7    0    18     Mención

Hola a todos! Tienen los resultados de Paraguay? Desde ya gracias

[quote="uglysolutions"]Parabéns Matheus!!!  :D

Resultados de Argentina:

ARG1 (Germán): 7 7 6 7 7 3 (37)

ARG2 (Iván): 7 5 4 6 6 1 (29)

ARG3 (Miguel): 7 7 0 7 7 4 (32)

ARG4 (Juan): 4 7 1 7 7 4 (30)

Los cortes fueron 24, 32 y 38, así que son 2 bronces, 1 plata bronceada y 1 plata dorada xD.[/quote]

Se dio una modificación de los cortes y quedaron así:
Bronce 24
Plata 31
Oro 38

[quote="uglysolutions"]Otra cosa: ¿se sabe qué países propusieron los problemas? Vi que el 3 era de El Salvador, pero me gustaría saber los demás también.

Muchas gracias![/quote]

PROBLEMA 1     MEXICO
PROBLEMA 2     MEXICO
PROBLEMA 3     EL SALVADOR    :roll:
PROBLEMA 4     MEXICO
PROBLEMA 5     ARGENTINA
PROBLEMA 6     ARGENTINA

En la pagina http://www.oei.es/cienciayuniversidad/spip.php?article645 se pueden encontrar los problemas y soluciones, y un informe muy completo sobre el desarrollo de la Olimpiada


Can we find four different primes p ,q , r ,s as great as we want and four naturel numbers a ,b ,c ,d such that:
(p^a)(q^b) - (r^c)(s^d) = 1.
Same qUestion with two primes and two naturel numbers.
How about two coprime numbers.

for the question about two coprime numbers the solution is yes and there are an infinity of solution of the form
(n+1) - n = 1
since they are coprime.
for the second question i realised that my question is catalan conjecture.AND THE RESPONSE IS NO;
remain the first question.It's probably possible because of third question.It's a kind of adjustion the terms in the left member of the equation.??


What is your predicted number correct?
I got 8 of them I think.

5 and a half  :|

result of one english class of working on the problems and one hour before midnight writing them.

Well, only one more day waiting to see if we get in or not, for those of us who applied early. I wonder how many applicants they'll get compared to last year.

[quote="Klebian"][quote="Aegor"]I third that.  I did not get 8, and I missed half of the solutions for 6.  So more like 8 and a half, if all others are correct.[/quote]
8 was very much like an AMC10 problem a few years ago. It was in an AMC practice packet that I had been working on earlier in the day and I thus found it pretty simple.[/quote]
Wait, you got number ten, but missed number 8?
I personally thought 8 was the easiest of the entire test, because it's just setting up an equation.
Then again, I did bad on the test overall...

So much confusion. :P  I got 8 problems correct.  I did not get 10, I DID get 8, and I got half of 6.  So, 8 fully correct, and 1 partially correct, and one not so much.

Eh, I seemed to have misinterpreted "Round Robin" and the 5 people thing.  I thought Round Robin meant each person plays each other person some equal number of times, and the two games each as each person playing two games against everyone each.  so...my answer was a bit different.

If you count that one wrong, I have 8.  I totally didn't solve #10.

Ahah, sure.

And yeah, maybe 7 at least, 8 at most.

[quote="Aegor"][quote="stupidityismygam"]pretty sure that i got the 9 i answered, i definitely did not get #10 though[/quote]

I don't think that many did (at least from this forum).  I drew the diagram and then gave up; mainly because I had spent all weekend doing the other problems and it was 10:30 at night.  They weren't that bad, all of them (save number 10) were easier than the hardest AMC problems, even if you got the same amount of time to do both.[/quote]
Huh, SplashD got number 10.

[quote="bpms"]
Huh, SplashD got number 10.[/quote]

Good job to him!  Did anyone start working on them as soon as they could?  I was planning to attend a local math program, but then someone who had sent their child to Awesomemath recommended it to me.  So, on Friday night, I decided to apply.  :P   Needless to say, my weekend was very busy.

by the way, the solutions are up for eveybody interested


geometry
3D geometry
tetrahedron
IMO 1965
prism
IMO

Given the tetrahedron $ABCD$ whose edges $AB$ and $CD$ have lengths $a$ and $b$ respectively. The distance between the skew lines $AB$ and $CD$ is $d$, and the angle between them is $\omega$. Tetrahedron $ABCD$ is divided into two solids by plane $\epsilon$, parallel to lines $AB$ and $CD$. The ratio of the distances of $\epsilon$ from $AB$ and $CD$ is equal to $k$. Compute the ratio of the volumes of the two solids obtained.

