trigonometry
geometry
trig identities
Law of Sines
cyclic quadrilateral
exterior angle
similar triangles

In convex quadrilateral ABCD, $\angle CAD=3\angle BAC$, $\angle ABD=\angle ADB$, and $\angle ACB=\angle BDC$.

Find the numerical value of $\angle ACB$.

[hide="Is it"]$270/11$?[/hide]

Actually, that's probably wrong. :(

That's what I thought. :D

[hide="Then is it"]
$36$?
If not then maybe $540/11$? :P
[/hide]

[hide]i think that ACB is equal to 45 degrees. ACB is equal to 90-2BAC. because ACB=BDC, and triangle ADB is isosceles, it is soon found that BAC=22.5 degrees and ACB=45 degrees.[/hide]

[hide]
First I drew a circle. Then I (somewhat randomly) placed the points $A,B,C,D$. I connected the points to form the given figure.

Let $\angle ACB = \theta$. Now because the quad. is cyclic, we know that $\theta = \angle ADB = \angle ABD = \angle BDC = \angle CAB$. Therefore $\angle DAC = 3\theta = DBC$. Therefore $180 =\angle CAB + \angle ABD + \angle DBC + \angle ACB = \theta + \theta + 3\theta + \theta \rightarrow \theta = 30^{\circ}$
[/hide]

Yes, $30^\circ$ is the correct answer :)

Yeah. I learned how to use cyclic quadrilaterals! :)

hey chess , when you made a circle and in it made the given figure , means you are actually taking it given that A, B,C,D are concyclic whereas this is NOT given .

so please clarify to how you assumed or proved A, B, C, D are CONCYCLIC

[quote="frt"]Does anyone have a trig solution? :)
(Although it's longer, my original solution uses some trig)[/quote]

hey, can you post your solution?

Ok, here's my solution.

[hide]Let $\angle BAC=x$, $\angle ABD=y$ and $\angle ACB=z$.

$\angle ABD=\angle ADB=x$, so $\triangle ABD$ is an isosceles triangle.
Let $AB=AD=1$ (length doesn't matter here because we are dealing with angles). Then $BD=2\cos y$.
Looking at the angles of $\triangle ABD$, we have $3x+x+y+y=180^\circ$, or $2x+y=90^\circ \Longleftrightarrow y=90^\circ -2x$.

Applying Law of sines to $\triangle ABC$, we obtain $\frac{1}{\sin z}=\frac{BC}{\sin x}$, which gives us $BC=\frac{\sin x}{\sin z}$.

$\angle ACD=180^\circ -\angle CAD -\angle ADC=180^\circ-(3x+y+z)$
$\angle BCD=\angle ACB+\angle ACD=180-(3x+y)=90^\circ -x$

Applying Law of sines to $\triangle BCD$, $\frac{2\cos y}{\sin (90^\circ -x)}=\frac{BC}{\sin z}$

Therefore, $BC=\frac{2\cos y\sin z}{\cos x}=\frac{2\cos (90^\circ -2x\sin z}{\cos x}=\frac{2\sin 2x\sin z}{\cos x}=4\sin x\sin z$

$\angle BAD=4x$, so $0^\circ<4x<180^\circ\Longleftrightarrow 0^\circ<x<45^\circ$

$4\sin x\sin z=\frac{\sin x}{\sin z}$ and $\sin x\ne 0$, so $\sin^2 z=\frac{1}{4}\Longleftrightarrow \sin z=\pm\frac{1}{2}$

Since $0^\circ<2z<180^\circ\Longleftrightarrow 0^\circ<z<90^\circ$, $z=\boxed{30^\circ}$

[/hide]

And I also had the same question as eminem... chess64, can you explain how you knew that $ABCD$ is cyclic?

:blush: I think I just misread "convex" to be "cyclic"...

It shouldn't be too hard to prove it though... I guess it must be true b/c my answer was right. Can someone do a proof? I just learned about cyclic quads so I don't know what I'm talking about yet. :D

I'm trying to prove it, but it's not easy... (although it does look simple) :(

Maybe $ABCD$ does not have to be cyclic. What I mean by that is, the "cyclic case" is just one possible case. There are possibly other quadrilaterals that are [i]not[/i] cyclic and yet still satisfy the given condtions.

I just got AoPS vol 2. Let me go check the chapter about Cyclic Quads. :)

EDIT:

In a cyclic quadrilateral, $\angle ABD = \angle ACD$. (in the figure in this problem)

"If points C and D lie on the same side of segment AB such that $\angle ACD = \angle ADB$, then the four points A, B, C, and D are concyclic."

how do you know $\angle ABD = \angle ACD$?, that comes from the knowledge that it is cyclic, doesn't it?

this is my attempt at showing that we do not have enough information... just skip to the end if you do not feel like reading it...
let $\angle ADB = y = \angle ABD$
$\angle BAC = x$ then $\angle CAD=3x$
let the intersection of the diagonals be M
by exterior angle theorem, $\angle AMB = 3x +y$ and $\angle AMD = x +y$
since they are supplementary, $4x + 2y = 180$
using vertical angles, we get $\angle CMB = x + y$
using the similar triangles $\triangle DCB \sim \triangle CMB$ (verify this for yourself)
$\angle CMB = \angle DCB = x +y$
so $\angle DAB + \angle DCB = 4x +x +y$
using, $4x + 2y = 180$, this means that
$\angle DAB + \angle DCB = 180 +x -y$
so for it to be cyclic, $x=y$ i.e $\angle CAB = \angle DBA$ must be true, but we cannot show that, we just do not have enough given info, we have only 1 given congruency for that section of that triangle, and the other givens do not give any help either

[quote="fourierseries"]Yes that's correct.  This quadrilateral doesn't have to be cyclic![/quote]

...The AoPS book said it did... :huuh:

[quote="AoPS Volume 2"]If points C and D lie on the same side of segment AB such that $\angle ACD = \angle ADB$, then the four points A, B, C, and D are concyclic.[/quote]

[quote="chess64"]If points C and D lie on the same side of segment AB such that $\angle ACD = \angle ADB$, then the four points A, B, C, and D are concyclic.[/quote]
Isn't it $\angle ACB=\angle ADB$? :?

yes it may say that, but you do not know that $\angle ABD = \angle ACD$ (we know that $\angle ABD = \angle ADB$) and there are no other angles that allow you to use that theorem, you do not have 2 angles described as the theorem states, look at the given information again, to show that those angles are congruent, you had to use the knowledge that it is cyclic, circular reasoning does not prove anything

yes frt you are right but we know that $\angle ABD = \angle ADB$


What is the atomic weight of a metal that forms an oxide, M2O3, that is 17.29% oxygen by weight?

1)57.4

2)115

3)76.6

4)229 g/mol

5)38.3

answer is
[hide]2
48g is 17.29 then you can figure out what is the percentage of M (82.71) then do a simple ratio . and the result follows.[/hide]


inequalities
Cauchy Inequality
inequalities unsolved

Let a, b, c > 0, a + b + c = 1. Show that

  $ \frac {a^2+b}{b+c} +\frac{b^2+c}{c+a}+ \frac{c^2+a}{a+b} \ge 2$

hmm.. lets see what I can do:

we rewrite the ineq as: $ \sum ( \frac{a^{2}+b \cdot 1}{b+c} -a) \geq \sum a$ but this becomes: $ \sum ( \frac{a^{2}+b(a+b+c)}{b+c} -a) \geq \sum a$  and after a few computations we get $ \sum \frac{a(a+b)}{b+c} \geq \sum a$. Now  we rewrite the ineq as: $ \sum( \frac{a^{2}+ab}{b+c} -a) \geq 0$ and this is equivalent to: $ \sum \frac{a(a-c)}{b+c} \geq 0$. Lets make a substitution: b+c=x, c+a=y and a+b=z, thus we get: $a=\frac{y+z-x}{2}$,  $b=\frac{z+x-y}{2}$ and $c=\frac{x+y-z}{2}$. The inequality becomes: $ \sum \frac{(y+z-x)(z-x)}{x} \geq 0$, clearing the denominators we get $ \sum (x^{2}z^{2}+x^{3}z) \geq 2 \sum xyz^{2}$ which is true by Muirhead. that is all

Is this okay?

Lagrangia, I have checked your computations and I think that after clearing the denominators, in your last expression, we get

$ x^2y^2 +y^2z^2+z^2x^2 + xy^3+yz^3+zx^3 \geq 2xyz(x+y+z)$

which cannot be proved by AM-GM but by Muirhead.

Thank you manlio, I was silly to write such a thing!  :blush:  :blush:  I wrote something on paper and another thing here.. I hope it won't happen again!

thank you again!

hi!
here's another solution, with a similar idea:
fisrt rewrite the ineq : $\sum a+\sum (\frac{a(1-b-c)+b}{b+c}) \geq 3$, but clearly this is equivalent to $\sum \frac{a+b}{b+c} \geq 3$ wich follows from AM-GM, since $\frac{\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}}{3} \geq (\frac{(a+b)(b+c)(c+a)}{(b+c)(c+a)(a+b)})^{1/3} = 1$  :lol:

I think that still that inequality can be proved with AM-GM, but with very smart actions :)

Well,in fact,we can have clearly:
[tex] \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{c+a} +1= \frac{a}{b+c}
+\frac{b}{a+c}+\frac{c}{a+b} [/tex]
So our inequality is equal to:
 [tex] \frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} \geq 3 [/tex]
and it is easy from now. :)

Oh,sorry.I don't know that RaMIaF have posted the solution already.I just see that my solution is different from Lagrangia's and I... :blush:
Again, sorry... :blush:  :blush:  :)
PS:Can anybody tell me what is AM-GM?Is that BCS,Cauchy inequality or something else??? :maybe:  :)

At last,I really have another solution,a new one :lol: .By that way,we can prove that:
   [tex] \frac {a^2+3b}{b+c}+\frac{b^2+3c}{c+a}+\frac{c^2+3a}{a+b} \geq 5 [/tex]
which can be illative Mr.Namdung's inequality.So everybody,why don't you try it,at least one time?? :)

AM _GM means the inequality between arithmetical mean and geometrical mean.

have fun!

No one want to solve my problem?? :(
So,I will post it as a proposed & own problem. :)


Prove that there exist an infinite number of ordered pairs $(a,b)$ of integers such that for every positive integer $t$, the number $at+b$ is a triangular number if and only if $t$ is a triangular number.

At first, we see that $(a,b)=(9,1)$ works:
Let $T$ denote the set of triangular numbers.
If $t \in T$, let $t=\frac{n(n+1)}2$ and $m=3n+1$. Then $9t+1=\frac{9n^2+9n+2}2 = \frac{(3n+1)(3n+2)}2=\frac{m(m+1)}2 \in T$.
Conversely, if $9t+1=\frac{m(m+1)}2$, we look $\mod 3$ to get that $m \equiv 1 \mod 3$, thus there is an integer $n$ with $m=3n+1$. The same identity like before shows that $t=\frac{n(n+1)}2 \in T$.
So we proved $t\in T \iff 9t+1 \in T$.

To get infinitely many pairs $(a,b)$, we just see them as linear maps and convolute them. Thus if $t \in T \iff at+b \in T$ and $t \in T \iff ct+d \in T$, we also have $t \in T \iff a(ct+d)+b=act+(ad+b) \in T$. It's clear that our pair $(9,1)$ generates infinitely many pairs because the first paramter $a$ always increases by this convolution.

[quote="Peter"]Prove that there exist an infinite number of ordered pairs $ (a,b)$ of integers such that for every positive integer $ t$, the number $ at\plus{}b$ is a triangular number if and only if $ t$ is a triangular number.[/quote]

 Let $ \mathcal{T}$ be the set of triangular numbers:
\[ \mathcal{T}\equal{}\left\{\frac{n(n\plus{}1)}{2}\;\vert\; n\in\mathbb{N}_{\geq 0}\right\}.\]
We prove the following proposition.

 Suppose that $ d$ is an odd integer (so that $ d^{2}\equiv 1\; (mod\; 8)$). Let $ t$ be a positive integer. Then,  $ t\in\mathcal{T}$  if and only if $ d^{2}t\plus{}\frac{d^{2}\minus{}1}{8}\in\mathcal{T}$.

{Proof} $ (\Rightarrow)$ Assume that $ t\in\mathcal{T}$ so that $ t \equal{}\frac{\alpha(\alpha\plus{}1)}{2}$ for some $ \alpha\in\mathbb{N}_{\geq 0}$. Set $ l \equal{}\frac{d (2\alpha\plus{}1)\minus{}1}{2}$. Since $ d$ is odd, we see that $ l\in\mathbb{N}_{\geq 0}$. One may easily check that
\[ d^{2}t\plus{}\frac{d^{2}\minus{}1}{8}\equal{}d^{2}\frac{\alpha(\alpha\plus{}1)}{2}\plus{}\frac{d^{2}\minus{}1}{8}\equal{}\frac{l(l\plus{}1)}{2}\in\mathcal{T}.\]
$ (\Leftarrow)$ Assume that $ d^{2}t\plus{}\frac{d^{2}\minus{}1}{8}\in\mathcal{T}$ so that
\[ d^{2}t\plus{}\frac{d^{2}\minus{}1}{8}\equal{}\frac{l(l\plus{}1)}{2}\]
for some $ l\in\mathbb{N}_{\geq 0}$. It follows that
\[ (8t\plus{}1) d^{2}\equal{}(2l\plus{}1)^{2}\]
so that $ 8t\plus{}1$ is a square of an odd integer $ 2\alpha\plus{}1$, where $ \alpha\in\mathbb{N}_{\geq 0}$. Therefore, $ t \equal{}\frac{\alpha(\alpha\plus{}1)}{2}\in\mathcal{T}$.

