I seem to remember there being a discussion about the possibility of this, but I can't find it anymore...

Anyway, will there be one this year?

I second this request/inquiry/demented scream of pain (ok, the last part probably is unnecessary (hey, I spelled it right! (can I use parenthesis like this (oh well)))).

I move that we talk to Mr. Fulton about it, as I'm overmuchly busy to try to deal with it myself right now.

Jerry,

Your signature is outdated.

Please rectify the situation.

Thank you.

I'll be happy to host one. The bad part of that is that it would be in Cobb County (not the middle of town) and I'm not sure how much I can help answering the problems. But I can dig up some old problems. And I should have a copy of Jack's "Field Guide to the AIME".

Let's talk about it at the Northview tournament.

Was the talking at the Northview Tournament (a.k.a. the tournament I didn't go to because I had an SAT subject test to do) fruitful? What did you decide?


Let $ABC$ be a non isosceles triangle wit h$\angle A = 90$. The incircle of $ABC$ touches $AB$ and $AC$ at $P$ and $Q$ respectively. Line $PQ$ intersects line $BC$ in $M$. The line tangent to the circumcircle of triangle $ABC$ at point $A$, intersects $BC$ in $N$ and $PQ$ in $U$. Prove that $\triangle MUN$ is isosceles.

Haven't tried it, they say it's the easiest geometry problem ever.

Daniel

Let $S$ and $R$ be the points of intersection between line $PQ$ and the circumcircle.
Then, according to [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=113059[/url], $BI$ and $PQ$ concur in $S$, and $CI$ and $PQ$ concur in $R$.

Then $\measuredangle MBS = 90^{\circ}+\measuredangle ACB+\frac{\measuredangle ABC+\measuredangle ACB}{2}$.

Hence $\measuredangle SMB = 45^{\circ}-\measuredangle ACB = \measuredangle AUQ = \measuredangle MUN$.

[quote="darij grinberg"][quote="dhernandez"]Haven't tried it they say it's the easiest geometry problem ever.[/quote]

I agree. But why do you want A to be 90° ?

  darij[/quote]

Hey! I don't want anything :) It's just how the problem went in the IbMO :)

Daniel

[quote="darij grinberg"][quote="dhernandez"]Haven't tried it they say it's the easiest geometry problem ever.[/quote]

I agree. But why do you want A to be 90° ?

  darij[/quote]I agree that A doesn't need to be 90°.


geometry
IMO
perimeter
inradius
area of a triangle
Heron\u0027s formula
IMO 1964

A circle is inscribed in a triangle $ABC$ with sides $a,b,c$. Tangents to the circle parallel to the sides of the triangle are contructe. Each of these tangents cuts off a triagnle from $\triangle ABC$. In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a,b,c$).

Let $(I,r)$ the inscribed circle of $\triangle ABC$

The tangent parallel to the side $BC$ intersects the sides $AB,AC$ at the points $B_1,C_1$, forming in this way a triangle $T_1=\triangle AB_1C_1$.

Since $B_1C_1 \parallel BC$, we have $\triangle AB_1C_1 \sim \triangle ABC$, with ratio of similarity $m_1=\frac{B_1C_1}{BC}$

Similarly, the other formed triangles $T_2=\triangle A_2BC_2, \ \ T_3 = \triangle A_3B_3C$ are also similar to $\triangle ABC$, with ratio $m_2$ and $m_3$ respectively.

The distance of the parallel lines $BC,B_1C_1$ is $2r$

Let $AD$ the altitude of $\triangle ABC$, and $AD_1$ the altitude of $\triangle AB_1C_1$
We bring a parallel line $L$ from $A$ to $BC$.

From the parallel lines $L, B_1C_1,BC$ we get:

$\frac{AD_1}{AD} = \frac{AB_1}{AB} = m_1 \Rightarrow$

$AD_1 = m_1 AD$

But $DD_1$ is the diameter of $(I,r)$, so $DD_1=2r\Rightarrow AD=AD_1+2r$

$AD = m_1 AD + 2r \Rightarrow$

$(1-m_1)AD = 2r \Rightarrow$

$\frac{AD}{r} = \frac{2}{1-m_1} \ \ \ (1)$

Let $s = \frac{a+b+c}{2}$, the semi-perimeter of $\triangle ABC$

The area of $ABC$ can be expressed in two different ways:

$(ABC) = \frac{AD \cdot a}{2} = s \cdot r \Rightarrow$
$\frac{AD}{r} = \frac{2s}{a} \ \ \ (2)$

$(1),(2) \Rightarrow$

$\frac{2}{1-m_1} = \frac{2s}{a}$

So we find that $m_1 = \frac{s-a}{s}$, and similarly $m_2 = \frac{s-b}{s}, m_3 = \frac{s-c}{s}$

The area of $(I,r)$ is $\pi \cdot r^2$.

Each of the triangles $T_1,T_2,T_3$ is similar to the triangle $ABC$, so the incircle of $T_k$ is similar to $(I,r)$ with the same ratio of similarity $m_k$. So we have:

Inradius of $T_1 = m_1 \cdot r$
Inradius of $T_2 = m_2 \cdot r$
Inradius of $T_3 = m_3 \cdot r$

Hence, the areas of their inscribed circles are $\pi {m_1}^2 r^2,\ \ \pi {m_2}^2 r^2,\ \ \pi {m_3}^2 r^2$

So we want the sum $S = \pi \cdot r^2 \cdot (1+{m_1}^2+{m_2}^2+{m_3}^2)$

The problem is already solved, because $m_1,m_2,m_3$ are functions of $a,b,c$.

But I'll try to find another expression, not containing $m_1,m_2,m_3$

${m_1}^2 = (\frac{s-a}{s})^2 = (1-\frac{a}{s})^2 = 1-2\frac{a}{s}+\frac{a^2}{s^2}$

${m_2}^2 = (\frac{s-b}{s})^2 = (1-\frac{b}{s})^2 = 1-2\frac{b}{s}+\frac{b^2}{s^2}$

${m_3}^2 = (\frac{s-c}{s})^2 = (1-\frac{c}{s})^2 = 1-2\frac{c}{s}+\frac{c^2}{s^2}$

$\Rightarrow$

${m_1}^2 +{m_2}^2 +{m_3}^2 = 3 -2\cdot \frac{a+b+c}{s} +\frac{a^2+b^2+c^2}{s^2}$

${m_1}^2 +{m_2}^2 +{m_3}^2 = 3 -2\cdot 2 +\frac{a^2+b^2+c^2}{s^2} = -1 +\frac{a^2+b^2+c^2}{s^2}$

Hence,

$1+{m_1}^2 +{m_2}^2 +{m_3}^2 = \frac{a^2+b^2+c^2}{s^2}$

Finally,

$S = \pi \cdot r^2 \cdot \frac{a^2+b^2+c^2}{s^2}$

Obviously the three smaller triangles are similar to triangle $ABC$.

We employ barycentric coordinates with reference triangle $ABC$. Since $I=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)$, the ratios of the side lengths to the side lengths of $ABC$ are
\[\frac{a+b-c}{a+b+c},\frac{a-b+c}{a+b+c},\frac{-a+b+c}{a+b+c},\frac{a+b+c}{a+b+c}.\]
Summing the squares, the ratio of the final area to the area of the inscribed circle of triangle $ABC$ is just
\[\frac{4(a^2+b^2+c^2)}{(a+b+c)^2}.\]
So our answer is just
\[\frac{\pi(a^2+b^2+c^2)(a+b-c)(a-b+c)(-a+b+c)}{(a+b+c)^3}.\]


geometry
3D geometry
rectangle
perimeter
arithmetic sequence
algebra
factorization

Sprint State 2004
The terms x, x+2, x+4,...x+2n form an arithmetic sequence with x an integer. If each term in the sequence is cubed, the sum of the cubes is -1197. What is the value of n if n>3?

A nonsquare rectangle has integer dimensions. If the perimeter equals the area then what is the area? How do you solve [b]without[/b] guess and check.

The second one:

[hide]xy=2x+2y

y=2+2y/x

1=2/x+2/y

1/2=1/x+1/y

1/2=1/3+1/6

the dimensions are 3 and 6[/hide]

Or...

[hide]
\begin{eqnarray*}
xy &=& 2x+2y\\
0 &=& xy-2x-2y\\
4 &=& xy-2x-2y+4\\
4 &=& (x-2)(y-2)\\
\end{eqnarray*}
Because it's non-square, $x \neq y$, so the only pair of factors left for $4$ is $4 \cdot 1$, so $x-2=4 \rightarrow x=6$, $y-2=1 \rightarrow y=3$. $\textbf {3 x 6}$ are the dimensions, as sirorange said.
[/hide]

[quote="pgpatel"]The terms x, x+2, x+4,...x+2n form an arithmetic sequence with x an integer. If each term in the sequence is cubed, the sum of the cubes is -1197. What is the value of n if n>3?[/quote]

I remember this! :roll:
[hide]
Because the sum of cubes is negative, $x$ is negative, and there are more negative than  positive terms. How can we get $1197$ as a sum of cubes? $5^3+7^3+9^3=1197$. Hey...that was pretty lucky! :D I'm serious!! I just typed some random amount of numbers, and voila! ... Well, normally you probably wouldn't be that lucky...

Anyways, We can have $(-9)^3+(-7)^3+(-5)^3=-1197$, but then $n=3$. So we need to somehow add numbers, but not change the value. Hmmm...how about adding positives and negatives together? $(-9)^3+(-7)^3+(-5)^3+[(-3)^3+(-1)^3+1^3+3^3]=-1197$, so to find $n$, $-9+2n=3 \rightarrow n=\fbox {6}.$
[/hide]


what means n=2,3 of natural numbers? u meant n\in N,n\geq 2??
is the third condition good?

I edited my post. The third condition is correct.

[quote="orl"]A sequence $\{a_{n}\}$ with n = 2,3,... of positive integers satisfy the following conditions:
(1) For each natural number $m \geq 1$: $a_{2^{m}}= \frac{1}{m}$.
[/quote]
A sequence of positive integers? $\frac{1}{m}$???

[quote=orl]A sequence $\{a_{n}\}$ with n = 2,3,... of positive integers satisfy the following conditions:

(1) For each natural number $m \geq 1$: $a_{2^{m}}= \frac{1}{m}$.

(2) For each natural number $n \geq 2$: $a_{2n-1}a_{2n}= a_{n}$.

(3) For all natural numbers $n,m$ with $2^{m}> n \geq 1$: $a_{2n}a_{2n+1}= a_{2^{m}+n}$.

Determine $a_{2000}$. Without proof you can assume that by (1), (2) and (3) a sequence $\{a_{n}\}$ with $n = 2,3$ of positive integers is defined.

[i]Edited by Myth[/i][/quote]

Lemma: For $n\geq0$ we have $\prod_{k=2^n+1}^{2^{n+1}}a_k=1$.

Proof by induction. By (1) we have $a_2=1$. So the claim holds for $n=0$.

Assume the claim holds for $n$. By condition (2) we have $1=\prod_{k=2^n+1}^{2^{n+1}}a_k=\prod_{k=2^n+1}^{2^{n+1}}a_{2k-1}a_{2k}=\prod_{k=2^{n+1}+1}^{2^{n+2}}a_k$. So the claim holds for $n+1$.

By using condition (3) a lot, condition (1) twice and the lemma at the end we get
\begin{align*}
a_{2000}=a_{1952}a_{1953}=a_{1856}a_{1857}a_{1858}a_{1859}=\prod_{j=1664}^{1671}a_j=\prod_{j=1280}^{1295}a_j=\prod_{j=512}^{543}a_j=\frac{1}{9}\prod_{j=513}^{543}a_j=\frac{1}{9}\prod_{j=2}^{63}a_j=\frac{1}{9a_{64}}=\frac{2}{3}
\end{align*}


calculus
integration
real analysis
real analysis unsolved

let $ (f_{n})$ and $ f$ are integrable
prove that if $ f_{n}\rightarrow f$  a.e  and ${ \int|f_{n}|\rightarrow \int |f|}$  then $ \int f_{n}\rightarrow \int f$

thanks for your help

Apply Fatou's lemma to both $ |f_n| \minus{} f_n$ and $ |f_n| \plus{} f_n$.

SORRY, it should be

let $ (f_{n})$ and $ f$ are integrable
prove that if $ f_{n}\rightarrow f$  a.e  and ${ \int|f_{n}|\rightarrow \int |f|}$  then $ \int |f_{n}- f|\rightarrow 0$

thanks for your help

Oh, then apply Fatou's lemma to $ |f| \plus{} |f_n| \minus{} |f \minus{} f_n|$.


Let $b_n=3a_n-1$. It's easy to show tht the recurrence relation for $a_n$ is equivalent to $b_1=8,\ b_{n+1}=\frac{b_n^2}2$, so $b_n=2^{x_n}$ s.t. $x_1=3,\ x_{n+1}=2x_n-1$. This means that $x_n=2^n+1$, so $b_n=2^{2^n+1}$.

The conclusion is equivalent to $3^{n+1}|b_{3^n}+1\Leftrightarrow 3^{n+1}|2^{2^{3^n}+1}+1$. I didn't show this yet, but I don't think it's that hard.

