[quote="shaaam"]Given that $ a,b,c \geq 0$, Prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq \frac {3}{2} \plus{} 2\sum_{cyc}\left(\frac {a \minus{} b}{a \plus{} b \plus{} 2c}\right)^2
\]
:)[/quote]
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b} \equal{} \frac{3}{2} \plus{} \frac{1}{2}\sum \frac{(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)}$and
$ (a\plus{}c\plus{}b\plus{}c)^2\ge 4(a\plus{}c)(b\plus{}c)$ and we are done :wink:

Yes, that was really easy one. How about the following, find the maximum $ k$ (best constant), so that, the following inequality holds:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq \frac {3}{2} \plus{} k\sum_{cyc}\left(\frac {a \minus{} b}{a \plus{} b \plus{} 2c}\right)^2
\]

 :wink:

The following inequality also holds.
For $ a,b,c$ positive reals , prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\]
\[\ge\frac{3}{2}+ \frac{1}{4(a+b+c)}(\frac{(b-c)^2}{b+c}+\frac{(c-a)^2}{c+a}+\frac{(a-b)^2}{a+b})\ge\frac{3}{2}.\]
\[\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\]
\[\le\frac{1}{2}+ \frac{3}{2}(\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ab}{(b+c)(c+a)})<2.\]
《Mathematics research in high school 》（China Nanchang）No.12(1999)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=555965]Iranian math comp:[/url]For positive real $a,b,c$ prove that $ \sum{\frac{a}{b+c}} \geq \frac{3}{4}+\frac{1}{2}\sum{\frac{a^2+b^2}{(a+b)^2}} $

[quote="shaaam"]Yes, that was really easy one. How about the following, find the maximum $ k$ (best constant), so that, the following inequality holds:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq \frac {3}{2} \plus{} k\sum_{cyc}\left(\frac {a \minus{} b}{a \plus{} b \plus{} 2c}\right)^2
\]
[/quote]
$2$ is the best value of $k$. :wink:

[quote=iliyash][quote=sqing]For $ a,b,c$ positive reals , prove that
$$\frac{2bc}{(c+a)(a+b)}+\frac{2ca}{(a+b)(b+c)}+\frac{3ab}{(b+c)(c+a)}\geq \frac{5}{3}$$
Old?
Where?[/quote]
@sqing
When is the equality case?[/quote]
$c=2a=2b$

[quote=sqing]For $ a,b,c$ positive reals , prove that
$$\frac{2bc}{(c+a)(a+b)}+\frac{2ca}{(a+b)(b+c)}+\frac{3ab}{(b+c)(c+a)}\geq \frac{5}{3}$$
Old?
Where?[/quote]

[img]http://s1.sinaimg.cn/middle/006ptkjAzy7bDu1mfWo40&690[/img]

[quote=sqing]For $ a,b,c$ positive reals , prove that
$$\frac{2bc}{(c+a)(a+b)}+\frac{2ca}{(a+b)(b+c)}+\frac{3ab}{(b+c)(c+a)}\geq \frac{5}{3}$$
Old?
Where?[/quote]
$$\iff a\left((2b)^2+c^2\right)+b\left((2a)^2+c^2\right)+c(a^2+b^2)\ge10abc$$
By AM-GM,...

[quote=sqing]For $ a,b,c$ positive reals , prove that
$$\frac{2bc}{(c+a)(a+b)}+\frac{2ca}{(a+b)(b+c)}+\frac{3ab}{(b+c)(c+a)}\geq \frac{5}{3}$$
Old?
Where?[/quote]
For $ x,y,z$ positive reals , prove that

\[2\,{\frac {x}{y+z}}+{\frac {y}{z+x}}+{\frac {z}{x+y}}\geq -1+2\,\sqrt {2}\]

when equality hold?

[quote=iliyash][quote=sqing]For $ a,b,c$ positive reals , prove that
$$\frac{2bc}{(c+a)(a+b)}+\frac{2ca}{(a+b)(b+c)}+\frac{3ab}{(b+c)(c+a)}\geq \frac{5}{3}$$
Old?
Where?[/quote]

We need to prove:
$$\frac{2(b+c)bc+2(a+c)ac+3(a+b)ab}{(a+b)(b+c)(c+a)}\ge\frac{5}{3}$$
Or
$$6(b+c)bc+6(a+c)ac+9(a+b)ab\ge5(a+b)(b+c)(c+a)$$
Or
$$6\sum a^2b+6\sum a^2c+3ab(a+b)\ge5\sum a^2b+5\sum a^2c+10abc$$
Or
$$\sum a^2b+\sum a^2c+3ab(a+b)\ge10abc$$
Which is obviously true by AM-GM![/quote]
Of course.
But have to be reassembled.

[quote=sqing]For $ a,b,c$ positive reals , prove that
$$\frac{2bc}{(c+a)(a+b)}+\frac{2ca}{(a+b)(b+c)}+\frac{3ab}{(b+c)(c+a)}\geq \frac{5}{3}$$[/quote]
From:
For $ a,b,c$ positive reals , prove that
$$\frac{bc}{(c+a)(a+b)}+\frac{ca}{(a+b)(b+c)}+\frac{ab}{(b+c)(c+a)}\geq \frac{3}{4}$$
$$\iff a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)\ge 6abc$$

[quote=shaaam]Given that $ a,b,c \geq 0$, Prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq \frac {3}{2} \plus{} 2\sum_{cyc}\left(\frac {a \minus{} b}{a \plus{} b \plus{} 2c}\right)^2
\]
:)[/quote]

By computer,we have
$$LHS-RHS=\frac{1}{2}\sum_{cyc}{\frac{(a-b)^4}{(a+b+2c)^2(b+c)(c+a)}}\ge{0}$$


AMC
AIME
geometry
3D geometry
tetrahedron
floor function
conics

Ok here is the AIME I wrote. The rules are standard:
1. no calculators
2. answers are integers 0-999
3. 3 hour time limit

Of course we are going by the honor system. You answers should be sent to me through a PM. I don't really intend on having a deadline for the contest, so just send me your answers whenever you feel like it. Also I would appriciate it if you maybe gave a like 1 sentence description of how you solved it so I can see other ways of solving the problems I wrote. Also I am considering writing a (kalva-type) solutions document after a week or two. (It is in pdf format and it is posted below).

Anyway, hope you guys like it!  :)  If you have any questions, please either post or PM them.

[hide="The Test"]

$ {\Large Mock Aime 2006} \qquad By: Alex Anderson (Altheman) $

1. A 5-digit number is leet if and only if the sum of the first 2 digits, the sum of the last 2 digits and the middle digit are equal. How many 5-digit leet numbers exist?

2. Qin Shi Huang wants to count the number of warriors he has to invade China. He puts his warriors into lines with the most people such that they have even length. The people left over are the remainder. He makes 2 lines, with a remainder of 1, 3 lines with a remainder of 2, 4 lines with a remainder of 3, 5 lines with a remainder of 4, and 6 lines with a remainder of 5. Find the minimum number of warriors that he has.

3. $T_1$ is a regular tetrahedron. Tetrahedron $T_2$ is formed by connecting the centers of the faces of $T_1$. Generally, a new tetrahedron $T_{n+1}$ is formed by connecting the centers of the faces of $T_n$. $V_n$ is the volume of tetrahedron $T_n$. $\frac{V_{2006}}{V_1}=\frac{m}{n}$ where $m$ and $n$ are coprime positive integers, find the remainder when $m+n$ is divided by $1000$.

4. Let $P(x)=\sum_{i=1}^{20}(-x)^{20-i}(x+i)^i$. Let $K$ be the product of the roots. How many digits are does $\lfloor K \rfloor$ have where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$?

5. A parabola $P: y=x^2$ is rotated $135$ degrees clockwise about the origin to $P'$. This image is translated upward $\frac{8+\sqrt{2}}{2}$ to $P''$. Point $A: (0,0)$, $B: (256,0)$, and $C$ is in Quadrant I, on $P''$. If the area of $\triangle ABC$ is at a maximum, it is $a\sqrt{b}+c$ where $a$, $b$ and $c$ are integers and $b$ is square free, find $a+b+c$.

6. Define a sequence $a_0=2006$ and $a_{n+1}=(n+1)^{a_n}$ for all positive integers $n$. Find the remainder when $a_{2007}$ is divided by $1000$.

7. $f(x)$ is a function that satisfies $3f(x)=2x+1-f(\frac{1}{1-x})$ for all defined $x$. Suppose that the sum of the zeros of $f(x)=\frac{m}{n}$ where $m$ and $n$ are coprime positive integers, find $m^2+n^2$.

8. $R$ is a solution to $x+\frac{1}{x}=\frac{ \sin210^{\circ} }{\sin285^{\circ} }$. Suppose that $\frac{1}{R^{2006}}+R^{2006}=A$ find $\lfloor A^{10} \rfloor$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.

9. Zeus, Athena, and Posideon arrive at Mount Olympus at a random time between 12:00 pm and 12:00 am, and stay for 3 hours. All three hours does not need to fall within 12 pm to 12 am. If any of the 2 gods see each other during 12 pm to 12 am, it will be a good day. The probability of it being a good day is $\frac{m}{n}$ where $m$ and $n$ are coprime positive integers, find $m+n$.

10. Define $S= \tan^2{1^{\circ}}+\tan^2{3^{\circ}}+\tan^2{5^{\circ}}+...+\tan^2{87^{\circ}}+\tan^2{89^{\circ}}$. Find the remainder when $S$ is divided by $1000$.

11. $\triangle ABC$ is isosceles with $\angle C= 90^{\circ}$. A point $P$ lies inside the triangle such that $AP=33$, $CP=28\sqrt{2}$, and $BP=65$. Let $A$ be the area of $\triangle ABC$. Find the remainder when $2A$ is divided by $1000$.

12. There exists a line $L$ with points $D$,$E$,$F$ with $E$ in between $D$ and $F$. Point $A$, not on the line is such that $\overline{AF}=6$, $\overline{AD}=\frac{36}{7}$, $\overline{AE}=\frac{12}{\sqrt{7}}$ with $\angle AEF > 90$. Construct $E'$ on ray $AE$ such that $(\overline{AE})(\overline{AE'})=36$ and $\overline{FE'}=3$. Point $G$ is on ray $AD$ such that $\overline{AG}=7$. If $2*(\overline{E'G})=a+\sqrt{b}$ where $a$ and $b$ are integers, then find $a+b$.

13. $\triangle VA_0A_1$ is isosceles with base $\overline{{A_1A_0}}$. Construct $A_2$ on segment $\overline{{A_0V}}$ such that $\overline{A_0A_1}=\overline{A_1A_2}=b$. Construct $A_3$ on $\overline{A_1V}$ such that $b=\overline{A_2A_3}$. Contiue this pattern: construct $\overline{A_{2n}A_{2n+1}}=b$ with $A_{2n+1}$ on segment $\overline{VA_1}$ and $\overline{A_{2n+1}A_{2n+2}}=b$ with $A_{2n+2}$ on segment $\overline{VA_0}$. The points $A_n$ do not coincide and $\angle VA_1A_0=90-\frac{1}{2006}$. Suppose $A_k$ is the last point you can construct on the perimeter of the triangle. Find the remainder when $k$ is divided by $1000$.

14. $P$ is the probability that if you flip a fair coin, $20$ heads will occur before $19$ tails. If $P=\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by $1000$.

15. A regular 61-gon with verticies $A_1$, $A_2$, $A_3$,...$A_{61}$ is inscribed in a circle with a radius of $r$. Suppose $(\overline{A_1A_2})(\overline{A_1A_3})(\overline{A_1A_4})...(\overline{A_1A_{61}})=r$. If $r^{2006}=\frac{p}{q}$ where $p$ and $q$ are coprime positive integers, find the remainder when $p+q$ is divided by $1000$.

[/hide]

if you want to discuss any of the problems, feel free to pm me!

[quote="zanttrang"]Just as a point of clarification, don't we number the mock AIMEs and use the letter system for the AMCs?[/quote]

i did not notice that before...changed

[quote="Silverfalcon"]Another note.. Since it's 2006, shouldn't we start renumbering them? (i.e. AIME 1, AIME 4)[/quote]
I think we are numbering in terms of the academic, rather than calendar, year.

[quote="Zellex"]Typo?

[hide]
[quote]If any of the 2 gods see each other during 12 pm to 12 am, it will be a good day.[/quote]

Do you mean if any of the 3 gods see each other?[/hide][/quote]

[hide]i don't know if it is clear, i mean that if at least any pair of them sees each other it will be a good day[/hide]

[quote="boxedexe"]You have a latex problem in number 13.[/quote]

fixed...i forgot a "}", the pdf version did not have the problem so i did not notice...

