geometry
3D geometry
sphere
ratio
MATHCOUNTS

[quote="13375P34K43V312"]State/National level. If interest is high enough, I will post it within about a week. No problems will intentionally be based on either the school or chapter tests. No countdown.

Takers?[/quote]
Sure, count me in too.

Yeah. We PM you the answers, right?

I'm in

Are questions from team round included?
Count me in.

I'm in.

FYI, my parents disconnected the internet yesterday (right now I'm supposed to be doing science fair :roll: at my friends house), so I might not be able to do the test right when it comes out.

sure, i'll do it

im gonna pown (lol)

-jorian

[quote="frost13"]you totally stole #25 from me.  The Anti Keegan problem!!!!!!!!!!!![/quote]
Any comments on difficulty?

[quote="frost13"]you totally stole #25 from me.  The Anti Keegan problem!!!!!!!!!!!![/quote]

I threw in a twist.

#2 means S*U*P*E*R

[quote="13375P34K43V312"]It's being posted right now.

At least the sprint is. I whipped this up in about 90 minutes, so don't be surprised if there are any typos and/or ambiguities. I'm checking my answer key with Mr. Frost. Let me know of any typos and/or ambiguities in this thread ONLY IF they don't give away ANYTHING about the answers.

Enjoy.

Edited once for #4 and #7.[/quote]
On #7, question is only talking about a circle and a square. Then where is the sphere coming from? Did you mean to say, "Find the ratio of the area of the circle to the area of the square"?

[quote="13375P34K43V312"]It's being posted right now.

At least the sprint is. I whipped this up in about 90 minutes, so don't be surprised if there are any typos and/or ambiguities. I'm checking my answer key with Mr. Frost. Let me know of any typos and/or ambiguities in this thread ONLY IF they don't give away ANYTHING about the answers.

Enjoy.

Edited once for #4 and #7.[/quote]
I think even after the editing, the question #7 is still incorrect. Did you mean to say, "Find the ratio of the area of the circle to the area of the square"?

#7 does still need fixing to area of circle to area of SQUARE

#20 is the same as #5

I believe there are two others that are either impossible or absurdly non-simplifyable(good word) that were not intended.

#22 Lines don't have vertices, so even though I am not sure what he wants, this can't have an answer

#29 [hide]I think you need a number in here, this is not one you can just pick a number for because it would all cancel out anyway.[/hide]

I liked it on he ones that worked though.  I tried reaching LEET with my points, but haven't heard in a bit, so he may be gone for the night.

Frost

1337, can you make corrections and then repost it before I take it? I don't want to waste my 40 minutes going on the forum asking for clarity.

I am editing right now. New version should be up in around five minutes.

Maybe next time I write a mock I'll proofread...[/i]

I seriously doubt they would have a test anywhere close to thist in MathCounts, because it was nothing but computation... and really really hard computation too, that doesn't test any skill and is time wasting. The concepts were like chapter/state level it's just that for every problem you needed to multiply 4 digit numbers or find a bunch of useless information and simplify it in random ways etc...

Yeah I noticed that, but it should be better next time.

Well thanks anyway for taking the time for putting together a mock!

One more important thing: None of the problems had any details and I wasted a lot of time trying to figure out what I should solve for. Many problems needed details such as "two [i]distinct[/i] integers" when it only said "two integers", etc.


If $ (a\plus{}1)(b\plus{}1)(c\plus{}1)\equal{}8$ prove that $ 1\geq abc$

Here is my proof:

$ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \equal{} abc \plus{} ab \plus{} ac \plus{} bc \plus{} a \plus{} b \plus{} c \plus{} 1 \equal{} 8$
$ abc \equal{} 7 \minus{} (ab \plus{} bc \plus{} ac \plus{} a \plus{} b \plus{} c)$

$ 1\geq 7 \minus{} (ab \plus{} ac \plus{} bc \plus{} a \plus{} b \plus{} c)$ than $ ab \plus{} bc \plus{} ac \plus{} a \plus{} b \plus{} c\geq6$

$ (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \minus{} abc\minus{}1\geq6$ than $ \minus{} abc\geq \minus{} 1$

and finally $ 1\geq abc$

Edit: Found my mistake

[quote="sinajackson"][hide="HeRe"]kind of easy! :)
By using AM-GM 3 times:

$ 8 \equal{} (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)\geq2\sqrt {a}\cdot2\sqrt {b}\cdot2\sqrt {c}$

and Done![/hide][/quote]
You can't use it when a,b,c are negative numbers :wink: .But the problem will be wrong if it doesn't have that condition. So you're right :lol:


Hello

Let ABCD be a cyclic quadrilateral.  Diagonals AC and BD intersect at E.  Let P be a point inside the quadrilateral.  Let $ O_{1}$, $ O_{2}$, $ O_{3}$, and $ O_{4}$ be the circumcetners of triangles ABP, BCP, CDP, and DAP, respectively.  PRove that lines $ O_{1}O_{3}$, $ O_{2}O_{4}$, and $ OE$ are concurrent.

I tried searching for this, but I couldn't find anything.

I am sorry for posting this in the wrong section, could someone please move this to the geometry forum.  Thanks

Take a look at: http://www.mathlinks.ro/viewtopic.php?t=187214  :D


number theory
relatively prime
recurrence relation
counting
Frobenius
IMO Shortlist

An eccentric mathematician has a ladder with $ n$ rungs that he always ascends and descends in the following way: When he ascends, each step he takes covers $ a$ rungs of the ladder, and when he descends, each step he takes covers $ b$ rungs of the ladder, where $ a$ and $ b$ are fixed positive integers. By a sequence of ascending and descending steps he can climb from ground level to the top rung of the ladder and come back down to ground level again. Find, with proof, the minimum value of $ n,$ expressed in terms of $ a$ and $ b.$

[hide]
I claim the answer is $ a\plus{}b\minus{}\gcd(a,b)$. First we will cover the case with $ a$ and $ b$ being relatively prime, so we wish to prove that the answer is $ a\plus{}b\minus{}1$.

Assume there are $ n$ rungs, with $ n<a\plus{}b\minus{}1$. Then there are $ n\plus{}1$ total possible spots for the mathematician, (we include the "ground floor" as a spot). below the rungs. Now suppose there exists a cycle for this mathematician, such that he can start at one rung and end up back at that rung, and consider the smallest such cycle. If he makes $ x$ ascending steps and $ y$ descending steps, then $ xa\equal{}yb$. We assume $ a$ and $ b$ are relatively prime here, so the minimum possible value of $ x$ is $ b$, and the minimum possible value of $ y$ is $ a$. Thus this cycle includes at least $ a\plus{}b$ steps. If $ n<a\plus{}b\minus{}1$, then there are at most $ a\plus{}b\minus{}1$ possible spots for the mathematician. Hence he must have visited one of these places twice, but that means there is a smaller subcycle within his path, but this subcycle cannot exists as we are assuming this cycle is minimal, a contradiction.

So we must have $ n\ge a\plus{}b\minus{}1$. Now I will show that it is indeed possible to have $ n\equal{}a\plus{}b\minus{}1$. On a ladder with $ a\plus{}b\minus{}1$ rungs, there are $ a\plus{}b$ possible spots for the mathematician (again, this includes the ground floor). It is always possible for the mathematician to make a move no matter what rung he is on; to see this, suppose he is on the $ i^{th}$ rung. If $ i\le b\minus{}1$, then he can ascend $ a$ rungs, and if $ i\ge b$, then he can descend $ b$ rungs. Since he can move indefinitely, there must be some cycle he can make. As proven above, there cannot be a cycle within at most $ a\plus{}b\minus{}1$ spots, so this cycle must include the ground floor and the top $ (a\plus{}b\minus{}1)^{th}$ rung. This path is the one we wished to find.

Now suppose $ a$ and $ b$ are not relatively prime. Then we know that $ \gcd(a,b)|n$, and we can think of each set of $ \gcd(a,b)$ rungs as a single rung, so it is like we have $ \frac{n}{\gcd(a,b)}$ total rungs, and each ascending step is $ \frac{a}{\gcd(a,b)}$ rungs and each descending step is $ \frac{b}{\gcd(a,b)}$ rungs. Then the minimum number of these "big rungs" is $ \frac{a}{\gcd(a,b)}\plus{}\frac{b}{\gcd(a,b)}\minus{}1$. To find the total number of rungs, we just multiply back the $ \gcd(a,b)$ to get $ a\plus{}b\minus{}\gcd(a,b)$ as the minimum number of rungs.
[/hide]

[quote=yunustuncbilek][quote="tjhance"]
Now suppose there exists a cycle for this mathematician, such that he can start at one rung and end up back at that rung, and consider the smallest such cycle. If he makes $ x$ ascending steps and $ y$ descending steps, then $ xa\equal{}yb$. We assume $ a$ and $ b$ are relatively prime here, so the minimum possible value of $ x$ is $ b$,
[/quote]
But, what if such a cycle doesn't exist?[/quote]

Eventually,if he wants to get back

Solved with [b]Aayuu[/b]

We claim that the answer is $\boxed{a+b-\gcd (a,b)}$.\\

By bezout's lemma we may find $x,y \geq 0$ with $ax-by=\gcd(a,b)$. Then we have $b(y+1)-a(x-1)=a+b-\gcd(a,b)$, and it is clear that we can perform such a sequence of moves (if we are ever "stuck" then we would have $ka<b$ which is obviously impossible). This shows that the answer works.\\

Suppose now that we have $n<a+b-1$. Considering all the moves in total, we have $ax=by$ for some $x,y \in\mathbb{Z}$, which implies that we move down atleast $a$ times and similarly move up atleast $b$ times. Since $\gcd(a,b)=1$, $b \cdot k$ (for $k \in \{0,1,\dots,a-1\})$ forms a complete residue system modulo $a$, so we would necessarily have the height be $b-1 \pmod a$ at some point. But since $n<a+b-1$, we have no legal move at this point which is obviously a contradiction.\\

This completes the proof and we are done. $\blacksquare$

The answer is $a + b - \gcd(a,b)$.
We will only solve the case where $\gcd(a,b) = 1$ because the general answer follows by a suitable homothety on the ladder. In other words, we will assume $\gcd(a,b)=1$ and show the answer is $a+b-1$.
To prove $n=a+b-1$ is possible, we use the following algorithm to ascend to the top of the ladder (the descent is symmetric). [list] [*]The mathematician goes up by $a$ if possible. [*]Otherwise he goes down by $b$; this must be possible since there are at least $n-(a-1)=b$ rungs below him. [*]To show this works, note that if the mathematician ever visits the same rung twice, then he must have had at least $a$ descents (and at least $b$ ascents). Since $\gcd(a,b)=1$, this means the mathematician visited a rung in every residue class modulo $a$. But if they had visited the rung which was $b-1 \pmod a$, then they reach the top of the ladder. [/list]
We now prove that $n < a+b-1$ isn't possible by contradiction. Given a trip with total displacement $0$, again, this means that the mathematician took at least $a$ descents and $b$ ascents, and hit every residue class modulo $a$ or $b$.
WLOG, let's assume $a \ge b$, and consider the rung $n-a+1$. It is the only rung which is $n-1 \pmod a$, so it must have been visited but there are no legal moves from this rung, a contradiction.


is this a name reaction ...find the products
cyclohexanone+ sodium ethoxide ----->B+ Nitro Methane

Is B the product from cyclohexanone and sodium ethoxide?

Then B is the enolate from cyclohexanone, or B is the product from attack of that enolate to another molecule of cyclohexanone: an alpha,beta unsaturated ketone. Then addition of nitromethane (still in the presence of base) results in a Michael Addition type reaction.


overall, how hard is it to get into top 500, and subsequently top 20? Would it be considerable to making USAMo, then MOP?
Also, is it common for a bio honors student to make it?

I have heard the difficulty is a joke compared to the math olympiads


How many selections of at least one red ball can be made from 4 red balls and 3 green balls if balls of the same colour are different in size?

My attempt:

C(4,1) [ C(6,0) + C(6,1) + C(6,2) + C(6,3) + C(6,4) + C(6,5) + C(6,6) ]

= 2^8.

But thats not the answer!

Well, there are 7 objects so only $ 2^7$ subsets to choose, so we can throw out any answer larger than that right away :)

Here's your mistake: $ \binom{4}{1} \cdot \binom{6}{1}$ counts the choice "choose the smallest red ball, then choose the largest red ball," and it counts separately the choice "choose the largest red ball, then choose the smallest red ball."  Similarly, your $ \binom{4}{1} \cdot \binom{6}{6}$ is supposed to be the number of ways of choosing 7 balls from among 7.  This should give 1, not 4.

(What you've calculated is the number of ways of choosing some number of balls from that collection with a specially selected red ball.  This is different than what the question asks.)

