ARML geometry ratio AMC USA(J)MO USAMO induction

Once the awards have been handed out and people have survived their bus trips back home, I'd like to use this thread here for people to share their thoughts on this year's contest. Too hard? Too easy? Too [i]awesome?[/i]  :wink:

Good luck to all y'all's this weekend, and remember (it's noted elsewhere) that that sites might not be perfectly in sync throughout this year's ARML, so wait until Saturday evening, after the awards, to do any discussin'  :)

Partial results:

Teams:
1. Lehigh Fire, 215
2. Exeter Red, 201
3. San Francisco Bay Area A1, 197
4. New York City A1, 196
5. Montgomery A1, 196
5. Exeter Red, 192
6. Southern California A1, 191
7. North Carolina A1, 186
8. San Diego Surf, 182
9. Indiana Gold, 182
Georgia: South Carolina A1, 173
Iowa: Chicago A1, 177
Las Vegas: Washington Gold, 166
Penn: Thomas Jefferson A1, 180

Division B:
1. Thomas Jefferson B1, 156
2. Andover B1, 150
3. Lehigh Ice, 144
4. Central Jersey B1, 136
5. San Diego Sun, 130
6. Utah B1, 129
7. Maine B1, 126
Georgia: Grissom B1, 115
Iowa: Michigan Naturals, 112
Las Vegas: Colorado B1, 104
Penn: NYC B3, 124

International:
1. Macau

(Site awards for teams were not all announced)

Individuals:
1. Alex Song, Ontario West
2. Warut Suksimpong, Exeter
3. Qin Xuan Pan, Montgomery
4. Junu Bae, Western Penn
5. Chong Gu, NYC
6. David Yang, SoCal
7. Bayley Wang, SFBA
8. Zhifan (Ivan) Zhang, SoCal
9. Aki Hogge, Montgomery
10. Robert Yang, Utah
Georgia site: John Berman, NC
Iowa site:
Las Vegas site: Sam Keller, Washington Gold
Penn site: Ofir Nachum, Western Mass A1

The top three individuals scored 10, with the other winners at 9.

[color=green][Edited for Exeter/Andover correction.][/color]

It was really a great competition this year.  The Team Contest was harder than usual which is sometimes enjoyable. The Power Round, I thought, was pretty easy aside from such problems as #10 and, according to my teammates, 5(b).  The Individual round I failed completely =.= I got 1, 2, 3, 5, 6, and 9.  For 4, I was just being stupid and tried to solve for the actual legs of the triangle rather than the sum of their reciprocals (of course, in the last 30 seconds, I realized my stupidity and got the correct set of ratios, but was just a little too late).  #7 I forgot the 00, 00 case, and so got 41 instead of 42.  #8 I didn't really try (as I had a [correct] feeling that my answer to #7 was missing something...) and #10 I was close to getting, but the time constraints wouldn't allow for it.  So, while I should have gotten an 8 easy, I am instead left with a 6. Disappointing a little, but the problems were really fun and I had a great time.

By the way, how do you do Team #9?

The Power Round and Individual Questions were far easier this year compared to last year. Our team, Chesapeake A1, got a 44 on the Power Round after a more disappointing score last year and losing our best seniors. As for the individual questions, I got the first eight and should have gotten nine had I not wasted time trying to bash #10 and just trying to check #9, where I had messed up the surface area of a cylinder.  :mad: Excluding 10, none of the problems seemed that difficult. I would have liked to make tiebreakers; I think I would have been one of the few non-USAMO qualifiers to do so had I made tiebreakers. I would have gotten the question quickly too; I guessed the correct answer by virtue of having absolutely no idea on how to proceed and dividing the two numbers given.  :rotfl:

The Team Round seemed harder, but I think it was because our proctor messed up our timing and we had a couple minutes of consternation trying to figure out exactly how much time we had left. We failed badly on Relays.

I liked this ARML more than last year's ARML. The Song Contest at Penn State never ceases to delight.

[quote="jmerry"]Partial results:

Teams:
1. Lehigh Fire, 215
2. SFBA A1, 197
3. New York City A1, 196
4. Montgomery A1, 196
5. Exeter Red, 192
6. Southern California A1, 191
7. North Carolina A1, 186
8. San Diego Surf, 182
9. Indiana Gold, 182

Division B:
1. Exeter B1, 159
2. Thomas Jefferson B1, 156
3. Lehigh Ice, 144
4. Central Jersey B1, 136
5. San Diego Sun, 130
6. Utah B1, 129
7. Maine B1, 126

International:
1. Macau

Individuals:
1. Alex Song, Ontario West
2.
3.
4.
5.
6. David Yang, SoCal A1
7. Bayley Wang, SFBA A1
8. Zhifan (Ivan) Zhang, SoCal A1
9. , Alabama
10. Robert Yang, Utah B1
Georgia site:
Iowa site:
Las Vegas site: Sam Keller, Washington Gold
Penn site:

The top three individuals scored 10, with the other winners at 9.[/quote]
I remember from PA site:
2. someone from Exeter
3. Qinxuan Pan, Montgomery
4. someone from NY?
5. Junu Bae, Western PA

yeah that didnt really help :P

The ones that jmerry is missing are:

2. Warut Suksimpong, Exeter
3. Qin Xuan Pan, Montgomery
4. Junu Bae, Western Penn
5. Chong Gu, NYC
9. Aki Hogge, Montgomery

Georgia site winner: John Berman, NC

I believe that Montgomery is Montgomery County, Maryland. Not Montgomery Alabama. For the 1st time in a while the normal Alabama team didn't come. There was a Grissom, AL team that was the Georgia B site winner.

My team (AAST Mu A) answered everything on power, and tied for third. I think we got a few points off for style/handwriting/stuff like that, I dunno. Well, we did use group theory. Or maybe we ACTUALLY FAILED but I do not wish to believe that.

[quote]The Power Round, I thought, was pretty easy aside from such problems as #10 and, according to my teammates, 5(b). [/quote]
Guess what the only problem I did the whole time period was! Oddly enough, I recently found [url=http://mathworld.wolfram.com/EulerZigzagNumber.html]this[/url], and the answer was on [url=http://www.research.att.com/~njas/sequences/A000111]this[/url], but of course I forgot it. I made up my own much more complicated formula ($ a_1 \equal{} 1$ and $ a_{n \plus{} 2} \equal{} \sum_{k \equal{} 1}^{\frac {n \plus{} 1}{2}} {n \plus{} 1 \choose 2k \minus{} 1} a_{2k \minus{} 1} a_{2n \minus{} 2k \plus{} 2}$, or something like that.) I just proved that formula's validity and bashed it out to find $ a_5$ and $ a_9$. Proud that I got it, I assigned it to a minion who wasn't doing anything. And he got it wrong, which resulted in me doing the fastest computations in my life to change the answer in the last minute or so.

In the team round, I was assigned to work on #9 the whole time, and therefore did not contribute to the team round at all. And since I only worked on 5 for power, I can't tell you anything about those.

I failed individual, I don't really want to talk about it. Same with relay.

The only thing I'll say is that I used complementary counting once and it screwed me over. And then I didn't use it and I got screwed over again. Individual #4 and relay 1 #3, respectively.

I must say I was fond of #8. It was a fun little problem. I wrote #10 in terms of complex numbers (a tool on recently learned here), and ended up with a cubic, and ran out of time.

Overall, my team failed relay. One member read 12 as 21.

Overall, my team failed to place at all. Not even in the singing contest.

By the way, I was the brave volunteer who held up the microphone.

[quote]6. David Yang, SoCal A1 [/quote]
SURPRISE!!!!1111111

[quote="J.Mayers"]
By the way, how do you do Team #9?[/quote]
Say that $ a\plus{}b\equal{}c$, then you know that $ b\equal{}c\minus{}a$.  You can substitute this in, expand, multiply the first equation by 8, and then set the coefficients of each $ a$ term equal to each other.  Then you solve for $ c$.  Sorry--I don't have a copy of the problem, and nor did I get it in the round, but this is the solution our coach gave us afterwards.

Actually, the third place team should read "New York City A1: Murph and the Magictones"

[quote="jmerry"]
Individuals:
1. Alex Song, Ontario West
[/quote]
Is this the same Alex Song from Waterloo, Ontario who qualified for USAMO last year as a 5th grader? I refuse to believe a 6th grader won individuals....

[quote="buzzer11"]I used induction for power round #10. How did everyone else do it?[/quote]

What was the problem again? I didn't look at it during the round.

Comments:

Individual Round:
This round seemed a lot easier this year. 10 was hard but a lot of people i know bashed it and got a quadratic in x^2 but got -8 root 3 instead of +8root3.
Power Round:
Ehh, I only looked at 2 problems, so I might be wrong, but the problems I looked at weren't very hard and according to my team 10 was the only hard question?
Team Round:
This round was really good... it was also harder than normal. My team (SoCal A1) missed 4 problems on this.....
Tiebraker #1: I got this by looking at a case where everything seemed to work, but I missed that the triangle was degenerate... arg I didn't solve this legitly but at least I got the right answer...

Was there a "smart" way to do individual #10?

For #10, divide both sides by $ 2$. Now take the tangent of both sides. Tangent of $ \frac{\pi}{2}$ is undefined, so the denominator of the tangent of the left side is $ 0$. If we have $ \tan \left(a\plus{}b\plus{}c\plus{}d\plus{}e\right)$, the denominator is $ e_0\minus{}e_2\plus{}e_4$, so set that equal to $ 0$ and you have a quadratic in $ x^2$.

[quote="Ravi B"][quote="Team Round #9"]Let $ a$ and $ b$ be real numbers such that
\[ a^3 \minus{} 15 a^2 \plus{} 20a \minus{} 50 \equal{} 0 \quad \textrm{and} \quad 8 b^3 \minus{} 60 b^2 \minus{} 290 b \plus{} 2575 \equal{} 0.
\]
Compute $ a \plus{} b$.[/quote]
Here's a cheapo method for solving Team Round #9.  By the form of the question, it seems likely that the roots of the first equation are some constant $ c$ minus the roots of the second equation.  Let's make that assumption.  Hence the sum of the roots of the first equation is $ 3c$ minus the sum of the roots of the second equation.  By Vieta's formula, the sum of the roots of the first equation is $ 15$ and the sum of the roots of the second equation is $ \frac {15}{2}$.  So we get the equation $ 15 \equal{} 3c \minus{} \frac {15}{2}$.  Solving for $ c$ gives $ c \equal{} \boxed{\frac {15}{2}}$..[/quote]
Here's another cheapish method, found by grn_trtle:

Multiply the first equation by $ 8$ and add it to the second, to obtain $ 8b^3\minus{}60b^2\minus{}290b\plus{}8a^3\minus{}120a^2\plus{}160a\plus{}2175 \equal{} 0.$
Now, we guess that this factors as $ (a \plus{} b \minus{} c)(Aa^2 \plus{} Bb^2 \plus{} Cab \plus{} Da \plus{} Eb \plus{} F) \equal{} 0$, where $ c$ is the desired answer to this problem.
Comparing coefficients of $ a^3$ and $ b^3$, we have $ A \equal{} B \equal{} 8$.
Looking at the coefficients of $ a^2$ and $ b^2$, we get $ \minus{} 8c \plus{} D \equal{} \minus{} 120$ and $ \minus{} 8c \plus{} E \equal{} \minus{} 60$.  Subtracting, we obtain $ D \minus{} E \equal{} \minus{} 60$.
Looking at the coefficients of $ a$ and $ b$, we have $ \minus{} Dc \plus{} F \equal{} 160$ and $ \minus{} Ec \plus{} F \equal{} \minus{} 290$.  Subtracting, we have $ \minus{} c(D \minus{} E) \equal{} \minus{} 450$.
Substituting $ D \minus{} E \equal{} \minus{} 60$, we obtain $ c \equal{} \frac {15}{2}$.

[quote="ra5249"]#3 in relay round, had the question "...has T+1 colors of finger polish..." implying T is an integer. was passed back "10^(1+rt5)/4". i was like... uh okay let's just pretend he meant to pass back 5. probably should have guessed power of 10 = 10 but whatever[/quote]

The final answer is $ (T\plus{}1)T^4\minus{}(T\plus{}1)T$. R1-2 is a log-base-10 equation, so my thinking was that if the person in position #2 messed up or had a wrong answer to begin with, they would pass back a power of 10. This, coupled with the fact that $ (T\plus{}1)T^4$ is a royal pain to compute for all but a few values of $ T$ (like 2, 3, and 10) leads me to believe that a 2-man relay team would have stood a good chance at arriving at the correct final answer. If position 3 receives some goofy non-integer power of 10, their first (and only) guess for the intended $ T$ should be 10.

