geometry
circumcircle
incenter
3D geometry
tetrahedron
inradius
function

[url=http://www.mathlinks.ro/Forum/download.php?id=1614]Click to here get the PDF Version[/url]

Click on each grade to see the problems, and click on the problem number to see the topic of discussion with solution.

[hide="7th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28991]Problem 1[/url]
Prove that for all $a\in\{0,1,2,\ldots,9\}$ the following sum is divisible by 10:
\[ S_a = \overline{a}^{2005} + \overline{1a}^{2005} + \overline{2a}^{2005} + \cdots + \overline{9a}^{2005}. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28992]Problem 2[/url]
Let $ABC$ be a triangle and let $M$ be the midpoint of the side $AB$. Let $BD$ be the interior angle bisector of $\angle ABC$, $D\in AC$. Prove that if $MD \perp BD$ then $AB=3BC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28993]Problem 3[/url]
We denote with $m_a$, $m_g$ the arithmetic mean, respectively geometrical mean, of the positive numbers $x,y$.

a) If $m_a+m_g=y-x$, determine the value of $\dfrac xy$;

b) Prove that there exists exactly one pair of different positive integers $(x,y)$ for which $m_a+m_g=40$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28994]Problem 4[/url]
In the triangle $ABC$ let $AD$ be the interior angle bisector of $\angle ACB$, where $D\in AB$. The circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $BCD$. Prove that $AC^2 = AD\cdot AB$.
[/hide]
[hide="8th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28995]Problem 1[/url]
Let $M$ be the set of the positive rational numbers less than 1, which can be expressed with a 10-distinct digits period in decimal representation.

a) Find the arithmetic mean of all the elements in $M$;

b) Prove that there exists a positive integer $n$, $1<n<10^{10}$, such that $n\cdot a - a$ is a non-negative integer, for all $a\in M$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28996]Problem 2[/url]
Let $ABCD$ and $ABEF$ be two squares situated in two perpendicular planes and let $O$ be the intersection of the lines $AE$ and $BF$. If $AB=4$ compute:

a) the distance from $B$ to the line of intersection between the planes $(DOC)$ and $(DAF)$;

b) the distance between the lines $AC$ and $BF$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=28999]Problem 3[/url]
Prove that if the circumcircles of the faces of a tetrahedron $ABCD$ have equal radii, then $AB=CD$, $AC=BD$ and $AD=BC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29001]Problem 4[/url]
Prove that no matter how we number the vertices of a cube with integers from 1 to 8, there exists two opposite vertices in the cube (e.g. they are the endpoints of a large diagonal of the cube), united through a broken line formed with 3 edges of the cube, such that the sum of the 4 numbers written in the vertices of this broken lines is at least 21.
[/hide]
[hide="9th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29004]Problem 1[/url]
a) Prove that if $x,y>0$ then
\[ \frac x{y^2} + \frac y{x^2} \geq \frac 1x + \frac 1y. \]

b) Prove that if $a,b,c$ are positive real numbers, then
\[ \frac {a+b}{c^2} + \frac {b+c}{a^2} + \frac {c+a}{b^2} \geq 2 \left( \frac 1a + \frac 1b + \frac 1c \right). \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29005]Problem 2[/url]
Let $ABC$ be a triangle inscribed in a circle of center $O$ and radius $R$. Let $I$ be the incenter of $ABC$, and let $r$ be the inradius of the same triangle, $O\neq I$, and let $G$ be its centroid. Prove that $IG\perp BC$ if and only if $b=c$ or $b+c=3a$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29006]Problem 3[/url]
Let $ABC$ be a non-right-angled triangle and let $H$ be its orthocenter. Let $M_1,M_2,M_3$ be the midpoints of the sides $BC$, $CA$, $AB$ respectively. Let $A_1$, $B_1$, $C_1$ be the symmetrical points of $H$ with respect to $M_1$, $M_2$ and $M_3$ respectively, and let $A_2$, $B_2$, $C_2$ be the orthocenters of the triangles $BA_1C$, $CB_1A$ and $AC_1B$ respectively. Prove that:

a) triangles $ABC$ and $A_2B_2C_2$ have the same centroid;

b) the centroids of the triangles $AA_1A_2$, $BB_1B_2$, $CC_1C_2$ form a triangle similar with $ABC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29007]Problem 4[/url]
Let $\{a_k\}_{k\geq 1}$ be a sequence of non-negative integers, such that $a_k \geq a_{2k} + a_{2k+1}$, for all $k\geq 1$.

a) Prove that for all positive integers $n\geq 1$ there exist $n$ consecutive terms equal with 0 in the sequence $\{a_k\}_k$;

b) State an example of sequence with the property in the hypothesis which contains an infinite number of non-zero terms.
[/hide]
[hide="10th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29008]Problem 1[/url]
Let $a,b>1$ be two real numbers. Prove that $a>b$ if and only if there exists a function $f: (0,\infty)\to\mathbb{R}$ such that

 i) the function $g:\mathbb{R}\to\mathbb{R}$, $g(x)=f(a^x)-x$ is increasing;

ii) the function $h:\mathbb{R}\to\mathbb{R}$, $h(x)=f(b^x)-x$ is decreasing.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29009]Problem 2[/url]
Find the functions $f:\mathbb{Z}\times \mathbb{Z}\to\mathbb{R}$ such that

a) $f(x,y)\cdot f(y,z) \cdot f(z,x) = 1$ for all integers $x,y,z$;

b) $f(x+1,x)=2$ for all integers $x$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29011]Problem 3[/url]
Let $O$ be a point equally distanced from the vertices of the tetrahedron $ABCD$. If the distances from $O$ to the planes $(BCD)$, $(ACD)$, $(ABD)$ and $(ABC)$ are equal, prove that the sum of the distances from a point $M \in \textrm{int}[ABCD]$, to the four planes, is constant.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29012]Problem 4[/url]
Let $n\geq 3$ be an integer. Find the number of functions $f:\{1,2,\ldots,n\}\to\{1,2,\ldots,n\}$ such that
\[ f(f(k)) = f^3(k) - 6f^2(k) + 12f(k) - 6 , \ \textrm{ for all } k \geq 1 . \]
[/hide]
[hide="11th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29014]Problem 1[/url]
Let us denote with $H$ the set of all the square matrices of order $n\geq 2$ with elements positive integers and with $P$ the set of matrices in $H$ such that the sum of the elements of each line and each column is equal with 1.

a) Prove that if $A\in P$ then $\det A \in \{-1,1\}$;

b) Prove that if $A_1$, $A_2$, $\ldots$, $A_p$ are in $H$, and $A_1A_2\cdots A_p \in P$ then $A_k\in P$ for all $k=\overline{1,p}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29016]Problem 2[/url]
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for all $a,b\in \mathbb{R}$, $a<b$, with $f(a)=f(b)$, there exists a $c\in(a,b)$ with $f(a)=f(b)=f(c)$. Prove that the function $f$ is monotonous on $\mathbb{R}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29017]Problem 3[/url]
a) Let $A,B\in\mathcal{M}_3(\mathbb{R})$ such that $\mathop\textrm{rang} A> \mathop\textrm{rang} B$. Prove that $\mathop\textrm{rang}{(A^2)} \geq \mathop\textrm{rang}{(B^2)}$;

b) Determine the non-constant polynomials $f$ with real coefficients, with the property that for any matrices $A,B\in\mathcal{M}_4(\mathbb{R})$ with $\mathop\textrm{rang}A > \mathop\textrm{rang}B$ we have
\[\mathop\textrm{rang}{f(A)} \geq \mathop\textrm{rang}{f(B)}. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29018]Problem 4[/url]
Let $f:\mathbb{Q}\to\mathbb{Q}$ be a monotonous one-to-one onto (bijective) mapping.

a) Prove that there exists an unique continuous function $F:\mathbb{R}\to\mathbb{R}$ such that $F(x)=f(x)$ for all $x\in\mathbb{Q}$;

b) Give an example of a non-one-to-one (non-injective) polynomial function $G:\mathbb{R}\to\mathbb{R}$, such that $G(\mathbb{Q}) \subset \mathbb{Q}$ and its restriction $G |_{\mathbb{Q}}$ is one-to-one.
[/hide]
[hide="12th Grade"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29019]Problem 1[/url]
Let $A_1$, $A_2$, $\ldots$, $A_n$, $n\geq 2$ be $n$ finite sets with the properties

 i) $|A_i| \geq 2$, for all $1\leq i \leq n$;

ii) $|A_i\cap A_j| \neq 1$, for all $1\leq i<j\leq n$.

Prove that the elements of the set $\displaystyle \bigcup_{i=1}^n A_i$ can be colored with 2 colors, such that all the sets $A_i$ are bi-color, for all $1\leq i \leq n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29020]Problem 2[/url]
Let $f:[0,1]\to\mathbb{R}$ be a continuous function and let $\{a_n\}_n$, $\{b_n\}_n$ be sequences of reals such that
\[ \lim_{n\to\infty} \int^1_0 | f(x) - a_nx - b_n | dx = 0 . \]

Prove that:

a) The sequences $\{a_n\}_n$, $\{b_n\}_n$ are convergent;

b) The function $f$ is linear.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29021]Problem 3[/url]
Let $(G,\cdot)$ be a group and let $F$ be the set of elements in the group $G$ of finite order. Prove that if $F$ is finite, then there exists a positive integer $n$ such that for all $x\in G$ and for all $y\in F$, we have
\[ x^n y = yx^n. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29022]Problem 4[/url]
Let $(A,+,\cdot)$ be a finite unit ring, with $n\geq 3$ elements in which there exist [b]exactly[/b] $\dfrac {n+1}2$ perfect squares (e.g. a number $b\in A$ is called a perfect square if and only if there exists an $a\in A$ such that $b=a^2$). Prove that

a) $1+1$ is invertible;

b) $(A,+,\cdot)$ is a field.
[/hide]
.

Too many linear algebra :? . And the whole set for 11th and 12th grades seems quite strange for me, [i]even if I realize its purpose[/i].

[quote="Valentin Vornicu"][quote="mna851"]It's interesting that the grade 12 test has calculus on it.[/quote]Of course. All grade 12th (and 11th for that matter) have calculus (sometimes even much harder than most putnam problems) on them ;)[/quote]
But are they so nice as many Putnam problems? ;)

It is just my opinion realated to olympiad traditions which vary from country to country.

[quote="Myth"]Too many linear algebra :? . And the whole set for 11th and 12th grades seems quite strange for me, [i]even if I realize its purpose[/i].

But are they so nice as many Putnam problems? ;)[/quote]Sometimes they are ;) :P I like many linear algebra problems :D Usually there are more calculus problems, but this year is an exception :)

[quote="Mamat"]Inequalities for 9th grade are nice Second one can be solved by using first  :)[/quote]
The corresponding [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29004]topic[/url] titled as "easy and classical inequality" ;)


If m,n,p are positive integers such that m/(n-p)+n/(p-m)+p/(n-m)=2008,
prove that at least one of the fractions m/(n-p), n/(p-m), p/(n-m) can be reduced.

It is easy to prove that $ \gcd(m\minus{}n,n\minus{}p)\equal{}\gcd(m\minus{}n,p\minus{}m)\equal{}\gcd(n\minus{}p,p\minus{}m)\equal{}d$
Then exist $ (d,x,y,z)$ such that
$ m\minus{}n\equal{}dx,n\minus{}p\equal{}dy,p\minus{}m\equal{}dz$ where $ (x,y,z)$ are pairwise relative prime .
Suppose that each of fractions can not be reduce. It means that $ \gcd(p,dx)\equal{}1$
From the condition we have :
$ dxyz|(mxz\plus{}nxy\plus{}pyz)$
$ \Rightarrow xyz|(mxz\plus{}nxy\plus{}pyz)$
$ \Rightarrow x|pyz$
Because $ \gcd(x,yz)\equal{}1$ so $ x|p$
Because $ \gcd(p,x)\equal{}1$ ,it follows that $ |x|\equal{}1$
Siminar $ |x|\equal{}|y|\equal{}|z|\equal{}1$
But $ x\plus{}y\plus{}z\equal{}0$ it gives contradiction .


From the top of a building 10m tall, the angle of elevation to the top of a flagpole is 11degrees. At the base of the building the angle of elevation to the top of the flagpole is 39 degrees. Find the height of the flagpole. (answer is 13.2)

(please explain/show the picture is possible)

i have the solution  but  how to  upload image can u tell me

Basic precalc problem.

[hide="Solution"]
 [geogebra]7258204595d418284e1f6693eb433be6b47c0ad9[/geogebra]

Let the distance between the building and the flagpole be $ k$, and let the height of the flagpole be $ 10\plus{}x$.

We have $ \tan 39\equal{}\frac{10\plus{}x}{k}$ and $ \tan 11\equal{}\frac{x}{k}$

Equating, we have $ 10\equal{}k(\tan39\minus{}\tan11)\implies k\equal{}16.2495$

Since $ \tan 11\equal{}\frac{x}{16.2495}$, we have $ x\approx 3.2$

Thus, the height of the flagpole is $ 10\plus{}3.2\equal{}13.2$

Answer: 13.2
[/hide]


Let be $ a,b\in \mathbb{R}$ and let be $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ a\cdot f(x)\plus{}b\cdot f(1\minus{}x)\equal{}x\ (1)\ ,\ (\forall)x\in \mathbb{R}$ .

i)Show that $ f(x)\plus{}f(1\minus{}x)\equal{}\frac{1}{a\plus{}b}\ ,\ (\forall)x\in \mathbb{R}\ ,\ a\neq \minus{}b$ ;

ii)If $ a\equal{}b$ , prove that $ (\nexists)f$ satisfying $ (1)$ ;

ii)If $ a\neq b$ , find $ f$ .

[b]i)[/b]$ a\cdot f(x)\plus{}b\cdot f(1\minus{}x)\equal{}x\ .......(1)$
$ a\cdot f(1\minus{}x)\plus{}b\cdot f(x)\equal{}1\minus{}x ........(2)$
(1)+(2) and the result follows.