Another way to this is to observe that for some number of red and green balls respectively, which we denote R and G, each selection with decrease the sum $ R\plus{}G$ by one. Additionally, for each selection there are two possibilities: either a red or green ball is selected.  Note that if we disregard the order and specifics of each selection, after we select R+G balls we have no balls left. I completed the problem by drawing a tree graph starting with the # of balls remaining $ n\equal{}7$ and the # of balls selected $ s\equal{}0$. Hence for $ (n,s)\equal{}(7,0)$ there is one possibility, no balls. I then can pick a predicted ball or unpredicted ball; $ (6,1)\equal{}2$. Again I can pick a predicted or unpredicted ball; $ (5,2)\equal{}4$. From there we see that where $ n\equal{}7\minus{}s$ one can pick $ s$ balls in $ 2^s$ ways. Because there are $ 7$ balls and we pick all of them (every way includes at least one ball), there are $ 2^7$ ways to pick 7 balls.

Not exactly a standardized or reliable method, but it works.


For all positive integers $ n$, let $ f(n) \equal{} \log_{2002} n^2$. Let
\[ N \equal{} f(11) \plus{} f(13) \plus{} f(14)
\]
Which of the following relations is true?

$ \textbf{(A)}\ N < 1 \qquad \textbf{(B)}\ N \equal{} 1 \qquad \textbf{(C)}\ 1 < N < 2 \qquad \textbf{(D)}\ N \equal{} 2 \qquad \textbf{(E)}\ N > 2$

[hide]We can rewrite the function as $ f\left(n\right) = 2\log_{2002} n$. Then:
\begin{align*}N & = 2\log_{2002} 11 + 2\log_{2002} 13 + 2\log_{2002} 14 \\
& = 2\log_{2002} \left(11 \cdot 13 \cdot 14\right) \\
& = 2\log_{2002} 2002 \\
& = 2\cdot 1 \\
& = 2 \end{align*}
I'm guessing the most explicit relation is $ \boxed{D}$, but why wouldn't $ A$ work as well?[/hide]

[quote="The Zuton Force"]I bet it should read $ \textbf{(A)} \; N < 1$[/quote]It even reads $ N>1$ in the official AMC 12 2002A PDF. must be a mistake...

[quote="123456789"][quote="The Zuton Force"]I bet it should read $ \textbf{(A)} \; N < 1$[/quote]It even reads $ N > 1$ in the official AMC 12 2002A PDF. must be a mistake...[/quote]This problem in the PDF is generated from my post above.

TZF, if you catch a typo when I mass-post problems from a contest, PM me as I usually don't follow the (large quantity of) threads. Nice catch, though.

[quote="123456789"][quote="worthawholebean"]This problem in the PDF is generated from my post above.[/quote]You typed up the problems for the MAA's Official PDFs on disk?
Because this problem doesn't appear on AoPS's contests section.[/quote]Oh, gotcha. Forgot I hadn't put this in the contests section yet. Yeah, that's where the typo came from, since I just get the problems from those PDFs.

On the official MAA PDF, choice A says N>1 not N<1.

HELLO HERE IS MY "SOLUTION" THANK YOU VERY MUCH PLEASE UNFORM ME OF ANY AND ALL AND ANY ERORS

\[N=\log_{2002} 11^2+\log_{2002} 13^2 +\log_{2002} 14^2\]

SINCE WE KNOW THAT

\[\log_a b + \log_a c = \log_a bc\]

WE THEN HAVE

\[N=\log_{2002} (11\cdot 13\cdot 14)^2=\log_{2002} (2002)^2\]

THUS

\[N=2\]


[hide="solution"]we can se that none of them are $ 0$ and we can assume that all are positive(we just add the negative solutions later)
from the second equation we have that $ \frac{(xy)^{2}}{(zt)^{2}}=\frac{x^{2}+y^{2}}{z^{2}+t^{2}}=1$
squaring the first equation and subtracting $ 4x^{2}y^{2}$ gives us
$ (x^{2}-y^{2})^{2}=(z^{2}-t^{2})^{2}$ so either
$ x^{2}-y^{2}=z^{2}-t^{2}$ or $ x^{2}-y^{2}=t^{2}-z^{2}$ so either
$ x=z$ and $ y=t$ or $ x=t$ and $ y=z$.
so the solutions are $ x,y,z,t$ are nonzero integers such that the sets $ (|x|,|y|)$ and $ (|z|,|t|)$ are identical.[/hide]


Thirteen boys and girls wait to take their seats in the same row in a movie theater.  The row is 13 seats long.  They decide after the first person sits down, the next person has to sit next to the first.  The third sets next to one of the first two and so on until all 13 are seated.  In other words, no person can take a seat that separates him from at least one other person.  How many different ways can this be accomplished, assuming that the first person can choose any of the 13 seats.

let $ a_n $ be the number of ways to do this for $ n $ people and $ n $ seats.  observe that, as a consequence of the rules, for the case of $ n $ people the last of them to sit must sit in either the first seat or the last, and then for each of these possibilities there are $ a_{n-1} $ ways in which the first $ n - 1 $ people could have sat down.  thus $ a_n = 2a_{n-1} $, and since $ a_1 = 1 $, by induction we then have that $ a_n = 2^{n-1} $.  in particular, $ a_{13} = 2^{12} = 4096 $.  if it matters *who* sat in *what* seat, then the answer is $ 4096 \cdot 13! $.


geometry circumcircle Euler symmetry trigonometry parameterization ratio

Given is an acute triangle $ ABC$ and the points $ A_1,B_1,C_1$, that are the feet of its altitudes from $ A,B,C$ respectively. A circle passes through $ A_1$ and $ B_1$ and touches the smaller arc $ AB$ of the circumcircle of $ ABC$ in point $ C_2$. Points $ A_2$ and $ B_2$ are defined analogously.

Prove that the lines $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ have a common point, which lies on the Euler line of $ ABC$.

??? I deleted my solution , and please don't participate me. thanks

Let $ (O)$ be the circumcircle of $ ABC$ with circumcenter $ O$
Let $ (O_1)$ the circle that is tangent to $ (O)$ at $ A_2$ and similarly we denote the $ (O_2),\ (O_3)$ .Then the common tangent of $ (O_1)$ ,$ (O)$ and the lines $ B_1C_1$ , $ BC$ are concurrent at the radical center of the circles $ (O_1)$ , $ (O)$ and the circumcircle of $ BCB_1C_1$ .
Let K this point .Similarly the common tangent at $ C_2$ ,$ AB$ , $ A_1B_1$ are concurrent at $ L$ and the the common tangent at $ B_2$ ,and the lines $ AC$ , $ A_1C_1$ are concurrent at $ M$ .
Now since $ KB\cdot KC\equal{}KC_1\cdot KB_1$ we have that $ K$ lies on the radical axis of the circumcircle of $ ABC$ and the circumcircle of $ A_1B_1C_1$ . Similarly for $ L,M$ .So $ K,L,M$ lie on the on the radical axis of the circumcircle of $ ABC$ and the circumcircle of $ A_1B_1C_1$ so these points are collinear .Let $ H$ the orthocenter of the triangle $ ABC$
Then from the complete quadrilateral $ AC_1HB_1BC$ we take that it's diagonals are cut at harmonic conjugate points ,ie $ (K,B,A_1,C)\equal{}\minus{}1$ (this can be done also by Ceva and Menelaos) .From the previous fact that  $ (K,B,A_1,C)\equal{}\minus{}1$ and the fact that $ KA_2$ is tangent to the circumcircle of the $ ABC$ we conclude that $ A_2A_1$ is the polar of $ K$ with respect to the circumcircle $ (O)$ of $ ABC$ .Similarly $ B_2B_1$ and $ C_2C_1$ are the polars of $ M,L$ respectively.Let $ A_1A_2$ and $ B_1B_2$ meet at $ Q$ .Then $ Q$ lies both to the polars of $ K,M$ so $ KM$ is the polar of $ Q$ . But $ KM$ passes through $ L$ so $ L$ belongs to the polar of $ Q$ so $ Q$ belongs of the polar of $ M$ .So $ C_1,Q,C_2$ are collinear so the lines $ A_1A_2,B_1B_2,C_1C_2$ are concurrent.
Now since the line $ M,K,L$ (let $ l$ this line) is the polar of $ Q$ we have that $ OQ\bot l$  . But $ l$ is also the radical axis of the circumcircles of $ ABC$ and $ A_1B_1C_1$ . So if $ O'$ is the circumcenter of $ A_1B_1C_1$ (this is also the nine-point center) we have that $ OO'\bot l$ so $ OO'$ passes through $ Q$ so $ Q$ belongs to the Euler line and we are done .

$ I$ $ CASE:$Exist two sides of triangle that are equals, $ WCGL$ $ AB\equal{}CA$->Then, $ A \in A_1A_2$, By the symmetry $ B_1B_2$ and $ C_1C_2$,have a common point,which the line $ A_1A_2$,But the line of euler of $ ABC$, is the same that $ A_1A_2$, then the lines $ A_1A_2$,$ B_1B_2$ ,$ C_1C_2$ have a common point, which lies on the Euler line of $ ABC$.$ \blacksquare$

$ II$ $ CASE:$ Dont exist two sides of triangle that are equals.
If $ \alpha$ is circumcircle of $ ABC$ and $ \beta$ is circumcircle of $ A_1B_1C_1$ , if $ \mu$ is Radical axis of $ \alpha$ and $ \beta$.
$ LEMMA:$ If $ A_3\equal{}B_1C_1 \cap BC$, $ B_3\equal{}A_1C_1 \cap AC$, $ C_3\equal{}A_1B_1 \cap AB$, and $ H$:Orthocenter, $ O$: Circuncenter.Then $ A_3$,$ B_3$,$ C_3$ are collinear,and $ OH \perp A_3B_3$.
$ DEM$.
Know we that: $ <BC_1C\equal{}<BB_1C\equal{} \pi /2$, Then-> $ A_3B.A_3C\equal{}A_3C_1.A_3B_1$
then $ A_3 \in \mu$, Similarly $ B_3 \in \mu$, $ C_3 \in \mu$. If $ O_9$ center of $ \beta$, know we that $ O_9O \perp \mu$(By Radical Axis), but $ O$, $ H$, $ O_9$ lies on the euler line of $ ABC$. Then $ OH \perp \mu$. $ \blacksquare$.

In the problem, know we that the circumcircle of $ B_1C_1A_2$, is tangent to $ \alpha$ in $ A_2$, if $ X_1$ the line tangent to $ \alpha$ in $ A_2$, then $ X_1$ is Radical axis of circumcircle of $ B_1C_1A_2$ and $ \alpha$. $ BC$ is Radical Axis of $ \alpha$ and circumcircle of $ B_1CB$, and $ C_1B_1$ is radical axis of circumcircle of $ B_1C_1A_2$ and circumcircle of $ B_1C_1B$, but $ A_3\equal{}B_1C_1 \cap BC$ then $ A_3 \in X_1$(By Radical Axis).
Now respect of $ \alpha$,we have $ A_1A_2$ is the polar of $ A_3$( because know we that $ A_3$, $ B$,$ A_1$,$ C$ points harmonics and $ A_3A_2$ is  tangent to $ \alpha$), similarly $ B_1B_2$ is the polar of $ B_3$ and $ C_1C_2$ is the polar of $ C_3$, now, the polos are collinear then the polars have a common point, its $ Z$ ($ Z \in A_1A_2$,$ Z \in B_1B_2$, $ Z \in C_1C_2$ , $ Z$ Is the polo of $ \mu$, know we that : si $ \phi$ is polar of $ Y$, then $ OY \perp \phi$, then $ OZ \perp \mu$, but by the LEMMA $ OH \perp \mu$. Then $ Z \in OH$ , then $ Z$ belong to the line of euler of $ ABC$.$ \blacksquare$

Our overall aim is to show that all the lines pass through the pole $ E$ (with respect to the circumcircle) of the radical axis of the circumcircle and the nine-point circle (which, for want of a better name, I call the Euler axis).

Let the circumcircle and nine-point circle of $ ABC$ be denoted by $ \omega_1$ and $ \omega_2$ respectively, let the circle $ B_1C_1A_2$ which is given as tangent to $ \omega_2$ at $ A_2$ be denoted by $ \gamma_1$, and let the circle on diameter $ BC$ (which clearly contains the points $ B_1,C_1$) be denoted by $ \gamma_2$.

[b]Step 1[/b]: These four circles all have a common radical point, implying that the lines $ B_1C_1$,$ BC$ and the tangent at $ A_2$ are concurrent on the Euler axis at a point we denote by $ P$.

Proof: $ \omega_2, \gamma_1, \gamma_2$ are coaxal, all containing the points $ B_1$ and $ C_1$, and thus sharing $ B_1C_1$ as a radical axis. Now intersect $ B_1C_1$ with the Euler axis, the radical axis of $ \omega_1, \omega_2$, to get the point $ P$. By the radical axis theorem applied to $ \omega_1, \omega_2, \gamma_1$, the tangent at $ A_2$ (being the radical axis of $ \gamma_1, \omega_1$) passes through $ P$, and by the radical axis theorem applied to $ \omega_1, \omega_2, \gamma_2$, $ BC$, the radical axis of $ \gamma_2, \omega_1$, passes through $ P$.

[b]Step 2[/b]: $ P$ and $ A_1$ are harmonically conjugate with respect to $ BC$, and, therefore, conjugate with respect to $ \omega_1$.

