function
inequalities
calculus
derivative
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inequalities proposed

Suppose $ a,b,c$ are real numbers, with $ a\plus{}b\plus{}c\equal{}3$. Prove that $ \frac{1}{5a^2\minus{}4a\plus{}11}\plus{}\frac{1}{5b^2\minus{}4b\plus{}11}\plus{}\frac{1}{5c^2\minus{}4c\plus{}11} \le \frac{1}{4}$.

[quote="Raja Oktovin"]Suppose $ a,b,c$ are real numbers, with $ a \plus{} b \plus{} c \equal{} 3$. Prove that $ \frac {1}{5a^2 \minus{} 4a \plus{} 11} \plus{} \frac {1}{5b^2 \minus{} 4b \plus{} 11} \plus{} \frac {1}{5c^2 \minus{} 4c \plus{} 11} \le \frac {1}{4}$.[/quote]
Let $ f(t) \equal{} \frac {1}{5t^2 \minus{} 4t \plus{} 11}$ we have $ f"(t) \le\ 0 \rightarrow f(t)$ is inconvex fuction

Follow Jensen : $ f(a) \plus{} f(b) \plus{} f(c) \le\ 3 f(\frac {a \plus{} b \plus{} c}{3}) \equal{} \frac {1}{4}$

We have done!

[quote="lenhamquy"][quote="onlylove_math"][
Let $ f(t) \equal{} \frac {1}{5t^2 \minus{} 4t \plus{} 11}$ we have $ f"(t) \le\ 0 \rightarrow f(t)$ is inconvex fuction
![/quote]
please, I don't think so[/quote]
But i think so  :lol:
$ f"(t)\equal{}\frac{\minus{}10t^2\minus{}120t\minus{}78}{(5t^2 \minus{} 4t \plus{} 11)^3} \le\ 0$  :wink:

[quote="onlylove_math"]
$ f"(t) \equal{} \frac { \minus{} 10t^2 \minus{} 120t \minus{} 78}{(5t^2 \minus{} 4t \plus{} 11)^3} \le\ 0$  :wink:[/quote]
I got $ 150x^2\minus{}120x\minus{}78$ instead of $ \minus{} 10t^2 \minus{} 120t \minus{} 78$

That's getting pretty funny indeed! Because I found...
$ f''(t) \equal{} \frac { 50t^2 \minus{} 120t \minus{} 78}{(5t^2 \minus{} 4t \plus{} 11)^3}$ :huh:

....and by checking $ discriminant> 0$ we cannot conclude for all $ t$, $ f''(t) \leq 0$.

[quote="Raja Oktovin"]Suppose $ a,b,c$ are real numbers, with $ a \plus{} b \plus{} c \equal{} 3$. Prove that $ \frac {1}{5a^2 \minus{} 4a \plus{} 11} \plus{} \frac {1}{5b^2 \minus{} 4b \plus{} 11} \plus{} \frac {1}{5c^2 \minus{} 4c \plus{} 11} \le \frac {1}{4}$.[/quote]
The $ pqr$ technique helps here.  :)
Denote $ q \equal{} ab \plus{} bc \plus{} ca$ and $ r \equal{} abc$ then
\[ \sum {\left( {5a^2 \minus{} 4a \plus{} 11} \right)\left( {5b^2 \minus{} 4b \plus{} 11} \right)} \equal{} \minus{} 90r \plus{} 25q^2 \minus{} 264q \plus{} 1089
\]

\[ \prod {\left( {5a^2 \minus{} 4a \plus{} 11} \right)} \equal{} 125r^2 \minus{} \left( {100q \plus{} 814} \right)r \plus{} 275q^2 \minus{} 1694q \plus{} 5324
\]
Hence we have the inequality equalivents to
\[ f\left( r \right) \equal{} 125r^2 \minus{} \left( {100q \plus{} 454} \right)r \plus{} 175q^2 \minus{} 638q \plus{} 968 \ge 0
\]
Actually we have $ f'\left( r \right) \equal{} 250r \minus{} 100q \minus{} 454 \le 0$ and $ r \le \frac {{q^2 }}{9}$
\[ \Rightarrow f\left( r \right) \ge f\left( {\frac {{q^2 }}{9}} \right) \equal{} \frac {{\left( {q \minus{} 3} \right)\left( {125q^3 \minus{} 525q^2 \plus{} 8514q \minus{} 26136} \right)}}{{81}} \ge 0
\]
We are done. Equality holds if and only if $ a \equal{} b \equal{} c$.

Finally, for the funtion $ f\left( t \right) \equal{} \frac{1}{{5t^2  \minus{} 4t \plus{} 11}}$ we have $ f''\left( t \right) \equal{} \frac{{150t^2  \minus{} 120t \minus{} 78}}{{\left( {5t^2  \minus{} 4t \plus{} 11} \right)^3 }}$.  :D

Hello Honey_S! Akashnil got the same too.
Can you show the elementary steps of obtaining $ f''(t)$? I am still obtaining $ 50t^2$ instead of $ 150t^2$. :maybe:

It is just a simple calculation about derivative :) The result's honey is correct $ f''(x)\equal{}honey$ :D

Before transforming the expression into pqr we can prove this one very easily by the following way:
\[ \Delta\equal{}(100q\plus{}454)^2\minus{}4\times 125(175q^2\minus{}638q\plus{}968)\equal{}\minus{}4(19375q^2\minus{}102450q\plus{}69471)<0,\forall q\in \mathbb{R}\]

yes, Honey_S is right
to onlylove_math but $ f'' < 0\Longleftrightarrow 0\le t\le 1.224621125$

[quote="Raja Oktovin"]Suppose $ a,b,c$ are real numbers, with $ a \plus{} b \plus{} c \equal{} 3$. Prove that $ \frac {1}{5a^2 \minus{} 4a \plus{} 11} \plus{} \frac {1}{5b^2 \minus{} 4b \plus{} 11} \plus{} \frac {1}{5c^2 \minus{} 4c \plus{} 11} \le \frac {1}{4}$.[/quote]
There exists a nice solution with Cauchy Schwarz. Can someone try this? ;) :)

Here's my solution. It's simple.
WLOG, we can assume $ a\equal{}max(a,b,c)$
Lemma: if $ x\leq \frac{9}{5}$ we have:  $ \frac{1}{5x^2\minus{}4x\plus{}11} \leq \frac{1}{24}(3\minus{}x)$ (because it's equivalent to $ (9\minus{}5x)(x\minus{}1)^2 \geq 0$)
Case 1:
 $ a \leq \frac{9}{5}$, applying the lemma, we get $ \sum{\frac{1}{5a^2\minus{}4a\plus{}11}} \leq \sum{\frac{1}{24}(3\minus{}a) \equal{}\frac{1}{4}}$
Case 2:
$ a > \frac{9}{5}$, it's obvious that $ \frac{1}{5a^2\minus{}4a\plus{}11} < \frac{1}{20}$
Note that: $ 5t^2\minus{}4t\plus{}11\equal{}5(t\minus{}\frac{2}{5})^2\plus{}\frac{51}{5} \geq \frac{51}{5}$, therefore
$ \frac{1}{5b^2\minus{}4b\plus{}11} \plus{}\frac{1}{5c^2\minus{}4c\plus{}11} \leq \frac{10}{51}$
So,  $ \sum{\frac{1}{5a^2\minus{}4a\plus{}11}} < \frac{1}{20}\plus{}\frac{10}{51} < \frac{1}{4}$

But where is the equality case considered when $ a>\frac{9}{5}$? You are obtaining a strict ineq.

[quote="Sunkern_sunflora"]But where is the equality case considered when $ a > \frac {9}{5}$? You are obtaining a strict ineq.[/quote]
There isn't a equality case with $ a > \frac {9}{5}$ :wink:

But we should obtain the same condition as required by the problem everywhere! :o

[hide="Sudhu Akashnil er jonno"]Tomar number theory er problem topic e r kono post koroni keno?
:) [/hide]

[quote="Raja Oktovin"]Suppose $ a,b,c$ are real numbers, with $ a \plus{} b \plus{} c \equal{} 3$. Prove that $ \frac {1}{5a^2 \minus{} 4a \plus{} 11} \plus{} \frac {1}{5b^2 \minus{} 4b \plus{} 11} \plus{} \frac {1}{5c^2 \minus{} 4c \plus{} 11} \le \frac {1}{4}$.[/quote]

your ineq is equivalent to :
$ \sum{\frac {x^2}{x^2 \plus{} 51}} \geq \frac {9}{20}$ with $ x \plus{} y \plus{} z \equal{} 9$
by cauchy-shwarz : $ LHS ( \sum{ (x^2 \plus{} 51)(x \plus{} 1)^2} ) \geq ( \sum{ x(x \plus{} 1)} )^2$
so it remains to prove that  $ 20 ( \sum{ x(x \plus{} 1)})^2 \geq 9 ( \sum{ (x^2 \plus{} 51)(x \plus{} 1)^2} )$
let $ f(x,y,z) \equal{} LHS \minus{} RHS$ , note that   $ f(x,y,z) \minus{} f(x,t,t) \geq 0$ where $ t \equal{} \frac {y \plus{} z}{2}$ and $ x \equal{} max(x,y,z) \geq 3$
so it sufficies to prove the inequality when two variables are equal
but  $ f(x,x,9 \minus{} 2x) \equal{} (x \minus{} 3)^2(31x^2 \minus{} 144x \plus{} 261) \geq 0$ , our inequality is proved

[quote="Raja Oktovin"]Suppose $ a,b,c$ are real numbers, with $ a \plus{} b \plus{} c \equal{} 3$. Prove that $ \frac {1}{5a^2 \minus{} 4a \plus{} 11} \plus{} \frac {1}{5b^2 \minus{} 4b \plus{} 11} \plus{} \frac {1}{5c^2 \minus{} 4c \plus{} 11} \le \frac {1}{4}$.[/quote]
Here is my second solution.  :)
First we will prove the follow inequality.  :oops:
\[ 179\sum {a^4 } \plus{} 855\sum {a^2 b^2 } \minus{} 406\sum {ab\left( {a^2 \plus{} b^2 } \right)} \minus{} 222abc\sum a \ge 0
\]
for $ a,b$ and $ c$ are real numbers. It is obviously true with $ m \equal{} 179,n \equal{} 855,p \equal{} g \equal{} \minus{} 406$
Then, assume $ a \plus{} b \plus{} c \equal{} 3$ and $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$ then we obtain the follow result from the above inequality
\[ \Rightarrow 1074 \minus{} 748q \plus{} 150q^2 \minus{} 180r \ge 0
\]
Now come back to the problem, using the Cauchy Schwarz inequality we have
\[ \sum {\frac {1}{{5a^2 \minus{} 4a \plus{} 11}}} \le \frac {1}{4}
\]

\[ \Leftrightarrow \sum {\frac {{5a^2 \minus{} 4a \plus{} 1}}{{5a^2 \minus{} 4a \plus{} 11}}} \ge \frac {1}{2}
\]
\[ \left( {\sum {\frac{{5a^2  \minus{} 4a \plus{} 1}}{{5a^2  \minus{} 4a \plus{} 11}}} } \right)\left[ {\sum {\left( {5a^2  \minus{} 4a \plus{} 1} \right)\left( {5a^2  \minus{} 4a \plus{} 11} \right)} } \right] \ge
\]
\[ \ge \left( {\sum {5a^2  \minus{} 4a \plus{} 1} } \right)^2
\]
Hence we need to prove
\[ 2\left( {\sum {5a^2 \minus{} 4a \plus{} 1} } \right)^2 \ge \sum {\left( {5a^2 \minus{} 4a \plus{} 1} \right)\left( {5a^2 \minus{} 4a \plus{} 11} \right)}
\]

\[ \Leftrightarrow 1074 \minus{} 748q \plus{} 150q^2 \minus{} 180r \ge 0
\]
which is known before,  :).

[quote="Math pro"][quote="Erken"]lol:
I am pretty sure Jensen is useless here.But I'm afraid that using derivatives is necessary,refering to [b]arqady[/b]'s solution,which he gave almost one year ago...
Searching is a virtue :wink: http://www.mathlinks.ro/viewtopic.php?p=968464#968464[/quote]
I don't think so because if $ f(x)$ is not concave at any point,the ineq is not true,I really think Jensen is very strong! :wink:[/quote]
Huh? :huh:
Dang...
OK,you can think whatever you want,but please speak more clearly,I am not telling anyone that Jensen is not strong,I just said that Jensen inequality here is not helpful and that's all...
Probably you mean that this function is concave or convex,but this function on the given segment simultaneously has "concavity" and "convexity" points...

