geometry
parallelogram
trigonometry
angle bisector
geometry proposed

Let BC=a, AC=b, AB=c and $p=\frac{a+b+c}2$.
Let $\{ R \} =AX \cap BC$.
BKL and BMN are isosceles triangles and it is easy to get:
AM=BL=BK=p-c
KC=p-b
CN=p-a
KN=KC+CN=c
ML=p-c+c+p-c=a+b
$\angle BLK=\frac{\angle ABC}2$
$\angle BMN=90^o-\frac{\angle ABC}2$
So $MX \bot LX$.

KNN' is isosceles, so KN'=c and $\angle NKN'=2\cdot \angle XKN=\angle ABC$. So KN'||AB and KN'=AB=c, so ABKN' is a paralelogram, so AN'||BC and AN'=BK=p-c.

AN'||NR and X is the middle of NN' implies AN'RN is a paralelogram, so NR=AN'=p-c. So CR=CN+NR=p-a+p-c=b. So ACR is isosceles and CX is the median, so CX is the angle bisector of ACR.

Amfulger, what is the point $N'\ ?$[hide="Here is a short metrical solution."]
$LB=AM=p-c\ ;\ KB=p-c\ ;\ KC=p-b\ ; CN=p-a\ ;$
$KN=KC+CN=c\ ;\ LM=LB+BA+AM=a+b\ ;\ I_bM=I_bN=r_b\ .$
$m(\widehat {CI_b N})=\frac C2\ ;\ m(\widehat {CI_b M})=180^{\circ}-B-\frac C2\ .$ Denote $Y\in MN\cap CI_b\ .$

$\blacktriangleright$ Apply the Menelaus' theorem to the transversal $\overline {LKX}$ and the triangle $BMN\ :$
$\frac{LB}{LM}\cdot \frac{XM}{XN}\cdot \frac{KN}{KB}=1\Longrightarrow \frac{p-c}{a+b}\cdot \frac{XM}{XN}\cdot \frac{c}{p-c}=1\Longrightarrow \boxed {\frac{XM}{XN}=\frac{a+b}{c}}\ \ (1)\ .$

$\blacktriangleright$ In the triangle $MI_bN$ apply to the cevian $I_bY$ a well-known relation :
$\frac{YM}{YN}=\frac{I_bM}{I_bN}\cdot \frac{\sin \widehat {MI_bY}}{\sin \widehat {NI_bY}}=\frac{\sin \widehat {MI_bC}}{\sin \widehat {NI_bC}}=\frac{\sin (180^{\circ}-B-\frac C2)}{\sin \frac C2}=\frac{\sin \left(B+\frac C2\right)}{\sin \frac C2}=$
$\frac{2\cos \frac C2\sin \left(B+\frac C2\right)}{2\sin \frac C2\cos \frac C2}=\frac{\sin A+\sin B}{\sin C}=\frac{a+b}{c}\Longrightarrow \boxed {\frac{YM}{YN}=\frac{a+b}{c}}\ \ (2)\ .$

$\blacktriangleright$ From the relations $(1)$, $(2)$ results $X\equiv Y$ because $\{X,Y\}\subset (MN),$
$\frac{XM}{XN}=\frac{YM}{YN}\ .$ Therefore, the ray $[CX$ bisects the angle $\widehat {ACN}\ \ \mathrm {q.e.d.}$[/hide]

I have a nice synthetic proof based on the [i]lemma[/i]:

[i]Let $ABC$ be a triangle and $D$, $E$, $F$ the points where the incircle cuts $BC$, $CA$, respective $AB$ and $P \in BI \cap EF$. Thus, $BP \perp CP$.[/i]

Denote $L \in EF \cap BC$, but the triangles $\triangle{BPE}$ and $\triangle{BPD}$ are congruent, so $\angle{BPL}=\angle{BPD}$. $(1)$

On other hand $(LBDC)=-1$, so from $(1)$ we have $BP \perp CP$.

Returning to the initial problem, it is easy to see that $\angle{ALK}=\frac{180-\angle{LBK}}{2}=\frac{B}{2}$, so $BI \| LK$. $(2)$.

Similar, $I_{a}I_{c}\| QR$, where $Q,R$ are the points where the incircle cuts $AC$, respective $BC$.

Now from the lemma, $BP \perp CP$, where $P \in BI \cap QR$, thus $\angle{BAP}=90-\frac{\angle{B}}{2}$.

But in the isosceles triangle $\triangle{BMN}$, we have $\angle{BMN}=90-\frac{\angle{B}}{2}$, so $MN \| AP$, and it is clear now that $LX \perp MX$.

Now denote $X' \in LK \cap I_{a}I_{b}$ and from the lemma, it results that $LX' \perp MX'$, but $LX \perp MX$, so $X \equiv X'$, i.e $X \in I_{a}I_{b}$, so $CX$ bisects $ACN$.

[quote="Darij Grinberg"][color=darkred]The $A$-excircle of a triangle $ABC$ touches the side $BC$ at the point $K$ and the extended side $AB$ at the point $L$. The $B$-excircle touches the lines $BA$ and $BC$ at the points $M$ and $N$, respectively. The lines $KL$ and $MN$ meet at the point $X$. Show that the line $CX$ bisects the angle $ACN$.[/color][/quote]
[color=darkblue][u][b]Proof I (resumption).[/u][/b] Suppose w.l.o.g. that $a>b$. Denote $\{\begin{array}{c}S_{1}\in MN\cap LK\\\\ S_{2}\in MN\cap CI_{b}\\\\ V\in BA\cap CI_{b}\end{array}$. Observe that : $\{\begin{array}{c}LB=MA=KB=p-c\\\\ KN=KC+CN=(p-b)+(p-a)=c\\\\ LM=LB+BM=(p-c)+p=a+b\end{array}$

and $\frac{VA}{b}=\frac{VB}{a}=\frac{c}{a-b}$ $\implies$ $VB=\frac{ac}{a-b}$ $\implies$ $VM=VB-MB=\frac{ac}{a-b}-p=\frac{(a+b)(p-a)}{a-b}$.

$1\blacktriangleright$ Apply the Menelaus' theorem to the transversal $\overline{LKS_{1}}$ in the triangle $BMN$ :

$\frac{LB}{LM}\cdot\frac{S_{1}M}{S_{1}N}\cdot\frac{KN}{KB}=1$ $\implies$ $\frac{p-c}{a+b}\cdot\frac{S_{1}M}{S_{1}N}\cdot \frac{c}{p-c}=1$ $\implies$ $\boxed{\frac{S_{1}M}{S_{1}N}=\frac{a+b}{c}}\ \ (1)$.

$2\blacktriangleright$ Apply the Menelaus' theorem to the transversal $\overline{VS_{2}C}$ in the same triangle $BMN$ :

$\frac{VM}{VB}\cdot\frac{CB}{CN}\cdot\frac{S_{2}N}{S_{2}M}=1$ $\implies$ $\frac{\frac{(a+b)(p-a)}{a-b}}{\frac{ac}{a-b}}\cdot\frac{a}{p-a}\cdot\frac{S_{2}N}{S_{2}M}=1$ $\implies$ $\boxed{\frac{S_{2}M}{S_{2}N}=\frac{a+b}{c}}\ \ (2)$.

From the relations $(1)$ and $(2)$ results that $\frac{S_{1}M}{S_{1}N}=\frac{S_{2}M}{S_{2}N}$ $\implies$  $S_{1}\equiv S_{2}\equiv S\in MN$, i.e. the lines $LK$, $CI_{b}$ are concurrently in the point $S\in MN$ for which $\frac{SM}{SN}=\frac{a+b}{c}$.[/color]

[color=darkred][b][u]Lemma.[/u][/b] Let $XYZ$ be a right triangle for which $XY\perp XZ$. Denote the projections $U$, $V$ of the points $Y$, $Z$ respectively to a mobile line $d$ for which $X\in d$.

Define the point $W$ for which $WU\perp XY$ and $WV\perp XZ$. Then $W\in YZ$.[/color]

[color=darkblue]Indeed, $\{\begin{array}{c}x=m(\widehat{UXY})\\\\ W_{1}\in YZ\ ,\ UW_{1}\perp XY\\\\ W_{2}\in YZ\ ,\ VW_{2}\perp XZ\end{array}\|$ $\implies$ $\{\begin{array}{c}W_{1}Y=YZ\sin^{2}x\\\\ W_{2}Z=YZ\cos^{2}x\end{array}\|$ $\implies$ $W_{1}Y+W_{2}Z=YZ$ $\implies$ $W_{1}\equiv W_{2}\equiv W\in YZ$.

Otherewise, denote $\{\begin{array}{c}R\in UW_{1}\cap XY\\\\ S\in VW_{2}\cap XZ\end{array}\|$ $\implies$ $\triangle UXY\sim\triangle VZX$ $\implies$ $\frac{UY}{UX}=\frac{VX}{VZ}$. Therefore,

$\{\begin{array}{c}\frac{W_{1}Y}{W_{1}Z}=\frac{RY}{RX}=(\frac{UY}{UX})^{2}\\\\ \frac{W_{2}Y}{W_{2}Z}=\frac{SX}{SZ}=(\frac{VX}{VZ})^{2}\end{array}\|$ $\implies$ $\frac{W_{1}Y}{W_{1}Z}=\frac{W_{2}Y}{W_{2}Z}$ $\implies$ $W_{1}\equiv W_{2}\equiv W\in YZ$.

[b][u]Proof II.[/u][/b] Apply the above lemma to the right triangle $BI_{a}I_{b}$ and the line $d\equiv BA$.[/color]

Another solution of this problem, has already been posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=164092/post#9

kostas vittas.


Given, $S=x^{3}y^{2}z$
and $x+y+z=1$
Determine the greatest value of S.
How do we start this type of question?
I started with
$S=xxxyyz$ where there are six terms;
used AM-GM and found that x=y=z, I know it's wrong, but why?
Then I think it occurs because there's no limits of them.
Therefore, I sub $z=1-x-y$
So $S=(1-x-y)(x^{3}y^{2})$
But there's no clear solution for me.
What should I do next?

Yes. It is correct.
But, why is my method totally unworkable?


This is what I have. When I compile, it says "ERROR: \begin{align*} allowed only in paragraph mode. What can I do to get align to work?
\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\begin{document}
\section{Problem 1}
\subsection{Part a}
$ \$$\begin{align*}
x^2+2x+9&=&x^2+2x+1-1+9\\
&=&(x+1)^2+8
\end{align*}$ \$$

Get rid of the dollar signs -- you don't need them in documents, only on the forum.  ("Paragraph mode," as opposed to "math mode.")

You also have too many alignment characters: it should be
\begin{align*}
x^2+2x+9 & = x^2+2x+1-1+9\\
&= (x+1)^2+8
\end{align*}


Hey did any other schools do team mandelbrot today? Gunn did. I thought it was pretty interesting.
Our A team had ccy, jungii, krustyteklown, and brian zhang (crazy pro people...), B team was me and some other people you won't know about ;) I thought it was interesting, with the mods and stuff although I sucked and failed. Jungii managed to write a proof for number 6 so they got perfect supposedly, we didn't get number 6 so we failed.

Our A team faildeathed. B team probably beat A team and C team probably tied A team. In our defense, we were all pretty sleep-deprived at the time. Hopefully we'll all be a little less silly next time.


number theory
greatest common divisor
modular arithmetic
number theory unsolved

Find all triples ($a,b,n$) such that:
$(a^3+b^3)^n=4(ab)^{1995}$

I assume $a$, $b$ and $n$ are integers satisfying the equation

$(1) \; (a^3 + b^3)^n = 4(ab)^{1995}$.

Observe that $ab=0$ iff $a^3 + b^3 = 0$ by (1), implying $ab=0$ iff $a=b=0$.  Hence  $(a,b,n)=(0,0,n)$, where $n \neq 0$, is a solution of (1).
Assume $ab \neq 0$.
If $|a^3 + b^3| = 1$, then $|(a^3 + b^3)^n| = 1$ and $|4(ab)^{1995}| \geq 4$.  Hence $|a^3 + b^3| \neq 1$ by (1), implying $|a^3 + b^3| \geq 2$, which according to (1) implies $n>0$.
Let $d=(a,b)$.  Then there exist two nonzero coprime integers $s$ and $t$ s.t. $a=sd$ and $b=td$, which inserted in (2) give us

$(2) \: d^{3n}(s^3 + t^3)^n = 4d^{3990}(st)^{1995}$.

[b]Case 1:[/b] $n \geq 1330$.  Then by (2)

$d^{3(n-1330)}(s^3 + t^3)^n = 4(st)^{1995}$.

The fact that $GCD(st,s^3+t^3)=1$ implies $(s^3+ t^3)^n$ is a divisor of 4.  If $|s^3 + t^3| > 1$, then $|(s^3+ t^3)^n| \geq 2^{1330}$ since $n \geq 1330$.  Hence $|s^3 + t^3| = 1$, i.e. $|s + t| \cdot |s^2 - st + t^2| = 1$.  Therefore $|s + t| = |s^2 - st + t^2| = 1$, implying $1 = |(s+ t)^2 - 3st| = |1 - 3st|$, meaning $st=0$ or $3st=-2$, which both are impossible.  Hence (1) has no solution in this case.

[b]Case 2:[/b] $n < 1330$.  Then by (2)

$(3) \; (s^3 + t^3)^n = 4d^{3(1330-n)}(st)^{1995}$.

Consequently $st$ is a divisor of $(s^3 + t^3)^n$, implying $st = \pm 1$ due to the fact that $GCD(s^3+t^3,st)=1$.  Therefore $|s| = |t| = 1$.  If $s \neq t$, then $s^3 + t^3 = (\pm 1)^3 + (\mp 1)^3 = 0$, implying $st=0$.  This contradiction implies $s = t = \pm 1$, meaning $st=1$ and $s^3 + t^3 = \pm 2$, which inserted in (3) result in

$(4) \; (\pm 2)^n = 4d^{3(1330-n)}$.

Thus $d = 2^k$ for a non-negative integer $k$, which inserted in (4) gives

$(5) \; (\pm 2)^{n-2} = 2^{3k(1330-n)}$.