I told this to a fellow coach, who said agreed, but who had heard several people in position #2 say they solved and got $ \log x\equal{}1$ and passed back $ x\equal{}1$  :blush: It's hard to use at least 3 different colors of fingernail polish when you only have one color  :P

On relay 1-3, I was passed the correct answer, $ 10$, I still didn't get the problem right! I got that there were $ (T \plus{} 1)T ^ 4 \equal{} 110000$ ways for Kay to paint her fingernails such that no two consecutive fingernails have the same color, but not necessarily using at least three colors. I then attempted to subtract the number of combinations that used fewer than three colors, but I thought that there were $ 2 ^ 5 \cdot \binom{11}{2}$ of those combinations, instead of $ 2 \cdot \binom{11}{2}$, because I didn't realize that I had to subtract combinations where no two fingernails are the same color!  :blush:

It was pretty fun for my first ARML. I got a 5 on my B1 with two mistakes, but it was a very fun experience. In the team round we did pretty bad, with a 2, but I checked and corrected #1 and got #3 for our second problem. The power round was mostly a failure for us. I thought (and many of my teammates) thought it focused to much on terminology and much less on actual mathematical ability. I had to check back multiple times just to see what some of the terms meant. In result, we got a 23/50 there. Our relay and super relays were pretty much disasters. On the first our third person (I was 2nd) couldn't get it for 5 or so minutes. On the 2nd, the first guy gave me a coordinate, not a number. So I guessed. I was wrong.

On the super relay, I was at position 7. Unluckily, the lines on both sides stopped at around 3 and 10. We never were close to finishing.

I thought Team #9 was ridiculously contrived - as soon as you start solving the cubics (depressing them) the answer just appears out of thin air. Once you remembered your trig and had the insight to see the denominator had to be 0, Individual #10 was an algebra bash. Overall, I felt the problems on Individual were sort of ugly this year.

I had a cleanish solution to individual #10 using complex numbers, which I'll post later when I have time.

@archimedes1: So you finally got that system of coefficient equations to work out :D

Here's my solutions:
[hide="No Complex Numbers"]
Let $ x \equal{} \tan(A) \Rightarrow A \equal{} \tan^{ \minus{} 1}(x)$ and $ 3x \equal{} \tan(B) \Rightarrow B \equal{} \tan^{ \minus{} 1}(3x)$.

We are trying to solve $ 6A \plus{} 4B \equal{} \pi \Leftrightarrow 3A \plus{} 2B \equal{} \frac {\pi}{2} \Rightarrow \tan(3A)\tan(2B) \equal{} 1$.

Using triple and double angle formulas (Which aren't too hard to derive during the 10 mins):

$ \tan(3A) \equal{} \dfrac{3\tan(A) \minus{} \tan^3(A)}{1 \minus{} 3\tan^2(A)} \equal{} \dfrac{3x \minus{} x^3}{1 \minus{} 3x^2}$

$ \tan(2B) \equal{} \dfrac{2\tan(B)}{1 \minus{} \tan^2(B)} \equal{} \dfrac{6x}{1 \minus{} 9x^2}$

Thus, we have:
$ \dfrac{3x \minus{} x^3}{1 \minus{} 3x^2} \cdot \dfrac{6x}{1 \minus{} 9x^2} \equal{} 1$

$ (3x \minus{} x^3)(6x) \equal{} (1 \minus{} 3x^2)(1 \minus{} 9x^2)$

$ 18x^2 \minus{} 6x^4 \equal{} 27x^4 \minus{} 12x^2 \plus{} 1$

$ 33x^4 \minus{} 30x^2 \plus{} 1 \equal{} 0$

Using the quadratic formula yields:
$ x^2 \equal{} \dfrac{ \minus{} ( \minus{} 30) \pm \sqrt {( \minus{} 30)^2 \minus{} (4)(33)(1)}}{2(33)} \equal{} \dfrac{15 \pm 8\sqrt {3}}{33}$

Now since clearly $ 0 < A < B$, we need $ \frac {\pi}{2} \equal{} 3A \plus{} 2B > 5A \Rightarrow 0 < A < \frac {\pi}{10}$, and thus, $ 0 < \tan(A) < \tan(\frac {\pi}{10}) < \frac {1}{2}$

Thus, the desired solution is $ x^2 \equal{} \boxed{\dfrac{15 \minus{} 8\sqrt {3}}{33}}$.[/hide]

[hide="Complex Numbers"]
Note that $ arg(1\plus{}xi) \equal{} \tan^{\minus{}1}(x)$, and $ arg(1\plus{}3xi) \equal{} \tan^{\minus{}1}(3x)$

Then, the equation becomes:
$ 6arg(1\plus{}xi) \plus{} 4arg(1\plus{}3xi) \equal{} \pi$

$ 3arg(1\plus{}xi) \plus{} 2arg(1\plus{}3xi) \equal{} \frac{\pi}{2}$

$ arg\left[(1\plus{}xi)^3(1\plus{}3xi)^2\right] \equal{} \frac{\pi}{2}$

$ (1\plus{}xi)^3(1\plus{}3xi)^2 \equal{} Ki$ for some $ K \in \mathbb{R^\plus{}}$

$ Re[(1\plus{}xi)^3(1\plus{}3xi)^2] \equal{} 0$

After some tedious expansion, this becomes:
$ Re[(33 x^4\minus{}30x^2\plus{}1)\plus{}(9x^5\minus{}46x^3\plus{}9x)i] \equal{} 33 x^4\minus{}30x^2\plus{}1 \equal{} 0$

Then proceed the same way as in the non-complex numbers method
[/hide]

Yeah, that's the second solution in the official packet.. the complex numbers one is the first.

That was my solution as well, though I didn't think to divide by 2 while taking the actual test :blush: Time pressure...

At least I saw the complex numbers behind it, so I'm happy with myself...

Complex numbers is essentially what I did, except I didn't divide by 2, thus making it even more bashy. I somehow ended up with a cubic in $ x^2$ instead. Probably made an algebra mistake somewhere.

Probably shouldn't have spent so much time checking 9.

EDIT: Wow I somehow didn't see grn_turtle's post.

hey guys... check the national rankings again!

OMG really? That stinks lol.

Yeah, this year there were many embarrassing  mistakes made by the ARML staff. Including telling my team that we were one of the top 5 teams...and then thirty minutes later, they tell us that they were kidding lol.

Team contest problem $ 9$ is an old idea with different numbers...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268928

The tiebreaker problem 1 was quite pretty!

Overall, the test looks great. Too bad my ARML days are over :(

the official solution to tiebreaker one was weird....I think it had some typos?

It took 5 lines to show that $ \angle ADE \equal{} \angle CBE$ when it could have just said $ \angle ADE \equal{} 180 \minus{} \angle EDC \equal{} \angle CBE$ since $ BCDE$ is cyclic.

Also, the angle bisector through me off when I worked this since I think it is irrelevant to the problem?

I just saw this problem as:

 since $ BPC\sim EPD$ and $ ABC\sim ADE$ (since $ BCDE$ is cyclic),

\[ \dfrac{BP}{PE} \equal{} \dfrac{BC}{DE} \equal{} \dfrac{AC}{AE} \equal{} 3.\]


What is the value of $ 1\minus{}2\plus{}3\minus{}4\plus{}5\minus{}6\plus{} \ldots \minus{}98\plus{}99\minus{}100$?

Alternatively, its the sum of the first $ 50$ odd integers minu the sum of the first $ 50$ even integers. The sum of the first $ n$ odd integers is $ n^2$. The sum of the first $ n$ even integers is $ n(n\plus{}1)$. So, $ 50^2\minus{}50(51)\equal{}50(\minus{}1)\equal{}\boxed{\minus{}50}$


Two circles $C_1$ and $C_2$ intersect at $S$.
The tangent in $S$ to $C_1$ intersects $C_2$ in $A$ different from $S$.
The tangent in $S$ to $C_2$ intersects $C_1$ in $B$ different from $S$.
Another circle $C_3$ goes through $A, B, S$.
The tangent in $S$ to $C_3$ intersects $C_1$ in $P$ different from $S$ and $C_2$ in $Q$ different from $S$.
Prove that the distance $PS$ is equal to the distance $QS$.

As an angle between a chord and a tangent in the circle $C_3$, the angle < BSP is equal to the angle < BAS. Also, as an angle between a chord and a tangent in the circle $C_1$, the angle < ASB equals the angle < SPB. In other words, we have < BSP = < BAS ans < SPB = < ASB. Hence, the triangles BSP and BAS are similar, and it follows that PS : BS = SA : BA. In other words, PS : BS = AS : AB. Thus, $PS \cdot AB = AS \cdot BS$. Similarly, $QS \cdot AB = AS \cdot BS$. Hence, PS = QS as intended to show.

  Darij

[b][u]Remark.[/u][/b] A intersting metric relation :

$C_k=C(O_k,r_k),\ k\in \overline {1,2},\ \triangle SAQ\sim \triangle BSP\sim \triangle BAS\Longrightarrow \underline {\overline {\left| \frac {AQ}{BP}=\left( \frac {r_2}{r_1}\right) ^2.\right| }}$

Ancient topic, but it in Contest Collections, so...
With $T$ - midpoint of $AB$, $TS$ - middle line of trapezoid $AQPB$. Done.

Let $\angle SAB=\alpha,\angle SBA=\beta,\angle BSA=\gamma$. Then, by the tangents, we have $$\angle PSB=\alpha,\angle SPB=\gamma,\angle QSA=\beta,\angle SQA=\gamma.$$ Hence, we also have $$\angle PBS=\beta,\angle QAS=\alpha.$$ Hence, by Law of Sines spam, $$SQ=PS\cdot\frac{\sin \gamma}{\sin \beta}\cdot\frac{\sin \beta}{\sin \alpha}\cdot \frac{\sin\alpha}{\sin\gamma}=PS,$$ so we are done.


limit
calculus
derivative
real analysis
real analysis unsolved

Let $a_0>0$ and $a_{n+1} = \arctan a_n$ for $n \ge 0$.
Show that $\lim_{n \to 0} a_n = 0$ ([i]easy[/i]).
Then find numbers $L>0, \, \alpha$, possibly depending on $a_0$, such that $\lim_{n \to \infty}\frac{a_n}{Ln^\alpha} = 1$, or prove that no such numbers  exist.

classical method:
$arctan(x)=x(1-2x^2+o(x^2))$ thus $arctan(x)^a=x^a(1-2ax^2+o(x^2))$. So if one chooses $a=-2$,
$arctan(x)^{-2}=x^{-2} +2 +o(1)$.
So $u_{n+1}^{-2}-u_n^{-2}$ tends towards $2$. This is why $u_n$ is equivalent to $(2n)^{-\frac{1}{2}}$

[quote="alekk"] $arctan(x)=x(1-2x^2+o(x^2))$ thus $arctan(x)^a=x^a(1-2ax^2+o(x^2))$. [/quote]
Why?

Edit- I figrd it out. Stuid me!

[quote="alekk"]classical method:
$\arctan(x)=x(1-2x^2+o(x^2))$[/quote]
$\arctan(x)=x(1-x^2/3+o(x^2))$, giving a different constant. But I'd still like to see a proof that goes beyond formal asymptotics :)

[quote="alekk"]what do you mean ? (i.e what is formal asymptotics  :)  )[/quote][i]Formal [/i] asymptotics is asymptotics without proving that the asymptotic statement is valid ("Expand in powers of $\epsilon$, compare coefficients, the result is ...")  :huh:

What I was looking for is a proof that makes sense to a first year college student or advanced high school student. The problem can be stated at this level, so there should be a solution not too far from this level.


linear algebra
matrix
induction
algebra
polynomial
quadratics

Consider the matrix $ A\equal{}\left(\begin{array}{cc}6 & 5\\ \ \minus{}3 & \minus{}2\end{array}\right).$ Compute $ A^n, n \in \mathbb{N}^*.$

Applying Cayley-Hamilton's theorem, we get $ A^2 \plus{} 4A \plus{} 3I \equal{} O$. ($ I$ is the unit matrix, and $ O$ is all-zero matrix)

It can be rewritten as
$ A(A \plus{} I) \equal{} \minus{} 3(A \plus{} I)\cdots\boxed{1}$
$ A(A \plus{} 3I) \equal{} ( \minus{} 1)(A \plus{} 3I)\cdots\boxed{2}$

By induction,

$ A^n(A \plus{} I) \equal{} ( \minus{} 3)^n(A \plus{} I)\cdots\boxed{1}'$
$ A^n(A \plus{} 3I) \equal{} ( \minus{} 1)^n(A \plus{} 3I)\cdots\boxed{2}'$

$ \frac {1}{2}\cdot(\boxed{2}' \minus{} \boxed{1}')$, we get $ A^n \equal{} \frac {( \minus{} 1)^n \minus{} ( \minus{} 3)^n}{2}A \plus{} \frac {3( \minus{} 1)^n \minus{} ( \minus{} 3)^n}{2}I,\quad{n\in{N}}$

Another method;

Let's think about a polynomial identity.
$ x^n \equal{} (x \plus{} 1)(x \plus{} 3)Q_{n}(x) \plus{} p_{n}x \plus{} q_{n}$ We can define $ p_{n},q_{n}$ by substitute $ x \equal{} \minus{} 1, \minus{} 3$.
And then, substitute $ x$ to matrix $ A$, we get the same answer. (Think $ 1, 3$ to $ I, 3I$.)