[quote="shalex"]I'm eager to see the solution applying Cauchy-Schwarz Inequality which, I guess, can_hang2007 loves very much and does well in :wink:[/quote]
Ok, here is the Cauchy Shwarz solution of mine. :)
It is easy to see that there exist $ a,b$ such that $ (a\minus{}1)(b\minus{}1) \ge 0$, then $ a^2\plus{}b^2 \le 1\plus{}(a\plus{}b\minus{}1)^2 \equal{}c^2\minus{}4c\plus{}5$. Now, rewrite our inequality as
\[ \left( 5 \minus{}\frac{51}{5a^2\minus{}4a\plus{}11}\right) \plus{}\left( 5\minus{}\frac{51}{5b^2\minus{}4b\plus{}11}\right) \ge \frac{51}{5c^2\minus{}4c\plus{}11} \minus{}\frac{11}{4}\]
or
\[ \frac{(5a\minus{}2)^2}{5a^2\minus{}4a\plus{}11}\plus{}\frac{(5b\minus{}2)^2}{5b^2\minus{}4b\plus{}11} \ge \frac{ 83\plus{}44c\minus{}55c^2}{4(5c^2\minus{}4c\plus{}11)}\]
By Cauchy Schwarz Inequality, we have
\[ \frac{(5a\minus{}2)^2}{5a^2\minus{}4a\plus{}11}\plus{}\frac{(5b\minus{}2)^2}{5b^2\minus{}4b\plus{}11} \ge \frac{(5(a\plus{}b)\minus{}4)^2}{ 5(a^2\plus{}b^2)\minus{}4(a\plus{}b) \plus{}22} \ge \frac{ (5c\minus{}11)^2}{5c^2\minus{}16c\plus{} 35}\]
It suffices to prove that
\[ \frac{ (5c\minus{}11)^2}{5c^2\minus{}16c\plus{} 35} \ge \frac{ 83\plus{}44c\minus{}55c^2}{4(5c^2\minus{}4c\plus{}11)}\]
which is just equivalent to
\[ (c\minus{}1)^2( 775c^2 \minus{}2150c\plus{}2419) \ge 0\]
which is obviously true. :)

[quote="Math pro"][quote="Erken"]lol:
I am pretty sure Jensen is useless here.But I'm afraid that using derivatives is necessary,refering to [b]arqady[/b]'s solution,which he gave almost one year ago...
Searching is a virtue :wink: http://www.mathlinks.ro/viewtopic.php?p=968464#968464[/quote]
I don't think so because if $ f(x)$ is not concave at any point,the ineq is not true,I really think Jensen is very strong! :wink:[/quote]
You didnt understand all of the idea of Jensen ineq, Okey ,you can try the ineq in  Poland 1991 to find you are wrong (the function is not convex with all $ x$ but the ineq is still true, :lol: )and of course,If it had been able to be killed by only Jensen(simply),it would have appeared in China TST prblem
Your proof is very nice,Can_hang :o  :o

To [b]MathPro[/b]:
It is really annoying me to dispute with you about who of us is right...
In order to clarify the point I would like to reply,but for the last time.
There was some kind of discussing whether Jensen solves this problem or not.Some member said that it does and he/she gave some calculation of second derivative,but as it was observed later, there was a mistake.
I said that this problem can not be solved by the use of Jensen inequality and in order to show the solution(I must admit that this solution is really good) I put a link.That's all.
Then,I don't know why,you were starting to make some senseless replies,like "I really think that Jensen is very strong".OK,I agree with you,but what is the point of telling a such things?
You said:"I don't think so"
Did I say anything about concavity in my first post?
Did I say that Jensen is "b*llsh*t"?
Man...Do you understand how senseless you replies look like?And if really you don't want to be a spamer then stop doing a such
things.

it's really nice can_hang2007, and I think you are always "playing" Cauchy-Swartz Inequality very well!

[quote="kunny"][url]http://www.mathlinks.ro/Forum/viewtopic.php?t=175288[/url][/quote]
What about to have a look at the first page of this thread?
Edited:
No,you misunderstood me,I guess,I meant that I have already meantioned it before you,in the first page of this thread.

Ah, sorry, you have already posted it.  :blush:

[quote="Erken"]lol:
I am pretty sure Jensen is useless here.But I'm afraid that using derivatives is necessary,refering to [b]arqady[/b]'s solution,which he gave almost one year ago...
Searching is a virtue :wink: http://www.mathlinks.ro/viewtopic.php?p=968464#968464[/quote]

[quote="Vasc"]Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac 1{9 \minus{} 10a \plus{} 9a^2} \plus{} \frac 1{9 \minus{} 10b \plus{} 9b^2} \plus{} \frac 1{9 \minus{} 10c \plus{} 9c^2} \le \frac 3{8}$.[/quote]It's really nicer in respect that

$ \sum{\frac {4}{11 \minus{} 4a \plus{} 5a^2}}\leq\sqrt [3]{\frac {8}{3}\sum{\frac {1}{9 \minus{} 10a \plus{} 9a^2}}}\leq 1$

hold for $ a \plus{} b \plus{} c \equal{} 3$.

[quote="Vasc"]I think this is nicer.

Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac 1{9 \minus{} 10a \plus{} 9a^2} \plus{} \frac 1{9 \minus{} 10b \plus{} 9b^2} \plus{} \frac 1{9 \minus{} 10c \plus{} 9c^2} \le \frac 3{8}$.[/quote]

my method that I used to the original inequality works here  :wink:

[quote="Vasc"]I think this is nicer.

Let $ a,b,c$ be real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ \frac 1{9 \minus{} 10a \plus{} 9a^2} \plus{} \frac 1{9 \minus{} 10b \plus{} 9b^2} \plus{} \frac 1{9 \minus{} 10c \plus{} 9c^2} \le \frac 3{8}$.[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-120-m-rozenberg.html

Suppose $ a,b,c$ are real numbers, with $ a+b+c=3$. [url=https://math.stackexchange.com/questions/509580/how-prove-this-frac12a2-6a9-frac12b2-6b9-frac12c2-6c9-le-f?rq=1]Prove that[/url] $$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}{5}$$


geometry
circumcircle
geometric transformation
reflection
symmetry
trapezoid
angle bisector

The points $A, B, C, D$ lie in this order on a circle $o$.  The point $S$ lies inside $o$ and has properties $\angle SAD=\angle SCB$ and $\angle SDA= \angle SBC$. Line which in which angle bisector of $\angle ASB$ in included cut the circle in points  $P$ and $Q$. Prove $PS =QS$.

Let $O$ be the circumcenter of the quadrilateral $ABCD$. The triangles $\triangle SDA \sim \triangle SBC$ are directly similar, having 2 internal angles equal. Reflect one of these triangles in the line $SO$, say the triangle $\triangle SBC$ into the oppositely congruent triangle $\triangle SB'C'$. The reflections $B', C'$ of the points $B, C \in (O)$ in the line $SO$ passing through the circle center also lie on the circle $(O)$. Thus we obtain a cyclic quadrilateral $AC'B'D$ with the triangles $\triangle SDA \sim \triangle SC'B'$ oppositely similar. This implies that the common vertex $S$ of these 2 triangles is the diagonal intersection of the quadrilateral $AC'B'D$, i.e., the points $D, S, C'$ are collinear and the points $A, S, B'$ are also collinear. Since the symmetry axis $SO$ bisects one angle $\angle BSB'$ formed by the lines $AB' \equiv SB'$ and $SB$, the normal to this symmetry axis at the point $S$ bisects the complemetary angle $\angle ASB$ formed by these 2 lines. This normal is thus the defined angle bisector and it meets the circumcircle $(O)$ at points $P, Q$. The line $SO$ passing through the circle center $O$ and perpendicular to the chord $PQ$ cuts this chord at its midpoint $S$, hence, $PS = QS$.

Here's my a little bit long proof.

Let $P$ is on arc $AB$, and $Q$ is on arc $CD$. and $X$ is the intersection of $AD$ and $BC$.

WLOG $X$ is on right side of $CD$

It's self-confident that $ASCX$, $BXDS$ is cyclic respectively.

so we can know that the three common cords of circles $AC, BD, XS$ is concurrent.

Let's denote $R$ by the intersection of three cords.

Here, $\angle CSX= \angle CAX=\angle CBD=\angle DSX$. So we get $\angle CSX=\angle DSX$.

Therefore, $P, Q$ are on line $SX$.

And here, we can find that from $\angle CAX=\angle XSD$, $ASRD$ is cyclic. similarly, $BSCR$ is also cyclic.

Now, we almost get it!

$XQ\times XP=XD\times XA= XS\times XR$ $\cdots (Y)$

$XR\times SR = AR\times RC= PR\times QR$ $\cdots (T)$

Now, Let $a=XQ, b=QR, c=SR, d=SP$.

Then, $(Y)$: $a(a+b+c+d)=(a+b)(a+b+c) \Leftrightarrow ad=b(a+b+c)$.

$(T)$: $c(a+b)=(c+d)b \Leftrightarrow ac=bd$.

from $(Y)$, we get $acd=bc(a+b+c)$ $\Rightarrow bd^2 = b(bd+bc+c^2 )$ (by $(T)$).

$\Leftrightarrow d^2- c^2 = b(c+d)$ $\Leftrightarrow d=b+c$

Therefore, $PS=QS$.

$Q.E.D.$

These solutions ([u]Yetti[/u]'s and [u]Chang Woo-Jin[/u]'s) are fine. I can add only my congratulations and a little remark: if $U\in AC'\cap PQ,\ V\in DB'\cap PQ$ then $SU=SV$ (the problem of "butterfly").

And here is my solution:

[color=blue][b]Problem.[/b] Let ABCD be a cyclic quadrilateral with the circumcircle k, and let S be a point inside this quadrilateral such that the triangles SBC and SDA are directly similar.
Let the internal angle bisector of the angle ASB meet the circle k at the points P and Q. Prove that PS = QS.[/color]

[i]Solution.[/i] We will work with directed angles modulo 180°.

Let W be the point of intersection of the lines BC and DA, let R be the point of intersection of the lines AC and BD, and let O be the center of the circle k.

Since the triangles SBC and SDA are directly similar, we have < SBC = < SDA, or, in other words, < SBW = < SDW. Hence, the point S lies on the circumcircle of the triangle BDW. Hence, < BWS = < BDS. Similarly, the point S lies on the circumcircle of the triangle ACW, and < CAS = < CWS. Thus,

  < RAS = < CAS = < CWS = < BWS = < BDS = < RDS.

Thus, the point S lies on the circumcircle of triangle RDA. Similarly, the point S lies on the circumcircle of triangle RBC.

[Note that thus, the point S is the Miquel point of the complete quadrilateral formed by the lines BC, DA, AC and BD.]

Since the point S lies on the circumcircle of the triangle ACW, we have < ASC = < AWC, and since the point S lies on the circumcircle of the triangle RBC, we have < BSC = < BRC. Thus,

  < ASB = < ASC - < BSC = < AWC - < BRC = < (DA; BC) - < (BD; AC)
            = (< (DA; AC) + < (AC; BC)) - < (BD; AC) = (< (DA; AC) - < (BD; AC)) + < (AC; BC)
            = < (DA; BD) + < (AC; BC) = < ADB + < ACB.

Since the points A, B, C, D lie on one circle (namely, the circle k), we have < ADB = < ACB, and since O is the center of this circle, the central angle theorem yields $\measuredangle AOB=2\cdot\measuredangle ACB$. Hence,

$\measuredangle ASB=\measuredangle ADB+\measuredangle ACB=\measuredangle ACB+\measuredangle ACB=2\cdot\measuredangle ACB=\measuredangle AOB$.

Thus, the point S lies on the circumcircle of triangle OAB. Similarly, the point S lies on the circumcircle of triangle OCD.

Since the point S lies on the circumcircle of triangle OAB, we have < ASO = < ABO and < OSB = < OAB. But since AO = BO (in fact, the point O is the center of the circle k which passes through the points A and B), the triangle AOB is isosceles, and thus < ABO = < OAB. Hence, < ASO = < OSB, so that the line OS bisects the angle ASB. From the arrangement of the points, it follows that the line OS is the external angle bisector of the angle ASB. Hence, the internal angle bisector of the angle ASB is perpendicular to the line OS (in fact, the internal angle bisector of an angle is always perpendicular to the external angle bisector). Thus, the triangles PSO and QSO both are right-angled at S. Also, these two right-angled triangles share a common catet (OS) and have the same hypotenuse (OP = OQ, since both points P and Q lie on the circle k with the center O). Thus, these two right-angled triangles are congruent, so that PS = QS, and the problem is solved.

  darij

Heeeere's my solution (?):

   Point S is on a circle built on diagonal AC and at the same time on a circle

   built on diagonal BD. Thus it is on their intersection.  Angle bisector of ASB

   is the radical axis of these two circles.  Their centers line is pependicular to radical

   axis. Thus line OS where O is the center of circle(ABCD) is perpendicular to PS

   and as diameter cuts chord PS in two equal segments.

   Looks very much like a Butterfly Theorem.

   Thank you.

   M.T.