[quote="chess64"]Ahh the numbers are too ugly, I'm getting frustrated. I could do about 10 of them if the numbers weren't so hideous, but I only feel like answering 5. :dry:

EDIT: Other than that, the problems were really good. :)[/quote]

if you find the best way to do the problems, there should not be any mess  :)

i will might make a solution thing on the weekend, but now i have a lot of tests to deal with, but i have typed out many solutions in pm's to people so once i have the time, it should not be that bad, the only thing that might take a while is if i want to make pictures, but you can pm me about specific problems now

This is a completely random question. Are there any Mock AMC 8s and if so, where? :huh:

search mock amc 8 on the forum to get...
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=111575#111575

you guys can discuss solutions now (not that anybody really cares...) and for the people who did not take the test, that is too bad, one of my problems was almost identical to the #23 on the AMC12B, just with different numbers, they stole my problem...  :mad:  :lol:, i might make a solution pdf based on how much time i have this weekend, but it is unlikely as i have a math team regional on Sat, so i will be gone for a while, also it depends on if i want diagrams and stuff, anyway i don't think i will post scores since not that many took it, but send me a pm telling me what you thought about it if you take it

[quote="Altheman"]not that anybody really cares...[/quote]I care but I did not have time to send you my answers. I liked your questions. Can you make more?

Anyone have solutions for 5,6,10, and 11. I couldn't solve those.

[quote="mysmartmouth"]Yes, a solution for number 11 would be greatly appreciated.  No diagram neccessary, unless it is neccessary. :P (Like if there is some secret extra line that you need to add, or something...)[/quote]
#11 is a variant of AMC 2006 B # 23 (though, to give credit, we saw it on the mock AIME first!), and so the solution is essentially the same for both (though what is asked for is slightly different).  Check it out here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433631#p433631

Can someone post there answer or at least pm it to me so I can see if I did it right. Im almost 99.9% sure I did it wrong but still would like to know the answer. Im not in the Contest. I only tried #4 because it was the only one I think I knew were to begin.

well if anybody cares, here are the answers:

[hide="answers"]
1. 330
2. 059
3. 908
4. 024
5. 594
6. 601
7. 005
8. 243
9. 015
10. 005
11. 505
12. 112
13. 270
14. 149
15. 642[/hide]

i am thinking about writing up solutions as the summer starts, but i will see if i have time to do that...if anybody wants to put up solutions as they go by the problems, that is fine, just make sure to hide them

here is an outline solution to #11

[hide]a common way to deal with points inside equilateral triangles or isosceles triangles is to rotate about a vertex, making a pair of equal sides match up

in this case you want to rotate 90 degrees about C, let P' be the image P
PCP' is an isosceles right triangle, with PP'=56

note that we then can get a closed triangle with sides 33,56,65 (noting that is a pythagorean triple), we can use the 90 degree angle and the 45 degree angle+ the law of cosines to find the side, and the answer follows[/hide]

[quote="shinwoo"]how did you do #5?

I thought the three points were
(0,(8+sprt)/2),(0,0),(256,0)
...[/quote]

this is not necessarily true...after you rotate it, the highest point is not the vertex...just draw it out

[hide="solution to #5"]consider after you rotate it, and a horizontal line (call this $l_1$) that is tangent to the parabola gives the max value (why?)

after rotating this back, it is a line ($l_2$) $y=-x+b$ tangent to $y=x^2$, use the quadratic discriminant to give one solution, making it tangent, you get $b=-\frac{1}{4}$,

now you need the height of the triangle (ignoring the $\frac{8+\sqrt{2}}{2}$, looking at the rotated parabola, it is the distance between $l_1$ and $y=0$, which rotated back is the distance from $l_2$ to $y=-x$, this is easily found by using similar triangles, or whatever... to be $\frac{\sqrt{2}}{8}$

then the area of the triangle is:

$\frac{1}{2}*\left(\frac{\sqrt{2}}{8}+\frac{8+\sqrt{2}}{2}\right)*256=512+80\sqrt{2}$...answer follows to be:

$512+80+2=\boxed{594}$[/hide]

[quote="4everwise"]Lol, Qin Shi Huang...

Originality points!  :D[/quote]

oh haha lol...i was actually studying chinese history in school during that time, he is the emperor who had the terra cotta army made to protect himself during the afterlife...

Let $sin^{2}x=x$

$\frac{1}{-x(x-1)}$ (Pythagorean identity, factoring out the negative one)

Use whatever partial fraction method you use.

$\frac{1}{\sin^{2}x}+\frac{1}{1-\sin^{2}x}=\frac{1}{\cos^{2}x\sin^{2}x}$

[quote="zhou_qiang"]How to split 1/cos^2sin^2 into partial fraction[/quote]

the trig summation is not a "pairing" problem, here is a hint:

[hide]find a polynomial with the roots $\tan^{2}1$, $\tan^{2}3$, ..., $\tan^{2}89$[/hide]

[quote="vvr1590"]In the tetrahedron problem, is the volume of the n+1 shape over volume of nth shape 1/27?

If so, how do you know?[/quote]

Look at the height. Since the point of the new tetra is at the centroid, the thing is 1/3 height (centroid is 1/3 of the way up), inverted.

[quote="zhou_qiang"]Sent me the solution for Mock AIME 4 question 10[/quote]

[hide="solution"](all in degrees)

We will construct a polynomial with the roots in the summation. Then use vieta's sums to find the sum of the roots and the desired answer. \[P(x)=(i\tan x+1)^{90}=\frac{1}{\cos^{n}x}*\exp 90x\] For $x_{0}=1,3,...,89$ we get that the real part of $P(x_{0})=0$. Then equate the real part of both sides (use the binomial theorem to expand and let $y=\tan x$). \[-y^{90}+\binom{90}{88}y^{88}-\binom{90}{86}y^{86}+...-...+1=0\] Now let $y^{2}=z$. The polynomial becomes degree 45.

Clearly, $\tan^{2}x$ for $x=1,3,5,...89$ are roots looking above. Since there are 45 distinct roots that must be solutions and the polynomial is degree 45, it must have those roots only, and no repeats.

So the desired sum (by vieta's sums) is: \[(-1)^{1}\cdot\frac{\binom{90}{88}}{-1}=\binom{90}{2}=45*89=4005\] it follows that the answer is $\boxed{005}$[/hide]

On 4, isn't the leading coefficient $-1$ and the constant $20^{20}$, making the product $-20^{20}$? :maybe:

EDIT: Never mind.

[quote="vvr1590"]On number 4, isn't K 20^20?[/quote]

[hide]no, the lead coefficient drops out (it is actually degree 19)[/hide]

[quote="zhou_qiang"]How do you come up with such a polymonial, please explain, also is the i here -1?[/quote]

did you read my solution above? the motivation is for you to figure out...i have seen this technique come up a few places, so that is how i know it

[quote="zhou_qiang"]Where have you seen such a technique (please sent me the websites), also , please explain why the given fuction have such roots, I really can't figure it out![/quote]

i have seen this idea while solving problems on aops...

to understand the solution fully, you first have to forget about how you would solve it, that can be done only when you see what is going on

so first, see that for x=1,3,5,..89 the expression is equal to 0

tan^2 of those are all different numbers (45 of them)

the resulting polynomial has 45 roots (since it is of degree 45)

therefore those are the only roots

What is the value of p such that 5x^2-5px+(66p-1)=0 has two integers solution?

From Singapore Mathematical Olympiad 2006 (Senior Section Round 1)

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3745700_2006_alex_anderson_mock_aime]here[/url]

enjoy  / start solving

USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url]
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)

PS. Answer key is [url=https://artofproblemsolving.com/community/c5h70988p522372]here[/url]


inequalities
logarithms
integration
inequalities proposed

Given:
$S_{n}=1+\frac12+\frac13+\frac14+...+\frac{1}{n}$
Show that:
$S_{p}+S_{q}-S_{pq}\leq 1$ Where $p,q$ are positive whole numbers

We're going to show
$S_{pq}-S_{p}\geqslant S_{q}-1$
denote
$a_{p}=S_{pq}-S_{p}=\frac{1}{p+1}+\frac{1}{p+2}+\cdots+\frac{1}{pq}$
notice that
$S_{q}-1=\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{q}=a_{1}$
So we just need to prove
$a_{p+1}\geqslant a_{p}\ (p=1,2,\ldots)$
and it's obvious that
$a_{p+1}-a_{p}\\ =\frac{1}{pq+1}+\frac{1}{pq+2}+\cdots+\frac{1}{pq+q}-\frac{1}{p+1}\\ \geqslant\frac{1}{pq+q}+\frac{1}{pq+q}+\cdots+\frac{1}{pq+q}-\frac{1}{p+1}=0$
:wink:

[b]Conjecture:[/b] If $a$ and $b$ are real sequences such that $(\log a_{1},\ldots,\log a_{n})\succ(\log b_{1},\ldots,\log b_{n})$ then
\[\sum_{j=1}^{n}\int_{0}^{a_{j}}\frac{1}{[x+1]}dx\geq\sum_{j=1}^{n}\int_{0}^{b_{j}}\frac{1}{[x+1]}dx.\]


Let$ a,b,c \ge 0$ such that $ ab \plus{} bc \plus{} ca\equal{}3$, Prove that :
$ \sum \frac {1}{a^2 \plus{} 1} \ge \frac {3}{2}$

[quote="tin_124"]such that $ ab \plus{} bc \plus{} ca$, [/quote]

what does this mean?

such that ab+bc+ca what? doesn't it lack sth

[quote="tin_124"]Let$ a,b,c \ge 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$, Prove that :
$ \sum \frac {1}{a^2 \plus{} 1} \ge \frac {3}{2}$[/quote]

We assume $ a \ge b \ge c$ then we have $ ab \ge 1$ and $ c \le 1$.
We have that

\[ \frac{1}{{a^2  \plus{} 1}} \plus{} \frac{1}{{b^2  \plus{} 1}} \ge \frac{2}{{ab \plus{} 1}} \Leftrightarrow \frac{{\left( {ab \minus{} 1} \right)\left( {a \minus{} b} \right)^2 }}{{\left( {ab \plus{} 1} \right)\left( {a^2  \plus{} 1} \right)\left( {b^2  \plus{} 1} \right)}} \ge 0
\]

Hence
\[ \frac{1}{{a^2  \plus{} 1}} \plus{} \frac{1}{{b^2  \plus{} 1}} \plus{} \frac{1}{{c^2  \plus{} 1}} \ge \frac{2}{{ab \plus{} 1}} \plus{} \frac{1}{{c^2  \plus{} 1}} \equal{} \frac{2}{{4 \minus{} c\left( {a \plus{} b} \right)}} \plus{} \frac{1}{{c^2  \plus{} 1}}
\]
But we have $ ab \plus{} bc \plus{} ca \equal{} 3 \Rightarrow a \plus{} b \plus{} c \ge 3$ or $ a \plus{} b \ge 3 \minus{} c$ therefore
\[ \frac{2}{{4 \minus{} c\left( {a \plus{} b} \right)}} \plus{} \frac{1}{{c^2  \plus{} 1}} \ge \frac{2}{{4 \minus{} c\left( {3 \minus{} c} \right)}} \plus{} \frac{1}{{c^2  \plus{} 1}} \equal{} \frac{2}{{c^2  \minus{} 3c \plus{} 4}} \plus{} \frac{1}{{c^2  \plus{} 1}}
\]
It suffices to show that
\[ \frac{2}{{c^2  \minus{} 3c \plus{} 4}} \plus{} \frac{1}{{c^2  \plus{} 1}} \ge \frac{3}{2} \Leftrightarrow \frac{{c\left( {1 \minus{} c} \right)\left( {c \minus{} 1} \right)^2 }}{{\left( {c^2  \plus{} 1} \right)\left( {c^2  \minus{} 3c \plus{} 4} \right)}} \ge 0
\]
Done. :)

Yes , you right , Loc .
P/S: Anh loc cho nick YH anh em minh noi chuyen cho vui ?

[quote="tin_124"]Let$ a,b,c \ge 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$, Prove that :
$ \sum \frac {1}{a^2 \plus{} 1} \ge \frac {3}{2}$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=164187
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176342


Stanford has a program called the Education Program for Gifted Youth Online High School.  Any experiences with this one?

Sorry for the late reply here.

I've heard a great deal about the EPGY OHS.  I'm not exactly sure what kind of opinions / responses you want, but most of the information you'll probably want to start out with can be found on their [url=http://epgy.stanford.edu/ohs/]website[/url].  If you have any specific questions about student life you can't find on the website, you can always pm me  :D

I attended part time last year, as a freshman. It was completely AWESOME. I'm really unhappy about not attending this year (I switched to full time at my normal school). I highly recommend you go.

No, it isn't. (Also, this thread is from 2010)

[quote=camacjo]hm, is this the same as Stanford OHS?[/quote]

yep ur a little late to the party there bud


Let $ M$ and $ N$ be the midpoints of the sides $ BC$ and $ AC$ of $ \triangle ABC$ ,$ BH$ be it's altitude.The line through $ M$,perpendicular to the bisector $ \angle HMN$ , intersects the line $ AC$ at point $ P$ such that $ HP \equal{} \frac{1}{2}(AB\plus{}BC)$ and $ \angle HMN \equal{}45^{\circ}$ Prove that $ \triangle ABC$ is isosceles.