Given the tetrahedron $ ABCD $ whose edges $ AB $ and $ CD $ have lengths $ a $ and $ b $ respectively. The distance between the skew lines $ AB $ and $ CD $ is $ d $, and the angle between them is $ \omega $. Tetrahedron $ ABCD $ is divided into two solids by plane $ \epsilon $, parallel to lines $ AB $ and $ CD $. The ratio of the distances of $ \epsilon $ from $ AB $ and $ CD $ is equal to $ k $. Compute the ratio of the volumes of the two solids obtained.

[size=150]Construction.[/size]
Through the point $ B $, we draw the line $ n $, $ n \parallel CD $, see picture 1.
Through the point $ C $, we draw the line $ m $, $ m \parallel AB $.
Then, plane $ (AB,n)$ $ \parallel $ plane $ (CD,m) $.
There are many lines, perpendicular to plane $ (AB,n)$ and plane $ (CD,m) $, there is only one line, perpendicular to $ AB $ and $ CD $, who is also crossing $ AB $ and $ CD $.
Suppose that $ EF $ is that special perpendicular. On $ EF $, we put the point $ M $, with the condition:
	\[\frac{\left|ME\right|}{\left|MF\right|}=k\]
On the picture, that ratio $ k=1 $.
Through the point $ M $, we construct the plane $ \epsilon $, parallel to lines $ AB $ and $ CD $, hence $ \epsilon \parallel (AB,n) \parallel (CD,m) $.
That plane $ \epsilon $ is crossing $ AD $ in the point $ P $, $ AC $ in the point $ Q $, $ BC $ in the point $ R $ and $ BD $ in the point $ S $. The points $ P,Q,R,S $ divide $ AD \ , AC \ , BC \ , BD $ with the same ratio $ k $. These points $ P,Q,R,S $ form a parallelogram and:
	\[AB \parallel PS \parallel QR \ \ ,\ \  CD \parallel PQ \parallel SR \]
The plane $ \epsilon $ divide our tetrahedron $ ABCD $ into two solids, two truncated prism $ (PS,QR,AB) $ (green on the picture 1 and rotated in picture 2) and $ (PQ,RS,CD) $ (red on the picture 1).
[size=150]Calculation.[/size]
An old theorem says: the volume of a truncated prism equals one third of the product of two factors: first factor is the sum of the lenght of the lateral edges, second factor is the surface of the perpendicular cross-section of these lateral edges.
If \[\frac{\left|ME\right|}{\left|MF\right|}=k\]  then \[\frac{\left|EF\right|}{\left|MF\right|}=k+1\] or \[\left|EF\right|=(k+1)\left|MF\right|\]
In $\triangle ABD$ and $\triangle ABC$, on the same way:
	\[\left|AB\right|=(k+1)\left|PS\right| \]and \[\left|AB\right|=(k+1)\left|QR\right|\]
The sum of the lenght of the lateral edges in the truncated prism $ (PS,QR,AB) $:
	\[\left|AB\right|+\left|PS\right|+\left|QR\right|=\left|AB\right| \cdot (1+\frac{2}{k+1})\]
In $\triangle ACD$ and $\triangle BCD$:
	\[\left|PQ\right|=\frac{k}{k+1}\left|CD\right| \]and \[\left|SR\right|=\frac{k}{k+1}\left|CD\right|\]
The sum of the lenght of the lateral edges in the truncated prism $ (PQ,RS,CD) $:
	\[\left|CD\right|+\left|PQ\right|+\left|RS\right|=\left|CD\right| \cdot (1+\frac{2k}{k+1})\]
Now the surface of the perpendicular cross-section of the truncated prism $ (PS,QR,AB) $ and $ (PQ,RS,CD) $; the cross-section of the truncated prism $ (PS,QR,AB) $ is perpendicular to $ AB $ and the cross-section of the truncated prism the cross-section of the truncated prism is perpendicular to $ CD $; the angle between them is $ \omega $; the ratio of these surfaces contains $ \cos \omega $. Because $\frac{\left|ME\right|}{\left|MF\right|}=k$, this ratio contains also $ k^{2}$.
The surface of the perpendicular cross-section of the truncated prism $ (PS,QR,AB) = k^{2} \cdot \cos \omega \cdot T $ ; $ T $ is the surface of the perpendicular cross-section of the truncated prism $ (PQ,RS,CD) $.
[size=150]Conclusion.[/size]
The volume of the truncated prism $ (PS,QR,AB)= \frac{1}{3}\left|AB\right|\frac{k+3}{k+1} \cdot k^{2} \cdot \cos \omega \cdot T $.\\
The volume of the truncated prism $ (PQ,RS,CD)= \frac{1}{3}\left|CD\right|\frac{3k+1}{k+1} \cdot T $.
The ratio of the volumes is:
	\[ \frac{a}{b} \cdot k^{2} \frac{k+3}{3k+1}\cdot \cos \omega \]