$Problem$: let $ABCD$ be a cyclic quadrilateral which is scribed in circle $(C)$.$S$ is a point in circle,inwhich:

$\angle SAD=\angle SCB$ and $\angle SDA=\angle SBC$.

the bisector of the angle $\angle ASB$,intersects circle $(C)$ in points $P$ and $Q$.

prove that:$SP=SQ$.

ok...

lets solve the problem by the most power full tool in geometry,$INVERSION$.

triangles $SAD$ and $SBC$ are simmilar to each other ,because:

$\angle SAD=\angle SCB$ and $\angle SDA=\angle SBC$ and $\angle ASD=\angle BSC$.

and the conclusion is :$SA.SB=SC.SD$.

consider an $INVERSION$ with the center $S$ and the radius $SA.SB$ or $SC.SD$.

lets invert the shape:

$A$ goes to $A'$ $(SA'=SB)$.

$B$ goes to $B'$ $(SB'=SA)$.

$C$ goes to $C'$ $(SC'=SD$.

$D$ goes to $D'$ $(SD'=SC)$.

$thorem$:the $INVERSION$ of a circle which doesnt go through the center of the $INVERSION$,is a circle.

then circle $(C)$,after invert changes to circle $(C')$,because it doesnt go through the center.

and $A'B'C'D'$ is a cyclic quadrilateral which is scribed in circle $(C')$.

Now we should use the fact:$SP$ is the bisector of $\angle ASB$ or $SQ$ is the bisector of $\angle DSC$.

pay attention to quadrilateral $A'B'C'D'$.$A',B',C',D'$, with a reflect to $PQ$,changes to $B,A,D,C,$, respectively.

then its clear that :the circles $(C)$ and $(C')$,also by reflecting to $PQ$,will change to each other.

and because of the smmetry of the shape,we can claime that:

circle $(C')$,goes through points $P$ and $Q$.

(note:dears ,this fact will be simply proved.and i think it doesnt need to disscus. ;).u can disscus it with ure self )

then,in the $INVERSION$,which we used the points $P$ and $Q$,are fixed.i mean $P=P'$and $Q=Q'$. then we

have:$SA.SB$=$SP.SP'$=$SP^2$

and also:

$SD.SC$=$SQ.SQ'$=$SQ^2$.

but:

$SA.SB$=$SD.SC$. (because the simmilaity of the triangles $SAD$ and $SBC$)

then:$SP^2=SQ^2$ and $SP=SQ$.

the probelm is really nice and u saw that,its proved very nice and simple by $INVERSION$.

i think the resourse of the problem is $INVERSION$.and the teacher created it by seeng and using inversive facts.

i say again.really a nice problem.

i can enjoy it during my life and up to the time my life stops. ;)

the blue circle is the inverted of circumcircle of$ABCD$,

and the green circle is the inversion circle!

i should say, this diagram is proposed by my dear friend$Amir.S$ :wink:

[quote="yetti"]Thus we obtain a cyclic quadrilateral $AC'B'D$ with the triangles $\triangle SDA \sim \triangle SC'B'$ oppositely similar. This implies that the common vertex $S$ of these 2 triangles is the diagonal intersection of the quadrilateral $AC'B'D$, i.e., the points $D, S, C'$ are collinear and the points $A, S, B'$ are also collinear.[/quote]Why? :huh:

First, the posted problem is trivial if S is the center of the given circle (O), so we assume that S is not the circle center. Extend AS, BS, CS, DS to meet the circle (O) again at B', A', D', C' (keepeing the notation of my figure). The cyclic quadrilaterals AC'B'D, A'CBD' have the common circumcircle (O), the same diagonal intersection S, the same angle between the diagonals (because $\triangle SBC \sim \triangle SDA$), and the same opposite sides AC' = A'C, B'D = BD' (because $\angle AB'C' = \angle A'BC,$ $\angle B'C'D = \angle BCD'$), therefore they are congruent. In particular, their diagonals are equal, AB' = A'B, CD' =  C'D, so that we obtain 2 isosceles trapezoids ABB'A' and CDD'C' with the same diagonal intersection S. The opposite sides of all these cyclic quadrilaterals (AC'B'D, A'CBD' and trapezids ABB'A', CDD'C') meet on the polar s of the diagonal intersection S with respect to the circle (O) and $s \perp OS.$ Because of symmetry, the opposite sides of the isosceles trapezoids ABB'A', CDD'C' meet at the foot P of the normal OS to this polar s. Thus these 2 isosceles trapezoids are symmetrical with respect to OS, so is the circumcircle (O), which means that A', B', C', D' ( originally obtained as the other intersections of BS, AS, DS, CS with (O)) are reflections of A, B, C, D in the line OS. Obviously, D, S, C' are collinear, A, S, B' are collinar, etc.

[quote="Virgil Nicula"]... Very nice your proof, Armpist ! In the my opinion, is the best...[/quote]

And of course, the purpose of opinions is to vary.

[quote="Virgil Nicula"][quote="armpist"][color=darkblue][b]Proof.[/b] Point $S$ is on a circle built on diagonal $AC$ and at the same time on a circle built on diagonal $BD$.  Thus it is on their intersection. Angle bisector of $\widehat{ASB}$ is the radical axis of these two circles. Their centers line is pependicular to radical axis. Thus line $OS$ where $O$ is the center of circle $(ABCD)$ is perpendicular to $PS$ and as diameter cuts chord $PS$ in two equal segments.

[b]Remark.[/b] Looks very much like a Butterfly Theorem.[/color][/quote]
Yes Armpist, I like much the Butterfly Theorem ! Very nice your proof, Armpist ! In the my opinion, is the best.
Please, forgive me if I could afford to write otherwise (more coloured) the your solution in "quote" ![/quote]Could any one clarify me why angle bisector of $\widehat{ASB}$ is the radical axis of these two circles, and why the line joining the centers is parallel to $OS$?

Thanks.

let $AD$ and $BC$ intersect at $T$ ;$AC,BD$ at $R$,then $ASCT;BSDT$ concyclic;hence in the 3 circles,equal-power axes $AC,BD,ST$are concurrent
so $PSRQT$ collinear.what follows is just a well-known conclusion in polar theory.

[quote=armpist]Thus line OS where O is the center of circle(ABCD) is perpendicular to PS
[/quote]

I still don't get it. Why $OS$ is parallel to line connecting centers of circles $ASC, BSD$?


function
inequalities
induction
algebra unsolved
algebra

$ f: N\to N$ satisfies i) and ii)

i) $ f(93)\equal{}116$
ii) $ f(f(n)) \plus{} 7f(n) \equal{} 8n \plus{} 207$

find $ f(n)$

[b]I will only apply property ii) to show $ \forall m: f(m)\equal{}m\plus{}23$.[/b]

Since $ f(f(m))>0$ we get $ f(m)\leq\frac{8m\plus{}206}{7}$ and after resubstitution $ 7f(m)\equal{}8m\plus{}207\minus{}f(f(m))\geq 8m\plus{}207\minus{}\frac{8f(m)\plus{}206}{7}$, which is equivalent to $ f(m)\geq\frac{56m\plus{}1243}{57}$.

Now define the sequences, $ \{a_n\},\{b_n\},\{c_n\},\{d_n\}$ via $ a_0\equal{}56,b_0\equal{}8,c_0\equal{}1243,d_0\equal{}206$ and recursively

$ a_{n\plus{}1}\equal{}8b_n,b_{n\plus{}1}\equal{}8a_n\plus{}7,c_{n\plus{}1}\equal{}207b_n\minus{}d_n,d_{n\plus{}1}\equal{}207(a_n\plus{}1)\minus{}c_n$

Assume we observe at some point the inequality (*) $ \forall m: \frac{ma_n\plus{}c_n}{a_n\plus{}1}\leq f(m)\leq\frac{m(b_n\plus{}1)\plus{}d_n}{b_n}$. Substitution into ii) yields $ \forall m: \frac{ma_{n\plus{}1}\plus{}c_{n\plus{}1}}{a_{n\plus{}1}\plus{}1}\leq f(m)\leq\frac{m(b_{n\plus{}1}\plus{}1)\plus{}d_{n\plus{}1}}{b_{n\plus{}1}}$.

[b]Claim[/b]: (1) $ \forall n>0: 22.3\leq\frac{c_n}{a_n\plus{}1}\leq 23\leq\frac{d_n}{b_n}\leq 23.2$
(2) $ \forall m: f(m)\equal{}m\plus{}23$

[i]Proof[/i]: As for (1) we verify the assertion for $ n\equal{}1$. For $ n>1$ using $ a_n,b_n>0$ and an induction argument yields $ \frac{c_{n\plus{}1}}{a_{n\plus{}1}\plus{}1}\equal{}\frac{207b_n\minus{}d_n}{8b_n\plus{}1}$ $ \leq \frac{207b_n\minus{}23b_n}{8b_n}\equal{}23$ and $ \frac{d_{n\plus{}1}}{b_{n\plus{}1}}\equal{}\frac{207(a_n\plus{}1)\minus{}c_n}{8a_n\plus{}7}$ $ \geq\frac{207(a_n\plus{}1)\minus{}23(a_n\plus{}1)}{8a_n\plus{}8}\equal{}23$. The inequalities for the outer bounds follow similarly taking into account the growth of $ a_n,b_n$.

For (2) note that $ f$ fulfills (*) for $ n\equal{}0$ and therefore for all $ n\in\mathbb{N}$. Due to (1) this implies for all $ m\in\mathbb{N}$ the inequality:

$ m\plus{}22.3\leq\limsup_{n\rightarrow\infty}\frac{ma_n\plus{}c_n}{a_n\plus{}1}\leq f(m)\leq\liminf_{n\rightarrow\infty}\frac{m(b_n\plus{}1)\plus{}d_n}{b_n}\leq m\plus{}23.2$ and with $ f(m)\in\mathbb{N}$ the assertion follows.


Let u_n:u_0=a>0;u_(n+1)=u_n/(1+(u_n) 2 ).
Find lim  u_n*\sqrt n.
This is a base problem for lim .

easily to have lim u_n=0
lim 1/(u_(n+1) 2 - 1/(u_n) 2
=lim (2+ u_n 2 ) = 2
By cesaro :
 lim 1/((u_n) 2 *n) =2
so lim u_n*  \sqrt n = 1/ \sqrt 2


If $ x$ is chosen at random from $ \{\minus{}4.\minus{}3,\minus{}2,\minus{}1,0,1,2,3,4\}$, what is the probability that it is a solution to $ x(x \minus{} 3)(x \plus{} 2) < 0$? Express your answer as a common fraction.

Or we could solve because we learn more that way

For $ x(x\minus{}3)(x\plus{}2)$ to be less than $ 0$, all of the factors must be negative or exactly one of them must be negative.  If all are negative, then the condition $ x\plus{}2<0$ suffices, so $ x<\minus{}2$.  There are $ 2$ values that work for this case.

If exactly one of them is negative, then it must be $ x\minus{}3$, so $ x>0$ and $ x\minus{}3<0\Rightarrow x<3$.  There are another $ 2$ values that work.

There are $ 9$ elements in the set, so $ \boxed{\frac{4}{9}}$


inequalities
geometry
circumcircle
inequalities solved

Show that in every triangle we have

1) (a^4b^4+b^4c^4+c^4a^2)/a^2b^2c^2 >= 9R^2 >= a^2 + b^2 + c^2

2) (r_a/a)^2 + (r_b/b)^2 + (r_c/c)^2 >=9/4

where a, b, c are sides, R is circumradius, r_a, r_b, r_c are escribed radii.

Namdung

9R^2 >= a^2 + b^2 + c^2

we know that sum(sin^2(A))=sum(1-cos^2(A))=3-sum(cos^2(A))=3-(1-2prod cos(A))=2(1+prod cos(A))<=2(1+1/8)=9/4
and we know that a/sin(A)=2R and replacing it in sum(sin^2(A)) we get the desired ineq!

cheers!

For the second one put a=y+z, etc and reduce it to sum (yz*(x+y+z)/(x(y+z)^2)>=9/4. Put m=yz, n=zx and p=xy. It becomes (mn+np+mp)(1/(m+n)^2+1/(n+p)^2+1/(m+n)^2)>=9/4, which holds by Iran (in fact another Crux) inequality.

Can you please write in detail how the inequality reduces to

sum yz*(x+y+z)/(x(y+z)^2)>=9/4.

Thank you very much.

well it's like so:

we know that r_a=S/(p-a)=sqrt((x+y+z)xyz)/x using the notation harazi just made: a=y+z; b=z+x and c=x+y.

I am very very very very stupid. I checked the first inequality and not  the second because I thought Harazi referred to:

(a^4b^4+b^4c^4+c^4a^2)/a^2b^2c^2 >= 9R^2

Thank you very much Lagrangia for your clarification.