Observe that $n$ odd implies $s=t=1$, i.e. $a,b>0$.  Futhermore (5) implies $n-2 = 3k(1330-n)$.  Hence

$(6) \; 3k + 1= \frac{2^4 \cdot 83}{1330 - n}$,

where $n>2$.  Hence

$(7) \; 1330 - n = 2^{4-u} \cdot 83^{1-v}$,

where $u \leq 4$ and $v \leq 1$ are non-negative integers, which inserted in (6) result in

$(8) \; 3k + 1 = 2^u \cdot 83^v$.

According to (8)

$1 \equiv 2^u \cdot 83^v \equiv (-1)^u \cdot (-1)^v = (-1)^{u+v} \!\! \pmod{3}$,

implying $u+v$ is even.  Combining this fact with (7) and (8) we end up with the following possibilities:

$\bullet \; u=0$, $v=0 \;\; \Rightarrow \;\; n = 2, \: k = 0  \;\; \Rightarrow \;\; (a,b,n)=(\pm 1, \pm 1, 2)$.

$\bullet \; u=2$, $v=0 \;\; \Rightarrow \;\; n = 998, \: k = 1  \;\; \Rightarrow \;\; (a,b,n)=(\pm 2, \pm 2, 998)$.

$\bullet \; u=4$, $v=0 \;\; \Rightarrow \;\; n = 1247, \: k = 5  \;\; \Rightarrow \;\; (a,b,n)=(2^5, 2^5, 1247)$.

$\bullet \; u=1$, $v=1 \;\; \Rightarrow \;\; n = 1322, \: k = 55  \;\; \Rightarrow \;\; (a,b,n)=(\pm 2^{55}, \pm 2^{55}, 1322)$.

$\bullet \; u=3$, $v=1 \;\; \Rightarrow \;\; n = 1328, \: k = 221  \;\; \Rightarrow \;\; (a,b,n)=(\pm 2^{221}, \pm 2^{221}, 1322)$.

Thus (1) only has these nine (non-trivial) solutions and the (trivial) solutions $(a,b,n)=(0,0,n)$, where $n \neq 0$.


modular arithmetic
induction
number theory
relatively prime

Prove that if $ k\ge 2$, then $ 16^k\minus{}1$ has a prime factor that is greater than $ 10$.

It suffices to show this for primes. For $ k\equal{}3,5,7$ verify by hand.

$ 16^k\minus{}1\equal{}(3*5\plus{}1)^k\minus{}1\equal{}...\binom{k}{2} (3*5)^2\plus{}k*3*5$ So if $ 3^2$ divides that, then $ 3|k$, $ k$ is prime, so $ k\equal{}3$. Similar with $ 5$.

Now we just need to show that there are no solutions to $ 16^p\minus{}1\equal{}15*7^m$.

I don't know...

Well, if you consider $ 16^p \minus{}1 \equal{} 15 \cdot 7^m$ in mod 7, then you get $ 2^p \equiv 1 \pmod 7$. And from there, you know that if there is a $ p$, it's divisible by 3...

My friend and i found this proof, dunno whether its right so plz check it :)
[hide="solution"]
Clearly $ 16^k \minus{} 1$ isnt divisible by 3 and 7 for all $ k \geq 2$ . So we need to prove that there exists another prime factor of it other than 5.
$ 16^k \minus{} 1 \equal{} (4^k \plus{} 1) \cdot (4^k \minus{} 1)  \implies$ it suffices to prove that $ 4^k \plus{} 1 \neq 5^m$ for $ k \geq 2, m \in \mathbb{N}$
To prove the claim:
We can clearly see that for $ 4^k \plus{} 1 \equal{} 5^m, k \equal{} 2t \plus{} 1$ if that is possible.
This yields $ 2^{2(2t\plus{}1)} \plus{} 1 \equal{} 5^m \implies 4 \cdot (2^t)^4 \plus{} 1^4 \equal{} 5$
By Sophie germain factorisation, $ (1 \plus{} 2 \cdot 4^t \plus{} 2 \cdot 2^t)(1 \plus{} 2 \cdot 4^t \minus{} 2 \cdot 2^t) \equal{} 5^m$
$ \implies$ we need to have both the factors (a and b where a and b are the two factors resp) as powers of 5 other than 1
Now take cases for t like
when $ t \equiv$ 1 or 2 $ \mod 4$, a $ \not \equiv 0 \mod 5$
when $ t \equiv$ 3 or 0 $ \mod 4$ ,b $ \not \equiv 0 \mod 5$

Hence the claim is proved :)

PS :Is there a better method for proving that there exists no soln for the eqn $ 4^k \plus{} 1 \equal{} 5^m$ for $ k \geq 2, m \in \mathbb{N}$ ?
[/hide]

I think I have got a relatively easy solution but I am not sure whether its right. Here it is, please check for any error.
Firstly for even 'k' the result is true.
And also note that its enough if we show that the result is true for prime 'k'.
For a prime value of 'k' , 2 or 7 cannot divide the given expression. Its clear that 2 cannot divide. I got that 7 cannot divide by writing down the powers of 16 modulo 7, keeping in mind that the exponent is a prime.
Also we can show that 25 or 9 cannot divide the expression by similarly writing the powers of 16 modulo 25 and 9 respectively. Now the proof is complete..(?) Is this right??

[quote="madness"]Is there a better method for proving that there exists no soln for the eqn $ 4^k \plus{} 1 \equal{} 5^m$ for $ k \geq 2, m \in \mathbb{N}$ ? [/quote]

You could cite [url=http://en.wikipedia.org/wiki/Mih%C4%83ilescu%27s_theorem]Mihailescu's theorem[/url].  :)

[quote="madness"]My friend and i found this proof, dunno whether its right so plz check it :)
[hide="solution"]
Clearly $ 16^k \minus{} 1$ isnt divisible by 3 and 7 for all $ k \geq 2$ .
[/hide][/quote]
It;s wrong with k=3 :wink: .
I think we need to divide into 2 cases:
Case 1:k is even.Then 16^k-1 is divisible by 17
Cse 2:k is odd. I haven't found out solution for this part. Hopefully I can do it tomorrow. :roll:

[quote]madness wrote:
My friend and i found this proof, dunno whether its right so plz check it Smile
solution

Clearly $ 16^k\minus{}1$ isnt divisible by 3 and 7 for all $ k\geq{2}$.

It;s wrong with k=3 Wink . [/quote]

I think he meant $ 4^k\plus{}1$ isn't divisible by 3 nor 7 for all $ k\geq{2}$, in which case his proof works out very well.

[quote="The QuattoMaster 6000"]Prove that if $ k\ge 2$, then $ 16^k \minus{} 1$ has a prime factor that is greater than $ 10$.[/quote]
[hide="must control impulse to use Catalan-Mihailescu"]$ (2^{k}\plus{}1)(2^{k}\minus{}1)(4^{k}\plus{}1)$
We can check k=2, 3 by hand with ease.
Let's induct or something.
We're assuming that k is odd, because if it's even then we can just break it into a smaller case.
So k is odd.
$ 4^{k}\plus{}1$ isn't a multiple of 2, 3, or 5, so now we need to observe it mod 7.
However, $ 4^{k}$ can only be 4, 2, or 1 mod 7, so clearly $ 4^{k}\plus{}1$ has no prime factors under 10.
So umm, we didn't really need induction I guess, but whatever.[/hide]

Nice Solution!  Here is my messier and longer solution:
[hide="Solution"]
First, change the term to $ 16^x\minus{}1$ instead of $ 16^k\minus{}1$ (sorry for the confusion.)

Now, if $ x$ is even, then we find that $ 16^x \minus{} 1\equiv ( \minus{} 1)^x \minus{} 1\equiv 1 \minus{} 1\equiv 0\bmod 17$, so $ 17|(16^x \minus{} 1)$ if $ x$ is even.  Now, if $ x$ is odd, then let $ x \equal{} 2k \plus{} 1$.  It follows that
\[ 16^x \minus{} 1 \equal{} 2^{4x} \minus{} 1 \equal{} (2^{2x} \minus{} 1)(2^{2x} \plus{} 1)
\]
Yet this is equal to
\[ (2^x \minus{} 1)(2^x \plus{} 1)(2^{4k \plus{} 2} \plus{} 2^{2k \plus{} 1}\cdot 2 \plus{} 1 \minus{} (2^{k \plus{} 1})^2)
\]
which becomes
\[ (2^{2k \plus{} 1} \minus{} 1)(2^{2k \plus{} 1} \plus{} 1)(2^{2k \plus{} 1} \plus{} 1 \minus{} 2^{k \plus{} 1})(2^{2k \plus{} 1} \plus{} 1 \plus{} 2^{k \plus{} 1})
\]
Now, we take pairs of factors and show that they are coprime.  $ \gcd (2^x \plus{} 1, 2^x \plus{} 1) \equal{} \gcd (2^x \minus{} 1, (2^x \plus{} 1) \minus{} (2^x \minus{} 1)) \equal{} \gcd(2, 2^x \minus{} 1)$, but $ 2$ is not a factor of $ 2^x \minus{} 1$, so $ 2^x \minus{} 1$ and $ 2^x \plus{} 1$ are coprime.  Also,
\[ \gcd (2^x \plus{} 1, 2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1) \equal{} \gcd (2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1, (2^x \plus{} 1) \minus{} (2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1)) \equal{} \gcd (2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1, 2^{k \plus{} 1})|2^{k \plus{} 1}
\]
but $ 2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1$ is odd, so the only factor of $ 2^{k \plus{} 1}$ which is odd is one, so $ 2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1$ and $ 2^{2k \plus{} 1}$ are coprime.  Similarly, $ 2^{2k \plus{} 1} \plus{} 2^{k \plus{} 1} \plus{} 1$ and $ 2^{2k \plus{} 1}$ are coprime.  Furthermore,
\[ \gcd (2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1, 2^{2k \plus{} 1} \plus{} 2^{k \plus{} 1} \plus{} 1) \equal{} \gcd (2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1, (2^{2k \plus{} 1} \plus{} 2^{k \plus{} 1} \plus{} 1) \minus{} (2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1)) \equal{} 2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1, 2^{k \plus{} 2})|2^{k \plus{} 1}
\]
and since $ 2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1$ is odd, we see that by similar reasoning as before, $ 2^{2k \plus{} 1} \plus{} 2^{k \plus{} 1} \plus{} 1$ and $ 2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1$ are coprime.  Finally, we see that $ \gcd (2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1, 2^{2k \plus{} 1} \minus{} 1)|\gcd (2^{2k \plus{} 1} \minus{} 1, 2(2^k \minus{} 1))$.  Yet $ (2^k \plus{} 1)(2^k \minus{} 1) \equal{} 2^{2k} \minus{} 1$, which is relatively prime with $ 2^{2k \plus{} 1} \minus{} 1$ since their difference is a power of $ 2$.  It follows that $ \gcd (2^{2k \plus{} 1} \minus{} 1, 2(2k \plus{} 1)) \equal{} 1$, so $ 2^{2k \plus{} 1} \minus{} 2^{k \plus{} 1} \plus{} 1$ and $ 2^{2k \plus{} 1} \minus{} 1$ are coprime as are $ 2^{2k \plus{} 1} \plus{} 2^{k \plus{} 1} \plus{} 1$ and $ 2^{2k \plus{} 1} \minus{} 1$.  We  conclude that all four of the factors are pairwise coprime.

Notice that when $ x \equal{} 1$, then $ 2^x \minus{} 1 \equal{} 1$, so there are only three factors after we take out the one.  In all other cases, though, we see that all factors are greater than one. It follows that when $ x\ge 2$, there are four factors, all of which are relatively prime, but none of them are divisble by $ 2$.  Since they are relatively prime, none of them share the same prime factor.  Yet, there are only three prime factors other than $ 2$ that are less than $ 10$, so at least one of these factors has a prime factor greater than or equal to $ 11$, as desired.   [/hide]

[quote="diophantient"][quote="The QuattoMaster 6000"]Prove that if $ k\ge 2$, then $ 16^k \minus{} 1$ has a prime factor that is greater than $ 10$.[/quote]
[hide="must control impulse to use Catalan-Mihailescu"]$ (2^{k} \plus{} 1)(2^{k} \minus{} 1)(4^{k} \plus{} 1)$
We can check k=2, 3 by hand with ease.
Let's induct or something.
We're assuming that k is odd, because if it's even then we can just break it into a smaller case.
So k is odd.
$ 4^{k} \plus{} 1$ isn't a multiple of 2, 3, or 5, so now we need to observe it mod 7.
However, $ 4^{k}$ can only be 4, 2, or 1 mod 7, so clearly $ 4^{k} \plus{} 1$ has no prime factors under 10.
So umm, we didn't really need induction I guess, but whatever.[/hide][/quote]
Very nice solution, diophantient! I just found out a similar solution 5 minutes ago. :wink:

Yeah sorry, i meant what jrshoch said, that is $ 4^k \plus{} 1$ isnt divisible by 3 or 7 for all $ k \geq 2$  :)


Vasc the first problem is a direct consequence of spanferkels inequality !

[quote="Albanian Eagle"]hey the inequality is true even if $\max{a,b,c,d} \le abcd$
:D[/quote]Of course, replace $a,b,c,d$ by $1/a,1/b,1/c,1/d$ respectively. :)


geometry
rhombus
trigonometry
AMC
AIME
geometric transformation
parallelogram

Quadrilateral $ABCD$ has $AB=BC=CD$, $\angle ABC=70^\circ$, and $\angle BCD=170^\circ$. What is the degree measure of $\angle BAD$?

$ \textbf{(A)}\ 75\qquad
\textbf{(B)}\ 80\qquad
\textbf{(C)}\ 85\qquad
\textbf{(D)}\ 90\qquad
\textbf{(E)}\ 95$

Draw equilateral triangle ABE with E close to C (i.e. inside the quadrilateral).  Then BCDE is a rhombus since it has three equal sides and adjacent angles add to 180.  Finally, draw DE to get isosceles triangle ADE.  The calculation of the unknown angles follows almost immediately.

[quote="the future"][quote="tenniskidperson3"]I just drew a picture with the protractor we were given...  :blush:[/quote]

lucky you had a protractor...

i only bring those items for the AIME but this would be a lot easier with a protractor..I was forced to use laws of sines (which techinically isnt required since this is the AMC 10)...and i got lucky that sin65=sin115 making one of those triangles isoscles (refer to above post)[/quote]
My school provided protracters and measuring devices...
 :lol:

So did mine...

infact a girl came up to me before the test and asked me if i needed a protractor or ruler...

i said no becuase i didnt have space left on my desk...

well i guess its a good thing i got it right becuase i wouldve killed myself i got it wrong...