[hide="Solution"][quote="anonymous111"]Let $ M$ and $ N$ be the midpoints of the sides $ BC$ and $ AC$ of $ \triangle ABC$ ,$ BH$ be it's altitude.The line through $ M$,perpendicular to the bisector $ \angle HMN$ , intersects the line $ AC$ at point $ P$ such that $ HP \equal{} \frac {1}{2}(AB \plus{} BC)$ and $ \angle HMN \equal{} 45^{\circ}$ Prove that $ \triangle ABC$ is isosceles.[/quote]
Suppose AB<BC, P lie on the opposite ray of CA. Let D symmetric to C wrt H, E symmetric to C wrt P.
Then, BD=BC, DE=AB+BC, $ \frac{EA}{ED}\equal{}\frac{AB}{BD}$
$ \Rightarrow AD\equal{}\frac{BC^2\minus{}AB^2}{BD}\Rightarrow AD.BC\equal{}CH^2\minus{}AH^2\equal{}AD.CD\Rightarrow AC\equal{}BC$[/hide]


[color=red]

here is an interesting problem, one that i am currently unable to do.

let S(x) denote the sum of the digits in the decimal representation of x.
prove that limit (S(2^n), n-> :inf: ) =  :inf:  .

the only idea i have had so far was to somehow bound the number of digits that are zero in 2^n, possibly inductively...

for instance, it is clear that if we have an n-digit number, all of whose digits are nonzero, the maximum number of zeros in 2n is the least integer greater than n/2 . however, i have played around with this fact for a bit and so far have not made substantial progress on the problem.

any comments would be much appreciated.

[/color]

Let n and m such as n >= 4m

2^n >= 16^m > 10^m so 2^n as at least one digit not null after m digits

If 2^n = M*10^m + r, r != 0, 2^m|r and r is the last m digits of 2^n : S(2^n) >= 1 + S(r)

Now we can play it again with m >= 4*p : 2^m has a digit not null >= p. so r has also a digit not null >= p.

If n >= 4^m, n has at least m digits not null : S(2^n) >= m >= [log_4(n)] -> +oo when n -> +oo

The minoration is *very* weak : S(2^10000) = 13561 and log_4(10000) ~ 5 !

I think it works also for 5^n :

If n >= 2*m, 5^n >= 25^m >= 10^m. If 5^n = M*10^m + r, r != 0, 5^m|r and so on.

S(5^n) >= m >= [log_2(n)]

HTH


HI PEOPLE...
THIS THE Morocco TEAM FOR IMO 2006 -SLOVENIA..

mar1:Nadir Gmira

mar2:Ali Chaer

mar3:Ali Ameziane

mar4:Yassine Meziani

mar5:Fayçal Yahyaoui

mar6:Youssef El Azbi


neshan dahid donbalei namotenahi az adade tabie'e manande $ a_{1},a_{2},\dots$ vojood darad be torike har adade tabie'e daghighan $ 1$ bar dar donbale zaher shavad va hamchenin har adade tabie'e daghighan yek bar dar donbaleye namotenahie $ |a_{1}\minus{}a_{2}|,|a_{2}\minus{}a_{3}|,\dots$ zaher shavad. :lol:

be nazar miad age hamintori kilooi pish berim moshkeli pish naiad! :D
yeni too har marhale $ a_n$ ro barabare koochiktarin adadi begirim ke too $ \{a_1,a_2,\ldots ,a_{n\minus{}1}\}$ naimade basheo hamchenin $ |a_n\minus{}a_{n\minus{}1}|$ too $ \{|a_2\minus{}a_1|,|a_3\minus{}a_2|,\ldots ,|a_{n\minus{}1}\minus{}a_{n\minus{}2}|\}$ naimade bashe. :maybe:
rahesh hamine?ya inke rahe sakhtariam dare?


find f:R+ ->R+ such that:
f(x).f(y)=f(x.y)+1/x+1/y

$f(1)^{2}= f(1)+2 \implies f(1) = 2,-1$
$f(1) f(y) = f(y)+1+\frac{1}{y}$

[b]Case:[/b]  $f(1) = 2$.  Then

$f(y) = 1+\frac{1}{y}$

Checking:

$f(x) f(y) = \left( 1+\frac{1}{x}\right) \left( 1+\frac{1}{y}\right) = 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}= f(xy)+\frac{1}{x}+\frac{1}{y}$

[b]Case:[/b]  $f(1) =-1$.  Then

$-2 f(y) = 1+\frac{1}{y}\implies$
$f(y) =-\frac{1}{2}\left( 1+\frac{1}{y}\right)$

But this solution does not satisfy the original equation.

[quote="t0rajir0u"]
[b]Case:[/b]  $f(1) =-1$.  Then

$-2 f(y) = 1+\frac{1}{y}\implies$
$f(y) =-\frac{1}{2}\left( 1+\frac{1}{y}\right)$

But this solution does not satisfy the original equation.[/quote]
This case is obviously impossible (from the assumption about $f$).


If the numbers  $ a$ , $ b$ , $ c$ are positive, then $ a \plus{} b \plus{} c\ \le\ \boxed{\ \sqrt {3\left(a^2 \plus{} b^2 \plus{} c^2\right)}\ \le\ \sum\ \frac {b^2 \plus{} c^2}{b \plus{} c}\ }\ .$

[hide="Attempt"]
The first inequality is, of course, well-known; I assume that's why you boxed the second part.

The inequality is homogenous; assume $ a\plus{}b\plus{}c\equal{}1$. Writing with common denominators and rewriting a bit, we have $ \sum (1\minus{}a^2)(1\plus{}a^2\minus{}b^2\minus{}c^2)\ge 2\sqrt{3}(a\plus{}b)(b\plus{}c)(c\plus{}a)$.[/hide]

I'm don't know where to proceed from here, though.

instead, does assuming (wlog) that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 0$ yield anything fruitful?

[hide="solution"]

Let $ a\plus{}b\plus{}c \equal{} 1$

Note that $ \sum\ \frac{b^2\plus{}c^2}{b\plus{}c} \equal{} \sum\ (b\plus{}c \minus{} \frac{2bc}{b\plus{}c}) \equal{} 2 \minus{} \sum\ \frac{2}{\frac{1}{b} \plus{} \frac{1}{c}} .$

Also, $ a^2\plus{}b^2\plus{}c^2 \equal{} (a\plus{}b\plus{}c)^2 \minus{} 2\sum ab \equal{} 1\minus{} 2\sum ab$

So, squaring our original inequality and substituting gives us

$ 3(1\minus{}2\sum ab ) \le (2\minus{}   \sum\ \frac{2}{\frac{1}{b} \plus{} \frac{1}{c}})^2$

This is equal to

$ \minus{}6\sum ab \plus{} 2 \sum\ \frac{2}{\frac{1}{b} \plus{} \frac{1}{c}} \le  (1\minus{} \sum\ \frac{2}{\frac{1}{b} \plus{} \frac{1}{c}})^2$.

We then prove that

$ \minus{}6\sum ab \plus{}2 \sum\ \frac{2}{\frac{1}{b} \plus{}\frac{1}{c}} \le 0$.

and we are done.
[/hide]


[quote="Silverfalcon"]Find the smallest positive integer whose leftmost digit is 1 and that has the property that if this digit is transferred to the end of the number, the number is tripled.[/quote]

[hide]
call the number after removing the 1, x
$ 10x\plus{}1\equal{}3(1*10....00\plus{}x)$
$ 7x\equal{}29999....9$

the lowest such x which fulfills these is 42857, which gives the RHS of equation 2 as 299999

so the integer is 142857

[/hide]


In a triangle $ PQR$, $ PQ \equal{} QR$. $ S$ is a point on $ PR$ such that $ PS \equal{} 15$ and $ SR \equal{} 21$. One circle touches $ PQ, QS$ and $ SP$ in $ X, Y$ and $ Z$. Another circle touches $ RQ, QS$ and $ SR$ in $ U, V$ and $ W$.

Find the length $ YV$

diagram ---> http://img522.imageshack.us/my.php?image=picture2uq0.png

This is a side-chase problem; repeated applications of the two-tangent theorem yields $ PQ \equal{} PX \plus{} XQ \equal{} PR \plus{} YQ \equal{} 13 \plus{} YQ \minus{} SY \equal{} 13 \plus{} VQ \minus{} VS \plus{} 2VY$ and $ QR \equal{} \cdots \equal{} 21 \plus{} VQ \minus{} VS$. From $ PQ \equal{} QR$, we find $ 2VY \equal{} 21 \minus{} 13 \Longrightarrow VY \equal{} 4$.

Hi kurt.math. It seems to me that the solution posted by azjps does not fit in with your diagram. I've chased the sides a number of ways and keep getting 3, and not 4. For example,
QP=QR
QX+XP=QU+UR
QY+PZ=QV+RW
QV+VY+15-SZ=QV+21-WS
VY+15-SY=21-SV
VY=6-(SY+YV)+SY
VY=3.
I hope I haven't made a mistake (as I tend to rush things without thinking.)

Oops. Now I get 2!
I think I wrote SY instead of SV so the penultimate line should read
VY=6-(SV+YV)+SY.
Then
2VY+SV-SY=6
3VY=6
VY=2.

(Perhaps this is wrong as well.)

Gah typo I wrote $ 13$ instead of $ 15$, it should be $ 3$.


geometry
parallelogram
analytic geometry

Let $ ABCD$ be a parallelogram and let $ X$ and $ Y$ be the points outside $ ABCD$ that make triangles $ BCX$ and $ CDY$ equilateral. Prove that triangle $ AXY$ is also equilateral.

Oh man this was in Challenging Problems in Geometry?
[hide]
Let ABC=x. ABX=60+x=ADY=YCX. AB=AD=DY=CY, BX=BC=AD=CX, so we have ABX congruent YCX congruent YDA, so AY=AX=XY, and we are done.[/hide]

what's loc?
this problem was way easy enough by simply subtracting an angle and doing sas
although i did this really cool extremely long and computational coordinate bash first  :D till i realized that there was a way easier way
but still, my coordiate bash was really awesome as it included using the distance formula about 4 times and huge 3 variable equations


calculus
integration
trigonometry
algebra
polynomial
derivative
complex analysis

How do I apply Cauchy's Theorem to $ f(z) \equal{} \frac{2z}{z\minus{}i}$ ;$ \Gamma$ is the circle $ |z\minus{}i|\equal{}3$?

I know $ \Gamma$ has to be a closed path, but what confuses me is the singularity of $ \minus{}i$ for $ f(z)$.

Thanks.

Hello mstrip.  $ \Gamma$ is a closed contour right.  It's the circle around $ \minus{}i$ that is $ |\minus{}i\plus{}z|\equal{}3$ but that's close enough since it will encircle the singularity which is at $ i$ and not $ \minus{}i$.  So Cauchy's Integral Formula requires:

$ \oint_{\Gamma}\frac{2z}{z\minus{}i}dz\equal{}2\pi i(2i)\equal{}\minus{}4\pi$

So does that mean if I had $ f(x)\equal{}\sin(\frac{1}{z})$ with $ \Gamma$ being the circle $ |z\minus{}(1\minus{}2i)|\equal{}1$ then

I could just apply Cauchy's Theorem to get $ \int_{\Gamma}f(z)dz \equal{} 0$ since the singularity of $ 0$ is NOT in $ \Gamma$?

Thanks again!

what about $ f(z) \equal{} (z\minus{}i)^2$ where $ \Gamma$ is the semicircle of radius $ 1$ about $ 0$ from $ i$ to $ \minus{}i$. How would I evaluate $ \int_{\Gamma} f(z) dz$ then? I understand it's analytic everywhere, but obviously Cauchy's Thm doesn't apply since $ \Gamma$ is a non-closed path.

Could I use $ z(t) \equal{} e^{it}$ and simply $ \int_{\minus{}\frac{\pi}{2}}^{\frac{\pi}{2}} f(z(t))\cdot z'(t) dt$?

If the contour went from $ i$ to $ -i$ then the limits would be from $ \pi/2$ to -$ \pi/2$ right?

Also, the polynomial and it's derivative are entire so the integral is path-independent.  Could just use antiderivatives:

$ \int_{\Gamma}(z-i)^2dz=\left1/3(z-i)^3\Biggr\vert_i^{-i}=8/3i$

or a simpler contour:

Let $ z=iy$ then:

$ \int_{\Gamma}(z-i)^2dz=\int_1^{-1}(-i)(y-1)^2dy=8/3i$

I understand, I really appreciate the help!


The ﬁgure shown is called a [i]trefoil[/i] and is constructed by drawing circular sectors about sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length $ 2$?
[asy]unitsize(1.5cm);
defaultpen(linewidth(.8pt)+fontsize(12pt));

pair O=(0,0), A=dir(0), B=dir(60), C=dir(120), D=dir(180);
pair E=B+C;

draw(D--E--B--O--C--B--A,linetype("4 4"));
draw(Arc(O,1,0,60),linewidth(1.2pt));
draw(Arc(O,1,120,180),linewidth(1.2pt));
draw(Arc(C,1,0,60),linewidth(1.2pt));
draw(Arc(B,1,120,180),linewidth(1.2pt));
draw(A--D,linewidth(1.2pt));
draw(O--dir(40),EndArrow(HookHead,4));
draw(O--dir(140),EndArrow(HookHead,4));
draw(C--C+dir(40),EndArrow(HookHead,4));
draw(B--B+dir(140),EndArrow(HookHead,4));

label("2",O,S);
draw((0.1,-0.12)--(1,-0.12),EndArrow(HookHead,4),EndBar);
draw((-0.1,-0.12)--(-1,-0.12),EndArrow(HookHead,4),EndBar);[/asy]$ \textbf{(A)}\ \frac13\pi\plus{}\frac{\sqrt3}{2} \qquad
\textbf{(B)}\ \frac23\pi \qquad
\textbf{(C)}\ \frac23\pi\plus{}\frac{\sqrt3}{4} \qquad
\textbf{(D)}\ \frac23\pi\plus{}\frac{\sqrt3}{3} \qquad
\textbf{(E)}\ \frac23\pi\plus{}\frac{\sqrt3}{2}$

This was just posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=251128]here[/url]...