It's actually not hard if we know that $3^{n+1}|2^{3^n}+1$. This was even proved on the forum a long time ago: I asked for a proof that $2$ is a primitive root modulo $3^n,\ \forall n\in N*$ and amfulger proved it. I'll use this without proof.

$3^n|2^{3^n}+1\Rightarrow 2^{3^n}+1=(2k+1)3^n\Rightarrow 3^{n+1}|2^{2^{3^n}+1}+1$, Q.E.D.


Is there anyway to find how many points each question is worth on past HMMT rounds?

Hm, much cooler would be 1223334444 for a total of 30, instead of what they do for 50.

Give too much weight to the backend, then either you can't really reward people for solving a really hard #10 or you're stuck making a bunch of medium diff questions (since four have same value) and you dont get really hard ones in


How much is the difference in the sale price if a 20% off coupon is used first and then a 25% off coupon is used versus using the 25% off coupon first and then the 20% off coupon?

Let's just say that the item was originally 100 dollars.

Case 1: 25 is used, then 20:

$ 100\times .75\times .80\equal{}60$

Case 2: 20 is used, the n25:

$ 100\times .80\times .75\equal{}60$

Both are equivalent to 40%. This is because of the [url=http://demonstrations.wolfram.com/TheCommutativePropertyOfMultiplication/]commutative property of multiplication.[/url]


Take the digits $ 1,2,3,4,5,6,7,8,9,0$. Use them all, and each one only once, to form a $ 10$ digit number such that it is the smallest such number divisible by all numbers from $ 1$ to $ 12$ inclusive.

Is the answer [hide]$ 1234759680$[/hide]?


Let x,y,z>0 such that x+y+z=1 prove that:
$ \frac{\sqrt{xy}}{\sqrt{xy\plus{}z}}\plus{}\frac{\sqrt{yz}}{\sqrt{yz\plus{}x}}\plus{}\frac{\sqrt{zx}}{\sqrt{zx\plus{}y}}\le\frac{3}{2}$


Got 177.. expected 190..
physics got 26 and 43
chem got 28 and 8
maths got 31 n 41..
individual percentiles phy got 98.1 n 98, total 98
maths got 99.1 n 99.7.. overall 99.4
chem got 85.6 n 62.5!! n overall some 82.somethin..
total percentile 98.9.......
Hope u ppl post urs too.. :D

157
97.7 overall

and as told before most contribution from physics 104 (99.9)
rest not meant to be remembered :P

math 81
physics 82
chem 81
percentile99.8

90 Percentile... :( Less than expected in everything so nuff to say.

Anyways Tarun you are pretty close to All India Topper 279.

92.2 percentile  :(
Didn't see it myself friend told me
idiotic server jammed  :rotfl:

Dont mind this post.
Srini if you have new private messages the private message icon will be highlighted .
Check often for pm i will be sending you.
I have sent one now. pl. reply

got 95.3 percentile and 135 in total chappli only told me same problem here da rohit server jammed :rotfl: does anyone here know how to rectify the problem??? :blush:

Incase you didn't noticee.

Highest Score: Physics Paper II - 77
Chem paper II 78
Maths Paper II - 81 (which is a full score)>

How the hell are people able to score this much???? Seriously it baffles me cause getting all 3 matches right is really difficult.

Plus this time a lot of solns in FIITJEE's soln sheet were wrong.


Outline a synthesis of $ \alpha\minus{}terpineol$ (look structure [url=http://en.wikipedia.org/wiki/Terpineol]here[/url]) from isoprene.

take one equivalent of isoprene and add HCl. The carbocation will be formed most stably at the tertiary center, so we should then have 2-chloro,2-methylbut-3-ene. React this with another equivalent of isoprene in a Diels-Alder reaction. Now substitute the chlorine with the hydroxyl group (S_N1).

That synthesis would require some harsh conditions, because the dienophile is not activated. Some high temperature would be required, and would the halide be stable? And from the addition of HCl to isoprene we get an allylic cation, don't forget, so more than one chloride can form.

Actually, you are allowed to use other compounds, if you like.

Do diels alder reaction with methyl vinyl ketone. :maybe:

chemrock has done the first part, so i only have to complete the synthesis :P
2. treat with ch3mgbr and water.
in the first step(the 1,4 product will predominate).

The first step of Chemrock is not completely correct. And what do you mean by "1,4 product will predominate"? What 1,4-product?

HERE BOTH THE REACTANTS ARE UN SYMMETRICAL, SO YOU CAN EXPECT TWO PRODUCTS
FROM WIKI
"In unsymmetrically substituted diene and dienophile, pseudo-ortho and para orientations in products are usually favored over meta orientation. A particular preference in location of substitutents in the product can, in some cases, be explained in terms of frontier orbital theory.The strongest interaction takes place between HOMO of diene and LUMO of dienophile. Carbons that have the highest coefficients in two frontier orbitals will begin to bond; therefore these carbons will direct the orientation of substituents and thus identity of major product of a DA reaction."
so, here pseudo-para product will predominate. why do you say that chemrock's step is not completely correct? :maybe:


The problem seems has a mistake. I guess $ \frac{1 + a + a^2 + ...... + a^n}{a + a^2 + ...... + a^{n - 1}}\ge \frac {n + 1}{n - 1}$
When a=1,It's true.
When a><1.We can prove that $ 1 + a^n \ge a^p + a^{n - p}$
proof.

$ (m \plus{} \frac {3 \plus{} 4p}{2})^2*(m \plus{} \frac {4k \minus{} 1 \minus{} 4 p}{2})^2 > (m \plus{} p \plus{} 1)(m\plus{}p\plus{}2)(m \plus{} 2k \minus{} p )(m\plus{}2k\minus{}p\minus{}1)$

kindly explain on both the proofs.....
unable to understand :oops:

Please check the first one.I think the problem has a small mistake.When a=1,……

[quote="anurags92"]Prove that $ \frac {1 \plus{} a \plus{} a^{2} \plus{} a^{3} \plus{} \cdots \plus{} a^{n}}{a \plus{} a^{2} \plus{} a^{3} \plus{} \cdots \plus{} a^{n \minus{} 1}} \ge \frac {n \plus{} 1}{n \minus{} 1},$

$ a$ is positive, $ n\ge2.$[/quote]$ (n \minus{} 1)\sum_{k \equal{} 0}^{n}{a^k} \minus{} (n \plus{} 1)\sum_{k \equal{} 1}^{n \minus{} 1}{a^k} \equal{} (1 \minus{} a)^2\sum_{k \equal{} 1}^{n \minus{} 1}{k(n \minus{} k)a^{k \minus{} 1}}\geq 0.$
[quote="anurags92"]Prove that

$ \left(m \plus{} \frac {3}{2}\right)\left(m \plus{} \frac {7}{2}\right)\cdots\left(m \plus{} \frac {4k \minus{} 1}{2}\right) > \sqrt {{\frac {(m \plus{} 2k)!}{m!}}}.$[/quote]$ \left(m \plus{} \frac {3}{2}\right)\left(m \plus{} \frac {7}{2}\right)\cdots\left(m \plus{} \frac {4k \minus{} 1}{2}\right) < \sqrt {{\frac {(m \plus{} 2k)!}{m!}}\left(1 \plus{} \frac {k}{4m^2}\right)}.$

for the second one we have (m+3/2)>√(m+1)(m+2)
because (m+1+m+2)/2>√(m+1)(m+2)....
we do the same with the oder factors. :) [/code]

[quote="Ji Chen"][quote="anurags92"]Prove that
$ \left(m \plus{} \frac {3}{2}\right)\left(m \plus{} \frac {7}{2}\right)\cdots\left(m \plus{} \frac {4k \minus{} 1}{2}\right) > \sqrt {{\frac {(m \plus{} 2k)!}{m!}}}.$[/quote]$ \left(m \plus{} \frac {3}{2}\right)\left(m \plus{} \frac {7}{2}\right)\cdots\left(m \plus{} \frac {4k \minus{} 1}{2}\right) < \sqrt {{\frac {(m \plus{} 2k)!}{m!}}\left(1 \plus{} \frac {k}{4m^2}\right)}.$[/quote]
$ \iff\prod_{i\equal{}0}^{k\minus{}1}\frac{(m\plus{}2i\plus{}1)\plus{}(m\plus{}2i\plus{}2)}{2\sqrt{(m\plus{}2i\plus{}1)(m\plus{}2i\plus{}2)}} \leq \sqrt{1\plus{}\frac{k}{4m^2}}$
just need to prove $ \frac{(m\plus{}2i\plus{}1)\plus{}(m\plus{}2i\plus{}2)}{2\sqrt{(m\plus{}2i\plus{}1)(m\plus{}2i\plus{}2)}} \leq \frac{\sqrt{4m^2\plus{}i\plus{}1}}{\sqrt{4m^2\plus{}i}}$
$ \iff\frac{((m\plus{}2i\plus{}1)\plus{}(m\plus{}2i\plus{}2))^2}{4(m\plus{}2i\plus{}1)(m\plus{}2i\plus{}2)} \leq \frac{4m^2\plus{}i\plus{}1}{4m^2\plus{}i}$
$ \iff\frac{1}{4(m\plus{}2i\plus{}1)(m\plus{}2i\plus{}2)} \leq \frac{1}{4m^2\plus{}i}$
that is obvious


geometry rectangle quadratics LaTeX inequalities

A question combined with geometry and algebra =P.

You're given a wire with length of 10m,
If you fold it into a rectangle,what is the maximum area you can get?

Determine the length of each side and the area covered by the wire. :)

Call a side of the rectangle $ s$. Thus we are looking for the maximum value of:
$ s\frac {10\minus{}2s}{2} 0 < s <5$.
Thus the area is $ 5s\minus{}s^2$. This is a quadratic with maximum point $ \frac {\minus{}5}{2(\minus{}1)} \equal{} \frac {5}{2}$
Thus, $ s \equal{} \frac{5}{2}$, $ 5\minus{}s \equal{} \frac{5}{2}$, area is $ {\frac {5}{2}}^2 \equal{} \frac{25}{4}$. Note that it is a square, this is true for all similar problems.

use the fact:
$ \sqrt[]{ab}\leq\((a+b)/2$
the arithmetic mean equals the geometric mean only when $ a=b$

[hide]$ 2x \plus{} 2y \equal{} 10\implies x \plus{} y \equal{} 5$

By AM-GM:
$ xy\leq\left(\frac {x \plus{} y}{2}\right)^2 \equal{} \frac {5^2}{2^2} \equal{} \boxed{\frac {25}{4}}$.

To be more correct, we should show that it's attainable, but I'm laaaazy.[/hide]

Edit: Thanks ThinkFlow for pointing out that little LaTeX mistake.

grn_trtle, you flipped an inequality sign. It should be
\[ xy \le \left( \frac{x \plus{} y}{2} \right)^2.
\]


Let $ 0\leq a,b,c \leq3$ and $ a\leq b \leq c$.Prove that:
$ (a\minus{}b)(c^2\minus{}9)\plus{}(b\minus{}c)(a^2\minus{}9)\plus{}(a\minus{}c)(b^2\minus{}9)\leq36$

[quote="Inequalities Master"]Let $ 0\leq a,b,c \leq3$ and $ a\leq b \leq c$.Prove that:
$ (a \minus{} b)(c^2 \minus{} 9) \plus{} (b \minus{} c)(a^2 \minus{} 9) \plus{} (a \minus{} c)(b^2 \minus{} 9)\leq36$[/quote]
We don't need $ 0 \le a,b,c \le 3$ here. Since
\[ \sum (a \minus{} b)(c^2 \minus{} 9) \equal{} \sum c^2(a \minus{} b) \minus{} 9\sum (a \minus{} b) \equal{} \sum c^2(a \minus{} b) \equal{}\minus{} (a \minus{} b)(b \minus{} c)(c \minus{} a) \le 0
\]
:)


solve \[ \lim\limits_{x\rightarrow 0}\frac{\sqrt{2(1-2x)}\sqrt{3(1-3x)}\ldots \sqrt{n(1-nx)}}{\sin{x}}\]

Something fishy here... For $ x\equal{}0$ the numerator is $ \sqrt{n!}$, while for $ x \to 0$ the denominator goes to zero. Why all this trouble to have a $ \pm \infty$ limit? (moreover, it depends from which side of $ 0$ comes $ x$; as such the limit does not exist, just lateral limits do).


geometry
circumcircle
rectangle
geometry proposed

Given a line d passs through the point M, the point A doesn't lie on d. Two mobile points B,C move on line d such that M is the midpoint of BC. Let D,E on AM such that AD=BD, AE=CE. BD cuts CE at F. Find the locus of F.

The midpoint of $ AB$ is $ L$, the midpoint of $ AC$ is $ N$ and $ AX\bot BC$

  The circumcircles of the triangles $ LBX$ and $ NCX$ intersect again at
the point $ F'$.  $ \{Y\}=F'X\cap LN$.

$ LN$ is the common tangent of the circumcircles of the triangles $ LBX$
and $ NCX$ $ \Rightarrow YL^{2}=YX\cdot YF'=YN^{2}\Rightarrow YL=YN$
$ \Rightarrow Y$ is the midpoint of $ LN\Rightarrow Y$ is the midpoint of
$ AM\Rightarrow XY$ is the median in the rectangle triangle $ AXM$.