Alternatively, just diagonalize it. We've seen the characteristic polynomial; the eigenvalues are $ 3$ and $ 1.$

For the eigenvalue $ 3,$ find the null space of $ \begin{bmatrix}3 & 5 \\
\minus{} 3 & \minus{} 5\end{bmatrix},$ getting $ \begin{bmatrix}5 \\
\minus{} 3\end{bmatrix}.$

For the eigenvalue $ 1,$ find the null space of $ \begin{bmatrix}5 & 5 \\
\minus{} 3 & \minus{} 3\end{bmatrix},$ getting $ \begin{bmatrix} \minus{} 1 \\
1\end{bmatrix}.$

Let $ P \equal{} \begin{bmatrix}5 & \minus{} 1 \\
\minus{} 3 & 1\end{bmatrix},$ so that $ P^{ \minus{} 1} \equal{} \frac12\begin{bmatrix}1 & 1 \\
3 & 5\end{bmatrix}.$

We have that $ P^{ \minus{} 1}AP \equal{} \begin{bmatrix} 3 & 0 \\
0 & 1\end{bmatrix} \equal{} D,$ or $ A \equal{} PDP^{ \minus{} 1}.$

We can then raise that to any power:

$ A^n \equal{} PD^nP^{ \minus{} 1} \equal{} \frac12\begin{bmatrix}5 & \minus{} 1 \\
\minus{} 3 & 1\end{bmatrix} \begin{bmatrix}3^n & 0 \\
0 & 1\end{bmatrix} \begin{bmatrix} 1 & 1 \\
3 & 5\end{bmatrix}$

$ \equal{} \frac12\begin{bmatrix}5 & \minus{} 1 \\
\minus{} 3 & 1\end{bmatrix}\begin{bmatrix}3^n & 3^n \\
3 & 5\end{bmatrix} \equal{} \frac12\begin{bmatrix}5\cdot3^n \minus{} 3 & 5\cdot3^n \minus{} 5 \\
\minus{} 3\cdot3^n \plus{} 3 & \minus{} 3\cdot3^n \plus{} 5\end{bmatrix}.$

[quote="kunny"][quote="J.Y.Choi"]Applying Cayley-Hamilton's theorem, we get $ A^2 \plus{} 4^A \plus{} 3I \equal{} O$. [/quote]

That's wrong.

[color=green][It was a typo. I've fixed it for him. - K.M.][/color][/quote]

Sorry, yet it is wrong. :(

The sum of each column of A is 3, therefore 3 is an eigenvalue with eigenspace spanned by (5,-3). Observe now that 6-5=1 and -3-(-2) = -1, so 1 is the other eigenvalue with eigenspace spanned by (1,-1). Hence,

$ A^n \equal{} \left[\begin{array}{cc}1 & 5 \\
\minus{} 1 & \minus{} 3\end{array}\right] \left[\begin{array}{cc}1 & 0 \\
0 & 3^n \end{array}\right]\left[\begin{array}{cc}1 & 5 \\
\minus{} 1 & \minus{} 3\end{array}\right]^{ \minus{} 1} \equal{} \frac12 \left[\begin{array}{cc}5 \cdot 3^n\minus{}3 & 5(3^n\minus{}1) \\ 3(1\minus{}3^n) & 5\minus{}3^{n\plus{}1}\end{array}\right]$.

That's correct.

Why don't you examine wheter the result is correct or not by substituting $ n \equal{} 0,\ 1$? :wink:

Answer:
\[ \boxed{\boxed{A^n \equal{} \frac {3^n \minus{} 1}{2}A \minus{} \frac {3^n \minus{} 3}{2}E}}\]

Yikes! I really screwed that up. The eigenvalues are $ 3$ and $ 1,$ not $ \minus{}3$ and $ \minus{}1.$ The characteristic polynomial is $ x^2\minus{}4x\plus{}3,$ not $ x^2\plus{}4x\plus{}3.$ And my diagonalization post was riddled with errors. I've tried to clean it up, but I won't swear I got everything.

Sorry,i made a mistake. (Applying Cayley Hamilton's theorem, we get $ A^2\minus{}4A\plus{}3I\equal{}O$.) But the approach was right, though.... :)

That's O.K. but you should edit it. :wink:

Anyway here is my solution:

The equation of the line passing through the points $ (1,\ 1),\ (3,\ 3^n)$ on $ y \equal{} x^n\ (n\geq 2)$ is $ y \equal{} \frac {3^n \minus{} 1}{2}(x \minus{} 1) \plus{} 1\Longleftrightarrow y \equal{} \frac {3^n \minus{} 1}{2}x \minus{} \frac {3^n \minus{} 3}{2}$, we can write $ x^n \equal{} (x \minus{} 1)(x \minus{} 3)Q(x) \plus{} \frac {3^n \minus{} 1}{2}x \minus{} \frac {3^n \minus{} 3}{2}$ where $ Q(x)$ is a monic polynomial with degree $ n\geq 2.$, thus by Caley-Hamilton theorem, we have $ A^2\minus{}4A\plus{}3E\equal{}\mathbb{O}$.

$ \therefore A^n \equal{} (A^2 \minus{} 4A \plus{} 3E)Q(A) \plus{} \frac {3^n \minus{} 1}{2}A \minus{} \frac {3^n \minus{} 3}{2}E$, yielding $ A^n \equal{} \frac {3^n \minus{} 1}{2}A \minus{} \frac {3^n \minus{} 3}{2}E\ (n\geq 0)$

By Caley-Hamilton thorem we have $ A^2 - 4A + 3E = \mathbb{O}\Longrightarrow A^n - 4A^{n - 1} + 3A^{n - 2} = \mathbb{O}\ (n\geq 2).$
we can write the equation in two ways as follows:

$ \left\{ \begin{array}{cc}{ll} A^n - A^{n - 1} = 3(A^{n - 1} - A^{n - 2}) & \quad \
A^n - 3A^{n - 1} = A^{n - 1} - 3A^{n - 2} & \quad \end{array} \right.$

$ \therefore \left\{ \begin{array}{{cc}{ll} A^n - A^{n - 1} = 3^{n - 1}(A - E})\ \cdots [1] & \quad \
A^n - 3A^{n - 1} = A - 3E \ \ \ \ \ \ \ \cdots [2] & \quad \end{array} \right.$

$ ([1]*3 - [2])\div 2$ gives $ A^n = \frac {3^n - 1}{2}A - \frac {3^n - 3}{2}E\ (n\geq 0)$

My penance for screwing this up: I led off my class today with exactly this example. We're studying eigenvalues, eigenvectors, and diagonalization.

My second lecture example was a formula for the powers of $ \begin{bmatrix}0&1\\1&1\end{bmatrix}.$

That's always a good example.

[quote="Kent Merryfield"]
My second lecture example was a formula for the powers of $ \begin{bmatrix}0 & 1 \\
1 & 1\end{bmatrix}.$

That's always a good example.[/quote]

Let $ \alpha ,\ \beta \ (\alpha < \beta)$ be the roots of the quadratic equation $ \lambda ^2 \minus{} \lambda \minus{} 1 \equal{} 0$,

Doing as my previous post, #10, we obtain $ A^n \equal{} \frac {\beta ^ n \minus{} \alpha ^n}{\beta \minus{} \alpha}A \plus{} \frac {\beta ^{n \minus{} 1} \minus{} \alpha ^ {n \minus{} 1}}{\beta \minus{} \alpha}E$, yielding $ \boxed{\boxed{A^n \equal{} \left( \begin{array}{cc} f(n \minus{} 1) & f(n) \\
f(n) & f(n \plus{} 1) \end{array} \right)\ (n\geq 0)}}$ for some $ f(n)$.

What's $ f(n)$? That's left for you. :D

I have more 4 solutions. :lol:

1.Diagonalization

2．Eigenvalue, Eigenvector

3. Reccurence

4. [color=blue]Spectrum Decomposition[/color]

I worked it into an explicit formula for the $ n$th Fibonacci number.

Yes, at first sight the problem, it occured to me.

Kent Merryfield, see my previous post, What's ''Spectrum'' in Spectrum Decomposition that appeares in Japanese University entrance examination?


What is the greatest prime factor of $ 55^{100}\plus{}55^{101}\plus{}55^{102}$?

Please excuse the revival, but what does the $ \nmid$ symbol mean?


calculus
derivative
function
real analysis
real analysis unsolved

Anyone can demonstrate how  to deduce  Faá di Bruno's formula for nth derivative of the composition f(g(t)),if f(t) and g(t) are functions for which all necessary derivatives are defined?

Please moderators Kent,blahblahblah,fedja,.., May you solve it?
thanks


The chart below was created from the results of a radio station survey. What percent
of the males surveyed listen to the station?
\begin{tabular}{| c | c | c | c|}
\hline &Listened to Station&Did not Listen to Station & Total\\ \hline
Male & $ 39\%$& $ 13\%$&$ 52\%$ \\ \hline
Female &$ 29\%$ & $ 19\%$ & $ 48\%$ \\ \hline
Total & $ 68\%$ & $ 32\%$ & $ 100\%$ \\ \hline
\end{tabular}

$ \begin{tabular}{| c | c | c | c|}
\hline &Listened to Station&Did not Listen to Station & Total\\ \hline
Male & 39\%&13\% & 52\%\\ \hline
Female & 29\%& 19\% &  48\%\\ \hline
Total & 68\% &  32\%& 100\% \\ \hline
\end{tabular}$

$ \frac{39}{52} \equal{} \boxed{75\%}$


geometry
3D geometry
sphere
topology
advanced fields
advanced fields unsolved

[quote="chungyc"]Show that the infinite-dimensional sphere $S^{\infty} = \cup_{n} S^{n}$ is contractible.[/quote]

How's the book reading going so far?

You seem to have lots of questions, do you ever come around to answering them?

I'm just curious by the way. I plan on studying Algebraic Topology sometime soon, like maybe in the Summer.

[quote="chungyc"]I would say that the book is suitable for beginners like myself because it has pictures and lots of descriptions to illustrate concepts. My lecturer isn't really helping much with the way he's teaching so I am struggling with the subject. Knowing the theory is one matter; being able to apply it to problems is another. I have been going through the exercise at the end of chapter 0, and so far I've only been able to solve about half of the questions. Those that I have posted are some questions that I've not been able to answer.[/quote]

Have you considered reading other Introductions to Algebraic Topology?

I find it better to have two texts. I follow one, and if it isn't clear at one point, I look it up in the other. That sometimes works, but then again there are plenty of books at the university library, which is where I assume you study.

If you are capable of answering half the questions, that's pretty good by my standard. The best thing to do is to just move on to Chapter 1, and come back to it later. Unless of course, you need to answer those questions for the next chapter.

I wish I can help, but like I mentionned, I'm just not there yet.

Evidently every homotopy group is trivial, to see this just map some n-sphere to the infinite-sphere.

On the other hand, you may construct a homotopy from the id to constant map.  For example, you slide the (n-1)-sphere along the n-sphere continuously to a point in the homotopy map F(x,t) when t belongs to [1-2^-(n-1),1-2^-n].

Or, you can just look at Example 1.B.3 in Hatcher.

It seems as if everyone uses Munkres for point-set topology, but no one uses the second half of the book for algebraic topology. Is it lacking in some way, or other books just better? I don't know since I haven't read any algebraic topology yet.


In the figure below, ABCD is a square piece of paper 6 cm on each side. Corner C is folded over so that it coincides with E, the midpoint of $ \overline{AD}$. If $ \overline{GF}$ represents the crease created by the fold, what is the length of $ \overline{FD}$? Express your answer as a common fraction.
[asy]import geometry_dev;
size(150);
pair A = (0,0), B = (0,1), C = (1,1), D = (1,0);
path square = A--B--C--D--cycle;
draw(square);
label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE);
pair E = midpoint(A--D);
line CE = line(C,E);
pair X = midpoint(C--E); line FG = perpendicular(X,CE);
pair[] intwithsquare = intersectionpoints(FG,square);
pair G = intwithsquare[0];
pair F = intwithsquare[1];
draw(F--G,dashed);
label("F",F,E);
label("G",G,W);[/asy]

since 6-x is the hypotenus (i killed that spelling) of the right triangle, shouldn't it be (x-6)^2[b]-[/b]9?

Yes, I think it should be.