I thought this was quite a nontrivial problem for the 10B, and found it harder than 1-23 on the 12B. I had no idea what it was doing nested before #25.

[hide="Solution"]
Let $ P$ be a point such that $ \triangle BPC$ is outside the quadrilateral and $ \triangle BPC$ is equilateral.  Thus, $ \angle ABP \equal{} \angle ABC \plus{} \angle CBP \equal{} 70 \plus{} 60 \equal{} 130$.  Also, $ \angle DCP \equal{} \angle BCD \plus{} \angle BCP \equal{} 170 \plus{} 60 \equal{} 230$, so the minor angle of $ \angle DCP$ is $ 130$.  Furthermore, $ AB \equal{} CD$, $ BP \equal{} CP$, and $ \angle ABP \equal{} 130 \equal{} \angle DCP$, so $ \triangle ABP$ is congruent to $ \triangle DCP$, which means that $ \triangle ABP\sim \triangle DCP$, so a spiral similarity maps $ \triangle PBC\sim \triangle PAD$, which implies that $ \angle PAD \equal{} \angle PBC \equal{} 60$.  Since $ AB \equal{} BP$ and $ \angle ABP \equal{} 130$, we see that $ \angle BAP \equal{} 90 \minus{} \frac {130}{2} \equal{} 25$.  Thus, $ \angle BAD \equal{} \angle PAD \plus{} \angle BAP \equal{} 85 \equal{} \boxed{C}$. [/hide]

EDIT to [b]RisingStar[/b]: http://mathworld.wolfram.com/SpiralSimilarity.html is a nice link to learn about it.  The way I meant it though is that if $ \triangle ABP\sim \triangle DCP$, then $ \triangle PBC\sim \triangle PAD$ (the link shows a nice animation of why that is true, but one could also prove it rigorously through SAS similarity.)

[quote="The QuattoMaster 6000"][hide="Solution"]
Let $ P$ be a point such that $ \triangle BPC$ is outside the quadrilateral and $ \triangle BPC$ is equilateral.  Thus, $ \angle ABP \equal{} \angle ABC \plus{} \angle CBP \equal{} 70 \plus{} 60 \equal{} 130$.  Also, $ \angle DCP \equal{} \angle BCD \plus{} \angle BCP \equal{} 170 \plus{} 60 \equal{} 230$, so the minor angle of $ \angle DCP$ is $ 130$.  Furthermore, $ AB \equal{} CD$, $ BP \equal{} CP$, and $ \angle ABP \equal{} 130 \equal{} \angle DCP$, so $ \triangle ABP$ is congruent to $ \triangle DCP$, which means that $ \triangle ABP\sim \triangle DCP$, so a spiral similarity maps $ \triangle PBC\sim \triangle PAD$, which implies that $ \angle PAD \equal{} \angle PBC \equal{} 60$.  Since $ AB \equal{} BP$ and $ \angle ABP \equal{} 130$, we see that $ \angle BAP \equal{} 90 \minus{} \frac {130}{2} \equal{} 25$.  Thus, $ \angle BAD \equal{} \angle PAD \plus{} \angle BAP \equal{} 85 \equal{} \boxed{C}$. [/hide][/quote]

Whats a spiral similarity?

[hide="Solution"]
I don't even know if this is mathematical but..

Say angle B is 60 deg. instead of 70 deg.

and that angle C is 180 deg (making BCD a line) instead of 170 deg.

You would then have a 30-60-90 triangle with angle A = 90 deg.

From that, increase angle B back to 70. That results in:

80 < angle A < 90

and from that, decrease angle C back down to 170 deg. That would still keep angle A within the limit mentioned above, and the only choice provided in the question is:

[b]C. 85[/b]

what a pity I never thought of this until _after_ I stepped out of the testing room.. :wallbash:
[/hide]

My solution involved drawing altitudes from $ B$ and $ C$ to the diagonals and then extending them, creating two kites. Then you can do some angle chasing and find that the two share an edge, and at the point it becomes easy.

[hide="translation"]
Translate CD so that C goes to B. Now, angle D'BC equals 180-170 = 10.  This means ABD' = 60, implying by SAS that ABD' is equilateral.  This in turn means AD'D is isosceles, because we know BCDD' is a parallelogram (and a rhombus, at that.)  From here it's just an angle chase.[/hide]

[quote="Moldytape"]For this problem, I made an educated guess that there is an isocoles triangle formed by the diagonals of the quadrilateral. So I got 85. :lol:[/quote]

Lots of people in my school did that.  But like everyone else said (maybe, I can't really understand) the real way to do it is with congruent triangles and equilateral triangles off the 60 degree angles between the diagonals.

Well, it might have helped me if i had read the problem correctly...
I thought m<BCD was 120...  :(  :mad:

[quote="lokesh"]This problem was stupid.[/quote]

That is a pointless comment. At least give a reason. If your reason is that the solution was too contrived, it's not; this is quite a common techinique (see USAMO 1996.5, maybe it was #4). It may be too hard for the AMC 10, however, as I had stated earlier. If your only reason is that you solved it incorrectly, then it's not a valid reason.

eh... I used law of sines twice to approximate the angle...  :blush:

I take it he said it was stupid because people used ruler+protractor to solve it. If that's whay he meant, then I agree that it is stupid that these are allowed, but the problem was good.

Yeah that is basically what my reasoning was behind that comment. The actual solution where you draw in an equilateral triangle and such was nice but frankly, if the MAA is going to ban calculators, they might as well ban protractors, rulers, and other tools that completely defeat the purpose of problem solving or just not include such problems on their tests.

A protractor and compass are very important in proof type geometry problems. So I would not expect them to be banned on the AIME (where the later problems have proof components) or the USAMO. However, I suppose that it is reasonable to ban them on the AMC 10/12 contests.

Yeah i guess that last statement was way too radical. They should just ban protractors and rulers from the AMC tests or not include such problems.

Hey, construction is a mathematical skill learned in geometry :P
...we spent like 4 weeks on it, constructing angles and stuff

But I dont really care if they ban protractors/compasses...since I never bring one, even to the AIME...

during the test i didn't know how to do it, and i was cleverly angle chasing (non trigonometrically) and the proctor announced like 1 minute left so i panicked and saw that i had written 85 for some angle and there were a butt-load of similar triangles so i wrote down 85 and bubbled it. i was almost sure that i got it wrong, but when i got home i used my handy construction skills to solve it and i realized that my answer was correct. well guessing this one correctly compensates for making a retarded mistake on #11.  :wink:

[hide="Trig Solution: How to do this Problem with a Minute left on the contest"]
We apply Quadrilateral Trig Ceva's Theorem, as we are tempted to draw in diagonals $AC, BD$:
$\frac{\sin \angle{DBC} \cdot \sin \angle BAC \cdot \sin \angle BDA \cdot \sin \angle ACD}{\sin \angle DBA \cdot \sin \angle CAD \cdot \sin \angle BDC \cdot \sin \angle BCA}=1$.

By simple angle chasing, we allow $\angle CAD=\theta$ and simplify the above equation to:

$\frac{\sin{5^\circ}\cdot \sin{55^\circ}\cdot \sin{\angle BDA}\cdot \sin{115^\circ}}{\sin{65^\circ}\cdot \sin{\theta} \cdot \sin{5^\circ}\cdot \sin{55^\circ}}=1$.
Since the total sum of the areas of the quadrilateral is $360^\circ$, we can easily see that $\angle{BDA}=60^\circ-\theta$.
We also utilize the fact that $\sin{\alpha}=\sin{180^\circ-\alpha}$ for some $\alpha$, and see that the above equation is simply equal to $\frac{\sin{60^\circ-\theta}}{\sin{\theta}}=1 \implies 60^\circ-\theta=\theta + 360^\circ k$ for some integer $k$. Then $60^\circ-360^\circ k=2\theta \implies \theta=30^\circ-180^\circ$. The only possible $\theta$ that works is when $k=0$, so $\theta=30^\circ$. So $\angle BAD=55^\circ+30^\circ=\boxed{\bold{(C)}}=85^\circ. \blacksquare$

[hide="Outline to Proof of Formula Used"]
Define $CA, BD$ to intersect at $K$. Then use LOS "symmetrically" on triangles AKD, DKC, CKB, KBA and combine. This will simplify to the formula Quadrilateral Trig Cevas provides. [/hide][/hide]

[quote=Altheman]Let $ P$ be the intersection of the angle bisector of $ \angle ABC$ and $ \angle  BCD$. By SAS congruency, it follows that $ \triangle APB\cong \triangle CPB\cong \triangle CPD$. Then $ \angle PBC\equal{}35$ and $ \angle PCB\equal{}85$. Then $ \angle APB\equal{}\angle BPC\equal{}\angle CPD\equal{}60$. Hence $ \angle APB\plus{}\angle BPC\plus{}\angle CPD\equal{}180$, so $ P$ lies on $ AD$. Then $ \angle PCB\equal{}\angle PAB\equal{}85$, so the answer is $ \boxed{C}$.[/quote]

[quote=Altheman]Comment: we can apply the same argument for any angles $ \alpha\equal{}\angle ABC$ and $ \beta\equal{}\angle BCD$ such that $ \alpha\plus{}\beta\equal{}240$. If we intersect $ AB$ and $ CD$ to intersect at $ E$, then we have $ \angle AED\equal{}60$. Then it becomes a classical problem.[/quote]

So the motivation for this solution was seeing that $\angle ABC + \angle BCD =240^{\circ} \implies \angle AED = 60^{\circ}$? I'm just curious as to what the motivation for drawing the angle bisectors was, or if it is a well-known ("classical"?) problem.

/bump sorry if I was unclear with my question. Let me rephrase it.

What was the motivation for drawing the angle bisectors? Was it simply the fact that this configuration is well known? I'm not sure what Altheman means by "classical problem." Does he just mean that the problem is pretty trivial after you find out $\angle ABC + \angle BCD =240^{\circ} \implies \angle AED = 60^{\circ}$?

[quote=perfect628][hide="Another, Worse, Way"]So, first connect $ AC$ and $ BD$, to obtain two isosceles triangles in $ \triangle ABC$ and $ \triangle BCD$. Let these lines interect at $ E$. Now, because those are isosceles, we see $ m\angle BAC\equal{}m\angle BCA\equal{}55^{\circ}$, and then $ m\angle CAD\equal{}m\angle CDA\equal{}5^{\circ}$. Now, we can thus see that $ m\angle ABE\equal{}65^{\circ}, m\angle AEB\equal{}m\angle CED\equal{}60^{\circ}$, and $ m\angle ECD\equal{}115^{\circ}$. Now, label the common side length given as $ x$. By the Law of Sines, we know $ \frac{\sin60^{\circ}}{x}\equal{}\frac{\sin115^{\circ}}{DE}$, and $ \frac{\sin60^{\circ}}{x}\equal{}\frac{\sin65^{\circ}}{AE}$. (from $ \triangle CED$ and $ \triangle AEB$ respectively) Now, since $ \sin65^{\circ}\equal{}\sin115^{\circ}$, we can see that $ AE\equal{}DE$, so $ \triangle ADE$ is isosceles. Since $ m\angle AED\equal{}120^{\circ}$, we know $ m\angle DAE\equal{}30^{\circ}$, so $ m\angle BAD\equal{}55\plus{}30\equal{}85^{\circ}$.[/hide]
Sorry, just wanted to add (one of) the trig methods.[/quote]

"Now, because those are isosceles, we see $ m\angle BAC\equal{}m\angle BCA\equal{}55^{\circ}$, and then $ m\angle CAD\equal{}m\angle CDA\equal{}5^{\circ}$."
I think you mean $ \angle CBD\equal{}\angle CDB\equal{}5^{\circ}$.
Also, this is my first time using code, so sorry if I did something wrong.

Good thing MAA banned protractors... right?

Another similar problem I've seen to this:
[quote = some mpfg problem]
Let $HEXAGON$ be a convex equilateral heptagon. If $\angle E = \angle O = 168^{\circ}$ and $\angle H = \angle N = 108^{\circ}$, find $\angle A$.
[/quote]

we proceed with guess and check :D

place A at origin, length = 1 for the 3 sides, D on the real axis

given $\angle BAD$ = $\theta$: $D = e^{i\theta}+e^{i(\theta-110)}+e^{i(\theta-120)}$

The $e^{i\theta}$ from going from A to B,
        $e^{i(\theta-110)}$ from B to C - you rotate the angle $\theta$ 110 down so that 180-110=70 is the angle of polygon
        $e^{i(\theta-120)}$ from C to D - rotate 10 down from $\theta-110$ so angle of polygon is 180-10=170

since D is on real axis, the real part of D equals 0, so $\sin\theta+\sin(\theta-110)+\sin(\theta-120)=0$

We now guess $\theta=85$ and try to verify

$\sin85+\sin-25+\sin-35=0$
$\cos5=\sin25+\sin35$
$\cos5=\sin(30-5)+\sin(30+5)$
$\cos5=2\sin30\cos5$
$\cos5=\cos5$

alternatively (slightly less complete guess):

$\sin\theta+\sin(\theta-110)+\sin(\theta-120)=0$
$\sin\theta+\sin(\theta-115+5)+\sin(\theta-115-5)=0$
$\sin\theta+2\sin(\theta-115)\cos(5)=0$
$\sin\theta-2\sin(\theta+65)\cos(5)=0$
$\sin\theta=2\sin(\theta+65)\cos(5)$

now guess $\theta=85$ so that the cos(5) is dealt with (and verify ofc this is guess and [b]check[/b])

alternatively alternatively expand out $\sin(\theta+65)$, divide by $\sin\theta$, and solve for $\cot\theta$, yielding $$\theta=\cot^{-1}\left(  \frac{\frac1{2\cos(5)}-\cos(65)  }  {\sin(65)} \right)$$ (use your inner calculator brain to evaluate it)

coordinates/vectors also works instead of complex #

[hide="messy algebra+trig solution"]
Let: point A be (0,0)
 D be (d,0)
AB=BC=CD=1
$\angle BAD=\theta$