$ \{D'\}=BF'\cap AM$ and  $ \{E'\}=CF'\cap AM$.

$ \angle D'AB=\angle LAX+\angle YAX=\angle LXA+\angle YXA=\angle LXY=
\angle F'BL=$
$ =\angle D'BA\Rightarrow\angle D'AB=\angle D'BA\Rightarrow$

$ \Rightarrow$ triangle $ D'AB$ is isosceles.

$ \angle E'AC=\angle NAX-\angle YAX=\angle NXA-\angle YXA=\angle NXY=
\angle F'CN=$
$ =\angle E'CA\Rightarrow\angle E'AC=\angle E'CA\Rightarrow$

$ \Rightarrow$ triangle $ E'AC$ is isosceles.

So  $ D'=D$ and $ E'=E$ $ \Rightarrow F'=F$ $ \Rightarrow F\in XY$.

The locus of $ F$ is $ (YX$.      :)


Hey. Maybe a Math Jam after the competition would be nice. To talk about the problems, talk about the solutions, the results, people's opinions, and the overall success of the competition.

I'm in favor of that, if there's enough interest. I should check with my fellow officers, though -- the Princeton Math Club is not a dictatorship, and even if it were, I'm not the dictator. ;)
What day and time would be good? If you're interested in this math jam, please indicate a good time for you that would be shortly after the contest (say, during the winter holidays). I will be traveling, but I am sure I can fit in a math jam. :D

Well, any day that is not Christmas, Christmas Eve or the other holidays.

Prefer evenings like the usual math jams 8pm to 9pm sounds cool. maybe the 30th of december? anytime rly. i can give up a party or two.

I'm up for a dose of math over Christmas.  I can't see the other officers caring very much one way or the other, so I'd say if there's interest, it should be fine.


linear algebra
matrix
quadratics
geometry
trigonometry
AMC
USA(J)MO

I thought it might be interesting to see some proofs you have done or seen that you found interesting. They could be from easy to ornate and vary in subject. Basically anything that you thought was interesting. Thanks.

Hardly original, but I think the proof of the determinant of the Vandermonde matrix is a pretty cool one.  Unfortunately the syntax is too annoying for me to post.

Also hardly original, the proof that $ \zeta(2)\equal{}\frac{\pi^{2}}{6}$ is quite nice.

[url=http://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity]Start here[/url].  :)

[hide="Problem"]What is the formula for the area of a regular polygon?[/hide][hide="Solution"]When a problem talks about the area of a polygon, it usually gives the number of sides. Let's call that $ n$. Also, we need the side length $ s$ to determine the area as well. Let the apothem be $ h$.

We can split the n-gon into n isosceles triangles with vertex angle $ \dfrac{360}{n}$, and base of length $ s$. We split it into two congruent right triangles of angles $ \dfrac{180}{n}$, $ 90$, and $ 90 \minus{} \dfrac{180}{n}$, and legs of length $ \dfrac{s}{2}$ and $ h$. Thus,
\[ \tan{\dfrac{180}{n}} \equal{} \dfrac{s}{2h}
\]

\[ h \equal{} \dfrac{s}{2\tan{\dfrac{180}{n}}}
\]
And the area of the right triangle is $ \dfrac{\dfrac{s}{2}\cdot h}{2} \equal{} \dfrac{s^2}{8\tan{\dfrac{180}{n}}}$. There are two of those triangles in each isosceles triangle, and $ n$ isosceles triangles. Thus,
\[ Area \equal{} \boxed{\dfrac{ns^2}{4\tan{\dfrac{180}{n}}}}.
\]
[/hide]

A proof of Bramagupta's formula (posted it sometime on aops). Then let $ d\equal{}0$ and you get heron's.

[quote]anything on the USAMO i cant prove, im just going to invent a formula for it and say because of this so and so such formula, this must be true, so we are done with the proof now.[/quote]
While this would be nice anytime we came to a problem we couldn't do, you would have to proof that your formula or method holds and is supported by logic.


inequalities
induction
inequalities unsolved

Prove that $(n-1)\cdot \left( \frac{1}{1-x_1}+\frac{1}{1-x_2}+ \ldots +\frac{1}{1-x_n} \right) \geq (n+1)\cdot \left( \frac{1}{1+x_1}+\frac{1}{1+x_2}+ \ldots +\frac{1}{1+x_n} \right)$,
where $x_1,x_2,\ldots,x_n \in (0,1)$ and have the sum $1$.

yptsoi dear , sorry ,but i don't understand you . Where is your solution ?

[quote="perfect_radio"][quote="Michal Marcinkowski"]
Let  $ f(x) = \frac{1-nx}{1-x^2} , x \in (0,1)$ - it is concave function. Jensen's inequality gives:
[/quote]
why is it so ?[/quote]
It is untrue. ;)

this means that the solution ain't good?

still, how can it be solved?

Ups... sorry, I've made a mistake  :(.  Second wrong solution, mayby third will be proper... :)

Michal

I think $(\frac{1-x_1}{n-1},...,\frac{1-x_n}{n-1})$ majorizes $(\frac{1+x_1}{n+1},...,\frac{1+x_n}{n+1})$
The sum of these two sequences are obviously equal.
For $1 \le i \le n-1$, we will prove that $\frac{1-x_1}{n-1} +...+\frac{1-x_i}{n-1} \ge \frac{1+x_1}{n+1} +...+ \frac{1+x_i}{n+1}$ by induction.
WLOG, assume $x_1 \le x_2 \le ... \le x_n$, then
i=1 case is equivalent to $x_1 \le \frac{1}{n}$, which is true.
Assume when i=k it's true, then we have $\frac{1-x_1}{n-1} +...+\frac{1-x_k}{n-1} \ge \frac{1+x_1}{n+1} +...+ \frac{1+x_k}{n+1}$, or equivalently $x_1+...+x_k \le \frac{k}{n}$, then consider,
\[ (x_1+...+x_k+x_{k+1}) + (x_1+...+x_k+x_{k+2}) + ... + (x_1+...+x_k+x_{n}) \]
The sum $=(n-k-1)(x_1+...+x_k) + 1 \le \frac{k(n-k-1)}{n} + 1 = \frac{(k+1)(n-k)}{n}$. Since $x_1+...+x_k+x_{k+1} \le x_1+...+x_k+x_{k+2} \le ... \le x_1+...+x_k+x_{n}$,
\[ x_1+...+x_k+x_{k+1} \le \frac{(k+1)(n-k)}{n(n-k)} = \frac{k+1}{n} \]
This completes the induction, as $ f(x) = \frac{1}{x}$ is convex, we hv the result by majorization inequality.

after seeing Armo's solution, i didn't enter this thread.

now i see there was another solution posted, but:

[quote="siuhochung"]no it's wrong.
I just realized my mistake.
However, it's already 3:14 ($\pi$) am here.
so i'd better sleep and work on it again tomorrow...[/quote]

i'm very sleepy right now and i can't see the mistake. could you please show me where is it?

(the main reason i'm posting now is that i'll forget about this problem til tomorow)

my numbers in the sequence is not ordered, so my majorization does not work.

[quote="cezar lupu"]Guys,this problem is  very easy.It follows from Chebyshef's inequality. ;)[/quote]

could develop your idea a bit, because i can't see how we can use chebyshev here

Just apply Chebyshev for the sequences $x_i$ and $\frac{2}{1-{x_{i}}^2} $.The conclusion follows imediately. ;)


[hide="Solution"] We seek an integer $ n$ such that $ n \equiv n^3 \bmod 100 \Leftrightarrow n(n \minus{} 1)(n \plus{} 1) \equiv 0 \bmod 100$.  We will solve $ \bmod 4, \bmod 25$ separately.

$ \bmod 4$ the LHS is $ 0$ if and only if $ n \equiv 0, 1, 3 \bmod 4$.

$ \bmod 25$, note that $ n \minus{} 1, n, n \plus{} 1$ are consecutive and therefore incongruous $ \bmod 5$, which means that the LHS is $ 0$ if and only if $ n \equiv 0, 1, 24 \bmod 25$.

By Chinese Remainder theorem, we conclude $ n \equiv 0, 76, 24, 25, 1, 49, 75, 51, 99 \bmod 100$.  (These are the [b]only[/b] numbers between $ 0$ and $ 99$ that work.)  The smallest two-digit solution is $ \boxed{24}$.  Indeed, $ 24^3 \equal{} 12824$. [/hide]

Yep, 24 was the answer.
Thanks,t0rajir0u

[quote="xpmath"]Find the smallest 2 digit number such that the ten's and unit's digit of the cube of the number are the same as the original number.
For example, the largest one is 99, since 970299, and the last 2 digits match the original number.

We did this in the Mathcounts PS, by trial and error, but is there a faster way?[/quote]
actually, the correct answer is -99. This problem was in a national mathcounts team question.

I think the original one said positive numbers.


A set S of positive integers less than or equal to 2004 satisfies the condition that no number in S is double in value of another number in S.What is the largest possible number of integers in S ?
 :?

Hey thanks --By the way , no straightforward formula,  right ??

I'm pretty sure this might come out as some complicated function, but I wouldn't suggest memorizing it anyways. It's just not helpful when you actually need to apply the math behind the formula.

My way doesn't take that long to do and is easy to memorize...

The logic behind this sum is very simple actually....you dont exactly need to remember loads of things.... :)

[quote="NoSoupForYou"]A number can be $S$:

1) If it a multiple of 1,
But not if it is a multiple of 2,
2) Unless it is a multiple of 4,
But not if it is a multiple of 8,
3) Unless it is a multiple of 16,
But not if it is a multiple of 32,
4) Unless it is a multiple of 64,
But not if it is a multiple of 128,
5) Unless it is a multiple of 256,
But not if it is a multiple of 512.
6) Unless it is a multiple of 1024,
But not if it is a multiple of 2048.

$1002$ integers fit the first criteria, $250$ fit the second, $62$ fit the third, $15$ fit the fourth, $3$ fit the fifth, and $1$ fits the sixths.

Thus there can, at most, be $\boxed{1333}$ integers in set $S$.[/quote]

I don't quite understand.  How can a number be a multiple of 4, but not 2.  A multiple of 16, but not 8.  A multiple of 64, but not 32.  And so on.

Hey G-Unit --
Try understanding it this way ---
A number can be in the set if it's a multiple of 4 but not 8.
what Nosoupforyou means is here -
There can be a 1 but not a 2.
There can be a 3 but not a six....
If you observe , there's no 2 in the set --
So 4 can still be there in the set even if it's a multiple of two...It still satisfies the set's conditions.....
If you still didn't understand , well , read the topic in detail again a few times .... you should get what he means...


Consider numbers written in base b where 2<=b<=9. Clearly indicate how you can tell quickly whether the number b is odd or even. Offer justification for your claim.

For $ b$ even is similar as the criteria for even in base $ 10$,
the last digit of the number written in base $ b$ has the same parity as the number
For $ b$ odd is similar as the criteria for $ 9$ in base $ 10$,
The sum of the digits written in base $ b$ has the same parity.


pove that :
$\sum a^{2}+3 \geq \sum a+\sum ab$   where $abc=1$ and $a,b,c >0$

it is a problem of[url=http://toantuoitho.nxbgd.com.vn/Forum/forum_posts.asp?TID=2301&PN=1]this[/url](viet nam ' web) :maybe:

[quote="sunchips"]Yes! I bashed it with lagrange Multipliers![/quote]
How?
And how do you bash with Lagrange Mulitpliers in general? :maybe:

[quote="rem"][quote="pohoatza"]turkevici for $\sqrt{a}, \sqrt{b}, \sqrt{c}, 1$.[/quote]
What's turkevici? :maybe:[/quote]

turkevici says : for $a,b,c,d$ pozitive real numbers we have : $a^{4}+b^{4}+c^{4}+d^{4}+2abcd\geq a^{2}b^{2}+a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}+c^{2}d^{2}$

now make the following substitutions: $a=\sqrt{x}$, $b=\sqrt{y}$, $c=\sqrt{z}$, $d=1$, and we obtain our ineq. :wink:


advanced fields
advanced fields unsolved

What is the next letter/number combination in this sequence?

S6
A10
R17
E21
H26
R30

I know that the next number is 33, what i cant figure out is the next letter

[color=red]i thought the number was 2[/color]

I don't know which language those month names are in, but these are dates 5 weeks apart in a calendar. The next number is 4, and the letter matches August.

Since this puzzle comes from http://www.neopets.com/games/conundrum.phtml , it may not even be mathematically related at all.  Some of the puzzles there have depended less on any logical or mathematical deduction than on intimate knowledge of the neopets site.


There are h gangs who are fighting . Some of them has 5n persons . The other k-n persons. If every person in a gang hits every person in every other gang , and the total hits were knh . Find the maximum number for each gang.

A complete solution is only acceptable.

Are you sure it is algebra and not combinatorics?
What the 'If' is for? Is it given that any person from any gang fight against any person from any other gang?
What is the question? You are looking for the maximum of what? Of gangs? Of members in any gang?
Moreover, I don't think it has to be post directly into the solved problems section...