$ (6 \minus{} x)^2 \minus{} 9 \equal{} x^2$

$ 36 \minus{} 12x \plus{} x^2 \minus{} 9 \equal{} x^2$

$ 27 \minus{} 12x \equal{} 0$

$ x \equal{} \boxed{\frac{9}{4}}$


What does $\wedge$ mean? Well I think if $D$ and $d$ are both $\in N$, then $d^2 \in N$. Since $D=d\sqrt{d}$, $D$ is in $N$, so the other side of the equation must be also in $N$, so $\sqrt{d}$ is also in $N$.

[quote="shobber"]What does $\wedge$ mean?[/quote]It means "and".

I think it is true, As shobber siad. So Bruno Bonagura Are we right? :huh:

$d$ is an integer so $\sqrt{d}$ [b]can't[/b] be a fraction. It is obvious but I am trying to prove it... This is a very difficult problem for me...

[quote="Bruno Bonagura"]I don't know  :|
I think shobber afirmation is false. Look...

$D = d^2\sqrt{d} \thickspace \Rightarrow \thickspace \sqrt{d} = \frac{D}{d^2}$

Both $D$ and $d^2$ are natural but their division may not result a natural too.

I'm really confused!  :?[/quote]

Since you KNOW $D=d^2\sqrt d$ and $D\in \mathbb{N}$ --> $d^2\sqrt{d}$ must also be Natural .  And if $d^2\sqrt{d}\in \mathbb{N}$ $\wedge$ $d\in \mathbb{N}$ --->$\sqrt d\in \mathbb{N}$ . So i guess shobber isnt wrong ....

[quote="Bruno Bonagura"][quote="shyong"]Since you KNOW $D=d^2\sqrt d$ and $D\in \mathbb{N}$ --> $d^2\sqrt{d}$ must also be Natural .  And if $d^2\sqrt{d}\in \mathbb{N}$ $\wedge$ $d\in \mathbb{N}$ --->$\sqrt d\in \mathbb{N}$ . So i guess shobber isnt wrong ....[/quote]

I also think it is right but I don't believe in "guess mathematics". I'm trying to prove it.

We know that for every $a,b \in \mathbb{N}$ we have $ab \in \mathbb{N}$ (field axiom). But a solution for this problem is to prove that: $a,b \in \mathbb{N} \thickspace \wedge \thickspace a = bc \thickspace \Rightarrow \thickspace c \notin \mathbb{N}$ It's true?  :?[/quote]

Lets see , if $ab\in \mathbb{N}$ , then is it always $a,b\in \mathbb{N}$ . Well , this is not true here since it can be $a=\frac 12$ and $b=10$ . But if $a\in \mathbb{N}$ , how would you say about $b$ ? Could $b$ always be natural ? No again since $b$ CAN be a fraction
.
Ok back to our question , now since $d^2\sqrt d\in \mathbb{N}$ and $d^2\in \mathbb{N}$ , $\sqrt d$ can be either Natural or Fraction . But you know that $d$ is Natural , and a square root of natural number wont give you a fraction (it can be either an integer or irrational number obvious) . So in conclusion , $\sqrt d \in\mathbb{N}$

[quote="Bruno Bonagura"]The proposition below is true or false? If it's true, then prove.

$D \in \mathbb{N} \thickspace \wedge \thickspace d \in \mathbb{N} \thickspace \wedge \thickspace D = d^2 \sqrt{d} \quad \Rightarrow \quad \sqrt{d} \in \mathbb{N}$[/quote]$D=d^2\sqrt{d}\Longrightarrow\sqrt{d}=\frac{D}{d^2}\in\mathbf{Q}$. It is well known, as shyong just said, that when $d\in\mathbf{N}$, $\sqrt{d}$ is either irrational or an integer. Since it is in $\mathbf{Q}$, it's not irrational, so $\sqrt{d}\in\mathbf{N}$. What's all the confusion about?

The fact $d\in \mathbb{N} \implies \sqrt{d}\notin\mathbb{Q-N}$.  So that means $\sqrt{d}$ is either natural or irrational.  Then $D=d^2\sqrt{d} \implies \sqrt{d}\in \mathbb{Q}$.  Thus, the intersection of these two implies that it is natural.

[quote="Bruno Bonagura"]Now it seems a little bit clearer... Thanks by the help!

But somebody can prove that?

$d \in \mathbb{N} \implies \sqrt{d} \in (\mathbb{R - Q}) \thickspace \vee \thickspace \sqrt{d} \in \mathbb{N}$

Please?[/quote]

I don't like the excessive use of symbols - words are nice too.  Anyways,

$d \in \mathbb{N} \implies \sqrt{d} \in (\mathbb{R - Q}) \thickspace \vee \thickspace \sqrt{d} \in \mathbb{Q}$

is a tautology, as soon as we know that square roots do exist (they do).

So we want to show that $\sqrt{d}\not\in(\mathbb{Q-N})$.  Assume otherwise.  Then we have $\sqrt{d}=m/n$ with $(m,n)=1$ and $n>1$, so $d=m^2/n^2$, and $(m^2,n^2)=1$, and $n^2>1$, so that $d\not\in \mathbb{N}$, a contradiction.

Well it was me the unblessed who formuled this question ( http://www.somatematica.com.br/forumsm/viewtopic.php?p=18843#18843 ).I based myself on that proposition to solve a very dificult question, but if that proposition is wrong, the solution to that question is wrong too. Bruno Bonagura participates of that forum and indicated me this forum.

Well, I will use Bruno Bonagura logic:
$\sqrt{d} = \frac{D}{d^2} \ \ \ \{D, d\} \in \mathbb{N}$

$D=d^2\sqrt{d} \ \Rightarrow \ d^2$ and $\sqrt{d}$ are divisors of $D$. If $d^2$ is a divisor of $D$, then $\frac{D}{d^2}$, which is equal to $\sqrt{d}$, is a natural number.

Is that right??

[size=75][i]edit: I liked that Latex code![/i][/size]  :D


algebra
polynomial
integration
algebra unsolved

[quote="Nbach"]Let be a natural number $ k$ and a polynomial $ P(x)\in R[x]$ such that $ P(2x) = 2^{k-1}(P(x)+P(x+\frac{1}{2}))$ $ \forall x\in R$ . Prove that :
        $ P(3x) = 3^{k-1}( P(x)+P(x+\frac{1}{3})+P(x+\frac{2}{3}))$ $ \forall x\in R$
Comment : This problem looks very nice ! It's the reason why I post it again .
and to N.T.Tuan : I'll try to corect my mistakes . :)[/quote]
Obviosly if $ P(x)$ is solution
(1) $ P_{k}(2x) = 2^{k-1}(P_{k}(x)+P_{k}(x+\frac{1}{2})$,
then $ aP(x)$ is solution too. Therefore we can consider only monic polinoms (with leader coefficient=1).
From (1) we get
(2) $ P_{k}'(2x) = 2^{k-2}(P_{k}'(x)+P_{k}'(x+\frac{1}{2})$ and
(3) $ P_{k}(0)(2^{1-k}-1) = P_{k}(\frac{1}{2})$.
From (2) and (3) we get unique monic polinom
$ P_{0}\equiv 1,\ P_{1}= x-\frac{1}{2},\ P_{2}(x) = x^{2}-x+\frac{1}{6},...P_{k}(x) = k\int_{0}^{x}P_{k-1}(y)dy-c,c =-\frac{k}{2-2^{1-k}}\int_{0}^{1/2}P_{k-1}(y)dy.$
Therefore $ P_{k}(x)$ is $ k-th$ polinom Bernulli.
Prove. Let $ f_{k}(x) = P_{k}(1-x)$, then $ f(2x) = P_{k}(1-2x) = 2^{k-1}(P_{k}(\frac{1}{2}-x)+P_{k}(1-x)) = 2^{k-1}(f_{k}(x)+f_{k}(\frac{1}{2}+x)$,
therefore $ P_{k}(1-x) = (-1)^{k}P_{k}(x)$ and $ P_{k}'(x) = kP_{k-1}(x)$. It give$ P_{k}(x)\equiv B_{k}(x)$.
By same method from $ P_{k}(3x) = 3^{k-1}(P(x)+P(x+\frac{1}{3})+P(x+\frac{2}{3})$ we get $ P_{k}(x)\equiv B_{k}(x)$ for monic polinoms.


limit
integration
calculus
calculus computations

It's been a couple years since I took Calc 1a, and my friend asked me a question that I couldn't figure out

$ \lim_{x \minus{} > \infty} \int_{x}^{x^2} e^{ \minus{} t^2} dt$

There has to be some sort of trick that I'm missing
I tihnk the answer is 0, but I have no idea if that is right nor how to show it

Any help is appreciated

Thanks

For $ x>1,$

$ 0<\int_x^{x^2}e^{-t^2}\,dt<\int_x^{\infty}e^{-t^2}\,dt$

$ \int_x^{\infty}e^{-t^2}\,dt<\int_x^{\infty}e^{-t}\,dt=e^{-x}$

And that goes to zero. We could make the estimate much tighter than this, but we don't need to in order to solve the problem.

Generalization: Suppose $ \int_0^{\infty}f(t)\,dt$ converges and suppose that $ g(x)<h(x)$ and $ \lim_{x\to\infty}g(x)=\infty.$ Then

$ \lim_{x\to\infty}\int_{g(x)}^{h(x)}f(t)\,dt=0.$

Thanks so much professor. That is easy to understand, I appreciate it.

[quote="ankur87"]Calculate $ L = \lim_{x\to\infty} \int_{x}^{x^2} e^{ - t^2}\ \mathrm {dt}\ .$[/quote]
Suppose w.l.o.g. that $ x > 1$ . Thus $ 1 < x < x^2$ and exists  $ c_x\in \left(x,x^2\right)$ so that

$ \int_{x}^{x^2} e^{ - t^2}\ \mathrm {dt} = \left(x^2 - x\right)\cdot e^{ - c^2_x}\ .$ Thus, $ x < c_x < x^2$ $ \implies$ $ \lim_{x\to\infty}c_x = \infty$ ,

$ 0 < \frac {x^2 - x}{e^{c^2_x}} < \frac {x^2 - x}{e^{x^2}}$ , $ \lim_{x\to\infty}\frac {x^2 - x}{e^{x^2}} =\lim_{x\to\infty}\frac {x^2-x}{x^2}\cdot\frac {x^2}{e^{x^2}}=0$ $ \implies$ $ L = 0\ .$

I also used mean value theorem of Integral, Virgil Nicula. :)


need help in this urgently

how do you prove the differentiability of the function a(x,y) if we know b(x,y) and b(x,y)*a(x,y) are both differentiable?

thanks

We don't: take $ b\equiv 0$ and choose $ a$ to be whatever you want. Of course, you didn't mean this case, so, in other situations, just recall that the quotient of two differentiable functions is differentiable when the denominator does not vanish.

Now as a moderator, I have to remark that this is [b]not[/b] a homework help hotline, so normally requests for "urgent help" with routine questions are ignored at best and result in not answering your posts at all at worst ;) (the fact that I answered now is due to a combination of my good mood and your being new to the forum :)).


[quote="cowpi"]9) Find the equation of the line which intersects the positive x and positive y-axis, contains the point $(-5/2,3)$ and the sum of whose intercepts is 4.

10) What is the slope of the line passing though the vertices of the graphs of the equations $y=x^2-4x+3$ and $y=-x^2-4x-1$?[/quote]
[hide="9"]
If y=y intercept and x=xintercept, we know that from the point given $\frac{3}{\frac{5}{2}+x}=\frac{y}{x}$ and $y+x=4$. Substituting $y=4-x$ into the first equation and solving for x, $x=\frac{5}{2}$ and then $y=\frac{3}{2}$, then finding the slope and yintercept the equation is $\boxed{y=-\frac{3}{5}x+\frac{3}{2}}$[/hide]
[hide="10"]
Vertex: 2 and -2
for the first one, $f(2)=-1$ and for the second, $f(-2)=3$. Points are (2,-1) and (-2,3) and the slope is $\boxed{-1}$. [/hide]

[hide="Number 10"]
I graphed the lines on my calculator and found the vertices are $(-2,3)$ and $(2,-1)$. Apply slope formula to get $\boxed {-1}$.[/hide]


AMC USA(J)MO USAMO MATHCOUNTS ARML percent factorial

The list is comprised of every state we have received information from with at least a 37 cutoff and the other option. I think Indiana or California will win the team.

I'm rooting for NY, although I have a feeling Texas is going to win again...
P.S. Is ESPN airing the countdown round?When?

I'm rooting for Indiana, and I have a feeling they will win

I doubt IN or TX will win again. I think Indiana's lost a lot of depth compared to last year's team, and TX has also lost its top scorers. When in doubt, vote Cal.

Michigan all the way.  We have a chance.

Thanks Kyyuanmathcount, I couldn't have said it better myself.

the maryland team will own in the competition.  THey are amazing. :D

Kansas is going to win :rotfl: !      ......I voted for Michigan.