Let $ x;y;z$ be positive real numbers .Find min value of:
$ T\equal{}\frac{x^3 y}{y^3 z\plus{}x^2 z^2}\plus{}\frac{y^3 z}{z^3 x\plus{}x^2 y^2}\plus{}\frac{ z^3 x }{x^3 y\plus{}y^2 z^2}$

[quote="chien than"]Let $ x;y;z$ be positive real numbers .Find min value of:
$ T \equal{} \frac {x^3 y}{y^3 z \plus{} x^2 z^2} \plus{} \frac {y^3 z}{z^3 x \plus{} x^2 y^2} \plus{} \frac { z^3 x }{x^3 y \plus{} y^2 z^2}$[/quote]
Is it your inequality, chien than? I think, I saw it before. :wink:


The infinite real sequence x1, x2, x3, ... satisfies |xi - xj| >= 1/(i + j) for all unequal i, j. Show that if all xi lie in the interval [0, c], then c >= 1.
my solution shows that c >= n-2/n for all n which implies c>=1.duz nebody have a different bound for c?
best regards

You can find a solution at
http://www.kalva.demon.co.uk/short/soln/sh02a2.html.


set theory
real analysis
real analysis unsolved

Hi,

I can't understand how to formulate what the author describes in the following text.
$ E\equal{}\bigcup P(E)$ is easy to prove.
But I can't formulate the "other order" that he talks about.
How can I formulate it ?

He's asking you to show how $ P(\cup E)$ compares to $ E$.

yes but that is $ P(\bigcup E)\equal{}P(E)$.
is it that simple ?

No: for example $ E \equal{} \{\{1, 2\}, \{2, 3\}\}$ implies $ \cup E \equal{} \{1, 2, 3\}$ and $ P(\cup E) \supsetneq E$.

but JBL,I thought elements of E are just numbers not other sets.
and how does $ \bigcup{E}\equal{}\{1,2,3\}$ if E contains sets.
For that matter what are the subsets of this strange set E ?
what is P(UE) in this case ?

The elements of the $ E$ I gave you are sets: $ E$ had two elements, $ \{1, 2\}$ and $ \{2, 3\}$.  (Also, in the context of set theory numbers [i]are[/i] sets, but that's irrelevant to this discussion.)  More importantly, I don't think you understand the notation you're using: $ \cup E \overset{\textrm{def}}{ \equal{} } \bigcup_{a \in E} a$.


projective geometry
geometry solved
geometry

Let ABC be arbitrary triangle and AA1, BB1 and CC1 be three concurrent lines such that A1, B1 and C1 belong to BC, CA and AB respectively. Let A1B1 and AB intersect at X, B1C1 and BC intersect at Y, A1C1 and AC intersect at Z. Prove that X,Y and Z are collinear.

Yes, I know that we should use Ceva and Menelaus, but I don't know the way. Can you pls post the exact solution?