[b]ii)[/b]If $ a\equal{}b$, then $ x\equal{}a(f(x)\plus{}f(1\minus{}x))\equal{}\frac{a}{a\plus{}b}$.
Assume that $ f$ satisfies (1) for all $ x$, then $ x\equal{}\frac{a}{a\plus{}b}$ which is fixed holds for all $ x$, contradiction, done.

[b]iii)[/b]If $ a \neq b$, then $ a \cdot f(x)\plus{}b \cdot (\frac{1}{a\plus{}b}\minus{}f(x))\equal{}x$
$ \Leftrightarrow f(x)\equal{}\frac{x}{a\minus{}b}\minus{}\frac{b}{a^2\minus{}b^2}$ for all $ x$ is the solution.


inequalities
function
vector
calculus
inequalities proposed

Yes, the inequalities are indeed trivial. Actually, the first one follows from

$ \frac {1}{2b \plus{} 1} \plus{} \frac {1}{2c \plus{} 1}\geq \frac {2}{b \plus{} c \plus{} 1}$;
$ \frac {1}{2c \plus{} 1} \plus{} \frac {1}{2a \plus{} 1}\geq \frac {2}{c \plus{} a \plus{} 1}$;
$ \frac {1}{2a \plus{} 1} \plus{} \frac {1}{2b \plus{} 1}\geq \frac {2}{a \plus{} b \plus{} 1}$

(this is because $ \frac {1}{2x \plus{} 1} \plus{} \frac {1}{2y \plus{} 1}\geq \frac {2}{x \plus{} y \plus{} 1}$ holds for any two positive reals x and y, what again is a consequence of $ \frac {1}{2x \plus{} 1} \plus{} \frac {1}{2y \plus{} 1} \minus{} \frac {2}{x \plus{} y \plus{} 1} \equal{} 2\frac {\left( x \minus{} y\right) ^{2}}{\left( 2x \plus{} 1\right) \left( 2y \plus{} 1\right) \left( x \plus{} y \plus{} 1\right) }\geq 0$). And the second inequality, hmm, well, it follows from Jensen applied to the convex function $ f\left( x\right) \equal{} \frac {1}{2 \plus{} 3x}$. But why one implies the other... Umm, if you call a limiting process elegant, then yeah, then it's right; in fact, if we define

$ f\left( x,y,z\right) \equal{} \frac {1}{2x \plus{} 1} \plus{} \frac {1}{2y \plus{} 1} \plus{} \frac {1}{2z \plus{} 1}$

as a function of the vector (x, y, z), then the first inequality states that $ f\left( x,y,z\right) \geq f\left( \frac {y \plus{} z}{2},\frac {z \plus{} x}{2},\frac {x \plus{} y}{2}\right)$. But since the iterative application of the transformation $ \left( x,y,z\right) \rightarrow \left( \frac {y \plus{} z}{2},\frac {z \plus{} x}{2},\frac {x \plus{} y}{2}\right)$ produces a convergent sequence with limit $ \left( \frac {x \plus{} y \plus{} z}{3},\frac {x \plus{} y \plus{} z}{3},\frac {x \plus{} y \plus{} z}{3}\right)$, and since the function f is continuous, we finally get

$ f\left( x,y,z\right) \geq f\left( \frac {x \plus{} y \plus{} z}{3},\frac {x \plus{} y \plus{} z}{3},\frac {x \plus{} y \plus{} z}{3}\right)$.

(This requires some easy calculus to prove, which I omit.) If we now set $ x \equal{} \frac {3}{4}a$, $ y \equal{} \frac {3}{4}b$, $ z \equal{} \frac {3}{4}c$, then this becomes

$ \frac {1}{2\cdot \frac {3}{4}a \plus{} 1} \plus{} \frac {1}{2\cdot \frac {3}{4}b \plus{} 1} \plus{} \frac {1}{2\cdot \frac {3}{4}c \plus{} 1}\geq 3\cdot \frac {1}{2\cdot \frac {3}{4}\cdot \frac {a \plus{} b \plus{} c}{3} \plus{} 1}$,

and this is actually equivalent to $ \frac {1}{2 \plus{} 3a} \plus{} \frac {1}{2 \plus{} 3b} \plus{} \frac {1}{2 \plus{} 3c}\geq \frac {3}{2 \plus{} a \plus{} b \plus{} c}$. So the first inequality implies the second one. But I hardly think this is the elegant way you intended, cuz it's killing flies with machine guns...

And yes, Harazi, I completely agree with you, I'm sick of median inequalities...

[b]EDIT:[/b] My above limit argument is very similar to the tactics discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=113248 .

  Darij

[quote="harazi"]Below are two trivial inequalities:
   1) For $ a,b,c>0 $ we have $\frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}\geq\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} $.
   2) For any positive a,b,c we have $\frac{1}{2+3a}+\frac{1}{2+3b}+\frac{1}{2+3c}\geq\frac{3}{2+a+b+c} $.
   The non-trivial (I hope) question is: which one implies the other? I hope you can find an elegant solution, because there is one.[/quote]

(1) implies (2): Let $x=\frac{3}{4}a$, etc. Then:

$\frac{1}{2+3a}+\frac{1}{2+3b}+\frac{1}{2+3c} = \frac{1}{2}\left(\frac{1}{2x+1}+\frac{1}{2y+1}+\frac{1}{2z+1}\right)\geq\frac{1}{2}\left(\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\right) \geq \frac{1}{2}\left(\frac{9}{2(x+y+z) + 3}\right) = \frac{3}{2+a+b+c} $

Both solutions are splendid, but I propose to you another challenge (in fact, this was the initial inequality I found, but, as usual, I made things easier with a stupid question):
   Prove that for positive numbers $ a,b,c$ we have:
  $ \sum\frac {1}{2a \plus{} 1} \minus{} \sum\frac {1}{a \plus{} b \plus{} 1}\geq\frac {1}{3}\cdot (\sum\frac {1}{2 \plus{} 3a} \minus{} \frac {3}{2 \plus{} a \plus{} b \plus{} c})$.
   Even more generally, let us consider $ a_i$ positive real numbers. Define $ A_k \equal{} C(n,k)\cdot\sum\frac {1}{ka_i \plus{} k \minus{} 1} \minus{} n\sum\frac {1}{a_{i_1} \plus{} a_{i_2} \plus{} ... \plus{} a_{i_k} \plus{} k \minus{} 1}$ where the second sum is taken after all subsets with k elements of the set {1,2,...,n} and $ C(n,k)$ is the number of these subsets.  Then prove that $ A_2 \minus{} A_3 \plus{} A_4 \minus{} ...$ is nonnegative (the sum is until n).


[quote="xp"]Δινεται κανονικο $ n$-γωνο $ A_1A_2...A_n$ εγγεγραμενο στον μοναδιαίο κυκλο και μεταβλητο σημειο $ M$ του επιπέδου του .
Να βρεθεί το $ [\sum_{i \equal{} 1}^{n}{(MA_i)}]_{min}$[/quote]

Πολύ όμορφη άσκηση :first:

Δηλαδή, ο μέσος όρος των αποστάσεων ενός σημείου από τις κορυφές ενός κανονικού n-γώνου προκύπτει μεγαλύτερος ή ίσος από την ακτίνα του περιγεγραμμένου κύκλου!

[hide]Δοκιμάστε την και με μιγαδικούς![/hide]
Cheerio,

Durandal 1707

[quote="silouan"]Είναι ειδική περίπτωση αυτής http://www.mathlinks.ro/viewtopic.php?search_id=1258401600&t=168828

Μεταφράζοντας κάποιος την πρώτη λύση μπορεί να βρει μια αμιγώς γεωμετρική λύση[/quote]

Κάτι δεν μου αρέσει στην πρώτη λύση. Στην περίπτωση που ο ν είναι περιττός, δεν καταλαβαίνω πως ζευγαρώνει τα υπόλοιπα σημεία. (Μπορώ βασισμένος στην ίδια ιδέα να την λύσω διαφορετικά, αλλά δεν είμαι σίγουρος αν γίνεται ή όχι το ζευγάρωμα.)

[quote="silouan"]Αυτό που κάνει είναι το εξής . Όταν είναι άρτιος ζευγαρώνει τα αντιδιαμετρικά και όταν είναι περιττός λέει ότι υπάρχει σημείο έστω $ A_j$ που είναι τέτοιο ώστε $ MA_j\geq R \equal{} 1$  . Και μετά ζευγαρώνει πάλι τα αντιδιαμετρικά  :)[/quote]

Μα όταν είναι περιττός δεν υπάρχουν αντιδιαμετρικά!

ooops  :blush:  Με συγχωρείς ... Πάλι όμως δεν μπορούμε να τα πάρουμε ανά δύο ώστε να κάνουμε τη δουλειά μας αλλά δεν είμαι σίγουρος και δεν έχω και χρόνο να το κοιτάξω . Πάντως η τελευταία λύση στο λινκ τουλάχιστον δίνει λύση  :)

Για γεωμετρική λύση (στην περίπτωση που ο ν είναι περιττός) αρκεί να δείξουμε ότι το Μ πρέπει να βρίσκεται πάνω στην ευθεία που ενώνει μια κορυφή με το μέσο της απέναντι πλευράς. (Άσκηση  :) )

Η αλγεβρική λύση υπάρχει στο link.

Sil δεν σε παρακολουθούσα παντα στο αγγλικό.....
Ειναι γεωμετρικη μεταφορα κλασικης ανισωσης μιγαδικων .
H λυση αλγεβρικα (ιδια) λιγο πιο αναλυτικα απ το Link....
Δηλαδη οι εικονες του ν-γωνου ειναι εικόνες λυσεων της εξισωσης $ z^n=a$ οπου $ a$ κατάλληλος μιγας και γνωριζουμε οτι $ \sum{z_i}=0$ οποτε $ \sum{|z-z_i|\geq{\sum{|z-(1/z^{-}_i)|}}....\geq{|n-z\sum{z_i}}|={n}}$


Welcome, x-and-y ! I see that you have made your first post. You can learn everything there is to know about the Art of Problem Solving website by exploring. Try the Getting Started in the Highschool forum.

[quote="x-and-y"]hi i am roien in germany.
i have been in germany for about 2 years.
i am just talking my evening high school.
The school is totally in german.
me really really bad in german langauge.
everything would be new for me, i don't know i can get through it.
anyways i love design i mean digital designing. something like website and photomanuplations.

i love martial arts(kung fu)

that is all i know about me. :lol:[/quote]

only u love artial arts?

which kind of them do u play?

thanks for you nutz_for2.718281828

[quote="Ashegh"]

only u love artial arts?

which kind of them do u play?[/quote]

well i love to do tiaku , it is a kind of martial art which was create by a local master.
anyways i love it.


[size=150][color=red]Το μετέτραψα σε εμφανές πεδίο , διότι περιέχει πολλές ωραίες ασκήσεις
  που οι νεότεροι δεν μπορούν πια να βρουν  ψάχνοντας στο ελληνικό φόρουμ!

 Καλή μελέτη !!![/color][/size]

[b][color=blue]ΧΡΟΝΙΑ ΠΟΛΛΑ  ΚΑΙ ΚΑΛΗ ΧΡΟΝΙΑ !!![/color][/b]

[b] [color=red]ΑΣΚΗΣΗ 21[/color] [/b]

  Δίναιται κύκλος με κέντρο $ O$  και ένας άλλος κύκλος που περνάει από το $ O$ και τον τέμνει στα σημεία $ A , B$. Μια ευθεία που περνάει από το $ O$ τέμνει τους δύο κύκλους με τη σειρα στα σημεία $ C, O , D , Z$ (To $ CD$  είναι δηλαδή διάμετρος). Να αποδειχθεί ότι το σημείο  $ D$  είναι έγκεντρο του τριγώνου $ ABZ$.

[u] Μπάμπης[/u]

[color=red]ΑΣΚΗΣΗ 21 -ΛΥΣΗ[/color]

Έστω $K$ το  σημείο τομής της $CZ$ με την $AB$ .
Αφού $BO=OA$ η $ZO$ είναι διχοτόμος της γωνίας $\widehat{AZB}$

Επιπλέον είναι γνωστό ότι τα σημεία $C,\ K,\ D,\ Z$ σχηματίζουν αρμονική τετράδα (αν δεν το ξέρετε πείτε μου να βάλλω απόδειξη) και ο κύκλος με διάμετρο $CD$  είναι ο Απολλώνιος κύκλος του τριγώνου $AKZ$ ως προς τη γωνία $\widehat{AKZ}$ .Συνεπώς οι ευθείες $AD,\ AC$ είναι η εσωτερική και η εξωτερική αντίστοιχα διχοτόμος της γωνίας $\widehat{AKZ}$
άρα η $AD$ διχοτόμος της $\widehat{BAZ}$
και άρα το $D$ είναι το έγκεντρο του τριγώνου $BAZ$

[i]Τέτοια λύση περίμενα περίπου  (με αρμονικά ), διότι αυτή είχα διαβάσει όταν πρωτοείδα την  άσκηση. Σήμερα , ωστόσο , που ήμουνα Αθήνα , και μη έχοντας τίποτα άλλο να κάνω το μεσημέρι , είπα  να την ξαναρίξω μια ματιά , γιατί θυμόμουνα την εκφώνηση. Μοιράζομαι μαζί σας τις σκέψεις μου , που όπως θα διαπιστώσετε ,είναι τελείως στοιχειώδεις: [/i]

  Φέρνουμε την $OA$ . Τότε :

   $\angle BAD = \angle OAD-\angle OAB = \angle ODA-\angle OZB = \angle ODA-\angle OZA = \angle DAZ$  ,

 διότι από το τρίγωνο $ADZ$ παίρνουμε

 $\angle ODA= \angle OZA+\angle DAZ$ , δηλαδή

  $\angle ODA-\angle OZA = \angle DAZ$ .

   Επομένως η $AD$ είναι διχοτόμος της γωνίας  $BAZ$  , και επειδή η $ZO$ είναι διχοτόμος της γωνίας  $AZB$  (διότι  $OA = OB$)  , προκύπτει ότι το  $D$   είναι έγκεντρο στο τρίγωνο  $ABZ$.