Pierre.

I did not answer the problem myself but I found it in an old book without a solution and I translated it into English and sorry for putting the problem in wrong place :D


algebra
polynomial
calculus
calculus computations

Find the third degree Maclaurin polynomial for $ y\equal{}e^{2x}$.

Use this to approximate $ e^{.2}$.

How close is your answer.  (Use a remainder thereom, don't just subtract the two).

I'm not sure what you mean here.

$ e^{x} \equal{} \sum_{n\equal{}0}^{\infty} \frac{x^{n}}{n!}$

So, $ e^{2x} \equal{} \sum_{n\equal{}0}^{\infty} \frac{(2x)^{n}}{n!} \equal{} 1 \plus{} 2x \plus{}2x^{2} \plus{} \frac{4}{3}x^{3}$ to the third degree.

Putting $ x\equal{}\minus{}1$ gives $ \frac{1}{3}$ which is a long way from $ e^{\minus{}2} \equal{} 0.13533528....$.

Have I misunderstood the question?

Sorry, my eyesight was the problem.

Substituting $ x \equal{} 0.1$ we get $ 1.2213333333333...$ instead  of $ 1.221402758...$. So it is fairly close.

For estimating that remainder, we can use Lagrange's form:

(there is a number $ \zeta \in \left(0,\frac15\right)$ such that:)

$ R_n(x) \equal{} \frac {f^{(n \plus{} 1)}(\zeta)}{(n \plus{} 1)!} (x \minus{} a)^{n \plus{} 1}$, where $ R_n(x)$ is equal to $ f(x) \minus{} T_n(x)$, where $ T_n(x)$ is the $ n$th degree Taylor polynomial around $ a$.

So now we have $ a \equal{} 0$, $ n \equal{} 3$ and get:

$ R_3\left(\frac15\right) \equal{} \frac {e^{2\zeta}}{4!}\, \left(\frac25\right)^4$.

As we have $ \zeta \in \left(0,\frac15\right)$ and $ e^{2 x}$ is increasing on this interval can estimate $ e^{\frac25}$ by $ e$ and get:

$ R_3\left(\frac15\right) \leq \frac {2\, e}{3 \cdot 5^4} \approx 0.00289950062$.


calculus
integration
trigonometry
calculus computations

* Use substitution [b]x = sin v[/b] to find the exact value of [url=http://img40.imageshack.us/my.php?image=27314496.jpg][img]http://img40.imageshack.us/img40/1835/27314496.jpg[/img][/url]

$ x \equal{} \sin v$
$ dx \equal{} \cos vdv$
$ \int^{\frac {1}{2}}_0 \frac {dx}{(1 \minus{} x^2)^{\frac {3}{2}}}$
$ \equal{} \int^{\frac {\pi}{6}}_0 \sec^2v dv$
$ \equal{} \tan \left (\frac {\pi}{6}\right) \minus{} \tan 0$
$ \equal{}\frac{\sqrt 3}{3}$

isnt it ( sec x)^3 inside the integral??

if u do not want the substitution there is an alternative way put x = 1/y and proceed


algebra polynomial function articles calculus derivative

Solve:
$ f(x) \equal{} x^3\minus{}x^2\plus{}4x\plus{}1\equal{}0(mod 125)$
$ 0\le x <125$

I have found solution $ x\equal{}94$ by brute force ( first look at mod5,then mod25,...),but is there any simpler way?
Thanks.

Rewrite the function as $ x^3 \minus{} x^2 \minus{} x \plus{} 1 \equiv 0 \mod 5$. Now it's factorable, so
$ (x \minus{} 1)^2(x \plus{} 1) \equiv 0 \mod 5$. From this $ x \equal{} {5a \plus{} 1,5b \plus{} 3, or 5c \plus{} 4}$, so what's left would be finding all these from $ 0$ to $ 125$..unless you'd accept the answer in this form then it would be $ x \equal{} {5a \plus{} 1,5b \plus{} 3, or 5c \plus{} 4}$.

$ 1,3,4,6,8....121,123,124$.

Having a solution $ \bmod 5$ doesn't guarantee that it'll continue to be a solution $ \bmod 125$ or even $ \bmod 25$.  What you need after that is [url=http://en.wikipedia.org/wiki/Hensel%27s_lemma]Hensel's lemma[/url].

well...i think that would be to complecated...don't even know what it is...wikipedia article is too technical for me :wallbash:

there is probably a way to solve it with factoring, I think

The problem with factoring in non-prime moduli is that there can be more roots than factors, which is why you use Hensel's lemma instead.  The basic idea is this:  once you know that $ x \equal{} \pm 1 \bmod 5$ (by the way, $ 3$ is definitely not a root), you know that the roots $ \bmod 25$ must be of the form $ x \equal{} 5a \pm 1$, so you substitute:

$ (5a \pm 1)^3 \minus{} (5a \pm 1)^2 \plus{} 4(5a \pm 1) \plus{} 1 \equiv 0 \bmod 25$

and when you divide by $ 5$ you'll end up with some polynomial $ \bmod 5$ that you can then find the solutions of.  Then you repeat $ \bmod 125$.  (This is where the formal derivative comes in:  if it evaluates to zero then this step fails.)


geometry
geometric transformation
rotation
function
linear algebra
matrix
trigonometry

[quote="symmetry"]What is the image of point (1, 0) after a clockwise rotation of 30 degrees?[/quote]
Image is just a fancy word meaning the output of a function. In this case the function is the clockwise rotation matrix.

$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} \cos 30^{\circ} & -\sin 30^{\circ} \\ \sin 30^{\circ} & \cos 30^{\circ} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

The image of $(1,0)$ under $30^{\circ}$ rotation is $(\cos 30^{\circ}, \sin 30^{\circ})$


function
trigonometry
limit
integration
algebra unsolved
algebra

Let $ f : [0, \plus{}\infty) \to \mathbb R$ be a differentiable function. Suppose that $ \left|f(x)\right| \le 5$ and $ f(x)f'(x) \ge \sin x$ for all $ x \ge 0$. Prove that there exists $ \lim_{x\to\plus{}\infty}f(x)$.

I am not sure I get it... if $ g(x)\equal{}f^2(x)\plus{}2\cos(x)$, then $ g$ is increasing because $ g'\equal{}2(f'(x)f(x)\minus{}\sin(x)) \ge 0$, and $ g$ is bounded because $ f$ and $ \cos(x)$ are bounded, thus $ g$ has a limit at $ \infty$. If $ f$ has a limit too, it implies that $ 2\cos(x)$ also has a limit - contradiction.

What about $ \left|f(x)\right|\le 5 $? :maybe:

What about it? [b]bambaman[/b] offers that the function $g(x) = 2\int_0^x (f(t)f'(t) - \sin t) \textrm{d} t + (f(0)^2 + 2) =$ $ (f(t)^2 + 2\cos t)_0^x + (f(0)^2 + 2)=$ $ (f(x)^2 + 2\cos x) - (f(0)^2 + 2) + (f(0)^2 + 2) =$ $ f(x)^2 + 2\cos x$ defined on $[0,\infty)$ is non-decreasing, therefore $\lim_{x\to \infty} g(x)$ does exist. Since $|g(x)| \leq |f(x)|^2 + 2|\cos x| \leq 27$, it follows that $g(x)$ is bounded, thus the above limit is finite, be it $L$. Now, if $\lim_{x\to \infty} f(x)$ were to exist, be it $\ell$, then $\lim_{x\to \infty} \cos (x) = \frac {1} {2} (L - \ell^2)$ would also exist, absurd.

[quote="mavropnevma"]What about it? [b]bambaman[/b] offers that the function $g(x) = 2\int_0^x (f(t)f'(t) - \sin t) \textrm{d} t + (f(0)^2 + 2) =$ $ (f(t)^2 + 2\cos t)_0^x + (f(0)^2 + 2)=$ $ (f(x)^2 + 2\cos x) - (f(0)^2 + 2) + (f(0)^2 + 2) =$ $ f(x)^2 + 2\cos x$ defined on $[0,\infty)$ is non-decreasing, therefore $\lim_{x\to \infty} g(x)$ does exist.[/quote]

if   $g(x)$ is non decreasing why  $\lim_{x\to \infty} g(x)$ exists?

P.S.
sorry i got my mistake

The statement $g(x)$ is non-decreasing means that for arbitrary $0\leq x_1< x_2$ we have $g(x_1)\leq g(x_2)$. Therefore $\lim_{x\to \infty} g(x) = \sup\{g(x) \mid x\geq 0\}$, which exists, either a finite real number, or else $+\infty$.


linear algebra
matrix
vector
Gauss
algebra
system of equations

Write the system of linear equations as a matrix equation Ax=b, where the matrix A (3x3) is defined by the coefficients of the linear equations, x is the vector of unknowns and b is the vector of constant terms. Use then Gauss to solve the matrix equation. The answer is 14, if I haven't made an arithmetical mistake.

it is gauss elimination
i think
[url=http://en.wikipedia.org/wiki/Gaussian_elimination]try this[/url]

Define M as a matrix representing the variables:
$ M \equal{} \left( \begin{array}{ccc} 2 & 4 & 9 \\
2 & \minus{} 10 & \minus{} 5 \\
3 & \minus{} 8 & \minus{} 18 \end{array} \right)$
Use Cramer's rule:
$ x \equal{} \left| \begin{array}{ccc} 9 & 4 & 9 \\
51 & \minus{} 10 & \minus{} 5 \\
\minus{} 46 & \minus{} 8 & \minus{} 18 \end{array} \right|/{\det(M)}$
$ y \equal{} \left| \begin{array}{ccc} 2 & 9 & 9 \\
2 & 51 & \minus{} 5 \\
3 & \minus{} 46 & \minus{} 18 \end{array} \right|/{\det(M)}$
$ z \equal{} {\left| \begin{array}{ccc} 2 & 4 & 9 \\
2 & \minus{} 10 & 51 \\
3 & \minus{} 8 & \minus{} 46 \end{array} \right| }/{\det(M)}$

This means:
$ x \equal{} \minus{} 4$
$ y \equal{} \minus{} 44/5$
$ z \equal{} 29/5$

Which means $ x^2 \plus{} 2y \plus{} z \equal{} 16 \minus{} 88/5 \plus{} 29/5 \equal{} \boxed{21/5}$

Oh, and for reference:
http://en.wikipedia.org/wiki/Cramer%27s_rule

this is all weirded out i just use simultaneous equations
can someone please explain gaussian elimination/Cramer's rule to me? thanks

Gaussian elimination is an efficient way to solve systems of equations, by doing three things:
1. Swap two rows
2. Multiply a row by a nonzero constant
3. Replace one row with the sum of two rows.
Using these repeatedly and in combination, you can get a system of equations down to one variable, then solve.

Example:
$ 2x\plus{}3y\equal{}8$
$ 3x\plus{}y\equal{}5$
We can multiply the bottom row by -3 so that we have:
$ 2x\plus{}3y\equal{}8$
$ \minus{}9x\minus{}3y\equal{}\minus{}15$
We can add the bottom row to the top row and replace the bottom row:
$ 2x\plus{}3y\equal{}8$
$ \minus{}7x\plus{}0y\equal{}\minus{}7$
Now the bottom row is solvable, and we know that $ \boxed{x\equal{}1}$
Now just substitute into either equation:
$ 3(1)\plus{}y\equal{}5\Rightarrow{\boxed{y\equal{}2}}$

Cramer's rule is a way to solve systems using matrices. It works like this:
Say we have a system:
$ ax \plus{} by \equal{} c$
$ dx \plus{} ey \equal{} f$
The matrix that represents this is:
$ \left[ \begin{array}{cc} a & b \\
d & e \end{array} \right] \equal{} \left[ \begin{array}{c} c \\
f \end{array} \right]$
Then, it can be solved like this:
$ x \equal{} \left| \begin{array}{cc} c & b \\
f & e \end{array} \right|\left/\left| \begin{array}{cc} a & b \\
d & e \end{array} \right|\right.$

$ y \equal{} \left| \begin{array}{cc} a & c \\
d & f \end{array} \right|\left/\left| \begin{array}{cc} a & b \\
d & e \end{array} \right|\right.$

Wikipedia explains Cramer's rule much better...

Simplified explanation of Cramer's...

Write the equations as an equation with matrices. If you want to solve for a variable, substitute the column containing that variable's coefficients with the solutions. Take the determinant of that, and divide it by the determinant of the original matrix. That's the variable.

i just tried it out, and i don't know matrix's and stuff, so i just went the old hard way.
if you add all the three equations you get: -14y-7z=14
from which z=-2-2y
and then you get these three equations
2x-14y=27
-14y-5x=55
3x+28y=-82
now: 4x-28y=54
3x+28y=-82
add them up: 7x=-28 so x=-4
and: 2x-14y=27 means y=35/-14=-5/2
and z=3
and it verifies. is it ok?