[quote="Namdung"]Show that in every triangle we have

[b]1[/b]) $ \frac {b^2c^2}{a^2} \plus{} \frac {c^2a^2}{b^2} \plus{} \frac {a^2b^2}{c^2} \geq9R^2;$

[b]2[/b]) $ \frac {r_{a}^{2}}{a^2} \plus{} \frac {r_{b}^{2}}{b^2} \plus{} \frac {r_{c}^{2}}{c^2}\geq\frac {9}{4},$

where $ a, b, c$ are sides, $ R$ is circumradius, $ r_a, r_b, r_c$ are escribed radii.[/quote]A counterexample $ a \equal{} 3,b \equal{} c \equal{} 2$ of the first one :

$ \frac {b^2c^2}{a^2} \plus{} \frac {c^2a^2}{b^2} \plus{} \frac {a^2b^2}{c^2} \equal{} \frac {178}{9} < \frac {144}{7} \equal{} 9R^2.$

The following inequality

$ \left(\frac {r_{a}^{2}}{a^2} \plus{} \frac {r_{b}^{2}}{b^2} \plus{} \frac {r_{c}^{2}}{c^2}\right)\left(\frac {b^2c^2}{a^2} \plus{} \frac {c^2a^2}{b^2} \plus{} \frac {a^2b^2}{c^2}\right)\geq\frac {81}{4}R^2$

holds for all triangle.

The second one see also here : http://www.mathlinks.ro/viewtopic.php?t=130999

let:p-a=x,...
we need to prove that:
$ \frac{x\plus{}y\plus{}z}{xyz}(\sum\frac{x^2y^2}{(x\plus{}y)^2}) \geq \frac{9}{4}$
but:
$ LHS \geq \frac{x\plus{}y\plus{}z}{xyz}*\frac{(\sum xy)^2}{\sum (x\plus{}y)^2}$
remember that:
$ \sum(x\plus{}y)^2 \leq \frac{4}{3}(x\plus{}y\plus{}z)^2$
so $ LHS \geq \frac{x\plus{}y\plus{}z}{xyz}*\frac{3}{4}*\frac{(\sum xy)^2}{(\sum x)^2}\equal{}\frac{3}{4}*\frac{(\sum xy)^2}{xyz(x\plus{}y\plus{}z)} \geq \frac{9}{4}$


Hi! I need help with the following problem:

A ship is steaming parallel to a straight coastline, distance D offshore, at speed V. A coastguard cutter, whose speed is u<V sets out from a port to intercept the ship.
a)Show that the cutter must start out before the ship passes a point a distance {D(V^2-u^2)^1/2}/u back along the coast.
b)If the cutter starts out at the latest possible, where and when does it reach the ship?

I just found the solution  :lol: .It was not really difficult but i didn't see something obvious.
For the part b) the solution  is:
Distance:
Du/{(V^2-u^2)^1/2}
Time:
DV/u{(V^2-u^2)^1/2}

Sorry but i don't know Latex :blush:


calculus
integration
derivative
trigonometry
calculus computations

Let $ f,g$ be differentiable in an interval containing $ 0$ where $ g > 0$, and let $ f(x) = o(g(x))$ as $ x \rightarrow 0$.  As $ x \rightarrow 0$ prove or give a counterexample to:

(i) $ \int_{0}^{x}f = o \left( \int_{0}^{x}g \right)$

(ii) $ f'(x) = o(g'(x))$

You can prove (i) using L'Hôpital's rule.

Here's an alternative argument: Given $ \epsilon>0,$ there exists $ \delta>0$ such that $ |f(x)|<\epsilon g(x)$ for $ 0<x<\delta.$ Then $ \left|\int_{0}^{x}f\right| <\epsilon\int_{0}^{x}g,$ which proves what we want.

Counterexample for (ii):

$ g(x)=x,\ f(x)=x^{2}\sin\left(\frac1{x^{10}}\right).$

No fair because $ f$ isn't positive? We can make similar examples even for positive $ f.$ I'm still working on the details, but it can be done.


To insert a figure you use the \begin{figure} \end{figure} command.  Is it possible to have it automatically label stuff like "Figure 1.1", and so I can maybe reference it later, like the \begin{equation} and \end{equation} environment??

add \caption and you can label the picture whatever you like.


trigonometry
function
algebra
polynomial
inequalities
analytic geometry
conics

From important to easy
Trig Functions
Graph and Inverse of Trig Funtions
Applications of Trig
Trig Identities and Equations
Polunomial Functions
Inequalities and Linenear Programming
Exponential and Logarithmic Functions
Polar coordinates and Complex Numbers
Conic Sections
Matrices and Vectors
Sequences and Series
Probability

[b]Very important[/b]
Conic Sections
Trig Functions
Graph and Inverse of Trig Funtions
Trig Identities and Equations
Applications of Trig

[b]Important[/b]
Polynomial Functions
Polar coordinates and Complex Numbers
Exponential and Logarithmic Functions

[b]Slightly important[/b]
Inequalities and Linear Programming
Sequences and Series
Matrices and Vectors
Probability

Importance for the list is basically what will be covered in calculus. The easiest topics are under the slightly important list. (in my opinion, anyways)

Are there any good precalc book that I should read before doing precalc in the fall?

If you're doing precalc in the fall, I highly doubt you need to read about it beforehand. You could, but you don't need to.

The book I recommend is Lial Hornsby Schneider Precalculus Textbook. It goes over everything there and some more, and implies a bit of the AoPS Learning of math.

But as davidyko said you really don't need it because you are taking precalculus in the fall.

if you are trying to teach ur self precal over summer, what book will be the best?

Personally, I'd move up everything in the "slightly important" category to "important" except Linear Programming and Inequalities.


Without a calculator, find the foci of the ellipse $ 4x^2\plus{}y^2\equal{}4$

[hide="Solution"]
We put the equation $ 4x^2+y^2=4$ into standard form of ellipses to figure out its foci.
\begin{align*}
4x^2+y^2&=4\\
\frac{x^2}{1^2}+\frac{y^2}{2^2}&=1\\
\end{align*}
So we have $ a=1$ and $ b=2$. $ b>a$, so the ellipse has a vertical major axis. We know that the ellipse's center is at the origin due to the fact that $ h$ and $ k$ are both $ 0$. Now we have to find $ c$ to find out the coordinates of the foci.
\begin{align*}
c&=\sqrt{b^2-a^2}\\
c&=\sqrt{4-1}\\
c&=\sqrt{3}\\
\end{align*}
So the foci are $ (h,k\pm c) \Longrightarrow (0,0\pm \sqrt{3}) \Longrightarrow \boxed{(0,\sqrt{3}),(0,-\sqrt{3})}$[/hide]

[hide="Solution using definition of focus"]
The foci are points $ A,B$ such that for any point $ X$ on the ellipse, $ AX\plus{}XB \equal{} c$ for some $ c$.

Easily see that if $ X$ lies on the y-axis, then we must have $ A(0,f)$ and $ B(0,\minus{}f)$, so that $ AX\plus{}XB\equal{}(2\plus{}f)\plus{}(2\minus{}f)\equal{}4$

Now, let $ X$ be the point $ (1,0)$ on the ellipse. We have that $ 2 \sqrt{1\plus{}f^2} \equal{} 4 \implies f \equal{} \pm \sqrt{3}$

Thus, the foci are $ (0, \pm \sqrt{3})$[/hide]


algorithm
number theory
Diophantine equation

This problem is actually on the final challenge of the class "Introduction to Number Theory" given on here!

"Given any solution to the Diophantine equation 41x+53y=12, find the residue of x in modulo 53."

Enjoy!

-interesting_move

wanna have fun try proving that little fact

you wouldnt believe how much is behind the "simple statement"

a|bc, (a,b)=1 => a|c.

its crazy.

I don't see why that's so hard to prove. Let b = p1_e1*p2^e2*p3^e3... and c = q1_f1*q2^f2*q3^f3*..., where ps and qs are primes. Since (a,b)=1, none of the ps are factors of a. Thus, if a != 1, then a must be divisible by some of the qs, and then we're done -- that is, if the Fundamental Theorem of Arithmetic can be proved without this fact, which may or may not be possible.

thats exactly the problem.

you need it to prove the fundamental theorem of arithmetic.

Basically you need to go

WOP -> Div. Algorithm -> Linear Diophantine Eq -> thm i just stated -> fundamental thrm of arithmetic.

Linear Dio Eq:
Given a,b integers
 ax + by = 1 is solvable in integers x,y iff (a,b) = 1.


Can any Mathlinks forum member who is going to take part as a contestant in IMO 2004 in Athens leave a message here? :) It would be nice to see who will be going.

wow, finally someone admitted that he/she is gonna be present at IMO :D ;)

The post above was by me...hm, how did I get logged out :?

[quote="Valentin Vornicu"][quote="Anonymous"]I suppose this will end up being a huge "looking for MathLinks people" session :)  Maybe the Romanian members should coordinate it ;)

It'll probably be pretty obvious to spot me, being a girl and all...  :blush:[/quote] :) yeah well if I manage to go too (afterall) I will probably be wearing a nice MathLinks t-shirt :P and who knows maybe I'll bring some spares :D[/quote]
Question :
HOW many girls have username on mathlinks and how many of them will go to IMO this year ?  :lol: Don't be  :blush: !!!

valiowk is a girl

Arne, iura, dust aren't

so that makes a  75%-25%  balance for now... not bad :)

I feel so old and tired... :)  I cannot go to IMO and so I will never participate at this competition. Too bad. But I wish to all of you goooooood luck.

[quote="Valentin Vornicu"][quote="pbornsztein"]I won't go too :(  but I agree to wear the mathlinks t-shirt at home :D
Hey Valentin, do not forget to send some t-shirts with you book when it will be available.

Pierre.[/quote] well the t-shirt idea depends whether I can find a sponsor or not :P but hopefully I will.[/quote]
I WANT A ML T-SHIRT TOO !!!! And what book is Pierre refering to ???... :?

ok, let's not deviate too much from this IMO topic, ok? :) people going to IMO should post here :P

alex please do not use caps locks anymore, ok? :) he is probably talking about the translation (and add-ups) of my book in English ... which I didn't have time to complete yet :( ... but it will be done in the near future :P :)

i'm leaving tomorrow morning(just a 2 hours flight :D ). hope to see you as well.

Peter

Amazingly, we (moldavian team) are going also tommorow, but we have a 3 :?  hours flight, and Moldova is closer to Greece than Germany. I guess you  will be flying on something very fast :D:)

hmm...maybe it were 3 hours, but anyway not too long(we're taking a normal plane, nothing special :D ).

Peter

Well, our plain departs in Frankfurt at 12:55, and we arrive in Athens at 16:40. Taking into account the time difference of 1 hour, the flight dures 2.45 h.

  Darij

I'm at the airport in Athens right now.  Our flight arrived one hour early (!!), so I have a lot of time to burn while waiting for the UK and Israel teams to arrive.

Unfortunately there is also a 10-minute Internet limit and the connection is really slow :(

To those who haven't gotten on the plane yet, have a safe trip!

I am leaving in 7 hours only. We're one of the last teams to arrive. :?

Vietnamese team arrived on 8 July.
I hope i'll meet the European Champions :D :D

[quote="pbornsztein"]Do not forget to take some pictures of all of you...

Pierre.[/quote]

Yeah, I completely agree with Pierre.  Something like the IMO2001 page with lots of contestants' pictures and some vita/preferences/interests stuff would be very nice.  :)

BTW will Valentin distribute any AOPS/ML T-Shirts at the IMO ?  :D

Hi Valentin,

I would like to have a mathlink T-shirt, large size.

Thx oubi

So any pictures of all of you?

Some are available here (french team) :
 http://membres.lycos.fr/imo04/Photos%20Antony/

Pierre.

it looks as if there's an Asian in French team. Which country is he from?

I think you are talking about Antony Lee. As far as I know, his family comes from Taiwan.

Pierre.

we (romanian team) will also post some pictures later :)

yeah, I'm from Taiwan, and actually, I'm now in Taiwan (for vacation)


On one of the proof writing tutorial pages, there was a problem to find S(S(S(4444^4444))) where S(x) is the sum of the digits of x.  I don't understand the proof for why S(S(S(4444^4444))) is congruent 7 mod 9.  Can someone please explain this to me?

Also, is there any place I can find a copy of all the 2005 AIME questions?

Thanks!

Can't help you with your problems, but I can tell you where to find this year's AIME.  If you click "math jams" at the top of the page, you'll find a list of upcoming math jams.  Math jams are where the admins give a list of problems from a test, ie. AIME, and walk people through detailied solutions.  Click "trancripts" to find transcripts of old math jams.  Both AIMEs of this year are there.

[quote="alekk"]compute $4444^4444$ modulo $9$ ..[/quote]

alekk, Ithink you forgotr to put {} around your 4444; it should look like this:4444^{4444}

The problem is actually IMO 1975 problem 4... See http://www.mathlinks.ro/Forum/viewtopic.php?t=31409 .

  darij


Euler
geometry
circumcircle
geometry proposed

(V.Protasov, 9--10) The Euler line of a non-isosceles triangle is parallel to the bisector of one of its angles. Determine this
angle (There was an error in published condition of this problem).