[hide="solution"]there are four sectors, each with central angle of $ 60$ and radius $ 1.$

$ \frac {60\cdot4}{360} \cdot1\cdot(\pi^2)$

$ \frac23\cdot\pi^2$

$ (\text B)$

[/hide]

EDIT: didnt see maths post, sorry.


function
algebra solved
algebra
functional equation

Does there exist a function $f:\mathbb N\to\mathbb N$ such that
\[f(f(n-1)) = f(n+1) -f(n) \] for all $n \geq 2$?

f(n)=0 verifies it... but shoul it be a nonzero constant function?

cheers! :D

You will be laugh: in Russia and Belarus N={1,2,3,4,...}

actually everywhere besides Romania N means positive integer, 1,2,3 ...

so I guess that's why they say Bucharest is the Little Paris :D


Prove that for every n,n is natural,the numbers 18^n and 18^n+2^n have the same number of digits.(For example if n=1 then 18^1=18 has 2 digits and 18^1+2^1=20-again 2 digits.)

[hide]The claim will be proven by contradiction.

Assume that for some natural number $ n$, the number $ 18^n \plus{} 2^n$ has a different number of digits than $ 18^n$.  It is obvious that it must have more digits, not less.  Let $ d$ be the number of digits in $ 18^n\plus{}2^n$.  It follows that

$ 18^n\plus{}2^n>10^d>18^n$
$ 2^n(9^n\plus{}1)>2^d*5^d>2^n*9^n$
$ 9^n\plus{}1>2^{d\minus{}n}*5^d>9^n$

If $ d\leq{n}$ then $ 5^d<9^n$ and $ 2^d\leq{2^n}$ and $ 2^d*5^d<2^n*9^n$, contradiction; thus, $ d$ must be greater than $ n$.  However, this means that $ 2^{d\minus{}n}*5^d$ is an integer, which contradicts the fact that it is greater than the integer $ 9^n$ and less than the integer $ 9^n\plus{}1$.  By contradiction, the claim is proved.[/hide]


maybe you guys have already discussed this.. buuut..
i'm having trouble finding a safety school that i would be happy to go to.  it must have strong math and physics departments + solid english department + music and dance.  and it must not be too hot, because i melt =b

i'm looking for something a little less selective than uchicago/harvey mudd/carleton

help?

Hey :D IM SO HAPPY TO SEE YOU HERE  :welcome:

hmm.. for safeties, I applied to georgia tech (tho I got rejected  :oops: ), University of Illinois at Urbana Champaign. They do have a strong science/engineering program, but not sure about english/music/dance tho.

I wish i was more helpful...

[quote="saracampbellsoup"]i'm having trouble finding a safety school that i would be happy to go to.  it must have strong math and physics departments + solid english department + music and dance.  and it must not be too hot, because i melt =b

i'm looking for something a little less selective than uchicago/harvey mudd/carleton
[/quote]

Most of your description sounds just like [url=http://www.stolaf.edu/]St. Olaf College[/url] in Northfield, Minnesota, very near to [url=http://www.carleton.edu/]Carleton College[/url]. The math is very strong and credible nationally at St. Olaf, music is in the school's DNA, the climate is Minnesotan and thus not hot during the school year, and the English department should still be well reputed too. (My late paternal grandmother was an English major there, and my dad was a chemistry major there half a century ago.) You can read about how strong the math department is at St. Olaf in [url=http://www.amazon.com/exec/obidos/tg/detail/-/082183455X/learninfreed]A Mathematician's Survival Guide: Graduate School and Early Career Development[/url] by Steven G. Krantz.

It's very hard to take a private college or an out-of-state college as a safety, but your odds increase as your statistics rank higher and higher in a school's general pool of applicants. What I like for a safety school is a state university that admits in-state students strictly by the numbers and still has a decent program. That would be the University of Minnesota for my oldest son, and possibly for all of my children, depending on what they desire to study.

Good luck in your applications.

[quote="manuel"]Geehon whay did u get accepted in mit and rejected in georgia tech??? thats nonsense[/quote]

not necessarily. i have read that if a school knows you are their "safety school" (which is kind of obvious if you are applying to a lot of top tiers as well and have a very good scholastic record) they may reject you on the grounds that you are "over qualified" and unlikely to attend their school. you are just taking up space on their list over people that really honestly want to go to their school and would make a good fit.

[quote="tokenadult"]Most of your description sounds just like St. Olaf College in Northfield, Minnesota, very near to Carleton College.[/quote]
tokenadult, St. Olaf sounds like an excellent school.  Do you think students who are not religious would feel out of place there?

[quote="wild octagon"]tokenadult, St. Olaf sounds like an excellent school.  Do you think students who are not religious would feel out of place there?[/quote]

I think there are plenty of secular (that is, not religious) students there. There are DEFINITELY non-Lutheran students there, even though it was founded as a Lutheran college in a state that has a huge Lutheran population. I went to a state university (in the same state) myself, so I don't have personal experience of that issue, but I think that St. Olaf is like any other college these days in desiring "diversity," including religious diversity, on campus.

thank you everyone for your suggestions!!
i wish we had a nice in-state school... not university of WOOOO BEER

[quote="saracampbellsoup"]i wish we had a nice in-state school... not university of WOOOO BEER[/quote]
i know the feeling.


inequalities
function
logarithms
inequalities proposed

I found that forall $a,b,c>0$ and $n\geq1$ there is
$\left(\frac{a}{b+c}\right)^{n}+\left(\frac{b}{c+a}\right)^{n}+\left(\frac{c}{a+b}\right)^{n}\geq\left(\frac{3}{2}\right)^{n}\frac{a^{n}+b^{n}+c^{n}}{(a+b+c)^{n}}.$
It can be solved with Karamata Inequality. But are there other opinions about it?

I have:
  $\sum(\frac{a}{b+c})^{n}\geq (\frac{3}{2})^{n}\frac{a^{n}+b^{n}+c^{n}}{(a+b+c)^{n}}$
  $\leftrightarrow \sum a^{n}(\frac{a}{b+c}+1)^{n}\geq (\frac{3}{2})^{n}\sum a^{n}$
Let $a\geq b\geq c$
I have:
  $\frac{a}{b+c}\geq \frac{b}{a+c}\geq \frac{c}{a+b}$
therefore,accord chebyshev's inequality :
 $\sum a^{n}(\frac{a}{b+c}+1)^{n}\geq \frac{1}{3}\sum a^{n}\sum(\frac{a}{b+c}+1)^{n}$
Finally,Use convex function,I have:
$\sum(\frac{a}{b+c}+1)^{n}\geq \frac{1}{3}(\sum\frac{a}{b+c}+1)^{n}\geq (\frac{3}{2})^{n}$

Thank you, a nice one! :lol:
but I think the last one
[quote="silver1989"]
$\sum(\frac{a}{b+c}+1)^{n}\geq \frac{1}{3}(\sum\frac{a}{b+c}+1)^{n}\geq (\frac{3}{2})^{n}$[/quote]
should be
$\frac{1}{3}\sum\left(\frac{a}{b+c}+1\right)^{n}\geq\left[\frac{1}{3}\sum\left(\frac{a}{b+c}+1\right)\right]^{n}\geq\left(\frac{3}{2}\right)^{n}$

Actually the inequality
$\sum\left(\frac{a}{b+c}\right)^{n}\geqslant\left(\frac{3}{2}\right)^{n}\frac{\sum a^{n}}{\left(\sum a\right)^{n}}$
is true for $n\geqslant\log_{2}3-1$


Two integers in the range${2, 3,..., 99}$are secretly drawn. The
sum of the two integers is given to Dr. S, and the product is
given to Dr. P. Dr. S and Dr. P, both of whom are excellent
mathematicians, then engage in the following short dialogue:

Dr. P: I do not know the two numbers.
Dr. S: I knew that you didn't know the two numbers.
Dr. P: Now I know the two numbers.
Dr. S: Now I know the two numbers.

What were the two numbers drawn?


Let f(x) be an integer polynomial of the form: f(x)=x^2s +... . We know, that there are an infinit number "x" for witch f(x) is square. Prove, that f(x)=(g(x))^2, for some integer polinomial g(x).

Could you explicit the form of the polynomial $ f(x)$? Is it perhaps $ f(x)=x^{2s}+a_{2s-1}x^{2s-1}+a_{2s-2}x^{2s-2}+...+a_{1}x+a_{0}$, its coefficients $ a_{i}$ being integers?
Then, if $ f(t)$ is a perfect square for infinite values of $ t$, should I think that there is at least a value $ t_{0}$ for which $ f(t_{0})$ is not a square?

Thank you...

[quote="Ponnamperuma"]should I think that there is at least a value $ t_{0}$ for which $ f(t_{0})$ is not a square?[/quote]

Well, then the problem would be false, wouldn't it?

Yes, of course: in fact I was thinking to a typo... Then could you explain to me what's the request? I don't get it...

[quote="t0rajir0u"]We are given an integer monic polynomial with even degree that takes on integer square values infinitely often.  We wish to prove that it is the square of another integer polynomial.

[hide="I suspect..."] That bounding will prove useful. [/hide][/quote]

Yes this is. Thans t0rajir0u.


[quote="Ronald Widjojo"]For $ a,b,c \ge 0$
Prove that :

$ \sum \sqrt {\frac {a}{b \plus{} c}} \ge \frac {3}{\sqrt {2}}$[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$ :wink:

Hey Arqady, i've edited it..

Sorry for the mistake..

I saw this problem at Hojoo Lee's book..

[quote="Ronald Widjojo"]For $ a,b,c > 0$
Prove that :

$ \sum \sqrt {\frac {a}{b \plus{} c}} \ge \frac {3}{\sqrt {2}}$

Edited : I'm sorry..[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0.07.$ :wink:

Oh, i've got my mistake..
It must be :
$ \sum_{cyc} \frac {a}{\sqrt{b \plus{} c}} \ge \sqrt {\frac {3}{2}(a\plus{}b\plus{}c)}$

When does equality hold ?

Put $ a \plus{} b \plus{} c \equal{} 3$
 $ f(x) \equal{} \frac {x}{\sqrt {3 \minus{} x}}, x\in(0,3)$ and the inequality result
 $ \frac {f(a) \plus{} f(b) \plus{} f(c)}{3} \geq \frac {1}{\sqrt {2}} \equal{} f(1) \equal{} f(\frac {a \plus{} b \plus{} c}{3})$ $ \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (A)$
since
 $ f'(x) \equal{} \frac {6 \minus{} x}{2(\sqrt {3 \minus{} x)^3}}$
 $ f''(x) \equal{} \frac {12 \minus{} x}{4(\sqrt {3 \minus{} x})^5} > 0, x\in(0,3) \Longrightarrow$ $ \minus{} \minus{} \minus{} \minus{} \minus{} \minus{}$ for the (A) applying Jensen's inequality we have
$ \sum_{cyc} \frac {a}{\sqrt {b \plus{} c}} \ge \sqrt {\frac {3}{2}(a \plus{} b \plus{} c)}$

_____
 :wink:

[quote="Ronald Widjojo"]Oh, i've got my mistake..
It must be :
$ \sum_{cyc} \frac {a}{\sqrt {b \plus{} c}} \ge \sqrt {\frac {3}{2}(a \plus{} b \plus{} c)}$

When does equality hold ?[/quote]
Applying Horder Inequality , we obtain :
$ (\sum\frac{a}{\sqrt{b\plus{}c}})^{2}(\sum a(b\plus{}c)) \geq (a\plus{}b\plus{}c)^{3}$
$ \equal{}(a\plus{}b\plus{}c).(a\plus{}b\plus{}c)^{2}\geq
3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$
$ \Leftrightarrow  \sum \frac {a}{\sqrt {b \plus{} c}} \ge \sqrt {\frac {3}{2}(a \plus{} b \plus{} c)}$
;)

[quote="Ronald Widjojo"]Oh, i've got my mistake..
It must be :
$ \sum_{cyc} \frac {a}{\sqrt {b \plus{} c}} \ge \sqrt {\frac {3}{2}(a \plus{} b \plus{} c)}$

When does equality hold ?[/quote]

NOPE! the problem is ABSOLUTELY correct just as in my post.
It is correct:
$ \sum_{cyc} \sqrt {\frac {2a}{b \plus{} c}}\le 3$

[hide]
 I am substituting:
$ x \equal{} \sqrt {\frac {b}{a}}$ ; $ y \equal{} \sqrt {\frac {c}{b}}$, $ z \equal{} \sqrt {\frac {a}{c}}$
Hence, as usual, $ xyz \equal{} 1$
Making the substitution, we have to prove that ::
$ \sqrt {\frac {2}{1 \plus{} x^2}} \plus{} \sqrt {\frac {2}{1 \plus{} y^2}} \plus{} \sqrt {\frac {2}{1 \plus{} z^2}}\le3$

We assume that $ x\le y\le z$ which implies that $ xy\le1$, $ z\ge1$
Now, we get from the RMS AM ineq;
$ (\sqrt {\frac {2}{1 \plus{} x^2}} \plus{} \sqrt {\frac {2}{1 \plus{} y^2}})^2 \le 2(\frac {2}{1 \plus{} x^2} \plus{} \frac {2}{1 \plus{} y^2})$  $ \equal{}$  $ 4(1 \plus{} \frac {1 \minus{} x^2y^2}{(1 \plus{} x^2)(1 \plus{} y^2)})$
                                                   $ \le 4(1 \plus{} \frac {1 \minus{} x^2y^2}{(1 \plus{} xy)^2}) \equal{} \frac {8}{1 \plus{} xy} \equal{} \frac {8z}{z \plus{} 1}$ (as $ xyz \equal{} 1$)

So $ \sqrt {\frac {2}{1 \plus{} x^2}} \plus{} \sqrt {\frac {2}{1 \plus{} y^2}}\le 2\sqrt {\frac {2z}{z \plus{} 1}}$ So it is only required to prove that$ 2\sqrt {\frac {2z}{z \plus{} 1}} \plus{} \sqrt {\frac {2}{1 \plus{} z}}\le 3$.