The y coordinate of D is: $\sin\theta+\sin(\theta-110^\circ)+\sin(\theta-120^\circ)=0$

$\sin\theta+\sin(\theta-115^\circ+5^\circ)+\sin(\theta-115^\circ-5^\circ)=0$
$\sin\theta+2\sin(\theta-115^\circ)\cos(5^\circ)=0$
$\sin\theta-2\sin(\theta+65^\circ)\cos(5^\circ)=0$
$\sin\theta=2\sin(\theta+65^\circ)\cos(5^\circ)$

Let: $\phi=\theta+65^\circ$

$\sin(\phi-65^\circ)=2\sin\phi\cos(5^\circ)$
$\sin\phi\cos(65^\circ)-\cos\phi\sin(65^\circ)=2\sin\phi\cos(5^\circ)$
$\cos(65^\circ)-\cot(\phi)\sin(65^\circ)=2\cos(5^\circ)$
$\cot(\phi)(\sin(60^\circ)\cos(5^\circ)+\cos(60^\circ)\sin(5^\circ))=(\cos(60^\circ)\cos(5^\circ)-\sin(60^\circ)\sin(5^\circ)) -2\cos(5^\circ)$
$\cot(\phi)(\frac{\sqrt{3}}{2}\cos(5^\circ)+\frac{1}{2}\sin(5^\circ))=(\frac{1}{2}\cos(5^\circ)-\frac{\sqrt{3}}{2}\sin(5^\circ)) -2\cos(5^\circ)$
$\cot(\phi)(\frac{\sqrt{3}}{2}\cos(5^\circ)+\frac{1}{2}\sin(5^\circ))=(\frac{-3}{2}\cos(5^\circ)-\frac{\sqrt{3}}{2}\sin(5^\circ))$
$\cot(\phi)(\frac{\sqrt{3}}{2}\cos(5^\circ)+\frac{1}{2}\sin(5^\circ))=-\sqrt{3} (\frac{\sqrt{3}}{2}\cos(5^\circ)+\frac{1}{2}\sin(5^\circ))$
$\cot(\phi)=-\sqrt3$

$\phi=150^\circ$

$\theta=150^\circ-65^\circ=85^\circ$

[b]$\angle BAD = 85^\circ$[/b]


Let $ A$ is a natural number.Change position of the digits of $ A$ we have $ B$
Suppose that $ B \minus{} A \equal{} 11..1(n$ ditgit $ 1)$
Find min value of $ n$

[hide]
Well the difference of a permutation is divisible by $ 9$ so $ 111.. \equiv 0 \mod 9$
Also the if a number is divisible by $ 9$ then the sum of it's digits is also divisible by $ 9$
so $ n_{min} \equal{} 111111111$ (with $ 9$ ones)[/hide]


modular arithmetic
number theory
greatest common divisor
relatively prime
combinatorics unsolved
combinatorics

Suppose that necklace $\, A \,$ has 14 beads and necklace $\, B \,$ has 19. Prove that for any odd integer $n \geq 1$, there is a way to number each of the 33 beads with an integer from the sequence \[ \{ n, n+1, n+2, \dots, n+32 \}  \] so that each integer is used once, and adjacent beads correspond to relatively prime integers. (Here a ``necklace'' is viewed as a circle in which each bead is adjacent to two other beads.)

Obviously $ (m,m\plus{}1)\equal{}(m,1)\equal{}1$. Make a $ 33$ bead arrangement with the numbers in order. We can easily verify that $ (n,n\plus{}32)\equal{}(n,32)\equal{}1$ since $ n$ is odd. Hence this is a good necklace. Now lets remove $ 13$ beads from this larger necklace, and connect the two parts into the two satisfactory necklaces. Say the first number that we remove is $ n\plus{}k$, then the end of that $ 14$ chain is $ n\plus{}k\plus{}13$. We only have to find $ k$ so we have the two new adjacent beads have relatively prime labels: $ (n\plus{}k,n\plus{}k\plus{}13)\equal{}1$ and $ (n\plus{}k\minus{}1,n\plus{}k\plus{}14)\equal{}1$. Note that $ k\in[0,20]$. Let $ a\equal{}n\plus{}k$.

The condition is $ (a,a\plus{}13)\equal{}(a,13)\equal{}1$ and $ (a\minus{}1,a\plus{}14)\equal{}(a\minus{}1,15)\equal{}1$. Note that $ a\minus{}1$ ranges between $ 21$ consecutive integers. Choose $ a\minus{}1$ even. Then $ (a\minus{}1,15)\equal{}1$ and $ a$ ranges from at least $ 10$ consecutive even integers. Now if $ (a,13)\equal{}1$, we are done. If $ (a,13)\equal{}13$, then $ (a\minus{}2,13)\equal{}(a\plus{}2,13)\equal{}1$ and since $ a$ is among at least $ 10$ consecutive even integers, $ a\minus{}2$ and/or $ a\plus{}2$ is also among these integers. Hence such $ a$ exists and we are done.

Sorry for bumping this, but I have some questions about altheman's proof. First, shouldn't k range from 1 to 18, not 0 to 20? (if the 14-chain is to be in between the two ends). Also, while trying to find a range of suitable  a-1's such that the gcd (a-1, 15) = 1, choosing a-1 as even does not guarentee that the two are relatively prime.

I think you need to prove (as seshadri said) that in any set of 18 consecutive numbers s = n+k (k ranging from 1 to 18) there exists at least one number such that it is not 1 mod 3, 1 mod 5, and 0 mod 13:

In the set n+1, n+2, ... n+18, where n is odd, we can put each number in modulo 3, and despite the shift of n the distribution of the residues will be the same (because of symmetry). So there are 6 numbers 1 mod 3.

Writing in mod 5, the most frequent residue appears 4 times so there are at most 4 numbers 1 mod 5. By the same logic, there are at most 2 numbers 0 mod 13.

At the most we could rule out 12 numbers (maybe less because of overcounting mod 15), yet we have 18 to choose from therefore a suitable n + k exists.

If I just got confused and didn't really understand I apologize and help would be appreciated =p.

[quote=Arslan]Is the oddness of $n$ used in the solutions above? It seems no need for that. Or do I miss something?[/quote]

Because if n is even then by Pigeonhole Principle you always have 2 adjacent even beads so it does not work

Consider the 3 constructions
A: n, n+1, n+2, ... , n+13, B: n+14, n+15, n+16, ... , n+32
A: n+19, n+20, n+21, ... , n+32, B: n, n+1, n+2, ... , n+18
A: n+2, n+3, n+4, ... , n+15, B: n, n+1, n+28, n+27, n+26, ... , n+16, n+29, n+30, n+31, n+32

First one fails if n = 1 (mod 3) or n = 0 (mod 13)
Second one fails if n = 0 (mod 3) or n = 7 (mod 13)
Third one fails if n = 2 (mod 3) or n = 10 (mod 13)

Since no n can fail all 3 constructions, at least 1 of them work.

If you are viewing this a few years in future the full proof would be linked at [url]https://infinityintegral.github.io/contest-collections[/url]

[quote=seshadri]we start with a simple observation: if we line up all the integers $n,n+1,\ldots,n+32$ in that order circularly,(so that $n$ has neighbors $n+1,n+32$, $n+1$ has neighbors $n,n+2$ and so on) then this is a 'good' arrangement(i.e. any two consecutive are relatively prime), since $n$ is odd.
to get the required arrangement, we could 'break and fuse' this arrangement to create two good circular arrangements. suppose we form the necklaces as $n+k,n+k+1,\ldots,n+k+13$ and $n+k+14,\ldots,n+32,n,\ldots,n+k-1$ then in order that this works we need that $(n+k,n+k+13)=1,(n+k-1,n+k+14)=1$. this is equivalent to saying that $r=n+k$ satisfies
$r\not\equiv 1\pmod 3,r\not\equiv 1\pmod 5,r\not\equiv 0\pmod {13}$ and also $k\le 18$.
it is now a simple exercise to see that this can be achieved quite easily(i am not going to prove this, but i think it should be pretty straightforward, since among $18$ numbers there can be atmost $2$ of them that are divisible by $13$ but lots more that satisfy the other two).[/quote]

Finishing your proof, of the 18 numbers at most 2 are multiple of 13, 6 are multiple of 3, and at most 4 are multiples of 5. Even in worst case they dont coincide we still have 6 ok numbers.


How it should look like:

[quote="Franz Joseph"]$1*2*3*..............*99*100*101= 2006$

Replace the asterisks with $+$ and $-$. Is equality possible for the above expression?[/quote]

:lol:  I must have been sleeping.........

Nothing prevents you from editing the original post, after which we can erase these subsequent three as unnecessary :)

[quote="buzzer11"][quote="SplashD"]It's impossible because the LHS will always be odd while the RHS will always be even[/quote]

Is this because there are an odd number of odd terms?[/quote]
Basically, yes.

$1\pm2\pm3\pm ... \pm 100\pm101\equiv1\pm0\pm1\pm ... \pm0\pm1\equiv1\pmod2$

left=odd, right=even for reasons given by lotrgreengrapes


geometry
geometric transformation
reflection
trigonometry
algebra
linear equation

Let $\triangle ABC$ be an equilateral triangle and let $P$ be a point on $\left[AB\right]$.
$Q$ is the point on $BC$ such that $PQ$ is perpendicular to $AB$. $R$ is the point on $AC$ such that $QR$ is perpendicular to $BC$. And $S$ is the point on $AB$ such that $RS$ is perpendicular to $AC$.
$Q'$ is the point on $BC$ such that $PQ'$ is perpendicular to $BC$. $R'$ is the point on $AC$ such that $Q'R'$ is perpendicular to $AC$. And $S'$ is the point on $AB$ such that $R'S'$ is perpendicular to $AB$.
Determine $\frac{|PB|}{|AB|}$ if $S=S'$.

It is a nice question ... the answer is [hide]$\frac 27$, and the most straightforward way to get it is to keep using 30-60-90 triangles all the way around and solving a linear equation for the ratio.[/hide]

I feel like there should also be an argument based on reflection of some sort, but I don't know what it is.

Nice problem ! Suppose w.l.o.g. $AB=1$ and denote $PB=x$.
Then  $AS=2[1-2(1-2y)]$ and $AS'=\frac 12\left[1-\frac 12\left(1-\frac x2\right)\right]\ .$
Therefore, $S\equiv S'\Longleftrightarrow AS=AS'\Longleftrightarrow x=\frac 27\ ,$ i.e. $\frac{PB}{AB}=\frac 27\ .$[hide="Here is an easy extension."]
[color=darkblue][b]An easy extension.[/b] Let $ABC$ be an equilateral triangle

and let $P$ be a point on $[AB]$. Define the points

$\{Q,Q'\}\subset [BC]$, $\{R,R'\}\subset [CA]$, $\{S,S'\}\subset [A,B]$ so that

$\triangle BPQ\sim\triangle CQR\sim\triangle ARS\sim\triangle BQ'P'\sim\triangle CR'Q'\sim\triangle AS'R'\ .$

Prove that if the point $S$ coincides with the point $S'$, then

$\frac{PB}{AB}=\frac{t}{t^2+t+1}$, where $t=\frac{\sin\widehat {BPQ}}{\sin \widehat {BQP}}\ .$
[u]A particular case :[/u] $t=2\ \vee \ t=\frac 12\ ,$ then $\frac{PB}{AB}=\frac 27\ .$
[u]Remark.[/u] If denote $f(t)=\frac{t}{t^2+t+1}\ ,$ then $f(t)=f\left(\frac 1t\right)$ for any $t\ne 0\ .$[/color][/hide] [hide="Here is a similar problem."]
[color=darkred][b]A similar problem.[/b] Let $ABC$ be an acute triangle.

Define the points $M\in (BC)$, $N\in (CA)$, $P\in (AB)$ so that

$MN\perp BC$, $NP\perp CA$, $PM\perp AB\ .$ Prove that :

$1.\blacktriangleright \frac{MN}{a}=\frac{NP}{b}=\frac{PM}{c}=\frac{4S} {a^2+b^2+c^2}\ ,$ where $S=[ABC]\ .$

$2.\blacktriangleright\frac{MB}{MC}\cdot \frac{NC}{NA}\cdot \frac{PA}{PB}\ge 8$ and $MN^2+NP^2+PM^2\le \frac 13 (a^2+b^2+c^2)\ .$

$3.\blacktriangleright (a^2+b^2+c^2)\sqrt{[MNP]}=4S\sqrt S$ and $3[MNP]\le S\ .$[/color][/hide]

In the my last reply I added an easy extension and a similar problem.

How do you find $AS=2[1-2(1-2y)]$ and $AS'=\frac 12\left[1-\frac 12\left(1-\frac x2\right)\right]\ ?$

Could anyone clarify this problem ? The only solution given is quite complicate honestly. I really need more details or maybe a simpler solution please.


On the plane are given $n\geq 3$ points which are not all collinear. For an arbitrary point $A$, let $S(A)$ be the sum of the distances from $A$ to the $n$ points. A point $A$ is called good if for any other point $B$ in the plane, $S(B)\geq S(A)$. Show that there can be at most one good point.

From my friend
[hide="hint"]If $M$ is midpoint of $[AB]$ then $XM\leq\frac{XA+XB}{2}\;\forall X$.[/hide]


Suppose that a,b,c are real numbers such that for any real positive numbers x_1,..,x_n we have

(A^a)*(B^b)*(C^c)  \geq 1

where

A=sum(i=1,..,n)x_i/n

B=sum(i=1,..,n)(x_i)^2/n

C=sum(i=1,..,n)(x_i)^3/n

Prove that the vector (a,b,c) has the form p(-2,1,0) + q(1,-2,1) for some positive real numbers p and q.