Let $ a = (2l+1)^2, b= \frac{l(l+1)}{2}$
Whenever $at + b$ is triangular, say $k(k+1)/2$,
By trivial calculations, $t = \frac{(\frac{k-l}{2l+1})(\frac{k-l}{2l+1} + 1)}{2}$
Since $t$ is an integer, we know that $\frac{k-l}{2l+1}$ is also an integer. And hence, $t$ is triangular.

Similarly in the reverse case, we may express any integer $ m$ as $\frac{k-l}{2l+1}$ and hence the result follows.


topology
real analysis
real analysis unsolved

Let $ E$ be the set of all $ x\in [0, 1]$ whose decimal expansion contain only 4 and 7. Prove that $ E$ is perfect.

----
Work So Far: I know that $ E$ is compact, this would imply that $ E$ is closed since $ E$ is a compact subset of the metric space $ \mathbb{R}$. How would I prove that if $ p\in E$, then $ p$ is a limit point of $ E$?

Write down explicitly the sequence of points that tends to a given point.


Depends on the year.  The very early AHSMEs are pretty easy (but they make for some interesting history - note the relative lack of counting/discrete math topics in those pre-CS years).  The last few before they started the AIME are, I think, the hardest.  Most of the late 80s are a little harder (perfect scores were very rare then) than current AMC->12.

Yeah, the ones in the 1950's were of a completely different format, and the problems were significantly easier than those in the 80's onward.


number theory
greatest common divisor
number theory proposed

Let $ p$ prime number with $ p^2|2^{p\minus{}1}\minus{}1$.Prove that,for every $ n\in\mathbb{N}$ the number $ (p\minus{}1)(p!\plus{}2^n)$ has at least three different prime divisors

Sorry for my English :blush:

If $ p\minus{}1$ is not power of 2, then it is obviosly. If $ p\minus{}1$ is power of 2, then $ p\equal{}F_n\equal{}2^{2^n}\plus{}1$.
But \[ 2^{p\minus{}1}\minus{}1\equal{}\prod_{k\equal{}0}^{2^n\minus{}1}F_k\] and $ (F_k,F_j)\equal{}1$ if $ i\not \equal{}j$. Therefore only for k=n $ p|F_n\equal{}p$ and not $ p^2|2^{p\minus{}1}\minus{}1$.

Rust wrote:
[quote]If  is not power of 2, then it is obviosly. [/quote]
Why?Can you explain this case?I think the first case more harder than the second :maybe:

if $ p \minus{} 1 \equal{} 2^{t}(2s \plus{} 1), s > 0$:
$ gcd(p \minus{} 1,p! \plus{} 2^{n}) \equal{} (p \minus{} 1, 2^{n}) \equal{} 2^{min(t,n)}$
2 and the prime divisors of (2s+1) are at least 2 different prime divisors of the expression. if $ 2s \plus{} 1 \equal{} q^{m}, q$ prime:
if we assume that the expression has only 2 prime divisors, then -$ p! \equal{} 2^{n'} \minus{} 2^{n}$ (because of the gcd), and.... (EDIT: I missed that case earlier, now thinking why it can't be)

In the other case, (2s+1) has more than 1 prime divisor and we're done.

[quote]and $ (F_i,F_j)\equal{}1$ if $ i\not \equal{}j$[/quote]

why is this true?
otherwise i get it

Let $ i>j$, then \[ F_j|F_i\minus{}2\equal{}2^{2^i}\minus{}1\equal{}F_{i\minus{}1}F_{i\minus{}2}...F_0\equal{}F_ja_j\]. Therefore $ (F_i,F_j)\equal{}(F_ja_j\plus{}2,F_j)\equal{}(2,F_j)\equal{}1$.


A polynomial
\[ P(x) \equal{} c_{2004}x^{2004} \plus{} c_{2003}x^{2003} \plus{} ... \plus{} c_1x \plus{} c_0
\]has real coefficients with $ c_{2004}\not \equal{} 0$ and $ 2004$ distinct complex zeroes $ z_k \equal{} a_k \plus{} b_ki$, $ 1\leq k\leq 2004$ with $ a_k$ and $ b_k$ real, $ a_1 \equal{} b_1 \equal{} 0$, and
\[ \sum_{k \equal{} 1}^{2004}{a_k} \equal{} \sum_{k \equal{} 1}^{2004}{b_k}.
\]Which of the following quantities can be a nonzero number?

$ \textbf{(A)}\ c_0 \qquad \textbf{(B)}\ c_{2003} \qquad \textbf{(C)}\ b_2b_3...b_{2004} \qquad \textbf{(D)}\ \sum_{k \equal{} 1}^{2004}{a_k} \qquad \textbf{(E)}\ \sum_{k \equal{} 1}^{2004}{c_k}$

[hide="Solution"]
$x = 0$ is a root, so $c_0 = 0$.  Because the coefficients are real, $\sum b_k = 0 \implies \sum a_k = 0 \implies c_{2003} = 0$.  Now, $\sum c_k = P(1)$, so this is zero iff $1$ is a root of the equation, and if it is then $b_2 b_3 b_4 ... b_{2004} = 0$ - hence the answer is $\boxed{E}$ (clearly some $b_p = 0$ but the corresponding $a_p \neq 1$ so $P(1) \neq 0$).
[/hide]

[quote="t0rajir0u"][quote="Hikaru79"][quote="t0rajir0u"]Because the coefficients are real, $\sum b_k = 0 \implies \sum a_k = 0 \implies c_{2003} = 0$.[/quote]

Is this because for all the coefficients to be real, every complex root must have a conjugate somewhere else in the polynomial, hence the two imaginary parts must be negatives of one another to cancel each other out?

I wasn't quite sure why you said this and that seems to be the best explanation I can think of. Am I right?[/quote]

Yes.  In addition, I should probably add that because each complex root must be paired with another one but you know $x = 0$ is a root of the polynomial, there must be at least one other real root (so some other $b_k = 0 \implies \prod_{i=2}^{2004} b_i = 0$) which gives the answer straightforwardly without my reasoning, which takes advantage of the fact that only one of the answers are correct.  :)[/quote]

Yes, I arrived at that conclusion myself :) Your solution was a lot of fun to understand on my own, since it came with minimal explanation. I feel I learned quite a bit pondering it. Thank you!  :)

What is the motivation for the fact that complex roots appear in conjugate pairs?

Conjugation preserves addition and multiplication and fixes $ \mathbb{R}$, so if $ P(x)$ is a polynomial with real coefficients then $ P(\bar{z}) \equal{} \overline{P(z)}$, hence one is zero if and only if the other is zero.

Alternately, the minimal polynomial of $ a \plus{} bi$ over $ \mathbb{R}$ is $ x^2 \minus{} 2ax \plus{} a^2 \plus{} b^2$, which is equivalent to the Fundamental Theorem of Algebra.

Sorry to revive this post again,

but I don't quite get the reasoning behind why answer choice E) can not equal 0. You said something about:

clearly some $ b_{p} \equal{} 0$ but the corresponding  $ a_{p}\neq 1$ so $ P(1)\neq 0$

Can you please explain it a bit further?

Thank you so much!  :oops:

The reason is that all solutions to a polynomial equation are conjugate pairs. Since we have $ b_{1}= 0 $, there must be another $ b_{p}= 0 $, since all conjugate pairs are in form $ a+bi $ and $ a-bi $.

Hope this helps.

This problem is hard,someone please name which AMC problem it is...

[quote="basketball9"][quote="w squared parabola"]I actually have a question of my own: why$\sum_{k}^{2004}a_{k}=0$ ?[/quote]

rEAD the first solution, a few times..Now do you understand?[/quote]

i don't understand since the solution is short...

So this was on AMC 12 2004 A 23, I think. I understand why $ \sum_{k}^{2004}b_{k}=0 $ due to conjugate pairs, but I still do not get why $ \sum_{k}^{2004}a_{k}=0 $.

why the problem classify as introductory?

Sorry, I am being very stupid, $ \sum_{k}^{2004}a_{k}=0=\sum_{k}^{2004}b_{k}$ was stated in the problem itself. Did a stupid mistake. :oops:

Someone please explain why choice c have to be eliminate?

By the Conjugate Pairs Theorem.  We are given one of the zeros which is $0$.  This is real.  So if we do polynomial division by $x$ then we should get the polynomial $c_{2004}x^{2003} + c_{2003}x^{2002} + ... + c_1$ with degree $2003$.  Since non-real roots come in conjugate pairs when we have a polynomial with real coefficients, a.k.a if $a + bi$ is a root then $a - bi$ must also be a root, there must be one root left which is real since the degree of the polynomial is odd, so that root's $b_k$ is $0$ and the product $b_2b_3...b_{2004} = 0$.

Sorry for the bump, but does that mean $0$ is a double root? I thought the question said the roots are [b]distinct[/b]

Professordad is basically saying that there are two values for $b_i$ that can be zero. The corresponding values of $a_i$ do not have to be zero however, so they are still distinct.


A foreman noticed an inspector checking a $ 3"$-hole with a $ 2"$-plug and a $ 1"$-plug and suggested that two more gauges be inserted to be sure that the fit was snug.  If the new gauges are alike, then the diameter, $ d$, of each, to the nearest hundredth of an inch, is:

$ \textbf{(A)}\ .87 \qquad \textbf{(B)}\ .86 \qquad \textbf{(C)}\ .83 \qquad \textbf{(D)}\ .75 \qquad \textbf{(E)}\ .71$

Restatement:
A circle of radius 3 with center O has a diamater AB. AP and PB are the diameters of two circles of radius 1 and 2, respectively, internally tangent to each other and externally tangent to circle O. A third circle Q is tangent to all three circles. Find the radius of circle Q.
[hide="Solution"]
Let R be the center of the circle with diameter AP and S be the center of circle with diamter PB. Let T be the point of tangency of circle Q with the outer circle. Let Q have radius x. $ AR\equal{}RP\equal{}PO\equal{}OS\equal{}1/2$. Also, $ RQ\equal{}x\plus{}1$ and $ QS\equal{}x\plus{}2$. Note that O,Q and T are collinear, so $ OQ\equal{}OT\minus{}TQ\equal{}3\minus{}x$. By Stewart's on $ \triangle RQS$, we have
$ (x\plus{}1)^2\plus{}2(x\plus{}2)^2\equal{}(3\minus{}x)^2 \cdot 3 \plus{} 6$. Solving, $ x\equal{}\frac{6}{7}$, so the answer is B.
[/hide]

[quote=skrublord420]This problem is popping up everywhere! Insane! 1-2-3 circles seem to be all over the place, in 10 different competitions at 10 different times. I'm gonna watch out for this, because it might pop up again next year.[/quote]

Watch Problem 25 on the AMC 10B be pretty much this same problem except in 4D (making my [url=http://artofproblemsolving.com/community/c5h1380350p7652834]prediction[/url] true) but with 12 tangent hyperspheres :ninja:


trigonometry
geometry
trapezoid
rectangle
ratio
trig identities
Law of Cosines

In quadrilateral $ ABCD$, it is given that $ \angle A \equal{} 120^\circ$, angles $ B$ and $ D$ are right angles, $ AB \equal{} 13$, and $ AD \equal{} 46$. Then $ AC \equal{}$

$ \textbf{(A)}\ 60 \qquad \textbf{(B)}\ 62 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 65 \qquad \textbf{(E)}\ 72$

[hide]First find $BD$ by Law of Cosines: if $d=BD$, then $d^2=13^2+46^2-2*13*46(-1/2)=2883\Rightarrow d=31\sqrt{3}$.

Let $\theta$ be the measure of $\angle ADB$. By Law Of Sines, $\frac{\sin \theta}{13}=\frac{\frac{\sqrt{3}}{2}}{31\sqrt{3}}=\frac{1}{62}\Rightarrow \sin \theta=\frac{13}{62}$.

Angles $ABC$ and $ADC$ are both right angles, so they are supplementary and quadrilateral $ABCD$ can be inscribed in a circle. Angles $ADB$ and $ACB$ both intercept arc $AB$, so they are equal. Then $\sin \angle ACB=\sin \theta$.

If $x$ is the length $AC$, then $\frac{13}{x}=\sin \theta=\frac{13}{62}\Rightarrow x=62\Rightarrow \boxed{B}$.[/hide]

[hide="Answer"]Drawing 30-60-90 triangle $ABE$, with $E$ outside $ABCD$ and $AB$ the hypotenuse, we have $AB=13$, $AE=\frac{13}{2}$, and $BE=\frac{13\sqrt{3}}{2}$. Now, we extend $AD$ and $BC$ until they meet at $F$, forming 30-60-90 $\triangle CDF$ and $\triangle BEF$. $EF=\frac{39}{2}$, so we have $DF=46+\frac{13}{2}+\frac{39}{2}=72$, so $CD=24\sqrt{3}$. Therefore, $AC=\sqrt{(24\sqrt{3})^2+46^2}=62\Rightarrow \boxed{B}$.[/hide]

[hide="Another Approach"]Extend $CD$ and $BA$ to point $E,$ forming $30^\circ-60^\circ-90^\circ$ triangles $EDA$ and $EBC.$ Note that $EA=2DA=92,$ so $EB= EA+AB=105.$

It follows that $BC=EB\sqrt{3}=35\sqrt{3}.$ Applying the pythagorean theorem on right triangle $ABC,$ $AC=\sqrt{AB^{2}+BC^{2}}=\sqrt{13^{2}+(35\sqrt{3})^{2}}=62.$[/hide]

[quote="mihail911"]i don't know but i got this wrong.
i thought that quadrilateral ABCD was a trapezoid and angle D was 60 degrees
so extending A to DC and forming a perpendicular i formed a 30-60-90 triangle.
so i got that BC= 23*sqrt 3
then i used pythagorean theorem to find AC=2*sqrt439

where did i go wrong?[/quote]

Um, ABCD is not a trapezoid. The problem said that $\angle{B}=\angle{D}=90$, so we can only say that it is cyclic.

are you sure it can't be a trapezoid?
consider this, there is a geometric theorem/postulate that says if a line segment (BC) is perpendicular to each of two other line segments, then those two ling segments (AB & CD) are parallel. and its impossible for AD and BC to be parallel therefore quad ABCD only has one pair of parallel sides and is a trapezoid.

im probably wrong.