Now, as $ (1 \plus{} z)^2\le 2(1 \plus{} z^2)$ implies that $ \sqrt {\frac {2}{1 \plus{} z^2}}\le \frac {2}{1 \plus{} z}$

So it suffices to prove that $ 2\sqrt {\frac {2z}{z \plus{} 1}} \plus{} \frac {2}{1 \plus{} z} \le 3$, which is equivalent to $ 1 \plus{} 3z \minus{} 2\sqrt {2z(1 \plus{} z)}\ge 0$,
which, in turn, is equivalent to : $ (\sqrt {2z} \minus{} \sqrt {1 \plus{} z})^2 \ge 0$, which is obviously TRUE!!!!!!!! :D  :lol:

[b]Equality holds if and only if[/b]  $ a\equal{}b\equal{}c$!! :lol:
Isn't this correct?
[/hide]
PS: Do you like my solution? :)

[quote="Potla"] Do you like my solution? :)[/quote]
No! Try $ a\rightarrow\plus{}\infty.$  :wink:


geometry
trigonometry
function
quadratics
geometry theorems

In triangle $ ABC$. $ T$ is any point.Denote $ TM,TN,TP$ are the feet from point $ T$ to $ BC,CA,AB$.
Prove that $ [MNP] \equal{} (\frac {1}{2})(R^2 \minus{} OP^2)\sin A \sin B \sin C$
$ \equal{} \frac {R^2 \minus{} OT^2}{4R^2}([ABC])$

Anyone? anyone can solve this problem? help me please :)

I have a solution. It is quite surprizingly but this problem doesn't require any calculations.

Let $ S(T)$ be an area of corresponding triangle $ MNP$. Note that $ S(T)\equal{}\frac{1}{2}(TN\cdot TN\sin \alpha\plus{}TN\cdot TP\cdot \sin \beta\plus{}TP\cdot TM\sin \gamma)$ (a sum of 3 triangles). Consider a diameter $ XY$ containing p.$ T$. Lengths $ TN$, $ TM$ and $ TP$ are linear functions of $ T$ on $ XY$, so $ S(T)$ is a quadratic function on $ XY$, but $ S(X)\equal{}S(Y)\equal{}0$ and $ S(O)\equal{}\frac{1}{4}[ABC]$. Thus the only such quadratic function is $ S(T)\equal{}\frac {R^2 \minus{} OT^2}{4R^2}([ABC])$.

[quote="HelloWorld"]In triangle $ ABC$. $ T$ is any point.Denote $ TM,TN,TP$ are the feet from point $ T$ to $ BC,CA,AB$.
Prove that $ [MNP] \equal{} (\frac {1}{2})(R^2 \minus{} OP^2)sinAsinBsinC$
$ \equal{} \frac {R^2 \minus{} OP^2}{4R^2}([ABC])$[/quote]
$ \equal{} \frac {R^2 \minus{} OT^2}{4R^2}([ABC])$


probability
geometry
3D geometry
group theory

Somebody scrambles a Rubik's cube by randomly turning 5 different faces of the cube. Adrian then tries to solve the cube. He can complete the solve if there are only two moves left, but everything until then is done randomly. What is the probability of Adrian solving the cube in just 5 moves (assume there is only one correct sequence of moves)?

um 1/(18^3)?
[hide="solution"]each move he has 3 planes to move the cube. but he can move it forwards and back wards so thats 6. but then, there are 3 diferent places on a side of the cube that he can move. so thats 18 moves. since there is only 1 way, he will have to move 3 ways until he has two moves left. on each move, he has a 1/18 probability of doing the corect one. so the answer is 1/(18^3)[/hide]

[quote="i_like_pie"]That is correct. In reality, however, that would not be correct because more than one sequence of moves can solve the cube.[/quote]

It's not really necessary to tell people that their answers are correct in every thread...

But to come up with the actual solution, you would have to do some casework and/or group theory, it looks like.


let $ABC$ be triangle such that :
 $\angle ABC=115^0$and, $\angle BAC=55^0 ,$
   the side $CD$ such that $ACD=DCB$,$M$ on side $CD$ such that$\angle MAC=25^0$.
        find $\angle MBC$

[quote="tranthanhnam"][quote="tranthanhnam"]let $ABC$ be triangle such that :
 $\angle ABC=115^0$and, $\angle BAC=55^0 ,$
   the side $CD$ such that $ACD=DCB$,$M$ (CD is bisector of $\angle C$)on side $CD$ such that$\angle MAC=25^0$.
        find $\angle MBC$[/quote][/quote]
1)You can use the edit button,can't you?
2)[hide="My solution"]
From [color=blue]Ceva's theorem in sine form [/color]result that:
$\frac{\sin\widehat{ECA}}{\sin\widehat{ECB}}.\frac{\sin\widehat{EAB}}{\sin\widehat{EAC}}.\frac{\sin\widehat{EBC}}{\sin\widehat{EBA}}=1$
$\Longrightarrow\frac{\sin\widehat{EAB}}{\sin\widehat{EAC}}.\frac{\sin\widehat{EBC}}{\sin\widehat{EBA}}=1$
$\Longrightarrow\frac{\sin 30^o}{\sin 25^o}.\frac{\sin\widehat{EBC}}{\sin\widehat{EBA}}=1$
$\Longrightarrow\frac{\sin 30^o}{\sin 25^o}=\frac{\sin\widehat{EBA}}{\sin\widehat{EBC}}$[color=red](1)[/color]
Otherwise,$\frac{\sin 30^o}{\sin 25^o}=\frac{1}{2\sin 25^o}=\frac{\cos 25^o}{\sin 50^o}=\frac{\sin 65^o}{\sin 50^o}$[color=red](2)[/color]
From [color=red](1)[/color] and [color=red](2)[/color] we have:
$\frac{\sin\widehat{EBA}}{\sin\widehat{EBC}}=\frac{\sin 65^o}{\sin 50^0}$[color=red](3)[/color]
We also have:$\widehat{EBA}+\widehat{EBC}=\widehat{ABC}=115^o=65^o+55^o$[color=red](4)[/color]
From [color=red](3)[/color] and [color=red](4)[/color] we have $\widehat{EBA}=65^o,\widehat{EBC}=50^o$[/hide]

[hide="one more solution"]
$\widehat{CAB}<\widehat{ABC}\Longrightarrow CA> CB$
Let $E$ be a point on the side $CA$ such that $CE=CB\Longrightarrow EB\bot MD$ at $J$
$\widehat{JDB}=\widehat{BAC}+\widehat{DCA}=60^o\Longrightarrow \widehat{JBD}=30^o\Longrightarrow\widehat{JED}=30^o$[color=red](1)[/color]
Otherwise,$\widehat{DMA}=\widehat{MAC}+\widehat{MCA}=30^o$[color=red](2)[/color]
Denote that $MA$ intersects $DE$ at $I$
From [color=red](1)[/color] and [color=red](2)[/color] we have:$E,I,J,M$ is concyclic points $\Longrightarrow\widehat{EIM}=\widehat{EJM}=90^o\Longrightarrow ED\bot MA$
Otherwise,$\widehat{DAM}=\widehat{DMA}\Longrightarrow DA=DM\Longrightarrow EM=EA$
$\Longrightarrow \widehat{CEM}=2\widehat{EAM}=50^o\Longrightarrow \widehat{MBC}=50^o$[/hide]
In fact, this is a very,very easy problem,am I right [b][i]tranthanhnam[/i][/b]? :P


When I try to visit  a website,I see a message
[i]"Service Temporarily Unavailable
The server is temporarily unable to service your request due to maintenance downtime or capacity problems. Please try again later.

Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request.[/i] "
I don't know what happend? Was it attacked by some hackers? could anyone give me some suggestions to log on it normally.
thanks all in advance.

You said it yourself.

[quote]Please try again later.[/quote]


logarithms
inequalities
calculus
calculus computations

Determine all $ k$ for which

$ \sum_{n=1}^\infty \left( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n) }\right)^{k}$

converges.

Since

$ \frac{1}{2\sqrt{n}}<\frac{1\cdot 3 \cdots (2n-1)}{2\cdot 4\cdots (2n)}<\frac{1}{\sqrt{2n}}$, $ n\geq 1$,

we get that your series converges for $ k>2$ and diverges otherwise.

Let $ P_{n}$ be the product inside the parentheses: $ P_{n}=\prod_{k=1}^{n}\left(1-\frac1{2k}\right).$

Take the logarithm. Since for large $ k,\ \ln\left(1-\frac1{2k}\right)$ is comparable to $ -\frac1{2k}$ we conclude that $ ln(P_{n})$ is comparable to $ -\sum_{k=1}^{n}\frac1{2k},$ meaning in turn that $ \ln(P_{n})=-\frac12\ln n+O(1).$

Exponentiate this to get that there exists constants $ c_{1}$ and $ C_{2}$ such that $ \frac{c_{1}}{\sqrt{n}}\le P_{n}\le\frac{C_{2}}{\sqrt{n}}.$

That means we have a two-sided comparison between $ \sum_{n=1}^{\infty}(P_{n})^{k}$ and $ \sum_{n=1}^{\infty}\frac1{n^{k/2}}.$

Hence we have convergence for $ k>2$ and divergence for $ k\le 2.$

[color=green][Ah, didi beat me to it by typing faster. Same argument.][/color]

thanks those are nice solutions.

Kent's is more like the way I approached it.  Didi, why did you use that inequality, is it obvious somehow?  I can prove it but I wouldn't have thought of using that.

For example:

Let $ A = \prod_{k=1}^{n}\frac{2k}{2k-1}$ and $ B = \prod_{k=1}^{n}\frac{2k+1}{2k}$. We have that $ A > B$, since $ \frac{2k}{2k-1}> \frac{2k+1}{2k}, \, \forall k \in \overline{1,n}$. Hence, $ A^{2}> AB = 2n+1$, from where we get that $ A > \sqrt{2n+1}$. Hence, $ \frac{1 \cdot 3 \cdot \ldots \cdot (2n-1)}{2 \cdot 4 \cdot \ldots \cdot (2n)}= \frac1{A}< \frac1{\sqrt{2n+1}}$.
In a similar manner, we can obtain $ A < 2 \sqrt n$, from where $ \frac{1 \cdot 3 \cdot \ldots \cdot (2n-1)}{2 \cdot 4 \cdot \ldots \cdot (2n)}= \frac1{A}> \frac1{2 \sqrt n}$.


[quote="math92"]What are the odds of rolling a pair of ones on two standard, six-sided dice?[/quote]

[hide]

The odds of an event are generally given by $\frac{q}{p}$, where $q$ is the probability the event didn't occur; $p$ is the probability that it did occur.

That said, the probability of "snake eyes" is $p=\frac{1}{36}$ ; $q=1-p=\frac{35}{36}$

The odds are therefore $35:1$
[/hide]

AHA! it asked for odds not probability.

I thought it looked a little too obvious :P

Important life lesson there  ;)

I was thinking probability at first as well.

bah notation shmotation.  1/36 conveys the same info as 35:1  :lol:

EDIT:[hide]
$\frac{35}{36}$
therefore the odds are 35:1[/hide]

[quote="gr8sk8r"][hide]
$\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}$[/hide][/quote]

I don't think you just read what was going on. It asked for odds...

woops sorry.nice observation Dr. No i didn't see that at all. :oops:

[quote="yankeefan13"]wise is 236factorial[/quote]

No, it's Dr. No. (That sounds funny  :lol: )
By the way, that grammar was awful.


A gaseous mixture of dioxygen and Krypton has a density of 1.104 g/L at 435 torr and 300 K. What is the mole percent dioxygen in the mixture?