Source :IRAN 2000 mathematical olympiads

first, set n = 1 , then we get x1a*(x12)b*(x13)c = x1a+2b+3c \geq 1, for x1 > 0, particularly this holds for x1 < 1 and x1 > 1, so we must have that a+2b+3c = 0, then if we set p = b + 2c, and q = c we have p(-2,1,0) + q(1,-2,1) = (-2p+q,p-2q,q) = (-2b-3c,b,c) = (a,b,c), and to complete the problem we must show that p = b+2c \geq 0 and q=c \geq 0.

p \geq 0:  set n = 2, x1 = 1 and x2 = e, and define f(e) = ((1+e)/2)a*((1+e2)/2)b*((1+e3)/2)c which is \geq 1
as e -> 0, f(e) -> 1/2a+b+c implying that 1/2a+b+c \geq 1, then a+b+c \leq 0, and p = b+2c = (a+2b+3c) - (a+b+c) \geq 0

q \geq 0: set n = k+1, x1 = x2 = ... = xk= 1-e, and xk+1 = 1+ke, k is a positive integer and e is a real number in the interval (0,1), using the definition from the problem, we see that:
A = k(1-e) + (1+ke) = k+1
B = k(1-e)2 + (1+ke)2 = (k+1)(1+ke2)
C = k(1-e)3 + (1+ke)3 = (k+1)(1+3ke2 + (k2-k)e3)
define g(k,e) = 1a*(1+ke2)b*(1+3ke2 + (k2-k)e3)c, and by the inequality, we see that g(k,e) \geq 1 for all k,e.  Now take e = k-1/2, and g(k,k-1/2) = 2b*(4+(k-1)/ \sqrt k)c \geq 1 for all positive integers k, but since (4+(k-1)/ \sqrt k) can be made arbitrarily large for sufficiently larger k, c must be non-negative, and q = c \geq 0

so p and q are non-negative as desired.


For any positive integer $ x$, let $ S(x)$ be the sum of the digits of $ x$, and let $ T(x)$ be $ |S(x \plus{} 2) \minus{} S(x)|$.  For example, $ T(199) \equal{} |S(201) \minus{} S(199)| \equal{} |3 \minus{} 19| \equal{} 16$.  How many values $ T(x)$ do not exceed 1999?

For what values of $ x$? 'Cause any $ x$ with units digit 0-7 is going to make $ T(x)\equal{}2$...

[quote="Brut3Forc3"]For what values of $ x$? 'Cause any $ x$ with units digit 0-7 is going to make $ T(x) \equal{} 2$...[/quote]

Right, so the units digit of $ x$ will be 8 or 9. This is a problem from the AIME Problem Series B message board but I'm confused by the one solution someone provided. Here it is:

[quote="timwu"][hide]Let $ d_i$ denote the $ 10^{i}$ digit of $ x$ , and $ f$ as the function on those digits after the number added by 2. Also denote that $ T_i \equal{} d_i \minus{} f(d_i)$.

Let's start with the unit digit. If $ 0\leq d_0\leq 7$, no carrying over occurs, so plainly $ T(x) \equal{} 2$, and we are done.
If $ d_0 \equal{} 8,9$, $ f(d_0) \equal{} 0, 1$ respectively, so $ T_0 \equal{} 8$. But then we encounter carrying over.
Now, if $ 0\leq d_1\leq 8$, $ f(d_1) \equal{} d_2 \plus{} 1$, $ T_2 \equal{} \minus{} 1$, and we are done. Else $ d_1 \equal{} 9$, $ f(9) \equal{} 0$, $ T_1 \equal{} 9$, and we have to go to $ d_2$, and etc.
Suppose this recursion goes up to $ n \plus{} 1^{th}$ digit, then
$ T(x) \equal{} |\sum_{i \equal{} 0}^{n \plus{} 1}{T_i}| \equal{} |8 \plus{} 9 \plus{} ... \plus{} 9 \minus{} 1| \equal{} 9n \plus{} 7$.
Because $ T(x)\leq 1999$, $ 9n \plus{} 7\leq 1999$, $ 0\leq n\leq 221$. Therefore, including $ 2$, in total there are $ 223$ different values of $ T(x)$.

[/hide][/quote]

I posted this in response but I'm still waiting:
[hide]
[quote="RisingStar"]Could OP or someone else explain this? The notation confuses me.

[quote="timwu"]
Now, if $ 0\leq d_1\leq 8$, $ f(d_1) \equal{} d_2 \plus{} 1$, $ T_2 \equal{} \minus{} 1$, and we are done.
[/quote]

Here I think it was a typo, where it should be $ f(d_2)$ rather than $ f(d_1)$. But I still don't get it. If $ 0\leq d_1\leq 8$, then why would $ d_2$ always be 1 larger as a result?

[quote="timwu"]
Else $ d_1 \equal{} 9$, $ f(9) \equal{} 0$, $ T_1 \equal{} 9$, and we have to go to $ d_2$, and etc.
[/quote]
Why is $ f(9) \equal{} 0$?[/quote]
[/hide]


I would appreciate any help on this problem. Also, I was unsure which forum to post this in. I apologize if this question is not appropriate for this forum.

Problem:

Consider the following statement: "For any two lines l and m there exists a one to one correspondence between the set of points lying on l and the set of points lying on m." Prove that this statement is independent of the axioms of incidence geometry."

[quote="-fs-"]Consider the following statement: "For any two lines l and m there exists a one to one correspondence between the set of points lying on l and the set of points lying on m." Prove that this statement is independent of the axioms of incidence geometry."[/quote]

What are the axioms, for purposes of this problem?


Determine the rightmost five digits of the number $2005^{2006}$


linear algebra
matrix
linear algebra unsolved

Let U be the set of the (n,n) real matrices A=[A(i,j)] s.t. A²=n*A. If A is in U then let m(A)=inf(A(i,j)) over all (i,j).
    Find sup(m(A)) over all A in U.

Since there are matrices in $U$ with nothing but positive elements, $\sup\{m(A)\}>0.$

Suppose that $m(A)=r>0.$ Then $a_{i,j}\ge r$ for all $(i,j).$ It follows that each element of $A^{2}$ is greater than or equal to $nr^{2}.$

Since the miminum element of $A^{2}$ is just $n$ times the minimum element of $A,$ the value of that element must be $nm(A)=nr.$ Hence $nr\ge nr^{2}$ from which we derive that $r\le 1.$

But we can have $r=1.$ Just take the matrix $J,$ all of whose entries are $1.$ Then $J^{2}=nJ$ and $m(J)=1.$

So $\sup\{m(A)\}=1.$

---

My first-glance reaction to seeing this problem was that $J$ had to be the extremal case - which turned out to be the correct intuition.


IMO 1998 Shortlist.

http://www.kalva.demon.co.uk/short/soln/sh98a1.html


Who do you think he will hire? We can pretty much assume Foreman is a lock. So the people left are:

13 (Chick possibly with Huntington's)
Amber (Cutthroat Biatch)
Taub (Bald Plastic Surgeon)
Kutner (Cole [Black Mormon]'s former buddy)

Then, there's Chase (Scottish dude) and Cameron (Scottish dude's gf now blonde), who are working at the hospital but not with house.

I personally think Taub gets fired next episode, then the episode after that, 13 and Kutner get fired to be replaced by Chase and Cameron. The episode after that, Cutthroat Biatch will get pissed that she has to work with House's former team, and she'll be gone too.

Predictions?

Who knows maybe House will decide that the dream team is his old team!


analytic geometry
trigonometry
quadratics

If we negate the first equation and then add, we get
$ a^{2}\minus{}b^{2}\equal{}12$
You can then choose which one to solve for in terms of the other.  Let's just say $ b$:
$ b\equal{}\pm\sqrt{a^{2}\minus{}12}$
Now just plug in to get $ a$.

ya that is true. Jrav that is the first steps. can you show how u find answers plz

Use the first equation:

$ a^{2}\minus{}12\minus{}2a\sqrt{a^{2}\minus{}12}\equal{}\frac{117}{4}$

Now move the terms so we get
$ a^{2}\minus{}\frac{117}{4}\equal{}2a\sqrt{a^{2}\minus{}12}$

Square both sides to get:
$ a^{4}\minus{}\frac{234}{4}a^{2}\plus{}\frac{13689}{16}\equal{}4a^{4}\minus{}48a^{2}$

Combine like terms and solve.

I encountered  these equations when trying to solve another problem:

Given three parallel lines L1, L2, and L3 where L2 lies between L1 and L3. The distance from L1 to L2 is $ 1$ and that from L2 to L3 is $ \frac{11}{2}$. Three points A, B, and C are on L1, L2, and L3, respectively, such that ABC form an equilateral triangle. Find the length of this triangle.

Setting coordinates and you will end up with something like those messy equations. Any other good idea??

$ a\sin x\equal{}1; \; \; a\sin (60^\circ \minus{}x)\equal{}\frac {11}{2}$. From this $ \frac {\sin (60^\circ \minus{}x)}{\sin x}\equal{}\frac {11}{2}$   or   $ \frac{sin 60^\circ \cos x \minus{}cos 60^\circ \sin x}{\sin x}\equal{}\frac {11}{2}$.

$ \frac {\sqrt 3}{2 \tan x} \equal{}6 \Rightarrow \tan x \equal{} \frac {1}{4 \sqrt 3}\Rightarrow \sin x \equal{} \frac {1}{7} \Rightarrow a \equal{} 7$

hello, write the system in the form
$ \frac{b^2-2ab}{a^2-2ab}=\frac{117}{165}$ and from here we get
${ \frac{\frac{b}{a}-2}{\frac{a}{b}-2}}=\frac{117}{165}$. By setting $ z=\frac{a}{b}$ you will get a quadratic equation.
Sonnhard.


ratio
geometry
circumcircle
trigonometry
inequalities
email
geometry solved

Here's a very brief sketch:

Let $U,V,T$ be the projections of $P$ on $BC,CA,AB$ respectively, and try to show that precisely the same relations hold, except that you replace every occurence of $D,E,F$ with $U,V,T$ respectively. The rest is, as Darij's signature says, angle chasing :).

Probably I'm blind but I don't understand how you do this:

[quote="grobber"]try to show that precisely the same relations hold, except that you replace every occurence of $D,E,F$ with $U,V,T$ respectively.[/quote]

Of course, as soon as this is done, the rest is angle chasing indeed...

  Darij

Oh, I seem to have made a mistake about that... :blush:

Still unsolved, sad. :(
Let's try to solve this guys. I spent 3 hours on this and still no result.

Because of cyclic quadrilaterals we have $|<AEP| = |<BFP|= |<CDP|$. WLOG suppose that $<AEP \leq 90$. Now label $<PAB=A_1, <PAC=A_2$ etc. Let $K$ be a point on $AE$ such that $KPE$ and $BDP$ are similar. Now $AE=EK$, and it follows from the cosine rule that $PA \leq PK$. Hence $A_2 \geq <AKP = B_1$, and since $ABC$ is acute, $\sin A_2 \geq \sin B_1$. Similarly $\sin C_2 \geq \sin A_1$ and $\sin B_2 \geq \sin C_1$. Hence $\sin A_2 \sin B_2 \sin C_2 \geq \sin A_1 \sin B_1 \sin C_1$, with equality IFF $<AEP = 90$. But of course we know that equality holds; each side is the product of the perpendiculars from $P$ divided by $PA.PB.PC$. Hence $DEF$ is the pedal triangle from $P$, and the result follows easily.

Your proof is cool Fiachra. Thank you. :)

[quote="Fiachra"]Because of cyclic quadrilaterals we have $|<AEP| = |<BFP|= |<CDP|$. WLOG suppose that $<AEP \leq 90$. Now label $<PAB=A_{1}, <PAC=A_{2}$ etc. Let $K$ be a point on $AE$ such that $KPE$ and $BDP$ are similar. Now $AE=EK$, and it follows from the cosine rule that $PA \leq PK$. Hence $A_{2}\geq <AKP = B_{1}$, and since $ABC$ is acute, $\sin A_{2}\geq \sin B_{1}$. Similarly $\sin C_{2}\geq \sin A_{1}$ and $\sin B_{2}\geq \sin C_{1}$. Hence $\sin A_{2}\sin B_{2}\sin C_{2}\geq \sin A_{1}\sin B_{1}\sin C_{1}$, with equality IFF $<AEP = 90$. But of course we know that equality holds; each side is the product of the perpendiculars from $P$ divided by $PA.PB.PC$. Hence $DEF$ is the pedal triangle from $P$, and the result follows easily.[/quote]

Fiachra, your solution is interesting, but it uses inequalities. There are indeed some geometry problems which require using inequalities (an example of such a problem is the Steiner-Lehmus theorem), but this problem is not one of them. In other words, there exists a solution without using inequalities. The advantage of such a solution is that it is sure to work not only for acute-angled triangles ABC and points D, E, F lying inside the sides BC, CA, AB, but also - with respective sign conventions and with some minor modifications - for arbitrary triangles ABC and points D, E, F lying arbitrarily on the lines BC, CA, AB, while a solution which uses inequalities can completely fail in such cases. In our case, this means that the original problem:

[quote="warut_suk"]In the acute-angled triangle $ABC$, the points $D$, $E$, $F$ lie on the sides $BC$, $CA$, $AB$, respectively. The circumcircles of the triangles $AEF$, $BFD$, $CDE$ meet at the point $P$. Prove that if $\frac{PD}{PE}=\frac{BD}{AE}$, $\frac{PE}{PF}=\frac{CE}{BF}$, $\frac{PF}{PD}=\frac{AF}{CD}$, then the lines $AD$, $BE$, $CF$ are the altitudes of triangle $ABC$.[/quote]

can be extended to arbitrary triangles as follows:

[color=blue][b]Extended problem.[/b] Let D, E, F be any three points on the sidelines BC, CA, AB of an arbitrary triangle ABC. The circumcircles of the triangles AEF, BFD, CDE meet at a point P. Prove that if $\frac{PD}{PE}=\frac{BD}{EA}$, $\frac{PE}{PF}=\frac{CE}{FB}$ and $\frac{PF}{PD}=\frac{AF}{DC}$, then the lines AD, BE, CF are the altitudes of triangle ABC.

Hereby, all segments on the lines BC, CA, AB, PD, PE and PF are directed as follows:

[b]1.[/b] The lines BC, CA, AB are directed such that the segments BC, CA, AB are positive (and thus, the segments CB, AC, BA are negative).
[b]2.[/b] The line PD is directed such that the segment PD is positive if the point P lies in the same halfplane with respect to the line BC as the point A, and negative if the points P and A lie in different halfplanes with respect to the line BC.[/color]

And here is my solution for this problem. It is explicitely restricted to the case when the points D, E, F lie inside the sides BC, CA, AB of triangle ABC, but it needs just some minor modifications to hold in any other case (what cannot be said about Fiachra's solution, which really makes use of the condition that triangle ABC is acute-angled).