[quote="mihail911"]are you sure it can't be a trapezoid?
consider this, there is a geometric theorem/postulate that says if a line segment (BC) is perpendicular to each of two other line segments, then those two ling segments (AB & CD) are parallel. and its impossible for AD and BC to be parallel therefore quad ABCD only has one pair of parallel sides and is a trapezoid.
[/quote]

The situtation you describe does describe a trapezoid, except your description doesn't follow what was stated in the problem. I think your confusion is stemming from misreading the problem, angle D is right, not angle C.

The theorem is correct but the vertices of any polygon must go in clockwise or counterclockwise order A, B, ...
You can't switch the vertices around to benefit you (although it would be awesome if it actually works :) ).

Also, it seems like you're trying to make a new condition when it's not stated in the problem. The only way for ABCD to be a trapezoid is as you said, for $AB\parallel CD$ and that's impossible. So there are no parallel sides.

An approach not yet discussed:

[hide]
Using the Law of Cosines (as has been suggested) we get $BD=31\sqrt{3}$. Draw point I on BC and drop the altitude J from I to DC. Then we have rectangle AIJD and 30-60-90 triangles ABI and ICJ. Using 30-60-90 triangle ratios we get $BC=35\sqrt{3}$ and $DC=24\sqrt{3}$.

Since B and D are right angles, quadrilateral ABCD is cyclic. Thus we have the theorem $(AB)(DC)+(AD)(BC)=(BD)(AC)$. The only remaining unknown is AC, which we therefore find to be $\boxed{62}$.

For proof of that theorem see AOPS Vol. II Chapter 4 (I think).
[/hide]

[hide]Using law of cosines as above solutions, $BD=31\sqrt3.$
Since $B$ and $D$ are right angles, $ABCD$ is cyclic, and the midpoint of $AC$ is the center, $O$, of the circle. Let the radius of the circle be $r.$ Since $B,D$ lie on the circle, $BO=DO=r.$ Since $\angle BOD=120,$ as $O$ is center of arc $BCD$ of the circle, we can use 30-30-120 triangle $BOD$ to find $r,$ and thus $AC.$

[/hide]

Hi, on this problem I was trying to do some auxillary lines... why couldn't i have done AC as bisector of angles dab and dcb? wouldn't that make a 30 60 90 triangle? but it doesnt work ....

when do I extend lines!?!!? its so confusing to know when

@above: Towards a contradiction, suppose AC was a bisector of angle DAB and angle DCB. Then  triangle ADC and triangle ABC would be congruent by having congruent angles and sharing a side. Since corresponding parts of congruent triangles are congruent, it follows that AD = AB. But we are given AB = 13 and AD = 46. So 13 = 46, leading to contradiction.

[hide="Solution w/o trig"]Just extend BA to a point P such that the lines PD and AB are perpendicular. Extend line PD until to a point E such that the lines CE and PD are perpendicular. We now have a rectangle. PBCE. Using the fact that angles DAP and BAD are supplementary we get that triangle ADP is a 30-60-90 right triangle. Therefore since AD is hypotenuse is 46 AP is 23 and PB is 23+13=36. We also know that DP is 23 rt3.  We now also know that CE has to be 36 since it is a rectangle. From there we also know that <EDC=6- degrees because the transversal CD intersects the two parallel lines CB and EP and since angle DCB is 60 degrees so must angle EDC be. Now we have another 30-60-90 right triangle CDE and it has 36 opposite 60 degrees so DE opposite 30 degrees must be 12rt3. DE+DP=EP=BC=12rt3+23rt3=35rt3. BC=23rt3 and BA is 13  so using pythagorean theorem on right triangle ABC we get 13^2+(35rt3)^2=AC^2=3844. SQRT(3844)=62 and your done.[/hide]


Euler
geometry
perimeter
projective geometry
geometry proposed

Let $A'B'C'$ and $A''B''C''$ be the orthic and median triangle of $ABC$, $B'C''\cap C'B''=D$ similarly we have $E,F$, prove that $D,E,F$ are colinear and do you recognize the intersection of this line with Euler line?

The points D, E, and F are collinear by Pascal's theorem, but I do not recognize the intersection of this line with the Euler lie.

We denote as $N,$ the center of the nine point circle $(N),$ of $\bigtriangleup ABC$ and as $B_{1},$ $C_{1},$ the midpoints of the segments $BH,$ $CH$ respectively, where $H,$ is the orthocenter of $\bigtriangleup ABC.$

Because of the points $B_{1},$ $N,$ $B'',$ are collinear, as the diameter of $(N)$ and the same about the points $C_{1},$ $N,$ $C'',$ by Pascal’s theorem $($ from non-convex cyclic hexagon $B'C''C_{1}C'B''B_{1}$ $),$ we conclude that the points $D\equiv B'C''\cap C'B'',$ $H\equiv C_{1}C'\cap B'B_{1}$ and $N\equiv C''C_{1}\cap B''B_{1},$ as the intersection points of it’s opposite sidelines, are collinear.

So, the point $D,$ lies on the Euler’s line of the triangle $\bigtriangleup ABC$.

Similarly about the points $E,$ $F$ and so, we conclude that their segment line, is coincided with the Euler’s line of the triangle $\bigtriangleup ABC$ and the proof is completed.

Kostas Vittas.

 PS. The same as in a convex hexagon, we say as opposite two side segments of a non-convex one, when between the end of the one of the them and the start of the other, with respect to the order of hexagon’s perimeter, two it’s adjoined side segments are inserted.

So, in the configuration of the above non-convex hexagon $B'C''C_{1}C'B''B_{1},$ the side segments $B'C'',$ $C'B'',$ are opposite, because of between $C''$ $($ the end of $B'C''$ $)$ and $C'$ $($ the start of $C'B''$ $),$ are inserted the adjoined side segments $C''C_{1}$ and $C_{1}C'.$


function
algebra
domain
analytic geometry
polynomial
limit
inequalities solved

Let positive reals [b]x,y,z[/b] satisfy the condition [b]x+y+z=1[/b].
Find all extremes of the function
$f(x.y.z)=x^{2}z+xy^{2}+yz^2$

The domain is a triangle with vertices on the coordinate axes. There are three kinds of points for us to deal with:

1. Critical points in the interior of the triangle.

By Lagrange multipliers, there is only one such point: $\left(\frac13,\frac13,\frac13\right).$

The value of $f$ at that point is $\frac19.$

2. Critical points of the function restricted to the (interiors of) the edges.

We get three points this way: $\left(\frac13,\frac23,0\right),\left(0,\frac13,\frac23\right)$ and $\left(\frac23,0,\frac13\right).$

The value of $f$ at each of those points is $\frac4{27}.$

3. The vertices: $(1,0,0),(0,1,0)$ and $(0,0,1).$

The value of $f$ at each of those points is zero.

This is a continuous function on a compact domain; it assumes its maximum and minimum. The maximum value is $\frac4{27},$ achieved once on each edge; the minimum value is $0,$ achieved at each vertex. The critical point at the center of the domain is neither a maximum nor a minimum, not even locally.

Note that I worked the problem as if you meant for $x,y,z$ to be nonnegative reals. If you restrict them to being postive reals, then $f$ does not achieve its extreme values on the domain, but the supremuman infininum are as listed in the previous post.

One can find [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2549]here[/url] a generalization of this problem and an elementary proof. BTW, I tried to solve the generalization with Lagrange Multipliers, and I couldn't do it. Does anyone think it's possible?

[quote="Bourne"]Let nonnegative reals $ x,y,z$ satisfy the conditions $ x \plus{} y \plus{} z \equal{} 1$ and $ yz \plus{} zx \plus{} xy > 0.$ Prove that

$ f(x,y,z) \equal{} \frac {1}{x \plus{} y^2} \plus{} \frac {1}{y \plus{} z^2} \plus{} \frac {1}{z \plus{} x^2} > \frac {13}{2}.$[/quote]Without losing the generality, we may assume $ x \equal{} \min\{x,y,z\}.$

The domain is a triangle with vertices on the coordinate axes. There are three kinds of points for us to deal with:

[b]1.[/b] Critical points in the interior of the triangle.

By Lagrange multipliers, there are two such points: $ \left(\frac13,\frac13,\frac13\right)$ and $ \left(t_3,t_1,t_2\right),$

where $ t_3 \equal{} 0.1814495\cdots,t_1 \equal{} 0.4656039\cdots,t_2 \equal{} 0.3529465\cdots$ are three real roots

(the least, the greatest and the middle, respectively) of the following irreducible polynomial over $ \mathbb{Q}:$

$ 774144t^{27} \minus{} 6967296t^{26} \plus{} 32901120t^{25} \minus{} 106334208t^{24} \plus{} 270355968t^{23}$

$ \minus{} 583690752t^{22} \plus{} 1147745664t^{21} \minus{} 2127342528t^{20} \plus{} 3708075024t^{19} \minus{} 5927352480t^{18}$

$ \plus{} 8496262296t^{17} \minus{} 10770428544t^{16} \plus{} 11979589941t^{15} \minus{} 11621888361t^{14}$

$ \plus{} 9773328369t^{13} \minus{} 7065521081t^{12} \plus{} 4359747743t^{11} \minus{} 2288408071t^{10}$

$ \plus{} 1044657625t^9 \minus{} 448146989t^8 \plus{} 208533820t^7 \minus{} 109307100t^6 \plus{} 55399246t^5$

$ \minus{} 23049242t^4 \plus{} 6935253t^3 \minus{} 1346057t^2 \plus{} 146935t \minus{} 6767.$

$ f\left(\frac13,\frac13,\frac13\right) \equal{} \frac {27}{4} \equal{} 6.75,$

$ f\left(t_3,t_1,t_2\right) \equal{} 6.7970257\cdots$ is the unique real root of the following irreducible polynomial

$ 4227136m^9 \minus{} 51889328m^8 \plus{} 194004672m^7 \minus{} 239831432m^6 \minus{} 119857988m^5$

$ \plus{} 400521285m^4 \plus{} 44318448m^3 \minus{} 279928224m^2 \minus{} 117441792m \minus{} 13886208.$

[b]2.[/b] Critical points of the function restricted to the (interiors of) the edges.

We get only one point: $ \left(0,y_0,z_0\right),$ where $ y_0 \equal{} 0.64508152274 \cdots$ is the unique real root of the following irreducible polynomial

$ y^7 \minus{} 6y^6 \plus{} 16y^5 \minus{} 22y^4 \plus{} 22y^3 \minus{} 16y^2 \plus{} 8y \minus{} 2,$

$ z_0 \equal{} 0.35491847725\cdots$ is the unique real root of the following irreducible polynomial

$ z^7 \minus{} z^6 \plus{} z^5 \minus{} 3z^4 \plus{} 9z^3 \minus{} 9z^2 \plus{} 5z \minus{} 1.$

$ f\left(0,y_0,z_0\right) \equal{} 6.5175776615\cdots$ is the unique real root of the following irreducible polynomial :

$ 12k^7 \minus{} 136k^6 \plus{} 436k^5 \minus{} 436k^4 \plus{} 612k^3 \minus{} 1760k^2 \minus{} 900k \minus{} 125;$

[b]3.[/b] $ \lim_{yz\to 0}f(0,y,z) \equal{} \plus{} \infty.$

The minimum value $ f\left(0,y_0,z_0\right) > \frac {13}{2}.$

See also here : http://www.mathlinks.ro/viewtopic.php?t=150085


What are the center and the radius of a a circle whose equation is (x-A)² + (y-B)²= C?

And, Line k is drawn so that it is perp to two distinct planes, P and R. What must be true about the planes?

Planes P and R are skew
Planes P and R are parrallal
Planes P and R are perp
Plane P intersects plane R but it is not perp to plane R

another one-

Lines k and k2 intersect at E. Line m is perp to lines k and k2. What is always true?

Linkes k and k2 are perp
Line M is parralal to the plane determined by lines k and k2
Line m is perp to the plane determined by lines k and k2
Line m is coplaner with the lines k1 and k2

#1- center = $ (A,B)$
radius = $ \sqrt{C}$

#2- Planes P and R are paralell.

#3- Line $ m$ is perp to the plane determined by lines $ k$ and $ k_2$.

I just learned that circle thing a few weeks ago in Geometry!

I'm also in geometry, but we're starting on surface area, volume, etc.


geometry
circumcircle
parallelogram
geometry solved

Triangle $ABC$ with $AC=BC$ have an interior point $P$ such that $\angle PAB=\angle PBC$.
If $M$ is midpoint of $AB$.