Find out for yourself. :wink:

Plug in the solution you got into the equations. Does everything work out? If it does, you're right. If not, check your work. It's always a good idea to plug in the variables, if you think your answer might be wrong.

it is alright, i just checked again. I just solved my first problem :D


Define $ \delta$ as a mathematical operation using a combination of only addition and multiplication of the two input numbers. Given that $ 2 \sim \delta \sim 4 \equal{} 24$, $ 4 \sim \delta \sim 6 \equal{} 60$, $ 2 \sim \delta \sim 6 \equal{} 48$ and $ 6 \sim \delta \sim 2 \equal{} 16$, what is the value of $ 6 \sim \delta \sim 4$?

you can notice that $ x\sim \delta\sim y\equal{}(x\plus{}y)*y$.

therefore, $ 6\sim \delta\sim 4\equal{}10*4\equal{}40$

answer : 40

The mathematical operation for $ a \delta b$ is $ (a \plus{} b)(b)$. (We find this using some guesswork. For each example, it works.)

So, substituting $ 6$ and $ 4$, we get:

$ (6 \plus{} 4)(4)$

$ (10)(4)$

$ \boxed{40}$

EDIT: Beaten to it.

Use

[code]$\boxed{x}$[/code]

which becomes

$\boxed{x}$.

There is a LaTeX forum for these kinds of questions.


AMC
AIME
Putnam
USA(J)MO
USAMO
AMC 10
function

What's the worst experience you've ever had in a math competition? Not experiences in which you "did bad", but experiences in which outside factors affected your performance.
Examples: You didn't sleep all night before a competition, a fire alarm went off during the test, etc....

Mine was last year just before the AIME (2006 II) . Three days before the test, I got [b]really [/b]sick. Really sick.The day before the AIME, I had a $103.5^\circ$ fever, I was nauseous, I had a really bad cough, and my throat hurt a lot. The morning of the AIME, I felt worse than ever. After I struggled out of bed and got dressed, I took two aspirins and a painkiller and got my temperature down to $102.5^\circ$ just before the test. I also went to school with a lot of cough drops.  But I still felt very miserable taking the test, and I just couldn't think clearly while doing the problems. There were some problems I just read over five times without really reading. Some other problems I would've usually gotten I also missed. Then I also made a lot of stupid mistakes checking my work, which I think I would have avoided if I was healthy (or maybe not, I'm usually really careless  :lol: ).

After the test, I went straight home. Later that day, my temperature went up to $105^\circ$.

So needless to say, I'm hoping I won't feel like that during the AIME this year.  :D

So about 3 weeks ago, I fell asleep during an entire test and got a 0.

Don't really care that much though.

Ooh.  I love blaming my bad performances on outside influences.

Actually, nothing really has happened to me.  At MC State 05, I nearly choked to death on a cough drop.  But it was nothing life-threatening.

oh math competitions, in that case, nothing other than nerves and stuff

One time in 9th grade I was at a math contest at another school and the person sitting directly in front of me pissed his pants, so he left the room but he left his puddle in the seat so it smelled like piss the entire test

But I still got 1st so I dunno maybe pee actually makes me do better?

My worst was sleeping through the first hour and a half of Putnam 2006.  Fortunately, I went on to have what I believe was the best hour and a half of mathematics that I've ever had.  I solved three problems, one in the last five minutes.  But I still think there's a good chance I could have gotten at least one more, not to mention a much better writeup on the last one, if I had had the extra hour and a half.

AIME 2006 I by far, even though no external circumstances were against me.
Due to my super fast and illegible handwriting, I misread my own answers and misbubbled 3 answers in my AIME I 06, 2 of which were correct, and missing those two cost me USAMO qualification.
Stupidity made me miss #1 and #2.
However, the dumbest mistake I made regarding that AIME was trying to pin the blame got my own messiness to my proctor for time pressure.

Of course, I wouldn't have gotten anywhere with USAMO so whatever.

[quote="drunner2007"][quote="bubka"]Last year, throughout the camp, I underslept, and my had was full of other things, and I kept on getting horrible marks.[/quote]

which camp. did you go to mop?[/quote]

I'm thinking the Indian mathematics camp, since he lives in India?

[quote="kyyuanmathcount"]Ugh, I had a fever before AMC10 last year.  went on moltrin and was dizzy the whole time, so I made a dumb mistake.  137 -_-.

Hm, taking a test in school uniform is a bit uncomfortable, so I slip out of dress code for the few minutes and the proctors thankfully don't care.[/quote]
School uniform? WTF?

Lol, yeah, it's not that bad.  Ask Neil Gurram, he goes to my school.  It's only bad when you try to play basketball in it.

[quote="kyyuanmathcount"]Ugh, I had a fever before AMC10 last year.  went on moltrin and was dizzy the whole time, so I made a dumb mistake.  137 -_-.

Hm, taking a test in school uniform is a bit uncomfortable, so I slip out of dress code for the few minutes and the proctors thankfully don't care.[/quote]

So you take math tests somewhat close to naked?

What the...

[quote="kyyuanmathcount"]Na, the uniforms aren't THAT uncomfortable.  I just take off the tie and the sweater.[/quote]
Tie...and...sweater?

I would not want to go to your school...

[quote="jhredsox"]i think his amc 8 pic has this in it???

i know they're not that bad, but kankersores can be somewhat annoying (i have one now)

-jorian[/quote]

Lol, yeah, Allen Yuan w/out the sweater.

lol...our uniforms require us to have a white polo shirt and navy pants.  we can wear a sqeatshirt if we want.

the only bad part is on thursdays when we have chapel.  we have to where a tie and a sqeater.  ugh  :mad:

Dang... school uniforms from hell.  I go to a public school where there is no dress code... but they're considering implementing one because there's this one girl who walks around campus dressed in a miniskirt and a tube top every day.  Even when it rains.  Had to sit in back of her one semester in French.  It was the most traumatizing experience I've ever had.

[quote="Piggypi314"]Dang... school uniforms from -.  I go to a public school where there is no dress code... but they're considering implementing one because there's this one girl who walks around campus dressed in a miniskirt and a tube top every day.  Even when it rains.  Had to sit in back of her one semester in French.  It was the most traumatizing experience I've ever had.[/quote]

DID U TAP THAT JUNK?

Dude, shut up.  I think Piggypi314 is a girl if my stalker powers remember correctly...

I went to a school last year that didn't have uniform.

To Jli: christian school?

[quote="Piggypi314"]Dang... school uniforms from -.  I go to a public school where there is no dress code... but they're considering implementing one because there's this one girl who walks around campus dressed in a miniskirt and a tube top every day.  Even when it rains.  Had to sit in back of her one semester in French.  It was the most traumatizing experience I've ever had.[/quote]
Your life is hard.

To contribute I had a linear algebra exam about a year ago which I managed to sleep about 2 hours for. I had just moved away from my parents and I was nervous in front of the test. Still, I don't think it really affected me on the test. You can take a night without sleep if you've been sleeping the days before it. This semester my analytic function theory exam was 8 am on a saturday, and needless to say a party started in the floor under me almost the second I went to bed at 11 pm. Still wasn't a big deal, though. You don't need sleep.

Hm, I DID get a full 4 hrs of sleep before chapters last year.  Oh well, I did well anyways.

[quote="Piggypi314"]Dang... school uniforms from -.  I go to a public school where there is no dress code... but they're considering implementing one because there's this one girl who walks around campus dressed in a miniskirt and a tube top every day.  Even when it rains.  Had to sit in back of her one semester in French.  It was the most traumatizing experience I've ever had.[/quote]

Bet you enjoyed that huh?

lol I was in sixth grade and it was winter. My parents made me wear this retarded sweater that made your body temperature shoot up 7 degrees. So I had a math test second period and then for some reason the heating was over done. So the temperature in that  class  room was easily over 80 degrees....No one could think and we nearly all failed.

[quote="sapphyre571"]Georgetown HS in TX hosted a MC invitational. During target, we had to to find THAT if HA*AH=THAT. They didn't tell us HA and AH meant the two-digit numbers HA and AH until there were 30 seconds left in the round, and we were all confused. But I got to countdown, and the answer key was plagued with wrong answers. So I got knocked out early despite getting 2 answers (which the audience said were correct but the key did not) before my first opponent...[/quote]

Sorry about that.

I don't think I have any worst experiences.

Just messing up.

I had this stupid SAT, well the one everyone takes in March or so, the one where they give percentiles in Science, Math , etc. in middle school.

In 2005, I think I was in 6th grade. During the OLSAT, I was really sick, and I fell asleep during the test, and didn't finish 39 of the problems out of 80 or so.

I made 65 percentile, compared to 99 in all the years before. In 7th grade, I took it and made a 99 again. Proves that Sickness is EVIL!!!!!!!!!!

[quote="nutz_for2.718281828"]My SAT. 4 hours of torture in hard chairs that really make your bones hurt, only 3 breaks, and I had to wait another 30 minutes to get home!!!!

I got 2 on the essay.  :(  NO WONDER!!!! :P[/quote]

I got a 3 on the essay in 7th grade =P I took it in January 2006.

I think 560 writing score.

[quote="asdf2345"]I once went without food for three days, and I had a really important math test the next day. I ended up eating my test. I'm not joking. It was really embarrassing.[/quote]

Haha. No you didn't.  :lol:

You're healthy.

Hehe You people have it lucky...

I have an assortment of crazy things happen: (Though these happened over many competitions)

1)Heard a proctor next door call out the answers to a test we were currently taking.

2) Had 4 hours of sleep the night before

3) Calculator died/almost lost it

4) Pencil Ran out of lead/Broke Tip

5) Did not get a trophy due to getting one extra question wrong (Note: This has happened to me 4 times)

6) Lost due to tiebreaker/sudden death TWICE

7)No Clock and Instructor forgot to give us warnings

8)Went up 15 flights of stairs 8 times including once where I had to go up the flights in under a minute and a half...

ive never had anything too bad but...

1) last year at chapters, our mathcounts team [which i wasnt on, i was like on 2nd team, i made state by individual] all got sick and were throwing up.  we got 1st in florida in 04 and 05, and last year we got 3rd at chapter and didnt even make state.

2) i got 1 hr of sleep before a competition, where i quickly took the test in like 30 minutes and went to sleep, and i got a perfect score.

3) we went on like 40 minutes of sleep to a competition this year. i got 3rd.

[quote="Treething"][quote="perfectnumber628"]this year on the day of the AIME, i was like "i'm gonna make pancakes for breakfast!" (there was some logical connection between the AIME and pancakes that made sense at the time...)  and so i got out a bowl and poured the mix in it, and then there was this spider on the inside edge of the bowl and it was scary!  and then i got somebody to come and throw the spider in the sink.  and then i made pancakes.

this had no effect on my AIME score.[/quote]

Sir, you are a genius.[/quote]

i agree

[quote="Nienna"]I was casted out once because I have neat and pretty nice handwriting and professor thought I was cheatin on the exam. He told me that no one had such beautiful handwriting and I had brought the paper with written answers. He didn't want to hear that I didn't know the questions and I didn't write answers at home and that I have neat handwriting, so I had to quit the test and to come back in next term. Since then I don't like complex analysis very much.
[/quote]
 :o
that happened to you on a college?  :o

[quote="deimos"]that happened to you on a college? [/quote]
Unfortunately yes. That was one of my last exams and I knew theory very well so written part of exam wasn't supposed to be a problem...Or at least I thought so...

once i took a test and i lost my manhood. no joke.

I've got it! It was a driving test, you were a man, then you had an accident, and died. This is your second(or up) life.

I'm sort of surprised that no one has mentioned the 2006 ARML individual at the "Paradise Campus" at UNLV. Or do you all just think better when it's 105°?

No one believes me but this is true.

My best experience:

Our math teacher was a very easy-going man. In two summers we went to Lapland for a week to calculate mathematics. We studied so much that the test was hold on the last night of the camp at 8 pm to 6 am, when we went back to Helsinki. I was little tired and I just asked if I can go for a swim during the exam. Teacher said that I can go for a sauna if I want. Then I just went to a sauna and swimming and after that I calculated the rest of the problems of that test.


Are there any computer software that can help us calculate non-commuting symbols?

For example, if we have a rule ba = ab+a, I would like the computer to output:
b^2a = ab^2+ 2ab +a
by commuting the symbols to a particular order.

This will certainly be useful to deal with nonabelian group or ring.

I'm pretty sure [url=http://www.cs.unm.edu/~mccune/prover9/]Prover9[/url] can do things like this (it's been two years since I used it and I didn't use it in this way).


The number 236! ends in how many zeros when expressed as an
integer?

To have 0's, we need tens. We've got plenty of 2's, so we just need to count the number of 5's. So:

$ 236 \div 5 \approx 47$

and

$ 236 \div 5^2 \approx 9$

and finally

$ 236 \div 5^3 \approx 1$

We add up the numbers, getting a result of $ \boxed{57}$.

Note izzy:
To clarify, it should be the floor function and not the approximation function.

Also, instead of dividing by 5^2 and 5^3, you could divide 47/5, then 9/5.