Let $ O,H,N$ be the circumcenter, orthocenter and 9-point center of $ \triangle ABC.$ Assume that $ OH$ is parallel to the internal bisector of $ \angle BAC.$ Which implies that $ HO$ bisects $ \angle BHC,$ since $ HB,HC$ and the external bisector of $ \angle BAC$ bound an isosceles triangle with apex $H.$ Hence if $ U,V$ denote the midpoints of $ HC,HB,$ then the quadrilateral $ HUNV$ is cyclic, where $ N$ is the midpoint of the arc $ UV$ of its circumcircle. By simple angle chase we obtain then

$ \angle UNV \plus{} \angle BHC \equal{} 180^{\circ} \Longrightarrow  2\angle BHC \plus{} \angle BHC \equal{} 180^{\circ}$

$ \Longrightarrow  \angle BHC \equal{} 60^{\circ} \Longrightarrow  \angle BAC \equal{} 120^{\circ}.$

easy problem.just see that $ HO$ bisects $ BHC$ and $ BO\equal{}CO$ so $ BHCO$ is cyclic.so $ 360\minus{}2A\plus{}180\minus{}A\equal{}180$ so $ A\equal{}120$


Pascal\u0027s Triangle
combinatorics unsolved
combinatorics

Consider the coefficient of $x^{n}$ in the expansion

$(1+x)^{2n}= \left( (1+x)^{n}\right)^{2}$

And the result follows immediately.

Alternately, a path-counting solution down Pascal's triangle to ${2n \choose n}$ is equivalent.

Finally, using ${n \choose i}={n \choose n-i}$, consider the union $U$ of two disjoint sets $A, B$ of $n$ elements each.  Then count the number of ways to create $n$-subsets of $U$ by selecting $k$ elements from $A$ and $n-k$ elements from $B$.

Hmmm... just posted about 2 days ago most recently...

(you might as well have proven Vandermonde's)


hello orl, i have found this inequality, do you meant
${ \frac{1}{2}(x+y)<\frac{x-y}{2\arctan\left(\frac{x-y}{x+y}\right)}}<\left(\frac{1}{2}(x^2+y^2)\right)^{\frac{1}{2}}<\left(x^xy^y\right)^{\frac{1}{x+y}}$?
Is this true?
Sonnhard.


For a given integer $ n\geq 2,$ determine the necessary and sufficient conditions that real numbers $ a_{1},a_{2},\cdots, a_{n},$ not all zero satisfy such that there exist integers $ 0<x_{1}<x_{2}<\cdots<x_{n},$ satisfying $ a_{1}x_{1}\plus{}a_{2}x_{2}\plus{}\cdots\plus{}a_{n}x_{n}\geq 0.$

[quote="math10"]Perhaps this is a simple matter.
an application by a simple format of equality Abel.
I really want to have a solution in 2003.2004(Chinese also).
here is an image.[/quote]

As I can not understand the language(either its Chinese or Japanese(kanji), please post your solution in english.

Remark: The scanning was unclear.! :D
I will try understanding it after you or any other member translates it.. :maybe:

What I think it's saying:

The necessary and sufficient condition is that there is some $ k$ for which $ \sum_{i\equal{}k}^n x_i>0$.   For necessity, note that we can write
\[ \sum a_ix_i \equal{} x_1(a_1\plus{}\dots\plus{}a_n)\plus{}(x_2\minus{}x_1)(a_2\plus{}\dots\plus{}a_n)\plus{}(x_3\minus{}x_2)(a_3\plus{}\dots\plus{}a_n)\plus{}\dots\plus{}(x_n\minus{}x_{n\minus{}1})a_n.\]
Since the differences of the $ x_i$ are by assumption positive, at least one of the sums of the $ a_i$ must be positive.

For sufficiency, let us suppose that the statement holds for some $ k$.  Then let $ x_1\equal{}1, x_2\equal{}2, \dots x_{k\minus{}1}\equal{}k\minus{}1, x_k\equal{}m\plus{}k, x_{k\plus{}1}\equal{}m\plus{}k\plus{}1, \dots x_n\equal{}m\plus{}k\plus{}n$, where $ m$ is to be determined later.

Then $ \sum a_ix_i$ has the form $ m(x_{k}\plus{}x_{k\plus{}1}\plus{}\dots\plus{}x_n)\plus{}c$, where $ c$ is some constant not depending on $ m$.  For large enough $ m$, this is positive.


linear algebra
matrix
linear algebra solved

Prove that two similar matrix have the same rank.

the two matrices represent the same endomorphism (in different basis), sothey have same rank.

Suppose $A$ is similar to $B$ and $A$ has rank $k.$ $A$ had rank $k$ iff there exists an invertible matrix $E$ such that $EA=M,$ where $M$ is in row echelon form and has $k$ pivots, hence exactly $k$ nonzero rows.

There exists an invertible $P$ such that $A=PBP^{-1}.$

Then $EA=EPBP^{-1}=M$

$(EP)B=MP$

So $B$ can be row reduced to $MP$, but if we examine the definition of matrix multiplication, we see that $MP$ has no more than $k$ nonzero rows. That means that the rank of $MP$ and hence the rank of $B$ is no more than $k.$

Thus $\text{rank}(B)\le\text{rank}(A).$

Since similarity is a symmetric relation, we have by the same argument that

$\text{rank}(A)\le\text{rank}(B).$

Actually, it's pretty well-known that multiplication with an invertible matrix does not change the rank: whenever $T$ is invertible, $r(AT)\le r(A)$ on the one hand, and, on the other hand, $r(A)=r(AT\cdot T^{-1})\le r(AT)$.

i  think that this question was giving in the class  :D

[quote="grobber"]Actually, it's pretty well-known that multiplication with an invertible matrix does not change the rank: whenever $T$ is invertible, $r(AT) = r(A)$.[/quote]

Is there an elementary proof of this?

See Kent Merryfield's post in this thread, and modify appropriately.


Download [url=http://www.mathlinks.ro/Forum/download.php?id=2268]PDF Version[/url] of the problems.

[b]Day 1[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=242547]Problem 1.[/url]
Let $x$, $y$ be two positive integers such that $3x^2+x=4y^2+y$.
Prove that $x-y$ is a perfect square.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=242550]Problem 2.[/url]
Two right angled triangles are given, such that the incircle of the first one is equal to the circumcircle of the second one.
Let $S$, respectively $S'$, be the areas of the first triangle and respectively of the second triangle.

Prove that $\displaystyle \frac{S}{S'}\geq 3+2\sqrt{2}$.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=242552]Problem 3.[/url]
In an international meeting of $n \geq 3$ participants, 14 languages are spoken. We know that:

- Any 3 participants speak a common language.
- No language is spoken more that by the half of the participants.

What is the least value of $n$?

[b]Day 2[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=243242]Problem 4.[/url]
Define $[u]$ as the integer part of any real $u$. Let $X$ be a non empty subset of $\mathbb{N}^*$.
Suppose that for all $x \in X$, $4x \in X$ and $[\sqrt{x}] \in X$.

Prove that $X=\mathbb{N}^*$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=243246]Problem 5.[/url]
Let $ABC$ be a triangle such that $BC=AC+\frac{1}{2}AB$. Let $P$ be a point of $AB$ such that $AP=3PB$.

Show that $\widehat{PAC} = 2 \widehat{CPA}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=243253]Problem 6.[/url]
Let $P$ be a polynom of degree $n \geq 5$ with integer coefficients:
\[ P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_0 \quad \textrm{ with } \ a_i \in \mathbb{Z},\ a_n \neq 0.  \]
Suppose that $P$ has $n$ different integer roots: $0,\alpha_2,\ldots,\alpha_n$. Find all integers $k \in \mathbb{Z}$ such that $P(P(k))=0$.

Is it a day with 21 points Igor?  :P

Pierre.

No ... I expect (12-)13-14.

Tomorrow will be a 21 point day  :P

Wow, so a total of 28-30 points. I don't know if it's good or bad in the French TST :) These ones seem a little bit more difficult than last year's tests, isn't that true Pierre? :P :)

I will be something like 2nd (with a lot of luck) or 4th-5th (with no luck) I think ...

That's good :) I have edited your first post to include the problems and the pdf version of them! Nice choice of problems there!

Thanks !

By the way, the working time is 4.5 hours and not 4 hours  ;)

Oh OK :) I will rebuild the .pdf file :P :)


[hide="Hint"]What happens when you square it and subtract 2?[/hide]

[hide][quote="daermon"]evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}$[/quote]

2
just keep subtracting and minusing 2[/hide]

[quote="moogra"][hide][quote="daermon"]evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}}$[/quote]

2
just keep subtracting and minusing 2[/hide][/quote]
 :huh: Your solution doesn't make too much sense. Can you clarify?
[hide="a hint"]Solve this like an infinite summation[/hide]
[hide="a solution"]$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}=x$
Squaring gives you $2+x=x^{2}$
$x^{2}-x-2=0$
$(x-2)(x+1)$
-1 is clearly extraneous :|
therefore the answer is 2.[/hide]
[hide="a hint for the VERY unlikely case that this is a target problem seeing as a version of this was on a sprint before"]If by some unlikely case this is a target problem then you can find the answer by punching in the first five or so sqrts and see what it tends to.[/hide]

[quote="daermon"]evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}$[/quote]

[hide="solution"]
Let $x$ = $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}$
$x^{2}=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}$
$x^{2}=2+x$
$x^{2}-x-2=0$
$(x+1)(x-2)=0$
principal sq rt cannot be negative
$x=2$
[/hide]
Edit-(I know ur gonna ask for generalization...)
[hide="Generalization"]
for $a \geq-\frac{1}4$
$x^{2}-x-a=0$
$x^{2}-x+\frac{1}4=a+\frac{1}4$
$x-\frac{1}2=\sqrt{a+\frac{1}4}$
$x-\frac{1}2=\frac{\sqrt{4a+1}}2$
if $a>0$ then
$x=\frac{1+\sqrt{4a+1}}2$
Elseif ($a\leq 0$) $AND$ ($a\geq-\frac{1}4$) then
$x=\frac{1+\sqrt{4a+1}}2$ or $x=\frac{1-\sqrt{4a+1}}2$
Else ($a\leq-\frac{1}4$) then
No Solution
[/hide]

[hide="solution"]We can call that $\omega$. Therefore,

$\omega=\sqrt{2+\sqrt{2+\sqrt{2+\cdot\cdot\cdot}}}$

We notice that

$\omega=\sqrt{2+\omega}$

So we can solve:

$\omega^{2}=2+\omega$
$\omega^{2}-\omega-2$
$(\omega-2)(\omega+1)=0$
$\omega=[2,-1]$

Since we want the positive value, $\boxed{2}$ is the answer.[/hide]

Do you want a generalzation? :gleam:

[quote="ragnarok23"][hide="Generalization"]
\begin{eqnarray*}\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{\ldots}}}}&=&x \\ \sqrt{n+x}&=& x \\ n+x&=&x^{2}\\ x^{2}-x-n&=&0 \\ \frac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(n)}}{2(1)}&=&\boxed{\frac{1\pm \sqrt{1+4n}}{2}}\\ \end{eqnarray*}
Of course, we are taking the positive value.[/hide][/quote]For this generilazation is protestanT's distributed form better or is ragnarok's?

oops i forgot to rationalize denominator


analytic geometry
graphing lines
slope
calculus
derivative
calculus computations

I was assigned a problem by my Calc teacher that requires me to model a ski jump first with a single curve and then with multiple.

I have been working out the details for further along in the process, but I got stuck at the very first part. I have deduced the following:

the curve at the takeoff point, (0,0), must have a slope (derivative) of tan(10.5)=.1853
the curve must intersect the point (144, 100) (slope does not matter there)
at some point (x, y) the curve must have slope (derivative)= tan(39)=.809784

I can take it from there, but I am struggling with this part. Any ideas?

[quote="aveld"]I was assigned a problem by my Calc teacher that requires me to model a ski jump first with a single curve ...
[/quote]

What single curve are you planning on using?

How about a Lorentz, 1/(1 + x*x) ?


[quote="easyas3.14159..."]For a natural number n, n^2 gives a remainder of 4 when divided by 5, and n^3 gives a remainder of 2 when divided by 5. What is the remainder when n is divided by 5?[/quote]

[hide]i got
[size=200]2[/size][/hide]

Well actually, the answer is [hide]3[/hide]

[quote="robinhe"][hide]One way of doing this is to substitute real numbers into $n$.  One number that fits the conditions is 3.  So, the answer is $3$ because $3 \div 5=0 \ R3$.[/hide][/quote]

can you solve without guess and check? >.> i tried without guess and check but i got the answer wrong

[quote="easyas3.14159..."]For a natural number n, n^2 gives a remainder of 4 when divided by 5, and n^3 gives a remainder of 2 when divided by 5. What is the remainder when n is divided by 5?[/quote]

[hide="Answer"]

  $n^2 = 4$ (mod 5)

  $n^3 = 2$ (mod 5)

  Substitue the value of $n^2$ into the second equation:

  $n^3=(n^2) \cdot n=4\cdot n=4n=2$ (mod 5)

  $4n=2=12$ (mod 5)

  $n=3$ (mod 5)

  Hence, $n$ yields a remainder 3 on division by 5. [/hide]


AMC
AIME
USA(J)MO
USAMO
algebra unsolved
algebra

[quote="admire9898"]$ a_{1} \equal{} 1,a_{2} \equal{} 2,a_{n}a_{n \plus{} 2} \equal{} 1 \plus{} a_{n \plus{} 1}^{3} (n\geq1)$
(1)   proof that all $ a_(n) \in Z^{ \plus{} }$
(2)   find $ a_(n) (n\geq1)$[/quote]

I think it is not a real olympiad or olympiad preparation problem.