Prove that $\angle APM+\angle BPC=\pi$

edit: Sorry wrong place..
Can anyone move this topic?
Thx..

[quote] Triangle $ABC$ with $AC=BC$ have an interior point $P$ such that $\angle PAB=\angle PBC$.
If $M$ is midpoint of $AB$.Prove that $\angle APM+\angle BPC=\pi$ [/quote]

It's clear.
$\triangle CAP\cong \triangle CBP$ and $\angle PAB=\angle PBA$.Since $C,P,M$ are collinear, $\angle APM=\angle BPM$.Also, $\angle BPM$and $\angle BPC$ are supplementay angles.So, $\angle APM$and $\angle BPC$ are supplementary angles.

I think the problem when $\angle PAB=\angle PBC$ not $\angle PAB=\angle PBA$

So your solution is wrong..

Can anyone help me?
Thx..

Lets draw circumcircle of the $ \triangle ABC$.Lines $ CP,BP$ intersect with the circle at $ K,L$ respectively. If $ m(BAP)\equal{}x,m(CAP)\equal{}y,m(ABP)\equal{}m(BCP)\equal{}a$, $ m(ACP)\equal{}m(CBP)\equal{}b$ then, $ m(BKC)\equal{}m(BLC)\equal{}m(BAC)\equal{}x\plus{}y$, $ m(LCA)\equal{}m(LBA)\equal{}m(KAB)\equal{}a$, and $ m(KBA)\equal{}m(KCA)\equal{}m(CAL)\equal{}b$.We will prove $ m(BPM)\equal{}b\plus{}y$.
Easily we see that, $ m(KAL)\equal{}x\plus{}y\plus{}a\plus{}b,m(BPC)\equal{}x\plus{}y\plus{}a\plus{}b$, $ m(PKA)\equal{}a\plus{}b\equal{}m(PLA)$.That is, $ AKPL$ is a parallelogram! Therefore, $ AP$ bisect $ KL$.Other hands, $ \triangle KLP~\triangle BPC$ (A.A.A) and $ PN,PM$ are medians of the similar triangles.$ m(KPN)\equal{}m(BPM)\equal{}b\plus{}y$. Since $ m(APC)\equal{}180^{o}\minus{}b\minus{}y$, $ \angle APC$ and $ \angle BPM$ are supplementary angles.


Let $\ell$ be a line, and let $A$, $B$ and $P$ be points on $\ell$ (where $B$ is between $A$ and $P$). Let $a$ and $b$ be the lines passing through $A$ and $B$, perpendicular to $\ell$. Let $m$ be a line passing through $P$ which is not parallel or perpendicular to $\ell$, and let $m$ intersect $a$ and $b$ in $Q$ and $R$ respectively. Let $S$ be the point on $a$ such that $AR \perp BS$. Let $U$ be the intersection of $AR$ and $BS$. Let $T$ be the point on $b$ such that $BQ \perp AT$. Let $V$ be the intersection of $BQ$ and $AT$.

a) Show that $P$, $S$ and $T$ are collinear.
b) Show that $P$, $U$ and $V$ are collinear.


Julie is picking peas from her garden. she picks x pea pods. each pod has an average of y peas in it. she cooks and eats x peas, then makes 3x peas for a potluck dinner at her church.  she is left with 200 peas left, which is 1/2 the number of peas she started out with.  find the coordinates for (x,y) and find its distance from the origin (0,0).

[hide]
From what you said,
$xy-x-3x=xy-4x=200$
$xy=400$
Substitute $xy=400$ into the first equation to get $x=50$. Then $y=8$.
So the answer is $\sqrt{2500+64}=\sqrt{2564}=2\sqrt{641}$[/hide]

[hide="My answer"]If she has 200 peas left, then she started with 400 peas, and did stuff with 200 peas. She ate x peas and cooked 3x peas, so x + 3x = 200, 4x=200, and x=50. She picked x or 50 pea pods, which means there were an average of eight peas per pod (400 total peas), which is y. So, the coordinates are (50,8) and the distance from the origin is 2 * sqrt641. I don't think that can be simplified. [/hide]

[hide]4x=200 peas, xy=400 the 400 peas total. Solving, x=50, y=8, distance from (x,y) to origin is $\sqrt{2564}=2\sqrt{641}$.[/hide]

[hide=" :P "]You figured the puzzle out!!! YAY!
[hide="solution"]There are 8x peas to start with, so y=8 and x=400/8=50. (x,y)=(50,8) and the distance is $2\sqrt{641}$[/hide][/hide]

[hide]I get equations to be
$xy-x-3x=xy-4x=200$
$xy=400$
Then by substitution you get $x=50$.

So $y=8$.

So the coordinate is $(50,8)$ and the distance is $\sqrt{50^2+8^2}=2\sqrt{641}$.

$(50,8)$
$2\sqrt{641}$[/hide]


[quote="JAS"]EVALUATE
(1/2Πi)   ∫ ezt/(z2 +1)2 dz where c is the circle│z│=3

(1/2 pi i) integration of e raise to power zt divided by z square plus 1 and its sqaure dz[/quote]

I think no one wants to mess with this because it's annoyingly ambiguous.  It's either $ \oint \frac{e^{zt}}{(z^2\plus{}1)^2}dz$ or $ \oint e^{\frac{zt}{(z^2\plus{}1)^2}}dz$.  The first version is easy.  The poles are second order, diffferentiate the appropriate quotient to get the residues and it's done.  The second one however, I think is quite a bit more challenging.

it has been solved . no need to reply

thanks.


function
limit
inequalities
trigonometry
calculus
calculus computations

Discuss the continuity of $f(x)=\lim_{n \to \infty}\sqrt[n]{1+x^{2n}}$
Happy new Year!

$x$ is real, i think. Occur of cases:

If $|x|\leq 1$ then $1<\sqrt[n]{1+x^{2n}}<\sqrt[n]{2}$ -->f(x)=1.

If $|x|>1$ then $x^{2}<\sqrt[n]{1+x^{2n}}<\sqrt[n]{2}.x^{2}$ follow $f(x)=x^{2}$. Now, problem is easy!  :wink:

[quote="math_music"]

Maybe it's kind of obvious but when i have a inequalitie, i can apply any function like $\sqrt$, $\sqrt[n]{}$, $\sin$, $\frac{1}{x}$ . But how i deal with the signs?  I need to know when the function is strictly (increasing) or decreasing to (un)change the sign?
Thanks[/quote]

That is my experienced , i think  :)

I'm sorry if I  misunderstand. English of mine is poor  :(

My writing is not good too, sorry. Anybody knows other way to solve it?

[quote="math_music"]Much Thanks N.T.TUAN, of course x is real. Let me see if i understood it correctly (I really need some advise on inequalities):
($|x|\leq 1$):

for $|x|\leq 1$ we have $|x|^{2n}=|x^{2n}|\leq 1$ and therefore $x^{2n}\leq 1$ since $x^{2n}=(x^{n})^{2}\geq 0$. Joining the two inequalities  we get $0\leq x^{2n}\leq 1$  by adding one to the last inequality we get $1 \leq 1+x^{2n}\leq 2$ and finally taking n-root we achieve $1 \leq \sqrt[n]{1+x^{2n}}\leq \sqrt[n]{2}$. When $n \to \infty$ by squeeze theorem and the last inequality $\lim_{n \to \infty}\sqrt[n]{1+x^{2n}}=1$

($|x|> 1$):

Analogous computations gives $x^{2n}>1$ and therefore $1<x^{2n}<x^{2n}+1<x^{2n}+x^{2n}=2x^{2n}$ since $x^{2n}>1$. Taking n-roots lead us to $1<x^{2}<\sqrt[n]{x^{2n}+1}<\sqrt[n]{2}x^{2}$ and by squeeze theorem  $\lim_{n \to \infty}\sqrt[n]{1+x^{2n}}=x^{2}$

Maybe it's kind of obvious but when i have a inequalitie, i can apply any function like $\sqrt$, $\sqrt[n]{}$, $\sin$, $\frac{1}{x}$ . But how i deal with the signs?  I need to know when the function is strictly (increasing) or decreasing to (un)change the sign?
Thanks[/quote]

That was beautiful!

As for the inequalities, say we have $a\leq b\leq c$, then you may apply a any positive non-decreasing function f(x), you may say that

\[f(a)\leq f(b)\leq f(c)\]

now if you multiply $a\leq b\leq c$ by a negative, then you reverse the order of it to get $-c\leq-b\leq-a$. Here is a [url=http://en.wikipedia.org/wiki/Inequality]link that will help[/url] you with the rest.

[quote="Kalle"][quote="math_music"]Benorin, much thanks for the words! I was wondering why that kind of property does not appear in Calculus I textbooks (correct me if i'm wrong) since it will be used so many times in limits computations. Thanks for the link, is the first time i see a text who cover the "apply function to both sides".[/quote]
The fact that such a thing holds is the very DEFINITION of non-decreasing function.[/quote]

I think you right Kalle, but why say that if $0 < a < b \Rightarrow \frac{1}{b}< \frac{1}{a}$ if we can say that we are applying a decreasing function on that interval? I mean why not do the general case? Sorry for the english and the off-topic.


Carcul why arent u here ???????? :rotfl:  :rotfl:  :(

Isn't expected that the C=O will be reduced to CH2? The carboxyl groups should remain untouched.

Under alcaline conditions the carboxyl groups are converted into the carboxylate anions, which are unreactive towards nucleophilic addition.

is it due to resonance
also my book say s that clemmensen reduction also does the same thing why is taht

Yes, the carboxylate anion (and also the protonated carboxylate group) is stabilised by ressonance.


I don't know if I'm just being over-paranoid, but I sent my solutions for Round 3 via snail mail on Monday, January 3rd, and it has been over a week with still no check mark.

In the event of the solutions being lost in the mail, is there any way to make up for it, or is all for nothing?  Sorry for my impatience..

Send email to usamts@usamts.org, and be sure to include your USAMTS ID#.

Don't get paranoid (and don't send email) until Thursday.  We haven't finished logging all the papers (we received 10 today, for example, that were postmarked on the due date).


function
topology
induction
set theory
superior algebra
superior algebra theorems

I don't quite understand the axiom of choice, is it, "For any set {a}, set {a} can be represented uniquely by a random member of set {a}"? Or am I off?

This does not belong here in "Intermediate Topics".

Here's what the Axiom of Choice actually says:

Given a collection of nonempty sets $X_{\alpha},$ where $\alpha$ belongs to some set of indices $A,$ is it possible to find a set which selects one element from each $X_{\alpha}.$ Another way to state this is that it is possible to find a function $f: A\mapsto\bigcup_{\alpha}X_\alpha$ such that for each $\alpha,\, f(\alpha)\in X_{\alpha}.$

Yet another paraphrase is that it is possible to find an element of the cartesian product of all of the nonempty sets $X_{\alpha}.$ You must understand that no restriction is being placed on the index set $A$ - it can be infinite, even (especially) uncountable.

I have heard the Axiom of Choice described as "obviously true" by someone who turned around to describe the Well-Ordering Principle as "obviously false". (The Well-Ordering Principle states that any set may be given a total order in which every nonempty set has a least element.) Of course, the Well-Ordering Principle is logically equivalent to the Axiom of Choice. Both are independent of the usual previous axioms of set theory - they may be neither proved nor falsified within those axioms.

Arguments with make use of the Axiom of Choice or any of its equivalents (Zorn's Lemma, the Hausdorff Maximal Principle, the Well-Ordering Principle, transfinite induction) are notorious for providing "constructions" that cannot be visualized. This is abstract mathematics at its least concrete.

It's quite important in case the given set is infinite.
None of us are able to choice the number of infinite.
It takes infinitely many times.
However, it may exist.
Axiom of Choice shows that the aspect of that.
In addtion, Godel and Cohen proved 'AC' is indepedent to the established system.


function
search
algebra unsolved
algebra

Let $ X$ be the set of all positive integers greater than or equal to $ 8$ and let $ f: X\rightarrow X$ be a function such that $ f(x\plus{}y)\equal{}f(xy)$ for all $ x\ge 4, y\ge 4 .$ if $ f(8)\equal{}9$, determine $ f(9) .$

[quote=Brut3Forc3]We have \[ f(8)\equal{}f(4\plus{}4)\equal{}f(16)\equal{}f(8\plus{}8)\\\equal{}f(64)\equal{}f(16\plus{}4)\equal{}f(20)\equal{}f(5\plus{}4)\equal{}f(9).\][/quote]

Quicker process:
f(9) equals f(8 plus 1) equals f (8) equals 9.and done!!!!

[quote=ac9991]
Quicker process:
$f(9)=f(8+1)=f(8)=9$.and done!!!![/quote]

[quote=makar]$ f(x\plus{}y)\equal{}f(xy)$ for all $ x\ge 4, y\ge 4$[/quote]

Plug \[y=4\]

Then, \[f(x+4) = f(4x) = f(2x+2x) = f(4x^2) = f(x.4x) = f(5x) = f(x+5)\]

From here, we get that f is a constant function. And we are done!!

[quote=subham1729]Actually generalization to this one came as one of TST problem this year, dunno what happened to me at hall.... the solution which I got at home just now is..... $f(2hx)=f(h^2x^2)=f(h^2x+x)=f(x+h^2+1)=f(x+2h)$ for all $x\geq h$ so, $f(x)=f(x+(h-1)^2)\implies f(x+2h-1)=f(x+2h+1)\implies f(x)=f(x+2)$ for all $x\geq h^2+5h$ now as we can take $k,l$ such that $2k+(h+2)^2=2l+1+(h+1)^2\implies f(x)=f(x+1)$ for all $x\geq h^2+5h$ now taking a number greater equals to $h$ and keep multiplying with that number we get $f(x)$ is constant for all $x\geq h$ and that's what we were asked to show.[/quote]

Could you share that problem?