Ohio has a 37 cutoff, but seeing as I am the only one in the state and I already voted for another, it isn't going to get any votes...

[quote="kyyuanmathcount"]Yes for Michigan.

Who else voted for Michigan, I thought Neil and I were the only "Michiganders" (I think).

Definitely many teams have a chance, but, though the Michigan cutoff wasn't that high, we have NEIL GURRAM, that says it all.

Mostly what matters is the team round, so as long as the team can work together and has good strategy, you should get a 10 on the team round, and with it, the national championship.[/quote]

Team round isn't going to be the determining factor, it's going to be the team's average.

Top 5 I think will be Texas, California, Indiana, Michigan, and somewhere else.

I don't think Texas will win this year =\ California has Kevin Yang and generally high state scores obviously maybe 2nd or 3rd? Indiana I think will win, their team is still pretty good, Michigan I think will also get 2nd or 3rd just missing Indiana :P

[quote="kyyuanmathcount"]Who else voted for Michigan, I thought Neil and I were the only "Michiganders" (I think).
[/quote]

MICHIGAN PRIDE!!!

You thought wrong :D

I voted for MI because I'm from MI, but I still think the MI team lost everything when Alan Huang left.

But we have gained stuff, and don't think that we will not do well just because Alan Huang is gone.

Obviously South Carolina will win it all.  They've got me...lol.

I thought the answer was unfair at the time, but don't they start the timer right a fter you buzz in anyways?

[quote="janieluvsmusic"][quote="Treething"]hey, I won kinda fairly...[/quote]

i'm not saying you didn't, from what i heard, you just used it to your advantage.[/quote]

see, loopholes are for winners :D

neal wu could've beat me 4-0 if he used that to his advantage, Andrew could've beaten me 4-2 probably

but nathan benjamin was funny either way

[quote="Fanatic"]Virginia was 1/4 point ahead of Washington, Indiana was 1/4 point behind Washington, Texas was far back (2-4 points?)[/quote]

heh. 2.5

[quote="Ignite168"][quote="janieluvsmusic"]what was the cutoff for team awards?[/quote]

I think you need like top 3 teams, and Pulak definitely could've improved 11 points for TX to get third. Pulak's score, let's just say, was within 3 points of half of Neal Wu's score.[/quote]

35 would've put me in 14th place, which I wasn't really expecting anyway, so I doubt I could've improved that much myself.

Congratulations to the Virginia kids for their showing at Nationals.  They did a great job and deserve all their kudos!!!

Also, congratualtions to the 4 Texas kids.  4th in the nation is fantastic.  I am proud of everything that you did to prepare this year.  A special congratulations for Kevin Chen for finishing 3rd in the Nation as a 7th grader. 29th as a 6th grader and 3rd as a 7th grader..WOW!  A great improvement!!

Also congratualtions to everyone that went to Nationals.  No matter how you did you are all STATE CHAMPIONS!!!!  No one can ever take that away from you.  You are all the best in your state!

 :lol:  :lol:  :lol:  :lol:  :lol:

Jeff Boyd
Texas Coach

yay

now i just have to start preparing for next year

[quote="samath"]I have no next year :( goodbye mathcounts.
I'll probably be back as some sort of assistant coach though ;)[/quote]
But, you did make countdown, and you did do pretty well.


real analysis
topology
advanced fields
advanced fields unsolved

Given $ \varepsilon > 0$,  construct an open set $ E\subset [0,1]$ that is dense in $ [0,1]$ such that $ m(E) \equal{} \varepsilon$ ($ m(E)$ is the Lebesgue measure)

I thought of a set which seems to work but is cumbersome to write down: Take open intervals of lenghts $ \frac {\varepsilon}{2}$, $ \frac {\varepsilon}{4}$, ...
Divide the first into two and put the parts at the ends of the intervals, then place the second centred around $ \frac {1}{2}$ and so on.

This looks something like:

$ \left(0,\frac {\varepsilon}{4}\right)\cup\left(1 \minus{} \frac {\varepsilon}{4},1\right)\cup\left(\frac {1}{2} \minus{} \frac {\varepsilon}{8},\frac {1}{2} \plus{} \frac {\varepsilon}{8}\right)\cup\left(\frac {1}{4} \plus{} \frac {\epsilon}{32},\frac {1}{4} \plus{} \frac {3\varepsilon}{32}\right)\cup\left(\frac {3}{4} \minus{} \frac {3\epsilon}{32},\frac {3}{4} \minus{} \frac {\varepsilon}{32}\right)\cup\ldots$

Any ideas of a more elegant solution?
Thanks!

choose an enumeration $ q_1,q_2,...$ of the rationals in $ [0,1]$ and choose balls around the $ q_n$ of radius $ \epsilon 1/2^{n\plus{}1}$ (or something like that).

That gets you an open set with measure strictly less than $ \epsilon$, since the balls overlap.

The construction in the first post works. Another idea is the complement of a Cantor-type set:
Let $ a_n$ be a sequence of numbers in $ (0,1)$ with $ \prod_n (1\minus{}a_n)\equal{}1\minus{}\epsilon$. Repeat this procedure: at step $ n$, remove the open middle $ a_n$ part of each remaining interval. After infinitely many steps, we are left with a closed set with empty interior and measure $ 1\minus{}\epsilon$. Its complement is the desired dense open set of measure $ \epsilon$.

Thanks jmerry

[quote="jmerry"]Repeat this procedure: at step $ n$, remove the open middle $ a_n$ part of each remaining interval.[/quote]

I guess you mean "remove the open middle parts of all remaining intervals so as total length removed is $ a_n$"

Also, is there a better way to write down my original construction?

No. I mean my $ a_n$ to be the proportion removed at each step. The original Cantor set uses $ a_n\equal{}\frac13$, so its measure is $ \prod_{n\equal{}1}^{\infty}(1\minus{}\frac13)\equal{}0$.


logarithms
calculus
calculus computations

(1) Solve $ x\frac {dy}{dx} \plus{} y \equal{} 0$.

(2) Solve $ x\frac {dy}{dx} \plus{} y \equal{} x\ln x$ in using (1).

hello, let's solve your equation (1), writing your equation in the form
$ \frac{dy}{y}\equal{}\minus{}\frac{1}{x}dx$ from here we get all solutions in the form
$ y(x)\equal{}\frac{C}{x}$ with $ x\neq0$.
For $ x\equal{}0$ we have $ y(0)\equal{}0$.
Sonnhard.

hello, to your second problem, $ y(x) = \frac {C}{x}$ is a solution of the homogenius part of the equation $ y^'+\frac {y}{x} = \ln(x)$,by the ansatz $ y(x) = \frac {C(x)}{x}$ we will find a special solution of the inhomgenius part, we have $ y^'(x) = \frac {C^'(x)x - C(x)}{x^2}$ inserting this in the equation we get $ C^'(x) = x\ln(x)$ and from here $ C = \frac {1}{2}x^2\ln(x) - \frac {1}{4}x^2$, finally we get
$ y(x) = \frac {1}{2}\ln(x)x - \frac {1}{4}x + \frac {C}{x}$.
Sonnhard.


floor function
number theory proposed
number theory

Let $ x_{1},x_{2},\cdots,x_{n}$ be real numbers such that the identity

$ \left\lfloor mx_{1}\right\rfloor\plus{}\left\lfloor mx_{2}\right\rfloor\plus{}\cdots\plus{}\left\lfloor mx_{n}\right\rfloor \equal{}\left\lfloor m\left(x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{n}\right)\right\rfloor$

holds for all positive integers $ m$. Prove that at most one of the $ x_{k}$'s ($ 1\leq k\leq n$) can be non-integer.

[hide="Hint (zero and one non-integer)"]Using that $ \lfloor z\plus{}r\rfloor\equal{}z\plus{}\lfloor r\rfloor$ with $ z\in\mathbb Z$ and $ r\in\mathbb R$ it is easy to prove that it is true for zero or one non-integer[/hide]
[hide="Hint 2 (Simplification for the two or more non-integer)"]Using that if $ x_{1},x_{2},\cdots,x_{n}$ works then $ x_{1}\minus{}z_{1},x_{2}\minus{}z_{2},\cdots,x_{n}\minus{}z_{n}$ also works with $ z_{i}\in\mathbb Z$ we just need to prove that $ f_{1},\cdots,f_{n}$ does not work for ($ 0< f_{i}< 1$) and ($ 2\leq n$)[/hide]


algebra
polynomial
search
Vieta
number theory
prime numbers
AMC

For each positive integer $ n$, let $ f(n)\equal{}n^4\minus{}360n^2\plus{}400$. What is the sum of all values of $ f(n)$ that are prime numbers?

$ \textbf{(A)}\ 794\qquad
\textbf{(B)}\ 796\qquad
\textbf{(C)}\ 798\qquad
\textbf{(D)}\ 800\qquad
\textbf{(E)}\ 802$

[hide="hint"]Try to factor $ f(n)$ as a difference of squares.  See Problem 1 at [url]http://www.princeton.edu/~ploh/docs/math/b1-polynomials.pdf[/url] [/hide]

[hide="Solution"]Factor $ f(n) \equal{} n^4 \minus{} 360n^2 \plus{} 400 \equal{} (n^2 \plus{} 20)^2 \minus{} 400n^2 \equal{} (n^2 \minus{} 20n \plus{} 20)(n^2 \plus{} 20n \plus{} 20)$. WLOG take $ n\geq0$. Then we must have $ n^2 \minus{} 20n \plus{} 20 \equal{} 1$ (or $ (n\minus{}19) (n\minus{}1) \equal{} 0$) and $ n^2 \plus{} 20n \plus{} 20$ prime. Both $ n \equal{} 1$ and $ n \equal{} 19$ satisfy the primality condition, producing $ f(n) \equal{} 41$ and $ f(n) \equal{} 761$, respectively. Adding, we have $ 41 \plus{} 761 \equal{} 802$.[/hide]

[quote="360_Fan"]This might be a stupid question, but what's WLOG?[/quote]

Without Loss of Generality.

[hide="Terribly Stupid Solution"]Check $ f(1)$. It's prime. $ f(2)$ and a few afterward are negative. Because we've memorized our squares, we see that $ f(n)\equal{}n^2(n^2\minus{}360)\plus{}400$ becomes positive again at or before $ f(19)$. $ f(18)$ is still negative, so we test $ f(19)\equal{}761$. Since any remaining output is higher than any of the answer choices, there can be no more primes, and $ 41$ alone is too small. So the answer is $ 41\plus{}761\equal{}\boxed{802}$.[/hide]

Sorry to revive.
[hide="Solution"]Let $f(n)=p$ for some prime number $p$. Then, $n^4-360n^2+400-p=0$. It only remains for us to check perfect squares. To make the search easier, note that the sum of the roots must be $0$ by Vieta's Formulas. Thus, if $n$ is a positive integer, the polynomial can be factored as $(n^2-a^2)(n^2-b^2)$ for some positive integers $a$ and $b$. Let $a^2=m$ and $b^2=n$. Then, $m+n=360$ and $mn=400-p$. Moreover, we only need to check $(m,n)=(-1,-359)$ and $(m,n)=(1,-361)$, which result in $p=41$ and $p=761$, respectively. Both $41$ and $761$ are prime, and $41+761=(D) \boxed{802}$.[/hide]

amctrivial grind: This factors as $(n^2-20n+20)(n^2+20n+20)$. This can only be prime when one of these values is $1$. If the first value is $1$, then $n^2-20n+19=0$, so $n = 1, 19$. If the second value is $1$, then $n^2+20n+19=0$, so $n = -1, -19$. Only the first two values are positive, and plugging in gives $41, 761 \implies \boxed{802}$.


geometry
circumcircle
incenter
geometry solved

Let ABC be a triangle and P a varying point on the arc BC of the circumcircle of ABC. Prove that the circle through P  and the incenters of PAB and PAC pass through a fixed point independent of P.

I have the feeling this was posted before.

[color=red][Moderator edit: Yes, http://www.mathlinks.ro/Forum/viewtopic.php?t=555 .][/color]

I don't know but somehow I have the feeling that paul_mathematics likes to repost problems which are already on the forum.


find all function f:$R\to R$  and $f$ continuous such that :
 $f(f(f(x))) +f(x)=2x$  for all x in real numbers.

Look around on the first page of this section and you'll find the answer.


geometry
incenter
angle bisector
geometry unsolved

ABC a triangle, I the incenter of ABC, U, V, W the symmetrics of I wrt BC, CA, AB, A'B'C' the orthic triangle of ABC,
Y the point of intersection of BC and B'C', Z	the point of intersection of AB and A'B', By, Bz the internal bisectors wrt <C'YB, <BZA', P the point of intersection of Bz and BC, R the point of intersection of By and AB, P' the point of intersection of UV and BC, R'	the point of intersection of VW and AB.
Conclusion :	PR is parallel to P'R'

Here is my solution.