               [u]Μπάμπης[/u]

[color=blue] Την άσκηση την είχε προτείνει ο Virgil Nicula στο Ρουμάνικο forum. Η απόδειξη που δόθηκε από κάποιον μαθητή βασίστηκε στη σχέση $OD^{2}= OK \cdot OZ$  που αποδείχτηκε με το θεώρημα τεμνομένων χορδών. Αλλά επειδή το $O$ είναι μέσο του $CD$ ,  η παραπάνω σχέση εξασφαλίζει ότι τα σημεία  $C , K , D , Z$ αποτελούν αρμονικά τετράδα κλπ. [/color]

Είναι ένα από τα αγαπημένα μου :lol:
http://www.mathlinks.ro/Forum/viewtopic.php?t=108097

[b][color=red]ΑΣΚΗΣΗ 22 [/color][/b]

Δίνεται τραπέζιο  $ABCD , AB // CD$ , το σημείο τομής $O$ των διαγωνίων του  ,  η παράλληλη από το $O$ προς τις βάσεις , που τέμνει την $AD$  στο $E$ και τη $BC$ στο $F$ ,  το σημείο τομής $G$  των $DA , CB$ και η προβολή $I$  του $G$  στην $EF$ . Να αποδείξετε ότι η $IE$ είναι διχοτόμος της γωνίας  $AID$ και  η $IF$ είναι διχοτόμος της γωνίας  $BIC$.

 [u]ΜΠΑΜΠΗΣ [/u]

Επειδή δεν έχω χρόνο γιατί ετοιμάζομαι να φύγω για Θεσ/νικη και λίγο βιαστικά που την είδα..

Έστω $Z$ σημείο στην $DC$ τ.ω. $GD=GZ$ και $GI\cap DZ=M$.. Θα βγει το $\triangle DIZ$ ισοσκελές , $EI\parallel DZ$ και τελικά προκύπτει το ζητούμενο..

Stelio δεν εξασφάλισες όμως ότι τα $A,\ I,\ Z$ είναι συνευθειακά  :wink:  (είναι βέβαια αλλά δεν το αναφέρεις καν και από οτι βλέπω δεν είναι πολύ εύκολο να αποδειχθεί (ούτε όμως πολύ δύσκολο  :wink: )).

Σωστά Σιλουανέ.. παρασυρόμενος από το σχήμα δεν απέδειξα ότι τα $A$, $I$, $Z$ είναι συνευθειακά..

Λοιπόν, αν δεν έχω λάθος στο συλλογισμό μου:

Έστω $DI\cap GZ=N$. Στο $\triangle AIN$ , $\angle AIN=90^\circ$. Στο $\triangle DZI$ , $\angle DIZ=90^\circ$. Άρα οι γωνίες είναι κατά κορυφήν, άρα τα $A$, $I$, $Z$ προκύπτει εύκολα ότι είναι συνευθειακά.

Αλλιώς.. το $ANZD$ είναι ισοσκελές τραπέζιο και $I$ το σημείο τομής (γιατί?) των διαγωνίων του.

[b][size=117][color=red]ΑΣΚΗΣΗ  23.[/color][/size][/b]

 Με βάσεις τις πλευρές $AB,$ $AC,$ τριγώνου $\bigtriangleup ABC$ κατασκευάζουμε τα όμοια ορθογώνια τρίγωνα $\bigtriangleup ABD,$ $\bigtriangleup ACE$ $($ $\angle BAD = \angle CAE = 90^{o}$ και $\angle ABD = \angle ACE$ $)$ προς το εξωτερικό μέρος αυτού.

Δια του $A,$ φέρνουμε ευθεία παράλληλη προς την ευθεία που συνδέει τα μέσα των $BD,$ $CE$ και έστω $M,$ $M',$ τα σημεία στα οποία τέμνει αντιστοίχως, τις ευθείες $CD,$ $BE.$ Αποδείξτε ότι $AM = AM'.$

Κώστας Βήττας.

[b][size=117][color=red]ΑΣΚΗΣΗ  24.[/color][/size][/b]

Με βάσεις τις πλευρές $AB,$ $AC,$ τριγώνου $\bigtriangleup ABC,$ κατασκευάζουμε τα όμοια ισοσκελή τρίγωνα $\bigtriangleup B'AC,$ $\bigtriangleup C'AB,$ προς το εξωτερικό μέρος αυτού.

Από το κέντρο $K,$ του περιγεγραμμένου κύκλου του $\bigtriangleup ABC,$ φέρνουμε κάθετες ευθείες επί των $AC',$ $AB',$ όι οποίες τέμνουν αντιστοίχως τις ευθείες $BC'$ $CB',$ στα σημεία $D$ και $E.$ Αποδείξτε ότι $DE\parallel BC.$

Κώστας Βήττας.

[b][size=117][color=red]ΑΣΚΗΣΗ  25.[/color][/size][/b]

Δίδεται τετράπλευρο $ABCD$ και έστω $E,$ το σημείο τομής των διαγωνίων του $AC,$ $BD.$ Δια των $B,$ $C,$ φέρνουμε τισ παράλληλες ευθείες προς τις πλευρές $CD,$ $AB$ αντιστοίχως, οι οποίες τέμνονται στο σημείο έστω $Z.$ Αποδείξτε ότι $EZ\parallel MN,$ όπου $M,$ $N,$ είναι τα μέσα αντιστοίχως, των $BC,$ $AD.$

Κώστας Βήττας.

[quote="silouan"]Βρήκα μια λύση για την 23 σε δύο γραμμες   :wink:
[hide]Θεωρούμε μετασχηματισμό (σύνθεση στροφής ομοιθεσίας κέντρου  $A$  ,.....[/hide][/quote]

Σιλουανέ, πάντα με γοήτευαν οι γεωμετρικοί μετασχηματισμοί και ποτέ δεν κατάφερα να εξοικειωθώ αρκετά με αυτούς. Η λύση που έχω υπόψη μου, είναι ορθή όπως πιστεύω, αλλά αρκετά κοπιαστική και θα είναι πολύ ενδιαφέρον αν μας δώσεις περισσότερες λεπτομέρειες.

Βασισμένος στο πρόβλημα αυτό, μπορεί να αποδείξει κανείς, ένα άλλο πρόβλημα που έχω βάλει στο αγγλικό φόρουμ.

http://www.mathlinks.ro/viewtopic.php?t=115173

Κώστας Βήττας.

[b][color=red]ΑΣΚΗΣΗ 26( Για Αρχιμήδη)[/color][/b]

[i]Ξέρετε , ποτέ δε δίνω πολύ δύσκολες ασκήσεις. Τέτοιες ,  όποιος θέλει , ας διαβάσει από το IMO COMPENDIUM .Είναι το καλύτερο! Προσπαθώ να βρίσκω προβλήματα  διδακτικά που να λύνονται σχετικά εύκολα και να αρέσουν.[/i]

 Δίνεται  σημείο $M$  στην πλευρά $AB$ ενός τετραγώνου $ABCD$ με κέντρο $O$. Η ευθεία $CM$  τέμνει την ευθεία $DA$ στο σημείο  $S$. Η  $SO$ τέμνει την  $DM$ στο σημείο $E$. Να αποδείξετε ότι $AE \perp DM$.

[b]Μια άλλη διατύπωση[/b]

 Ένα μεταβλητό σημείο $M$ κινείται στην πλευρά $AB$ ενός τετραγώνου $ABCD$ με κέντρο $O$. Η ευθεία $CM$  τέμνει την ευθεία $DA$ στο σημείο  $S$. Η  $SO$ τέμνει την  $DM$ στο σημείο $E$. Να αποδείξετε ότι ο περιγεγραμμένος κύκλος του τριγώνου $AED$ διέρχεται και από τρίτο σταθερό σημείο.

[b][color=red] ΑΣΚΗΣΗ 27[/color][/b]
[i]
  Λοιπόν , χθες μου έστειλε ένας Ρουμάνος συνάδελφος αρκετά βιβλία από τη Ρουμανία , καθώς και μαθηματικά περιοδικά. Για να τα ξεφυλλίσω μόνο θέλω μια εβδομάδα ! Ωστόσο ,  στο τεύχος 8 / 2006 του GM βρήκα τυχαία την παρακάτω ωραία άσκηση. Δεν πρέπει να έχουμε ...πρόβλημα, διότι αν δε βρεθεί λύση , θα σας στείλλω την επίσημη  λύση. Είναι πραγματικά άσκηση επιπέδου Αρχιμήδη , αλλά και πολύ διδακτική(γι αυτούς που θα δουν αργότερα τη λύση). Χαρείτε την ! [/i]

Δίνεται ισοσκελές τρίγωνο $ABC, AB= AC$ εγγεγραμμένο στον κύκλο $(C)$.Ένας  άλλος κύκλος εφάπτεται στις προεκτάσεις των πλευρών $CA , CB$ και εξωτερικά του κύκλου $C$ στο σημείο $M$ ( To $M$ είναι το σημείο επαφής των δύο κύκλων) . Να αποδείξετε ότι $MC = 2MA$

 [u]Μπάμπης[/u]
 [i]( Μέχρι τον Αρχιμήδη ,  δε θα προτείνω άλλη άσκηση Γεωμετρίας, για να μη σας βγάζω από τους ρυθμούς σας ! )[/i]

Με αναλογα ποσά δε το κάνεις;Εμβαδά, ύψη κτλ....

Ναι, αν έχεις χρόνο βάλε και τη λύση σου, έτσι για να υπάρχει.

Ευχαριστούμε  :)

mia apli pu 8imi8ika tora...
$.............34.............$

or8ogonio trigono $ABC$   $A=90$ kai o egegramenos kiklos tu,efaptete stin ipotinusa sto simio $D$.
ndo..$(ABC)=DB*DC$

 :D

Κλασικό....Μου το βάλανε και στο σχολείο!!! :D  :D

Δε βάζω λύση γιατί δεν έχω συνηθίσει με το Latex και γράφω αργά....δεν είναι πολύ δύσκολη όμως...

[b][color=red]ΑΣΚΗΣΗ 34 [/color][/b]

Μου τη στείλανε από τη Χιλή.Δε φαίνεται δύσκολη. Ζητάνε λύση χωρίς τριγωνομετρία , αν γίνεται.

ali8ia ...ine 8ema diagonismu +zitane lisi xoris trigonometria??mu fenete periergo.. :huh:
pantos ipologizontas mon ena imitono mias gonias..vgeni eykola...
Hint...

[hide]$AB=AC=2\sqrt{6}$[/hide]

 :D

[b][size=109][color=red]ΑΣΚΗΣΗ  34 - ΛΥΣΗ[/color][/size][/b]

Η σχέση $(ABC) = 3\sqrt{15}$ $,(1)$  είναι καθοριστική για τη λύση χωρίς τριγωνομετρία.

Εάν $K,$ $K',$ είναι αντιστοίχως οι προβολές των $B,$ $C,$ επί της διαμέσου $AM,$ τότε έχουμε $BK = CK'$ και έτσι εύκολα προκύπτει ότι $(ABC) = (BK)\cdot (AM)$ $,(2)$

Από $(1),$ $(2)$ $\Longrightarrow$ $BK = \sqrt{15}$ $,(3)$ $($ γιατί $AM = 3$ $)$

Από ορθογώνιο τρίγωνο $\bigtriangleup BKM,$ $\Longrightarrow$ $(KG)^{2}= 16-15 = 1$ $($ $BG = 4$ $)$ $\Longrightarrow$ $KG = 1$ $,(4)$ όπου $G,$ είναι το βαρύκεντρο του $\bigtriangleup ABC.$

Από $(4)$ συμπεραίνουμε ότι το $K,$ είναι μέσο του $AG$ γιατί $AG = 2$. Επομένως το τρίγωνο $\bigtriangleup BAK$ είναι ισοσκελές και άρα $AB = BG = 4$ $,(5)$

Από ορθογώνιο τρίγωνο $\bigtriangleup BKM,$ και επειδή $KM = 2$  $\Longrightarrow$ $(BM)^{2}= 15+4$ $\Longrightarrow$ $BM = \sqrt{19}$ $\Longrightarrow$ $BC = 2\sqrt{19}$ $,(6)$

Με βάση το πρώτο θεώρημα των διαμέσων στο $\bigtriangleup ABC,$ μπορούμε τώρα να υπολογίσουμε την πλευρά $AC.$

$(AB)^{2}+(AC)^{2}= 2(AM)^{2}+\frac{(BC)^{2}}{2}$ $\Longrightarrow$ $AC = 2\sqrt{10}$ $,(7)$

Τέλος, πάλι με το ίδιο θεώρημα, έχουμε ότι $(AC)^{2}+(BC)^{2}= 2(CP)^{2}+\frac{(AB)^{2}}{2}$ από όπου προκύπτει το ζητούμενο $CP = 3\sqrt{6}.$

Ελπίζω ότι δεν έχω κάνει λάθος στις πράξεις. Θα το ξανακοιτάξω πάλι.

Κώστας Βήττας.

Ευχαριστώ !

 Αυτός που το ζήτησε δίνει εισαγωγικές  στο Πανεπιστήμιο. Για ποιο λόγο  δε θέλει τριγωνομετρία , δε μου το είπε , αν και τον ρώτησα.

  Θα τον παραπέμψω στις λύσεις σας για να τις δει.
 Μπάμπης

[quote="RAKAR"]Ontos...kati dn pai kala..ta numera pu vrika dn taftizonte me ayta tis lisis tu vittasko...to mono pu dn katalava sti lisi...meta  tin (4) pos simperenume oti to K ine meso tu AM?( lete epidi $AM=2$..ala i $AM=3$...)
Dn ksero mpori na min katalava ego kati...
i dikia mu lisi pantos ine i eksis..(dn tin egraya prin gt opos ipa ipologizo to imiotno+to sinimitono mia gonias..)
ine $AG=2$ $GM=1$ $BG=4$ $GN=2$...
Omos $(ANG)=\frac{(ABC)}{6}$ opote  8etontas $x=AGN$
$\frac{1}{2}*2*2*sin(x)=\frac{1}{6}*3*\sqrt{15}$
opote $sin(x)=\frac{\sqrt{15}}{4}$ kai ara $cos(x)=\frac{1}{4}$
meta apo n sinimitonon sto ANG perno $AN=\sqrt{6}$opote $AC=2*\sqrt{6}$
Meta perno nomo sinimitono sta $ABG$ k $BGM$ kai perno antistixa
$AB=2*\sqrt{6}$ kai $BC=2*\sqrt{15}$
opote perno meta ton tipo tis diamesu kai vrisko $(m_{c})^{2}=36$ diladi $m_{c}=6$
8a to ksanado pantos...