Since Kent's series #4 was solved so quickly, here is a nice series:

$\sum_{n=1}^N \frac{n}{a^n}$ with $a>0$

There are at least 3 or 4 methods I have been shown for this series

A harder problem (one I cannot do)
$\sum_{n=1}^N \frac{n^k}{a^n}$ with $a>1$ and $k>0$

[quote="Kent Merryfield"]I'd ask this question for infinite series rather than finite sums - that is,

$\sum_{n=1}^{\infty}\frac{n^k}{a^n}$ with the restriction that $a>1.$

Perhaps that's what you meant anyway.[/quote]

I did mean $a>1$

I was originally intersted in the finite summation for a problem in class.

However I am happy for the limit case as N goes to infinity aswell. (After all it is a computations and tutorials forum - I am not looking for the answer, rather meaning to give others some interesting questions)


function
algebra
polynomial
algebra unsolved

Find all functions $ f: R \rightarrow R$ such that $ f(f(x))\plus{}xf(2x)\equal{}f(x)^2\plus{}8f(\minus{}x)\plus{}35x$

[quote="Dumel"]Find all functions $ f: R \rightarrow R$ such that $ f(f(x)) \plus{} xf(2x) \equal{} f(x)^2 \plus{} 8f( \minus{} x) \plus{} 35x$[/quote]

There is a trivial polynomial solution $ f(x) \equal{} 2x \minus{} 5$ but I think it's nearly impossible to show if this is the unique one.
I'm not sure that  the professor who created this problem has the solution.
Where is this problem from ?

[quote]Where is this problem from ?[/quote]I don't know. Someone posted this problem on "matematyka.pl" (polish forum), I asked him, but he didn't reply.


Let $ ABC$ be a triangle and let $ \left\|\begin{array}{c}
\{Y_1,Z_2\}\subset (BC)\\\\
\{Z_1, X_2\}\subset (CA)\\\\
\{X_1,Y_2\}\subset (AB)\end{array}\right\|$ be any points for which exists

$ P\in X_1X_2\cap Y_1Y_2\cap Z_1Z_2$ . Prove that $ \frac {PX_1}{PX_2}\cdot\frac {PY_1}{PY_2}\cdot\frac {PZ_1}{PZ_2}\equal{}\frac {AX_1}{AX_2}\cdot\frac {BY_1}{BY_2}\cdot\frac {CZ_1}{CZ_2}$ .


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $S(\mathbb N)$ be the group of permutations of $\mathbb N$.

What are the normal subgroups of $S(\mathbb N)$? (there realy are only a few of them :))

P.S.

Once this is solved, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112250]this[/url] problem will follow almost immediately, at least in the case when $X$ is countaby infinite.

I know two: {generated by all 2-cycles}, and {generated by all 3-cycles}. Can't think of any other.  :huh:


Let ABC be a triangle with orthocenter H, and let M and K denote the midpoints of AB and Ch. Prove that the angle bisectors of <CAH and <CBH meet at a point on the line MK

I'll change the notations a bit, to make them more comfortable (for me :)).

Let $M,K$ be the midpoints of $BC,AH$ respectively. Let $T$ be the intersection of the bisectors of $\angle ABH,\angle ACH$. We have to show that $T\in MK$.

Notice that $BT\perp CT$, so $T$ is on the circle of diameter $BC$. A trivial angle-chase in  the right triangle $BCT$ shows that $\angle CMT-\angle BMT=2(\angle B-\angle C)$ (in my picture $ABC$ is acute; this might be important, but the solution must, of course, be basicly the same in the other case). We must thus compute the angle between $KM$ and $BC$. It's well-known that $AH=2OM\Rightarrow AK=OM$, and $AK||OM\perp BC\Rightarrow$ $KM||AO$, so the angles are easy to compute and the equality between $\angle (KM,BC)$ and $\angle (TM,BC)$ is easy to check.


algebra
polynomial
inequalities
modular arithmetic
arithmetic sequence
number theory proposed
number theory

Find all polynomial $P(x)\in Z[x]$.such that  for each $p$ is prime ,there exist $q$ is prime  so $P(p)=q^{m}$

Is the $m$ given or does it change as $p$ changes ?

I will assume $m$ depending on $p$, since otherwise it's just easier (the inequality part is not necessary). But the following is not really elementary (Dirichlet is used):

Assume there are different primes $p,q$ and different primes $s,t$ with $p \nmid P(p)=s^m , q \nmid P(q)=t^n$.
Then by the Chinese Remainder Theorem there is an $c$ with $c \equiv p \mod s$ and $c \equiv q \mod t$ and $c$ coprime to $st$.
So by Dirichlet's theorem on primes in arithmetic progression there is even a prime $c$ with that property. This prime fulfills $P(c) \equiv P(p) \equiv 0 \mod s$ and analogously for $t$, thus $st|P(c)$, giving a contradiction.

Now let $n$ be the degree of $P$. Then $x^{n-1} < P(x) < x^{n+1}$ for all big enough $x$

Case 1: there are $\infty$ many primes $p$ such that $P(p)$ is a power of $p$.
Thus $P(p)=p^n$ for all big enough primes $p$ with this property by the mentioned inequality.
Now a polynomial of degree $n$ is uniquelly defined by $n+1$ values, thus we just take $n+1$ large enough primes to see that the polynomial is $x^n$ with no other choice.

Case 2: Otherwise, for all sufficiently large $p$ we have that $P(p)$ is not a power of $p$. From now, associate to all primes the term "big enough". Then for any two primes $p<q$ we have that $P(p), P(q)$ are powers of the same prime $s$ (otherwise, we get the contradiction above). When $P$ isn't constant, we get also $P(p)<P(q)$, so when $P(p)=s^m$ then $P(q)=s^n$ with $n>m$. By that, $p_k^{n+1} > P(p_k) > s^{k-c}$ for $p_k$ the $k$-th prime and $c$ some constant.
And this is impossible to be archieved: there are more primes than squares, thus $k^2 > p_k$, so $k^{2n+2} > s^{k-c}$, which is wrong for big $k$.

So either $P(x)=x^n$ or $P$ is constant (then $P$ must be the power of some prime).

[quote="Vor"]other way : use Dirichle 's theorem[/quote]
Well, I used it... How do you do it then¿

If $p\#q$ then $P(p+tq^m)-P(p)$ divise by $p^m$. Thus there exist infinite $p$ is prime so $P(p)=q^m$

ZetaX's solution can be a little bit simplified as we don't need this "analytic" part. If we have that $ P(p) \equal{} s^m$ for all sufficiently big $ p$ and suitable $ m$ then note that we can take infinitely many primes $ q$ such that $ q \equiv p \pmod{s^{m \plus{} 1}}$. Then $ f(q) \equiv f(p) \equal{} s^m \pmod{s^{m \plus{} 1}}$, so $ f(q) \equal{} s^m$ for all such $ q$. Hence $ f(x) \equal{} s^m$.


Given are a$\triangle ABC$ with side a,b,c. Prove that with any point M in plane then:
$(b+c)MA+(c+a)MB+(a+b)MC > ab+bc+ca$
 :)  :)

We have $LHS = \sum (MA + MB).c \geq a^2 + b^2 +c^2 \geq ab + bc +ca$

 The equality can't happen since $M$ can't be on $BC, CA, AB$ at the same time.

 Hence $\sum (b + c)MA > ab + bc +ca$


calculus
integration
calculus computations

hello , some one can help me

evaluate :
$ \Large \int_{0}^{1}\frac{(x\minus{}1)e^{x}}{(x\plus{}1)^{3}}dx$

hello, we consider the integral $ \int\frac{x\minus{}1}{(x\plus{}1)^3}e^x\,dx$, setting $ e^x\equal{}t$ we have $ dx\equal{}\frac{1}{t}\,dt$ and so our integral is $ \int\frac{\ln(t)\minus{}1}{(\ln(t)\plus{}1)^3}\frac{1}{t}\,dt$,substituting again we set $ u\equal{}\ln(t)\plus{}1$ so we get $ \frac{1}{t}\,dt\equal{}du$ and now our integral is $ \int\frac{u\minus{}2}{u^3}\,du$.
Sonnhard.

$ \int\frac{x\minus{}1}{(x\plus{}1)^3}e^x\,dx\equal{}\int(\frac{1}{(x\plus{}1)^2}\minus{}\frac{2}{(x\plus{}1)^3})e^x\,dx\equal{}\frac{e^x}{(x\plus{}1)^2}\plus{}c$

Recall that $ \int(f(x)\plus{}f'(x))e^x\,dx\equal{}e^{x}f(x)\plus{}c$

Hey,
[quote="pankajsinha"]
Recall that $ \int(f(x) \plus{} f'(x))e^x\,dx \equal{} e^{x}f(x) \plus{} c$
[/quote]
Do you have a proof of this equation? Furthermore are there any restrictions on this equation (for example restrictions on what $ f(x)$ can be)?

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715][u][b]APSStudent[/b][/u][/url]

Hey,
[quote="jmerry"]
If you want to verify a proposed antiderivative, differentiate it.
[/quote]
Perhaps my choice of words could have been better, however besides simply differentiating it I was looking more for a derivation or other unique proof.

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715][u][b]APSStudent[/b][/u][/url]

It's integration by parts, just written in the way that naturally suggests that it's a consequence of the product rule.

[quote="Dr Sonnhard Graubner"]hello, we consider the integral $ \int\frac {x \minus{} 1}{(x \plus{} 1)^3}e^x\,dx$, setting $ e^x \equal{} t$ we have $ dx \equal{} \frac {1}{t}\,dt$ and so our integral is $ \int\frac {\ln(t) \minus{} 1}{(\ln(t) \plus{} 1)^3}\frac {1}{t}\,dt$,substituting again we set $ u \equal{} \ln(t) \plus{} 1$ so we get $ \frac {1}{t}\,dt \equal{} du$ and now our integral is $ \int\frac {u \minus{} 2}{u^3}\,du$.
Sonnhard.[/quote]There's a mistake, Sonnhard. It's in the first substitution. Your substitutions are pretty much redundant, in my opinion.

hello, yes, i have made a mistake, after setting $ e^x\equal{}t$  we get the integral $ \int\frac{\ln(t)\minus{}1}{(\ln(t)\plus{}1)^3}\,dt$ which is $ \frac{t}{(\ln(t)\plus{}1)^2}\plus{}C$.
Sorry!
Sonnhard.


Are archives for problem series class stay long as subject classes?

I'm thinking of taking two problem series and want to make sure they are. I don't see why they won't but I just want to make sure. Thanks.

[quote="MCrawford"]Currently we never revoke access to archives after you take a class.

It may be that one day we put a system in place to allow access for something reasonable like four years, but that's a long way off.[/quote]

Yay :)


calculus HCSSiM algebra function domain AMC USA(J)MO

Which AP exams are you taking this May?

I was going to make a poll, but only 10 poll options are allowed, and many people may be taking more than one AP...

For me, AP Calc BC and AP Biology.

It would be interesting to find out which APs are most popular among AoPSers.  BC Calculus is probably fairly popular.

BC, Chem, and American History.  I am soooo stressed out.  I need 5's on BC and Chem, just cause I'm like that, and I haven't been paying attn at all this year in history, cause we Americans are so boring.  I don't know any other juniors in my school taking 3 aps though, so at least I'll get into college for my efforts (if I pass the damn tests).

Edit:  Lol, it replaced, let's say "darn" with the -.

World Hist, Bio, Stats, Comp Sci. Calc'll be next year. ^^;;

Is World History fun/worthwhile?  I might take it next year (as my only AP :( )

[Edit] Oh, never mind.  Saw your comment.

-.-" I'm a she. And the comment isn't referring to one on this site, I think.

I'm very sorry on the gender mix up.

How many girls are on this site? </derail>

Oof.  Well, gender mix-ups are common here, especially concerning Tare.  I don't think 'Fiery' ever mistook 'ripply' for a boy, though.

[quote="Magnara"]How many girls are on this site? </derail>[/quote]

Quite a few.  I'd estimate that there are about 100 girls registered on this site, but only 15-20 are active posters, where an active poster is defined as a person with more than 10 posts.  Namely, rippledance, MathFiend, aalindsey, BHorseMath, hikari, bleumoose, Caroline, and others are girls.

Rippledance is right - the comment is on another site.  But I'm still not sure how

[quote]He only meant he'll be taking calc next year, but he's in world now. That's a period, not a comma.[/quote]

follows from my asking about World History.  Even if I had mistaken the period for a comma, that would only call into question the part about calculus; it wouldn't change whether I believed Hann was taking World History.

I suppose I'm just carrying on a pointless topic, though.  Thank you for reading this, and good night.

this is only for people who are in high school now, right?

Well, if you're a middle schooler taking an AP, then say something.  But generally speaking, APs are taken by (select) high schoolers.

Just AP government this year. Pending scheduling worries, Calc, AP Chem, AP Bio, and Macroecon next year.

No APs for me this year; I guess I'll be taking AP Calc exam at the end of next year.

[quote="mathfanatic"]Quite a few. I'd estimate that there are about 100 girls registered on this site, but only 15-20 are active posters, where an active poster is defined as a person with more than 10 posts. Namely, rippledance, MathFiend, aalindsey, BHorseMath, hikari, bleumoose, Caroline, and others are girls. [/quote]

I think there's a thread on fun and games about girls doing math... but yes, there are definitely girls on this site, though not nearly as many as there are guys ;)

just AB, but maybe i'll take the american history one for fun.