Let $ D, E$ be intersections of $ IU$ and $ BC$ ; of $ IW$ and $ AB$.

$ C'R$ and $ YR$ are 2 angle bisectors of triangle $ C'YA'$ => $ A'R$ is the angle bisector of $ \angle{C'A'B}$.
Similarly, $ C'R$ is the angle bisector of $ \angle{A'C'B}$
Then $ A'R$ and $ C'R$ are 2 angle bisectors of triangle $ BA'C'$.

Since, $ AI$ is the line bisector of $ VW$, after a few deductions, we have $ \angle{EIR'} \equal{} \angle{EIA}$
Therefore, $ ER'.EA \equal{} DP'.DC \equal{} r^2$

From this, we can position the points $ P'$ and $ R'$. Then I use the algebraic transformation to deduce : $ \frac{BP}{BP'} \equal{} \frac{BR}{BR'}$
(I 'm just lazy to write all of this here :D )

(QED)

Can you explain your "algebraic transformation"?
Thank you
Jean-Louis

"Algebraic transformation" means calculating all the segments.

We can calculate $ BP, BP', BR, BR'$ (based on $ a,b,c$ - the 3 sides of triangle $ ABC$) and then all that we have to do is proving an identity.

I have found a proof based on "Two parallels (2) and (3)".
I will put it on my web site which I will open soon.
Jean-Louis


Um, there's already a topic about this.  :maybe:

Oh well.

Are you Eric Yang the guy who scored an 85 last year?

The results come up tomorrow morning.  :lol:

I guess results won't be up till later today.  :maybe:

oops i won

edit> roflrofl

Vestavia won

and we only had people compete in three divisions even tho all five  counted

that's hilarious

haha actually only four counted.

A 12th grader at our school got a 50. 11th grader highest was 60, 10th grade highest at our school 60, and 9th grade highest at our school 60.

So those four add up to 230, the final score. Yeah I got a 34 lol bet you can't guess which one I am.

Vestavia only entered in 10th 11th and 12th.
Scores for the three added up to 240.  :rotfl:

The scores were a bit down this year...

ARGH HOW DID I GET A 52 MAYBE THESE PEOPLE CAN'T GRADE
or maybe:
1) Wrong answer key given
2) Teachers can't grade

MAN I AM SO MAAAAAAAAAAAAAAAADDDD! :cursing:

lol that happened to me on the fall startup.

I was told I had a higher score than the posted score that came up a few weeks later.

oh well I need to speed up lolz in non geometry ciphering

two of our top people's grades dropped from what our teacher expected (wayne zhu and evan zhao both went from 88 to 84)

surprisingly mine didn't drop so I guess I missed different problems

you probably did harder problems that had a definite answer. maybe the answer key given had a few typos or something.  :lol:

mine didn't drop either. Ahhhh Hopefully AoPS Vol. 2 will help me do more of these hard problems, or at least learn some stuff.

MINE DROPPED BECAUSE MY TEACHER DIDN'T SUBMIT MY ANSWER FORM!!!!!!!!!!!!!!!!!!!!


Let $a$, $b$, $c\ge-\frac{7}{15}$ such that $a+b+c=1$. Prove that
\[\frac{1}{1+a+a^2}+\frac{1}{1+b+b^2}+\frac{1}{1+c+c^2}\ge\frac{27}{13}\]

[edit]
 :blush: Sorry, I messed up the constant. It should now work.

We have \[\frac{1}{1 + a + a^2} \geq \frac{162 - 135a}{169}\] and the similar inequalities, since the above inequality is equivalent to \[(15a + 7)(3a - 1)^2 \geq 0.\]
Adding these gives the result.


1) Let ABC  be a fixed triangle . With compass and ruler draw a line which divides the area of the triangle into two equal parts.

2) With compass and ruler draw a line wich divides the area of the triangle into two parts such that the ratio of the areas of the parts be equal to k (k is the length of a given line segment).

Maybe you want some extra restrictions? Like the line passing through some given point? Because otherwise it's easy.. :?

Indeed, there is a fixed point. Not so many time ago I wrote a solution for a problem, where cases $k=1$ and $k=2$ asked (actually we were asked to divide triangle into three parts of equal area). But the idea can be easily applied for your problem.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=28947 .

hi

as you have said , i have checked the question agian

yes it needs a fixed point

sorry for the mistake

and mr.myth thank you for your link
it was intersting


function
complex analysis
complex analysis unsolved

If a function $ f$ statisfies $ f(z^2)\equal{}f(z)\minus{}z$, can you conclude that $ f$ has a singularity in the point $ z\equal{}1$ and $ z\equal{}\minus{}1$. Justify your answer ! Thanks in advance.

Assuming you mean a [i]meromorphic[/i] function $ f$ on the whole complex plane--you really need to be specific about these things--then it seems that no such function exists.  So you can conclude whatever you want.

Specifically, repeated application of the equation for $ z\in \mathbb{C}$, $ |z| < 1$, gives $ f(z^{2^n})\equal{}f(z) \minus{} \sum_{i\equal{}1}^n z^i$, so $ f(0)\equal{}f(z) \minus{} \frac{z}{1\minus{}z}$.  Then $ f(z^2)$ would have a pole at $ \minus{}1$, while $ f(z)\minus{}z$ does not, contradiction.

Somehow I saw the complicated answer to this question first.  :roll:

You can also just set $ z\equal{}1$ if $ f$ has no singularity at $ 1$, then you get $ 0\equal{}\minus{}1$, a contradiction.  Then, if $ f$ has no singularity at $ \minus{}1$, the right hand side has a finite limit as $ z\to \minus{}1$, but the left hand side does not, contradiction.


Let $ A$, $ M$, and $ C$ be digits with
\[ (100A \plus{} 10M \plus{} C )(A \plus{} M \plus{} C ) \equal{} 2005.
\]What is $ A$?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$

[hide="Click for solution"]
The prime factorization of $ 2005$ is $ 5 \times 401$ so $ 100A\plus{}10M\plus{}C\equal{}401$ and $ A\plus{}M\plus{}C\equal{}5$.  Because they are digits, $ A\equal{}4$, $ M\equal{}0$, $ C\equal{}1$ or $ \boxed{\textbf{(D)}}$.
[/hide]

Solution: [hide]Seeing the multiplication shows that 2005 = 5 * 401. The LHS is essentially a 3-digit number times the sum of the digits. We see that 401  * 5 works. Therefore A = 4 and the answer is D[/hide]


Let G be a graph with n vertices with no three cycles, yet contains an x-cycle where x is odd and >3. Prove that there exists a vertex in G with degree at most 2n/5.

Bomb


calculus
integration
trigonometry
geometry
3D geometry
sphere
analytic geometry

$ \int\int(3x\plus{}4y^2)dxdy$ where R is bounded by $ y\equal{}\sqrt{4\minus{}x^2}$,$ y\equal{}\sqrt{1\minus{}x^2}$,$ y\equal{}0$

$ \int\int\int(x^2)dxdydz$ where Q is above $ z\equal{}\sqrt{3x^2\plus{}3y^2}$ and inside $ z\equal{}\sqrt{25\minus{}x^2\minus{}y^2}$

Volume of a cylinder $ (x\minus{}1)^2\plus{}(y\minus{}1)^2\equal{}2$ between $ z\equal{}0$ and $ z\equal{}10\minus{}x^2\minus{}y^2$

The first one I found the limits of x to be -4 to -1, and y from 0 to ?
The second and third I'm not really sure about. Thanks in advance.

[quote="xerosmythe"]$ \int\int(3x + 4y^2)dxdy$ where R is bounded by $ y = \sqrt {4 - x^2}$,$ y = \sqrt {1 - x^2}$,$ y = 0$

$ \int\int\int(x^2)dxdydz$ where Q is above $ z = \sqrt {3x^2 + 3y^2}$ and inside $ z = \sqrt {25 - x^2 - y^2}$

Volume of a cylinder $ (x - 1)^2 + (y - 1)^2 = 2$ between $ z = 0$ and $ z = 10 - x^2 - y^2$

The first one I found the limits of x to be -4 to -1, and y from 0 to ?
The second and third I'm not really sure about. Thanks in advance.[/quote]

[hide]$ 1)$ $ \int\int_R (3x + 4y^2) ~dA$ is bounded by the two semicircles above the x-axis with radii $ r = 1$ and $ r = 2$.

$ R = \{ (x,y) \in \mathbb{R}^2 | ~~ - 2 < = x < = 2, ~~\sqrt {1 - x^2} < = y < = \sqrt {4 - x^2} \}$.  But we also can convert this to polar so:
$ R = \{ (r,\theta) \in \mathbb{R}^2 | ~~0 < = \theta < = \pi, ~~1 < = r < = 2 \}$

$ \int\int_R (3x + 4y^2) ~dA = \int_{ - 2}^{2}\int_{\sqrt {1 - x^2}}^{\sqrt {4 - x^2}}(3x + 4y^2)~dy~dx$
$ = \int_0^{\pi}\int_1^2 (3r\cos \theta + 4r^2 \sin^2 \theta) r~dr~d\theta$

$ 2)$ $ \int\int\int_Q x^2 ~dV$ where $ Q$ is bounded by the cone $ z = \sqrt {3x^2 + 3y^2}$ and the hemisphere $ z = \sqrt {25 - x^2 - y^2}$.

The easiest way to do this one is using spherical coordinates.  Obviously the radius of the sphere is $ 5$ so, $ \rho = 5$.  Obviously $ 0 < = \theta < = 2\pi$.  Using $ z = \sqrt {3x^2 + 3y^2}$, and substituting what we know $ x,y,$ and $ z$ to be in spherical coordinates, we see $ 0 < = \phi < = \frac {\pi}{6}$.

$ Q=\{ (x,y,z) \in \mathbb{R}^3 | ~~ -\frac{5}{2}<=x<=\frac{5}{2}, ~~-\sqrt{\frac{25}{4}-x^2}<=y<=\sqrt{\frac{25}{4}-x^2}, ~~\sqrt{3x^2+3y^2}<=z<=\sqrt{25-x^2-y^2} \}$
$ Q=\{ (\rho,\phi,\theta) \in \mathbb{R}^3 | ~~0<=\rho<=5, ~~0<=\phi<=\frac{\pi}{6}, ~~0<=\theta<=2\pi \}$

So $ \int\int\int_Q x^2 ~dV = \int_0^{2\pi}\int_0^{\frac {\pi}{6}}\int_0^5 \rho ^2 \sin ^2 \phi \cos ^2 \theta ~\rho ^2 \sin \phi ~d\rho ~d\phi ~d\theta$

$ 3)$ $ V=\int\int\int_E 1 ~dV$ where

$ E=\{ (x,y,z) \in \mathbb{R}^3 | ~~1-\sqrt{2}<=x<=1+\sqrt{2}, ~~1-\sqrt{2-(x-1)^2}<=y<=1+\sqrt{2-(x-1)^2}, ~~0<=z<=10-x^2-y^2 \}$

$ E=\{ (x,y,z) \in \mathbb{R}^3 | ~~-\sqrt{2}<=x<=\sqrt{2}, ~~\sqrt{2-x^2}<=y<=\sqrt{2-x^2}, ~~0<=z<=10-(x-1)^2-(y-1)^2 \}$
$ E=\{ (r,\theta,z) \in \mathbb{R}^3 | ~~0<=\theta<=2\pi, ~~0<=r<=\sqrt{2}, ~~0<=z<=8-r^2+2r(\cos \theta + \sin \theta) \}$

$ V=\int\int\int_E 1 ~dV=\int_0^{2\pi}\int_0^{\sqrt{2}}\int_0^{8-r^2+2r(\cos\theta+\sin\theta)} r~dz~dr~d\theta$[/hide]

1. Draw a picture of the region. After that, you'll see that if you change to polar coordinate system, 1 <= r <= 2 and 0 <= theta <= Pi. The calculation is much easier using polar system.
2. Q is bounded by a sphere(top) and an elliptic cone(bottom). Draw a graph. Do the integration in this order instead: dz, dy, dx. For z, the limit is from sqrt(3x^2+3y^2) to sqrt(25-x^2-y^2). For y, the limit is from -sqrt(25/4-x^2) to sqrt(25/4-x^2). For x, the limit is from -5/2 to 5/2.
   You can also change to spherical coordinate system to simplify your calculation. In this case, r is from 0 to 5. theta is from 0 to 2PI, and phi is from 0 to Pi/6.
3. The Region is bounded by the cylinder, z=0 and the elliptic paraboloid. This is pretty much the same as 2. Do the integration in the order of dz, dy, dx. The limits will be clear if you draw a graph.


trigonometry
function
quadratics
vector
algebra
real analysis
real analysis unsolved

I am just beginning differential equations, and wanted to find the solution to $y'' + y = 0$.  I know the solution is $A\cos{t}+B\sin{t}$, but I had trouble in the following (after setting up the characteristic equation).