 :D[/quote]

 Πριν λίγο είχα ένα μάθημα και βρήκα λίγο χρόνο να το κοιτάξω και εγώ.Δεν είχα δει ότι έστειλε λύση και ο Rakar , οπότε είπα να την κοιτάξω και μόνος μου. Τριγωνομετρικά ,  με τον ίδιο τρόπο που  δούλεψε και ο Ράλλης , βρήκα τα ίδια νούμερα. Αλλά βλέπω ότι και εγώ θεώρησα ότι η γωνία $BGM$ είναι οξεία , ενώ βλέπω ότι στο σχήμα του Vittasko η γωνία αυτή είναι αμβλεία.Βάζοντας αμβλεία τη γωνία αυτή , τότε συμφωνώ με τον Κώστα και βρίσκω $BC =2 \sqrt 19$ και όχι $BC =2 \sqrt 15$ που αρχικά βρήκα.
 Κώστα , δες λίγο το σχήμα με τον υπολογιστή και εξέτασε μήπως πρέπει να πάρουμε δύο περιπτώσεις ή μήπως η μία αποκλείεται.
 Μόλις τελειώσω το μήνυμα , θα επαληθεύσω όλα τα αποτελέσματα , να δω αν τα τρίγωνα δίνουν το ίδιο εμβαδόν.
 [u]Μπάμπης [/u]

[quote="vittasko"][b][size=109][color=red]ΑΣΚΗΣΗ  34 - ΛΥΣΗ[/color][/size][/b]

Η σχέση $(ABC) = 3\sqrt{15}$ $,(1)$  είναι καθοριστική για τη λύση χωρίς τριγωνομετρία.

Εάν $K,$ $K',$ είναι αντιστοίχως οι προβολές των $B,$ $C,$ επί της διαμέσου $AM,$ τότε έχουμε $BK = CK'$ και έτσι εύκολα προκύπτει ότι $(ABC) = (BK)\cdot (AM)$ $,(2)$

Από $(1),$ $(2)$ $\Longrightarrow$ $BK = \sqrt{15}$ $,(3)$ $($ γιατί $AM = 3$ $)$

Από ορθογώνιο τρίγωνο $\bigtriangleup BKM,$ $\Longrightarrow$ $(KG)^{2}= 16-15 = 1$ $($ $BG = 4$ $)$ $\Longrightarrow$ $KG = 1$ $,(4)$ όπου $G,$ είναι το βαρύκεντρο του $\bigtriangleup ABC.$

Από $(4)$ συμπεραίνουμε ότι το $K,$ είναι μέσο του $AM$ γιατί $AM = 2$. Επομένως το τρίγωνο $\bigtriangleup BAK$ είναι ισοσκελές και άρα $AB = BG = 4$ $,(5)$

Από ορθογώνιο τρίγωνο $\bigtriangleup BKM,$ και επειδή $KM = 2$  $\Longrightarrow$ $(BM)^{2}= 15+4$ $\Longrightarrow$ $BM = \sqrt{19}$ $\Longrightarrow$ $BC = 2\sqrt{19}$ $,(6)$

Με βάση το πρώτο θεώρημα των διαμέσων στο $\bigtriangleup ABC,$ μπορούμε τώρα να υπολογίσουμε την πλευρά $AC.$

$(AB)^{2}+(AC)^{2}= 2(AM)^{2}+\frac{(BC)^{2}}{2}$ $\Longrightarrow$ $AC = 2\sqrt{10}$ $,(7)$

Τέλος, πάλι με το ίδιο θεώρημα, έχουμε ότι $(AC)^{2}+(BC)^{2}= 2(CP)^{2}+\frac{(AB)^{2}}{2}$ από όπου προκύπτει το ζητούμενο $CP = 3\sqrt{6}.$

Ελπίζω ότι δεν έχω κάνει λάθος στις πράξεις. Θα το ξανακοιτάξω πάλι.

Κώστας Βήττας.[/quote]

 [b]Είναι τα περίεργα που κρύβει η επιστήμη ! [/b]

 Μόλις πριν λίγο εξέτασα τα αποτελέσματα . Όλα είναι σωστά ! Απλά υπάρχουν δύο τέτοια τρίγωνα. Η πλάκα είναι ότι και τα δύο είναι αμβλυγώνια και παρόλα αυτά έχουν ίσα εμβαδά.
Επομένως προκύπτει ότι

[color=red]'' Αν δύο τρίγωνα έχουν δύο διαμέσους ίσες και ίσα εμβαδά , τότε δεν είναι κατάγκην ίσα '' ! [/color]

 Ή.... κάνω κανένα λάθος :maybe:  ;
 [u]Μπάμπης[/u]

[quote="RAKAR"]Ontos...kati dn pai kala..ta numera pu vrika dn taftizonte me ayta tis lisis tu vittasko...to mono pu dn katalava sti lisi...meta  tin (4) pos simperenume oti to K ine meso tu AM?( lete epidi $AM=2$..ala i $AM=3$...) :D[/quote]

Ευχαριστώ RAKAR, για την παρατήρηση. Ήδη το διόρθωσα. Είναι $AG = 2$ και το $K,$ μέσον του $AG,$ πως φαίνεται στο σχήμα.

Κώστας Βήττας.

[quote="stergiu"]...

 [b]Είναι τα περίεργα που κρύβει η επιστήμη ! [/b]

 Μόλις πριν λίγο εξέτασα τα αποτελέσματα . Όλα είναι σωστά ! Απλά υπάρχουν δύο τέτοια τρίγωνα. Η πλάκα είναι ότι και τα δύο είναι αμβλυγώνια και παρόλα αυτά έχουν ίσα εμβαδά.
Επομένως προκύπτει ότι

[color=red]'' Αν δύο τρίγωνα έχουν δύο διαμέσους ίσες και ίσα εμβαδά , τότε δεν είναι κατάγκην ίσα '' ! [/color]

 Ή.... κάνω κανένα λάθος :maybe:  ;
 [u]Μπάμπης[/u][/quote]

Αν θεωρήσουμε ένα ισοσκελές τρίγωνο $\bigtriangleup ABC$ $($ $AB = AC$ $),$ με διάμεσο $AM = 3$ και εμβαδόν $(ABC) = 3\sqrt{15},$ τότε παίρνουμε $BC = 2\sqrt{15}$ και άρα $BM = CM = \sqrt{15}.$

Από το ορθογώνιο τρίγωνο $BMG,$ προκύπτει $(BG)^{2}= 15+1$ $\Longrightarrow$ $BG = 4,$ που επαληθεύει το $BN = 6$ $($ και $CP = 6$ $)$ και άρα όντως υπάρχουν δύο λύσεις του προβλήματος.

Κώστας Βήττας.

[quote]opote $sin(x)=\frac{\sqrt{15}}{4}$ kai ara $cos(x)=\frac{1}{4}$ [/quote]
edo kano to la8os +xano tin deyteri lisi :wallbash:  gt 8eoro au8ereta oti i gonia ine oksia...
pantos pernontas+deyteri periptosi oti i gonia ine amvlia,me ton idio akrivos tropo,(3 fores n sinimitonon), prokiptun i lisis tu vitasko..+ontos ine +i dio dektes+dn aporiptete kamia,gt plirun+i dio tis proipo8esis tis ekfonisis..
[quote]Μου τη στείλανε από τη Χιλή.Δε φαίνεται δύσκολη.
[/quote]
na pu telika omos ekrive pagida.. :wink:

 :D

[b][color=red]ΑΣΚΗΣΗ 35[/color][/b]

[b]Ας συνεχίσουμε με συνθετική γεωμετρία
 Α! Καλή επιτυχία στον αυριανό Αρχιμήδη ![/b]

 Έστω $AB, AC$ οι εφαπτόμενες προς έναν κύκλο και $D$ ένα σημείο στην προέκταση της $AB$ προς το  to $B$. Ο περιγεγραμμένος κύκλος του τριγώνου $ADC$  τέμνει τον αρχικό κύκλο στο σημείο $P$.Ας είναι $Q$ η προβολή του $B$ στην ευθεία $CD$ .Να αποδειχθεί ότι $\angle DPQ = 2 \angle ADC$.


Find the angles $ \alpha , \beta , \gamma$ of a triangle if they satisfy the relation :

$ \sin(\frac{\alpha \minus{}\beta}{2})  \plus{}  \sin(\frac{\alpha \minus{}\gamma}{2})  \plus{} \sin(\frac{3\alpha}{2})  \equal{} \frac{3}{2}$

[quote="guru_bhai"]Find the angles $ A$ , $ B$ , $ C$  of a triangle $ ABC$ if they satisfy the relation $ E\equiv \sin \frac {A \minus{} B}{2} \plus{} \sin\frac {A \minus{} C}{2} \plus{} \sin\frac {3A}{2} \equal{} \frac {3}{2}$ .[/quote]

[b][u]Proof[/u].[/b] Denote $ x \equal{} 90 \minus{} \frac {A \minus{} B}{2}$ , $ y \equal{} 90 \minus{} \frac {A \minus{} C}{2}$ , $ z \equal{} \frac {3A}{2} \minus{} 90$ . Observe that $ x \plus{} y \plus{} z \equal{} 180^{\circ}$ .

If $ A\ge 60$ , then  $ \{x,y,z\}\subset (0,180)$ and $ \cos x \plus{} \cos y \plus{} \cos z\le \frac 32$ . Hence $ x \equal{} y \equal{} z \equal{} 60\implies$

$ A \equal{} 100$ , $ B \equal{} C \equal{} 40$ . Prove easily that if $ A < 60$ , then $ E < \frac 32$ .


we have: (i will use f instead of f(x))
2f-2ff=x-f => (x-f)=2(f-ff) => 2| (x-f)
now replacing x with f in 2f-2ff=x-f we get that
2ff-2fff=f-ff =>4ff-4fff=2f-2ff=x-f => 4|(x-f)
Continuing you get that 2^n|(x-f) for any n =>x-f=0 =>f=x
so f(x)=x

That looks good.

Could you do it if, instead of the integer condition, you knew f was continuous?

f(x)=x is a solution, yes, but how would you show that it is the only solution (if it is) in this alternate case?

if you take out the integer condition and just say f is continuous, already there is one more solution at least.
that is f(x)=x/2, as f(f(x))=x/4 and
3f(x)-2f(f(x))=3x/2-x/2=x.
i'll think if there are any other solutions. :)


IMO Shortlist
combinatorics
dissection
Tiling

For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with non-overlapping trominos?  When it is possible, what is the minimum number of trominos needed?

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

Let $n=2t+1$. Then there are $(t+1)^{2}$ black squares in a row that is an even number of rows away from the bottom row (i.e. in the first,third,fifth... row counting from the bottom). But each of these squares must be covered by a different tromino- so at least $(t+1)^{2}= \frac{(n+1)^{2}}{4}$ trominos are required. If $n=3,5$, then this means that the task is impossible since there is not enough room on the board for this many trominos.

For $n=7$ an arrangement can be found (split the board into 3 3x4 rectangles, tile them in the obvious fashion, and then fill in all but a remaining white square). It uses 16 tiles. (Notice that $16=\frac{(7+1)^{2}}{4}$).

Now suppose that there is a tiling for $n=k$ ($k$ odd) that uses $\frac{(k+1)^{2}}{4}$ tiles. Then add to this a $2$x$(k+2)$ and a $2$x$k$ rectangle. Each of these can be covered (its black squares covered) by cutting a $2$x$3$ rectangle off one end, putting 2 trominos on it and then a line of trominos covering the 2 black squares in each $2$x$2$ square. Between these two rectangles, we have used an extra $k+2$ trominos. So a $(k+2)$x$(k+2)$ rectangle can be tiled (have its black squares covered) using $\frac{(k+1)^{2}}{4}+k+2=\frac{(k+3)^{2}}{4}$ tiles.

So by induction, for all odd $n\geq 7$, we can tile the black squares of an $n$x$n$ board using $\frac{(n+1)^{2}}{4}$ trominos. We have also shown this to be the minimum.

Let $n=2k+1$. Color the black squares red and blue so that blue squares are in rows with $k+1$ black squares and red squares are in rows with $k$ black squares.

A tromino can't cover multiple blue squares. There are $(k+1)^2$ blue squares. So, there are at least $(k+1)^2$ trominos. For $k=1, 2$, $3(k+1)^2>(2k+1)^2$, so the space the trominoes need to take up is bigger than the board, so it's impossible.

We will prove by induction that for $k\geq 3$ it's possible to cover the black squares with $(k+1)^2$ blue squares. We can check that this is possible for $k=3$.

For the inductive step, we basically need to show that we can cover the outer $2$-wide $L$ shape with $2k+1$ trominoes, since the rest of the board is a square with side length $2k-1$ which can be tiled with $k^2$ trominoes.

We can also prove this claim by induction. By putting a tromino at each tip of the $L$ shape, we reduce it to an $L$ shape with side length $2k-1$, which can be covered with $2k-1$ trominoes. We can use a base case of $n=5$ where we can cover the legs with $2\times 3$ grids and cover the corner with $1$ tromino.

Call a black square [i]even[/i] if there are an even number of squares from this black square to any side of the chessboard. For example, all black squares on the edge are black squares. Note that no two even squares may be covered by the same tromino and that there are exactly $\left(\frac{n+1}{2}\right)^2$ even squares. Thus, we need at least $\left(\frac{n+1}{2}\right)^2$ trominos.

Note that this also means that $n=3$ and $n=5$ do not work because we need at least $4$ and $9$ trominos, covering $12$ and $27$ squares respectively. We claim that al black squares can be covered in $\left(\frac{n+1}{2}\right)^2$ trominos for all $n\ge 7.$ We will proceed with induction.