See http://www.research.att.com/~njas/sequences/A003818

It seems nobody yet gave there a closed formula for this sequence ...

Maybe admire9898 may give us the origin of this problem, and how does he know there is a solution.

[quote="pco"]

I think it is not a real olympiad or olympiad preparation problem.

Maybe admire9898 may give us the origin of this problem, and how does he know there is a solution.[/quote]

this problem is from a chinese contest like AIME in USA.BUT NOT equivalent to USAMO
the (1) is only a lamma to solve that problem
but (2) is my guess to this problem :P   :D  I'm interested to this sequence if it really have a the formula of general term

[quote="admire9898"]this problem is from a chinese contest ... [/quote]

Thanks for this answer. So we now are sure there is a general closed formula for this sequence.

Dont hesitate to post this formula when the solutions of the contest will be published.

Did the solutions of the contest publish ?


function
induction
search
complex analysis
real analysis
real analysis unsolved

Easy one but interesting:
let $f$ be an increasing (no continuity etc ..) function on $\mathbb{R}$.
Prove that there is a power series $g(x)=a_0+a_1x+..+a_kx^k+..$, with infinite radius of convergence that satisfies:
$g(x)>f(x)$ for all $x \in \mathbb{R}$

What's your solution ? Mine is kind of tedious to write: on each interval $[n,n+1]$ (where $n \in \mathbb{Z}$), $f$ is bounded. Note $M_{n} \in \mathbb{Z}$ one of its upper bounds on this interval. Now we construct a power series $S(x) = \sum a_{n}\frac{x^{2n}}{n!}$ with coefficients $(a_{n})$ as follows:
$\forall n \in [0,M_{1}],a_{n}=M_{0}$.
 $\forall n \in [M_{1},M_{2}+T_{2}],a_{n}=M_{1}$, where $T_{2}$ is large enough in order to guarantee that $min(S^{*}(x),x \in [1,2]) \geq M_{1}$, where $S^{*}$ has the coefficients previously defined on $[0,M_{1}]$ and coefficients stationary to $M_{1}$ for $n \geq M_{1}$.
And so on by induction on $[M_{2}+T{2},M_{3}+T_{3}]$, etc ... Using also the parity of $S$ and its relation to $e^{x^{2}}$, the radius of convergence is clear and the majoration too.

it is easy to check that the problem can be reduced to the following:
Let $b_0,b_1,..,b_k,..$ be arbitrary real numbers
Show that there exists an even entire function $f$ (i.e. $f(x)=f(-x)$) that satisfies $f(n) \geq b_n$ for $n \geq 0$.

We will search $f$ under the shape $f(x)=a_0+a_1x^{p_1}+..+a_nx^{p_n}+..$ where $a_k > 0$, $p_{n+1}>p_n$, $p_i$ is an even integer.
*To assure the convegence, we will take $(a_n, p_n)$ such that $|a_nx^{p_n}| < \frac{1}{n^2}$ for all $x \in [0;\sqrt{n}]$.
*to assure the hypothesis we will also take $(a_n, p_n)$ such that $a_nn^{p_n} > b_n$

It suffices that : $a_nn^{\frac{p_n}{2}}<\frac{1}{n^2}$ and $a_nn^{p_n} > b_n$.
i.e. $a_n<\frac{1}{n^{\frac{p_n}{2}+2}}$ and $a_n>\frac{b_n}{n^p_n}$.

But for $p_n$ large: $\frac{b_n}{n^{p_n}}<\frac{1}{n^{\frac{p_n}{2}+2}}$. So we can find $a_n$.
Conclusion follows.


\[\text{Day I}\]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=182624#p182624]Problem 1 Discussion[/url]
Let $x,y$ be real numbers satisfying the condition:
\[x-3\sqrt{x+1}=3\sqrt{y+2}-y \]
Find the greatest value and the smallest value of:
\[P=x+y \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=182627#p182627]Problem 2 Discussion[/url]
Let $(O)$a fixed circle with the radius $R$.Let $A$ and $B$ be fixed points in $(O)$ such that $A,B,O$ are not collinear.
Consider a variable point $C$ lies on $(O)(C\neq A,B)$.Construct two circles $(O_{1}),(O_{2})$ passed through $A,B$ and tangent $BC,AC$ at $C$,respectively.The circle $(O_{1})$ intersect the circle $(O_{2})$ in $D(D\neq C)$.Prove that:
a)
\[CD\leq R \]
b)The line $CD$ passed through point independent of $C$(i.e there exist fixed point on the line $CD$ when $C$ lies on $(O)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=182628#p182628]Problem 3 Discussion[/url]
Let $A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}A_{8}$  be convex 8-gon (no three diagonals concruent).
The intersection of arbitrary two diagonals will be called "button".Consider the convex quadrilaterals formed by four vertices of $A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}A_{8}$ and such convex quadrilaterals will be called "sub quadrilaterals".Find the smallest $n$ satisfying:
We can color n "button" such that for all $i,k \in\{1,2,3,4,5,6,7,8\},i\neq k,s(i,k)$ are the same where $s(i,k)$ denote the number of the  "sub quadrilaterals" has $A_{i},A_{k}$ be the vertices and the intersection of two its diagonals is "button".
\[\text{Day II}\]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=183462#183462]Problem 4 Discussion[/url]
Find all function $f: \mathbb R\to \mathbb R$ satisfying the condition:
\[f(f(x-y))=f(x)\cdot f(y)-f(x)+f(y)-xy \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=183463#183463]Problem 5 Discussion[/url]
Find all triples of natural $(x,y,n)$ satisfying the condition:
\[\frac{x!+y!}{n!}=3^{n}\]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=183465#183465]Problem 6 Discussion[/url]
Let $\{x_{n}\}$ be a real sequence defined by:
\[x_{1}=a,x_{n+1}=3x_{n}^{3}-7x_{n}^{2}+5x_{n}\]
For all $n=1,2,3...$ and a is a real number.
Find all $a$ such that $\{x_{n}\}$ has finite limit when $n\to+\infty$ and find the finite limit in that cases.

[quote="chavez"]kunny: why?  :D[/quote]

It's secret. :D

Please, everyone which posted usefull information in this thread, repost these in the first topics (the ones in which the problems are discussed). They will be deleted if you do not do that... Thanks!

Wow! You are the fastest one today,Pigfly! :D
The exam has just taken this morning and in the evening, I see it'problem here
How do you do in teh exam? :D

I have just posted 3 problems come from Day II :) .See #1
@Minh Khoa:How did you do in the exam? :)

Wow!You did 5 problems
Xin chuc mung
I did'nt solve problem 3 and problem 6 in case 0<a<1

[quote="TienVinh"]In proplem 6, you should divice (0,1/3) and (1/3,1)[/quote]
Please, next time send your comments into the corresponding topic ;) :)
http://www.mathlinks.ro/Forum/viewtopic.php?p=183465#183465

Does anyone know where Vietnam 2004 problems can be found (in english)
Between it seems that this year problems are easier then previous year problems.

After making a lot of effort I've succeeded to build a *.pdf with the problems (since I haven't yet seen it on the forum). Here it is.

Thank you very much, Xixas!

But there are some reasons to improve your $\LaTeX$ skills. ;) :)
Just post the tex-sorce here.

I know, my skills should be improved, since it's the first *.pdf, I've compiled.

The source:

\documentclass[11pt]{article}
\pdfpagewidth 8.5in
\pdfpageheight 11in
\topmargin 0in
\headheight 0in
\headsep 0in
\textheight 8in
\textwidth 6.5in
\oddsidemargin 0in
\evensidemargin 0in
\headheight 40pt
\headsep 0.2in
\begin{document}
\pagenumbering 0
\Huge{Vietnam Mathematical Olympiad 2005}\\
\\
\large{Day I} \\
\\
1. Let $x,y$ be real numbers satisfying the condition:\\
$\displaystyle{x-3\sqrt {x+1}=3\sqrt{y+2} -y}$\\
Find the greatest value and the smallest value of:\\
$P=x+y$\\
\\
2. Let $(O)$ a fixed circle with the radius $R$. Let $A$ and $B$ be fixed points in $(O)$ such that  $A,B,O$ are not collinear. Consider a variable point $C$ lies on $(O)$ $(C\neq A,B)$. Construct two  circles $(O_1)$, $(O_2)$ passed through $A,B$ and tangent $BC,AC$ at $C$, respectively.The circle $(O_1)$  intersect the circle $(O_2)$ in $D$ ($D\neq C$). Prove that: \\
a) $CD\leq R$ \\
b) The line $CD$ passed through point independent of $C$ (i.e there exist fixed point on the line $CD$ when $C$ lies on $(O)$). \\
\\
3. Let $A_1A_2A_3A_4A_5A_6A_7A_8$ be convex 8-gon (no three diagonals concruent).
The intersection of arbitrary two diagonals will be called "button".Consider the convex quadrilaterals formed by four vertices of $A_1A_2A_3A_4A_5A_6A_7A_8$ and such convex quadrilaterals will be called "sub quadrilaterals". Find the smallest n satisfying: \\
We can color $n$ "button" such that for all $i,k \in\{1,2,3,4,5,6,7,8\}$, $i\neq k,s(i,k)$ are the same where $s(i,k)$ denote the number of the "sub quadrilaterals" has $A_i,A_k$ be the vertices and the intersection of two its diagonals is "button".\\
\\
\large{Day II}\\
\\
4. Find all function $f:\mathbb R\to \mathbb R$ satisfying the condition:\\
${\displaystyle{f(f(x-y))=f(x)\cdot f(y)-f(x)+f(y)-xy}}$\\
\\
5. Find all triples of natural $(x,y,n)$ satisfying the condition:\\
\\
$\displaystyle{\frac{x!+y!}{n!}=3^n}$\\
\\
6. Let $\{x_n\}$ be a real sequence defined by:\\
\\
$\displaystyle {x_1=a,x_{n+1}=3x_n^3-7x_n^2+5x_n}$\\
\\
For all $n=1,2,3...$ and a is a real number. Find all a such that $\{x_n\}$ has finite limit when $n\to +\infty $ and find the finite limit in that cases.
\end{document}

Oh,I think the problems for this year this very very easier than that last year!


group theory
abstract algebra
superior algebra
superior algebra solved

Let  G be a finite group, P sylow subgroup.
If H is a subgroup of G s.t. N(P) is contanained in H is contained in G, Then
[G:H]=1 modp

By Sylow III, $[H: N(P)]$ and $[G: N(P)]$ are congruent to 1 mod p. So we are done.


[quote="bibobeo"]Find x,y integer satisfy
$19^{19}= x^{3}+y^{4}$[/quote]

Work modulo 13.

The exponents $3, 4$ both have the property that they divide $\varphi(13) = 12$.  This means that there are only $4, 3$ values that each can take, respectively, because we know that

$a^{12}\equiv 1 \bmod 13$

From which it follows that

$(a^{4})^{3}\equiv (a^{3})^{4}\equiv 1 \bmod 13$

So next time, we just mutilply the exponent of x and y then find x that has $\varphi(x) =$ that number. then work with modulo x. Is it right?

Of course, such a modulus may not exist.  The next best thing is a modulus $m$ such that the exponents of $x, y$ either both divide or both have common factors with $\varphi(m)$.


In an office at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box.  When there is time, the secretary takes the top letter off the pile and types it.  There are nine letters to be typed during the day, and the boss delivers them in the order 1, 2, 3, 4, 5, 6, 7, 8, 9.

While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing.  The colleague wonder which of the nine letters remain to be typed after lunch and in what order they will be typed.  Based upon the above information, how many such after-lunch typing orders are possible?  (That there are no letters left to be typed is one of the possibilities.)

My answer seems a little too simplistic to fit the problem difficulty, which leads me to believe that it is incorrect, but I'll post anyway. [hide="Just maybe"]Each letter she could or could not have typed, because she could, say, type each letter right after the boss puts it down, or delay until he puts the next one down and not type the previous. The fact that 8 has been typed doesn't say anything about whether or not any of the other 8 letters have been typed, so each of the other eight can or can not be typed, so $2^8=\boxed{256}$ different possibilities.[/hide]

Hokkage, you left out a key fact:

[hide] Letter 9 might not have even been deposited yet, so it might be done out of order.  If n letters from 1 through 7 remain to be typed, 9 could be done in one of n+2 ways:  the n+1 spots between the n other letters, or not at all (it was typed in the morning).  We can thus go across   the seventh row of Pascal's Triangle, muliplying entry k by k+2, to get

1*2+7*3+21*4+35*5+35*6+21*7+7*8+1*9=704.