Let $ H$ be the orthocentre of $ ABC$. Considering $ ABCH$ as a quadrangle, it has diagonal point triangle $ A_1B_1C_1$, and since $ P \equal{} B_1C_1 \cap BC$, the result is an immediate consequence of the harmonic properties of any complete quadrangle (any range of four points in the complete quadrangle configuration is harmonic: in this case we consider the range lying along the line $ BC$).

[b]The Final Push[/b]
Here if we talk of a pole, polar or a conjugacy, we always mean with respect to $ \omega_1$.

Now, since $ PA_2$ is a tangent to $ \omega_1$ at $ A_2$, $ A_2$ lies on the polar of $ P$. Since $ A_1$ is conjugate to $ P$, $ A_1$ also lies on the polar of $ P$, so the line $ A_1A_2$ is the polar of $ P$. But $ P$ lies on the Euler axis, so the pole $ E$ of the Euler axis must lie on the polar of $ P$, so $ E$ lies on the line $ A_1A_2$.

By a similar argument relabeling letters appropriately, $ E$ also lies on the lines $ B_1B_2$ and $ C_1C_2$, so these lines are indeed concurrent at $ E$.

Furthermore, the pole of a line $ l$ with respect to a circle centre $ O$ must lie on the line through $ O$ perpendicular to $ l$. But the line through the centres of two circles is perpendicular to their radical axis. It follows that $ E$ lies on the line containing the centres of the circumcircle and the nine-point circle, which is the Euler line.

:)

(Step 2 Original Proof): We use unnormalised areal (barycentric) co-ordinates (whose basic workings are outlined in my PDF at http://www.bmoc.maths.org/home/areals.pdf or one of the references given therein).

It is well known that the points $ A_1,B_1,C_1$ have areal co-ordinates $ (0,\tan B, \tan C), (\tan A, 0, \tan C), (\tan A, \tan B, 0)$ respectively (to prove this, we remark that $ BA_1 \equal{} 2R\sin C\cos B, A_1C \equal{} 2R\sin B\cos C$ by easy trigonometry, and so $ BA_1: A_1C \equal{} \tan C : \tan B$, whence the co-ordinates $ (0,\tan B, \tan C)$, and similarly for the other two points). Now, $ P$ is the intersection of $ B_1C_1$ with $ BC$. The line $ B_1C_1$ has equation (it may be easily checked that both $ B_1$ and $ C_1$ occupy this line)
\[ \frac {x}{\tan A} \equal{} \frac {y}{\tan B} \plus{} \frac {z}{\tan C}
\]
Intersecting this with the line $ BC$ with equation $ x \equal{} 0$ gives $ \frac {y}{\tan B} \plus{} \frac {z}{\tan C} \equal{} 0$, which gives the co-ordinates of $ P(0, \tan B, \minus{} \tan C)$.
Applying a parameter $ t$ to the line $ BC$ as $ (0, \frac {\tan B}{\tan C}, t)$, we see $ t \equal{} 0,\infty$ at $ B,C$ and $ t \equal{} 1, \minus{} 1$ at $ A_1, P$, so the cross ratio $ (A_1,P;B,C)$ evaluates to $ \minus{} 1$. This is well known in projective homogeneous co-ordinates (of which unnormalised areals are a special case) to characterise a harmonic separation, so we are done.
(for the results used on projective co-ordinates, see E.A. Maxwell, or more recently Christopher Bradley's 'Algebra of Geometry', published by Highperception).

[b]Result quoted in given Step 2 Proof[/b]: Any range in a harmonic quadrilateral is harmonic.
The theorem is easy to prove with projective homogeneous co-ordinates. Let us prove the result needed for the question, and its generality is obvious. We are using projective co-ordinates, so may choose four points without loss of projective structure, so we pick $ A(\minus{}1,1,1), B(1,\minus{}1,1), C(1,1,\minus{}1), H(1,1,1)$. This forces the diagonal point triangle (by intersecting lines like $ AH: y\equal{}z$ with lines like $ BC: y\plus{}z\equal{}0$), to be $ A_1(1,0,0), B_1(0,1,0), C_1(0,0,1)$. Finally, to give the complete quadrangle, we intersect every line present in the current diagram with every other line as much as possible. For example, intersecting $ B_1C_1$ and $ BC$ gives a point $ P(0,1,\minus{}1)$, so $ P,B,A_1,C$ form a range. If we apply a parameter to the line $ BC (1, t, \minus{}t)$ we have $ P,A_1$ corresponding to $ t\equal{}\infty,0$ and $ B,C$ corresponding to $ \minus{}1,1$, so these pairs of points are separated harmonically. We can apply the same logic to any range in the configuration.

Assume firstly the triangle is not isosceles.
Denote $ \{A'\} \equal{} B_1C_1\cap BC$. Define $ B'$ and $ C'$ similarly. Let $ \Gamma(O,R),\Omega(\omega,R/2)$ be the circumcircle, $ 9$-point Circle of $ ABC$, respectively and let $ \rho(X,C)$ denote the power of point $ X$ wrt circle $ C$.

Let's firstly show that $ A',B'$ anc $ C'$ are collinear. Indeed, note that $ BCB_1C_1$ is a cyclic quadrilateral, hence $ \rho(A',\Gamma) \equal{} A'B\cdot A'C \equal{} A'C_1\cdot A'B_1 \equal{} \rho(A',\Omega)$. By similar arguments, we get that $ A',B',C'$, lying outside $ \Gamma$ have the same power wrt $ \Gamma$ and $ \Omega$ thus are collinear lying on the radical axis $ d$ of $ \Gamma$ and $ \Omega$. An important observation is that $ d\bot O\omega$, or $ d\bot OH$. By a very-well known property (which is also very easy to prove), circles of diameters $ OA',OB',OC'$ are concurrent in some point $ Y$. Since we have $ OY\bot YA'$, $ OY\bot YB'$ and $ OY\bot YC'$, and $ A',B',C'\in d$, we obtain $ Y\in d$. Since $ d\bot OH$, $ Y \equal{} OH\cap d$.

Let $ \Delta$ be the circle passing through $ B_1,C_1$ and internally tangent to $ \Gamma$ in $ A_2$. Let $ X$ be the intersection point of the tangent in $ A_2$ to $ \Gamma$ (and hence to $ \Delta$) with $ BC$ (the point exists, since the triangle is not isosceles). Then $ \rho(X,\Gamma) \equal{} XB\cdot XC \equal{} XA_2^2 \equal{} \rho(X,\Delta)$, hence $ X\in d$. Since $ d\cap BC \equal{} \{A'\}$, we get $ X \equal{} A'$.

Consider the inversion $ I_1(A',k)$ with $ k \equal{} \rho(A',\Gamma) \equal{} \rho(A',\Delta) \equal{} \rho(A',\Omega)$. Circles $ \Gamma,\Omega,\Delta$ remain invariant under this inversion. Let $ M\in\Omega$ be the midpoint of $ BC$. Note that $ OM\bot MA'$ and $ OA_2\bot A_2A'$. So $ O,M,A_2,A'$ are inscribed into some circle $ L$, which, inverted becomes a line perpendicular to the diameter $ OA'$. This line is $ A_1A_2$ since $ I_1(M) \equal{} A_1$ and $ I_1(A_2) \equal{} A_2$. So, $ A_1A_2\bot OA'$.

Consider now the [i]beautiful[/i] inversion $ I(O,R^2)$. $ I(A_2) \equal{} A_2$, so $ I(L)$ is a line perpendicular to $ OA'$ (diameter of $ L$). According to the above, this line must be $ A_2A_1$. Since circles of diameters $ OA',OB'$ and $ OC'$ are concurrent in $ Y$, lines $ A_1A_2,B_1B_2$ and $ C_1C_2$ are concurrent in $ I(Y)$. As we have seen, $ OH\bot d$, and $ Y \equal{} OH\cap d$. Clearly then $ I(Y)\in OH$, hence lines $ A_1A_2,B_1B_2,C_1C_2,OH$ are concurrent.

[color=#0000BF]Since $ \angle C_1BB_1\equal{}\angle C_1CB_1\equal{}90^\circ\minus{}A$, it follows that $ BCB_1C_1$ is a cyclic quadrilateral.

$ B_1C_1$ is the radical axis of $ C(B_1C_1A_2)$ and $ C(B_1C_1BC)$

$ BC$ is the radical axis of $ C(ABC)$ and $ C(B_1C_1BC)$

$ A_2A_2$ is the radical axis of $ C(ABC)$ and $ C(B_1C_1A_2)$

(where $ A_2A_2$ is the common tangent of $ C(ABC)$ and $ C(B_1C_1A_2)$).

It follows that there exist a point $ M$ such that $ M$=$ B_1C_1\cap BC\cap A_2A_2$ (M is called the radical centre of the $ C(B_1C_1A_2)$, $ C(ABC)$ and $ C(B_1C_1BC)$).

From Menelaus' theorem we conclude that:
$ \frac{CB_1} {B_1A} \cdot \frac{AC_1} {C_1B} \cdot \frac {MB} {MC} \equal{}1$

From Ceva's theorem we conclude that:
$ \frac{CB_1} {B_1A} \cdot \frac{AC_1} {C_1B} \cdot \frac {BA_1} {A_1C}\equal{}1$

It follows that $ \frac {MB} {MC} \equal{} \frac {A_1B} {A_1C}$, so the division $ (B,C,M,A_1)$ is harmonic. Since $ A_2$ is the tangent point from $ M$ to circle $ C(ABC)$, we conclude that $ A_1A_2$ is the polar of $ M$ relative to circle $ C(ABC)$.

Let $ A'$ be the midpoint of $ (BC)$ and $ {S}\equal{}MH \cap AA'$. Since the division $ (B,C,M,A_1)$ is harmonic, it follows that $ A_1B \cdot A_1C\equal{}A_1A' \cdot A_1M$. But $ A_1B \cdot A_1C\equal{}A_1H \cdot A_1A \Rightarrow \frac {A_1A'} {A_1A}\equal{}\frac {A_1H} {A_1M} \Rightarrow$ tan$ A_1AA'$=tan$ A_1MH$ $ \Rightarrow \angle A_1AA'\equal{}\angle A_1MH \Rightarrow \angle ASM \equal{} \angle AA_1M\equal{} 90^\circ \Rightarrow$ $ MH \perp AA'$ $ (1)$

Let $ O$ be the circumcenter of triangle $ ABC$ and $ G$ be the centroid of triangle $ ABC$. $ {N}\equal{}A_1C_1 \cap AC$ and $ {P}\equal{}A_1B_1 \cap AB$. It is well known that $ H\minus{}G\minus{}O$ is the Euler line of $ ABC$. $ B'$ is the midpoint of $ AC$.

From $ (1)$ $ \Rightarrow MH \perp AA'$ and $ NH \perp BB'$. $ U \in MH \cap AA', V \in NH \cap BB'$. Quadrilaterals $ AMA_1U$ and $ BNB_1V$ are inscribed in the circles of diametres $ AM$, $ BN$, respectively.

So, $ MH \cdot HU\equal{}HA \cdot HA_1$, $ HN \cdot HV\equal{}HB \cdot HB_1$. But $ HA \cdot HA_1\equal{}HB \cdot HB_1$. It follows that $ HM \cdot HU\equal{}HN \cdot HV$, which means that the quadrilateral $ MVUN$ is cyclic. But the quadrilateral $ HGUV$ is also a cyclic one, so it follows that $ MN$ is the image of circle $ C(HGUV$) under the inversion of pole $ H$. As $ HG$ is diameter in $ C(HGUV)$ , we conclude that $ HG \perp MN$ $ (2)$.

From $ (2)$ $ \Rightarrow MP \perp HG \Rightarrow N\minus{}M\minus{}P \perp H\minus{}G\minus{}O$

Let $ N\minus{}M\minus{}P$ be the polar of $ X$ relative to the circle $ C(ABC)$. It follows that $ OX \perp N\minus{}M\minus{}P \Rightarrow X \in H\minus{}G\minus{}O$.