The main element of my solution is to show that the point P lies on the line BE. This can be done as follows:

We want to show that the point P lies on the line BE. This is equivalent to showing that the point B lies on the line EP. In other words, we have to show that the lines EP, FB and DB are concurrent (indeed, this implies that the point of intersection of the lines FB and DB lies on the line EP; in other words, the point B lies on the line EP).

Applying the trigonometric version of the Ceva theorem to triangle EFD, we see that, in order to prove that the lines EP, FB and DB are concurrent, it is enough to establish the equation

$\frac{\sin\measuredangle DEP}{\sin\measuredangle PEF}\cdot\left(-\frac{\sin\measuredangle EFB}{\sin\measuredangle BFD}\right)\cdot\left(-\frac{\sin\measuredangle FDB}{\sin\measuredangle BDE}\right)=1$,

where angles are not directed, and the two minus signs are due to the fact that the lines FB and DB divide the angles EFD and FDE externally.

Well, proving the equation

$\frac{\sin\measuredangle DEP}{\sin\measuredangle PEF}\cdot\left(-\frac{\sin\measuredangle EFB}{\sin\measuredangle BFD}\right)\cdot\left(-\frac{\sin\measuredangle FDB}{\sin\measuredangle BDE}\right)=1$

is not so difficult. At first, since the point P lies on the circumcircle of triangle CDE, we have < DEP = < DCP, and since the point P lies on the circumcircle of triangle AEF, we have < PEF = < PAF. Thus,

$\frac{\sin\measuredangle DEP}{\sin\measuredangle PEF}=\frac{\sin\measuredangle DCP}{\sin\measuredangle PAF}=\frac{\sin\measuredangle BCP}{\sin\measuredangle PAB}$.

On the other hand, since the point P lies on the circumcircle of triangle AEF, we have < EFA = < EPA, and since the point P lies on the circumcircle of triangle BFD, we have < BFD = < BPD. But we know that $\frac{PD}{PE}=\frac{BD}{EA}$; in other words, $\frac{EA}{PE}=\frac{BD}{PD}$. By the Sine Law in triangle APE, we have $\frac{EA}{PE}=\frac{\sin\measuredangle EPA}{\sin\measuredangle EAP}$, and by the Sine Law in triangle BPD, we have $\frac{BD}{PD}=\frac{\sin\measuredangle BPD}{\sin\measuredangle PBD}$; thus, the equation $\frac{EA}{PE}=\frac{BD}{PD}$ becomes $\frac{\sin\measuredangle EPA}{\sin\measuredangle EAP}=\frac{\sin\measuredangle BPD}{\sin\measuredangle PBD}$. In other words, $\frac{\sin\measuredangle EPA}{\sin\measuredangle BPD}=\frac{\sin\measuredangle EAP}{\sin\measuredangle PBD}$. Since < EFA = < EPA and < BFD = < BPD, we thus have

$\frac{\sin\measuredangle EFA}{\sin\measuredangle BFD}=\frac{\sin\measuredangle EPA}{\sin\measuredangle BPD}=\frac{\sin\measuredangle EAP}{\sin\measuredangle PBD}=\frac{\sin\measuredangle CAP}{\sin\measuredangle PBC}$.

Since < EFA = 180° - < EFB, we have $\sin\measuredangle EFA = \sin\measuredangle EFB$, so that we get

$\frac{\sin\measuredangle EFB}{\sin\measuredangle BFD}=\frac{\sin\measuredangle CAP}{\sin\measuredangle PBC}$.

Similarly,

$\frac{\sin\measuredangle EDB}{\sin\measuredangle BDF}=\frac{\sin\measuredangle ACP}{\sin\measuredangle PBA}$,

so that

$\frac{\sin\measuredangle FDB}{\sin\measuredangle BDE}=\frac{\sin\measuredangle BDF}{\sin\measuredangle EDB}=1: \frac{\sin\measuredangle EDB}{\sin\measuredangle BDF}$
$=1: \frac{\sin\measuredangle ACP}{\sin\measuredangle PBA}=\frac{\sin\measuredangle PBA}{\sin\measuredangle ACP}=\frac{\sin\measuredangle ABP}{\sin\measuredangle PCA}$.

Altogether, we have found the following equations:

$\frac{\sin\measuredangle DEP}{\sin\measuredangle PEF}=\frac{\sin\measuredangle BCP}{\sin\measuredangle PAB}$;
$\frac{\sin\measuredangle EFB}{\sin\measuredangle BFD}=\frac{\sin\measuredangle CAP}{\sin\measuredangle PBC}$;
$\frac{\sin\measuredangle FDB}{\sin\measuredangle BDE}=\frac{\sin\measuredangle ABP}{\sin\measuredangle PCA}$.

Thus, the equation that we have to prove:

$\frac{\sin\measuredangle DEP}{\sin\measuredangle PEF}\cdot\left(-\frac{\sin\measuredangle EFB}{\sin\measuredangle BFD}\right)\cdot\left(-\frac{\sin\measuredangle FDB}{\sin\measuredangle BDE}\right)=1$

becomes almost trivial:

$\frac{\sin\measuredangle DEP}{\sin\measuredangle PEF}\cdot\left(-\frac{\sin\measuredangle EFB}{\sin\measuredangle BFD}\right)\cdot\left(-\frac{\sin\measuredangle FDB}{\sin\measuredangle BDE}\right)$
$=\frac{\sin\measuredangle BCP}{\sin\measuredangle PAB}\cdot\left(-\frac{\sin\measuredangle CAP}{\sin\measuredangle PBC}\right)\cdot\left(-\frac{\sin\measuredangle ABP}{\sin\measuredangle PCA}\right)$
$=\frac{\sin\measuredangle BCP}{\sin\measuredangle PAB}\cdot\frac{\sin\measuredangle CAP}{\sin\measuredangle PBC}\cdot\frac{\sin\measuredangle ABP}{\sin\measuredangle PCA}$
$=\frac{\sin\measuredangle CAP}{\sin\measuredangle PAB}\cdot\frac{\sin\measuredangle ABP}{\sin\measuredangle PBC}\cdot\frac{\sin\measuredangle BCP}{\sin\measuredangle PCA}=1$,

where the last equality sign follows from the trigonometric version of the Ceva theorem, applied to the triangle ABC and the concurrent lines AP, BP, CP.

This completes the proof of the fact that the point P lies on the line BE. Similarly, the point P lies on the lines CF and AD. Now, the remaining part of the solution is very easy:

From now on, we work with directed angles modulo 180°.

Since the point P lies on the lines BE and CF, we have < BPC = < EPF. On the other hand, since the point P lies on the circumcircle of triangle AEF, we have < EPF = < EAF. Thus, < BPC = < EAF. On the other hand, if H is the orthocenter of triangle ABC, the lines BH and CH are altitudes in triangle ABC, and thus they are perpendicular to the lines CA and AB, respectively, so that < (BH; CA) = 90° and < (AB; CH) = 90°. Consequently,

  < BHC = < (BH; CH) = < (BH; CA) + < (CA; AB) + < (AB; CH)
             = 90° + < (CA; AB) + 90° = 180° + < (CA; AB) = < (CA; AB) = < EAF.

Comparing this with the equation < BPC = < EAF, we see that < BPC = < BHC. Thus, the point P lies on the circumcircle of triangle BHC. Similarly, the point P lies on the circumcircles of triangles CHA and AHB. But the circumcircles of triangles BHC, CHA and AHB have only one common point, namely the point H. Hence, the point P must coincide with the point H, i. e. with the orthocenter of triangle ABC. Thus, the lines AD, BE, CF, passing through the point P, must be the altitudes of triangle ABC. And the problem is solved.

[b]PS.[/b] I did not really expect to be able to access ML during this IMO training seminar, but anyway, being given this possibility for a few hours right now, I'm posting solutions to some of the ML problems I solved within the last couple of days. I will not have any time to reply to most of the PMs and emails I got recently.

  darij


kewl they made a forum for mathpath
anybody else going? :D

did you already send in your thing(or went there last year)? i doubt my solutions are smooth enough when i send mine.

*cries*
okay well i'll send mine in the first week of march..

you dont have to send it in early i just wanted to be able to plan my summer :wink:

[quote="funcia"]are you going this year? or have you gotten into mathcamp....[/quote]

I applied to AwesomeMath. I think I learn of their decision in like 3 weeks.


[quote="7h3.D3m0n.117"]Two questions about the AP final for chemistry (my instructor never informed us):

What type of questions can I expect on the final? Are they more conceptual and theory based? Or are they plug-n-chug questions?

And do any of you know any good websites for practice/past exams?

Thanks![/quote]

All the math questions from what i've seen are plug into the formula.  The multiple choice is a lot of concepts and some of them get pretty hard.  After preparing a ton for usnco none of them really seem all that hard anymore.


Give triangle $ ABC $ which have three sides $ a,b,c $. Prove that:
[tex]  \frac{(a\cdot B-b\cdot A)^2+(b\cdot C-c\cdot B)^2+(c\cdot A-a\cdot C)^2}{(a+b+c)^2} \leq \frac{{\pi}^2}{4} [/tex]

http://www.mathlinks.ro/Forum/viewtopic.php?t=15100

Much better: http://www.mathlinks.ro/Forum/viewtopic.php?t=15786

  darij


function
real analysis
real analysis unsolved

Give an example f s.t  f(x+y)=f(x)+f(y) and also f isn't continuous and linear.

Define $f$ on a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$. You can define it however you want. Since $H$, the hamel basis is uncountable there will be a point $a$ in the closure of $H$. So, you can define the function such that $f$ is not continuous in $a$. The function is welldefined.(simple exercise).

Ha Orif javob qoniqarlimi? Agar tushungan bo'lsang shu erning o'ziga tushuntirib ber!


inequalities
pigeonhole principle
algebra proposed
algebra

Prove that for any $ a;b;c;d \in R$, we always choose two numbers $ x;y$ satisfying
\[ 0\le \frac{x\minus{}y}{1\plus{}xy}\le 1
\]

hint: [hide]what's the tangent addition formula?[/hide]

2nd hint: [hide]this should really be in the combinatorics forum[/hide]

[quote="Allnames"]Prove that for any $ a;b;c;d \in R$, we always choose two numbers $ x;y$ satisfying
\[ 0\le \frac {x \minus{} y}{1 \plus{} xy}\le 1
\]
[/quote]
Denote $ (a,b,c,d)\equal{}(tanx,tany,tanz,tant)$.
From now, you shall get the result..... :lol:

@ zeb : $ (x,y)$ is chosen from $ (a;b;c;d)$  :) .And I think it is a Algebra problem   .Moderator will decide that. :D .
@mathVnpro. Continue, guy!.

[quote="mathVNpro"][quote="Allnames"]Prove that for any $ a;b;c;d \in R$, we always choose two numbers $ x;y$ satisfying
\[ 0\le \frac {x \minus{} y}{1 \plus{} xy}\le 1
\]
[/quote]
Denote $ (a,b,c,d) \equal{} (tanx,tany,tanz,tant)$.
From now, you shall get the result..... :lol:[/quote]
Countinue... hence $ x,y,z,t \in ( \minus{} \frac {\pi}{2},\frac {\pi}{2})$.
Divide $ ( \minus{} \frac {\pi}{2},\frac {\pi}{2})$ into $ 4$ sections:
$ ( \minus{} \frac {\pi}{2}, \minus{} \frac {\pi}{3}),...,(\frac {\pi}{6},\frac {\pi}{2})$
Therefore, according to the [b]Dirichlet principle[/b], there exist $ 2$ numbers such that:
$ x,y$ belong to $ 1$ section,which implies that $ 0 < x \minus{} y < \frac {\pi}{3}$ ($ WLOG: x > y$)
$ \Longrightarrow 0 < tan(x \minus{} y) < \sqrt {3}$.....OK, I am sorry for my carelessness, I have proved a weaker inequality than yours :D

[quote="Allnames"]Prove that for any $ a;b;c;d \in R$, we always choose two numbers $ x;y$ satisfying
\[ 0\le \frac {x \minus{} y}{1 \plus{} xy}\le 1
\]
[/quote]

I'm very sorry but everybody seems to understand this problem except me : what is the link between $ x,y$ and $ a,b,c,d$ in this problem.

Any $ x\equal{}y$ matches the equation $ 0\le \frac {x \minus{} y}{1 \plus{} xy}\le 1$

It seems we miss some part of the problem but mathvnpro, allnames, zeb all know the missing part. Am I mad ?

[quote="pco"]Am I mad ?[/quote]Yes. But it's treatable!

Just kidding. $ x, y$ are two of the four numbers $ a, b, c, d$, but you can't choose the same number twice (so $ (x,y) \equal{} (b, b)$ would be cheating). That make more sense?

Also - my second hint was in fact a hint. Not a suggestion to move the thread, but a hint about the problem (and how you should approach it). Stop thinking in terms of algebra, start thinking about four points on a circle (or a semicircle), and how they could be arranged.

[quote="zeb"][quote="pco"]Am I mad ?[/quote]Yes. But it's treatable![/quote]
:)
[quote="zeb"]$ x, y$ are two of the four numbers $ a, b, c, d$, but you can't choose the same number twice (so $ (x,y) \equal{} (b, b)$ would be cheating). That make more sense? [/quote]

Ahhhh yes. That make quite more sense. Thanks.

Thank for your care , zeb. Your way is also a nice hint but my solution is rather short  :lol: .
 @pco: Of course, dear . If we choose $ a$ (for example) as $ x$ ,we can't choose $ a$ as $ y$ again.
By the way $ (a;b)$ when $ a\equal{}b$ is a right couple! .
@mathVNpro:
In my solution I use $ x,y,z,t\in [0,\pi]$
And I am Allnames  :D . I sure that  :lol:

[quote="Nora.91"]Just a query - does this problem have anything to do with the Pigeon Hole Principle  :?:[/quote]
[b]Dirichlet principle[/b] that mathVnpro used can also be understood as Pigeon Hole Principle .

ok. Thanks.

i didn't know that term for PHP   :oops:

[quote="mathVNpro"]
Divide $ ( \minus{} \frac {\pi}{2},\frac {\pi}{2})$ into $ 4$ sections:
$ ( \minus{} \frac {\pi}{2}, \minus{} \frac {\pi}{3}),...,(\frac {\pi}{6},\frac {\pi}{2})$
Therefore, according to the [b]Dirichlet principle[/b], there exist $ 2$ numbers such that:
$ x,y$ belong to $ 1$ section,which implies that $ 0 < x \minus{} y < \frac {\pi}{3}$ ($ WLOG: x > y$)
$ \Longrightarrow 0 < tan(x \minus{} y) < \sqrt {3}$.....OK, I am sorry for my carelessness, I have proved a weaker inequality than yours :D[/quote]

Is it just me, or are the four intervals that you used not obviously defined?
The width of $ ( \minus{} \frac {\pi}{2}, \minus{} \frac {\pi}{3})$ is $ \frac{\pi}{6}$ and the width of $ (\frac {\pi}{6},\frac {\pi}{2})$ is $ \frac{\pi}{3}$.