I found roots of the characteristic equation to be $i$ and $-i$.  So I said the solutions to the differential equation were $e^{it}$ and $e^{-it}$.  The real valued solution basis would then be $\cos{t}$ for $e^{it}$.  When I try to find the real solution basis for $e^{-it}$, I come up with the same.  However, it seems it should be sine instead of cosine (judging by the actual solution that I found online).

$z=e^{it}=\cos{t}+i\sin{t}\implies\textrm{Re}(z)=\cos{t}$

$z^{-1}=e^{-it}=\cos{t}-i\sin{t}\implies\textrm{Re}(z)=\cos{t}$

Why does $e^{-it}$ coorespond to the real valued solution $\sin{t}$?

Thanks for your help.

Alex

Since $e^{it} = \cos t + i\sin t$, and $e^{-it} = \cos t - i\sin t$, we have $\cos t = \frac{e^{it}+e^{-it}}{2}$ and $\sin t = \frac{e^{it}-e^{-it}}{2i}$.  So each of the pairs $\{\cos t,\sin t\}$ and $\{e^{it},e^{-it}\}$ can be expressed in terms of the other.

Suppose $L$ is a homogeneous linear differential operator with real coefficients, and suppose that $w(t)$ is a complex-valued function and $I$ an interval with no singular points of $L$ such that for $t\in I,$  $L(w)=0.$ Then (in all cases, for $t\in I$):

1. $L(\overline{w})=0.$

2. If $\text{Re}(w)=u$ and $\text{Im}(w)=v$, then $L(u)=0$ and $L(v)=0.$

3. If there exists a $t\in I$ such that $\frac{w'(t)}{w(t)}\in\mathbb{C}\setminus\mathbb{R},$ then $u$ and $v$ are linearly independent. If we now assume that $L$ is a second order linear differential operator, the complete general solution of $L(y)=0$ is $y=c_1u+c_2v.$

As for the proof:

1. Since the coefficients of $L$ are real, $\overline{L(w)}=L(\overline{w}).$

2. From (1), we have that $w$ and $\overline{w}$ are solutions of the homogeneous linear differential equation. Hence, so also are any linear combinations of $w$ and $\overline{w},$ notably $u=\text{Re}(w)=\frac{w+\overline{w}}2$ and $v=\text{Im}(w)=\frac{w-\overline{w}}{2i}.$ Note that this is exactly what rgep did above.

3. We have $w=u+iv$ where $u$ and $v$ are real. Then

$\frac{w'}w=\frac{u'+iv'}{u+iv}=\frac{(u'+iv')(u-iv)}{(u+iv)(u-iv)} = \frac{(uu'+vv')+i(uv'-vu')}{u^2+v^2}.$

The condition that this take on a non-real value is precisely the condition that $uv'-vu'\ne0$ at some point. But this is simply the Wronskian of $u$ and $v$, and if the Wronskian is ever nonzero on $I,$, then $u$ and $v$ are independent and, for second order equations, span the solution space.

---

Putting this to practical use:

Consider the differential equation $y''+2y'+5y=0.$ Is we assume solutions of the form $y=e^{rt},$ we get the quadratic equation $r^2+2r+5=0,$ which has roots $r=-1\pm2i.$

Let's just focus on the solution $w=e^{(-1+2i)t}=e^{-t}e^{2it}=e^{-t}(\cos 2t+i\sin 2t).$

Then $u=\text{Re}(w)=e^{-t}\cos 2t$ and $v=\text{Im}(w)=e^{-t}\sin 2t.$

These are independent, and the general solution can be written as

$y(t)=c_1e^{-t}\cos 2t+c_2e^{-t}\sin 2t.$

You may worried about justifying the line "assume solutions of the form ...." The justification comes after the fact. However we came up with that, we did find solutions, and having found a two-dimensional vector space of solutions (in particular, we can solve for all possible initial conditions), the existence-uniqueness theorem tells us we have found all solutions.


linear algebra
matrix
linear algebra unsolved

Let $ A$ be a $ n\times n$ matrix with real entries.

Prove that we cannot have
\[ A^2 \equal{} \text{diag}(\alpha_1,\ldots,\alpha_n)
\]
where $ \alpha_i<0$ for $ 1\leq i\leq n$.

$ A\equal{}\begin{bmatrix}0&\minus{}1\\1&0\end{bmatrix}.$ $ A^2\equal{}\begin{bmatrix}\minus{}1&0\\0&\minus{}1\end{bmatrix}.$

Now, what's the real question supposed to be?

How about "$ n$ is odd" as an additional hypothesis? Then it would be true, since $ A$ would have to have at least one real eigenvalue and hence $ A^2$ would have at least one nonnegative eigenvalue.

[quote="blahblahblah"]actually I copied the question correctly; there must have been an error.  That is a nice argument for the odd case though.[/quote]Transfer to a different university.

[quote="blahblahblah"]That is a nice argument for the odd case though.[/quote]

A better one would be $ \alpha_1\cdot\alpha_2\cdot ...\cdot\alpha_n\equal{}\det\left(A^2\right) \equal{} \left(\det A\right)^2\geq 0$. ;)

  darij


superior algebra
superior algebra theorems

In a ring with identity and with proper ideals, there always exist maximal ideals. Is the statement true for rings with a nontrivial multiplication and with no identity?

[quote="-oo-"]http://sierra.nmsu.edu/morandi/notes/NoMaxIdeals.pdf, Theorem 4[/quote]
Good link, thank you very much!  :wink:


inequalities
calculus
inequalities proposed

I solve this problem today , and I think the inequality is very strict :D

$ \sqrt {\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc}} \geq 1 \plus{} \frac {3}{5} \left( 1 \minus{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \right)$

with the condition $ a,b,c$ are positive.

P.S CAS solution is welcome.  :)

It's obviously beacause: $ (a\plus{}b)(b\plus{}c)(c\plus{}a)(a\plus{}b\plus{}c)^2 \geq\ 24abc(a^2\plus{}b^2\plus{}c^2)$
(Next, assume $ a^2\plus{}b^2\plus{}c^2\equal{}3.$ Let $ a\plus{}b\plus{}c\equal{}t$, we are easily to complete the proof)
 :)

[quote="nguoivn"]It's obviously beacause: $ (a \plus{} b)(b \plus{} c)(c \plus{} a)(a \plus{} b \plus{} c)^2 \geq\ 24abc(a^2 \plus{} b^2 \plus{} c^2)$
(Next, assume $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Let $ a \plus{} b \plus{} c \equal{} t$, we are easily to complete the proof)
 :)[/quote]

Then I will change it to

$ \sqrt {\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc}} \geq 1 \plus{} \frac {17}{25} \left( 1 \minus{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \right)$

and your method does not work . Just check $ t \equal{} 2.9993$  :D

[quote="arqady"]$ \sqrt {\frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc}} \geq 1 \plus{} \frac {11}{16} \left( 1 \minus{} \sqrt {\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}} \right)$ is stronger.[/quote]

Yeah Arqady!
I think the best constant k is k = 0.6883 But I guess you have solved it already.

[quote="Gibbenergy"]
I think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]
I have got $ k_{max}\equal{}0.688300972...$

[quote="arqady"][quote="Gibbenergy"]
I think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]
I have got $ k_{max} \equal{} 0.688300972...$[/quote]

Arqady, Can you please show us your solution?

Thanks  :)

For $ k\equal{}\frac{11}{16}$, using my hint above and the old result:
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{abc} \geq\ 8\plus{}4\sqrt[]{2}\minus{} \frac{4\sqrt[]{2}(ab\plus{}bc\plus{}ca)}{a^2\plus{}b^2\plus{}c^2}$
$ (a\plus{}b)(b\plus{}c)(c\plus{}a)(a\plus{}b\plus{}c)^2 \geq\ 24abc(a^2\plus{}b^2\plus{}c^2)$
Consider 2 cases of t, we have done.
 :)

[quote="Gibbenergy"][quote="nguoivn"]For $ k \equal{} \frac {11}{16}$, using my hint above and the old result:
$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{abc} \geq\ 8 \plus{} 4\sqrt []{2} \minus{} \frac {4\sqrt []{2}(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2}$
$ (a \plus{} b)(b \plus{} c)(c \plus{} a)(a \plus{} b \plus{} c)^2 \geq\ 24abc(a^2 \plus{} b^2 \plus{} c^2)$
Consider 2 cases of t, we have done.
 :)[/quote]

[I  edited]

I think you did try to solve the problem. But there is a little mistake if you use  Hungkhtn's inequality  ? :)
Consider Arqady's case in which $ k \equal{} \frac {11}{16}$. Let $ t \equal{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}$

For the first  inequality, Hung's inequality is applicable if $ t > 2.933899575$

For the second inequality, it will fail if $ t > 2.877890984$[/quote]

What is the name of CAS program that you are using to determine such accuracies?

Very sorry. I was wrong in calulus  :oops:

[quote="shaaam"][quote="arqady"][quote="Gibbenergy"]
I think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]
I have got $ k_{max} = 0.688300972...$[/quote]

Arqady, Can you please show us your solution?

Thanks  :)[/quote]
It's a very ugly.
Let $ f(a,b,c) = \sqrt {\frac {(a + b)(b + c)(c + a)}{8abc}} - 1 - k \left( 1 - \sqrt {\frac {ab + bc + ca}{a^2 + b^2 + c^2}} \right),$ $ abc = 1$ and $ a = \max\{a,b,c\}.$
Hence, $ f(a,b,c) - f\left(a,\sqrt {bc},\sqrt {bc}\right)\geq$
$ \geq(\sqrt b - \sqrt c)^2\left(\frac {k(a^3 - 4a - 4)a^2}{2(a^3 + 2)^{1.5}\sqrt {2a\sqrt a + 1}} + \frac {a^2\sqrt a + 4\sqrt a + 1}{4(\sqrt {2(a^4 + a)} + \frac {a^{1.5} + 1}{\sqrt [4]a})}\right)\geq0.$
The last inequality is true for all $ 0 < k\leq0.7$ and $ a\geq1.$
Thus, we need to check  when $ f(a,1,1)\geq0,$ which gives
$ k\leq\min{\frac {\sqrt {a^2 + 2}(\sqrt {a^2 + 2} + \sqrt {2a + 1})}{2\sqrt a(\sqrt a + 1)^2} = 0.688300972...}$

[quote="arqady"][/quote][quote="shaaam"][quote="arqady"][quote="Gibbenergy"]
I think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]
I have got $ k_{max} = 0.688300972...$[/quote]

Arqady, Can you please show us your solution?

Thanks  :)[/quote][quote="arqady"]
It's a very ugly.
Let $ f(a,b,c) = \sqrt {\frac {(a + b)(b + c)(c + a)}{8abc}} - 1 - k \left( 1 - \sqrt {\frac {ab + bc + ca}{a^2 + b^2 + c^2}} \right),$ $ abc = 1$ and $ a = \max\{a,b,c\}.$
Hence, $ f(a,b,c) - f\left(a,\sqrt {bc},\sqrt {bc}\right)\geq$
$ \geq(\sqrt b - \sqrt c)^2\left(\frac {k(a^3 - 4a - 4)a^2}{2(a^3 + 2)^{1.5}\sqrt {2a\sqrt a + 1}} + \frac {a^2\sqrt a + 4\sqrt a + 1}{4(\sqrt {2(a^4 + a)} + \frac {a^{1.5} + 1}{\sqrt [4]a})}\right)\geq0.$
The last inequality is true for all $ 0 < k\leq0.7$ and $ a\geq1.$
Thus, we need to check  when $ f(a,1,1)\geq0,$ which gives
$ k\leq\min{\frac {\sqrt {a^2 + 2}(\sqrt {a^2 + 2} + \sqrt {2a + 1})}{2\sqrt a(\sqrt a + 1)^2} = 0.688300972...}$[/quote]

I appreciate  :)

(P.S why do you assume $ abc = 1$)

[quote="shaaam"]

 why do you assume $ abc \equal{} 1$[/quote]
Because the original inequality is homogeneous and $ abc\equal{}1$ simplifies this inequality.

[quote="arqady"][/quote][quote="shaaam"][quote="arqady"][quote="Gibbenergy"]
I think the best constant k is k = 0.6883 But I guess you have solved it already.[/quote]
I have got $ k_{max} = 0.688300972...$[/quote]

Arqady, Can you please show us your solution?