1) Show that there is no positive integer $n$ such that $121$ divides $n^{2}+3n+5$

2) Let $X$ be a finite set with $n$ elements. Find the number of ordered pairs $(A, B)$ of subsets of $X$ such that $A$ is not contained in $B$ and $B$ is not contained in $A$.

3) Find all positive integer solutions $x, n$ of the equation $x^{2}+615 = 2^{n}$

For problem number three one could rewrite the equation to 615 =(2^(n/2) + x )( 2^(n/2) - x ). And then primefactorize 615 and then compare factors. The end result should be that x = 59 and n = 12 which are the only solutions.

You have missed odd case that is easy though.

[hide="1st is evident"]$n^{2}+3n+5=(n-4)^{2}+11(n+1)$[/hide]
[hide="perhaps this approach to 3d is natural"]This number of coloring $n$ elements in 4 colores:
color 1 - element belongs to $A-B$
color 2 - element belongs to $B-A$
color 3 - element belongs to $A \cap B$
color 4 - element does not belong to $A \cup B$
with restriction that both colors 1 and 2 must present.
Now it is covinient to see by inclusion-exlusing formula how much coloring does not satisfy the restriction...[/hide]


search
geometry
incenter
power of a point
radical axis
geometry proposed

It's an easy one.Let $ QH = r$ .Suppose that line $ QH$ intersect $ CD,(Q;r),(AB)$ at poits $ X,Y,Z$. We have
$ QY=QH=HZ=r$. Because $ CD$ is the radical axis of $ (Q;r) ,(AB)$ then $ P_{X}\(Q) = P_{X}\(AB)$
or $ XY.XH = XQ. XZ$
 $ \Leftrightarrow (XQ+r).XH = XQ.(XH+r)$
 $ \Leftrightarrow XQ=XH$ . QED

Sorry, I didn't get very well your solution, could you or someone else give another solution?

Thanks.

I post only the figure and I think it is enough.

Kostas vittas.

[quote="Multiple"][color=indigo]From a point $ Q$ in a circle of diameter $ \overleftrightarrow{AB}$, a perpendicular draws up to $ \overleftrightarrow{AB}$ with foot $ H$. Circle of diameter $ \overleftrightarrow{AB}$ and circle with center $ Q$ with radius $ \overline{QH}$ get cut in $ C$ and $ D$. Prove that $ \overline{CD}$ bisects to $ \overline{QH}$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote $ O$ be the middle point of $ AB,\,QP$ be the diameter of $ (O).$
And denote $ I,\,K$ be the intersections of $ CD$ with $ QH,\,QP,$ respectively.

In right triangle $ QPD,$ we have $ QD^{2}=QH^{2}=QK\cdot QP=2\cdot QK\cdot QO$
On the other hand, quadrilateral $ OKIH$ is cyclic, so $ QK\cdot QO=QI\cdot QH$

Hence $ QH^{2}=2\cdot QI\cdot QH\Rightarrow QH=2\cdot QI,$ i.e. $ I$ is the midpoint of $ QH\quad\mathbb{QED}$[/color]

See also problem :

 http://www.mathlinks.ro/viewtopic.php?search_id=1341443501&t=87126

 [u]Babis[/u]

[quote="Virgil Nicula"][color=darkred][b]An easy extension.[/b] Let $ ABC$ be a triangle for which $ a=\max \{a,b,c\}$. Denote : $ D\in (BC)\ ,\ AD\perp BC$ ;

the circle $ w'=\mathcal C(A,AD)$ ; the ciircumcirrcle $ w=\mathcal C(O)$ of the triangle $ ABC$ ; the points $ \|\begin{array}{c}\{E,F\}=w\cap w'\\\ P\in EF\cap AD\\\ \{D,R\}=AD\cap w'\\\ \{A,S\}=AD\cap w\end{array}$.

[b]Prove that[/b] : the points $ \{P,S\}$ are conjugate w.r.t. the points $ \{D,R\}$ and $ \boxed{\frac{PA}{PD}=\frac{AD^{2}}{DB\cdot DC}}$ ; the point $ D$ is the incenter of $ \triangle ESF$.[/color][/quote]

[color=darkblue][b]Proof.[/b] Observe that $ DA\cdot DS=DB\cdot DC$. The point $ P$ belongs to the radical axis of the circles $ w$ and $ w'$. Thus $ p_{w}(P)=p_{w'}(P)$ $ \Longleftrightarrow$

$ PA\cdot PS=PR\cdot PD$ $ \Longleftrightarrow$ $ PA\cdot (PD+DS)=(PA+AR)\cdot PD$ $ \Longleftrightarrow$ $ PA\cdot DS=DA\cdot PD$ $ \Longleftrightarrow$ $ \frac{PA}{PD}=\frac{DA}{DS}=\frac{DA^{2}}{DA\cdot DS}$ $ \Longleftrightarrow$

$ \boxed{\frac{PA}{PD}=\frac{AD^{2}}{DB\cdot DC}}$. Is well-known or prove easily that the division $ (R,P,D,S)$ is harmonically, i.e. $ \frac{DP}{DS}=\frac{RP}{RS}$.

Therefore,  $ \begin{array}{ccc}ED\perp ER & \implies & \widehat{DEP}\equiv\widehat{DES}\\\ FD\perp FR & \implies & \widehat{DFP}\equiv\widehat{DFS}\end{array}$, i.e. the point $ S$ is the incenter of $ \triangle ESF$.[/color]


algebra
polynomial
number theory proposed
number theory

a) $P(x),R(x)$ are polynomials with rational coefficients and $P(x)$ is not the zero polynomial. Prove that there exist a non-zero polynomial $Q(x)\in\mathbb Q[x]$ that \[P(x)\mid Q(R(x)).\]  b) $P,R$ are polynomial with integer coefficients and $P$ is monic. Prove that there exist a monic polynomial $Q(x)\in\mathbb Z[x]$ that \[P(x)\mid Q(R(x)).\]

\[Q(x)=\prod_{i}(x-R(x_{i})),\] were $x_{i}$ are roots of P(x).

Your solution does not work for b when $R(x)$ is not monic.
Try a new thing for the part b.

$Q(x)$ defined monic. Why it don't work?

You see any relation with your Old conjucture Omid Hatami ,Some of the Summer camp participants knew about this and you guys put it in the exam?
http://www.mathlinks.ro/Forum/viewtopic.php?t=18474

$Q(x)$ that you defined is monic but its coefficients are not integers when $R(x)$ is not monic.

Your conditions are incorrect, I edit them:
a) $P(x),R(x)$ are polynomials with rational coefficients and $P(x)$ is not the zero polynomial. Prove that there exist a non-zero polynomial $Q(x)\in\mathbb Q[x]$ that \[P(x)|Q(R(x))\] b) $P,R$ are polynomial with integer coefficients and $P$ is monic. Prove that there exist a monic polynomial $Q(x)\in\mathbb Z[x]$ that \[P(x)|Q(R(x))\] It is easy to check, that $R(x_{i})$ are algebraic integer numbers (for it don't needed, that R(x) - monic, sufficiently, that R(x) had integer coefficients). Therefore $Q(x)=\prod_{i}(x-R(x_{i}))$ monic and had integer coefficients.

[i]To Rust :[/i]
Yes, You are right. Your solution is correct. :blush: I was out of my mind.
[i]
To lomos_lupin :[/i]
I don't see any relations between these two problems.


What type of music do you listen to, and why?

although this is included under other, I think in this forum math/science songs deserve their own category

pesonally i listen to some of everything although my favorites are math/science songs, capitol steps, and queen

[quote="bleumoose"]i love how the makers of polls only consider the things they have heard of and then... "other"...

how about alternative, adult alternative, christian, country, r&b, celtic, world, reggae, new age, metal- is metal on there?,  big band, folk, punk... blues...
and that's only the ones i can remember at the moment.[/quote]

What the heck is adult alternative?

who can resist geeksta / nerdcore / cs rap?  :lol:

Wow, I was the first and only person to vote for Rap/Hip-Hop :o

My favorite is classical, but I can listen to and maybe even like any of the other ones except for rap.  Rap is more noise than music.

I usually listen to rock, though i dont mind listening to rap either

Let's see...
[b]Death Metal [/b](Cannibal Corpse, Morbid Angel, Deicide...)
[b]Oi![/b] (Cock Sparrer, Blitz, 4Skins, Angelic Upstarts...)
and of course [b]Depeche Mode  [/b]:lol:

[quote="zeb"]Anything except rap. Really. Except for punk. Or emo. Or pop. Or country.

For example, Nirvana and REM are some of my favorite bands.[/quote]

basically you can just scratch that entire first line, huh?

Does anyone remember the name of that death-metal orchestra?

i went and chose techno/trance. i just like most kinds of this genre, not all though

[quote="EUCLA"]i love rap...[/quote]
Music is like candy, it's best without (w)rappers.

lol, a friend of mine said that.

The list is diverse enough. If you like PUNK, that goes under ROCK. Anyway, Metal is also very dif from Rock, that should be mentioned. And theres no country, or reggae, which doesnt fit in any category. And Asian Pop isnt special. I'm sorry to burst ur... AZN-PR1D3, but.. no. Its POP.

Anyway, i mostly listen to rock.. lots of genres of it too, and a little bit of hiphop/rap.

Motown, of course :D  (especially from the 60's and 70's). My favorite groups are probably the Temptations, Marvin Gaye, and the Supremes. I like a few other oldies, too, and my favorite non-motown artists would probably be Beatles, Elton John, and Cat Stevens.

And when I'm in the mood for it, Hindi songs. But usually they're getting kind of trashy/ are just too repetitive wih no "soul", so I only like listening to a few songs.

[quote="WarpedKlown1335"]Wow...this is ridiculous.  Btw I think that death metal orchestra you're talking about was Dream Theater.  I'm not too sure.

Anyone here listen to Coheed & Cambria or The Mars Volta on here?[/quote]

Finally someone who's heard Coheed and Cambria...I've just heard "welcome home" and i love the heavy guitar and solos in there.I usualkly listen to A lot of rock like myfavourites [b]TheRasmus[/b],Creed,evanescence,Bonjovi(their new song welcometo...is really nice in terms of lyrics),linkinpark...and many more.Santana is nice as well...i only like the solos the old bands used to play,not their songs like Scorpion,Deeppurple,GnR(Yuck,but solos.).

And people saying rap is useless,i used to think the same till last yuear till someone told me 'bout eminem and as soon as i heard some songs of his like Loseyourself,withoutme,mynameis.....i fell in love with rap...I hate all the blackies in rap,pointless raaps they sing xcept 2pac.

 :rotfl:  :rotfl: 1234567890,since when do you listen to hindi songs?You Indian linked?I too have just 5 or six favourites...

Oh,i forgot,crazy frog is nice too,and lets get it started and my humps by Blackeyedpeas.

I'll chose...ummm......none right no,i can't make up my mind which genre to chose but i hate pop :censored:

[quote="shadysaysurspammed"]I hate all the blackies in rap,pointless raaps they sing.

Oh,i forgot,crazy frog is nice too,and lets get it started and my humps by Blackeyedpeas.
[/quote]

Oh man...the force is weak in this one.

And racism...oh man...

 :(

[quote="shadysaysurspammed"]i fell in love with rap...I hate all the blackies in rap,pointless raaps they sing
[/quote]

Your statement makes absolutely no sense considering rappers are  99% African Americans. I think this is a very racist comment and maybe you're just in love with Eminem, not rap :furious:

And try to post your thoughts in coherent sentences ;)

[quote="h_s_potter2002"][quote="shadysaysurspammed"]Oh,i forgot,crazy frog is nice too,and lets get it started and my humps by Blackeyedpeas.[/quote]

Those are some of the most annoying songs.  Ever.[/quote]
[img]http://www.csh.rit.edu/~tomp/YouWin.png[/img]

OOoopssss....I m not  racist or whtever,I just said tht the raps which most of the african americn people do justy don't suit me for various reasoms,not bcos they are BLaCK...I hate pop too,for what,Racial resons?Of course not.Didn't i mention i like  2pac too,he's blck too,I never discriminmte aginst anyone.

I just used some wrong languge bymistke,sorry to do that and thank you for all those comments.Nice and understnding users over here.

and buy the wy,I don't think I ridiculed anyone's choice here or I could hve started  whole new dictionry on wht i think apout some of the songs you guys listen too,I got warned about this once by Tokenadult .Every one has their own choice of music nd by the wy,Cray frog ws once U.K no.one...(NO,th'ts not the reason i like it)

BUT above all,sorry for using inpproprite language.[b]Peace[/b]

There are black people on AOPS btw.

Lol...I'm not mad though.

I like this song:
Daler Mehndi - Tunak Tunak Tun
really cool!

And I find this song and the music video attractive:
"Plan B" at http://gprime.net/
Is this hip-hop, rap?

And I like music of Seo Taiji. Seo Taiji is truely a musician, genius, semi-God of music.

HAHAHAHA!!! Tunak Tunak Tan, oh man, so weird yet funny. Totally cracked up first time I saw the video for it.