 [/hide]

Official Solution

[hide="Solution"]

At any given time, the letters in the box are in decreasing order from top to bottom. Thus the sequence of letters in the box is uniequely determined by the set of letters in the box. We have two cases: letter 9 arrived before lunch or it did not.

* Case 1: Since letter 9 arrived before lunch, no further letter will arrive, and the number of possible orders is simply the number of subsets of $ T \equal{} 1,2,...,6,7,9$ which  might still be in the box. In fact, each subset of $ T$ is possible, because the secretary might have typed letters not in the subset as soon as they arrived and not typed any orders. Since $ T$ has 8 elements, it has $ 2^8 \equal{} 256$ subsets (including the empty set).

* Case 2: Since letter 9 didn't arrive before lunch, the question is: where can it be inserted in the typing order? Any position is possible for each subset of $ U \equal{} \{ 1, 2, ..., 6,7 \}$ which might have been left in the box during lunch (in descending order).  For instance, if the letters in the box during lunch are 6,3,2, then the typing order 6,3,9,2 would occur if the boss would deliver letter 9 just after letter 3 was typed. There would seem to be $ k\plus{}1$ places at which letter 9 could be inserted into a sequence of $ k$ letters.  However, if letter 9 is inserted at the beginning of the sequence (i.e. at the top of the pile, so it arrives before any after lunch typing is done), then we are duplicating an ordering from Case 1. Thus, if $ k$ letters are in the basket after returning from lunch, there are $ k$ places to insert letter 9 (without duplicating Case 1 orderings). Thus we obtain

\[ \sum^{7}_{k\equal{}0} k \binom{7}{k} \equal{} 7 (2^{7\minus{}1}) \equal{} 448\]

new ordering in Case 2.

Combining these cases gives 256+448 = 704 possible typing orders. [/hide]

[quote="Silverfalcon"]\[ \sum^{7}_{k \equal{} 0} k \binom{7}{k} \equal{} 7 (2^{7 \minus{} 1}) \equal{} 448
\]
[/quote]

Um sorry about reviving the thread, but how did you get from the summation to 2^6? Shouldn't it be 2^7? Why did you subtract 1 from the power?

[quote="#H34N1"][quote="Silverfalcon"]
\[ \sum^{7}_{k \equal{} 0} k \binom{7}{k} \equal{} 7 (2^{7 \minus{} 1}) \equal{} 448
\]
[/quote]

Um sorry about reviving the thread, but how did you get from the summation to 2^6? Shouldn't it be 2^7? Why did you subtract 1 from the power?[/quote]
It's because $ \sum^n_{k\equal{}1}k\binom{n}{k}\equal{}n2^{n\minus{}1}$. Can you prove it?

Oh right. I forgot about that. I was thinking that the k was constant....may be that I wasn't thinking.

Anyway its really easy to prove. A combinatorial approach is something of the following: $ n$ committee possibilities, picking a committee of $ k$ team members and designating a leader. The RHS first fixes a leader $ n$ and then out of the $ n\minus{}1$ people left to pick, it goes through the possible subsets, namely $ 2^{n\minus{}1}$.

Solution:

Note that when we consider after-lunch sequences, we may have any of the first 7 in it, or not in it. Thus we must choose some of them, but then note that letter $9$ can be typed at any time, after any letter. Thus if we have $m$ letters that have yet to be typed, then letter $9$ has $k+1$ possibilities.

Thus, our answer is: $\sum_{i=0}^{7}(\dbinom{7}{i}\cdot (k+1) $) $=$ $704$

$Note$: Many get confused when they try to do casework based off of what letters arrive before lunch time. This is not efficient. If anything, do your casework based off of what $gets$ $done$ before lunch time. By doing casework on what $arrives$, it's easy to overcount the case when letter $9$ is done right after the break and when it is done before lunch time, as you might think that it's being there before lunch time implies it must be typed before she takes a break, which is not the case.

Lesson for this problem: Be smart and decisive about what you make your cases. In this problem, we don't need cases at all. Just a solid hand on where our letter $9$ can go.


linear algebra
matrix
vector
algebra
linear equation
complex numbers

I was just hoping someone could confirm this for me

If $A$[b]$x$[/b]=[b]$b$[/b] is a system of $n$ linear equation in $n$ unknowns were $x_1$ and $x_2$ are both solutions.

if $x_1$ is not equal to $x_2$ , does this mean this system has infinite solutions?

You have $Ax_1=b$ and $Ax_2=b$, where $A$ is an $n\times n$ matrix and $x_1,x_2,b$ are $n\times 1$ column vectors.

What can you say about $A(px_1+(1-p)x_2)$, where $p$ is an arbitrary real number?

Well, I apologize because I'm not sure where your going with that, but wht

No matter what $p$ is equal to $A(px_1+(1-p)x_2) = b$

I'm taking a huge stab in the dark here, but does this imply that A is singular

If $A$ is not singular, e.g, invertible, then the unique solution to $Ax=b$ is $x=A^{-1}b$.  So yes, it is clear that $A$ must be singular.

Anyways, you are correct - if $Ax_1=Ax_2=b$, then $A(px_1+(1-p)x_2)=p(Ax_1)+(1-p)(Ax_2)=pb+(1-p)b=b$.

But $px_1+(1-p)x_2$ is also a column vector which is distinct from $x_1,x_2$ as long as $x_1\not=x_2$ and $p\not\in\{0,1\}$ (well, that depends on the field you are working over, but we'll work over the real or complex numbers to keep everything nice)

So we do have infinitely many solutions, geometrically speaking, which lie on the line joining $x_1,x_2$ in $\mathbb{R}^n$ (or $\mathbb{C}^n$).

[quote="blahblahblah"]But $px_1+(1-p)x_2$ is also a column vector which is distinct from $x_1,x_2$ as long as $x_1\not=x_2$ and $p\not\in\{0,1\}$ (well, that depends on the field you are working over, but we'll work over the real or complex numbers to keep everything nice) [/quote]
That statement at least is true in any field; the only issue is that we need the field to be infinite in order to conclude that these give infinitely many solutions.


videos geometry calculus MATHCOUNTS AMC AIME articles

America seems to be more focused on sports than on academics.

[quote]math154:
I think the main problem is that the majority of students are only/mainly interested in athletics and don't have time/don't want to focus on school. People just don't see the value in academics. And also, this is kind of a cultural issue: no offense to anyone, but I think lots of the US is heavily geared towards sports, unlike, for example, Asia, where academics are heavily valued.

Also, gifted students almost get no special attention at all. People only recognize those people with disabilities. Of course, those people need more care and stuff, but for gifted students, there's basically nothing. Sure, many schools have "gifted" programs, but in most schools, like mines, about 30-50% of the school is in the "gifted" program! The standards have been set so low that, for example, the average 5th grader barely knows more than the average 2nd or 3rd grader in several other countries.

...anyway, my rant is officially over.
[/quote]

Does anyone think that the U.S. should stop labeling the gifted students as nerds and geeks (At my school, we define nerd as the ones who are interested in good grades and geeks as the ones who are completly obssesed with video games) and start making them equal to the jocks. Although at my school, Roosevelt Middle School in Eugene, Oregon, we do equate nerds with the jocks (Sometimes they overlap :lol: ). I think the problem is that No Child Left Behind has lowered our standards on grade levels. If we want to raise the standards, we need to improve No Child Left Behind to  increase standards and not alienate the gifted students.

On the part about 2nd and 3rd graders in other countries knowing more than a 5th in the U.S., I believe it maybe because of cultural values. A person, A, is from country X believes a C in the US is a fail. Another person,B, is from country Y believs that a D in the US is a pass. Let's say that both move to the US and go to the same school. Not surprisingly, A is a G & T (Gifted and Talented) student and B is failing all of his/her classes. When they go to college, A is accepted into an Ivy League school and B doesn't even go to college, because he/she was not accepted into anything, because of his/her grades. When they grow up, A becomes president and B doesn't have a job or works at a gas station. The worst part is A will be remembered forever (By forever, I mean as long as the human race lasts) :lol:  , but B leaves no trace of him/herself :( . This how much culture can influence you.

I am also sick of hearing people in my own math class praise how I am a genius. I am in Geometry and Algebra 2 right now. I was sick of hearing people praise me in 6th grade when I was in Algebra 1 that being a 6th grader in Algebra made me a genius, and then they forget that there were six 6th grader in the class. Now people praise me for being in Algebra 2 as well as Geometry, but there were four 7th graders in Algebra 2. Last year, there were four 8th graders in Pre-Calculus and one in Honore's Calculus plus a 6th grader who was in Geometry. If you want to praise someone for their ability in mathematics, you should praise them, not me. The only difference between everyone else in Algebra 1 last year, was that I made it to State in Mathcounts with all of the 8th graders in Pre-Calculus. I didn't even do well in the AMCs, I got 18 on the AMC 8, 72 on 10 and I could have taken the competition, but, I forgot. Now, I have to work hard to get into the Mathcounts team (Preferably the A team, the more advanced of the two teams) and qualify for the AMC 10 and 12.

Does anyone else have the same problem?

My response to the first question you gave is yes: I don't think students should be labeled in a negative way just because they are gifted in some sort of "extraordinary" way.  However, I think these labels are made in jealousy.

Yes, we should raise the standards.  I believe the US is 13th (?) in education standards, and as the United States is sort of like a "leadership country", we definitely need to step it up a notch.

Hmmm... for me, a B is fail.  In my culture (by the way, I'm Indian), we expect no less than perfect.  This may come from India's rigorous education (which let's you go straight to medical school if you want to become a doctor, by the way).

Obviously, you have some very smart people in your class.  However, in my school (private), they obviously have not seen anyone like me (please excuse me if you think I'm bragging, but I'm not).  So I can't really relate to this situation.

My answer for the question posed in the subject is, like ernie, yes. It is because of cultural reasons; just look at a standard movie. The people who earn good grades are labeled "nerds" or "geeks" and are portrayed as glasses-wearing, computer-savvy, short kids with big teeth and messy hair who lack social skills and stammer all the time. Meanwhile, the athletes are shown as strong, happy, kids with tons of friends.

In my school, people think I'm a genius, which is definitely far from the truth. I take Geometry in 7th grade, which I see nothing special about, because so many other AoPSers do the same. However, no one does it in my school, so everyone thinks I'm so great, which makes me uncomfortable. Anyway, there's nothing fantastic about taking math early - how good you are with problem solving is what truly matters.

[quote="isabella2296"]In my school, people think I'm a genius, which is definitely far from the truth. I take Geometry in 7th grade, which I see nothing special about, because so many other AoPSers do the same. However, no one does it in my school, so everyone thinks I'm so great, which makes me uncomfortable. Anyway, there's nothing fantastic about taking math early - how good you are with problem solving is what truly matters.[/quote]

Yes, I experience the same thing. If someone asks me, "Are you smart?" I usually reply no, but they still assume I am anyway. I am FAR from "smart" in terms of problem solving.

[quote="ernie"]
Hmmm... for me, a B is fail.  In my culture (by the way, I'm Indian), we expect no less than perfect.  This may come from India's rigorous education (which let's you go straight to medical school if you want to become a doctor, by the way).
[/quote]
Yes, I'm also Indian, and B is epic fail.

I consider anything less than a 100 an epic fail.

I'm Chinese and Polish, but since I was around 8 or so I was mostly self-pressured, rather than pushed by my parents.

Izzy's polish?
:p

B+ = phail
I have all A's, and a-'s, and my parents are mad for only have 2 A+'s...

Izzy, the thing you said about "people think ur genius, far from the truth" and the rest of that, that all applies to me also.

If there was more problem solving on standardized tests, first we would need to build a base up.  For example, maybe next years first grade curriculum could start a little, then the next year second grade, then 3rd, etc.  But they couldn't just start it, say, in 8th grade, when everybodys' already used to the old "exercises".

As I've said before, most kids aren't motivated enough, and there's no hope in trying to improve current education for older kids (say, middle school and above). I think the best way is to improve the fundamentals when children are really small, which would allow them to learn much more easily in the future. For example, today, my Algebra II teacher was trying to explain how $ a^{\minus{}n}\equal{}\frac1{a^n}$. Kids did not understand, so he didn't explain, and just told a random analogy. After everybody understood, he moved on to a more "complicated" problem, like $ \minus{}3a^{\minus{}n}$. The people in the class thought it was $ \frac1{3a^n}$...an example of the importance of fundamentals.

[quote="Yongyi781"]These experiences especially hold true for me. In Bedford, which is one of the more athletically-oriented schools in Massachusetts, almost every person I meet who is not my personal friend asks me "Are you a super genius? Do you take calculus at high school? What's 287 * 96?" I find absolutely no reason to continue the conversation, since the average person (athletic or not) views "smart at math" as "being able to do arithmetic fast."