Since $ M$ is on the polar of $ X$ relative to the circle $ C(ABC)$ $ \Rightarrow X$ is on the polar of $ M$ relative to the circle $ C(ABC)$$ \Rightarrow X \in A_1A_2$. Similar, $ X \in B_1B_2$ and $ X \in C_1C_2$.[/color]

First we have to show that lines $ A_{1}A_{2}$ ,$ B_{1}B_{2}$, and $ C_{1}C_{2}$ intersect at the same point.
Denote $ X$ as intersection between $ A_{1}A_{2}$ and $ B_{1}B_{2}$.
Denote $ H$ as orthocenter of $ \triangle ABC$
Denote $ C_{1}^\prime$ as intersection of line $ C_{1}H$ with circumcircle of $ \triangle ABC$
Note that $ |HC_{1}| = |C_{1}C_{1}^\prime|$ $ \leftarrow$  Reflection of orthocenter theorem, (or something like that :maybe: )
Denote circle $ \kappa_{1}$, with diameter $ |HC_{1}^\prime|$, and center at $ C_{1}$
Denote $ C_{2}^\prime$ as intersection of $ \kappa_{1}$ and line $ XC_{2}$, such that $ |XC_{2}^\prime|\geq|C_{2}^\prime C_{2}|$
Denote $ X^\prime$ as intersection of $ \kappa_{1}$ and line $ XC_{2}$, such that $ |XX^\prime|\leq |X^\prime C_{2}|$

Points $ X^\prime,H,C_{2}^\prime,$ and $ C_{1}^\prime$ are cocyclic.
$ \angle C_{1}^\prime X^\prime H = \angle C_{1}^\prime C_{2}^\prime H = \frac {\pi}{2}$ $ \leftarrow$ Thales theorem: as $ C_{1}^\prime H$ is diameter
$ \angle X^\prime C_{1}^\prime H = \angle X^\prime C_{2}^\prime H\leftarrow$ over same chord $ X^\prime H$
$ \angle C_{1}^\prime X^\prime C_{2}^\prime = \angle C_{1}^\prime H C_{2}^\prime\leftarrow$ over same chord $ C_{1}^\prime C_{2}^\prime$
$ \angle C_{1} C_{1}^\prime X^\prime = \angle C_{1} X^\prime C_{1}^\prime \leftarrow$ isosceles $ \triangle C_{1} C_{1}^\prime X^\prime$
$ \angle C_{1} X^\prime H = \angle C_{1} H X^\prime \leftarrow$ isosceles $ \triangle C_{1} X^\prime H$
$ \angle C_{1}^\prime X^\prime H = \angle C_{1}^\prime X^\prime C_{1} + \angle C_{1} X^\prime H = \angle C_{1} H C_{2}^\prime + \angle C_{1} H X^\prime = \angle X^\prime H C_{2}^\prime = \frac {\pi}{2}$
Hence $ X^\prime C_{2}^\prime$ is also diameter. Therefore $ C_{1}$ lies on that line.
But, because $ X^\prime$ and $ C_{2}^\prime$ lie on line $ X C_{2}$, point $ C_{1}$ lie on that line too.
Therefore $ X$ is also point on line $ C_{1} C_{2}$
Thus we proved that all $ 3$ lines intersect at same point.

Now we have to show that $ X$ lies on Euler's line of $ \triangle ABC$
We know that $ O$-circumcenter, $ H$-orthocenter, $ T$-centroid, and $ O_{e}$-center of $ 9$-point circle, are on that line.
Its enough to show that $ \angle XHO = \pi$
Denote $ E_{A},E_{B},E_{C}$ as Euler's points to the corresponding vertex of $ \triangle ABC$
Denote $ \alpha = \angle E_{A}HC_{1}$
Denote $ \beta = \angle C_{1}HX$
Denote $ \gamma = \angle XHE_{B}$
Denote $ \delta = \angle E_{B}HA_{1}$
Denote $ Y$ intersection between $ C C_{1}$ and $ E_{A} E_{B}$
Denote $ W$ intersection between Euler's line and $ E_{C} E_{A}$
Denote $ Z$ intersection between Euler's line and $ E_{A} E_{B}$

Note: $ \alpha + \beta + \gamma + \delta = \pi$
Because $ C_{1}, H, C$ are collinear $ \angle C_{1}, H, C = \pi \Rightarrow \angle A_{1} H E_{C} = \alpha$
Because $ E_{B} H B_{1}$ are collinear $ \angle E_{A} H B_{1} = \pi \Rightarrow \angle E_{A} H B_{1} = \delta$
$ \angle E_{A} E_{B} E_{C} = 2\pi - \frac {\pi}{2} - \frac {\pi}{2} - (\beta + \gamma + \delta) = \pi - \alpha - \beta - \gamma - \delta + \alpha = \alpha$
$ \angle E_{B} E_{A} E_{C} = 2\pi - \frac {\pi}{2} - \frac {\pi}{2} - (\alpha + \delta) = \pi - \alpha - \beta - \gamma - \delta - \alpha + \beta + \gamma = \beta + \gamma$
Note: $ \angle HYE_{B} = \frac {\pi}{2}$
$ \angle WE_{B}H = \pi - (\beta + \gamma) - \frac {\pi}{2} = \frac {\pi}{2} - \beta - \gamma$
$ \angle HZE_{B} = \pi - \angle WE_{B}H - \angle ZHE_{B} = \pi - \frac {\pi}{2} + \beta + \gamma - \gamma = \frac {\pi}{2} + \beta$
$ \angle E_{C}W H = 2\pi - \angle WE_{C}E_{B} - \angle E_{C}E_{B}Z - \angle HZE_{B} = 2\pi - \delta - \alpha - \frac {\pi}{2} - \beta =$   $ \frac {3\pi}{2} - \alpha - \beta - \delta - \gamma + \gamma = \frac {\pi}{2} + \gamma$
$ \angle E_{A}E_{C}Z = \pi - \angle E_{C}E_{A}Z - E_{C}ZE_{A} = \pi - (\beta + \gamma) - \frac {\pi}{2} = \frac {\pi}{2} - \beta - \gamma$ (Note:because $ Y$ lies on  line $ E_{A}E_{B}$ $ \angle E_{C}E{}_{A}{}_E{B} = \angle E_{C}E_{A}Z$)
$ \angle ZHE_{C} = \pi - \angle ZE_{C}H - \angle E_{C}ZH = \pi - \frac {\pi}{2} + \beta + \gamma - \frac {\pi}{2} - \gamma = \beta$ (Note:because $ Z$ lies on  line $ E_{A}E_{C}$ $ \angle E_{A}E_{C}E{B} = \angle ZE_{C}E_B$)
Since $ Z$ is on Euler's line $ \angle ZHE_{C} = OHE_{C}$

$ \angle XHO = \angle XHE_{B} + \angle E_{B}HA_{1} + \angle A_{1}HE_{C} + \angle E_{C}HO = \gamma + \delta + \alpha + \beta = \pi$
Thus points $ XHO$ are collinear, and because line $ HO$ is Euler's line point $ X$ lies on that line.

Let $ A_3$ be the intersection point of the lines $ B_1C_1$ and $ BC$, if this dont exist consider it as a point at infinity. Define similarly $ B_3$ and $ C_3$, $ w$, $ w_1$ are the circumcircle of ABC and the nine point circle of $ ABC$(with center $ Z$), finally define: $ w_a$ the circle passing throught $ B_1$, $ C_1$ and tangent to $ w$. Define similarly $ w_b$ and $ w_c$.
Consider $ w_a$, $ w$ and the circle passing throught $ B_1, C_1, B, C$, the three radical axis between these circles must concur, then $ B_1C_1$, $ BC$ and the tangent passing throught $ A_2$ to $ w$ concur at $ A_3$, now consider $ w_1$, $ w$ and $ w_a$, the 3 radical axis must concur at $ A_3$, then $ A_3$ belongs to the radical axis of $ w_1$ and $ w$, similar resoaning to $ B_3$ and $ C_3$.
Suppose $ A_3A_2$ and $ A_3A_4$ are tangent to $ w$ and $ R$ is the intersection point of $ A_2A_4$ and $ BC$, we have $ BA_4CA_2$ is an harmonic quadrilateral, so $ (A_3,R,B,C)\equal{}\minus{}1$, also B and C are the feet of the angle bisectors of the triangle $ A_3B_1A_1$, so $ (A_3,A_1,B,C)\equal{}\minus{}1$, then $ R\equal{}A_1$, this implies $ A_2A_1$ is the polar line(respect to $ w$) of $ A_3$, for duality as $ A_3,B_3,C_3$ are collinear then the polar lines concur at $ X$, and the polar line of $ X$ is the radical axis of $ w$, $ w_1$, then $ OX$ is perpendicular to the radical axis of $ w$, $ w_1$, finally $ OZ$ is perpendicular to the radical axis of $ w$, $ w_1$, then O, Z and X are collinear, this proves the result because $ OZ$ is the Euler line of triangle $ ABC$.

Let $ X \equal{} B_1C_1\cap BC$, $ Y \equal{} C_1A_1\cap CA$, $ Z \equal{} A_1B_1\cap AB$, and call $ H$ the orthocentre of triangle $ ABC$.

From the concurrency of $ AA_1$, $ BB_1$ and $ CC_1$, we conclude that $ (BCA_1X)$ is a harmonic division. Therefore $ A_1$ lies on the polar of $ X$ relative to the circumcircle $ \Gamma$ of triangle $ ABC$. $ (1)$

On the other hand, we have the quadrilateral $ BCB_1C_1$ is cyclic (because of $ \measuredangle BB_1C \equal{} \measuredangle BC_1C \equal{} 90^{\circ}$), which implies that $ X$ is the radical center of the circumcircles of triangles $ ABC$, $ B_1C_1A_2$ and $ BCB_1$. It means $ XA_2$ is tangent to the circle $ \Gamma$ at $ A_2$. In other word, $ A_2$ lies on the polar of $ X$ relative to the circle $ \Gamma$. $ (2)$

From $ (1)$ and $ (2)$ we conclude that $ A_1A_2$ is the polar line of $ X$ with respect to $ \Gamma$. Similarly, $ B_1B_2$ is the polar of $ Y$ and $ C_1C_2$ is the polar of $ Z$ relative to $ \Gamma$. Hence, to show that $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ concur, we will prove that points $ X$, $ Y$, $ Z$ are collinear.

Indeed, let's call $ O$ and $ N$ the circumcenters of triangles $ ABC$ and $ A_1B_1C_1$, respectively. From the cyclic quadrilateral $ BCB_1C_1$, we have $ XB\cdot XC \equal{} XB_1\cdot XC_1$, it implies that $ X$ lies on the radical axis of circles $ (O)$ and $ (N)$. Similarly, $ Y$ and $ Z$ also lie on the radical axis of circle $ (O)$ and $ (N)$. In conclusion, $ X$, $ Y$, $ Z$ lie on a line perpendicular to $ ON$. Notice that $ N$ is the nine-point center of triangle $ ABC$, so $ O$, $ N$, $ H$ are collinear, so we conclude that $ OH\perp\overline{XYZ}$ $ (*)$

From the collinearity of $ X$, $ Y$, $ Z$, we obtain that lines $ A_1A_2$, $ B_1B_2$ and $ C_1C_2$ concur at the pole of line $ \overline{XYZ}$ relative to the circle $ \Gamma$, let it's $ P$. It implies $ OP\perp \overline{XYZ}$ $ (**)$

Combining $ (*)$ and $ (**)$ we deduce that $ O$, $ H$ and $ P$ are collinear, i.e. lines $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ have a common point, which lies on the Euler line of $ \triangle ABC$, as desired.

Let $ B_{1}C_{1} \cap BC \equal{} A_{3}$, and so on.
Let's consider a three radical axis of circumcircle of $ ABC$ ( $ \equal{}w$), $ BCB_{1}C_{1}$ and $ B_{1}C_{1}A_{2}$.
Then we can show that $ B_{1}C_{1}$, $ BC$ and a line passing through $ A_{2}$ and tangent to $ w$ concur at a point and it should be a $ A_{3}$. Therefore if a line passing through $ A_{3}$ and tangent to $ w$ at arc $ BC$ was drawn, tangent point will be $ A_{2}$.
In addition, because $ A_{3}, B, A_{1}, C$ is harmonic, polar of $ A_{3}$ wrt $ w$ passing through $ A_{1}$.
So the line $ A_{2}A_{1}$ is the polar of $ A_{3}$ wrt $ w$. $ B_{3}, C_{3}$ also satisfy same property.

But by Desargue's Theorem, $ A_{3}, B_{3},C_{3}$ lie on a line. Let this line $ l$.
Now it is obvious that $ A_{1}A_{2}, B_{1}B_{2}, C_{1}C_{2}$ concur at a point. To show the rest part, It is enough to show that $ l \perp l_{e}$. ($ l_{e}$ denotes the Euler Line of $ ABC$) ( $ l_{e}$ contains $ O$, the circumcenter of $ ABC$)

Let $ H$ be a orthocenter. Let $ M, N$ be a point such that $ MB_{1} \parallel C_{1}B_{3}$, $ NC_{1} \parallel B_{1}C_{3}$, $ M \in AB$, $ N\in AC$.
Then $ \angle BMB_{1} \equal{} \angle BC_{1}A_{1} \equal{} \angle C \equal{} \angle AC_{1}B_{1}$. So $ B_{1}M \equal{} B_{1}C_{1}$.
By the same way, we get $ MB_{1} \equal{} B_{1}C_{1} \equal{} C_{1}N$.
Let $ O'$ be a symmetric point of $ O$ wrt $ AC$. Then $ O'$ is the circumcenter of $ AHC$. so $ OC\equal{}CO'\equal{}O'H$.
Furthermore, $ \angle A_{1}O'C \equal{} 2\angle HAC \equal{}\angle MB_{1}C_{1}$ and $ \angle OO'C \equal{} \angle B \equal{} \angle BB_{1}N$.
Now, $ B_{1}C_{1}MN \equiv O'CHO$. But $ NB_{1} \perp OO'$. Therefore $ MN \perp OH \equal{} l_{e}$.
And $ MN \parallel B_{3}C_{3}$ ( $ \frac{AM}{AC_{1}} \equal{} \frac {AB_{1}}{AB_{3}}$, $ \frac{AN}{AB_{1}} \equal{} \frac{AC_{1}}{AC_{3}}$)
Hence $ B_{3}C_{3}\perp l_{e}$, as desired.