[b]Base Case:[/b] When $n=7$ we can cover like this:
[center][img width=40]https://media.discordapp.net/attachments/944079414742892624/983571307868323890/Screen_Shot_2022-06-06_at_8.20.46_PM.png[/img][left]
[b]Induction Hypothesis:[/b] Suppose for $n=2k-1$ we can make a tiling.
[b]Induction Step:[/b] Then adding on something like this will do the job.
[center][img width=40]https://media.discordapp.net/attachments/944079414742892624/983572553798926376/Screen_Shot_2022-06-06_at_8.26.05_PM.png?width=1186&height=1170[/img][left]
Thus, we can always tile an $n\times n$ chessboard with $\left(\frac{n+1}{2}\right)^2$ trominos when $n\ge 7$ and furthermore, it will be optimal.

[hide = Solution] Write $n = 2k + 1$. I claim that you can tile all odd $n \geq 7$ and the minimum number of trominoes required is $(k + 1)^2$.

Construct two $k \times (k + 1)$ rectangles and two $(k + 1) \times k$ rectangles. Tile so that it leaves the center uncovered. We now have that if $k$ is odd, then there are $\left (\frac{k + 1}{2} \right)^2$ tetrominoes. Multiplying by $4$ gives us $(k + 1)^2$. For even $k$ it takes $\left (\frac{k}{2} \right)^2$ tetrominoes. Multiplying by $4$ gives us $(k + 1)^2$. However this proof would not get us many points on the real exam. We should use induction.

Assume for a $(2k - 1) \times (2k - 1)$ board it takes $k^2$ tetrominoes. Then for a $(2k + 1) \times (2k + 1)$ board we have an extra $8k$ squares which is made up from a $2 \times (2k + 1)$ rectangle and a $(2k - 1) \times 2$ rectangle. The least number of tetrominoes required to tile a $2 \times (2k + 1)$ rectangle is $k + 1$ and the minimum number of tetrominoes for a $(2k - 1) \times 2$ rectangle is $k$. With the induction hypothesis, the new minimum becomes $k^2 + (k + 1) + k = (k + 1)^2$, as desired. $\blacksquare$ [/hide]

This took me way too long.

Consider the $\left(\frac{n+1}{2}\right)^2$ black squares that are an even distance from all sides of the chessboard. Each tile can contain at most one of these but each square must be tiled, so we must use at least $\left(\frac{n+1}{2}\right)^2$ tiles. Then clearly this implies that $n=1,3,5$ fail by bounding. For $n=7$ consider the tiling attached below. For $n>7$ take a construction for $n-2$ and add on two rows and columns; note that we can tile these in just $n$ tiles by placing two $2\times 3$ rectangles in the corner (which takes up $4$ tiles) and adding on one more tile for every $2\times 2$ square (of which there are $n-2$). Then by induction we can reach $\left(\frac{n+1}{2}\right)^2$ for all $n\ge 7$. We are done. $\blacksquare$

[b]S[/b]olved with [b]NTGuy, everythingpi3141592[/b]

[hide = solution]
Such tilings are possible only for $n \ge 7$, and at least $\frac{(n+1)^2}{4}$ tiles are necessary and sufficient for these $n$. Remark that $n = 3$ and $5$ do not work due to size constraints. The corners are fixed and the middle elements of the $5 \times 5$ do not allow for the centre to be covered. Consider $n \geq 7$. A construction for $n = 7$ is provided below:
[img]https://i.imgur.com/LhTM8W4.png[/img]
Next imagine extending from a $n \times n$ board to a $n+2 \times n+2$ board by filling in the two $2 \times n$ boards with $\frac{n+1}{2}$ trominoes and the $2 \times 2$ board with a single one. This uses $n+2$ trominoes and inductively we use $\frac{(n+1)^2}{4} + (n+2) = \frac{(n+3)^2}{4}$ trominoes to filll in the $n+2 \times n+2$ board. Therefore, it suffices to show that these many trominoes are necessary. To prove this, consider the $\frac{(n+1)^2}{4}$ cells $(i, j)$ with both $i, j$ odd. Each trominoe covers at most one such cell, moreover each such cell is black; implying that we require at least $\frac{(n+1)^2}{4}$ trominoes, as required.
[/hide]


Consider a cyclic polygon with equal angles. Prove that if the number of the sides of polygon is odd than the polygon is regular.

There must be something wrong with this problem statement. First, the given is that it is cyclic, thus it is cyclic. Second, it is easily seen that $A_1A_2$ can be replaced by any $A_1'A_2'$ such that $\overline{A_1A_2} \parallel \overline{A_1'A_2'}$ where $A_1' \in \overline{A_nA_1}$ and $A_2' \in \overline{A_2A_3}$. But not all $A_1'A_2'A_3A_4\cdots A_n$ are cyclic.

Sorry for that idiot mistake, it is fixed now.And I agree with your second argument, I'll try to find the right problem.

Let $A_{0}A_{1}\ldots A_{2n}$ be the polygon. Then the triangles $A_{0} A_{1}A_{2}$ and $A_{1}A_{2}A_{3}$ are congruent. Indeed, $A_{1}A_{2}$ is a
common side; the angles $\angle A_{0}A_{1}A_{2}$ and $\angle A_{1}A_{2}A_{3}$ are congruent since the polygon is equiangular; and the angles $\angle A_{1}A_{0}A_{2}$ and $\angle A_{1}A_{3}A_{2}$ are also congruent (their measure is $\frac{1}{2}\overset{\frown}{A_{1}A_{2}})$. It follows that $A_{0}A_{1}=A_{2}A_{3}.$

In the same way we obtain
\[
A_{2}A_{3}=A_{4}A_{5}=\ldots=A_{2n-2}A_{2n-1}=A_{2n}A_{0}=A_{1}A_{2}%
=\ldots=A_{2n-1}A_{2n},
\]
and hence the polygon is regular.
P.S. Armo, check "Mathematical Olympiad Treasures", Equiangular polygons.


Two circumferences of radii $ r$ and $ R$ $ (R>r)$ are tangent internally. Find the radius of the circumference that is tangent to these two circumferences and the line that connects the two centers.

[b][u]Proof[/u].[/b] Let $ w_a(A,a)$ , $ w_b(B,b)$ , $ w_c(C,x)$ be three circles so that the circles $ w_b$ , $ w_c$ are internal tangent to the circle $ w_a$

and the circle $ w_c$ is external tangent to $ w_b$  and is tangent to $ AB$ . Denote $ D\in AB$ , $ CD\perp AB$ .  Observe that $ a\ >\ b$ ,

$ AB \equal{} a \minus{} b$ , $ BC \equal{} b \plus{} x$ , $ CA \equal{} a \minus{} x$ and $ CD \equal{} x$ . Using the [u][b]Heron[/b]'s relation[/u] obtain $ [ABC] \equal{} \sqrt {abx(a \minus{} b \minus{} x)}$ .

Therefore, $ 2\cdot [ABC] \equal{} AB\cdot CD$ $ \Longleftrightarrow$ $ 4abx(a \minus{} b \minus{} x) \equal{} (a \minus{} b)^2x^2$ $ \Longleftrightarrow$ $ \boxed {x \equal{} \frac {4ab(a \minus{} b)}{(a \plus{} b)^2}}$ .


Tamyra is making four cookies and has exactly four chocolate chips.  If she distributes the chips randomly into the four cookies, what is the probability that there are no more than two chips in any one cookie? Express your answer as a common fraction.

[hide]Prob=good/all
(order does't matter, because they aren't different cookies. They're just cookies)
ALL:
1,1,1,1
1,1,2,0
1,3,0,0
4,0,0,0
2,2,0,0
GOOD:
1,1,1,1
1,1,2,0
2,2,0,0
>>3/5<<[/hide]
Unless I'm missing something crucial...

Not 3/5.  This a really annoying handbook problem that I'm trying to find a shortcut to.

Treat the chocolate chips as well as cookies distinguishable.   Try again with this assumption.

Naga

[hide]That list looks good but are you forgetting to find how many permutations there are for each combination? If you do it that way it should be 19/35.[/hide]

As we saw in this problem, a common error is to count "good" over "all" without considering whether the possibilities you have counted are equally likely.

In this case, there are a lot more ways to distribute chips 3,1,0,0 than 1,1,1,1 (since with 1,1,1,1 you have to put one chip on each cookie -- but with 3,1,0,0 the 3 chips could be on any of the 4 cookies, and then the 1 remaining could be on any of the 3 remaining cookies).

[hide="Solution"]
There are 4^4=256 ways of distributing the chocolate chips (each of the 4 chips can go on any of the 4 cookies).  This count treats all chips and cookies as distinguishable so we must remember to do that in the rest of the problem, too.

Now we must find how many ways there are to have no more than two chips on any cookie.

In this problem, for me it's easier to find the complementary count (the number of "bad" ways, i.e., how many ways are there to have > 2 chips on at least one cookie).  There are two cases.

all 4 chips on one cookie -- 4 ways (you can choose any of the 4 cookies but then you have to put all 4 chips on it)

3 chips on one cookie, one on another -- there are 4 ways to choose which cookie gets only one chip, 4 ways to choose which chip that cookie gets, and then 3 ways to choose which cookie gets 3 chips.  4*4*3=48

In all 4*4*3 + 4 = 52 ways of having > 2 chips on at least one cookie.

Therefore 256-52 = 204 ways of having no more than 2 chips on any cookie.

Answer: 204/256 = 51/64.

(If the part about distinguishing chips and cookies is confusing, it may be clearer to think of the problem this way: I roll 4 four-sided dice.  What is the probability that no number occurs more than twice?  Here, the dice are the chips, and the number I roll corresponds to which cookie I put it on.)
[/hide]

Yakko

Good solution yakko. That is correct. :)


Forty-three percent of Americans have Type $ A$ molecules in their
blood, $ 15\%$ have type $ B$ molecules, and $ 46\%$ have neither
Type $ A$ nor Type $ B$ molecules. What percent of Americans have
both Type $ A$ and Type $ B$ molecules in their blood?

Since $ 46\%$ of Americans have neither type $ A$ or type $ B$ molecules. That leaves only $ 44\%$ with $ A$, $ B$, or both.

Then, since $ 43\%$ have $ A$ you do: $ 44\minus{}43\equal{}1\%$ which leaves only $ 1\%$ that have only $ B$ molecules.

Then, since $ 15\%$ of people have $ B$ molecules you do: $ 15\minus{}1\equal{}14\%$ that have both $ A$ and $ B$ molecules.

I'm fairly sure that's incorrect.

We can do $43+15+46$ to get a result of $104$ That means $4%
$ of the people have both.

You went wrong with $100%-46
$=44%$, whereas it really is 56%.

Let $x$ be the percent of people who have both type $A$ and $B$ molecules. Then, the percent of people who only have type $A$ molecules is $43 - x$, and the percent of people who only have type $B$ molecules is $15 - x$. Adding all four percentages gives us

\[(43 - x) + (15 - x) + x + 46 = 100\]\[x = \boxed{4}\]
Crumbzz's solution is correct.


For me, red.

Even though my favorite is just about the opposite of that.

EDIT: [sarcasm]I loved how the colors assigned in the poll matched the colors written SO well.[/sarcasm]

[quote="maybach"]I chose blue.  How did you get so many options on the poll?  When I tried to get 12 options on the "Where do you live?"  It wouldn't let me.  Why is that?[/quote]
When you edit your post, the poll disappears, so when you edit the post again, to add new poll results, the old results come up again. It's a strange glitch.

I like green, it's been my favorite color for a while :D

I chose pink.
It's been my fave color for 2 years. =]

I chose teal because when I think "color", I think "Game Boy Color", and my first GBC was teal.

As you can sort of tell from my avatar, I love anything between green and blue, and yet when I think color I think of red. Very weird.


If $ a\plus{}1\equal{}b\plus{}2\equal{}c\plus{}3\equal{}d\plus{}4\equal{}a\plus{}b\plus{}c\plus{}d\plus{}5$, then $ a\plus{}b\plus{}c\plus{}d$ is

$ \text{(A)}\ \minus{}5 \qquad
\text{(B)}\ \minus{}10/3 \qquad
\text{(C)}\ \minus{}7/3 \qquad
\text{(D)}\ 5/3 \qquad
\text{(E)}\ 5$

[quote="redcomet46"]If $ a \plus{} 1 \equal{} b \plus{} 2 \equal{} c \plus{} 3 \equal{} d \plus{} 4 \equal{} a \plus{} b \plus{} c \plus{} d \plus{} 5$, then $ a \plus{} b \plus{} c \plus{} d$ is

$ \text{(A)}\ \minus{} 5 \qquad \text{(B)}\ \minus{} 10/3 \qquad \text{(C)}\ \minus{} 7/3 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 5$[/quote]
[hide]Add $ a\plus{}1, b\plus{}2, c\plus{}3, d\plus{}4$, which is equal to $ 4 (a\plus{}b\plus{}c\plus{}d\plus{}5)$. Simplifying gives the answer as $ \minus{}10/3$.[/hide]

Or you can expand this in \[ a\plus{}b\plus{}c\plus{}d\equal{}d\minus{}1\] \[ a\plus{}b\plus{}c\plus{}d\equal{}c\minus{}2\] \[ a\plus{}b\plus{}c\plus{}d\equal{}b\minus{}3\] \[ a\plus{}b\plus{}c\plus{}d\equal{}a\minus{}4\] and you obtain B


calculus
integration
trigonometry
calculus computations

This should be in the calculus computations forum.

[hide="Beginning"]The curve C is parametrized by $(x, y) = \frac{a}{2}(1+\cos t, \sin t)$.  Arclength is simply given by $\frac{at}2$.  Then it's just a slog from there ... we have to integrate $\sqrt{1+\cos t}$ and keep track of the constants.[/hide]


Square ABCD has midpoints E, F, G, and H. AB = 15 centimeters. Find the area of the shaded interior square in square centimeters.