If the mole percent of dioxygen is $ x$, there are $ x$ moles of dioxygen in $ 100$ moles of mixture.
$ 100$ moles of mixture would occupy 2240 L at STP (273 K ant 760 torr).
At 435 torr and 300 K, they would occupy:
(2240 L)(760/465)(300/273) = 4300 L,
and would weigh:
(4300 L)(1.104 g/L) = 4750 g
Since the weight in grams of $ x$ moles of dioxygen is $ 32.0x$,
and the weight in grams of $ (100\minus{}x)$ moles of Krypton is $ 83.8(100\minus{}x)$,
the total weight in grams of the mixture is:
$ 32.0x\plus{}83.8(100\minus{}x) \equal{} 8380\minus{}51.8x$
Solving
$ 8380\minus{}51.8x \equal{} 4750$,
we find $ x \equal{} 70$


let $ a$ ,  $ b$ , $ c$ $ \in$ $ {\mathbb R}^{\plus{}}$ such that ; $ a\plus{}b\plus{}c\equal{}1$ , prove that :

$ \frac{a}{\sqrt{a\plus{}1}}\plus{}\frac{b}{\sqrt{b\plus{}1}}\plus{}\frac{c}{\sqrt{c\plus{}1}} \geq \frac{3\sqrt3}{2}$

[quote="y-a-s-s-i-n-e"]let $ a$ ,  $ b$ , $ c$ $ \in$ $ {\mathbb R}^{ \plus{} }$ such that ; $ a \plus{} b \plus{} c \equal{} 1$ , prove that :

$ \frac {a}{\sqrt {a \plus{} 1}} \plus{} \frac {b}{\sqrt {b \plus{} 1}} \plus{} \frac {c}{\sqrt {c \plus{} 1}} \geq \frac {3\sqrt3}{2}$[/quote]
must be : $ \frac{\sqrt{2}}{2} \leq \frac {a}{\sqrt {a \plus{} 1}} \plus{} \frac {b}{\sqrt {b \plus{} 1}} \plus{} \frac {c}{\sqrt {c \plus{} 1}} \leq \frac {\sqrt3}{2}$;
$ f(x)\equal{}  \frac{x}{\sqrt {x \plus{} 1}}$ convex then $ x>\minus{}1$ , so : $ f(1) \leq f(a)\plus{}f(b)\plus{}f(c) \leq 3f(\frac{1}{3})$;

$ f(x)\equal{}\frac{x}{\sqrt{x\plus{}1}}$ is concave !!!!

$ f"(x) \geq 0 \:   \Leftrightarrow  f \: is \: covexe$


1) Alkaline hydrolysis of methylisocyanate gives ???
2) Chloropicrin is obtained by?????
3) Alanine exists in acidic medium as ????
4) Furfural + NaOH gives???
5) Hoffmann mustard oil rxn is ????

 :( [b][size=150]Very Sorry, but i do not know how to post it in iitjee discussion thread.What ever i post comes here. Some one pls help!!!! [/size][/b] :maybe:  :( [/b]

instead of clicking on "post new topic" ,click on "reply to topic" :)

2.$ HNO_3 \plus{} CHCl_3$
5.Amines + CS2

i think the answer fo0r the 3rd question is zwitter ion....the base donates a pair of electrons or rather accepts a proton...a nd in the acidic medium...acid pat dissociates giving h+ions.......

please correct me  if i am worng

no
only the $ \minus{}NH_2$ grp is protonated
and $ COOH$ remains as such :D

as far as i kno...

1) isocyanates dont actually give alkaline hydrolysis, they only give acidic hydrolysis. Still if this is not the answer than it has to be a primary amine and carbonate as it resembles the last step of hoffman brommamide.

for the fifth question $ HgCl_2$ is also needed.

The fourth question is canizzaro reaction.


I have a nice problems,as follows:
Let $p$ denote a prime number $\geq$ 5.Does there exist positive number $n$ such that:
$d(p^{n-1}n)=n$ where $d(x)$ is number of divisors of $x$

$n=1$ and we have simply $d(p^{n-1}n)>n$ for every $n>1$. Or am I missing something?

He means (I think):
When $n>1$, then $d(p^{n-1} n) > n$ because of the divisors $1,p,p^2,...,p^{n-1}, n p^{n-1}$.


Hey, I'm stuck on this question while studying for my provincial exam:

If the function y = 3^X is expanded vertically by a factor of 9 to produce a new function, which of the following is an equation of the new function?

A) Y = 3^(2X)
B) Y = 3^(3X)
C) Y = 3^(X+2)
D) Y = 3^(X-2)

the function is 9 times larger, so the old function y, is multiplied by 9 to get the new function y', or

$y'=9*3^x=3^{x+2}$...$C$


[i]Let a be a positive constant[/i], [i]and the sequence [/i]( $ v_n$) [i]defined by :[/i]
     $ v_1 \equal{} a$

  $ v_{n \plus{} 1} \equal{} \frac {n.(v_n)^3 \plus{} 1}{(v_n)^{2}.\sqrt [3]{n^3 \plus{} n^2}}$ [i]with n= 1,2,3,....[/i]

 [u][i] Find[/i][/u] $ \lim_{n\rightarrow\infty}(\sqrt [n]{C_{2n}^{n}}.v_{n})$

 ( $ \ C_{n}^{k} \equal{} \frac {n!}{k!(n \minus{} k)!}$)
 :love: AN

the problem is equivalent to find the limite of $ U_n$ ,because we now tha $ lim(Vn)^{1/n}\equal{}limV_{n\plus{}1}/Vn$ , so $ lim [(2n)!/n!^2]^{1/n} \equal{}4$


geometry
incenter
geometric transformation
reflection
analytic geometry
circumcircle
perpendicular bisector

ABC is a triangle which is inscribed in circle O.  Bisector of angle ABC meet AC at F and O at G .
Bisector of BAC meet BC at D and O at E. Let P be a common point of lines GE and FD. Prove that AB || PI (I is a incenter of triangle ABC)

Since $ I$ is orthocenter of its circuncevian triangle, it follows that $ DE$ is the perpendicular bisector of $ AI.$ The tangent $ \mathcal{T}_a$ to $ (O)$ through $ A$ is then the reflection of the parallel $ \mathcal{L}_a$ to $ BC$ passing through $ I$ about $ DE.$ Hence, the problem is equivalent to show that $ V_bV_c , \mathcal{T}_a$ and $ \mathcal{L}_a$ concur. Using barycentric coordinates, we have

$ V_bV_c \equiv bcx \minus{} acy \minus{} abz \equal{} 0 \ , \ \mathcal{T}_a \equiv c^2y \plus{} b^2z \equal{} 0.$

$ \mathcal{L}_a$  passes through $ I(a: b: c)$ and the infinite point of $ BC$ $ (0: 1: \minus{} 1)$

$\Longrightarrow \mathcal{L}_a \equiv (b \plus{} c)x \minus{} ay \minus{} az \equal{} 0$

Therefore, $ V_bV_c , \mathcal{T}_a , \mathcal{L}_a$ concur at $ P \ (a(c \minus{} b): \minus{} b^2: c^2).$

Let $ EG\cap AB \equal{} \{L\}$
Appying [b]Menelaus's[/b] theorem for triangle $ IEG$ with line $ PDF$ we have:
$ \frac {PE}{PG}.\frac {FG}{FI}.\frac {DI}{DE} \equal{} 1$
$ \Rightarrow \frac {PE}{PG} \equal{} \frac {FI}{FG}.\frac {DE}{DI} \equal{} \frac {r}{AG.sin \angle \frac {B}{2}}.\frac{BE.sin \angle \frac {A}{2}}{r}\equal{} \frac {BE.sin \angle \frac {A}{2}}{AG.sin \angle \frac {B}{2}}$
$ \Rightarrow 1 \plus{} \frac {EG}{PE} \equal{} \frac {AG.sin \angle \frac {B}{2}}{BE.sin \angle \frac {A}{2}}$
On the other side, $ \frac {LE}{LG} \equal{} \frac {NE}{MG} \equal{} \frac {AE.sin \angle \frac {A}{2}}{BG.sin \angle \frac {B}{2}}$
$ \Rightarrow 1 \plus{} \frac {EG}{LE} \equal{} \frac {BG.sin \angle \frac {B}{2}}{AE.sin \angle \frac {A}{2}}$
$ \Rightarrow \frac {PE}{LE} \equal{} (\frac {BG.sin \angle \frac {B}{2}}{AE.sin \angle \frac {A}{2}} \minus{} 1): (\frac {AG.sin \angle \frac {B}{2}}{BE.sin \angle \frac {A}{2}} \minus{} 1)\equal{} \frac {BE}{AE}.\frac {BG.sin \angle \frac {B}{2} \minus{} AE.sin \angle \frac {A}{2}}{AG.sin \angle \frac {B}{2} \minus{} BE.sin \angle \frac {A}{2}}$
But $ \frac {BE}{AE} \equal{} \frac {EI}{AE}$ so $ IP//AB$ iff $ AG.sin \angle \frac {B}{2} \minus{} BE.sin \angle \frac {A}{2} \equal{} BG.sin \angle \frac {B}{2} \minus{} AE.sin \angle \frac {A}{2}$
$ \Leftrightarrow AG.sin \angle \frac {B}{2} \plus{} AE.sin \angle \frac {A}{2} \equal{} BG.sin \angle \frac {B}{2} \plus{} BE.sin \angle \frac {A}{2}$
$ \Leftrightarrow IG.sin \angle \frac {B}{2} \plus{} (AI \plus{} IE).sin \angle \frac {A}{2} \equal{} (BI \plus{} IG).sin \angle \frac {B}{2} \plus{} EI.sin \angle \frac {A}{2}$
$ \Leftrightarrow AI.sin \angle \frac {A}{2} \equal{} BI.sin \angle \frac {B}{2} \equal{} r$ (true!)
We are done!

Assume $ AC<BC$ ($ AC=BC$ is obvious). Apply Pascal Theorem to degenerate hexagon $ ACCBGE$ to get that $ P$ is intersection of $ GE$ and tangent to $ O$ at $ C$. We see that $ PC=PI$ and $ \angle PIC=\angle ICP=\angle B+\angle \frac{C}{2}$ and $ \angle CIG=\angle \frac{B}{2}+\angle \frac{C}{2}$ hence $ \angle GIP=\angle \frac{B}{2}=\angle GBA$.

Dear Mathlinkers,
1. let (O) the circumcircle of ABC, Y the second point of intersection of CI with (O), X the meet point of YE and BC, Tc the tangent to (O) at C.
2. According to a special case of Pascal's theorem applied to ACTcBGEA, Tc goes through P.
3. According to Pascal's theorem applied to CYEGBCTc, I, X and P are collinear.
4. Let Ty be the tangent to (O) at Y; we have Ty // AB.
5. According to a special case of Pascal's theorem applied to CYTyEABC, IX is parallel to AB
and we are done...
Sincerely
Jean-Louis


trigonometry
calculus
calculus computations

I keep differentiating this equation and getting a different answer from my book. Can somebody solve it so I know who made the mistake?

$y=e^{cx}(csinx-cosx)$

Thanks

well i get $\frac{dy}{dx}=e^{cx}(c^2+1)\sin x$


I think this is a description of an AoPS video, not an Alcumus problem.

Oh I just realized this was posted before I was born

[quote=GameBot]Combinations with Restrictions Part 3

We solve a hard problem involving combinations and restrictions.[/quote]

I thought GameBot is a robot… why is he posting for a video


AMC
USA(J)MO
USAJMO
USAMO
AIME
AMC 10
Support

[color=green]Note: This thread has been split from the announcement thread and the "What do you think?" thread per AMCDirector's request. -mod[/color]

With regard to red MOP: Supposing the top 25 scores on 1/J3 and 4/J6 are such that the cutoff is 8, but someone taking the USAMO gets a 070700, it looks like that person would not make it according to the guideline in 6, when they really should. The way I'm reading it USAMO 2/3/5/6 are only a tiebreaker; this does not seem right.

My personal fix to this would be to factor in the entire score for USAMO takers when finding that top 25, giving that group a deserved advantage.

Very interesting solution to the qualification problem you guys have been having.

Acknowledged!
We did consider that possibility, but (if the USAMO committee continues its usual excellent estimation of difficulty), scores like 070700 for 9th graders are unusual (but of course not impossible).  After some discussion, we decided that at this point it is probably better to try to specify an objective comparison criterion, even if it might have hypothetical flaws.
A year from now, we can measure whether the criteria accomplished what we intended and adjust.

An alternative is to trust the judgment of the graders about who is invited to MOSP on the basis of their MO scores.  Although I would trust the judgment process because the judgments are made by experienced, intelligent people, it is not a transparent process.  Unfortunately, it is hard for me to justify processes which are not transparent, even if they are fair or intelligent.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="perfect628"]Hmm well hopefully the freshmen taking the USAJMO will realize they need to start with the last problem...I imagine those talented enough who aren't in the loop about these things might just start with the first one, putting them at a severe disadvantage.[/quote]

Not so much start, but do exclusively. There isn't much incentive to do J1,J2,J4,J5 really, unless you want to say you are a junior winner (junior winner? how does [i]that[/i] sound?)

[quote="MellowMelon"]With regard to red MOP: Supposing the top 25 scores on 1/J3 and 4/J6 are such that the cutoff is 8, but someone taking the USAMO gets a 070700, it looks like that person would not make it according to the guideline in 6, when they really should. The way I'm reading it USAMO 2/3/5/6 are only a tiebreaker; this does not seem right.

My personal fix to this would be to factor in the entire score for USAMO takers when finding that top 25, giving that group a deserved advantage.