[quote="catcurio"][quote="TESoccer66"]Rap is more noise than music.[/quote]I couldn't agree more.  :D It amazes me how they can talk that fast though. :)[/quote]

Love rap, probably because I can talk that fast. :thumbup:

[quote="WarpedKlown1335"]Wow...this is ridiculous.  Btw I think that death metal orchestra you're talking about was Dream Theater.  I'm not too sure.[/quote]

Dream Theater is so far from death metal; it's mostly prog metal. That being said, I'm a huge DT fan; big prog/power metal fan in general...Dragonforce, DT, Ayreon, LTE, Mars Volta, ATDI, even the cheesy stuff like Falconer or Symphony X. Big fan of clubbin' music, like Sean Paul. Also love the virtuoso stuff; Bela Fleck, Steve Vai, Malmsteen, etc. Dunno what category I'd fit under.

Dude techno/trance is sweet.
Its the biggest genre if you look internationally and its in all the big dance clubs.
plus, whenever i listen to it i get a jolt of energy. that's how i managed to go through 5 aps as a sophomore.


calculus integration algebra polynomial quadratics

Prove that the number $ n^{2} \minus{} n \plus{} 2$ is not  dividiable by $ 7^{2}$

Edit: Sorry : it is not dividiable ..

HI! I wonder if there's any other polynomials with the same property as the other polynomial? And what kind of relation exist between the solution in this set?

$ n^2 \minus{} n \plus{} 2 \equiv 0 \bmod 49$ if and only if $ 4n^2 \minus{} 4n \plus{} 8 \equal{} (2n \minus{} 1)^2 \plus{} 7 \equiv 0 \bmod 49$, but clealry if $ 7 | 2n\minus{}1$ then $ 7^2 | (2n\minus{}1)^2$.  It is not hard to come up with other such examples:  for example, $ x^3 \plus{} 7 \equiv 0 \bmod 7^3$ also has no solutions.

The more interesting examples of polynomials of this form occur when $ 49$ is replaced by a prime rather than a prime power; for quadratic polynomials, a complete characterization of such polynomials can be obtained via [url=http://en.wikipedia.org/wiki/Quadratic_reciprocity]quadratic reciprocity[/url], and for higher-degree polynomials all irreducible polynomials $ \bmod p$ occur as a divisor of a polynomial of the form $ x^{p^n} \minus{} x \bmod p$ for some $ n$.


A photograph measuring 16 inches by 20 inches is reduced uniformly so that the greater measure becomes 5 inches. What is the number of inches in the perimeter of the reduce photo?

the greater measure is 20 inches, so to be reduced to 5 inches, you divide by 4
the other measure is 16 inches, so we divide that by 4 also, to get 4

now we have our side lengths (4 and 5), so we find the perimeter
$ 4\plus{}4\plus{}5\plus{}5\equal{}18$

our answer is 18

A less random sounding solution:

By "uniformly" they probably mean proportionally, so you're dealing with two rectangles which are the same shape but scaled down.

Call the big rectangle R and the little rectangle r.

The rule of proportionality would tell you: $ \frac{long side of r}{long side of R}\equal{}\frac{short side of r}{short side of R}$.

Numerically:
$ \frac{5}{20}\equal{}\frac{x}{16}$
$ \frac{1}{4}\equal{}\frac{x}{16}$
$ 4\equal{}x$
(Then the perimeter would be as above, 18)


algebra
polynomial
quadratics
search
integration
inequalities
inequalities unsolved

If the polynomial  $f(x) = ax^2 + bx +c$  has real roots , prove that  $a + b+ c \leq \frac{9}{4} max \{ a,b,c\}$

[u]Babis[/u]

I set the determinant of the quadratic equstion to greater than or equal to zero, and I can see that this is going to be fairly difficult.

Ah , now I remember that this ineq.was refered as MOSER inequality. Try to search in the web.I'm afraid , this a problem from old Amer. Math. Montnly Issues.I' ll try tis , too.

   [u]babis[/u]

The problem is not that hard, what is a bit tricky is to avoid too much computation in the solution.

[color=blue][b]Problem.[/b] Let a, b, c be three real numbers such that the polynomial $f\left(x\right)=ax^2+bx+c$ has real roots. Prove that $a+b+c\leq\frac94\cdot\max\left\{a;\;b;\;c\right\}$.[/color]

[i]Solution.[/i] For the sake of contradiction, assume that this is wrong, i. e. that we have $a+b+c>\frac94\cdot\max\left\{a;\;b;\;c\right\}$. Then, since $\max\left\{a;\;b;\;c\right\}$ is greater or equal to each of the numbers a, b, c, we have $a+b+c>\frac94a$, $a+b+c>\frac94b$ and $a+b+c>\frac94c$.

First we show that the numbers a, b, c are all positive. In fact, if one of the numbers a, b, c - say, a -, would be $\leq 0$, then we would have $a+b+c\leq b+c\leq\max\left\{a;\;b;\;c\right\}+\max\left\{a;\;b;\;c\right\}=2\cdot\max\left\{a;\;b;\;c\right\}$, what, compared with $a+b+c>\frac94\cdot\max\left\{a;\;b;\;c\right\}$, would yield $\frac94\cdot\max\left\{a;\;b;\;c\right\}<2\cdot\max\left\{a;\;b;\;c\right\}$, and thus $\max\left\{a;\;b;\;c\right\}<0$. Hence, it would follow that

$a+b+c\leq\max\left\{a;\;b;\;c\right\}+\max\left\{a;\;b;\;c\right\}+\max\left\{a;\;b;\;c\right\}$
$=3\cdot\max\left\{a;\;b;\;c\right\}<\frac94\cdot\max\left\{a;\;b;\;c\right\}$

(since $\max\left\{a;\;b;\;c\right\}<0$), what contradicts with $a+b+c>\frac94\cdot\max\left\{a;\;b;\;c\right\}$. Thus, it is clear that all numbers a, b, c must be positive.

Now, we can rewrite $a+b+c>\frac94b$ as $a+c>\frac94b-b=\frac54b$, and similarly we transform $a+b+c>\frac94a$ into $b+c>\frac94a-a=\frac54a$. Thus,

$a+c>\frac54b=\frac54\cdot\left(b+c\right)-\frac54c>\frac54\cdot\frac54a-\frac54c=\frac{25}{16}a-\frac54c$,

so that $c+\frac54c>\frac{25}{16}a-a$, or, equivalently, $\frac94c>\frac{9}{16}a$. Division by $\frac{9}{16}$ transforms this into 4c > a, so that 4c - a > 0. Similarly, 4a - c > 0. Thus, $25ac-4\left(a+c\right)^2=\left(4a-c\right)\left(4c-a\right)>0$, so that $25ac>4\left(a+c\right)^2$; hence, $81ac=\frac{81}{25}\cdot 25ac>\frac{81}{25}\cdot 4\left(a+c\right)^2=4\left(\frac95\left(a+c\right)\right)^2$. Now, since $a+c>\frac54b$, we have

$81ac>4\left(\frac95\left(a+c\right)\right)^2>4\left(\frac95\cdot\frac54b\right)^2=\frac{81}{4}b^2$.

Hence, we get $81ac>\frac{81}{4}b^2$, or, equivalently, $4ac>b^2$. Thus, $b^2-4ac<0$; hence, the polynomial $f\left(x\right)=ax^2+bx+c$ has a negative discriminant, so that it cannot have real roots. This is a contradiction to the condition of the problem. Hence, our assumption was wrong, and we indeed must have $a+b+c\leq\frac94\cdot\max\left\{a;\;b;\;c\right\}$, so the problem is solved.

[[b]EDIT:[/b] Corrected a mistake.]

  Darij

It is really a great work , Darij . Bravo !

 Thank you!

 [u]babis[/u]

[quote="draperp"][quote="zhaobin"]another proof:
I'll prove just $a+1$(I think the other case is $a=-1$)
we can assume $c \le a$(because c and a are symmetrial)
with $b^2-4c \ge 0$
if $b<1$
then  we have ${a+b+c \le \min\{1+b+\frac{b^2}4,1+b+1} \le \frac 9 4$
if $b \ge 1$
then we have  $a+b+c \le \min \{1+b+\frac{b^2}4,1+b+1 \} \le \frac 9 4 b$
so we have sloved the problem[/quote]

Do you mean a=1?
And why are a and c "symmetrical" and c<=a?[/quote]
because $c(\frac 1 x)^2+b\frac 1 x+a =0$,(we may assume x is not equal to 0,if not c=0,obviously)
also have a real root.
also you can understand it by the condition is $b^2-4ac \ge 0$
that  a and c "symmetrical".

Darij you said that
. Finally, $a+b+c>\frac94b$ yields $4\left(a+b+c\right)^2>4\left(\frac94b\right)^2=\frac{81}{4}b^2$.

I think that the following   is not true

$a+b+c>\frac94b$ yields $4\left(a+b+c\right)^2>4\left(\frac94b\right)^2$

since you cannot square because  you didn't know the sign of $b$.

If I missed something tell me.

[quote="silouan"]Darij you said that
. Finally, $a+b+c>\frac94b$ yields $4\left(a+b+c\right)^2>4\left(\frac94b\right)^2=\frac{81}{4}b^2$.

I think that the following   is not true

$a+b+c>\frac94b$ yields $4\left(a+b+c\right)^2>4\left(\frac94b\right)^2$

since you cannot square because  you didn't know the sign of $b$.[/quote]

Thanks, you are right. I corrected my proof (and simplified it a bit as well).

  Darij

A friend of mine gave me this solution
As darij said
[quote="darij grinberg"]. Thus, it is clear that all numbers a, b, c must be positive.
[/quote]

Now let $a$ the maximum of  the numbers  $a,b,c$.
We have $a+b+c=\int^{1}_{0}{(3ax^2+2ax+c)dx}\leq\int^{1}_{0}{(3ax^2+2ax+\frac{b^2}{4a})dx}\leq a\int^{1}_{0}{(3x^2+2x+\frac{1}{4})dx}=\frac{9}{4}a$

and we are done

[size=75]
The original question was published in
[1] [b]Moser L. and J.R.Pounder [/b], [color=blue][b] Problem 53 ,[/b][/color] Canad.Math.Bull., 5,70(1962)
There exists a generalization, namely published as
[2] [b]Dixon J.D. [/b], [color=blue][b] Polynomials with real roots ,[/b][/color] Canad.Math.Bull., 5, (1962) 259-263.

In [2] it's proved following
[b]THEOREM (Dixon)[/b][color=green][b] For all real polynomials $P(x)=\sum\limits_{k=0}a_kx^k$ of degree  $n$ , with only real zeros, we have
\[ \begin{array}{|rcl|} \hline \displaystyle P(1)= \sum\limits_{k=0}^{n}a_k &\le& \displaystyle \alpha_n\cdot\max\limits_{j}{a_j} \\ &&\\ \displaystyle \min\limits_{j}{a_j}&\le&\displaystyle \beta_n\cdot \max\limits_{j}{a_j}\; , \\ &&\\ \hline \end{array}  \]
where
\[ \left\{\begin{array}{lcl} \alpha_n &=&\displaystyle \frac{(n+1)^n}{{n\choose s}(n-s)^{n-s}(s+1)^s}\\ &&\\ \beta_n&=&\displaystyle {n\choose s}^{-1} \end{array}\right.  \]
with $s =\left[\frac{n}{2}\right] .$ The constants $\alpha_n$ and $\beta_n$ are the best possible.
[/b][/color]
These informations are from  the monograph by D.S.Mitrinovic "Analytic Inequalities",Springer  Verlag,Grundlehren,Band 165, 1970,p.217.
[/size]


calculus
integration
limit
trigonometry
calculus computations

When evaluating:

$ \int_{ - \infty}^{\infty}\frac {dx}{(1 + x^2)\sqrt {2 + x^2}}$

using residue integration, one is led to the limit:

$ \mathop\lim\limits_{\epsilon\to0}\int_{\sqrt {2} + \epsilon}^{\infty}\frac {dy}{(1 - y^2)\sqrt {y^2 - 2}}$

Is it easier to just evaluate the first directly?  I've done the second one (via trig sub) since that was the problem.  Haven't attempted the first directly though.  Anyway, though both were interesting integrals.

If you are interested in calculating the first integral here is a method of attack:

Let $ I$ be the value of your integral, then

$ I\equal{}2\int\limits_{0}^{\infty}\frac{dx}{(1\plus{}x^{2})\sqrt{2\plus{}x^{2}}}$. Using the substitution $ x\equal{}\frac{1}{t}$ we get that

$ I\equal{}\int\limits_{0}^{\infty}\frac{2tdt}{(1\plus{}t^{2})\sqrt{2t^{2}\plus{}1}}$.

$ t^{2}\equal{}u$ and you get that

$ I\equal{}\int\limits_{0}^{\infty}\frac{du}{(1\plus{}u)\sqrt{2u\plus{}1}}$.

Finally, the substitution $ \sqrt{2u\plus{}1}\equal{}z$ reduces the integral to an elementary one

$ I\equal{}\int\limits_{1}^{\infty}\frac{2dz}{z^{2}\plus{}1}\equal{}\frac{\pi}{2}$.

Thanks!  I used $ y\equal{}\sqrt{2}Sec(\theta)$ for the second one and it got a little messy with the expression $ ArcTan[Sin(\theta)]$.