[quote="GameBot"]The cost of 10 pencils is $ \$0.84$. At the same rate, what is the
cost, in dollars, of 25 pencils? Express your answer as a
decimal to the nearest hundredth.[/quote]

$ .84 \cdot \frac{5}{2}\equal{}1.68\plus{}.42\equal{}2.1$.


Solve for $\frac{1}{x}$ and $\frac{1}{y}$ instead, and then invert them.

[hide="method 1"]Add the equations and stuff cancels[/hide]
[hide="method 2"]Let $\frac{1}{x}=a$ and $\frac{1}{y}=b$[/hide]

[quote="linzk3"]if (1/x)-(1/y)=1/3 and (1/x)+(1/y)=1/4 solve for x and y...i am having trouble with getting started...any tips on how to solve?[/quote]
[hide]$\frac1x+\frac1y=\frac14;\,\frac1x-\frac1y=\frac13$

Add the equations to get $\frac2x=\frac7{12}$, which leads to $x=\frac{24}7$ and then $y=-24$.[/hide]
[hide]$\frac1x+\frac1y=\frac14;\,\frac1x-\frac1y=\frac13$

Let $a=\frac1x$ and $b=\frac1y$.

$a+b=\frac14;\,a-b=\frac13$

Again, add the equations to get $2a=\frac7{12}$, or $a=\frac7{24}$. Then, find $b=-\frac1{24}$.

Take the reciprocals of $a$ and $b$ to get $x=\frac{24}7$ and $y=-24$.[/hide]


Euler
trigonometry
algebra
polynomial
calculus
calculus computations

for the second one(the limit)
use the fact that e^x=1+x/1!+x^2/2!+....+x^n/n!

then the limit is  e

[quote="color"][quote="Carcul"]$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{2}\pi^{2}}\right) = \sinh(1)$.[/quote]
How  :| ?[/quote]

He used the infinite product formula.

You know the Euler-Wallis identity for $\sin \pi n$ but here it is for $\sinh$

$\frac{\sin x}{x}= \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)$.

Now consider x = i (or -i, which will do the same) and we get

$\frac{\sin (i)}{i}= \prod_{n=1}^{\infty}\left(1+\frac{1}{n^{2}\pi^{2}}\right)$.   (1)

Now, we know that $\sin \theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$.
Make the substitution $\theta = ix$ and we get a formula for the sine of a complex angle: $\sin(ix) =-\frac{\sinh x}{i}= i\sinh x$, and so

$\sinh x = \frac{\sin(ix)}{i}$.

For x = 1, and from expression (1), we get that

$\prod_{n=1}^{\infty}\left(1+\frac{1}{n^{2}\pi^{2}}\right) = \frac{\sin(i)}{i}= \sinh(1)$.

[quote="color"]Ok, thanks , but could you please
explain how the original product formula holds at all?
I mean:
$\frac{\sin x}{x}= \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)$.

(I googled "Euler-Wallis identity" and only got 2 results, which did not contribute.)[/quote]

The idea is that $\frac{\sin x}{x}$ is an infinite polynomial. (Look at its power series).

Then you can factor a polynomial if you know its roots.
The roots of $\frac{\sin x}{x}$ are $\pi n$ for $n\no = 0$.

[quote="color"]Aha I remember that from Euler's proof of the Basel problem.
Thanks  :lol: . Or am I right?[/quote]

Yeah you right! Euler so brilliant. But his solutions was not rigorous :)
(You should use Weierstrass Factorization Theorem).


[hide=" "][/hide][hide=" "][/hide][hide=" "][/hide][hide=" "][/hide][hide=" "][/hide]

There are no mods in this forum, so only admins can delete posts here. Just leave it; it'll sink down eventually.

Maybe if Astro becomes the mod he can delete this!


Find $ f(x)$ such that $ f(x)^2f(\frac{1\minus{}x}{1\plus{}x}) \equal{} 64x$ for $ x$ not $ 0,1,\minus{}1$.

Use the fact that $ g(g(x))\equal{}x$ with $ g(x)\equal{}\frac{1\minus{}x}{1\plus{}x}$.


[quote="nguoivn"]Here is my gift for my younger brothers: Honey_S (Loc) and tranquoc (Luat).
Given $ a, b, c > 0$. Prove that:
$ \sqrt {\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc}} \geq\ \frac {4}{3} (\sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}})$

PS: Think about Am-Gm, my brothers  :)[/quote]
The inequality is equivalent to
\[ mnp(m\plus{}n\plus{}p) \le \frac{3}{4}\]
where $ m\equal{}\sqrt{\frac{a}{b\plus{}c}}$ and $ n,p$ are similar.
As $ 3mnp(m\plus{}n\plus{}p) \le (mn\plus{}np\plus{}pm)^2$, we can deduce the inequality to
\[ mn\plus{}np\plus{}pm \le \frac{3}{2}\]
which is trivial as
\[ 2mn \equal{}2\sqrt{ \frac{bc}{(a\plus{}b)(a\plus{}c)}} \le \frac{b}{a\plus{}b}\plus{}\frac{c}{a\plus{}c}\]
Similar to $ np$ and $ pm$, and adding them up, we will get the result. :)


function
real analysis
real analysis unsolved

Does there exist an [b]onto[/b] map from $ R$ to $ R^2$?
Does there exist a [b]one-one[/b] and [b]onto[/b] map?
What is the map, if so? If not, why not?

It should be easy to show a mapping from $ \mathbb{R}^2$ onto $ \mathbb{R}^1.$ A simple projection will do.

One idea for constructing a mapping from $ \mathbb{R}^1$ onto $ \mathbb{R}^2$ is to express a real number as a decimal expansion and make an ordered pair out of the two numbers you get by taking only the even-index digits and only the odd-index digits, compressing each into a decimal expansion. (This isn't quite one-to-one because of the non-uniqueness of some decimal representations.)

By the Cantor-Schroeder-Bernstein theorem (look it up!) it follows that there exists a one-to-one correspondence between the two sets. That is, they have the same cardinality. Alternatively, you could try starting with the idea in my second paragraph and tinkering with it to fix it.

If you start imposing extra conditions on this function, such as continuity, then the story changes.


trigonometry
inequalities
inequalities solved

easy,i bet
find the maximum and minimum of  of
sinA+sinB+sinC+sinD+sinE+sinF
subject to the conditionA+B+C+D+E+F=2 \pi  and
 A,B,C,D,E,F  \varepsilon [0, \pi /2]

Are you sure it's so easy? I don't find an easy job to find the minimum value!

well i used karamata,and now that u bring it up i'll chek my proof,k

Do you consider Karamata easy? I think it is one of the most difficult inequalities. But I'm really interested in how you found the minimum using it.

What is Karamata's Inequality? >_< :blush:
Nevermind at all!!! I found out that this inequality was discussed a long time ago on this site!!! Link is below:
http://www.mathlinks.ro/viewtopic.php?t=430

this is what i did
f(x)=sin x is concave on [0, \pi /2]
and [\pi /2,\pi /2,\pi /2,\pi /2,0,0]>[A,B,C,D,E,F]>[2 \pi /6,2 \pi /6,2\pi /6,2 \pi /6,2 \pi /6,2 \pi /6]
now apply karamata's inequality
i never said proving karamata was easy.even i consider it pretty difficult.btw,is one required to prove it during contests
hope the proof is correct


The degree of (x2 + 1)^4(x3 + 1)^3 as a polynomial in x is

A. 5; B. 7; C. 12; D. 17; E. none

All of the coefficients are positive so nothing's go to cancel.  $ (x^2\plus{}1)^4$ has degree 8 and $ (x^3\plus{}1)^3$ has degree 9.  So their product has degree...


search
number theory unsolved
number theory

[b]a)[/b]Let $p$ be a prime number and define $\forall n\; 1<n<p-1\; A_{n}=\{k|1\le k\le p-1 \;\; ord_{p}k=n\}$. Find the cardinal of $A_{n}$.

[b]b)[/b]Find the number of all natural $n$ less than $54$ for which $(n, 540)=1,\; ord_{54}n=9$.

a) follows either directly from some proofs of cyclicity of $(\mathbb Z/ p \mathbb Z)^{*}$, or from the cyclicity itself; search the forum for the proofs.

b) follows after you solved a) with the Chinese Remainder Theorem.

[b]Basic problem[/b].
The number $k$ such that $ord_{p}{k}=d$ for $d|{p-1}$ is $\phi(d)$.
Solution.
In any book about number theory.


[quote="Virgil Nicula"]$ 1\blacktriangleright\left\|\begin{array}{c} \triangle\ ABC\  ,\ b\le 2c\ : \ M\in (BC) \\\
 \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \\\\
m(\widehat {BAM})\equal{}60^{\circ}\\\\
\frac {MB}{a \plus{} b \plus{} c} \equal{} \frac {MC}{2a \minus{} b \plus{} 2c}\end{array}\right\|\implies m(\widehat {BAM}) \equal{} \frac B2\ .$[/quote]

[quote="Virgil Nicula"]$ 2\blacktriangleright\left\|\begin{array}{c} \triangle\ ABC\  ,\ b\le 2c\ : \ M\in (BC) \\\
 \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \\\\
m(\widehat {BAM}) \equal{} \frac B2\\\\
\frac {MB}{a \plus{} b \plus{} c} \equal{} \frac {MC}{2a \minus{} b \plus{} 2c}\end{array}\right\|\implies A\equal{}60^{\circ}\ .$[/quote]

[quote="Virgil Nicula"]$ 3\blacktriangleright\left\|\begin{array}{c} \triangle\ ABC\  ,\ b\le 2c\ : \ M\in (BC) \\\
 \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \\\\
A\equal{}60^{\circ}\\\\m(\widehat {BAM}) \equal{} \frac B2\\\\
\end{array}\right\|\implies \frac {MB}{a \plus{} b \plus{} c} \equal{} \frac {MC}{2a \minus{} b \plus{} 2c}\ .$[/quote]

See http://www.mathlinks.ro/viewtopic.php?t=170435

I think that algebra helps solving your problem.

From $ \frac{MB}{a\plus{}b\plus{}c}\equal{}\frac{MC}{2a\minus{}b\plus{}2c} \equal{} \frac{MB\plus{}MC}{3a\plus{}3c} \equal{} \frac{MC\minus{}MB}{a\minus{}2b\plus{}c}$

If M lies between B and C then $ MB \equal{} \frac{a(a\plus{}b\plus{}c)}{3a\plus{}3c}$

If $ M \notin (B,C)$ then $ |MB| \equal{} \frac{a(a\plus{}b\plus{}c)}{a\minus{}2b\plus{}c}$


What are both primes $p>0$ for which $\frac{1}{p}$ has a purely periodic decimal expansion with a period 5 digits long? [Note: $\frac{1}{37}=0.\overline{027}$ starts to repeat immediately, so it's purely periodic.  Its period is 3 digits long.]

[hide="Solution"]Five-digit periodic numbers have the form ${k\over 99999}=\frac{k}{3^{2}\cdot 41\cdot 271}$, hence the desired numbers are $41$ and $271$: ${1\over 41}=0.\overline{02439},{1\over 271}=0.\overline{00369}$[/hide]


Prove that the remainder of a division of a prime number by 30 is either 1 or a prime number.

Denote $p$ is this prime then $p=30k+r$, where $r=0,1,2,...,29$ and ...


A set of 25 square blocks is arranged into a 5x5 square. How many different combinations of 3 blocks can be selected from that set so that no two are in the same row or column?

ps. is it 3600 or am i overcounting?


The owner of a movie theater was counting the money from 1 day's ticket sales. He knew that a total of 150 tickets were sold. Adult tickets cost $7.50 each and children's
tickets cost $4.75 each. if the total receipts for the day were $891.25, how many of each
kind of ticket were sold?

[quote="symmetry"]The owner of a movie theater was counting the money from 1 day's ticket sales. He knew that a total of 150 tickets were sold. Adult tickets cost 7.50 each and children's
tickets cost 4.75 each. if the total receipts for the day were $891.25, how many of each
kind of ticket were sold?[/quote]

[hide]Let x=child tickets, y=adult tickets

x+y=150
4.75x+7.5y=891.25

x=85, y=65. [/hide]

This is a two equations in two variables.
Thanks.


Determinati x,y intregi  daca:

3x - y = 56
[x;y] - 4(x;y) = 154

$\ [x,y]\cdot\ (x,y)=xy$
$\ [x,y]=d\Rightarrow\ x=dm,y=dn, (m,n)=1$


limit
trigonometry
logarithms
calculus
calculus computations

[u]Without L'Hopital's Rule or  Taylor Expansion[/u] find
$ \lim_{x \to 0^{\plus{}}}\frac{x^{(x^x)}\sin{x}}{\tan^2{2x}}$.

Same as the limit of $ \frac{1}{4}x^{x^x\minus{}1}$. The logarithm of $ x^{x^x\minus{}1}$ is $ \ln x(e^{x\ln x}\minus{}1)$, which is asymptotic to $ x(\ln x)^2$, which has limit $ 0$. Thus, the answer is $ \frac14$.

$ \lim_{x\to\infty}\ \frac {\ln x}{x} \equal{} 0\ \ \ \ ;\ \ \ \lim_{x\searrow 0}\ x^x\equal{}1$ .