French, US history, and calculus.  Not looking forward to the next few weeks, but it'll be nice when it's over.

I could never imagine taking a history AP for fun... especially with all those essays.  That's the test i'm worried the most about.

[quote="mathfanatic"]Wow, 9 APs?  That must be some kind of record.  I wonder, though: some of the exams are during the same time slot.  How are you going to take both Physics exams?

And what about test fees?  From what I hear, they're $75 each (although Macro and Micro only count for one; Comp. and US Government only count for one; and both Physics Cs only count for one).[/quote]

My girlfriend took 10 AP classes this year alone, and 7 last year... that is not even close to the record...

In florida, we do not have to pay for them, the state pays for us... lucky us :)

as for me... I have: calc bc, physics c mech, physics c E&M, english Lit, gov politics, gov comparative, micro econ, macro econ

[quote="mathnovice"]Is there any benefit to taking so many AP classes as a senior? (besides the joy of learning new things, of course  :D ) You get college credit and an AP award-- anything else?[/quote]

If for instance you were trying to get into let's say... MIT, Harvard, Stanford, CalTech, or any other top notch college, it is one of the most important things you can do...

[quote="jelyman"]Im quite dumbfounded at all of this. At my school, its almost standard practice to take 10+ ap's over the high school career. Ive taken 2 this year (soph), 6 scheduled for next year, and I have 6 scheduled for senior year, and I plan to study for physics myself (school not offering it anymore due to class limits). The number 1 in the junior class took 4 as a soph, followed it up with 6 this year, and is going to be take all ap's at my school and dual enrollment at a nearby college next year. My girlfriend took 7 this year as a junior, 2 as a soph, and is taking 6 next year.

I guess this is why we (at least the top 2 schools in my county, us and Stoneman) dont do as well on national competitions, because we overload (more like ignite up the reactor) by other standards.

Just to end my point, we have had only 2 usamo quals. 1 went to mop (back in 89) and one just made it twice (hes at harvard now). I plan to be the third :).[/quote]

You reference stoneman.... what school do you go to? My gf goes to stoneman, and I am good friends with a good deal of them... They are going to get crushed at MAO nationals this year though lol.

[quote="mathnovice"]Do you really need to take an AP English class? Most of the questions are similar to the verbal section of the SAT except that they use more literary terminology. Would it be worth it to take an AP English class where I would have to read and write a lot when most of this isn't really necessary? The exam can't possibly expect you to read every single book from the English literary canon, so why bother? I also happen to dislike reading and writing because they are both extremely time-consuming activities for me.[/quote]

WHAT!!!! lol English AP is NOTHING like the SAT, it's so much incredibly harder it's not even funny...

Why take them? If you wish to attend any top notch University... it's definitely required. Ivy-League won't even look at your app if you haven't taken Eng AP...

[quote="tokenadult"][quote="Spoon"]as for me... I have: calc bc, physics c mech, physics c E&M, english Lit, gov politics, gov comparative, micro econ, macro econ[/quote]

I have every reason to believe that that strengthened your college application.[/quote]

I have every reason to believe you are correct! And that is largly why I did it... I really didn't have an interest in taking the Gov classes, but I had an extra spot to fill and it was either Bio or Gov... I HATE bio, so I took gov...

It's really less for me than it looks... calc BC I didn't have a class for, I've known the math for quite some time and I didn't have to study or review any of it. Much of Micro I already knew as well, Macro is crazy, I don't like it. I swear to god my teacher in English has to be one of the hardest ever... but you know what, for someone who sucks badly at english, I think I got a 5 on the test the other day... Last year I got a 2 on English, so hmm...

[quote="Spoon"]Why take them? If you wish to attend any top notch University... it's definitely required. Ivy-League won't even look at your app if you haven't taken Eng AP...[/quote]

I refer you all to my many posts on the topic of "How will ____ help me get into ______, given ______?"

I took, over the course of my high school career, five AP exams, four  of which were half-credit exams.  I received a score of 5 on three of them.  I never took AP English.  I am at Harvard.  A good number of my friends have also not taken AP English.  It is very possible to get into Ivy League schools without AP exams.  Don't take classes that you aren't interested in and in which you won't do well -- taking a whole lot of AP classes and getting C's and 3's on the exams is not particularly impressive to a college, while taking only a few AP's that you happen to enjoy, taking other electives that look interesting, and doing well in all of them will quite possibly be seen in a better light.  Spoon is generalizing without any basis for it.

Also, please don't post 4 times in a row -- use the "edit" button to add additional information.

I second what JBL said. Many of the people that I went to Princeton with didn't take an outrageous number of AP exams, and neither did I. I went to a school that didn't officially call any of it's classes "AP", but I took the Calculus, Physics C, Chemistry, and US History exams.

I say if you have the time to study and you can pass them, go ahead and take as many exams as you want. But realize that it's not [i]absolutely[/i] necessary to take that many to get into a selective school. Many top Universities don't give credit for some of the AP exams that are offered, and most won't allow you to use more than a year's worth of those credits towards your degree (they have to keep you there as long as possible to make money off of you).

The more exams you can pass, that's great - it can never hurt. But don't feel like you have to take lots of APs just to keep up with anybody else. Just take classes that interest you and challenge you.

Will taking APs as a senior help you when you apply to a college since, by the time of the exams, you already have been accepted or rejected? It might look good if you take a lot of AP classes, but the college would never know if you are self-studying something.

Well I know that colleges take into consideration your senior year when you apply.  So unless you somehow let them know that you are independent studying something they will never know and thus it won' affect your chances of getting in.  I would,however,  don independent study if I thought I was going to be bored with my schedule because it does force you to do at least some studying depending on how well you know the subject.

On a different note, I would just like to say that I think it is amazing that some of you are doing the Physics C with such ease.  The reason that I say this is that at my school only two ppl in the AP physics class are in contention to pass and it is Physics B!  Ok, the reason that they are doing so horrible is that the teacher stinks and they are lazy.  Next year I know that I will pass the exam no matter how bad the teacher is(they just make excuses).

What do you mean by passing the exams? What score is "passing"?

I'm not really sure.  I'm just saying what I heard from them.  From what I gather though most of them probably won't get even a 3.  It's pretty sad really.

I self-taught stats and chem, I'm a sophomore... and a girl :P

this whole board is scary o.O 17 AP courses? no one can compete with that...

how many APs on average do you need to skip the freshman year of college? (I know, I know, depends on the school, but for anyone who has applied to college/is looking into colleges, how many did you need?)

and most schools accept 4's and 5's on exams, except for the "elite" and artsy ones, right?

What is the point of taking AP things if you go to school, and the colleges you're applying to doesn't accept credit for them? would it be better to just take courses at a local university?

Don't worry, not everyone on this board takes 17 AP classes  :wink: .  I am aiming to take about 7 min to 12 max before I graduate.  That is a lot for my school.  I take the AP classes because they are the hardest classes that my school offers.  I don't really want to go to my local university because it is about 30 minutes away and I'm not really sure that what they offer is much better than the courses at my school.  I don't think that you would be doing anything wrong if you did either.

hmm that trumps me nicely... I'm only going to have done:

BC Calc
AP Phys C
AP Comp Sci AB
(MAYBE AP Chem)

by the time I'm through. AAST is weird like that though. And I spend too much time on USACO / USAMO. :-\

Those classes that you are taking are what I would consider the only  AP classes that you really need.  I don't know where you are going or what you are majoring in but I can guess that it is somewhere in math or science.  The only way that I will take 7 or more is because of what I consider side AP classes like econ, gov, latin.  They are still AP but aren't really core.  This is all a matter of a person's views though.

I plan to major in math, minor (or second major) in a computer related field, so I think everything is in focus as far as that goes.

Of course, things can change, but I don't intend for them to anytime soon ;).


In how many distinct ways can a set of 6 different elements be divided into 3 subsets of 2 each?

I dont understand the question exactly.  Is this implying that {(a,b),(c,d),(e,f)} is different or the same as {(c,d),(a,b),(e,f)}?

What about from {(a,b),(c,d),(e,f)} and {(b,a),(c,d),(e,f)}

[quote="pleurestique"]you said "subsets", and sets are indifferent to the orders of their elements. on the other hand, the order of the subsets could matter, depending on your problem. if order doesn't matter, the answer is $ \frac {6!}{2!2!2!3!}$. (each of the 2!'s comes from not caring about the order of elements in a set, and the 3! from not caring about the order of the *sets*.) if order does matter, then the answer is $ \frac {6!}{2!2!2!}$ (same explanation).[/quote]

Thats exactly what I did- I did the $ \frac {{\binom{6}{2}}{\binom{4}{2}}{\binom{2}{2}}}{3!}$, which is 15 (also 6!/2!2!2!3!), but for some reason it gave the answer as 60.  Even if I multiply it by 3!, it still isnt 60.  Is the answer wrong?

Yes, the answer is 15: five choices to go with $ 1$, three choices to go with the smallest remaining number makes 15 choices.

Sorry but I am also having trouble with the definition of an ordered pair and set.

For an ordered pair, is (a,b) different from (b,a)?

And for a set is (a,b) the same from (b,a)?

For this problem, why is something like ((a,b),(c,d),(e,f)) not different from ((a,b),(e,f),(c,d))?  I understand why ((a,b),(c,d),(e,f)) and ((b,a),(c,d),(e,f)) would be considered the same (beucase they are subsets), but why is ((a,b),(c,d),(e,f))and ((a,b),(e,f),(c,d)) also considered the same (what in the problems gives that away)

Or is this problem a bit ambigious (I found this on some ARML practice website)

[quote="JBL"]As is suggested by the name, in an ordered pair (or triple, or whatever), order matters.  In a set, order doesn't matter.

For the second question, it is a slightly ambiguous question -- there's no language to specifically decide whether $ (ab)(cd)(ef)$ is different from $ (ab)(ef)(cd)$.  Typically, a partition of a set (in this case, into three subsets of size two) is an unordered partition.[/quote]

K thank you.  I also did find like 2 or 3 errors on this test that I got this problem from so Im sure on a formal test they would be more specific and better organized.


Prove that the expression $ \frac{(3n)!}{(3!)^n}$ is an integer for all $ n >\equal{} 0$.

[quote="guile"]Prove that the expression $ \frac {(3n)!}{(3!)^n}$ is an integer for all $ n > \equal{} 0$.[/quote] The first thing that comes to mind, is first let n be a positive integer i.e.$ 2r$ and then simplify (to get a nice expression). Then let n be an odd integer i.e. $ 2r\minus{}1$ (where $ r \ge 0$) and simplify again.

This can be be explained by a combinatorial proof.
Hint
[hide]
Consider the arrangement of 3n things in a straight line and 3n things in n rows of 3 things in each row.  :wink:
[/hide]

I forgot to mention that the first thing that comes to mind, always tend to be the wrong thing too! :P

(Not intentionally of course...  :ninja: )

$ \frac{(3n)!}{(3!)^n}\ \equal{}\ \left(\frac{1\cdot2\cdot3}{3!}\right)\left(\frac{4\cdot5\cdot6}{3!}\right)\left(\frac{7\cdot8\cdot9}{3!}\right)\ldots\left(\frac{(3n\minus{}2)\cdot(3n\minus{}1)\cdot(3n)}{3!}\right)$

In each group, the numerator is a product of three consecutive integers while the denominator is $ 3!\equal{}6$. The product of three consecutive integers is always divisible by 6 because three consecutive integers always contain exactly one multiple of 3 and at least one even number.

[hide="In general"]We have $ \frac{(mn)!}{(m!)^n}$ always an integer for $ m,n\in\mathbb N$. And it easily follows from the proof showed by Jane_Bennet.[/hide]

[quote="nayel"][hide="In general"]We have $ \frac {(mn)!}{(m!)^n}$ always an integer for $ m,n\in\mathbb N$. And it easily follows from the proof showed by Jane_Bennet.[/hide][/quote]

Is there a combinatorial meaning to that... I mean other than $ mn$ things in one row and $ m$ things in $ n$ rows


LaTeX
algebra
polynomial
algebra unsolved

[b]Problem[/b]: Does the set $\{\{n^2\alpha\}|n\in N\}$  dense in the interval $(0;1)$. Where $\alpha \notin Q$

It's too difficult for me. :blush:

@perfect_radio:
But $\alpha$ is a constant here.

[quote="perfect_radio"]And how do you make in  the other sign for diference of sets? (the backslash that is)[/quote]

'\setminus' or '\backslash'.
This yields $\setminus$, respectively $\backslash$.

@ raluca :yes, $\{x\}=x-[x]$. and $\alpha$ is a constant .

@ perfect_radio  : [code]\{\{n^2\alpha\}|n\in N\}[/code]

[quote="perfect_radio"]Look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=59918 . We get your problem by taking $\alpha = n a \in \mathbb{R} - \mathbb{Q}$.

And how do you make in $\LaTeX$ the other sign for diference of sets? (the backslash that is)[/quote]

What you have there doesn't work.

Anyways, Weyl proved that if $p(n)$ is a polynomial in $n$ with at least one irrational coefficient, then the sequence $(p(n))_n$ is equidistributed mod 1.  This also implies that it is dense.