[asy]pair A,B,C,D,E,F,G,H,W,X,Y,Z;
path I,J,K,L;
A = (0,2);  B = (2,2);  C = (2,0);  D = (0,0);
E = (2,1);  F = (1,0);  G = (0,1);  H = (1,2);
draw(A--B--C--D--A);
draw(A--E--C--G);
draw(H--D--F--B);
I = A--E;  J = G--C;
K = H--D; L = B--F;
W = IntersectionPoint(I,K);
X = IntersectionPoint(I,L);
Y = IntersectionPoint(J,L);
Z = IntersectionPoint(J,K);
fill(W--X--Y--Z--cycle);
label("A", A, NW);
label("B", B, NE);
label("C", C, SE);
label("D", D, SW);
label("E", E, E);
label("F", F, S);
label("G", G, dir(180));
label("H", H, N);[/asy]

i think the answer is 45, but i'm not sure how to get it

i just estimated that one of the triangles and one of the trapezoids add up to equal the area of the square

if we do that, there would be 5 squares

divide the area of the big square into 5 parts and you get 45

for a better solution, i think you would have to use similar triangles or something

Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=238174]this.[/url]

The problems are almost identical. Since this square has side length $ \frac34$ of the one in the link, our final answer is $ \left(\frac34\right)^2\cdot80\equal{}\boxed{45}$.


inequalities
geometry
geometric transformation
reflection
inequalities proposed

I proposed the following inequality in Mathematics Reflection along time ago,

If $ a,b,c$ are positive real numbers with sum $ 1$, then

\[ \frac{a}{\sqrt{a+2b}}+\frac{b}{\sqrt{b+2c}}+\frac{c}{\sqrt{c+2a}}\le \sqrt{\frac{3}{2}}.\]

Actually the solution to this one is not very hard. It is discussed in some Vietnamese Forum and also mathlinks, but no one shows a solution.

Maybe I'm wrong, but I can't figure out where. :blush:

we have (all sums are cyclic)

$ LHS=\frac{\sum a\sqrt{b+2c}\sqrt{c+2a}}{\sqrt{(a+2b)(b+2c)(c+2a)}}$

but $ \sum a\sqrt{b+2c}\sqrt{c+2a}\le \sqrt{\left(\sum a^{2}(b+2c)\right)\left(\sum c+2a\right)}$

$ =\sqrt{3\sum a^{2}(b+2c)}$

So we need to prove that

$ \sqrt{3\left(\sum a^{2}(b+2c)\right)}\le \sqrt{\frac{3}{2}(a+2b)(b+2c)(c+2a)}$

$ \iff 2\left(\sum a^{2}(b+2c)\right)\le (a+2b)(b+2c)(c+2a)$

$ \iff 2\left(\sum a^{2}b+2\sum ab^{2}\right)\le 9abc+2\sum a^{2}b+4\sum ab^{2}$

$ \iff 9abc\ge 0$.

$ LHS=\frac{\sum_{cyc}a\sqrt{b+2c}\cdot\sqrt{c+2a}}{\sqrt{(a+2b)(b+2c)(c+2a)}}$
but by Cauchy Schwarz inequality,
$ \sum_{cyc}\left(a\sqrt{b+2c}\right)\left(\sqrt{c+2a}\right)\le \sqrt{\left(\sum_{cyc}a^{2}(b+2c)\right)\left(\sum_{cyc}c+2a\right)}=\sqrt{3\sum_{cyc}a^{2}(b+2c)}$

So we need to prove that

$ \sqrt{\frac{3\sum_{cyc}a^{2}(b+2c)}{(a+2b)(b+2c)(c+2a)}}\le \sqrt{\frac{3}{2}}$

$ \iff 2\sum_{cyc}a^{2}(b+2c)\le (a+2b)(b+2c)(c+2a)$

$ \iff 2\left(\sum_{cyc}a^{2}b+2\sum_{cyc}ab^{2}\right)\le 9abc+2\sum_{cyc}a^{2}b+4\sum_{cyc}ab^{2}$

$ \iff 9abc\ge 0$.

Can you please specify exactly which part you don't understand? Then it would be easier for me to explain. :)

I think I just proved that the following inequality holds for $ k,a,b,c>0$ and $ a+b+c=1$
\[ \frac{a}{\sqrt{a+kb}}+\frac{b}{\sqrt{b+kc}}+\frac{c}{\sqrt{c+ka}}\le \sqrt{\frac{3}{k+1}}\]
Is it well known? I don't know.
hungkhtn, when does equality hold in your problem?

[quote="nayel"]I think I just proved that the following inequality holds for $ k,a,b,c>0$ and $ a+b+c=1$
\[ \frac{a}{\sqrt{a+kb}}+\frac{b}{\sqrt{b+kc}}+\frac{c}{\sqrt{c+ka}}\le \sqrt{\frac{3}{k+1}}\]
[/quote]
Actually, the following inequality holds for $ k,a,b,c>0$ :

\[ \frac{a}{\sqrt{a+kb}}+\frac{b}{\sqrt{b+kc}}+\frac{c}{\sqrt{c+ka}}<\sqrt{\frac{(k+1)(a+b+c)}{k}}\]

[quote="hungkhtn"]Oh... Nayel... You are really intelligent!!! Very very very nice solution. Thank you![/quote]

Thank you hungkhtn, but I am very stupid, I know that. So I am trying to learn as much as I can. I hope everyone will help me. :lol:

[quote="color"][quote="nayel"]

$ \iff 2\sum_{cyc}a^{2}(b+2c)\le (a+2b)(b+2c)(c+2a)$

$ \iff 2\left(\sum_{cyc}a^{2}b+2\sum_{cyc}ab^{2}\right)\le 9abc+2\sum_{cyc}a^{2}b+4\sum_{cyc}ab^{2}$
[/quote]

this part, please[/quote]

$ 2\left(a^{2}(b+2c)+b^{2}(c+2a)+c^{2}(a+2b)\right)\le (a+2b)(b+2c)(c+2a)$

$ \iff 2\left(a^{2}b+b^{2}c+c^{2}a+2ab^{2}+2bc^{2}+2ca^{2}\right)\le 9abc+2(a^{2}b+b^{2}c+c^{2}a)+4(ab^{2}+bc^{2}+ca^{2})$

$ \iff 9abc\ge 0$.

[quote="Vasc"]Actually, the following inequality holds for $ k,a,b,c>0$ :
\[ \frac{a}{\sqrt{a+kb}}+\frac{b}{\sqrt{b+kc}}+\frac{c}{\sqrt{c+ka}}<\sqrt{\frac{(k+1)(a+b+c)}{k}}\]
[/quote]

$ \sum\frac{a}{\sqrt{a+kb}}\le \sqrt{\frac{(k+1)(\sum a)\left(\sum a^{2}b+k\sum ab^{2}\right)}{(a+kb)(b+kc)(c+ka)}}$

Now $ \sqrt{\frac{(k+1)(\sum a)\left(\sum a^{2}b+k\sum ab^{2}\right)}{(a+kb)(b+kc)(c+ka)}}<\sqrt{\frac{(k+1)(a+b+c)}{k}}$

$ \iff k\left(\sum a^{2}b+k\sum ab^{2}\right)<(k^{3}+1)abc+k\sum a^{2}b+k^{2}\sum ab^{2}$

$ \iff (k^{3}+1)abc>0$.


It appears that the teenagers (young adults) (meaning high school and college students) always go to parties, and bars where they hang out in large groups all the time. When I look at the facebook website, I see some people have like 100 friends. If you do not like all these huge group gatherings and would like to go somewhere with only a couple of friends, and if you generally don't have that many friends, is that fine? I am not sure if my question makes any sense.

Obviously it's fine :) I mean, if you like smaller groups of friends, then so be it. As long as you and your friends enjoy it :lol:

I think its fine as long as you are happy with it. :)

I can only tell you this: I doubt that there are too many people with 100 [b]real[/b] friends ;)

Define friend  :P  If you consider everyone you've ever talked to a friend perhaps it's easy to get a hundred friends

About bars, I'm a devoted nonalcoholic and nondrug taker, so yeah, some people, especially like five years back when I was in high school , didn't respect that, sometimes people didn't really understand the meaning of no :|  :mad:
(For those in united states, I just went to united states and found myself under legal drinking age, but in Belgium, parties where fifteen year olds drink alhohol and smoke, either at home or at "cafés" are quite common, "pauvre jeunes...") :(

[quote="fredbel6"]About bars, I'm a devoted nonalcoholic and nondrug taker, so yeah, some people, especially like five years back when I was in high school , didn't respect that, sometimes people didn't really understand the meaning of no :|  :mad: [/quote]Strange, I never experienced that... I rarely drink, just because I don't like it. No one ever questioned that. They can find it strange you drink nothing at all, but as long as you just drink water they usually have no problem. Well, that was my group at least.

It depends , when you get older, the rest matures with you and that betters

Well, strange, if I would introduce the dosed injection of nitroglycerine, you would find it strange, some disgusting things just seem to accepted  :huh:

Keep in mind that Facebook friends can mean anything from best-friends-for-life to casual acquaintances.  It all depends on the person--a lot of people will friend-list just about anyone.

Yea, there are some people at our school that try to look cool and add everyone they see as a friend.  Like I didn't even know wh some of these people were untill they added me.  So they have like a 100 friends.

Yeah, I don't think hello was thinking about "blog friends" :)

Yeah, I mean normal friends. But the thing is, how many friends are normal to have? I mean is it fine if you just have 1 or 2 friends or is it too low? I dont seem to be very good at making friendships even though I really dont like to admit this in public. Also, I have no clue how to get gf's even though this was a really long topic before. Sorry if I am reinvigoring. I will try to look up if anyone in this site is from Philly area, maybe we can meet? Let me look..

hey.. theres no 'normal' and 'abnormal' amount of friends. u shouldnt judge how many friends u have, but the quality of friends. People can be the most popular people ever, with 100 people they are 'friendly' with - yet they arent happy. Find a few good friends, true friends, and ur set for life  :lol: .

As for gfs - we had a thread on this before? not since i came.. but anyway, theres probably someone there for you somewhere. Dont change who you are to try to appeal to someone else, cuz then it wont work out in the long run...

i have like four or five really good friends, then a whole bunch of just friends...

"About bars, I'm a devoted nonalcoholic and nondrug taker, so yeah, some people, especially like five years back when I was in high school , didn't respect that, sometimes people didn't really understand the meaning of no Neutral Mad
(For those in united states, I just went to united states and found myself under legal drinking age, but in Belgium, parties where fifteen year olds drink alhohol and smoke, either at home or at "cafés" are quite common, "pauvre jeunes...") Sad"

interesting...i'm in high school, but like a third of the school smokes or drinks or does something illegal and we only have freshmen and sophomores right now..but no one really seems to care. i always think, "what kind of idiot screws up their life at fifteen by doing drugs?!"

# of friends doesn't really matter, just how good they are...i used to be a loner :(

First thing, don't pay attention to how many friends facebook says you have.  Facebook is just like myspace and since I use myspace I know what it's like.  The majority of 'friends' on these websites consist of casual acquaintances that a person has just in case they ever need to say one little comment or something to them.

Second, don't try to change yourself to be more like someone else just so you can be their friend.  Be yourself.  If you're really rude and not at all friendly...then you'll want to try and change certain aspects of yourself (hypothetically speaking here).

Don't let your "friends" peer pressure you too much.  If it's something like going on a roller coaster where you might be a little timid to do this, their support can help you overcome your fear.  Just don't let your friends control your life or do anything that you consider wrong (from excessive drinking to theft, etc).  You just need to keep an open mind yet hold tight on your ethical and moral grounds.

Also, if you don't agree with drinking then going to a bar to meet friends probably wouldn't be the most logical thing to do :D  Try and meet friends at places that [i]you[/i] would normally go to.

Thanks for your replies, but, even though as I said, I hate continuing this topic so that all the public can see, but I am just trying to say what if you have the number of friends that can be counted on one hand. I mean friends that you go somewhere in general, what do you do, how do you get more friends if it seems that they are all too different? I really do hate posting on this topic, I just have no clue how to really improve.

Interesting. Peter, it is interesting how you imply that her wanting to have sex with you is by your physical strength (even though this is probably true).

But one of my most important questions, what can we talk to them about? I run out of topics very very quickly?

[quote="hello"]But one of my most important questions, what can we talk to them about? I run out of topics very very quickly?[/quote]Well, I really don't know what to help you with here. Try speaking in front of the mirror? :D Dunno :?

[quote="hello"]Interesting. Peter, it is interesting how you imply that her wanting to have sex with you is by your physical strength (even though this is probably true).[/quote]Most people with girlfriends I know are not athletes. But I don't know any 100-kg fatties or computer nerds spending 15 hours a day in front of their screen having one. I don't know you, but I think the social part will be a bigger deal usually.

[quote="hello"]But one of my most important questions, what can we talk to them about? I run out of topics very very quickly?[/quote]Yes... I used to have this problem as well. With some friends the conversations were limited to what new level we had reached in our favorite computer game... :?
I think there's only one solution to this: talk more. :) Especially to people who are pretty talkative themselves. And to girls, even if they have a bf already. That has done it for me at least... I have rarely ran out of subjects lately. :)
[ I think chatrooms etc may be a good start if you don't have any girls you know well. Local sports club or local <whatever> club may be a good place to meet new people too. ]

Valentin... I don't know, but isn't that usually the trick if you need to give a speech or something? Unless you have a talking mirror, that doesn't solve much here?

[quote="Peter VDD"]Valentin... I don't know, but isn't that usually the trick if you need to give a speech or something? Unless you have a talking mirror, that doesn't solve much here?[/quote]Dunno, I said I was just guessing. I have no problem with making (tons of) conversation :D, so I really don't know a solution for this ...  :(

Be spontaneous and just talk about anything that comes to your mind.  Try to avoid weather, though.   :)  And even weather works sometimes.  I remember talking for quite awhile about favorite seasons with a friend, and that led into something else.

Try asking the person questions about themselves and be a good listener.

Well, it appears that asking the other person about themselves is not a good idea, thats what it seems like to me.