$ \lim_{x\searrow 0}\ x\ln x \ \stackrel {\left(x: \equal{}\frac 1x\right)}{\equal{}}\  \lim_{x\to\infty}\ \frac 1x\cdot\ln\frac 1x \equal{} \minus{} \lim_{x\to\infty}\ \frac {\ln x}{x} \equal{} 0$ .

$ \lim_{x\searrow 0}\ x\ln^2x\ \stackrel {\left(x: \equal{}x^2\right)}{\equal{}}\ \lim_{x\searrow 0}\ x^2\ln^2x^2 \equal{} 4\cdot\lim_{x\searrow 0}\ (x\ln x)^2 \equal{} 0$ .

$ L\equiv\lim_{x\searrow 0}\ \left(x^x \minus{} 1\right)\ln x \equal{} \lim_{x\searrow 0}\ \frac {e^{x\ln x} \minus{} 1}{x\ln x}\cdot x\ln^2x \equal{} 0$ .

$ \lim_{x\searrow 0}\ x^{x^x \minus{} 1} \equal{} e^L \equal{} 1$ .

$ \lim_{x\searrow 0}\ \frac {x^{x^x}\sin x}{\tan^22x} \equal{} \lim_{x\searrow 0}\ \frac {\sin x}{x}\cdot\left(\frac {2x}{\tan 2x}\right)^2\cdot \frac x4\cdot x^{x^x} \equal{} \frac 14\cdot\lim_{x\searrow 0}\ x^{x^x \minus{} 1} \equal{} \frac 14$ .


An aircraft flies with a constant airspeed(speed of aircraft in still air) $200km/hr$ from position $A$ to position $B$,which is 100km northeast of $A$ and then flies back to $A$.Throughout the whole flight the wind velocity is $60km/hr$ from the west.Find the total time of flight from $A$ to $B$ ands back.

hey! I don't think there's a hitch in this problem.
assuming the aircraft to move along the same straight path from A to B and then back to A, the total distance travelled is
100+100=200km
the velocity of the aircraft will be 200-60=140kmph(given that both r constant)
hence the total time taken for flight from A to B and back should be
T=s/v;
i.e200/140=1.42hr(hope that helps)
(But its difficult to practically imagine the situation as it seems quite impossible that 60kmph wind velocity can affect an aircraft of huge mass)     ;)
 please reply and tell me whether I'm right.
as i'm a new user I would like to get a reply and hence keep the interaction going on.  :)
Sorry I haven't yet got the latex .[/quote]

well technically, yes
it does seem rather improbable that a feeble wind should change the course of
an aircraft, but half the assumptions in physics are improbable anyway.
    you could try the vector approach however. draw $AB$ as the resultant
flight vector inclined at $45$ degreed to the horizontal, and a $60 km/hr$
vector to the right i.e. in east-west direction. then join the tail of this vector to
$B$ and you have the true flight vetor, which is $200 km/hr$ then solve using cosine rule.
   apeksha, you can LaTeXify anything you want by putting it between two $--$ marks
like $(x^2 - 4y)$ look into top left bar for details.
madatmath


[img]http://i32.tinypic.com/znq0qh.jpg[/img]

[hide="my soln. but incorrect"]
a) $ \vec{D} \equal{} \vec{A} \plus{} \vec{B} \plus{} \vec{C}$

consider, $ |\vec{A}| \equal{} A_x \plus{} A_y$
$ A_x \equal{} 100cos30^o$
$ A_y \equal{} 100sin30^o$

///ly, $ B_x \equal{} \minus{} 80cos60^o$
$ B_y \equal{} \plus{} 80sin60^o$

$ C_x \equal{} \minus{} 40cos53^o$
$ C_y \equal{} \minus{} 40sin53^o$

$ D_x \equal{} A_x \plus{} B_x \plus{} C_x \equal{} 22.52$
$ D_y \equal{} 87.33$

$ |\vec{D}| \equal{} \sqrt {87.33^2 \plus{} 22.52^2}$
$ \equal{} 90.20 N$

b) $ tan\theta \equal{} \frac {87.33}{22.52}$
$ \equal{} 75.54^o$
???????
[/hide]

Your $ B_y$ should be positive, not negative.

[quote="bos1234"]
///ly, $ B_x \equal{} \minus{} 80cos60^o$
[/hide][/quote]

By is only 80 sin60 and not -80sin60, because if you resolve it, its component is in the the positive y direction only.

Your first answer turns out to be right because, though you've negated the value, since you square, mag remains the same.

But how abouot the second qn? To find the direction

After I change $ B_y$ to $ (\plus{})$ and add all the vectors
$ D_x\equal{}22.52
D_y\equal{}87.33$

But $ tan^{\minus{}1}\frac{87.33}{22.52}\equal{}75^o$

$ \vec{Fnet} \equal{} 22 \hat{i} \plus{} 87 \hat{j}$
So required force to counter this

$ \vec{Fnet} \equal{} \minus{}22 \hat{i} \plus{} \minus{}87 \hat{j}$

So this is obviously in the third quadrant. (because both x and y are negative)

So ans is tan(180+theta).. Theta is 75.

ah kkkk i understand now

thanks alot man!

You can use [url=http://noether.phas.ubc.ca/demo/demos/LamiTheorem/LamiTheorem.html]Lami's thorem[/url].


function HMMT Support calculus algebra polynomial derivative

I'm not sure if there is a nice solution to this problem, but it seems like a recursive approach might be useful.

Let $n$ be a positive integer. Find a closed form expression for

$\sum_{d|n}2^{d}$

Unfortunately, $2^{d}$ is not a multiplicative arithmetic function, so I doubt this has a nice solution.  Where is the problem from?

I came across it when I was working on another problem. actually, the problem it popped up in was the series question on the HMMT contest this year. but i'm pretty sure that the method i was using was not the best one for figuring out the problem.

[quote="t0rajir0u"]Hmm.  What was the original question?[/quote]

$w$ is a primitive $2007$th root of unity. Find $(2^{2007}-1) \sum_{k=1}^{2006}\frac{1}{2-w^{k}}$.

Uh, that answer is wrong

to avoid all that lame stuff (I mean calculus, that is)

[hide]

If you add  to the summation (so it goes from 1 to 2007), then the summation is the sum of hte roots of the polynomial

[/hide]

EDIT: Nevermind, my pairing argument unfortunately doesn't work...

[quote="Elemennop"]Uh, that answer is wrong

to avoid all that lame stuff (I mean calculus, that is)[/quote]

It's been fixed.  And I hardly see how expanding a geometric series is more advanced than the concept of complex roots of unity.

complex roots doesn't really play that much into my solution...I didn't really read your solution that much either, it just looked somewhat confusing

I was mainly pointing out how bad the official solution was (it was a calc test Q, and when I saw the solution i was like wtf this is pointless, why are you tkaing the derivative (btw what is a derivative)) so anything that doesn't use calculus on taht problem is probably better anyways

Hmm.  What's the official solution?  Yours is quite nice, so I'm not sure why the test-makers didn't consider that.

Edit:  [hide="Hmm, I think I know how it starts"] We consider the function

$f(x) = (x^{2007}-1) \left( \sum_{k=1}^{2006}\frac{1}{x-\omega^{k}}\right)$

Note that $x^{2007}-1 = \prod_{k=0}^{2006}(x-\omega^{k})$, so now the motivation for the problem becomes apparent; $f(x)$ is a polynomial.

In fact, we suspect from the other methods that $f(x) = 2005 \cdot x^{2006}+1$.  [/hide]

Edit 2:  [hide="How about integration"] Rather, consider the function

$g(x) = \sum_{k=0}^{2006}\frac{1}{x-\omega^{k}}$

(Note again the $0$ index.)  It's fairly clear that

$\int g(x) \, dx = \sum_{k=0}^{2006}\ln (x-\omega^{k})+C = \ln \left( \prod_{k=0}^{2006}(x-\omega^{k}) \right)+C = \ln (x^{2007}-1)+C$

Taking the derivative again,

$g(x) = \frac{2007x^{2006}}{ x^{2007}-1}$

Then proceed as in my previous solution. [/hide]

Their solution basically combined all the fractions in the summation, and then realized that the numerator is a derivative of the polynomial z^2006 + z^2005 + .... + z + 1.

Ah, I see.  That's basically my second solution.


calculus
integration
trigonometry
calculus computations

[quote="kunny"]Evaluate

\[\sum_{n=1}^{\infty} \int_{2n\pi}^{2(n+1)\pi} \frac{x\sin x+\cos x}{x^2}\ dx\ (n=1,2,\cdots)\][/quote]

I got $\frac{1}{{2\pi }}$.  It looked simple to me.  I used integration by parts for the second term in the integrand and first term gets cancelled in the process.

Is that right answer kunny?

Of course you are right,vidyamanohar. :)

Quotient rule:

$\frac{d}{dx}\left(\frac{-\cos x}x\right)=\frac{x\sin x+\cos x}{x^2}.$


geometry
parallelogram
angle bisector
geometry unsolved

Let ABCD be a parallelogram, M the midpoint of AB and N the intersection of CD
and the angle bisector of ABC. Prove that CM and BN are perpendicular iff AN is the angle bisector of DAB.

The angle bisector of BAD meet DC on K.

1) $ CM \perp BN \ \Longrightarrow \ K \equiv N$: clearly $ AN \perp BN$, so we must have $ AK \parallel CM$ and then $ \angle BMC \equal{} \frac{\angle DAB}{2}\equal{} 180 \minus{} \angle MBA \minus{} \angle BMC \equal{} \angle BCM$ and then $ \overline{AB}\equal{}2 \overline{BC}$. Call J the midpoint of CD, so BCJM and MJDA are rombus, so the angle bisector of $ \angle MAC$ and of $ \angle MAD$ pass through J (i.e. $ K \equiv N \equiv J$)

2) $ K \equiv N \ \Longrightarrow \ CM \perp BN$: we get $ \angle NBC \equal{} \angle BNC$ and $ \angle NAD \equal{} \angle AND$, from which N is the midpoint of DC, from which BCNM and MNDA are rombus, from which $ CM \perp BN$


algebra
polynomial
Ring Theory
superior algebra
superior algebra theorems

find all maximal ideals of C[x].then find all maximal deals of C[x1,x2,...,xn].         (C is complex number and C[x] is polynomials ring) :|

Any maximal ideal in $\mathbb{C}[x_{1},x_{2},...,x_{n}]$ is of the form $(x_{1}-z_{1},x_{2}-z_{2},...,x_{n}-z_{n})$ for some $z_{1},z_{2},...,z_{n}\in \mathbb{C}$:

Lety $I$ be our maximal ideal.Then by weak Hilbert Nullstellensatz $V(I) \neq \emptyset$, let $(z_{1},z_{2},...,z_{n}) \in V(I)$, then $(x_{1}-z_{1},x_{2}-z_{2},...,x_{n}-z_{n}) = I( \{z_{1},z_{2},...,z_{n}\}) \supset I(V(I)) = \sqrt{I}= I$, ($\sqrt{I}$ is the radical ideal of $I$, $I$ is maximal, thus prime and $\sqrt{I}=I$), so since our $I$ is maximal we indeed have that
$I=(x_{1}-z_{1},x_{2}-z_{2},...,x_{n}-z_{n})$ .


A bacterial colony grows exponentially, tripling every x years, $x>0$.  What is the ratio of the population after $\frac{x}{3}$ years to the population after $\frac{2x}{3}$ years?  (Assume that the initial population is greater than zero.)

[hide]Let $P(t)$ denote the popultation after $t$ years and $P_0$ the initial population.  From the given conditions in the problem, we have:

$P(x)=3(P_0)$
$P(2x)=9(P_0)=3^2(P_0)$

We can formulate the general formula for the problem as: $P(nx)=3^n(P_0).$ for some positive real value of $k$.

Thus, we have:
$P(\frac{x}{3})=3^{\frac{1}{3}}(P_0)$
$P(\frac{2x}{3})=3^{\frac{2}{3}}(P_0)$

The resired ratio is therefore:

$\frac{3^{\frac{2}{3}}(P_0)}{3^{\frac{1}{3}}(P_0)}=\boxed{3^{-\frac{1}{3}}}$[/hide]


Find all integers $ x$ and prime numbers $ p$ satisfying $ x^8 \plus{} 2^{2^x\plus{}2} \equal{} p$.

[hide="Hint"]Ever heard of that factorization for $ a^4\plus{}4b^4$?[/hide]

$ a^4\plus{}4b^4 \equal{} (a^2\minus{}2ab\plus{}2b^2)(a^2\plus{}2ab\plus{}2b^2)$

Now:
$ x^8 \plus{} 2^{2^x\plus{}2} \equal{} p$
$ x^8 \plus{} 4 \cdot 2^{2^{x}} \equal{} p$
$ (x^4 \minus{}2x^2 \cdot 2^{2^{x\minus{}2}} \plus{} 2\cdot 2^{2^{x\minus{}1}})(x^4 \plus{}2x^2 \cdot 2^{2^{x\minus{}2}} \plus{} 2\cdot 2^{2^{x\minus{}1}}) \equal{} p$

But in factoring we should notice that both factors are integers iff $ x \ge 2$, so we must test $ x\equal{}0 \implies p\equal{}8$ which is not prime, and $ x\equal{}1 \implies p\equal{}17$, a solution.  We can ignore $ x<0$ because all these will give $ 2^x\plus{}2 \in \mathbb{Q}$ and thus $ x^8\plus{}2^{2^x\plus{}2}$ will be irrational if $ x$ is a negative integer.