By Ceva:$\frac{BA_1}{A_1C}\frac{CB_1}{B_1A}\frac{AC_1}{C_1B}=1$
By Menelaus:$\frac{BA_1}{A_1C}\frac{CB_1}{B_1A}\frac{AX}{XB}=1$
$\frac{BY}{YC}\frac{CB_1}{B_1A}\frac{AC_1}{C_1B}=1$
$\frac{BA_1}{A_1C}\frac{CZ}{ZA}\frac{AC_1}{C_1B}=1$
So $\frac{BY}{YC}\frac{CZ}{ZA}\frac{AX}{XB}=1$
Then $X,Y,Z$ is collinear.

Come on, such a problem being given on an olympiad is :spam: . It is well-known itself and it is a particular case of the even more well-known [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30323]Desargues theorem[/url].

  darij


Will there be a Math Jam for Round 3? If so, when will it be?

oops...we totally forgot to schedule this!   :blush:

We're going to have it on Thursday, 1/11, at 7:30 Eastern (4:30 Pacific).  It'll be on the official schedule shortly.

By the way, we have our usual discussion embargo: please do not discuss Round 3 until after the Math Jam on Thursday.

[quote="jruzzy"][quote="bpms"][quote="purple plume"]of course you can. freedom of speech, unless the Bill of Rights has a flaw[/quote]
Actually, I pmed an admin and they said wait until after the math jam. About freedom of speech, the thing that prevents me from cheating on the USAMTS, is what is stopping me from discussing the round, so...[/quote]

The USAMTS runs on an honor system, so it follows "the thing" you mentioned is honor. If you find it goes against said honor to discuss problems before the Math Jam, go ahead and don't. Otherwise...[/quote]
Did you read my post? The admin specifically said don't discuss the third round until after the math jam.

[quote="bpms"][/quote][quote="jruzzy"][quote="bpms"][quote="purple plume"]of course you can. freedom of speech, unless the Bill of Rights has a flaw[/quote]
Actually, I pmed an admin and they said wait until after the math jam. About freedom of speech, the thing that prevents me from cheating on the USAMTS, is what is stopping me from discussing the round, so...[/quote]

The USAMTS runs on an honor system, so it follows "the thing" you mentioned is honor. If you find it goes against said honor to discuss problems before the Math Jam, go ahead and don't. Otherwise...[/quote][quote="bpms"]
Did you read my post? The admin specifically said don't discuss the third round until after the math jam.[/quote]

Jeez... no need to accuse people of failure to read posts.

I did, in fact, read your post (as I do with all posts I quote) and was commenting on the second half of it. I fail to see how you can equate outright cheating on the problems to simple discussion of the problems, but I digress...

The thing is in under 24 hours anyway, so this will become a non-issue soon enough.

[quote="jruzzy"][/quote][quote="bpms"][/quote][quote="jruzzy"][quote="bpms"][quote="purple plume"]of course you can. freedom of speech, unless the Bill of Rights has a flaw[/quote]
Actually, I pmed an admin and they said wait until after the math jam. About freedom of speech, the thing that prevents me from cheating on the USAMTS, is what is stopping me from discussing the round, so...[/quote]

The USAMTS runs on an honor system, so it follows "the thing" you mentioned is honor. If you find it goes against said honor to discuss problems before the Math Jam, go ahead and don't. Otherwise...[/quote][quote="bpms"]
Did you read my post? The admin specifically said don't discuss the third round until after the math jam.[/quote][quote="jruzzy"]

Jeez... no need to accuse people of failure to read posts.

I did, in fact, read your post (as I do with all posts I quote) and was commenting on the second half of it. I fail to see how you can equate outright cheating on the problems to simple discussion of the problems, but I digress...

The thing is in under 24 hours anyway, so this will become a non-issue soon enough.[/quote]
Oh sorry, my mistake. I didn't know what you were saying.

[quote="rrusczyk"]Towards the end of the month.[/quote]

Now that we are already reaching the end of the month, does anyone know of a more specific time that round 4 is coming out?

Probably at the end of the week -- Thurs or Fri.

[quote="vishalarul"]How about Round 3 scores? Is it also coming at the end of the week?

Can't wait... :jump:[/quote]

Last time it took about five weeks, probably not for another week at least.