Anyway, this problem appears to have been solved by several people, but no one has posted a solution. So here it goes:
[hide]
Let $ a \equal{} \tan(A)$, $ b \equal{} \tan(B)$, $ c \equal{} \tan(C)$, and $ d \equal{} \tan(D)$ where $ a,b,c,d \in \left(\minus{}\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Also, let $ x \equal{} \tan(X)$ and $ y \equal{} \tan(Y)$ with $ X,Y \in \left(\minus{}\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Also, WLOG, let $ A \le B \le C \le D$

Note that since switching $ x$ and $ y$ yields the negative of the original value, we can assume $ x \ge y \leftrightarrow X \ge Y$ and require that:

$ \minus{}1\le\frac{x\minus{}y}{1\plus{}xy}\le 1 \Longleftrightarrow \minus{}1\le\frac{\tan(X)\minus{}\tan(Y)}{1\plus{}\tan(X)\tan(Y)} \le 1$ $ \Longleftrightarrow \minus{}1 \le \tan(X\minus{}Y) \le 1$

Note that $ 0 \le X\minus{}Y < \pi$.

Thus, $ \minus{}1 \le \tan(X\minus{}Y) \le 1 \Longleftrightarrow 0 \le X\minus{}Y \le \frac{\pi}{4} \text{ \ OR \ } \frac{3\pi}{4} \le X\minus{}Y < \pi$

Suppose that it is impossible to do this. Then, $ \frac{\pi}{4} < X\minus{}Y < \frac{3\pi}{4}$.

$ (X,Y) \equal{} (B,A)$ must satisfy $ X\minus{}Y > \frac{\pi}{4}$. Thus, $ B > A \plus{} \frac{\pi}{4}$.

$ (X,Y) \equal{} (C,B)$ must satisfy $ X\minus{}Y > \frac{\pi}{4}$. Thus, $ C > B \plus{} \frac{\pi}{4} > A \plus{} \frac{\pi}{2}$

$ (X,Y) \equal{} (D,C)$ must satisfy $ X\minus{}Y > \frac{\pi}{4}$. Thus, $ D > C \plus{} \frac{\pi}{4} > A \plus{} \frac{3\pi}{4} \Rightarrow D\minus{}A > \frac{3\pi}{4}$

This is a contradiction. Thus, it is always possible to pick $ X,Y$ from $ \{ A,B,C,D \}$ such that $ \minus{}1 \le \tan(X\minus{}Y) \le 1$.
The conclusion follows directly from this.
[/hide]


How can we make the following conversion?

$ C_6H_5\minus{}CH_2\minus{}C \equiv N \longrightarrow C_6H_5\minus{}C \equiv N$

1.H3O+
2.SMIDT REACTION
3.NaNO2/HCl
4.K2Cr2O7/H+
5.HEAT WITH NH3/DCC
6.P4O10

Valeriummaximum's answer is correct. However, Chemrock's answer is incorrect because of step 3.

Another answer: amazingly, this conversion can be made simply by treating with sodium azide and a base. Can anyone provide a mechanism?

well, azide ion is a good nucleophile and will attack the carbon of CN and in presence of base C=N is converted to keto group( hydrolysis). n we know dat nitrogen is a good leaving group which leaves giving a nitrene or electron deficinent nitrogen to give hofmann bromamide product(kinda similar to schmidt reaction 'cept that a base is used)

No, that's wrong. Azide is not a good enough nucleophile to attack a cyanide carbon.

Hey Carcul
can you please explain your one step conversion using an azide??

[quote="valeriummaximum"]see if this method works..........

1.Hydrolyse to get the acid.
2.treat with SOCl2 and heat with ammonia to get the amide.
3.perform Hoffmann bromamide reaction.
4.Oxidise using CF3CO3H to get the nitro compound
5.Dehydrate to get the desired product..(some reagents you mentioned)

Is this correct :wink: ???????????????????[/quote]
Hey after hydrolysing it  to get acid you can convert it to CH3CH2COOAg and then decrease in homologous series by treating it with Br2 and now you have CH3COOH now convert it into Amide and the add P2O5 to get the required cyanide is this method also right??? :lol:

[quote="Srinitrs"]convert it to CH3CH2COOAg and then decrease in homologous series by treating it with Br2[/quote]

What kind of reaction is that?

I think srinitrs wanted to use the Hunsdiecker Borodine synthesis

First, you are probably referring to the Hunsdiecker Reaction, because the Simonini reaction is a special case of the first with a completely different outcome. Second, the Hunsdiecker Reaction is extremely hard to perform by using the silver salts of the acid, and instead mercury or lead salts are used. Third, the final outcome is

$ R\minus{}CO_2H \longrightarrow R\minus{}Br$

and not a chain reduced acid.

a confution of names :D

$ R\minus{}CO_{2}H \longrightarrow R\minus{}Br$
isn't a Carbon reduced here  :rotfl:

Of course, but not as the acid, which in this case makes all the difference. Look at Srinitrs post above.

i thought u read the question wrng and that i supported u :)

[quote="VARUNARASU"][quote="srinitrs"][quote="pardesi"]sorry i thought he wanted to decrease the chain :blush:
@srinitrs ne galli da :mad:[/quote]
hey i am  sorry but after you get that R-br you can add KCN to get the product no???
@pardesi:why me galli???[/quote]

what diff wud that make. you wud get back the reactant. :)[/quote]
OH thanks a lot now let me try the conversion
CONVERT2CN TO C6H5CH2C00H(AND ADD AgNO3) AND THEN DO A BORADINE HUNDSDIECKER REACTION TO GET C6H5CH2Br NOW OXIDISE IT USING KMnO4 AND YOU GET C6H5COOH NOW CONVERT IT TO C6H5CONH2 AND THEN ADD P2O5 TO REDUCE IT TO C6H5CN!!! NOW IS IT RIGHT???

[quote="Srinitrs"]NOW OXIDISE IT USING KMnO4 AND YOU GET C6H5COOH[/quote]

You could do that right from the start.


A 87.7 kg astronaut and a 1159 kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.121 m/s directly away from the shuttle. 7.38 s later the astronaut comes into contact with the shuttle. What was the initial distance drom the shuttle to the astronaut?

[hide]
conservation of momentum
momentum given to satellite = momentum of astronaut

1159 * .121 = 87.7 * speed of astronaut

speed of astronaut * time = distance to space shuttle


inequalities
trigonometry
calculus
derivative
function
inequalities proposed

If  $ A$ $ B$ and $ C$ are angles of triangle where $ B$ is obtuse or right angle, prove  the inequality:

$ \sin{A} \leq \cot \frac{B\minus{}C}{2}$

Is there any idea or solution (or disprovement) for this problem :?:

Inequality does indeed hold for right triangles, and it can be easily proven (using derivatives, for instance), but the inequality doesn't hold for obtuse triangles (i.e., counterexample: $ A \equal{} 1$, $ B \equal{} 2\pi/3$, $ C \equal{} \pi/3 \minus{} 1$). Interesting, $ LHS \minus{} RHS$ is negative for most $ A$s (if we consider it as a one-variable function (for a fixed obtuse angle $ B$), but not for all (it can be seen by drawing graphs of $ f(A) \equal{} LHS \minus{} RHS$ for various $ B$s).

Yes You are absolutely right [b]Hsiljak[/b]

Actually, this is an equivalent version of one old Crux problem which appeared in wrong form and before I posted this problem I hadn't seen it's correction that appeared in one of following issues.

And we find a solution for following inequality:

$ a^2 \plus{} b^2 \plus{}c^2 \plus{}abc \ge 4$, where $ a, b, c \ge 0$ and $ a\plus{}b\plus{}c\equal{}3$

[quote="nntien"]And we find a solution for following inequality:

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \ge 4$, where $ a, b, c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$[/quote]
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 1.5abc \ge 4.5$, where $ a, b, c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$ true too. :wink:

[quote="arqady"] $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 1.5abc \ge 4.5$, where $ a, b, c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$ true too. :wink:[/quote]

Nice arqady  :D !
Here is what i found :
Schur's inequality gives that $ 2(ab\plus{}bc\plus{}ca)\leq a^2\plus{}b^2\plus{}c^2\plus{}3abc$

So $ 2(a^2\plus{}b^2\plus{}c^2)\plus{}3abc\geq a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ca\equal{}9$ qed ... :)

Find the best value of $ k$ so that the inequality below is true for all $ a,b,c\ge 0$ and $ a\plus{}b\plus{}c\equal{}3$

$ a^2\plus{}b^2\plus{}c^2\plus{}kabc\ge 3\plus{}k$

[quote="quangpbc"]Find the best value of $ k$ so that the inequality below is true for all $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} kabc\ge 3 \plus{} k$[/quote]
Let $ b \equal{} c$
ineq $ \Leftrightarrow$
$ (a \minus{} b)^2(9 \minus{} \frac {3k}{2} \minus{} 3kc)\ge 0$
$ 3\equal{}a\plus{}2c\ge 2c$
$ \Rightarrow kmax \equal{} \frac{3}{2}$
 :blush:

[quote="silouan"][quote="arqady"] $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 1.5abc \ge 4.5$, where $ a, b, c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$ true too. :wink:[/quote]

Nice arqady  :D !
Here is what i found :
Schur's inequality gives that $ 2(ab \plus{} bc \plus{} ca)\leq a^2 \plus{} b^2 \plus{} c^2 \plus{} 3abc$

So $ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} 3abc\geq a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2bc \plus{} 2ca \equal{} 9$ qed ... :)[/quote]

The same way in my solution, we can use $ abc \ge (a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)$  :wink:


algebra
system of equations
complex numbers
number theory proposed
number theory

prove that there exist many infinitive odd perfect squares $n$ such that the system of equations
$n=a^2+b^2-ab=x^2+y^2+xy$ have the roots $(x,y,a,b)$ (poitive integers) satisfy the condition $(x,y)=(a,b)=1$

$(c^2-cd+d^2)^2 = \| c+ d \zeta_3 \|^2 = \| (c+ d \zeta_3 )^2\| = \| c^2 + 2cd \zeta_3 + d^2 \zeta_3^2\|=$
$= \| (c^2-d^2) + (2cd-d^2) \zeta_3 \| = (c^2-d^2)^2 - (c^2-d^2)(2cd-d^2) + (2cd-d^2)^2$
(with $\zeta_3$ a primitive $3$-rd root of unity and $\| \cdot \|$ the complex norm, so the square of the modulus).

Now
$\gcd(c^2-d^2 , 2cd-d^2) = \gcd(c(c-2d) , d(d-2c)) =$
$= \gcd(c-2d,d-2c) = \gcd(c-2d,3d)=\gcd(c-2d,3)$
as long as $\gcd(c,d) = \gcd(c,d-2c) = \gcd(d,c-2d) =1$.
So choose $c,d$ such that $\gcd(c,d)=1$ and $3 \nmid c-2d$.

So what the above shows is that there are $\infty$ many squares $s^2=(c^2-cd+d^2)^2$ such that we have coprime $m,n$ with $s^2=m^2-mn+n^2$.
Well, we have to handle that we can find coprime $x,w,t >0$ and $y<0$ with $x^2-xy+y^2 = s = t^2-wt+w^2$. Well, this can be archieved for sure if we look at
$c^2-cd+c^2 = | c+ d \zeta_3 \| = | c \zeta + d \zeta_3^2 \| =$
$\| (-d)+ (c-d) \zeta_3 \| = (-d)^2 - (-d)(c-d) + (c-d)^2$ (note that again $\gcd(-d,c-d)=1$).

Assume WLOG that we aren't done yet  for $w,t$ with $w=m,t=n$, so they have different signs (otherwise simply changing both sign solves the problem), WLOG after exchanging $t,w$ we have then $t>0>w$, but then $-w >0$ and $t-w >0$, so we are done with these values by the above.
$x,y$ are analogous now.

thanks zetax for the quick solution but how's about the elementary one?.

It's elementary, all the equations used can be checked without any complex numbers (just ignore everything between the first and last $=$ at these equations). I gave them just how to find the formulas.

i think i got it. But mine is totally different , it depends on the lemma: for every  prime $p=3k+1$ there exist 2 positive integers $x,y$ such that $p=3x^2+y^2$


Find the real roots of the equation
$ \sqrt{x\plus{}2\sqrt{x\plus{}2\sqrt{x\plus{}2\sqrt{x\plus{}...\plus{}2\sqrt{x\plus{}2\sqrt{x\plus{}2\sqrt{3x}}}}}}}$   =  x
Totally n radicles.

Not for guys who have aldready seen it.

Split 3x as x + 2x, Substitute for x & solve..

Ther real roots for these equatiuons are

x=0 and x=3 are the 2 real roots for the equation


inequalities
triangle inequality
inequalities proposed

Use the general case of triangle inequality for $n$.

$\sqrt[n]{(\sqrt[n]{a_1})^n+(\sqrt[n]{a_2})^n+...+(\sqrt[n]a_m)^n}+
\sqrt[n]{0+(\sqrt[n]{a_2})^n+...+(\sqrt[n]{a_m})^n}+...+\sqrt[n]{0+0+0+...+0+(\sqrt[n]{a_m})^n}\ge\sqrt[n]{(\sqrt[n]{a_1})^n+(2\sqrt[n]{a_2})^n+(3\sqrt[n]{a_3})^n+...+(m\sqrt[n]{a_n})^n} $

which is your problem mathx

What is the use of this kind of problem ?


hey i want to know if there is a faster way to do this other than guess and check.
Four children guess the height of their teacher to be 196cm, 163cm, 178cm, and 185cm. They all missed the exact height and they were off (in some order) by 1 cm, 6cm, 16cm and 17cm. What is the sum of the digits in the teacher's real height?