@blahblahblah and @raluca : Yes, you're both right: I wrote a stupidity... I wasn't thinking much when I was writing that post  :blush:

I read some time ago an advanced proof of "$\{ n \alpha \}$ is dense". It said there smth about Weil. And I understood nothing from it. blahblahblah, does the proof you're talking about also involve advanced stuff  :?     :(

(Almost) any proof of equidistribution will use the Weyl equidistribution theorem, yes.  It is entirely possible, however, that you can prove density, which is weaker, through elementary means.


calculus
derivative
function
integration
algebra
polynomial
number theory

here goes the problem
f'(x) + xf(-x) = 1 for all x belongring to the set of real numbers.
f is a continuous and differentiable function.f'(x) denotes the derivative :D
i am getting something like
f'(x)- f'(-x) = x*[{e^((-x^2)/2)}]. dont know if it's correct but am not able to go ahead.
ne1 with better ideas and a nice soln?
best regards

Ok I think I have a solution.The result is pretty ugly , though.

f'(x) = 1 - x f(-x) (1)
f'(-x) = 1 + x f(x) (2)

Let g(x) = f(x) + f(-x) => g'(x) = f'(x) - f'(-x)
By substracting (2) from (1) we get :

g'(x) + x g(x) = 0 . Since g(x)=g(-x) , there is a function h such that g(x)=h(-x^2/2). Then g'(x) = -x h'(-x^2/2) . So we have :

x (h(-x^2/2) - h'(-x^2/2)) = 0 , or h(-x^2/2) = h'(-x^2/2) . From here it follows easily that h(x)=e^x , for all x<0 (since it's defined only for negative numbers) , which means g(x) = e^(-x^2/2) for non-zero x. By continuity it's true for all x.

Now , multiply (1) by f(x) and (2) by f(-x) . Add them up to get :

f(x)f'(x) + f'(-x)f(-x) = f(x) + f(-x) = e^(-x^2/2)

So :

(f^2(x))' - (f^2(-x))' = 2 e^(-x^2/2) . Let F be a primitive of e^(-x^2/2). As you may know already , F cannot be expressed in terms of elemmentary functions, but it exists. You can integrate termwise the Taylor series expansion to get *something* , though .

Anyway , we get g(x)*(f(x)-f(-x)) = 2F + C => f(x)-f(-x) = (2F+c)/g
                                                                    f(x)+f(-x) = g

From here we can derive an expression for f. It's not nice though :D

I hope it's correct , I didn't have time to substitute in the original equation and see if it checks out.

hey even i tried the same approach and landed up with the same expression as u got .but i cudnt find the primitive of e^(-x^2/2) so dint attempt the problem further.,thanks for the help.
cheers mate!!

For Galois, it impossible to have an explicite primitive of

exp(-x^2) with usual functions.

Liouville Theorem: If  \int f(x).exp(g(x))dx is an elementary function,
where f(x) and g(x) are rational functions with degree(g)>0, then
\int f(x).exp(g(x))dx = h(x).exp(g(x)) where h(x) is a rational function.

Suppose that  \int exp(-t^2)dt = h(x).exp(-x^2) by derivation

(1)  h'(x) -2xh(x) = 1,

with (2) h(x) =p(x)/q(x), p,q are polynomials  gdc(p;q) = 1

(3) p'.q - p.q'-2xp.q = q^2

If q(x) = c constant polynomial (3)=> p'-2x.p = c impossible for polynomial.

Thus q is non-constant polynomial, let z_0 one of its complex roots
with multiplicity m,
(x-z_0)^m| q(x) and (x-z_0)^(m-1) | q'(x)

(3) => p.q' = (p'-2xp-q).q

so x-z_0 | p(x), wich contradicts gcd( p(x) , q(x)) = 1  :)

About the equation (1) f '(x) +xf(-x) = 1 where f is continuous and derivable

(1)=> (2) f'(x) = 1-xf(-x) => f ' is derivable

(3) f''(x) + f(-x) - xf'(-x) = 0

for x=/=0 (1) => (4) f(-x) = (1-f '(x))/x

change x as -x in (2) => (5) f'(-x) = 1+xf(x)

(3),(4),(5) give a differential equation of second order

(6) xf"(x) -f'(x) -x^2f(x) +1-x = 0

now you can try find a solution in term of power serie

f(x) =sum_{n>=0}a_n.x^n , recurrence on a_n, expression of

a_n in term of n and identify the function wich associate.  :D


algebra
polynomial
trigonometry
quadratics
algebra proposed

a) Show that you can divide an angle $\theta$ to three equal parts using compass and ruler if and only if the polynomial $4t^3-3t-\cos (\theta)$ is reducible over $\mathbb{Q}(\cos (\theta))$.

b) Is it always possible to divide an angle into five equal parts?

[b]a)[/b] This is the polynomial given by $\cos(3\arccos(t))-\cos(\theta)$. It is reducible iff it's roots are zeros of linear and quadratic equations which happens iff the desired angle can be constructed out from the given stuff.

[b]b)[/b] Surely not: the extension given by adjoining the $25$-th roots of unity to $\mathbb{Q}$ has degree $20$, which is not a power of two, thus by the known classification of constructible points the angle $14.4^\circ$ cannot be constructed by ruler and compass only (and even any angle-trisector will not help).


Let n be a positive integer greater than 5. Show that at most eight members of the set {n + 1, n + 2, ... , n + 30} can be primes

By the Inclusion-Exclusion Principle there are30/2+30/3+30/5-30/(2*3)-30/(2*5)-30/(3*5)+30/(2*3*5)=22

Whats the Inclusion-Exclusion Principle.Whats the general form of it?
Then, the rest of the solution is understandable.


[quote="Funcia"]I have no life.[/quote]
DUDE!!! That's a cool way to edit spam.

so who wants to collaborate on a list of aopsers with/without lives

we can clearly see where to put funcia

uhh you can just have a mod get rid of that post

I never said that next time check before you accuse me

edit:
Page 18 last post in Hurt and Heal it was mathalete not me who said that

and now Funcia
cause I will hunt you down and continue to do everything legal to make you pay for doing that

I won't be surprised if you set up the account to do that and accuse me as everyone hates me

I hope you get banned for attempting to get me banned

you guys are getting too personal. and ars is right, but still you should watch what you copy and paste.

If you personally don't like someone, use private messages. :wink:

And what was the point of starting this thread? It is just total spam. And what's the point of repeating the bad word again in a new thread?

And stop saying dude all the time, it is really annoying.

[quote="Ars"]dude I did not continue it

dude why are you saying dude

Dude if you noticed everything I said was appropriate

not like you dude

so just plea for my mercy dude

cause you are more annoying by the second dude

and I am not trying to do what lies you did dude just make you have no friends here dude

as its obvious you have no life if you continue accusing me after I proved you wrong dude

Mod can this get deleted as this is just a pityful attempt to make me look bad by the hordes of people who hate me.[/quote]

dude PM me i beg forgiveness and will edit every post made by myself in this thread
dude well i was going to say we need to make a list but you are so nice you finally helped me classify myself thank you so much ars!
funcia

Hey hey hey hey hey!!
Before this is locked I'll get this post in!

Dudes why are you fighting?
Omg this Ars dude is too quarrel prone!

Stop it dudes!
Dudes mustn't have hard feelings among themselves.
This is not the way for dudes to behave!

Funcia, dude, I can see more clearly than most people where to put you, for I have been watching you more carefully than most others!
If Ars isn't liking it, you shouldn't do this to a fellow dude.
If you do that dude, even if you are soundind cool, you are not being cool.

Ars, dude, common, you are older than funcia, can't you treat him with a large enough mind?
Look at me, dude, I could have picked so many quarrels on personal matters with so many people since I joined, but I never did!

Come on, dudes, forgive each other in the true AoPS spirit, and let's all be friends!

sadfljkdfijsfd;lkja

are you sure funcia is a guy?

Number of times "dude" was said in this topic so far:
[size=200][b]40[/b][/size]

[quote="1=2"]You all said Dude so much that I'm beginning to doubt that it's a word.[/quote]
dude 1=2 it's like the russians saying comrade.  that's in some of orwell's books too dude.

iamagenius
dude who thinks i am a guy i can't see there are so many post with dude in them

ars
i am glad you have forgiven me, it was my fault. i should stop posting random crap without proof.
i changed it so no one can see what i accused you of because you don't want some kid reading only the first post and jumping to a conclusion(i.e. funcia) about you again? i could change it back if you really really wanted me to. just pm me coz i better for me to stay out of g&ff dude or i will have made 8 of these posts

anyone can aim me at valkyriefuncia so that we can talk about the other topic(the non ars one) coz that's serious too

wow that's like i didn't say dude g2g

EDIT: bubka dude you seem nice but you don't understand that funcia has problems, pay more attention next time to all the edits and the upset victims

dudical dudes, dude the dude, dude.
duded dudical dude, duding the duded dudical dudes dudes the dude of dudes.
dude! dude.

[quote="Ars"]to the people like bubka:

first of all the conflict between funcia and I has been resolved by pm

second
Bubka you did not see what funcia posted and as a result do not talk as you do not know why I was forced to respond the way I was

finally
I am not prone to argument I got accused out of nowhere for something fairly bad

not only was I accused but the whole topic was made to try to show how bad of a person I was it even had my name in the title

so I tried to prove otherwise and have funcia admit that I was right

after I said it wasn't me but funcia decided that it was still my fault, that I had somehow continued the bad remarks, and funcia continued to use the topic to make me look bad though it was easy to check I didn't

Then I post again already get mad as this not only showed funcia was after me not the person who said that remark and that funcia was trying to regardless of the truth make look bad

now after I said how sad this was on Funcia's part Funcia changed the topic and all of the posts to make it look like I was making fun of Funcia over nothing

if you still believe it is my fault then you seriously must hate me for some unknown reason as all I did was normal behaivor in the conditions

As I said I have forgiven funcia

 but Bubka your attempts to always look good and help the person in need more often hurt the situation then help as you do not see all that has happened[/quote]

[quote="Ars"]I am not prone to argument [/quote]

 :rotfl:  :roll:

_________________________________________________

Hey I never said I hate you Ars! You don't quite see the point.
As one of my friends put it, friendship should transcend all barriers of having to say sorry, and that is what I feel too!
Best friends should be such that they continually abuse and accuse and belittle each other, without ever getting piqued.
Even now, as you write this post, you speak as if, out of your magnanimity, you have forgiven a great sinner.
That is the point I'm trying to address.
We are your friends here, Ars, and I'm trying to help you.
I admit I did not see the topic in its origninal form, and have little idea what it was all about.
But it does seem that you are always on the lookout for people trying to make you look bad. Get out of it, buddy!
Learn to see the world around you as a world full of friends as I do. I have been hurt many times because of that, but look at me, I'm still happy, happier than most of the people who hurt me.
People say I'm yet to get wise, that I'm too naive for my age. It may be true, but in matters such as human relations, I fell I'm far better off than most.
Learn to trust, Ars, learn to be warm. In doing so, even if you are swindled, you stay warm. I feel that from my heart.

I am not making an argument just a point unlike with your points there is a difference

you are attacking what I am doing (arguing)

I am defending (not arguing)

as I said cracking up again shows you are not trying to help

the point is over so you attack me saying that I am wrong

now I am sorry if you see this as arguing but your constant insistance that your behavior is better then mine makes everything you say hypocitical as you say that I am being arrogant while yourself saying you are better

Also you say you now human relations then you should now that sometimes there is a situation where people are wrongfully accused and serious problems come from it because

Another thing is that unlike you I have been accused several times for things I have not done and have had serious problems from that. Here I also was accused so I defended before it turned ugly. Now the reason I saw this was the thread  had my name in it a pretty obvious sign to look at it.

I may have seem arrogant but I believe that you coming in from nowhere would continue to defend funcia and your actions unless I explained it the way I saw it

Even funcia stated that but you don't seem to care and are still using this against me just as was trying to avoid in order to get me to change maybe you don't understand that the circumstances do matter in realtions more then you explain

Let me clear two things:
I am NOT trying to defend funcia
I am NOT against you

Unless you get this 'against me' thing out of your head, we can't proceed. This is the very problem I've been pointing out. I am trying to have a friendly discussion, and none of my posts now is against you. Okay?

Cheer up please!  :)


Expand and simplify $(a - x)(b - x) \cdots (z - x)$.

Heh, I did this problem on a math competition. Assuming it includes every letter of the alphabet, you'll have an $(x-x)$ in there, which is $0$. :roll:

Yea...  :)  By the way I didn't know brainteasers can come out in math competitions?

Not brainteasers, but trick questions.

A well known mathematical joke is:
Compute $\sin^2 39 + \cos^2 39$ to the nearest hundredth.

EDIT: BTW, this was in my algebra 2 textbook. :lol:

[quote="randomdragoon"][hide]1.00  :| [/hide]

I personally don't think trick questions (or at least common trick questions) should appear on math competitions - too much chance that someone's already seen it before![/quote]

Define a trick question. There was a great question with a trick in the Mathcounts forum: "Consider the number that is the product of the six possible numbers composed of one 1, one 3, and one 5. What is the greatest prime factor of it?" I would call that a trick question, et toi?