But Valentin, it is because you know how to do it that I am asking you, what is it that you talk to them about, the same with Peter and everyone else. Actually girls, you should respond to this because you are the ones that know best!

[quote="hello"]But Valentin, it is because you know how to do it that I am asking you, what is it that you talk to them about, the same with Peter and everyone else. [/quote]You can talk about anything (including weather) as long as you sound convincing, and not obviously hitting on them.

[b][i]Really, you've had tons of advice by now. If you can't or won't follow it, it's your problem. We can only give you advice, not go do it for you.[/i][/b]
[quote="hello"]I find that I can talk about something academic or politics related (otherwise, knowledge related and classes related). I can talk a bit about movies and some sports and thats it.[/quote]
If these are the only things you can talk about, maybe you should get yourself a girl who likes talking about these things. Or, as mentioned MANY times before in this thread and the other one you started earlier this year, get new interests and make yourself interesting to them. If you are boring, you will get a boring girl. Of course, if you like it better that way, that is fine.[/i]

[quote="hello"]I find that I can talk about something academic or politics related (otherwise, knowledge related and classes related). I can talk a bit about movies and some sports and thats it. I dont know what else I can talk about. It seems that bores them I think.[/quote]

Well, that sounds good to me---I think everyone I've dated has talked mostly about academics and politics.  (Maybe the problem is that you're talking to females who don't agree with your politics?  If there's too large of a gap in outlook, that can be a turnoff.)

[quote="smbelcas"]Well, that sounds good to me---I think everyone I've dated has talked mostly about academics and politics.[/quote]well... maybe that is because you (as the girl) showed interest in it. Most girls I know are completely not interested in it... so it's also a big task to find the right girl ;) [but I guess that's an obvious remark]

Okay I'll say this.
I have about 40 "friends" on facebook.  The majority of them, however, aren't real friends.  And my actual friends mostly don't go on facebook.  I don't do much on facebook: I just see it as a way for people to easily contact me.

Smbelacas, I think that Peter is right in what he is saying even though I much prefer that you were correct, or at least that I can find a girl that is how I described.

All I can say is that going into this with the attitude of "I am now going to find me a girlfriend" is the wrong way to even approach this. Assuming that you're not just looking for a one-night fling here, and you want a meaningful relationship, you don't just one day decide you're going to go find "the right person" and then magically she'll be there if you look hard enough. In general, these things happen _to_ you. Not only is actively seeking a girlfriend a bit of a turnoff (to girls), its also very unlikely that anyone you find in this manner will be the right person -- because if you're consciously searching, you're very very likely to hit upon a lot of 'false positives'.

When the right girl comes along, you won't need to worry about all the things you're worrying about :) Just go and give each other the chance to meet.


Got an exam coming up and ive been looking through some past papers.  There is one particular line, and I cant see how it was obtained...see if you can help:

b    /    ( ( b / d )+m x )        =          d ( 1 - (dm/b) x + ... )

In the context of the question, we neglect all x of order >1, which may be the reason for the '...' or it may be from the taylor series. let me know wat u think

The right hand side commes from the first terms of $ \frac{1}{1\plus{}x} \equal{} 1 \minus{}x \plus{}x^2\minus{}x^3\plus{}x^4\minus{}\plus{}... \equal{} \sum_{k\equal{}0}^\infty (\minus{}x)^k$.


geometry
geometric transformation
homothety
vector
angle bisector
power of a point
radical axis

Let u, v and w be three congruent circles pairwise intersecting each other.
Let the circles v and w intersect at the points A and A'.
Let the circles w and u intersect at the points B and B'.
Let the circles u and v intersect at the points C and C'.

Prove that the perpendiculars from the points A', B', C' to the lines BC, CA, AB concur at one point.

  darij

[quote="darij grinberg"]Let u, v and w be three congruent circles pairwise intersecting each other.
Let the circles v and w intersect at the points A and A'.
Let the circles w and u intersect at the points B and B'.
Let the circles u and v intersect at the points C and C'.

Prove that the perpendiculars from the points A', B', C' to the lines BC, CA, AB concur at one point.[/quote]

I was asked to post my solution of this problem, so I do now.

First we show an angle equality.

Let U, V, W be the centers of the circles u, v, w, and let r be their common radius. Then, UB = UC = VC = VA = WA = WB = r.

The line BB' is the common chord of the circles w and u, and thus it is perpendicular to the line WU which joins the centers W and U of these circles. Hence, it is the altitude in the triangle BWU from the vertex B. But this triangle BWU is isosceles (as WB = UB), and thus the altitude from its apex B must simultaneously be the angle bisector of its angle WBU. Hence, the line BB' is the angle bisector of the angle WBU, so that $ \measuredangle B^{\prime}BU \equal{} \frac {\measuredangle WBU}{2}$.

On the other hand, since UB = UC, the triangle BUC is isosceles, and thus its base angle < UBC equals $ \measuredangle UBC \equal{} \frac {180^{\circ} \minus{} \measuredangle BUC}{2} \equal{} 90^{\circ} \minus{} \frac {\measuredangle BUC}{2}$.

Hence,

$ \measuredangle B^{\prime}BC \equal{} \measuredangle B^{\prime}BU \minus{} \measuredangle UBC \equal{} \frac {\measuredangle WBU}{2} \minus{} \left(90^{\circ} \minus{} \frac {\measuredangle BUC}{2}\right)$
$ \equal{} \frac {\measuredangle WBU \plus{} \measuredangle BUC}{2} \minus{} 90^{\circ}$.

Similarly,

$ \measuredangle C^{\prime}CA \equal{} \frac {\measuredangle UCV \plus{} \measuredangle CVA}{2} \minus{} 90^{\circ}$;
$ \measuredangle A^{\prime}AB \equal{} \frac {\measuredangle VAW \plus{} \measuredangle AWB}{2} \minus{} 90^{\circ}$;

hence,

$ \measuredangle B^{\prime}BC \plus{} \measuredangle C^{\prime}CA \plus{} \measuredangle A^{\prime}AB$
$ \equal{} \left(\frac {\measuredangle WBU \plus{} \measuredangle BUC}{2} \minus{} 90^{\circ}\right) \plus{} \left(\frac {\measuredangle UCV \plus{} \measuredangle CVA}{2} \minus{} 90^{\circ}\right) \plus{} \left(\frac {\measuredangle VAW \plus{} \measuredangle AWB}{2} \minus{} 90^{\circ}\right)$
$ \equal{} \frac {\measuredangle WBU \plus{} \measuredangle BUC \plus{} \measuredangle UCV \plus{} \measuredangle CVA \plus{} \measuredangle VAW \plus{} \measuredangle AWB}{2} \minus{} 270^{\circ}$.

By the sum of the angles in the hexagon BUCVAW, we have < WBU + < BUC + < UCV + < CVA + < VAW + < AWB = 540°; thus,

$ \measuredangle B^{\prime}BC \plus{} \measuredangle C^{\prime}CA \plus{} \measuredangle A^{\prime}AB \equal{} \frac {540^{\circ}}{2} \minus{} 270^{\circ} \equal{} 90^{\circ}$.

This is an important identity we should keep in mind.

The radical axis of two intersecting circles is the line joining their two points of intersection. Hence, the lines AA', BB' and CC' are the pairwise radical axes of the three circles u, v and w, and thus concur at the radical center R of these three circles.

Now, let X, Y, Z be the orthogonal projections of the point R on the sidelines BC, CA, AB of the triangle ABC. We will show that the triangles XYZ and A'B'C' are homothetic.

Let the line B'C' meet the line BC at a point S. Then, by the sum of angles in triangle CSC', we have

< CSC' = 180° - < CC'S - < SCC' = (180° - < CC'S) - < SCC' = < B'C'C - < BCC'.

Since the points B, C, B' and C' lie on one circle (namely, on the circle u), we have < B'C'C = < B'BC, so this becomes

< CSC' = < B'BC - < BCC' = < B'BC - (< BCA - < C'CA) = (< B'BC + < C'CA) - < BCA.

Now, since < B'BC + < C'CA + < A'AB = 90°, we have < B'BC + < C'CA = 90° - < A'AB, so this becomes

< CSC' = (90° - < A'AB) - < BCA.

On the other hand, let the line YZ meet the line BC at a point T. Then, by the sum of angles in triangle CYT, we have

< CTY = 180° - < CYT - < TCY = (180° - < CYT) - < TCY = < ZYA - < BCA.

Since < AYR = 90° and < AZR = 90°, the points Y and Z lie on the circle with diameter AR, so that < ZYA = < ZRA, and in the right-angled triangle AZR, we have < ZRA = 90° - < RAZ. Thus, < ZYA = 90° - < RAZ = 90° - < A'AB. Hence,

< CTY = < ZYA - < BCA = (90° - < A'AB) - < BCA = < CSC'.

Hence, B'C' || YZ. Similarly, C'A' || ZX and A'B' || XY. Thus, we have shown that the triangles A'B'C' and XYZ are homothetic; hence, there exists a homothety mapping the points X, Y, Z to the points A', B', C', respectively. Let this homothety map the point R to a point Q. Then, since homotheties map lines to parallel lines, QA' || RX. But $ RX\perp BC$; hence, $ QA^{\prime}\perp BC$, and it follows that the point Q lies on the perpendicular from the point A' to the line BC. Similarly, we can show that the point Q also lies on the perpendiculars from the points B' and C' to the lines CA and AB. Hence, these three perpendiculars concur at one point.

Originally I intended to propose this problem for the QEDMO (a math fight of the QED community), but it turned out to be a bit too hard...

  Darij

This is a cool problem Darij !
 Last night, I tried this problem and obtained the following generalization
    Suppose $ABC$ and $A'B'C'$  are two orthologic triangles. Let $M, N, P$ be the points such that $AB'MC'$, $BC'MA'$, $CA'MB'$ are parallelograms. Then $ABC$ and $MNP$ are orthologic.
    Darij 's problem is a special case of this generalization when $A'B'C'$ is the triangle formed by the three centers of the circles.
    Of course this problem is not very unfamiliar because the idea is straightforward : Carnot 's theorem. However, it would be nicer if I use the following vector form of Carnot 's theorem (here by denoting $|a|$ I mean vector $a$ as I don't know how to write vector in Latex. Also, |a|.|b| is the dot product of the two vectors)
    Let $ABC$ and $A'B'C'$ be triangles. Then the sufficient condition for them to be orthologic is $|AA'|.|BC| + |BB'|.|CA| + |CC'|.|AB| = 0$
    This problem is easily proved
[hide="Hint"]Let $A_1, B_1, C_1$ be the midpoints of $BC, CA, AB$[/hide]
    Now the generalization is easily proved.  :)

What is orthology as affine transformation????

a short solution with vectors is also possible :)


Let $ f(n)\equal{}\left[\sqrt[4]{n}\right]$.
Compute $ f(1)\plus{}f(2)\plus{}...\plus{}f(2005)$.

Other way is consider the graph $ y \equal{} \sqrt [4]{n}$ and find the number of integer point.


I think you are missing something.

For instance, if c is negative, b is positive, a is positive, and the absolute value of c is greater than a and b, then the inequality fails.

I think
a,b,c are elements of positive real numbers

If the above people are correct, then here is my suggestion
[hide="hint"]AM-GM[/hide]

Clearly false as $a, b, c \to \infty$.  I think the original problem had the condition $abc = 1$, and in any case this problem is not appropriate for this forum.


Given nine points, with positions in a 3x3 square, go though all of them with a continuous path formed by four straigh segments.

I love this one...

[hide]Start from the upper left, go diagonal to the lower right (1), go up to the upper right and past it by one dot (2), go diagonally down and to the left so that you are one dot left of the bottom left dot (3), and finally, go right to connect the rest of the dots (4).[/hide]


Prove that $ 0.\bar{9}$ is equivalent to 1. (Source: Myself)

[hide="My Proof"]
We know that $ 0.\bar{9}$ is rational and we know that all rational numbers can be represented as a ratio of two integers.

We can let $ x\equal{}0.\bar{9}$. If we multiply $ x$ by $ 10$, we get $ 10x\equal{}9.\bar{9}$. If we subtract $ x$ from $ 10x$, we get rid of the recurring decimal. So we have:
$ 10x\minus{}x\equal{}9\\
9x\equal{}9\\
\fbox{x\equal{}1}$[/hide]

yes your answer is completely right :)
It can be proved also using infinite geometric sequence but that's not necessary.

Proved by an infinite geometric sequence? That sounds interesting  :P, mind explaining?

My proof: $ 1\minus{}0.99999...\equal{}0.000000...1$
But that last number $ 1$ will never appear because there are infinitely many nines. Little bit stupid...

Or if someone says that the pattern might not continue like that, just ask them to add 1/3and 2/3. :lol:


[quote="Hanako"]Let a, b, c  such that ab + bc + ca = abc. Prove that

$ (a \plus{} b \minus{} c \minus{} 1)(b \plus{} c \minus{} a \minus{} 1)(c \plus{} a \minus{} b \minus{} 1) \le 8$[/quote]

If $ {a,b,c}>0$ ,we let

$ a \equal{} \frac {x \plus{} y \plus{} z}{x},b \equal{} \frac {x \plus{} y \plus{} z}{y},c \equal{} \frac {x \plus{} y \plus{} z}{z},$

then the original inequality is equivalent  to
\[ 4\sum{y^3z^3(y \plus{} z)(x \minus{} y)(x \minus{} z)} \plus{} 8\sum{xy^3z^3(x \minus{} y)(x \minus{} z)}\]

\[ \plus{} [(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx) \plus{} xyz](x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\ge{0}.\]


linear algebra
matrix
algebra
polynomial
geometry
3D geometry
sphere

Let $A$ a matrix of order $3$ such that $A^3=I_3$. Prove that $tr(A)=0\iff tr(A^2)=0$ ;)

since X^3-1 =(X-1)(X-J)(X-J²)  so A is diagonalisable with the eingvalus of A  are in {1,j,j²} so let  theme  a,b and c ,if tr(A)=0 then it's clear that {a,b,c}={1,j,j²} and so tr(A²)=0 ,if tr(A²)=0 then it's also clear that tr(A)=0 (this is because  X²+X+1 is the minimal polynom of j and j² in IR[X]) , you need only some base in linear algebra :lol:

it's clear that detA=1....
we have  AA*=A*A=IdetA =>AA*=I=> A*=A square
Using Hamilton-Cayley, the conclusion follows

[quote="yassinus"]since X^3-1 =(X-1)(X-J)(X-J²) so A is diagonalisable with the eingvalus of A are in {1,j,j²}[/quote]
it is not true. if $A$ has all its eigenvalues $1$ then it is not necessary diagonalisable...
for example $\begin{array}{ccc}( 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1) \end{array}$.
It follows from what you wrote that the eigenvalues of $A$ are in the set $\{1,\varepsilon,\varepsilon^2\}$. The condition $Tr A=0$ or $Tr A^2=0$ gives you that the eigenvalues of $A$ are exactly those and the matrix is diagonalisable...