The original purpose of the No Child Left Behind act was to build a "bottom-up" foundation in academics, but what it ended up doing was that it left out the gifted people and didn't really improve the lower end. Even my little brother (who is in 6th grade) commented that the 10th grade MCAS was too easy. We need to, like someone above me said, "kick the standards up a notch." The hardest MCAS problem for 8th grade was a simplistic use of the Pythagorean Theorem.

What are your opinions on whether standardized tests should focus more on problem solving?[/quote]
omg, i can sooo relate to that! When i went to india for summer of 4th grade and i had started MC, people knew that i was pretty good at math (for my age). So, my friend asks me 2^10 and i yell at him for like 1 hour continiously and now he understands what math actually is btw, he's still my friend even though i yelled at him :)

And answering your poll, the average grade in my school is $ \approx 85\%$.

EDIT: How is that spam???

[quote="gauss1181"]how can you start MC in 4th grade?[/quote]
unofficially.

My problem with those people who label us (As in people in math competitions) as genii is that they asking me questions about geometry, when there are equally capable people out there who are in Geometry who can explain just as well as I do. As my username suggests, I am Chinese, and to me (And my parents), B is a fail. I am so sick of people saying that I am a genius, because I went the extra mile and competed in mathcounts. If you want genius, go find someone who understands Albert Einstein or Isaac Newton or Stephen Hawking's theories, including the mathematics.
I am also annoyed if someone asks me a (mediocre) aritmetic question. I always think that it is how much you can do, not how fast you do it that counts, although there is a limit to that rule. I know most of the time I comply and answer it, but I sometimes ignore the rediculous ones, such as 1037*4937 (This is an exageration and people actually rarely ask me). I fit perfectly with the classic nerd, I wear glasses (Sometimes), I have large front teeth, but the difference is that I don't obsess about computers, and I don't have braces. What does anyone think about people calling you a genius?

no he didn't... nobody patents actions. people only patent things.

[quote="1=2"][quote="Poincare"]And, 1=2, Angle Bisector Theorem is definitely not complex, try proving let's see.[/quote]

[hide="crucial step in here"]Try extending a segment in a two-column proof...[/hide][/quote]

Simple: just say something like "Point A' exists|Refer to diagram." It worked when I was in geometry, so... :rotfl:

[quote="AIME15"]I usually say:

Draw Point A such that... | Through any two lines there is at most one intersection[/quote]

Yeah, my geometry teacher told me this method was "proper".

[quote="Poincare"]And, another problem is that people think the education ends once school ends. For example, I was trying a medium difficulty problem about the extended Law of Sines and this eight grader looks at the diagram i am drawing and says I can do that that problem since I've done Geometry. So, I'm like sure, try it. And she stares at it and says "Wtf is the extended of law of sines?" I mean they could at least they could mention there is more interesting math than they teach in school??![/quote]

Couldn't have said it better myself.  If people think you're good at math, they immediately think you've remembered countless numbers of formulas.  Then when they see a [b]real[/b] hard problem, they think that they just haven't learned the right formulas yet.  Little do they know that many USAMO problems can be done with simple algebraic steps.  It's the creativity that matters, but that's a quality that just isn't valued in math class, or school in general for that matter.

Yeah, one thing that really distinguishes strong math countries from the USA is that they tend to emphasize the conceptual ideas and the big picture so that kids have a thorough understanding, as opposed to a surface understanding. This essentially expands students' flexibility and creativity, allowing them to do much more.

For the average joe, a USAMO problem may take a VERY long time to solve, if at all. Compared to the 1-second excercises that teachers have students do, USAMO problems, and any other contest problems (even AMC 12), take a much longer time. Teachers don't like this, because, by law, they have to teach to the curriculum. To teach to the curriculum, they have to complete the state standards for the class by the end of the year. Teachers want to be able to complete the standards, so they have no time giving creativity-requiring, meaningful problems. This is where the problem lies, not in the teachers, but in the constraints, such as standards and what-not, that the law imposes on them. The standards are self-destructive: while trying to learn so many things in so many specific and random ways (such as 2 column proofs.. my school had 3 column proofs agh), students are really not learning anything because:
1) The basics are not there to build anything upon
2) There is no mastery because of the lack of non-textbook excercises

[quote="ZzZzZzZzZzZz"]If people think you're good at math, they immediately think you've remembered countless numbers of formulas.  Then when they see a [b]real[/b] hard problem, they think that they just haven't learned the right formulas yet.  Little do they know that many USAMO problems can be done with simple algebraic steps.  It's the creativity that matters.[/quote]

This is so true; even with me! Last night I was in the WOOT classroom puzzling over [[2006 AIME I Problems/Problem 15]], and I wasn't making any progress on it, when somebody just squared both sides of the recurrence formula. I got it in a few seconds after that. It isn't that hard, it's just creativity that is needed.

I think what the problem is is that not enough time is spent explaining [i]why[/i] things are true, not just they're true.  It gives a broader understanding that allows a lot more promise of solving harder problems and comprehending harder material.

This is all that should be needed to re-explain everything we have talked about so far. My math teacher said the following today:
"(just decided the plan on what they should do on a geo problem)now that we have figured out how to do it, its going to be even harder to do it really!" Once you've figured out the right plan the problem's a breeze! THE EMPHASIS ON THE PROBLEM-SOLVING, NOT THE ARITHMETIC.

Perhaps not only the material, but the formality of the problems and methods given need to be improved.  In my high school my math teacher has a doctorate from Princeton, and he is $ A LOT$ more lacksidasical with how he teaches: he uses variables that have random symbols, likes trees or planes or something, and he uses words as point names, for example.  The problems and the solutions given and the ways in which math is taught in many high schools throughout the country make it a very "boring" subject to many when, in fact, it's easily the most beautiful subject of all.

The boring part comes from all the memorizing the teachers make you do. They don't really encourage you to find out why the rules work, or any of the logic behind it. Memorizing may be useful, but if you don't understand why the rule works, you won't be learning nearly as much.

And the annoying part about it is that it takes maybe one class at most to teach WHY something works, a class that will take away many questions that the kids have and, essentially, will make it a more efficient math class.

Please remember that this forum is for math teachers to discuss issues regarding mathematics. Students should use Classroom Math, High School Basics, or other appropriate forum to discuss these topics.

For those of you wishing better math instruction, I recommend going to a school for the highly gifted if your family can at all afford the tuition or if there is a good magnet school near you. AoPS also offers many classes which may be appropriate for you.

idea: move to round table. more efficient discussion there. :)


Let $\angle XOY = \frac{\pi}{2}$; $P$ is a point inside $\angle XOY$ and we have $OP = 1; \angle XOP = \frac{\pi}{6}.$ A line passes $P$ intersects the Rays $OX$ and $OY$ at $M$ and $N$. Find the maximum value of $OM + ON - MN.$

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

[hide="Remark"]
- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917
- second link
etc.
[/hide]

[b]Happy Problem-Solving![/b] :)

[quote="orl"]Let $\triangle XOY = \frac{\pi}{2}$[/quote]

Is this?

I think you mean $\angle XOY$ and $\angle XOP$, instead of $\triangle XOY$ and $\triangle XOP$.

Does anybody know how to solve this problem? It seems to have a very ugly answer based on calculus.

[b]Lemma 1:[/b] The inradius of $\triangle MON$ is $\frac{1}{2}(OM+ON-MN)$.

[b]Proof:[/b] Let the incircle centered at $I$ meet $MO,ON,MN$ at $A,B,C$ respectively. Note that $AOBI$ is a square of sidelength $r$. Also note $MA=MC$ and $NB=NC$, hence $OM+ON-MN=MA+OA+NB+OB-MC-NC=OA+OB=2r$. The lemma follows. $\square$

---------------------
From lemma 1 it suffices to maximize $2r$. Denote the line through $P$ as $\ell_P$. We use Cartesian Coordinates. Set $P=\left (\frac{1}{2},\frac{\sqrt{3}}{2}\right )$, and $I=(r,r)$.
------------------------------
[b]Lemma 2:[/b] We must have $r\leq\sqrt{3}$.

[b]Proof:[/b] Recall that $I$ must be inside $\triangle MON$. Let the point where $\ell_P$ intersects the line $y=x$ be $X$. clearly the furthest this point can be from the origin is $\sqrt{6}$ giving $r\leq\sqrt{3}$. $\square$

-----------------------------
Let $d(A,B)$ denote the shortest distance between a point on $A$ and a point on $B$. Now from definition of shortest distance between a point and a line we have $r=d(I,\ell_P)\leq d(I,P)$. Now we have

\begin{align*} d(I,P)&\geq r \\ \left(r-\frac{1}{2}\right)^2+\left (r-\frac{\sqrt{3}}{2}\right )^2&\geq r^2 \\ (2r-1)^2+(2r-\sqrt{3})^2-(2r)^2 &\geq 0 \end{align*}

Solving the quadratic inequality gives $2r\geq 1+\sqrt{3}+\sqrt[4]{12}\geq 4$ or $2r\leq 1+\sqrt{3}-\sqrt[4]{12}$. We have shown that $r\leq \sqrt{3}<2$ so the first case is impossible and thus the maximum value is $\boxed{1+\sqrt{3}-\sqrt[4]{12}}$ which is easily obtained through construction.

It is possible to achieve the quadratic without the use of coordinates in a slightly different manner. In particular define points as above. The $IP^2=IO^2+OP^2-2(IO)(OP)cos(15^\circ)$ or using that $IP\le r$ then $r^2\le1+2r^2-2\sqrt{2}rcos(15^\circ)$ and the above follows.
However an easier way to prove that $r\le\frac{\sqrt{3}}{2}$ is to let $R$ be the projection of $I$ on $OX$ and $Q$ be the projection of $I$ onto $OY$. Suppose that $r>\frac{\sqrt{3}}{2}$, then $P$ is contained inside $ORIQ$, which is obviously a contradiction.


My friend gave me this problem and I was able to solve it, but I thought I should post it here so that others can learn from it.

Evaluate $\sum_{k=1}^{\infty}\frac{k}{(k+1)(k+2)^2}$

The only snag on this problem is that you will have to know a reasonably well known sum.

[hide]

Using partial fractions, the equation becomes

The first pair of terms telescope leaving simply . The second, using the fact that , it becomes . The sum of these two is the answer, which is .[/hide]


geometry
circumcircle
incenter
angle bisector
geometry proposed

[b][size=100][color=DarkBlue]A triangle $ \bigtriangleup ABC$ is given and let $ A'$ be, the reflexion of $ A,$ with respect to the midpoint $ M$ of its side-segment $ BC.$ The line through $ A'$ and parallel to the internal angle bisector of $ \angle A,$ intersects the circumcircle $ (K)$ of the triangle $ \bigtriangleup BIC,$ where $ I$ is the incenter of $ \bigtriangleup ABC,$ at point so be it $ P.$ Prove that $ BD \equal{} DE \equal{} EC,$ where $ D\equiv AB\cap CP$ and $ E\equiv AC\cap BP.$[/color][/size][/b]

Kostas Vittas.

Assume  there exists the points $ D$ and $ E$ on $ BA$ and $ AC$ such that $ BD \equal{} DE \equal{} EC$ and $ P' \equiv BE \cap CD.$ We'll prove that $ P$ and $ P'$ are necessarily identical. It's easy to see that $ P'$ lies on $ \odot(IBC),$ due to

$ \angle DP'B \equal{} \angle EDC \plus{} \angle DEB \equal{} \frac {_1}{^2}(\angle AED \plus{} \angle ADE) \ \Longrightarrow$

$\angle DP'B \equal{} \frac {_1}{^2}(\pi \minus{} \alpha) \equal{} \pi \minus{}\angle BIC$ $ \Longrightarrow \ BIP'C$ is cyclic.

On the other hand, it is well-known that the locus of the intersection of the diagonals of the convex quadrilateral $ BD'E'C$ satisfying $ BD' \equal{} E'C$ is the A'-angle bisector of the antimedial triangle $ \triangle A'B'C',$ which is the parallel line to $ AI$ from $ A'.$ This was discussed in the topic [url=http://www.artofproblemsolving.com/viewtopic.php?t=300736]plane geometry 2[/url], hence $ P' \equiv P$ and the proof is completed.

[b]Remark:[/b]

[i][b][color=blue]If $ \widehat {BAC} \equal{} 60^\circ$, then $ P \equiv O$, where $ O$ is the circumcenter of $ \triangle ABC$[/color][/b][/i].

Best regards,
sunken rock


topology
advanced fields
advanced fields unsolved

Prove that two metric spaces are homeomorphic iff their [url=http://planetmath.org/encyclopedia/StoneVCechCompactification.html]Stone-Cech compactifications[/url] are homeomorphic.

The same is true for completely regular spaces, wich satisfy the first axiom of countability at all points.