Very interesting solution to the qualification problem you guys have been having.[/quote]

Could you please enlighten me about these "qualification problems"?  The way I see it, narrowing down the selection criteria from 6 problems to 2 makes the system more based on luck and unstable, and would result in Red MOP qualifiers who potentially are farther from the top 25 9th graders than the qualifiers by the old system.

A possible remedy to stabilize Red MOP would be to make the USAJMO J2/J5 = USAMO 1/4 and J3/J6 = 2/5 while introducing a new J1/J4 for the USAJMO that are not AIME type questions but instead easy proof questions.  This would provide a slightly higher level of difficulty, and would comply with mewto's suggestion that people would like to see something other than mostly calculation AIME style problems on the USAJMO, which is at this point basically AIME #s 16-21.
Red MOP could also now be calculated by 1/2/4/5 = J2/J3/J5/J6, which is twice as large a variety as 1/4 alone.

[quote="CircleSquared"]Could you please enlighten me about these "qualification problems"?  The way I see it, narrowing down the selection criteria from 6 problems to 2 makes the system more based on luck and unstable, and would result in Red MOP qualifiers who potentially are farther from the top 25 9th graders than the qualifiers by the old system.[/quote]
Qualification problems means the issue of taking the AMC10 being beneficial to qualifying for USAMO, which is the topic that has been hotly debated on this forum.

In practice, I believe the system will not really become more unstable because problems 2/3/5/6 on the USAMO have little effect on red MOP qualification. The change to a 2 problem system is actually a very small one, and saying the original is a 6 problem system is ludicrous.

EDIT: Somehow an extra negative crept in there...

[quote="MellowMelon"][quote="CircleSquared"]Could you please enlighten me about these "qualification problems"?  The way I see it, narrowing down the selection criteria from 6 problems to 2 makes the system more based on luck and unstable, and would result in Red MOP qualifiers who potentially are farther from the top 25 9th graders than the qualifiers by the old system.[/quote]
Qualification problems means the issue of taking the AMC10 being beneficial to qualifying for USAMO, which is the topic that has been hotly debated on this forum.

In practice, I do not believe the system will not really become more unstable because problems 2/3/5/6 on the USAMO have little effect on red MOP qualification. The change to a 2 problem system is actually a very small one, and saying the original is a 6 problem system is ludicrous.[/quote]

Ah, THOSE qualification problems.

Also, you have a very good point about how Red MOP is not often decided by anything other than 1 and 4.

Exception being this year, in my opinion.  Many Red MOPpers I know qualified not by 1 and 4 but by 2 and 4.  I think somehow including 2 and 5 might be a good idea to account for years like this one, where 2 is also a problem many people earn significant points on.

[quote]In practice, I do not believe the system will not really become more unstable because problems 2/3/5/6 on the USAMO have little effect on red MOP qualification. The change to a 2 problem system is actually a very small one, and saying the original is a 6 problem system is ludicrous.[/quote]

I'm not sure this is always true. This year, for example, I believe most of the points I earned came from #2, and I know someone who solved #2 and #6 and did no work on the others. (Both of us would have qualified for red MOP had we been freshmen.)

This is definitely an interesting solution and I'm looking forward to seeing how it works out. I'm a little uncomfortable with the continued incentive for underclassmen not to take both AMC-12s, but at this point it's probably just me being stubborn and thinking everyone should be like me (I've taken both AMC-12s since freshman year). I also believe from the above anecdotes that considering all six USAMO problems really can be noticeably different from and probably better than only considering the "easiest" two, but I don't have any constructive suggestions in that area.

I'm curious about whether there will be some sort of honor/recognition for top JMO scorers besides MOP qualification.

[quote="shtsxc12"]I think that this new system is flawed: for USAMO, a person with 700700 will make the "Red Mop" for sure, but a person with 700070 or 070700, etc. may not make "Red Mop", since 2/3 and 5/6 are only used in a tiebreaker. In this year, 1/4 were more difficult than those in previous years, I knew that somebody did better in 6 than 1 or 4.[/quote]

This was said in the announcement thread to be "rare" by the statistics, but as one of these people myself (failing to get #4 but getting significant points on #6), I understand this argument and I believe it is highly valid.

In essence, I think this provides an excellent, more worthwhile opportunity to those who aren't really seasoned olympiad problem solvers (e.g., 4th graders.) It'll give them things to work toward that aren't far too advanced for them, and they won't come walking back with 0's.

I do, however, find this a very weak selection process for freshmen taking USAJMO, as we're basically putting all our eggs in two baskets, rather than six. My personal opinion is that having four genuine Olympiad problems and two tiebreakers is more reasonable than two genuine Olympiad problems and four tiebreakers. In addition, I'm not at all comfortable with only one option for an Olympiad problem. While the majority of Red MOPpers get 1's and 4's, there are some who get other problems. This would, in a particular perspective, make those exceptions even more eligible (since they got problems "harder" than what most people qualified with), though under this system, they would not even qualify.

Take me, for instance. If I were forced to do 1 or 4, I wouldn't have made it to Red MOP. I'm 80% sure my distribution was 0/7/0/0/0/1, possibly 0/6/0/1/0/1, though I'm inclined to believe the former (since my friend did a lot more progress on 4 than I but ended up with an overall score of 0.)

While such a distribution "rare" and may support the majority, there's an unheard minority that needs to be accounted for. To say the least, woudn't it make sense to have less tiebreakers than genuine problems?

Other than that, I think this is a great idea. The only real flaw I see in this is the freshman selection process. If you're good enough, you don't need the 10 to qualify. If you're not good enough, you should probably just do USAJMO. I don't see any problems.

EDIT: Upon rereading my post, I found it unclear. I'm criticizing the format of the USAJMO, not the selection process for freshmen taking USAMO (though on hindsight, as timwu states below, that could be improved upon as well.)

[quote="AMCDirector"]6. Selection to "Red MOSP" will work in the following way (assuming funding for Red MOSP continues.)  Among all 9th graders taking the USAMO and JMO, the scores on problems J3 = 1 and J6 = 4 will be
totalled and ordered.  Then we will select the students with the top
25 scores on these two problems.  Ties among the top scores in each
category will be broken by using the scores from the other 4 problems
on each Olympiad.[/quote]

If there's a freshman wanting to make Red MOP that does not qualify for the USAMO but takes the JMO, wouldn't it mean this person would have to ignore 1, 2, 4, 5 and focus first on 3 and 6? This is a little weird. I certainly see it happening that someone takes the 10A, does well enough to make the JMO, but gets something like 120 on the 12B and 8 on the AIME, which would make him fall short of USAMO qualification.

Also I agree that although 1 and 4 are the major players in MOP qualification, it cannot be ignored that there should be other places in which kids should be able to show what they can do. I made red MOP in 2008, but didn't solve #1 this year; I got most of my day 1 points on #2. There was a student at red MOP last year who solved #3 on the USAMO and not #1 (although his understanding of graphs of polynomials was pretty questionable  :wink: ). Of course I have complete faith in the USAMO committee to order the problems properly, or else I wouldn't work on them in order, but everybody is different, and especially when #1 is a geometry problem you'll get a lot of people that will avoid it and make significant progress and/or solve a different one.

But I'm sure the committee will already have this mind and take extra care in addressing this. The Jr. USAMO is a very interesting idea and although it's not relevant at all to me anymore, I wish I would have had it back in 8th grade.

[quote="AMC Finance"]I believe if you look at the results and statistics, you will notice that the number of upperclassmen (11-12) vs underclassmen (10 and below) is about what the split for the qualification for the new USAMO/USAJMO.  Therefore there will be little to no dilution, nor disadvantage to anyone.  You will also notice that a great majority of the younger students score almost nothing on the USAMO making the possibility of scoring positively on the USAJMO a better use of their time and skills.[/quote]

I think the number of amc-12 qualifiers (upperclassmen + some underclassmen) vs the number of amc-10 qualifiers (underclassmen - those underclassmen) are a better split.

I also think Red MOP qualification process is highly flawed. Did the AMC conduct a survey of the points distribution of past Red MOPpers to conclude that getting scores on other problems is "rare"? A hypothetical situation: a 9th grader got 7/7/0/1/0/0 and the cutoff is 9 (7 on #1 and 2 on #4, e.g.). The said 9th grader will fall short by 1 point when he legitimately solved 2 problems!
An alternative proposal: take the two highest scores on the six problems by the USAMO 9th grader as his Red MOP index.

I like the ideas behind the new system, but this is a very major problem:

[quote="CatalystOfNostalgia"]If there's a freshman wanting to make Red MOP that does not qualify for the USAMO but takes the JMO, wouldn't it mean this person would have to ignore 1, 2, 4, 5 and focus first on 3 and 6? This is a little weird. I certainly see it happening that someone takes the 10A, does well enough to make the JMO, but gets something like 120 on the 12B and 8 on the AIME, which would make him fall short of USAMO qualification.[/quote]
Now, for freshmen trying to qualify for Red MOP (who will make up a significant portion of USAMO/JMO takers), the goal is no longer to get the best score they can on the test they're given, but to get the best score on two problems on the test.  I think most freshmen will spend all their time on the important problems and ignore the others unless they're 100% sure they've solved the important problems.

The upshot of this, in my opinion, is that despite the AMC's efforts to provide four more problems for younger test takers to tackle, the AMC has actually reduced the effective number of problems on the test to just two for a huge portion of the people taking it.

[quote="CatalystOfNostalgia"](although his understanding of graphs of polynomials was pretty questionable  :wink: ).[/quote]
I thought his new theory was pretty groundbreaking. :D

I'll put my alternative proposal for Red MOP qualification in the other thread here: for 9th-grade USAMOer, take the two highest scores on the six problems as his Red MOP index. This way 9th graders will try to solve as many problems as they can.

Also this current process will encourage 9th graders to spend time only on #1 and #4 (if they cannot get). This will not only cause people with the ability to solve other problems not make Red MOP but also will hide this flaw itself: as no one will even attempt to solve other problems, statistically most Red MOPpers in the future [b]will[/b] not solve other problems.

I don't think misinformation will be a problem for JMOers: just put the Red MOP qualification process in USAJMO manual and the test instruction.

[quote="not_trig"] Moreover, how would ties be broken when comparing between JMO and USAMO scores? If a freshman gets #2 on USAMO, is that better than solving #1 and #4 on JMO?
[/quote]

I"m getting the impression that many people commenting in this thread have not read the PDF attachment to AMC Director's initial post on this thread.  The format of the posts on this forum makes it easy to overlook such attachments, unfortunately.)

The CAMC memo attached to the first post in this thread makes clear that students who make significant progress on USAMO problems other than 1 and 4 would receive very strong consideration and priority before they even consider USAJMO students.

[quote="CAMC memo"]Currently we invite the approximately 25 top scoring 9th grade students on the USAMO to the MOSP.
Assuming continuing funding, the USAMO SubCommittee and the graders will still select about 25
top scoring 9th graders on the USAMO and USAJMO to the MOSP. The scores on the
common problems of the two exams will be compared first. Ninth grade students taking the
USAMO with significant scores on the remaining USAMO problems will be considered next. Say this is n
students. Then the top 25 – n 9th grade students taking the Junior Math Olympiad would be invited to
MOSP.[/quote]

I think I might be misreading something there.

So then if you do solve both #1 and #4, but are taking the USAMO, you won't be considered since you did not have "significant scores" on the other problems?

[quote="timwu"]An alternative proposal: take the two highest scores on the six problems by the USAMO 9th grader as his Red MOP index.[/quote]

Excellent suggestion. That was actually what I was about to say until I saw your post.

I also agree with everyone else that it should remain 330/160 rather than 270/230.

Another possibility would be to exclude USAJMO from red MOP, but there might not be enough 9th grade USAMO qualifiers to fill up MOP...

As many of the previous posters, I am somewhat worried about freshman USAMO participants who score very well on the USAMO while solving only one of 1 or 4. Every year, there are red MOPpers who got in with one of 1,4 and some other problem. It also concerns me that 1,4 may be on the easy side, and then the cutoff may be near 14, so that a hypothetical freshman who misses the blue cutoff by a point or two (this is entirely realistic) but had trouble with say #4 would not qualify. I think a more logical solution would be to base red MOP qualification on total score (for USAMO) or score on 3,6 (for USAJMO). While this places USAJMO participants at a disadvantage, that is a product of them choosing an easier test - in other words, the procedure should by default look for USAMO participants but also recognize particularly strong USAJMO participants who may not have been aware of their own capabilities (or the convoluted system). Furthermore, since red cutoffs are almost always less than 14, it would always be possible for USAJMO participants to qualify.

My other concern is that since even though the USAJMO would be offered to all students in grades 10 and below, only 9th graders could progress beyond it. This suddenly seems extremely arbitrary. While I realize that the 9th grade restriction for red MOP is a condition of Akamai, not the MAA, it would seem much more logical and consistent to offer red MOP to all USAJMO participants (and USAMO participants in corresponding grades), even if it means offering USAJMO only to freshmen-and-under.

But for now, all we can do is throw out thoughts. Although I won't be participating in either contest next year, I will be interested to see the first batch of data and develop more informed opinions. I do hope I won't see strong USAMO scores rejected in favor of weaker USAJMO participants.