If the minimal polynomial decomposes into all linear factors, then the matrix is diagonalizable. I think that was his justification for the claim.

but he does not know that from $A^3=I_3$.. it comes when you use $TrA=0$ or $TrA^2=0$.

Yes he does. $A^3 = I_3$ tells us that $x^3 - 1$ annihilates $A$, so the minimal polynomial of $A$ divides $x^3-1$. But $x^3 - 1$ already has only linear factors of multiplicity 1 (over the complex numbers, at least), so the minimal polynomial must also.

clearly A is digonlisable and has spectre in unite sphere so :
A²=A^(-1) and digonalistion shows that Tr(A^(-1))=conjugate(Tr(A)) , the equivalence follows immediately.

If $ a,b,c$ eigenvalues. They can take next values $ 1, (1 \plus{} \sqrt3i)/2, (1 \minus{} \sqrt3i)/2$, so $ conjugate(a) \equal{} a^2$, so if $ a \plus{} b \plus{} c \equal{} 0$ then $ conjugate(a \plus{} b \plus{} c) \equal{} a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$ and otherwise is also true.

since $x^{3}-1=(x-1)(x^{2}+1+x)$, and let $a_{1},a_{2},a_{3}$ be the eigenvalues of matrix A, so that $a_{1}+a_{2}+a_{3}=Tr(A)$,noting that the polynom $(x^{2}+x+1)(x-1)$is the  characteristic polynomial of matrix A, it is easy know that $ tr(A)=0\iff tr(A^{2})=0 $.


let $ (a_n)$ be an increasing sequence of positive reals then prove:

$ \frac{1}{a_1} \plus{} \frac{2}{a_1\plus{}a_2}\plus{}...\plus{}\frac{n}{a_1\plus{}a_2\plus{}....\plus{}a_n} < 4(\frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n})$

Actually the following stronger inequality holds too:
${ \frac {1}{a_1} + \frac {2}{a_1 + a_2} + \dots + \frac {n}{a_1 + a_2 + \dots + a_n} < 2(\frac {1}{a_1} + \frac {1}{a_2} + \dots + \frac {1}{a_n}})$.
And $ 2$ is best constant.
See here for a proof:
http://www.mathlinks.ro/viewtopic.php?t=169630


Announcement  --- SWIM, Princeton.

From June 11 to 19, 2009, Princeton University will host "Women in
Mathematics - Summer Workshop in Mathematics for high school
students".

This is a 9-day workshop for rising high school seniors interested in
mathematics, who would like to find out more about pursuing a math
major at Princeton.

Participants in the workshop will attend two mathematical courses, and
in the afternoon work in groups on an exploration topic related to a
course of their choice. Throughout the workshop students will meet
undergraduate students, graduate students and faculty of the
Mathematics Department.

In addition to the mathematical activities, various other activities
are planned: all the workshop attendees will participate in the "Women
in Science" seminar every afternoon, organized jointly with the "Women
and Mathematics" program at the Institute for Advanced Study
(http://www.math.ias.edu/wam); Princeton undergraduates will host a
discussion panel about applying to college; a mathematics-themed tour
of campus, ...

Workshop participants will be housed in one of the colleges on campus
(chaperoned by a Princeton female math major), and thus get a glimpse
of life as a Princeton undergrad. Every participant will receive full
support for travel and accommodation, to any reasonable extent.

More details can be found on http://www.math.princeton.edu/swim/

In particular, please encourage all mathematically talented female
students to apply to this program. Applications will be due at the end
of February 2009.


I am trying to figure out how to create a math based word document that contains a lot of latex use. However, I am not sure how to do this. Do I need to download a program involving latex for this to be possible?

Have a look at the AoPS guide starting [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url] which will tell you what to download and how to use the software.


Don inscribes a 120 degree central angle in circle Q with a radius of six inches.  This central angle intersects the circle at A and B.  Don cuts out sector AQB and rolls it to form a cone such that A meets B and there is no overlap.  What is the number of cubic inches in the volume of the cone?  Express your answer to the nearest cubic inch.

[hide]The slant height of the cone is 6 inches and the circumference is $4 \pi$ giving it a diameter of 4 and a radius of 2. Forming a right triangle with the slant height and the radius we find that the height of the cone is $\sqrt {36  - 4} = \sqrt {32} = 4 \sqrt 2$. Since the volume of a cone is $\frac {\pi r^2 h}{3}$ plugging in the values the volume of this cone is $\frac {4 \pi \cdot 4 \sqrt 2}{3} = \frac {16 \pi \sqrt 2}{3}$ which is approximately $24$.[/hide]

[hide]Folding into a cone, we see that the circumfrence of the cone's base is actually the length of the arc that was facing the inscribed angle, which was 120 degrees. The circumfence of the original circle was 12pi, and 120 is a third of 360 degrees, so the arc facing the inscribed angle must be 4pi.
Since the circumfence of the cone's base is 4pi, the radius of the cone must be 2.
The height can be found through the pythagorean thorem. The slant height is the original's circle's radius, which is 6, and is the hypotenuse of our imaginary triangle. The radius is one of the legs. using the thorem, we find that the height must be 4:sqrt: 2.

The formula for a cone's volume is 1/3(h)(r2pi), so substituting values, we get 23.695, or to the nearest whole number, 24 inches3[/hide]


Is the competition harder than previous years?

In my opinion, I think it just varies from year to year. Also, it may depend on a student's own judgment, particularly with subject matter, or whether he/she is better with conceptual creativity or computational style problems.

Our coach made up his own problems because he was afraid people would find the problems online and memorize the solutions or something.  The one he made up was a lot harder than the practice tests I did though.  :mad:

It was a lot harder than before last year, judging by the statistics, so I guess its either going to be the same or increase.

Wait, are you talking about AMC 8 or Mathcounts?

obviously it will probably get harder.
there were only 160 something perfect scores on amc 8 compared to 300 something last year

[quote="mathluver2.718"]I have heard that mathcounts has three difficulty levels which it cycles through every 3 years. If this is true, I am pretty sure that 2008 was the 'hard' year and 2009 was the 'easy' year, so , by process of elimination, 2010 will be the 'medium' year. That's my theory, anyway.[/quote]

Actually, the 2008 Mathcounts was really easy =/. The state coordinator also said that last year's contest was a lot harder than the previous ones; He said that Mathcounts was really 'stepping it up'.

[quote="TheDeity"]Actually, the 2008 Mathcounts was really easy =/. The state coordinator also said that last year's contest was a lot harder than the previous ones; He said that Mathcounts was really 'stepping it up'.[/quote]

Lolwut.  Last year's contest was much easier than the '08...judging by [b]scores[/b], not [b]opinions[/b].

09 was arithmetic intense, at all levels. That is why some people who generally suck at non-arithmetic-intense-bashy-type-math did better than normal.

What is the general trend when you look at the Mathcounts problems in the last 20 some years?
Is 90s harder than 80s and 2000s harder than 90s?

Or it just recycle everything two years (someone suggests every three years).

With more books, more websites (particularly aops website), students are getting better and better.
The problems have to get harder and harder.

Or Mathcounts just try to recycle emphasis:
from geometry to arithmetic to counting ...... each year?
Mewto suggests 2009 is the year of arithmetic.


I posted this in Advanced section/Number Theory/Number Theory Unsolved Problem early on Tuesday morning. It has had 58 views but no replies, so I though this forum might ber more appropriate.

Let m = (4^p  - 1)/3, where p is a prime number exceeding 3. Prove that 2^(m-1) has remainder 1 when divided by m.

Can anyone help?

Also, are solutions to (early) BMo round 2 questions available on the interent?

In reverse order,
No, there do not exist systematic sets of BMO2 solutions on the internet, or if there do, there shouldn't. It is BMOS policy for solutions to not be published on the internet, because this removes the temptation for people to look at them and 'revise' from solutions (which is a pretty useless activity unless one has actually solved the problem oneself).

Where have you gotten so far. Have you had a good look at the properties of m-1 yet?

Hmm, that hint may be a little obscure, and I should probably point out some useful stuff. My first deduction when I saw this (although I admit I did it at some point last year anyway) was

$ \text{Olympiad Question}\Rightarrow\text{Probably has a quite nice solution}$
Anything which involves actually substituting the given value of m-1 into $ 2^{m\minus{}1}$ immediately contradicts the "nice" criterion. If my assumptions about olympiad questions are true, we are therefore looking for some trick we can get by studying some select properties of $ m\minus{}1$.

That doesn't give it away, but should help.

[quote="probability1.01"]Or you can study relationships between 2 and m. For example, in general for odd $ m$, you only know $ 2^{\phi(m)} \equiv 1 \mod m$, but in this case, $ m$ is constructed from a power of $ 2$, so you can get extra information.[/quote]

Please could you be a bit more explicit?

Thanks

4^p = 4 (mod p) by Fermat's Little Theorem and since 3 and p are coprime,
m= ((4^p)  -1)/3 = 1 (mod p), i.e. p divides (m-1)
By parity observations, m-1 is even so, letting 2pk = m-1,
2^(m-1) = 2^(2pk)
= (4^p)^k
= (3m+1)^k
= 1 (mod m).


[quote="triplebig"]Prove:

$ a,b,c > 0\Longrightarrow\;\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2}\geq \frac {a \plus{} b \plus{} c}{3}$[/quote]
After expanding it gives:
$ 2(a^3 \plus{} b^3 \plus{} c^3) \geq a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b$ $ \boxed{1}$
But $ a^3 \plus{} b^3 \plus{} c^3 \geq 3abc$
So the $ RHS \equal{} a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc$
So our ineq is true by Schur...
The $ \boxed{1}$ we can also solve it by rearrangement...

$ \boxed{1}$ can be proved :

$ a^3\plus{}b^3\ge ab(a\plus{}b)\Leftrightarrow (a\plus{}b)(a^2\minus{}ab\plus{}b^2)\ge ab(a\plus{}b)\Leftrightarrow (a\minus{}b)^2\ge 0$

$ b^3\plus{}c^3\ge bc(b\plus{}c)$

$ c^3\plus{}a^2\ge ca(c\plus{}a)$

So the conclusion is obvious .

The inequality is equivalent to
$ a^3 \plus{} b^3 \plus{} c^3 \ge \frac13 (a\plus{}b\plus{}c)(a^2 \plus{} b^2 \plus{} c^2)$
and isn't it the chebyshev inequality? (QED)

[quote="TFPNH"]The inequality is equivalent to
$ a^3 \plus{} b^3 \plus{} c^3 \ge \frac13 (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$
and isn't it the chebyshev inequality? (QED)[/quote]
Yes it is...

hello, the given inequality is equivalent to $ \frac{2a^3\plus{}2b^3\plus{}2c^3\minus{}a^2b\minus{}ab^2\minus{}a^2c\minus{}ac^2\minus{}b^2c\minus{}bc^2}{3(a^3\plus{}b^3\plus{}c^3)}\geq0$. The numerator can be written as $ (a\minus{}b)^2(a\plus{}b)\plus{}(a\minus{}c)^2(a\plus{}c)\plus{}(b\minus{}c)^2(b\plus{}c)\geq0$ which is obviously true.
Sonnhard.

[quote="ElChapin"]Those inequalities can be proved using Cauchy.

If $ x_1, ..., x_n$ are positive numbers and $ m$ and $ l$ are real numbers then, by Cauchy:
\[ \dfrac{x_1^{m \plus{} l} \plus{} ... \plus{} x_n^{m \plus{} l}}{x_1^m \plus{} ... \plus{} x_n^m}\ge\dfrac{x_1^m \plus{} ... \plus{} x_n^m}{x_1^{m \minus{} l} \plus{} ... \plus{} x_n^{m \minus{} l}}
\]
[/quote]

This can be used to prove inequalities in the form of the one you posted triplebig.

:)

[hide][quote="triplebig"]Prove:

$ a,b,c > 0\Longrightarrow\;\frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2}\geq \frac {a \plus{} b \plus{} c}{3}$[/quote]

Let $ a^3 \ge b^3 \ge c^3$, therefore $ a^2 \ge b^2 \ge c^2$ and turn around.
Use Chebychev Ineq we proved: $ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \le 3(a^3 \plus{} b^3 \plus{} c^3)$

==> $ \frac {a^3 \plus{} b^3 \plus{} c^3}{a^2 \plus{} b^2 \plus{} c^2}\geq \frac {a \plus{} b \plus{} c}{3}$

:)[/hide]


How do you find the geometric mean of two segments?

Is it like this:

[img]http://img488.imageshack.us/img488/3065/geometricmeanmg5.png[/img]

$x=\sqrt{ab}$

Such construction in general case is impossible (using only straightedge and compass). Even ancient Greeks proved that $\sqrt[3]{2}$ can't be constructed.

I really could have banged myself in the head for missing a question on a math league about geometry means today. I simply forgot it. I saw it on 2003 grissom geometry test, yet I simply missed the formula by a bit. Well, now I will remember it forever after a disappointment.

so basically, if you want to find the geometric mean of $n$ elements, where
$a_{1},\;a_{2},\;a_{3}\;......\;,a_{n}$ are the elements, you would take the $n^{th}$ root of the product of the elements.

In symbols: $\sqrt[n]{a_{1}a_{2}a_{3}\cdots a_{n}}$

It isn't a formula, it is the definition.