Continuing on from our factorization:
$ x^4 \minus{}x^2 \cdot 2^{2^{x\minus{}2}\plus{}1} \plus{} 2^{2^{x\minus{}1}\plus{}1} \equal{} 1$ and $ x^4 \plus{} x^2 \cdot 2^{2^{x\minus{}2}\plus{}1} \plus{} 2^{2^{x\minus{}1}\plus{}1} \equal{} p$

$ x^4 \plus{} 2^{2^{x\minus{}1}\plus{}2} \equal{} p\plus{}1$ so (unless $ p\equal{}2$) $ x^4$ is even and therefore $ x$ is even.  But if this is the case $ x^4 \minus{}x^2 \cdot 2^{2^{x\minus{}2}\plus{}1} \plus{} 2^{2^{x\minus{}1}\plus{}1} \equal{} 1$ cannot be true because the LHS must be even and $ 1$ is not even.

So let $ p\equal{}2$, so we have $ x^4 \plus{} x^2 \cdot 2^{2^{x\minus{}2}\plus{}1} \plus{} 2^{2^{x\minus{}1}\plus{}1} \equal{} 2$.  Now the LHS is strictly increasing for $ x>0$ and we have $ x\equal{}2$ gives an LHS of $ 268$, so there are clearly no solutions here.

Thus, the only solution is $ (x,p)\equal{}(1,17)$.


function
induction
algebra proposed
algebra

The real-valued function $ f$ is defined on the reals and satisfies $ f(xy) \equal{} x f(y) \plus{} y f(x)$ and $ f(x \plus{} y) \equal{} f(x^{1993}) \plus{} f(y^{1993})$ for all $ x, y$. Find $ f(\sqrt{5753})$.

[hide="Solution"]-) $ y \equal{} 0: f(0) \equal{} xf(0)\forall x\Rightarrow f(0) \equal{} 0\Rightarrow f(x) \equal{} f(x^{1993})$
-) $ y \equal{} 1: f(x) \equal{} xf(1) \plus{} f(x)\forall x\Rightarrow f(1) \equal{} 0\Rightarrow f(x \plus{} 1) \equal{} f(x^{1993})$
$ \Rightarrow f(x \plus{} 1) \equal{} f(x)\Rightarrow f(n) \equal{} f(0) \equal{} 0$
-) $ y \equal{} x: f(x^2) \equal{} 2xf(x)\Rightarrow 0 \equal{} f(5753) \equal{} 2\sqrt {5753}f(\sqrt {5753})\Rightarrow f(\sqrt {5753}) \equal{} 0$[/hide]

$ y\equal{}x^n\Rightarrow f(x^{n\plus{}1})\equal{}xf(x^n)\plus{}x^nf(x)$ for all $ x\in\mathbb{R},n\in\mathbb{N}$.
Then by induction $ f(x^n)\equal{}nx^{n\minus{}1}f(x)$ for all $ x\in\mathbb{R},n\in\mathbb{N},n>1$.
And then $ f(x)\equal{}f(x^{1993})\equal{}1993x^{1992}f(x)$ gives $ f(x)\equal{}0$ for all $ x$, excpt maybe for $ \pm{1\over\sqrt[1992]{1993}}$.
Then $ f\equiv 0$ follows immediately. (for example take $ y\equal{}2$ in first equation)


ƒ(x) is defined on [0;1].Prove:
a)if  ƒ  is differentiable on (0;1) and  ƒ(0)= ƒ(1)=0, then there is  ξ in (0;1) s.t  ƒ′(ξ)+ƒ(ξ)=0
b)if  ƒ  is two time differentiable on (0;1) and  ƒ(0)= ƒ(1)= ƒ (1/2)=0,  then there is ξ in (0;1) s.t  ƒ ′′ (ξ)-ƒ(ξ)=0

a). Rolle theorem for $e^xf(x)$
b). Rolle theorem for $e^{-x}f(x)$ and intervals [0, 1/2] and [1/2, 1] then again Rolle for $e^x[f(x) - f'(x)]$  :blush:


AMC
AMC 12
AMC 12 B
AIME
geometry
3D geometry
tetrahedron

How did you do on the 2007 AMC12B?  (Answers are in the other thread).

I left 2 blank, so 141.

Assuming that I got all of mine correct, around 110.  I have never done well on these, for some odd reason.  However, I think that I finally qualified for the AIME....

I got a 109.5

16 right, 0 wrong.. Improved by a bit this year without any practice. *shrugs* I shoulda qualified for AIME last year (not quite sure what happened with my score, maybe a misbubble or something) so it's nice to finally do ok, heh.

Now, to get a decent AIME score..

I answered 16, and got 16 right...  I just hope that I bubbled everything correctly.  If I don't qualify because of that, I won't be happy. :P

[quote="randomdragoon"]So your poll assumes no one got 145.5?[/quote]

Yep, you shouldn't get that  :D  always guess!

I got 138- 23 right and two wrong.

One really stupid mistake, and one really lucky guess, so I guess they kinda cancel each other out.

121.5. (19,1,5)

After doing bad on the 12A, I tried to play it conservative and answer less than usual and check well, but I made a dumb mistake with the tetrahedron one (miscounted cases) and didn't even catch that when I checked. So that error pretty much negated my strategy.  :(

Both of my scores this year were worse than last year.   :o

141, 23 right and I left the last two blank.  hooray!

21 questions answered got them all...

132

112.5
yay!
i answered 17 but accidentally made a stupid mistake on the geometric series one because i (don't ask me why) assumed that it was a finite series not an infinite one

I got a 136.5! (22 right, 3 blank)
I answered 1-13, 15-22, and 24.

124.5...(3 blank 2 wrong)
all you genius 8th graders...

[quote="mathnerd314"]Oh, and the equilateral triangle one? The one where the point inside the triangle was 1, 2, and 3 units away from the sides? I had just proven a theorem in AoPS 1 exactly like that problem. But I blanked out and forgot the theorem (which states that the sum of the distances is equal to the altitude) and thought the sum of the distances was equal to the side length instead. I picked 6. It was $4\sqrt{3}$.

Presumably it's normal to do 20 points worse on the 12 than the 10. That's why the cutoff for the 10 is 20 points higher than that on the 12.[/quote]

Sum of areas:

$s(1+2+3)/2 = \frac{s^{2}\sqrt{3}}{4}$
Sounds like a pretty trivial theorem, if a one liner gives the same result.

[quote="fizix"]112.5...is it normal for people to do about twenty points worse on the 12 than they did on the 10 when they took it last year?  (This happened to two people at my school.)[/quote]

My freshman year was 144 on the 10 and 120 on the 12.  Sounds about right.  Depends a lot on the year, though.

$150$.  Took long enough  :)  Luckily, this week is our school's Midwinter Break, so I was able to get some sleep the night before.  Apparently, an extra few hours of sleep improves your score by $18$ points.

[quote="t0rajir0u"]Apparently, an extra few hours of sleep improves your score by $18$ points.[/quote]

I don't think so. It's more like $12$ :wink:.

My score was 112.5 (17 right, 1 wrong, 7 blank), which I'm more than satisfied with, considering that I only got 108 last year (with the easier scoring system).  My school only gives the B form, but I took the 12A for practice and got 100.5, just barely enough to qualify for the AIME, which had me worried.  I'm hoping that I had the highest score in my school this year, since this would be my second year in a row.

Questions 1 through 13 I did without too much trouble, though 14 took me a while (geometry isn't my best field).  I also managed to do 15 and 17, and 18 I got simply by guessing and testing, and on my first guess at that!  I guess the math gods were watching over me that day.  Since I wanted to answer 18 questions, I decided to guess at number 16, though for some strange reason my brain interpreted "tetrahedron" as "cube," which is why I incorrectly guessed the largest choice.

Overall, I'm more than happy with what I got, and congradulations to all of you geniuses who managed to get such great scores!

Man, missed #1, marked 867 instead of 876  :mad:. Ended up with 135 (22 right 1 wrong 2 blank).

[quote="h_s_potter2002"][quote="xingster14"]darn it , only 12% of poll say they got less than 100? that cutoff is getting greater and greater and greater......[/quote]

On the AMC 12 A poll, only 10% said they got less than 100, and that qualifying score was 97.5.[/quote]

Of course, using AoPS polls as an indicator of the general population's AMC scores is a little biased sampling... ;)

ahhh! 117... coulda gotten a 123 had I not marked 18 for #4 instead of 8. Yeah, reading your answer choices helps sometimes... you know, that little vertical bar on the left of the "8" makes a big difference.

109.5 :censored:
From what I have seen on here, I would have scored 130's on the A, this test had a lot of geometry, which I am incapable of doing. So I am really gonna have to power own the AIME to go on :|

Hey, xingster, what school do you go to?

[quote="fizix"]112.5...is it normal for people to do about twenty points worse on the 12 than they did on the 10 when they took it last year?  (This happened to two people at my school.)[/quote]

I'm not sure how normal it is, but I had a 32 point drop from 10th to 11th grade.

I believe there is also a substantial non-response bias among those who got less than 100 points on the AMC.

I got a 114, unofficially, taking it on my bed. (17, 0, 8)

(right,wrong,blank)

(17,2,6)

Therefore, I'm thinking I got a 111 if I didn't have a bubbling error.....

B was easier than the A. :|  No, I mean it.


Fibonacci numbers can be defined such that F[1] = F[2] = 1 and F[n] = F[n-1] + F[n-2].

This sequence looks like 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...

Find an expression for the sum of the first m Fibonacci numbers.  Explain why the relationship holds true.

i think the sum of the first m fibonacci numbers is F[size=59]m+2[/size]-1.

F[size=59]1[/size] + F[size=59]2[/size] +.....F[size=59]m[/size] = F[size=59]m+2[/size]-1, replacing m with k yields the same thing just with k, k=1 works, adding F[size=59]m+1[/size] to both sides yields the quantity F[size=59]m+2[/size]-1 + F[size=59]m+1[/size], or F[size=59]m+3[/size]-1, which is what was needed.

[quote="Barnacle"]i got the same relation.  is there something explicitly in terms of m? that'd be interesting  :?:[/quote]

There is, but that's a much more complicated issue.  There is a formula for the $n$th Fibonacci number called Binet's formula and it requires a bit higher level of algebra to derive.

If any students here are familiar with recursion or generating functions -- either of those tools would work.

Once the formula is known it can be proved by induction in a similar manner as the formula for the sum was proved above.


real analysis
inequalities unsolved
inequalities

I guess some of you just haven't seen it in Real analysis section

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=264907

Let $ f(x,y,z) \equal{} xy^2 \plus{} yz^2 \plus{} zx^2$

given $ c > 0$ (const.) find
\[ \sup_{x,y,z > 0\land x \plus{} y \plus{} z \equal{} c}f(x,y,z)
\]
you mustn't use Lagrange multipliers

I am not sure....

oh,I sorry,I misuse the wrong var.....

Why $ b(a\plus{}c)^2\leq \frac{1}{2}\left (\frac{2a\plus{}2b\plus{}2c}{3}\right )^3$?

:blush: I forgot the difference of Max and Sup...Maybe in analyze course.

[quote="MeKnowsNothing"]We can't choose $ a \equal{} 0$. We can't get the maximum but we can get the supremum by choosing (notation is in Jianxing113725's post so $ x$ is not defined.) $ x\in ]0,1[$, $ a \equal{} x , b \equal{} M/3, c \equal{} 2M/3 \minus{} x$ and letting $ x\to 0 \plus{}$. One should always be careful then dealing with maxs/sups.[/quote]
yes,
another thing - it's not sufficient to prove it in the case $ a\le b\le c$, the formula for $ f$ isn't symmetric...........  :(


number theory
Galois Theory
superior algebra
superior algebra unsolved

Let $ R$ be a ring of integers of a number field. Is every ideal class represented by a prime ideal?

Yes, but proving this is far away from being easy (it's similar to the standard proof for Dirichlet's theorem on primes in arithmetic progressions).

Well, I can tell you where to find it: J. Neukirch, "Algebraic Number Theory", Chapter 7, theorem 13.2.
I can post a sketch of the proof, but only if I have more time (so not now, but later^^).

I don't know how much this helps, but it might give some additional insight and make a connection to other results which have been mentioned on the forum: the statement can be proved by applying the [url=http://eom.springer.de/C/c130080.htm]Chebotarev Density Theorem[/url] to the [url=http://mathworld.wolfram.com/HilbertClassField.html]Hilbert class field[/url] $ L$ of your number field $ K$. The Frobenius map gives an isomorphism of the ideal class group of $ K$ onto the Galois group $ G\equal{}\mbox{Gal}(L/K)$, and Chebotarev's theorem says that the density of the set of prime ideals taken by the Frobenius map onto any one given element of $ G$ is $ \frac1{|G|}$. In particular, there are such prime ideals.