This method doesn't involve a lot of guessing: Since no two guesses differ by 1, the ones that are off by 16 and 17 are 196 and 163. Thus the teacher's height is either 179 or 180. Then 178 and 185 are off by 1 and 6, so the 185 is off by 6 and the 178 by 1, so the teacher's height is 179. The sum of the digits is 17. One question is whether the sum of the digits can be found without finding the actual height. I doubt that this is possible.


How would I modify the code for cse5 so that the constructs such as CR and MP accept a pen as a modifier?

CR doesn't draw anything, so it doesn't make sense to make a pen one of its arguments. As to MP, I admit that I was a bit stupid when I made it possible to modify the font size but not the font color. A quick fix would be to change the definition of MP to

pair MP(string s, pair A, int f=11,pair B=plain.S[color=red],pen p=black[/color])
{
if (fullbreak&&breakflag) return A;
if (A!=nullpair) label("$ \$$"+s+"$ \$$",A,B,fontsize(f)[color=red]+p[/color]);
return A;
}

This would allow you to use MP both with or without specifying the extra pen argument. So,

size(100);
pair A=(0,0),B=(2,1);
D(D(A)--D(B));
MP("A",A,19,NW,green);

will result in
[asy]import cse5; size(100); pair myMP(string s, pair A, int f=11,pair B=plain.S,pen p=black) { if (fullbreak&&breakflag) return A; if (A!=nullpair) label("$"+s+"$",A,B,fontsize(f)+p); return A; } pair A=(0,0),B=(2,1); D(D(A)--D(B)); myMP("A",A,19,NW,green);[/asy]
with big (size 19) letter $ A$ of green color.

I think that I'll ask Valentin to update the site version of cse5 in the near future and this (or equivalent) change in MP will be incorporated in the new version :).

Oops, almost forgot: don't forget to add the new pen argument to the definition of MarkPoint (long name of MP) at the end of cse5! If you don't do it, you'll get an error message.

I noticed that the \text{} command doesn't work within MP() in cse5. Is there another way to do this or some way to get it to recognize amsmath (I'd rather not get rid of the automatic math mode tags)?

Just use \mbox{} instead of \text{}:
[asy]import cse5; size(200); D(MP("\mbox{Let's call this point }A",D((0,0)),8,SW)--MP("\mbox{And this point }B",D((2,1)),8,SE));[/asy]

Thanks!  I also found it helpful to have cse5 call the olympiad package, as well as to define a short name for midpoint (useful for labeling a curve).  You might want to consider making these changes in the "official" version.  On second thought, that would pose a problem for people who don't have the olympiad package ...

I do not think that it is a good idea to include a call of olympiad.asy into the "standard version" of cse5.asy. On this site the olympiad package is included automatically, so you don't need to execute the import command. And if somebody gets tired of typing the multiple import commands in the beginning of each code file at home, he can just create once one file named all.asy (or something like that), which would consist of "include" commands for all packages he wants to use and put the [b]import all;[/b] line in the beginning of his code files.

As to short name for midpoint and labelling curves in general, it is a good idea and I will definitely implement it in the next version. Right now I'm thinking of two commands to add: PointonCurve() or PC()

pair PC(path P,real t=0.5)

which under normal circumstances would be equivalent to the waypoint command from the olympiad package but will enjoy a few foolproof features of other commands in cse5, and MarkCurve() or MC()

path MC(string s,path P,int f=pathfontsize,real t=0.5,pair B=plain.N,pen p=pathfontpen)

which would label the path P by string S at the point PC(P,t), the label being put in direction B from that point in font of size f using pen p, and which would return path P, so en-passe labelling will be possible for paths the same way it now is for points.

Is there anything else that would be useful for anybody?


integration
calculus
calculus computations

I think my teacher assigned this section by mistake because we've yet to cover this stuff but it's spring break so all I can do is try for the next few days! Can anyone help me with this problem so I can try the rest? Thanks so much!!!!

8(t+1) (dy/dt) - 5y = 15t
for t > -1 with y(0) = 16
y = ?

Thanks again =)

You can put this equation into standard linear form: $ y'\plus{}Py\equal{}Q$ where $ P\equal{}\minus{}\frac{5}{8(t\plus{}1)}$ and $ Q\equal{}\frac{15t}{8(t\plus{}1)}$. Multiply both sides by the integrating factor $ I(t)\equal{}\exp(\int P(t)dt)\equal{}(t\plus{}1)^{\minus{}5/8}$. Let the product rule work its magic: $ (Iy)'\equal{}IQ$. Integrate both sides and solve for $ y$.


Can someone prove that an irreducible markov chain on a finite state space $ \{0,1,2,...,m\}$ is not a martingale?

I do not know exactly what it means, but here is a thought:
so you have a process $ X_n$ that represents the evolution of your Markov chain, and you want to show that it is not a martingale, under any suitable filtration. What happens when you arrive at the level $ m$? If it were a martingale, $ E[X_{n\plus{}1}|X_n]\equal{}X_n$. So it means that once your Markov chain arrives at the level $ m$, it stays there. That is impossible since you supposed that it is irreducible. Is that what you meant?

Alekk, I think you are right. Moreover I want to define a stopping time so that the stopped process is a Martingale.

if your Markov chain goes up/down with equal probability when $ X_n \in \{1,2, \ldots, m-1\}$, then you can stop it the first time that it attains $ \{0,m\}$. The result will be a martingale.


function
modular arithmetic
number theory unsolved
number theory

Find all positive integer solutions $(a, b, c)$ to the function $a^{2} + b^{2} + c^{2} = 2005$, where $a \leq b \leq c$.

[quote="dblues"]Out of curiosity, let me ask another related qn. What is the smallest positive integer $k$ such that the eqn
${a_1}^2 + {a_2}^2 + \ldots + {a_k}^2$ = 2005 has infinitely many positive integer solutions in $a_1, a_2, \ldots, a_k$?[/quote]
I think that $\forall k\in \mathbb {N}$ there exists only finitely many solutions (if they exist at all), because $a_i<[\sqrt{2005}]$ and there are $< ([\sqrt{2005}]+1)^k$ k-tuples $(a_1,a_2,...,a_k)\in {\mathbb {R}}^k$ that has a chance to be a solution and hence the number of solutions is finite.

so sorry... tt was a dumb qn...  :blush:  :blush: of course there are finitely many solutions $\forall k$. Recently, I was actually working on a problem about the smallest positive integer $k$ such that the eqn ${a_1}^3 + {a_2}^3 + \ldots + {a_k}^3 = n $ has infinitely many positive integer solutions in $a_1, a_2, \ldots , a_k$. Either I cannot remember what the value of $n$ was in that qn, or I generalised some olympiad qn, but I was asking myself what if $n = 2005$ out of curiosity... Perhaps that was why i mixed it up... sorry...

What's the solution without brute force?  :?

[quote="bomb"]If you want number of solutions, apply Jacobi's formula (in terms of Dirichlet's class-number theirem) and you're done.

Bomb[/quote]

But that's not what we want...

[quote="Micchi"]wait! im so sorry... what is Jacobi's formula?[/quote]

But we want solutions, not number of solutions. I still don't know how this one is solved.

a^2 + b^2 = k is solvable iff, given $k = \prod p_i^{a_i}$, 4 does not divide $p_i^{a_i} + 1$ for all i.

http://mathworld.wolfram.com/SquareNumber.html

yeah, but that's trivial because of Fermat's 4n + 1 theorem (any prime of the form 4n+1 can be written as a sum of two squares), and the identity (a^2 + b^2)(c^2 + d^2) = (ad + bc)^2 + (ac - bd)^2

Can zhaoli or anyone post the official solution?  :)

I do not know the official solution, but here's my solution anyway.  :P

We consider $\mod 8$. Since all squares are $\equiv 0, 1$ or $4 \pmod{8}$ and $2005 \equiv 5 \pmod{8}$ it follows that $a^2, b^2, c^2$ are $\equiv 0, 1, 4 \pmod{8}$ in some order. WLOG, we ignore the order of $a,b,c$ (we drop the assumption of $a \leq b \leq c$) and assume $a^2 \equiv 0 \pmod{8}, b^2 \equiv 1 \pmod{8}, c^2 \equiv 4 \pmod{8}$. Let $a=4a_1, b=2b_1+1, c=4c_1+2$. We then have $a^2+b^2+c^2=16{a_1}^2+4{b_1}^2+4b_1+16{c_1}^2+16c_1+5 = 2005$ $\Leftrightarrow 4{a_1}^2+{b_1}^2+b_1 + 4{c_1}^2+4c_1 = 500$. Taking $\mod 4$, we have ${b_1}^2 + b_1 \equiv 0 \pmod{4} \Rightarrow b_1 \equiv 0$ or $-1 \pmod{4}$.

Case 1: $b_1 \equiv 0 \pmod{4}$. Let $b_1 = 4b_2$
We have $4{a_1}^2+16{b_2}^2+4b_2 + 4{c_1}^2+4c_1 = 500 \Leftrightarrow {a_1}^2+4{b_2}^2+b_2 + {c_1}^2+c_1 = 125$. Since $b_2 \in \mathbb{Z}^+$ and $4{b_2}^2 > 125 \ \forall \ b_2 > 5$, we have $1 \leq b_2 \leq 5$.
When $b_2 = 5$, we have ${a_1}^2 +{c_1}^2+c_1=20$. Note that ${c_1}^2+c_1 \equiv 0 \pmod{2}$, thus ${a_1}^2 = 4$ or $16$. We then have ${c_1}^2+c_1 = 16$ or $4$ respectively, which yield no solutions.
When $b_2=4$, we have ${a_1}^2 +{c_1}^2+c_1=57$. Note that ${c_1}^2+c_1 \equiv 0 \pmod{2}$, thus ${a_1}^2 = 1, 9, 25$ or $49$. We then have ${c_1}^2+c_1 = 56, 48, 32$ or $8$ respectively, which yield the only possible solution $c_1 = 7 \Rightarrow (a,b,c) = (4,33,30)$.
When $b_2=3$, we have ${a_1}^2 +{c_1}^2+c_1=86$. Similarly, ${a_1}^2$ can only be $4, 16, 36$, or $64$, which means ${c_1}^2+c_1 = 82, 70, 50$ or $22$ respectively, which yield no solutions.
When $b_2=2$, we have ${a_1}^2 +{c_1}^2+c_1=107$. Similarly, ${a_1}^2$ can only be $1, 9, 25, 49$ or $81$, which means ${c_1}^2+c_1 = 106, 98, 82, 58$ or $26$ respectively, which yield no solutions.
When $b_2=1$, we have ${a_1}^2 +{c_1}^2+c_1=120$. Similarly, ${a_1}^2$ can only be $4, 16, 36, 64$ or $100$, which means ${c_1}^2+c_1 = 116, 104, 84, 56$ or $20$ respectively, which yield the possible solutions $c_1 = 7, 4 \Rightarrow (a,b,c) = (32, 9, 30), (40, 9, 18)$.

Case 2: $b_1 \equiv -1 \pmod{4}$. Let $b_1 = 4b_2 -1$
We have $4{a_1}^2+16{b_2}^2-4b_2 + 4{c_1}^2+4c_1 = 500 \Leftrightarrow {a_1}^2+4{b_2}^2-b_2 + {c_1}^2+c_1 = 125$. Since $b_2 \in \mathbb{Z}^+$ and $4{b_2}^2 -b_2 > 125 \forall b_2 > 5$, we have $1 \leq b_2 \leq 5$
When $b_2 = 5$, we have ${a_1}^2 +{c_1}^2+c_1=30$. Similarly, by considering parity, we must have ${a_1}^2 = 4$ or $16$, which means ${c_1}^2+c_1 = 26$ or $14$ respectively, which yield no solutions.
When $b_2=4$, we have ${a_1}^2 +{c_1}^2+c_1=65$. Similarly, we have ${a_1}^2 = 1, 9, 25$ or $49$, meaning ${c_1}^2+c_1 = 64, 56, 40$ or $16$ respectively, which yield the only possible solution $c_1 = 7 \Rightarrow (a,b,c) = (12, 31, 30)$.
When $b_2=3$, we have ${a_1}^2 +{c_1}^2+c_1=92$. Similarly, we have ${a_1}^2 = 4, 16, 36$ or $64$, meaning ${c_1}^2+c_1 = 88, 76, 56$ or $28$ respectively, which yield the only possible solution $c_1 = 7 \Rightarrow (a,b,c) = (24, 23, 30)$.
When $b_2=2$, we have ${a_1}^2 +{c_1}^2+c_1=111$. Similarly, we have ${a_1}^2 = 1, 9, 25, 49$ or $81$, meaning ${c_1}^2+c_1 = 110, 102, 86, 62$ or $30$ respectively, which yield the possible solutions $c_1 = 5, 10 \Rightarrow (a,b,c) = (4, 15, 22), (36, 15, 42)$.
When $b_2=1$, we have ${a_1}^2 +{c_1}^2+c_1=122$. Similarly, ${a_1}^2$ can only be $1, 9, 25, 49, 81$ or $121$, which means ${c_1}^2+c_1 = 121, 113, 97, 73, 41$ or $1$ respectively, which yield no solutions.

Therefore, by considering the above $10$ cases, the only seven possible solutions are $(4,33,30), (32,9,30), (40,9,18), (12,31,30), (24,23,30), (4,15,22), (36,15,42)$.

As a side note, we can actually make use of what [i]Singular[/i] has mentioned regarding the solvability of $a^2+b^2 =k$ to reduce the number of cases, but this does not necessarily make the solution shorter.

Using the "Dirichlet Class-Number Theorem", or rather, finding the appropriate class numbers, to compute $\frac{r_3(2005)}{2^3 \cdot 3!} - \frac{r_2(2005)}{2^2 \cdot 2!} = 7$ is but an unnecessary complication. And besides, I haven't seen any olympiad problems that ask "Find the number of solutions..." instead of the usual "Find all solutions to...", so I doubt this theorem will be used that often.


Find the number of ordered triples (a,b,c) such that

4abc + a + b + c = 2ab + 2bc + 2ca + 500000