I'll try your method to do the second one.  Looks cleaner. :)

. . . kinda' messy too but still neat:

For the second, let $ y=\frac{1}{t}$.  This gives:

$ \int \frac{dt}{(1-t^2)\sqrt{1-2t^2}}$

Now let $ t^2=u$.  I get:

$ \frac{1}{2}\int \frac{du}{(1-u)\sqrt{1-2u}}$

Let $ z=\sqrt{1-2u}$ obtaining:

$ -\int\frac{dz}{1+z^2}=-ArcTan(z)=-ArcTan(\sqrt{1-2u})=
-ArcTan(\sqrt{1-2t^2})=-ArcTan(\frac{\sqrt{y^2-2}}{y})$

Alright then:

$ \mathop\lim\limits_{\epsilon\to 0}\int_{\sqrt{2}+\epsilon}^{\infty}\frac{dy}{(1-y^2)\sqrt{y^2-2}}=
\left-\mathop\lim\limits_{\epsilon\to 0}ArcTan(\frac{\sqrt{y^2-2}}{y})\right]_{\sqrt{2}+\epsilon}^{\infty}=
-\frac{\pi}{4}$

which when combined with everything else from the residue integration:

$ \int_{-\infty}^{\infty}\frac{dx}{(1+x^2)\sqrt{x^2+2}}=2\pi i(-i/2)-(-2)(-\pi/4)=\pi/2$

yields the desired answer.


Pretend you are Mr. Jones. Try to figure out way it was built.

Its okay but a little better control would be nice... what are you supposed to do anyway?

I beat it. It claims to have a virus, but w/e I scanned it. I'm actually pretty sure that you're not allowed to write programs that say YOUR COMPUTER IS NOW INFECTED.

i wrote a virus once, but all it did was create A LOT of txt files full of junk all over the place. but it was boring.

Well I mean this "virus" was a fake virus that just displayed YOUR COMPUTER IS INFECTED.

You can make some weird stuff with bat files lol

it is a fake virus. It is called a joke. no, it is not meant for mac

[quote="pacman2812"]Runtime error "11"

Division by zero

That's what I get :([/quote]
That is strange. Did you try to use cheat codes? Cheat code do not exist. I know this, because I built it.

@mathelete: I'm not a cheater. My comp is weird, so it might be it...

Sorry, I should not have called you a cheater.

You and mewto55555 just made me lose. :noo:

To play this, I think you need Windows XP or higher. Can you figure out how it was made?

[quote="ernie"]I lost "The Game". :noo:[/quote]
Look!  Up in the sky!  It's a bird!  It's a plane!  It's xkcd!

[img]http://imgs.xkcd.com/comics/anti_mind_virus.png[/img]


number theory
greatest common divisor
algebra
polynomial
Vieta
quadratics

if a,b and q=$ (a^2 \plus{} b^2)/(ab\plus{}1)$ are integers$ >\equal{}$0,the prove thet q=gcd$ (ab)^2$

the only solutions for this are $ (a^3 , a)$ or $ (a, a^3 )$ (the proof of this is well-known, using Vieta jumping).

$ \Rightarrow q \equal{} a^2 \equal{} [gcd(a,b)]^{2}.$ :juggle:

[quote="Bugi"]BTW, this is the harder version of IMO 88/6. The problem only wanted you to prove that q is a square.[/quote]

oh sorry, I thought the proof already covers the fact that the solutions are in the form I wrote earlier... :oops:

Vieta jumping is a technique in solving NT problems
Great article:
http://reflections.awesomemath.org/2007_5/vieta_jumping.pdf

I challenged myself that i will solve the problem from scratch. So how about this view in solving the problem.

This is one of those problems where you have a method of transforming
one solution into another.  In particular, if you have a solution of
the form

 (a, b, q)

where a, b, and q are all integers, then you can find another such
solution.  The trick here is thinking of this as a quadratic
polynomial in a:

 a^2 - (q*b)*a + (b^2 - q) = 0

Then this polynomial has two solutions (in a, for a given choice of b
and q) whose product is b^2 - q and whose sum is q*b.  For a
particular choice of b and q, those two solutions might not be
integers, or even rational, but if you have one solution in positive
integers, (a, b, q), then another solution will be (qb - a, b, q).

You can check this directly by substituting qb - a for a into the
equation

 a^2 + b^2 = q*(a*b + 1).

and checking that it simplifies to the same equation.

So what does this tell us?  Well, since the equation is also symmetric
in a and b, we can get from one solution (a, b, q) to either (b, a, q)
or (qb - a, b, q).  For example, we can start with (0, 2, 4) and get to

 (2, 8, 4),
 (8, 30, 4),
 (30, 112, 4),
 (112, 418, 4),
 (418, 1560, 4),
 (1560, 5822, 4),
 (5822, 21728, 4),
 (21728, 81090, 4),
 (81090, 302632, 4),
 (302632, 1129438, 4),
 (1129438, 4215120, 4),
 (4215120, 15731042, 4),
 (15731042, 58709048, 4),

and so on.  In this way, we can keep getting more and bigger
solutions.  But the value of q doesn't change, and since you want to
know about which q values you can get, these can all be considered as
a single solution.

But you notice that we can get that q value with extremely large a and
b values, which is annoying because it means that we can't rule out
very large a and b values when trying to decide which q values are
possible.  But there is something else we can do that's just as good:
 Suppose there is some solution (a, b, q).  Maybe a and b are very
large, but if so, then we can do the same method of getting from one
solution to another, but in the other direction, until we get to the
smallest solution in that chain (for that q value).

So we suppose that we have a solution in positive integers (a, b, q).
 If a < b, then we switch a and b, so that a >= b.  And if a >= b and
0 < bq - a < a, then we change a into bq - a and repeat.  Since the
positive integer a + 2b is getting strictly smaller at every step,
this can't go on indefinitely, which means that eventually the process
will stop at a solution which has both

 a >= b,   and
 either bq - a >= a or bq - a <= 0.

So we have proven that if there is a positive integer solution (a, b,
q) to the equation

 a^2 + b^2 = q*(a*b + 1),

then there is another positive integer solution with possibly
different a and b but which has the same q and also satisfies the
inequalities

 a >= b,   and
 either bq - a >= a or bq - a <= 0.

We might call such a solution a "minimal solution."  So if we want to
decide what q values are possible, then we only have to find all of
the (minimal) solutions in positive integers to

 a^2 + b^2 = q*(a*b + 1),
 a >= b,   and
 either bq >= 2a or bq <= a.

In the case where bq >= 2a, we have (since b and ab + 1 are both positive)

 (a^2 + b^2)/(ab + 1) = q >= 2a/b,
 b(a^2 + b^2) >= 2a(ab + 1).

Then since a >= b, we have a^2*b >= b^2*b = b^3, and therefore

 2*a^2*b = a^2*b + a^2*b >= a^2*b + b^3 = b(a^2 + b^2) >=
    2*a*(a*b + 1) = 2*a^2*b + 2*a

giving us 0 >= 2*a, contradicting that a is positive.  So all minimal
solutions must have bq <= a.  I will demonstrate that all minimal
solutions must have bq = a, because otherwise bq < a, so

 a >= bq + 1
 a^2 >= abq + a
 q(ab + 1) = a^2 + b^2 >= abq + a + b^2
 q >= a + b^2
 a^2 + b^2 = q(ab + 1) >= (a + b^2)(ab + 1)
      = a^2*b + a + ab^3 + b^2 > a^2 + a + a + b^2,

which is, again, impossible since a and b are positive.  So all
minimal solutions must have bq = a, which means that taking one more
transformation would take our solution to one with a = 0 (but that's
not positive, so we don't consider it a legal transformation).  Either
way, the point is that

 a^2 + b^2 = qab + q = a^2 + q,

and therefore q = b^2.

So the only possible values for q are perfect squares.  And you can
get all solutions in positive integers by starting with a solution of
the form (0, b, b^2) and then repeatedly applying the transformations

 (a, b, q) -> (b, a, q) or
 (a, b, q) -> (qb - a, b, q).


geometry
geometric transformation
dilation
geometry unsolved

Hey everyone, this one is paticularly driving me crazy:

Two circles are tangent internally at [i]P[/i], and a chord [i]AB[/i], of the larger circle is tangent to the smaller circle at [i]C[/i].  [i]PB[/i] and [i]PA[/i] cut the smaller circle at [i]E[/i] and [i]D[/i] respectively.  If [i]AB[/i]=15, [i]PE[/i]=2 and [i]PD[/i]=3, determine the length of [i]AC[/i].

note that a dilation through $P$ takes the small circle into the larger one. $\triangle PAB$ is therefore similar to $\triangle PDE$, and $DE \parallel AB$. this implies $AD: BE = 3: 2$, so let $AD = 3x, BE = 2x$; and let $AC = y$; then $BC = 15 - y$. by power of the point, we have

$y^2 = 3x(3x+3) = 9x(x+1)$;
$(15-y)^2 = 2x(2x+2) = 4x(x+1)$

dividing, we get $(\frac{y}{15-y})^2 = \frac94$; $\frac{y}{15-y} = \frac32$; $y = 9$.


[color=cyan]Tangents to circle A from a point P intersect the circle at M and N.  If PA=2 and MP=1, compute the area of the region bounded by segments MP and NP and minor arc MN.[/color]

Solution:

[hide]Draw radii MA, AN. By Pythagorean Theorem,

AM=AN=sqrt[3]

Thus, the area of the quadrilateral AMPN is sqrt[3]

NOw, arc MN is a 60 degree arc, so the area of AMN is pi/2.

So the area is question is sqrt[3] - pi/2

i think...[/hide]

[color=cyan]I think it's safe to say the area of that sector is not pi, because that would be larger than the area of the quadrilateral it is contained within, no?[/color]

My mistake; I drew a faulty diagram, of course.

Waiit, if the area is pi and the arc is 60 degrees, then isn't the corresponding sector of an area pi/6?

No, the circle has radius :rt3: and thus area 3:pi:, and so the sector has area :pi:/2.

Oh, my bad. Read the question wrong. Sorry.

The circle has area 3pi, but the central angle is 60, so the area of the sector is 1/6 the area of the circle, which is pi/2


Let $ a,\ b,\ x,\ y$ be positive numbers such that $ a\plus{}b\equal{}1$.
Prove that $ (ax\plus{}by)^n\leq ax^n\plus{}by^n$ for all positive integers $ n$.

Isn't this just weighted 2-variable power mean?


Facts:

1.  There are 5 houses (along the street) in 5 different colors:
      blue, green, red, white and yellow.
2.  In each house lives a person of a different nationality:
      Brit, Dane, German, Norwegian and Swede.
3.  These 5 owners
drink a certain beverage:
      beer, coffee, milk, tea and water,
smoke a certain brand of cigar:
      Blue Master, Dunhill, Pall Mall, Prince and blend,
and keep a certain pet:
      cat, bird, dog, fish and horse.
4.  No owners have the same pet, smoke the same brand of cigar, or drink the same beverage.

   Hints:

1.  The Brit lives in a red house.
2.  The Swede keeps dogs as pets.
3.  The Dane drinks tea.
4.  The green house is on the left of the white house (next to it).
5.  The green house owner drinks coffee.
6.  The person who smokes Pall Mall rears birds.
7.  The owner of the yellow house smokes Dunhill.
8.  The man living in the house right in the center drinks milk.
9.  The Norwegian lives in the first house.
10.  The man who smokes blend lives next to the one who keeps cats.
11.  The man who keeps horses lives next to the man who smokes Dunhill.
12.  The owner who smokes Blue Master drinks beer.
13.  The German smokes Prince.
14.  The Norwegian lives next to the blue house.
15.  The man who smokes blend has a neighbor who drinks water.

--------------------------------------------------------------------------------

The question is:  Who keeps fish?

bebinam in hamoon moamaye anishtain(nemidoonam)nist ke too rooznameye sharif chapesh kardan??
(2 sale pish)

salam
khaneye sabz mahi dare prince mikeshe ghahve mikhore va almany
khasti amare baghiye ye khanne ha ham behet bedam.
dar har soorat masaleye ghashangi bood.thanks aragon. ;)

Dear friend
I solved it at about 1 hour.

The German keeps fish.

If you have a doubt about it, I can explain that why this is true!!!!

By

kheili tekrarieh. kash masael yekam jadidtar basheh.


ABC is an equilateral triangle  and PQRS is a square inscribed in the circle APQBRCS.

If AB is parallel to PQ, prove that A, P, Q, B, R, C, S are some of the vertices of a regular 24-sided polygon

If they are points of a 24-gon, then they'll subtend segments of $\frac{360}{24}=15^{\circ}$ or multiples of that degree measure. Using subtended arcs and simple circle geometry you can quickly show that it's true.


Abundancy index $ I(x)=\frac{\sigma(x)}{x}$ where $ \sigma$ is the sum of divisors function. Note: $ I(x)=2$ means $ x$ is perfect.

I just came across a proof that the set of rational numbers that are abundancy indexes are dense in $ (0,\infty)$. Then I came across a proof that the set of rational numbers that are not abundancy indexes is dense in $ (0, \infty)$. Clearly this is a partition of $ \mathbb{Q}$.

Does anyone else have interesting examples of partitions of $ \mathbb{Q}$ in which all parts are dense in the same interval?

Odd numerator | even numerator (in lowest terms).


function
modular arithmetic
number theory
totient function
relatively prime

Can someone explain how euler's totient function works?

For example, use it to find the remainder when 3^2007 is divided by 1000.

$\phi(n) =$ Number of positive integers $k \le n$ such that $gcd(k,n) = 1$.

Euler's Totient Theorem states that:

If $a,m$ are relatively prime positive integers, then

$a^{\phi(m)}\equiv 1 \pmod{m}$.