Proofs of these results (like Chebotarev's theorem and the existence of the Hilbert class field) tend to be quite long and difficult. I'm sure a good algebraic number theory book will tell you a lot more than anyone could post here.

Neukirch needs the Dirichlet density of prime ideals in some ideal class to prove Chebotarev's theorem^^ (but I would not be astonished if one could skip it).
But I totally agree that reading some good books is much better than reading some posts.


Curious as to which books have been read.

I liked both of Brian Greene's books.  Only listened to The Briefer History of Time from an audio book. Read Penrose's book up to page seventy about since it involves alot of post graduate mathematics.  Smolin's book looked too cynical for me so, I did not read it.

Road to Reality & The Universe in a nutshell
I have a general dislike towards *Stephen Hawking* type books. I keep struggling with Penrose's book from time to time.

I have all of the Brian Greene books up there. I don't know if you count this as an Astrophysics but I am reading An atlas of the Universe by Sir Patrick Moore.

Yeah, I do not really like those "Stephen Hawking" type books either and not really his books in general.

hunter34 , The Atlas of the Universe looks quite interesting.

 :wink:  If anyone has other books not listed that are any good post it.

I don't know if they are available on the international market, but popular books by J.V. Narlikar (Seven Wonders of the Universe, A Cosmic Adventure etc.) are very good as poupular astrophysics books go.
Weinbergs 'The first three minutes' is also a classic.

Anyway, for those who really want to learn astrophysics, say high school onwards would do much better by using good astrophysics texts rather than popular astronomy books.

:wink:  Of course a book could never take the place of a good textbook.

i completely support shreyas that jayant narlikar books are a classic ,they just hav mind blowing concepts which clear ur fundamentals in astrophysics

The Elegant Universe was my first, struggled with the later portions.  Then I went onto Universe in a Nutshell, and Brief History of Time (then Briefer History)...now I am on some of Kaku's books (Physics of the Impossible being the current one I am on)

But I agree that these books cannot replace a good textbook.  :lol:

[quote="rt_08"]hi,
do u have the e book of the physics of the impossible.
regards,
romil[/quote]

Sorry, I checked it out of the library.

But I am certain they have audiobooks of it, because my friend is still using that.

And I have a complete Michio Kaku set seperate bought.

So anyone here manage to read through Road to Reality? I got to chapter 8 before I gave up. Might attempt it again this summer except that I'll get textbooks for all the math in there.

[quote="bnmanjunath"]Personally, I am a huge fan of Michio Kaku.[/quote]
 Yo that guy is awesome~

I read The Elegnat Universe by Briane Greene.
I also read Why does E=mc^2 by Brian Cox and Jeff Forshaw, as well as Beyond Einstein by Michio Kaku.

The Road to Reality is a good read for me, this can be treated as a textbook, others are for just Reading like a novel.


function
logarithms
algebra unsolved
algebra

Find all the functions $f:R\rightarrow R$ sitisfying :
$\left( \forall x\in R\right) :f\left( \sqrt{2}x\right) +f\left( \left( 4+3\sqrt{2}\right) x\right) =2f\left( \left( 2+\sqrt{2}\right) x\right) $

hé, dose anybody have a solution, I am really stuck with this problem  :?

please  :?  I would like that someone posts a solution for the problem. I really don't know how to solve it ...

erdos, please don't be so anxious. It's 2 hours 46 minutes after you posted your problem.

Erdos, are you sure this is the right statement? No mention of continuity?

Yes I am totally sure. this is the original statment. I hope that someone could fin a solution. it seems too difficult for me.

The satetement means that $f(ax)+f(bx)=2f(x\sqrt{ab})$. And it is reduces to the searching for a function satisfying $g(y-T)+g(y+T)=2g(y)$ for all $y\in\mathbb R$, $T$ is a fixed numbers. :? There are infinitely many such functions and no any nice rule for describing.
What is wrong?

Myth wrote:There are infinitely many such functions and no any nice rule for describing.
Yes!The answer :$f(x)=a$ for $x=0$ and $f(x)=g(\log_{\sqrt {2}-1}|x|)$ for $x\ne 0$(where $g(u)=\varphi (u) +uh(u),\forall u\in \mathbb R$ and $\varphi (u),h(u)$ are periodic functions with the period $1$)
@Myth:Your new avatar is nice ;)  :) !


search
function
number theory unsolved
number theory

Very trivial usage of Search Function...
Locked.

[quote="quangpbc"]Prove that the sequence defined by $ a_{1}=1$ and $ a_{n}=a_{n-1}+a_{[\frac{n}{2}]}$ contains infinitely many terms divisible by 7 :wink:[/quote]
Use function search with keyword ''Poland 1997'' you will have http://www.mathlinks.ro/Forum/viewtopic.php?search_id=209288072&t=4600 .

To ZetaX: He is a new member, I will hint him!  :wink:


[hide]HINT: Use the fact that $ e^{x} \equal{} \sum_{n\equal{}0}^{\infty} \frac{x^{n}}{n!}.$[/hide]

[hide]Start by noting that the given series converges, since the series with general terms $ \frac{n^3}{(n\minus{}1)!}$ and $ \frac{1}{(n\minus{}1)!}$ are both convergent (to see this, you can use, for example, the ratio test). To compute its value, let us now find four constants $ a,b,c,d$ such that

$ \frac {2n^{3} \plus{} 3}{(n \minus{} 1)!} \equal{} \frac {a}{(n \minus{} 1)!} \plus{} \frac {b}{(n \minus{} 2)!} \plus{} \frac {c}{(n \minus{} 3)!} \plus{} \frac {d}{(n \minus{} 4)!}$.

Performing the calculations we get a = 5, b = 14, c = 12, and d = 2. Therefore,

$ \sum_{n \equal{} 2}^\infty\frac {2n^{3} \plus{} 3}{(n \minus{} 1)!} \equal{} \sum_{n \equal{} 2}^\infty \frac {5}{(n \minus{} 1)!} \plus{} \sum_{n \equal{} 2}^\infty \frac {14}{(n \minus{} 2)!} \plus{} \sum_{n \equal{} 3}^\infty \frac {12}{(n \minus{} 3)!} \plus{} \sum_{n \equal{} 4}^\infty \frac {2}{(n \minus{} 4)!}$

$ \equal{} 5(e \minus{} 1) \plus{} 14(e) \plus{} 12(e) \plus{} 2(e)$

$ \equal{} 33e \minus{} 5$,

where the four series on the right hand side of the first equality are all convergent.[/hide]

Alternatively as mentioned, we can just consider $ e^x\equal{}\sum_{n\equal{}0}^\infty \frac{x^n}{n!}$.
For our own confinence, we first derive a formular, $ \frac{d}{dx} \big(e^x \phi (x) \big)\equal{}e^x\big(\phi (x) \plus{} \phi' (x)\big)$,
now we have $ xe^x \equal{} \sum_0^\infty \frac{x^{n\plus{}1}}{n!}$,
differentiate both sides and again multiply both sides by $ x$,  $ e^x(x^2\plus{}x)\equal{}\sum_0^\infty \frac{(n\plus{}1)}{n!}x^{n\plus{}1}$,
and so on, we get $ \sum_0^\infty \frac{(n\plus{}1)^3}{n!}x^n \equal{} e^x(x^3\plus{}6x^2\plus{}7x\plus{}1),$
substitute $ x\equal{}1$, we get desired sum.


Ring Theory
superior algebra
superior algebra unsolved

R is a commutative ring. M is R-module. $ \mathfrak{a}$ is an ideal of R.
let $ x\in\cap_{\mathfrak{a\subset m}}Ker (M\rightarrow M_\mathfrak{m})$ where $ \mathfrak{m}$ run all maximal ideals of R containing $ \mathfrak{a}$.

Prove that $ (x)\equal{}\mathfrak{a}(x)$.

Thks so much for helping me.

[quote="kidmath"]R is a commutative ring. M is R-module. $ \mathfrak{a}$ is an ideal of R.
let $ x\in\cap_{\mathfrak{a\subset m}}Ker (M\rightarrow M_\mathfrak{m})$ where $ \mathfrak{m}$ run all maximal ideals of R containing $ \mathfrak{a}$.

Prove that $ (x) \equal{} \mathfrak{a}(x)$.

Thks so much for helping me.[/quote]

Suppose $ \mathfrak{b}\equal{}\mathfrak{a} \plus{} \text{ann}_R(x) \neq R.$ Choose a maximal ideal $ \mathfrak{m} \supseteq \mathfrak{b}.$ So $ \mathfrak{m} \supseteq \mathfrak{a}.$ Thus $ x \in \ker(M \longrightarrow M_{\mathfrak{m}}),$ i.e. there exists $ s \in \text{ann}_R(x) \minus{} \mathfrak{m},$ which is impossible because $ \mathfrak{m} \supseteq \text{ann}_R(x).$

Thus $ \mathfrak{a} \plus{} \text{ann}_R(x) \equal{} R.$ Now multiply both sides by $ x$ to get $ \mathfrak{a}(x)\equal{}\mathfrak{a}x\equal{}(x). \ \Box$

thks very much. I did not find this way.


geometry
rectangle
trigonometry
3D geometry
pyramid
analytic geometry
ARML

Rectangle ABCD has an interior point E such that AE = 7, BE = 6, and CE = 4.  Find DE.

I can find a solution, but I suspect there is an extremely straightforward way.  This is I believe an ICTM problem.  My solution requires law of cosines and some cyclic quads.

Is it [hide]5[/hide]?  That was (almost) a random guess, so I have no idea if its right. :?

No, the answer is [hide]$\sqrt{29}$.  I showed that $x^2 + 6^2 = 4^2 + 7^2$.  But that might not be the best approach...[/hide]

Actually, that is the correct approach. Have you heard of the British Flag theorem?

It says that with any rectangle ABCD and any point, ANYWHERE in space, E, AE^2 + CE^2 = DE^2 + BE^2.

So, even if you have a rectangular-based pyramid, the sums of the squares of.. I guess you could say, "oppositely-paired" edges are equal.

Did that make sense?

Proving the theorem is pretty simple, especially if you use a coordinate system.

I guess my method is about as easy as it gets then, unless someone else comes along and shows a really easy way.  This is what I did:

[hide]Extend AE to X s.t. AXC is a right angle.  Also extend BE to Y s.t. BYD is a right angle.  Since A, B, C, and D are concyclic, and AC and BD are diameters, we have that A, B, X, and Y are also concyclic.  Then by power of a point (or whatever you call it), AE*EX = BE*EY.  Since EX = -CE*cos(AEC) and EY = -DE*cos(BED), we can say that

-AE*CE*cos(AEC) = -BE*DE*cos(BED)

Now AC = BD --> AC^2 = BD^2

Law of cosines yields

AE^2 + CE^2 - 2*AE*CE*cos(AEC) = BE^2 + DE^2 - 2*BE*DE*cos(BED)

Subtracting twice our first equation gives the result.  But considering that ICTM is a 50 minute, 20 question exam, I thought this was kind of hard.[/hide]

That's like what I did, except I didn't square the numbers.  Why is it called British Flag Theorem?

I think it's because the British flag has a cross on it.

Ah, I've never heard it called that before, but the way I prove it makes it like the british flag.
[hide]Draw lines parallel to the sides and passing through the point, then consider Pythagoras and its very simple.[/hide]

Hah yeah that's a bit easier than what I had.  I was afraid of all the variables.  Still imo a little bit hard for a fast contest.

i guess the contest was assuming that you'd know the formula, so it would be reduced to a 5 second problem? idk

[quote="probability1.01"]I guess my method is about as easy as it gets then, unless someone else comes along and shows a really easy way....[/quote]
If you are familiar with the median formula (or Stewart's theorem), then consider two triangles AEC and BED, they share the same median EO (where O is the center) and have same base lengths (AC=BD being diagonals).

This method works for any point, ANYWHERE in space, E as pointed out by [b]pkothari13[/b]

Using your new found knowledge, see if you can tackle this problem:

$\text{Pyramid EARLY has a rectangular base EARL, and apex Y, ER = } \sqrt{1105}, \text{ and EY}\perp \text{RY. If EA, AR, EY, AY, RY, and LY are distinct integers, compute the area of the smallest face of the pyramid.}$

ARML 2004 Team #10... fun stuff.

[hide]
At point $Y$, we know $AY^2+CY^2 = EY^2+RY^2 = 1105$ since $\triangle RYE$ is right. Also, $AE^2+AR^2 = 1105$.

Notice that $1105 = 33^2+4^2 = 32^2+9^2 = 31^2+12^2$.

Use the triangle inequality to arrange the lengths and I think you get a triangle with sides $9,4,12$. So it has area $\sqrt{\frac{25}{2} \cdot \frac{7}{2} \cdot \frac{17}{2} \cdot \frac{1}{2}} = \frac{5}{4}\sqrt{119}$.

Note that I had no idea how to do this problem when I was there last year ;).[/hide]

Looks good to me  :D ! yep, ARML 2004 Team 10! haha

if the answer is rt29 then i dont think its a difficult problem at all...something i think id see on a relay or something. if it is simply taking x^2+6^2=7^2+4^2...very similar to a theorem dealing with two chords of a circle.

That's funny... that is the exact theorem we had to prove in a recent assignment.