[quote]I think a more logical solution would be to base red MOP qualification on total score (for USAMO) or score on 3,6 (for USAJMO).[/quote]
I think this is the best modification of the proposed red MOP selection process I've seen so far. It means a reduction in the incentive for freshman USAMO takers to focus exclusively on 1-4 and (I think) appropriately favors USAMOers over USAJMOers.

I think the best way would be to have 3 regular (hard) problems in less time (maybe 1 1/3 hours?) on the first day and then 3 proofs on the 2nd being of 1,2, and 4 level.

What if someone makes a lot of progress on all of the problems, with something like 344344? Obviously, this person deserves to get into MOP, but will he? His score overall would be 22, which is impressive, but his score to get into MOP would be 6.

erm, 3's and 4's are extremely rare... and with a score of 22 one would definitely make blue. I like the idea of total USAMO score v. USAJMO 3/6.

[I'm sorry that my previous post didn't make much sense. Here I go again.]

Basically, on the USAJMOs, I think we need more actual problems than tiebreakers. Putting all our faith in the ability for freshmen to solve two problems, rather than six, while the other four are practically irrelevant doesn't sound so great to me. When I did my USAMO practices this year, I solved random combinations of 1's, 2's, and 3's, very rarely just 1's and 4's, perhaps because I chose the problems that best matched my preferred field of math. If I were restricted to one problem on each test, I would not have performed nearly as well as I had on the practice tests. Even this year, if I were restricted to 1 and 4, I wouldn't have made it to Red.

I like CircleSquared's idea of having J2/J5 = 1/4 and J3/J6 = 2/5, and having J1/J4 just being a fairly simple proof.

I also don't see why only 1 and 4 should be considered for USAMOs. The minority who get other problems should perhaps be even more qualified, because they solved "harder" questions, and yet through this system, they are completely ignored. Why can't all their points simply be considered?

[quote="mathcire"]Question - Mr. Dunbar, you wrote [quote]6. Selection to "Red MOSP" will work in the following way (assuming
funding for Red MOSP continues.) Among all 9th graders taking the
USAMO and JMO, the scores on problems J3 = 1 and J6 = 4 will be
totalled and ordered. Then we will select the students with the top
25 scores on these two problems. Ties among the top scores in each
category will be broken by using the scores from the other 4 problems
on each Olympiad.
[/quote]

So if I understand correctly, (and please correct me if I don't) but lets say two 9th graders - one taking the USAMO - the other, the JMO, both score the same on the 2 key selection problems.  Now you look at how they scored for the other 4 problems? If yes, then wouldn't it seem reasonable to assume that the student taking the presumably "less challenging" (my words) JMO be at a distinct ADVANTAGE over another 9th grader who took the USAMO? If so, this doesn't seem the most fair or logical method to me, as it would seem to potentially give a Red MOP (MOSP?) spot to a student who may actually have been the "lesser qualifier" using the old system and a student who took (and did well) on the AMC 12, may now lose out on a spot as a result of his choice (mistake??) to take the AMC 12 rather than only taking the AMC 10.  Am I correct? If not, please explain how this new tie-breaking system will be fair to those 9th grade students taking the AMC 12. Thank you, and I apologize up front if I've somehow misunderstood the new system.[/quote]

evidently they give preference to the real USAMO if you read the PDF. I'm not sure exactly how they do this though.

[u]Addressing this comment specifically to Mr. Dunbar:
[/u]I started to post my opinion, but I first read others' posts with my same thoughts and insights, and those before me stated them so eloquently, that I am hard pressed to improve upon. But just to add my "vote" to the mix, I whole- heartedly agree that your new system may be flawed and quite detrimental to next year's 9th graders, in particular.    Your statement of [quote][i]A year from now, we can measure whether the criteria accomplished what we intended and adjust.[/i][/quote] quite concerns me as it seems as though the AMC has decided (at least for next year) that it's new system is a "done deal" and is unwilling to even contemplate the idea that while the new system I'm sure was considered quite extensively, nonetheless, may still be quite flawed, particularly against potential Red Moppers. Since the USAMO system is not at all "transparent" as I believe is the term you've used, in that no one taking can ever really be "sure" of where they did or did not earn points, ([i]an issue which I'm sure could be a whole separate post!!) [/i]nonetheless, most students have a sense of where they understood or completed a problem, and I personally know several of our country's most promising young mathematicians who strongly believe they earned most of their points on problems other than the two that the "majority" got correct.  Frankly, I think that some of these "out of the box" thinkers who may get the more complex problems correct may also be the ones that you would ultimately want on our country's International Team, and yet, they may be the exact ones hurt by this new policy. I urge you to PLEASE reconsider going forward fully with your new system, and I urge anyone (coaches, students, teachers, professors, parents) agreeing with me and the many others who have already voiced concern or objection, to similarly speak up on behalf of ALL the students potentially involved with the USAMO.

I don't know if anybody has suggested this, but people have been joking around about White MOP. Should we make one for the top USAJMO participants and leave Red, Blue, and Black MOPs the way they were this year?

[quote="AIME15"]I'm not sure AMC would be able to get the funding for that.[/quote]

It's more a matter of redistributing funding - split the current Red MOP funding between Red and White.  At MOP, they could effectively be one group.  I like the idea.

[quote="dgreenb801"]If you can do a harder problem, shouldn't that be better? I'm confused. I think we may be trying to outsmart the process. I would get rid of the AMC10 and just have the AMC12 so you only need 100, so there is less luck involved and we can better distinguish between kids, I don't think this should discourage kids from participating. If our problem is we're not getting the right USAMO qualifiers, I would make two aimes: a harder one and an easier one instead, to not discourage kids. That way the kids that do very well on the amc could take an aime with problems around the 7-15 level to better determine who should really make USAMO. But USAMO is the last level, I don't know if making two of them is really a good idea.
Part of the problem, as I said before, is that we don't want to discourage kids by making it too hard at too early a stage, but we do have to take the selection process for the USAMO seriously. We could perhaps put some proof types problems at the very end of the AMC; consider for example, the Turkish, olympiad, with multiple choice proof-type questions. That way, more talented kids would stand out better, and it wouldn't be a competition of least mistakes.[/quote]

I agree. It seems that the distinction between AMC10 and AMC12 is causing far more problems than it is fixing. If getting beginning algebra students to take the AMC series is an issue, you can simply take out the material that requires more knowledge and let the AIME decide on that.

[quote="worthawholebean"]However, then the test ceases to be interesting. The AMC almost above all doesn't want the testing to be routine for anyone - then they have failed.[/quote]

I think the MAA's concern in separating the AMC10 and 12 is that people who have not taken Math Analysis might be discouraged by the presence of complex numbers and the like. I am here interpreting knowledge to be "basic understanding of material".

I'm not sure how "requiring less knowledge" leads to routine tests. It only limits the problem selection. Problems requiring Math Analysis are no more or less challenging than those that do not, unless the test taker does not know Math Analysis.

You might be interpreting knowledge to mean difficulty, but I can assure you that is not the case. This year's USAMO #3b holds its roots in Algebra 1, and I am sure you will agree that it is rather difficult. :)

[quote="Differ"][quote="AMCDirector"]For the intention, purpose, and history of development behind the AMC 10, the content in the AMC 10, and the scoring rules used in the AMC 10 please read the "Message from the AMC 10 Chair" on page 14 of the "[i]2008 Summary of Results and Awards[/i]".  That message eloquently explains it all .

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]

Unfortunately, some of us do not have access to this document. Where could we acquire a copy?[/quote]

The AMC10/12 archive pages on the AMC website have PDF files of the summary books for recent years, including 2008.

Here's the link to the 2008 Summary book:
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/Summary,2008HS.pdf

In the archive:
http://www.unl.edu/amc/e-exams/e6-amc12/archive12.shtml
...second to last row,
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/Summary,2008HS.pdf
...along with the Summary from 2007, 2006, 2005, and 2004.

Steve Dunbar
MAA Director, American Mathematics Competitions


Prove that the sum $1^{k}+2^{k}+3^{k}+...+n^{k}$ is a polinomyal in $n$ of degree $k+1$.

see http://mathworld.wolfram.com/PowerSum.html   :wink:


function
geometry
circumcircle
national olympiad

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=136052]Problem 1[/url]

Let $n$ be positive integers. Two persons play a game in which they are calling a integer $m\ (1\leq m\leq n)$ alternately. Note that you may not call the number which have already said. The game is over when no one can call numbers, if the sum of the numbers that the lead have said  is divisible by 3, then the lead wins, otherwise the the second move wins. Find $n$ for which there exists the way of forestalling.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=136045]Problem 2[/url]

Find all functions $f,$ defined on the positive real numbers and taking real numbers such that
\[f(x)+f(y)\leq \frac{f(x+y)}{2},\ \ \frac{f(x)}{x}+\frac{f(y)}{y}\geq \frac{f(x+y)}{x+y}\]
for all $x,\ y>0.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=136055]Problem 3
[/url]
Let $\Gamma$ be the circumcircle of triangle $ABC.$ Denote the circle which touches to the sides $AB, \ AC$ and touches to $\Gamma$ internally at $P$ by $\Gamma_{A},$ and the circle which touches to the sides $AB, \ BC$ and touches to $\Gamma$ internally at $Q$ by $\Gamma_{B},$ and the circle which touches to the sides $AC, \ BC$ and touches to $\Gamma$ internally at $R$ by $\Gamma_{C}.$ Prove that the lines $AP,\ BQ,\ CR$ are coccurent.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=136048]Problem 4
[/url]
On a plane, call the band with width $d$ be the set of all points which are away the distance of less than or equal to $\frac{d}{2}$ from a line.Given four points $A,\ B,\ C,\ D$ on a plane.If you take three points among them, there exists the band with width $1$ containing them. Prove that there exist the band with width $\sqrt{2}$ containing all four points.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=136049]Problem 5
[/url]
For real positive numbers $x,$ the set $A(x)$ is defined by
\[A(x)=\{[nx]\ | \ n\in{ \mathbb{N}}\}, \]
where $[r]$ denotes the greatest integer not exceeding real numbers $r.$Find all irrational numbers $\alpha >1$ satisfying the following condition.
Condition: If  positive real number $\beta$ satisfies $A(\alpha)\supset A(\beta),$ then $\frac{\beta}{\alpha}$ is integer.

Working Time: 4 hours.

Thank you very much  :)  This is version pdf  :wink:


function
trigonometry
calculus
calculus computations

Study the uniformly continuity of
\[ \sqrt [3]{{\sin \sqrt x }}
\]
on $ [0, \plus{} \infty)$.

I hope that some one can solve.

The composition of uniformly continuous functions is uniformly continuous.


In the AMC 10, some of the possible scores from 0 to 150 can be obtained in only one way. Some can be obtained only in two ways. There are three scores that can be obtained in exactly three ways. What is their sum?

(A) 175       (B) 179.5       (C) 182       (D) 188.5       (E) 201

6 points for every correct answer, 2.5 for every blank, and 0 for wrong answers I think.

Yes, that is. This was number 25 on a amc 10 test. I'm going to start trying numbers  :)

The answer is (D) 188.5. I'll show the solution after, but the three numbers are 60, 62.5, 66.

I know the answer. I know the numbers. I need a solution. ;)


geometry factorial counting distinguishability

Here are 2 problems i'm havin trouble with-

1) ABC is an acute-angled triangle with ∠A = 30degrees. H is the orthocenter and M is the midpoint of BC. T is a point on HM such that HM = MT. Show that AT = 2 BC.

2)There are 2 clocks.everyday,one clock increases its time by 2 minutes and the other decreaeses by 3 minutes.Find for an instant,when will both clocks show the correct time?
options-within-a)amonth,b)ayear,c)aweek,d)6months

For number one, how can HM=MT if T is on HM?

For number two, does AM and PM count? (I'm getting a weird answer anyway).

that's what i was thinking too. by the way, i don't think that AM and PM count on the clock. i worked out the problem and i couldn't get any of your answer choices for number 2. the answer i got was 360 days. does that count as a year?

The first clock is 2 minutes off everyday, meaning in 360 days, it will be correct. THe second clock is 3 mintues off everyday, meaning in 240 days it will be correct. That means every 720 days, they'll be correct. What's that supposed to be?  :?

236!'s answer seems correct. maybe the answer choices were typed wrong.

Factorial,the question says that one clock has its time increased every day,not decreased.(sorry if im interpretin ur post wrong)  :D

it doesn't matter if it's 360 or 720, it's not one of the choices!

yea, but, regardless of whether the clock increased or decreased, it would still display the same time after those days because AM and PM are indistinguishable on a clock.

hEY fACTORIAL OR H78REG,COULD YA TELL ME how u got 360 days and 240 days for the clock.i couldnt relly get u.
even if the ans is 360 days,the options are not exact,it says "within"

360 days is incorrect, 720 is the right answer. each clock will be correct after it's 12 hours off, or 720 minutes. the earliest time possible when both clocks are right is within 2 years, unless we misinterpreted something.

How did u get 720?please explain!!!
by the way,i hope u guys interpreted 1 day as 24 hrs,not 12.
h78reg,how will the clocks be right after 12 hours?

on a clock, the same time is displayed every 12 hours because clocks aren't 24 hours. as 236! said, one clock is 12 hours off after 360 days and the other is 12 hours off after 240 days. if a clock is 12 hours off, it displays the correct time, not taking into consideration AM and PM. therefore, the lowest common multiple of 360 and 240 is 720 days.