Try using this to solve your problem.

[quote="computer_nuke"]So...$3^{400}\equiv 1 \pmod{1000}$, and so $3^{2000}\equiv 1 \pmod{1000}$ according to the pattern of repeated mods, since $3^{0}\equiv 1 \pmod{1000}$, and therefore $3^{2007}\equiv 3^{7}\pmod{1000}$

Am I right?[/quote]

Yup. You may want to evaluate $3^{7}\pmod{1000}$ to get a final answer.

Ok thanks for the help. I'll also find a proof of it so I know how it works.

The proof I know is very similar to the one for Fermat's Little Theorem, which is fitting. So if your interested in trying to prove it on your own, you can approach it with that mindset


function
inequalities theorems
inequalities

Let $ f$ defined on the interval $ I$ be a convex and continuos function on $ I$.Let $ a,b,c\in I$ such that $ a\leq b \leq c$.We have the folowing ineguality :
\[ a( f(b)-f(c))+b(f(c)-f(a))+c(f(a)-f(b))\geq 0 \]

[quote="Vrajitoarea Andrei"][color=darkred]Let $ f$ defined on the interval $ I$ be a convex and continuos function on $ I$.Let $ a,b,c\in I$ such that $ a\leq b \leq c$.
We have the following ineguality : $ a\left[f(b)-f(c)\right]+b\left[f(c)-f(a)\right]+c\left[f(a)-f(b)\right]\geq 0$. [/color][/quote]
[color=darkblue][b]Proof.[/b] Denote the points $ A(a,f(a))$, $ X(x,f(x))$, $ C(c,f(c))$ which belong to the graph of the function $ f$, where $ a\le x\le c$.
The equation of the line $ AC$ is $ y\equiv d(x)=f(a)+\frac{f(c)-f(a)}{c-a}\cdot (x-a)$. Because the function $ f$ is convex obtain that for any
$ x\in (a,c)$ , $ f(x)\le d(x)$, i.e. $ f(x)\le f(a)+\frac{f(c)-f(a)}{c-a}\cdot (x-a)$ $ \Longleftrightarrow$ $ a\left[f(x)-f(c)\right]+x\left[f(c)-f(a)\right]+c\left[f(a)-f(x)\right]\ge 0$,
what is [b]well-known[/b] (it is a characterization of the convexity for a function $ f$). For $ x: =b$ obtain the required inequality.[/color]


Let $ p(x) \equal{} 2^x \minus{} 1$, where $ x$ is a prime number. What is the sum of the three smallest distinct values of $ p(x)$?

Try them out. Obviouly, negative and fractional values of x yield non-integers. So start with 0 and go up.

0-0
1-1
2-3
3-7
4-15
5-31

And 3 primes appear.
$ 3\plus{}7\plus{}31\equal{}41$

[quote="BOGTRO"]Huh? It says x is a prime number, not p(x).

Thus, $ x\equal{}2,3,5$, and $ p(x)\equal{}3,7,31$. Adding gives $ 41$.

Somehow you got the same answer, but I don't think your solution method is correct.[/quote]
Well, this formula is well-known to give prime outputs only for prime inputs. The first prime input that results in a non-prime output is $(11, 2047)$ as $2047 = 23*89$
Anyways, I digress. BOGTRO's interpretation is correct, however.


[hide]The number 5 must appear in either position 2 or 4, because if it didn't then there would have to be a number greater than it which is impossible. 1 and 2 must be in positions 1,3 or 5 because there are not two numbers which are less than 1 or 2. Therefore the possible candidates for numbers in positions 2 and 4 are (3,5) and (4,5).

Suppose that 3 appears in position 2 and 5 in position 4. The 4 cannot appear next to the 3 because it is larger than it, so the 3 must have 1 and 2 either side of it. There are two ways of doing this (1,3,2) , (2,3,1). Number 4 must go in the last remaining position. There are therefore 2 possible eyebrows when 3 is in position 2 and 5 is in position 4. For the same reasons, there are 2 eyebrows when 5 is in position 2 and 3 is in position 4. There are therefore [b]4[/b] eyebrows when the numbers in position 2 and 4 are (3,5)

Suppose that 4 appears in position 2 and 5 in position 4. The 1, 2 and 3 can go anywhere since 4 and 5 are greater than all of them. There 6 ways of arranging the numbers 1, 2 and 3. For the same reasons there are also 6 eyebrows when 5 is in position 2 and 4 is in position 4. There are therefore [b]12[/b] eyebrows when the numbers in positions 2 and 4 are (4,5)

The total number of eyebrows is therefore 12+4=[b]16[/b]

[b]There are 16 eyebrows.[/b][/hide]

These are more frequently referred to as [[alternating permutation]]s.  They're counted by [url=http://www.research.att.com/~njas/sequences/A000111]A000111[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

[quote="kurt.math"]An eyebrow is an arrangement of the numbers 1,2,3,4 and 5 such that the second and fourth numbers are each bigger than both of their immediate neighbours. For example (1,3,2,5,4) is an eyebrow and (1,3,4,5,2) is not.

The number of eyebrows is?[/quote]

We can not have the numbers 1 or 2 in the second or fourth places, so we have to try the other three numbers for those positions.

It doesn't matter whether the three is in the second place or the fourth place, we'll just multiply the number of eyebrow numbers in which three is the second number by two. If it is the second number then the one and the two have to be adjacent to it. There are two ways to arrange the two numbers. Once we have the first three numbers in, the five has to be in the fourth spot. So there are two possibilities in which the three is in the second spot. If we simply flip the order around then we have another two possibilities for a total of four possibilities.

Now that we have considered all the possibilities in which the three is one of the numbers in the second or fourth position, we now have to put the 4 and the 5 in the second and fourth spots. There are two ways to do this. Then there are 6 ways to arrange the other three numbers, so we have 12 ways.

12 + 4 = 16, so we have a total of sixteen eyebrow numbers.


function
calculus
derivative
integration
inequalities
real analysis
real analysis solved

Let $f(x)$ be a function with the second derivative $f''(x)$ and $r$ be a positive constant number.If $f''(x)<\frac{1}{r}$ holds for all real numbers $x$ and $f(0)=f'(0)=0$,then prove that a circle $x^2+(y-r)^2=r^2$ has no points of intersection with the graph of $y=f(x)$ except origin.

I think, some important information is missing.
Maybe $f(0)=f'(0)=0$ :?

Yes.You may be right.I will edit right now.Thank you,Myth.

kunny

Thank you for quick reply.

Consider $x\geq 0$.
Since $f'(0)=0$, we conclude $f'(x)=\int_0^xf''(t)dt\leq \frac{x}{r}$. Further, $f(x)=\int_0^{x}f'(t)dt\leq \int_0^x\frac{t}{r}dt=\frac{x^2}{2r}$.
Now we are to prove that $f(x)$ lies below circle for $x\in [0,r]$. Indeed, it reduces to following inequality:
\[f(x)\leq r-\sqrt{r^2-x^2},\]
using obtained inequality we obtain new inequality to prove:
\[\frac{x^2}{2r}\leq r-\sqrt{r^2-x^2}.\]
But it simply means that $\sqrt{r^2-x^2}\leq r-\frac{x^2}{2r}$ which is true by squaring.

That's right. :) I have no problems for you,as well as Kent Merryfield.
By the way,what kind of problems are you poor at?
I wonder this is a nonsense question for you,anyway thank you for your reply,Myth.

kunny

[quote="kunny"]By the way,what kind of problems are you poor at?[/quote]
I don't like high algebra (matrices, groups, etc.)  :)


advanced fields
advanced fields unsolved

Prove that every ultrametric space is totally disconnected.

in ultrametric spaces the open balls ar also closed and form a basis of the topology.


Suppose $ \cos x \equal{} 0$ and $ \cos (x \plus{} z) \equal{} 1/2$. What is the smallest possible positive value of $ z$?

$ \textbf{(A) } \frac {\pi}{6}\qquad \textbf{(B) } \frac {\pi}{3}\qquad \textbf{(C) } \frac {\pi}{2}\qquad \textbf{(D) } \frac {5\pi}{6}\qquad \textbf{(E) } \frac {7\pi}{6}$

[hide="Solution"]

By the cosine addition formula we have

\begin{eqnarray*} \cos(x+z) &=& \cos x\cos z-\sin x\sin z\\ &=& -\sin x\sin z\\ &=&1/2. \end{eqnarray*}

Also, if $\cos x=0$ then $\sin x=\pm 1.$  If $\sin x=1$ then $\sin z=-1/2$ and the smallest positive value for $z$ is $\frac{7\pi}7$ but if $\sin x=-1$ then the smallest positive value of $z$ is $\boxed{\frac{\pi}6}.$[/hide]

[hide]
On the unit circle, $\cos \frac{\pi}{2}$ and $\cos \frac{3\pi}{2}$ are equal to $0$
Also, $\cos \frac{\pi}{3}$ and $\cos \frac{5\pi}{3}$ are equal to $\frac{1}{2}$
We can see that the smallest difference is between $\frac{3\pi}{2}$ and $\frac{5\pi}{3}$
$\frac{5\pi}{3} - \frac{3\pi}{2}$
$= \frac{10\pi}{6} - \frac{9\pi}{6}$
$= \frac{\pi}{6}$
A
[/hide]

[quote=MOP_Janitor]Is this is a correct solution? I couldn't find this on the wiki or the real page so I wasn't sure if this thinking process was correct.

[hide=My Solution]For $\cos(x+z) = \frac12$ we need $x + z = 60$ degrees or $x+z = -60$ degrees. Also notice that for $\cos x = 0,$ $x$ has to be equal to $90$ or $-90$ degrees.

Since we want $z$ to be positive, we have $-90 + z = 60$ or $-90 + z = -60.$  Solving for $z$ we get the answer as $150$ or $30$ degrees. Since the problem asks for the smallest positive value of $z$ we find that $30$ degrees. Converting this into radians gives us $\boxed{\frac{\pi}{6}}$ which is answer option $\textbf{(A)}.$
[/hide][/quote]

Unnecessary bump.

I believe that you are wrong because a few weeks ago I bumped an old thread and people said it was ok as long as I provided a sol and didn't troll.

[url=https://artofproblemsolving.com/community/c5h1997705p17573387]Proof[/url]
I also believe that this bump isn't unnecessary because I am trying to understand if my thinking was correct.

I don't completely understand the hate against bumping. Yes, I know that bringing old threads to the top of the forum for no valid reason can get annoying, but when somebody actually posts a solution (that too, to a thread that is part of the [b]contest collection[/b] - refer to c&p posting rec #3) and is asking for verification from the community, I don't see the reasoning for why it should be characterized as unneeded or some form of misconduct. I mean to say, isn't the purpose of the AoPS community to engage in meaningful conversations about mathematics and problem solving in general? If a user is attempting to spark one by posting a solution to a contest math problem, in my opinion, they should be encouraged, not reprimanded.

$cos(\frac\pi2)=0$

$x=\frac\pi2$

$\cos(\frac{2\pi}{3})=\frac12$

$z=\frac{2\pi}3-\frac\pi2=\frac\pi6$

$\frac\pi6<\frac\pi3<\frac\pi2<\frac{5\pi}6<\frac{7\pi}6$

$\boxed{\frac\pi6}$


Let $ A_i\equal{}\lbrace c_i \plus{} za_i : z \in \mathbb{Z} \rbrace , \ a_i,c_i \in \mathbb{Z}, \ i\equal{}1,2,..n$. Prove that if:
$ \bigcup_{s\equal{}1}^{n} A_s \equal{} \mathbb{Z}  \wedge A_i \cap A_j\equal{}\emptyset$ then
$ \exists i \neq j : a_i\equal{}a_j$.

In another word, you ask to prove that if $ \mathbb{Z}$ is partitioned into finitely aritmetical sequences, then two of these sequences have the same ratio.
This is equivalent to prove that if, for a certain $ n$, you color the vertices of a regular $ n$-gon with finitely many colors such that, for each color $ c$, the vertices colored with $ c$ are the vertices of a regular polygon then there are at least two monochromatic isometric polygons.
This last problem is a famous russian one and already appeared in the forum here and there, and the usual proof deals with formal series.

Pierre.


function
quadratics
conics
parabola
algebra
quadratic formula

Let f(x) be a quadratic function such that the coefficient of the quadratic term is m, and the solution set of f(x)>-2x is (1,3).
(1) Find f(x) if there are 2 equal roots to f(x)+6m=0.
(2) If max{f(x)}>0, find the range of m.

[hide]Part A:
Realize that f(x)+2x=0 at 1 and 3 since a parabola can only curve downwards once.  Assume the equation is $mx^2+bx+c$.  Plug in x=0.  Since 0 isn't in the solution set, $c\not>0$.  Assume c is zero.  THen, at m(x-1)(x-3)=f(x)+2x, m is also 0 so the function isn't quadratic.  Therefore, c is negative.  After factoring the above equation and putting the 2x on the side with the m, c=3m b=-4m-2.  This means m is also negative.  Now, using Part A's requirement that f(x)+6m=0 has two equal roots, we redesign c to fit these new conditions.  c=9m.  SInce the roots are equal, by the quadratic formula, $16m^2+16m+4=36m^2$.  After combining like terms and dividing out the 4, $5m^2-4m-1=(m-1)(5m+1)=0$.  Since m is negative, m=-1/5.  If m=-1/5, then c=-3/5 and b=-6/5.  THerefore, if f(x) satisfies condition A, then $f(x)=-1/5(x^2+6x+3)$[/hide]