Thanks  :)[/quote][quote="arqady"]
It's a very ugly.
Let $ f(a,b,c) = \sqrt {\frac {(a + b)(b + c)(c + a)}{8abc}} - 1 - k \left( 1 - \sqrt {\frac {ab + bc + ca}{a^2 + b^2 + c^2}} \right),$ $ abc = 1$ and $ a = \max\{a,b,c\}.$
Hence, $ f(a,b,c) - f\left(a,\sqrt {bc},\sqrt {bc}\right)\geq$
$ \geq(\sqrt b - \sqrt c)^2\left(\frac {k(a^3 - 4a - 4)a^2}{2(a^3 + 2)^{1.5}\sqrt {2a\sqrt a + 1}} + \frac {a^2\sqrt a + 4\sqrt a + 1}{4(\sqrt {2(a^4 + a)} + \frac {a^{1.5} + 1}{\sqrt [4]a})}\right)\geq0.$
The last inequality is true for all $ 0 < k\leq0.7$ and $ a\geq1.$
Thus, we need to check  when $ f(a,1,1)\geq0,$ which gives
$ k\leq\min{\frac {\sqrt {a^2 + 2}(\sqrt {a^2 + 2} + \sqrt {2a + 1})}{2\sqrt a(\sqrt a + 1)^2} = 0.688300972...}$[/quote]

This is a nice and natural approach and I also had the same result.
@nguoivn: I actually like your sols, but it will be more convenient if you can write your sols more detail.
@Shaaam : I use Ti to find the root of equation.
  If you ask for CAS, I guess MatLab is #1, following by Mathcad, Maple, Mathematica, PolyMath,... etc


limit
function
integration
real analysis
real analysis solved

Let $\{a_i\},\{b_i\}$ be decreasing sequences of positive numbers s.t. $\displaystyle \sum_{n\ge 1}a_n=\sum_{n\ge 1}b_n=\infty$. Does it follow that $\displaystyle \sum_{n\ge 1}\sqrt{a_ib_i}=\infty$?

Note that sayng that $\min(a_i,b_i)\le \sqrt{a_ib_i}\le \max (a_i,b_i)$ doesn't suffice, because we can find sequences $\{a_i\},\ \{b_i\}$ which have the mentioned properties, but $\displaystyle \sum_{n\ge 1}\min(a_n,b_n)<\infty$.

Here's a concrete case: what if $b_n=\frac 1n$? Does it follow that $\displaystyle \sum_{n\ge 1}\sqrt{\frac {a_n}n}=\infty$?

I don't have the answer for the first question, though I think it is no, but for the second question the answer is yes, it follows:
  $ (\sqrt{\frac{a_1}{1}}+...+\sqrt{\frac{a_n}{n}})^2\geq\sum_{i=1}^n \sqrt{\frac{a_i}{i}}\cdot\sum_{j=1}^i {\sqrt{\frac{a_j}{j}}\geq\sum_{i=1}^n a_i }$

Yeah, I realized it wasn't that hard. I think that if $\{x_n\}$ is a decreasing sequence of positive numbers s.t. $\displaystyle \sum_{n\ge 1}x_n<\infty$ then $\displaystyle \lim_{n\to \infty}nx_n=0$. It's related to the following problem (posted and solved before on the forum: if $f : (0,\infty)\to (0,\infty)$ is a strictly decreasing function s.t $\displaystyle \int^{\infty}_0 f(x)dx<\infty$ then $\displaystyle \lim_{n\to \infty}xf(x)=0$. We can define a continuous function $f$ s.t. $f(n)=x_n$ and we're done.

We apply this result to $x_n=\sqrt{\frac {a_n}n}$. We get $\displaystyle \lim_{n\to \infty}n\sqrt{\frac {a_n}n}=0$. On the other hand, $a_n=\sqrt{\frac {a_n}n}\cdot n\sqrt{\frac {a_n}n}$, which gets really small when compared to $\sqrt{\frac {a_n}n}$, and if the latter has a convergent series, then so must $a_n$. If we look at it backwards, if $a_n$ has a divergent series, then so must $\sqrt{\frac {a_n}n}$.

[b] Moderator edit: Edited for readability (there was a problem with : in math mode in the original post)[/b]

if i'm right the lemma which is used in your solution come from vojtec could you refer me to proof Grobber?

Sorry, I can't seem to find it, although I distinctly remember me posting it and Moubinool giving a really nice solution. The one I know isn't that nice, and I don't really feel like posting it :). It was indeed proposed for Voytch Yarnk (if that's the correct spelling :D).

I can't forgot that magnific solution! Let $ F(x)=\int_0^x f(t) dt $. Then
 $ F(2x)-F(x) $ goes to zero, but $ F(2x)-F(x)\geq xf(2x)$.


Square $ABCD$ is inscribed in a circle. Point $X$ lies on minor arc $AB$ such that $[XCD]=993$ and $[XAB]=1$. Find $[XAD]+[XBC]$.

[hide]

Let r = radius of the circle
Then $\sqrt{2}r$ = the length of one side of the square
Let h = the height of $\triangle{XAB}$ = distance from X to AB
$[XCD] = \frac{1}{2}(\sqrt{2}r)(\sqrt{2}r+h) = 993$
$[XAB] = \frac{1}{2}(\sqrt{2}r)(h) = 1$
$r^2=992$
But since the height of XAD and XBC sum to $\sqrt{2}r$,
$[XAD]+[XBC] = \frac{1}{2}r^2 = 496$

[/hide]

[hide="Unfortunately, that is incorrect."]Your answer is off by a factor of 2...[/hide]

I think minsoens made a small error at the end. $[XAD]+[XBC] = \frac{1}{2}r^2 = 496$ should be $[XAD]+[XBC]=\frac{1}{2}\cdot(r\sqrt{2})^2=r^2=992$.

992 is correct.

http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=55093  :D

[quote="JavaMan"]I may sound stupid, but what's [XAD] etc mean?[/quote]I think that [XAD] indicates the area of the triangle XAD.

[hide="another sol"]
let side be $s$,
and perpendicular distance from $X$ to nearest side be $x$,
$\frac{1}{2}*s*(s+x)=993$
$\frac{1}{2}*s*x=1$
So,
$\frac{1}{2}*s^2=992$
We need to find:
$\frac{1}{2}*s*y+\frac{1}{2}*s*z$
but $y+z=s$,
So,
$\frac{1}{2}*s^2$
which is:
$\boxed{992}$
[/hide]


If one uses " in a description, it always gets converted to \". I haven't tested, but maybe other characters are influenced as well.

[quote="ZetaX"]If one uses " in a description, it always gets converted to ". I haven't tested, but maybe other characters are influenced as well.[/quote]

Yeah, and it is kind of annoying if you edit the description of a competition several times, e.g. I got \^{10}" several times. As I understand that is kind of escaping mechanism due to the coding but then there should be a post-running script trimming them off again.

In fact, the number of \ grows exponentially, causing a lot of fun  :D


function
college
integration
real analysis
real analysis solved

if f{x} is a real valued differentiable function with f{1} =1.If f{x} prime is equal to
1/{x^2+ f{x}^2},then proove that f{x}<1+pi/4 for all x.

can u please tell me a book in which these r found in plenty.

well I come from India and don't know what you r talking about.can u specifically tell me the source and please solve it also

Hmm. Maybe you can find it at [url]http://math.berkeley.edu/~desouza/pb.html[/url].
I've got a book of that.

I think the conclusion is $f(x)<1+\frac{\pi}{4}$ for $x\geq 1$.

$f'(x)>0$ so $f$ is strictly increasing, and $f(x)>1$ for $x> 1$.

$
f(x)-f(1) = \int_1^x f'(t)dt = \int_1^x \frac{dt}{t^2+f(t)^2} <\int_1^x \frac{dt}{t^2+1} < \int_1^\infty \frac{dt}{t^2+1} = \frac{\pi}{4}
$

The conclusion follows easily.

yup that is the answer
thanks lily for the link


Five students $ A, B, C, D, E$ took part in a contest. One prediction was that the contestants would finish in the order $ ABCDE$. This prediction was very poor. In fact, no contestant finished in the position predicted, and no two contestants predicted to finish consecutively actually did so. A second prediction had the contestants finishing in the order $ DAECB$. This prediction was better. Exactly two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so. Determine the order in which the contestants finished.

I don't know if this is correct or not, this is my first time doing this kind of logic problem...

From "two of the contestants finished in the places predicted, and two disjoint pairs of students predicted to finish consecutively actually did so," we see that

1) The two that finished in the correct place must be consecutive.

Proof:
Suppose they are not consecutive. Then they form at least a block of 3 (e.g. D_E..) which means that the other two numbers must satisfy the finishing consecutively condition. However, the number between them cannot be correct, which means a second consecutive pair cannot be formed. Contradiction.

2) The two that are correct and consecutive must be at the ends.

Proof:
If they are in the middle, then the other pair is fixed, making it correct also, leaving 4 correct places. Contradiction.

Then we test:

If DA is in the correct place, we have:
DACBE-- WRONG, E, C is in the correct place from the very first condition
DABEC-- WRONG, AB is a correct consecutive pair, violates first condition

If CB is in the correct place, we have:
AEDCB-- WRONG, A is in the correct place
EDACB-- Correct. Solution.

Very hard problem, might i ask what is the correct solution, i'm not discounting iniquitious' solution only i don't necessarily understand it all )=, i'm not used to seeing logic problems outside of the mensa workbooks and this seems 20x harder.

[hide="correct answer"]EDACB, nice job iniquitus[/hide]
[hide="lightentropy"]btw this was an IMO question, yea it was 1963, but it was still IMO[/hide]

[quote]btw this was an IMO question, yea it was 1963, but it was still IMO[/quote]
Yeah the level of difficulty has increased a lot since then :(

that makes sense still, i'm more at the AMC12/ and can do maybe 2-3 problems on AIME so it would make sense that i was unaware how to do this. I need to do more logic problems, they are interesting and fun.
cool post


geometry
analytic geometry
geometric transformation
reflection

got this from a middle school chinese mathbook.

Two points on a Cartesian plane, A and B, are located at (5,4) and (1,1) respectively. There lies a point P that is located on either the X or the Y axis. What are the coordinates of P so that AP + BP is minimized.

yeah, reflections are definitely the way because if you can place AP and a reflected point on a single line, it minimizes the value of AP + BP.


probability
percent
probability and stats

Seventy percent of the kids who visit a doctor have high temperature and 30% of the kids who have high temperature, also have smallpox. What's the probability that a kid who goes to the doctor's has a temperature and smallpox?

Just multiply. 21%.

As a bit of advice, please don't post multiple threads with the same title. We like to be able to tell threads apart. "Probability" is a very bad title for this forum, since it says almost nothing about the content.

Strictly speaking, unsufficient data: $ P(H|D)\equal{}0.7$, $ P(S|H)\equal{}0.3$, $ P(S\cap H|D)\equal{}?$. Some independence assumption is used.

Actually, it's an implicit assumption that "who visit the doctor" is the whole universe here.


Hi.

Hexagonal number nth term: $ f_6(n)\equal{}2n^2\minus{}n$
triangular number nth term: $ f_3(n)\equal{}0.5(n^2\plus{}n)$

Looking at a proof here to prove every hexagonal number is also triangular they did:

$ f_6(n) \equal{} f_3(2n\minus{}1)$ and continued from that point.

Can someone explain where did 2n-1 come from?

Thanks alot.

What this says is that the $ n^{th}$ hexagonal number is the same thing as the $ (2n \minus{} 1)^{th}$ triangular number.  Now, there are a couple of ways to understand this statement.

1.  If you look at those algebraic expressions, they're pretty similar; a linear and a quadratic term, no constant term.  There might be a way to turn one into the other.  Now, in $ f_3$ the linear and quadratic term are the same whereas in $ f_6$ one is twice the other, so perhaps we should look at $ f_3(2n) \equal{} 2n^2 \plus{} 2n$.  But the linear term is wrong, so we keep tweaking this until we arrive at the right answer.

2.  Can you figure out a way to transform hexagons into triangles?  (I think this is much more fun.  Play around with small cases, although make sure you understand the definition of a hexagonal number.)

I dont understand the goal nor the way to do it  :(

Are we trying to guess terms to put into $ f_3(n)$ so it becomes like the form of $ f_6(n)$ ?

Yes, pretty much, but I wouldn't advise you to think about the problem that way since it provides no motivation:  there's no reason to suspect that there is a way to do this at all, so I would prefer to see what happens if you try to do it geometrically (although I don't know if it works).

find m(n) such that
$ 2n^2 \minus{} n \equal{} 0.5(m^2 \plus{} m)
\\4n^2 \minus{} 2n \equal{} m^2 \plus{} m
\\m^2 \plus{} m \plus{} \dfrac{1}{4} \equal{} 4n^2 \minus{} 2n \plus{} \dfrac{1}{4}
\\(m\plus{}\frac{1}{2})^2 \equal{} (2n \minus{} \frac{1}{2})^2
\\m \plus{} \frac{1}{2} \equal{} (2n \minus{} \frac{1}{2})
\\m \equal{} 2n \minus{} 1$

we take the positive square root on the secondlast line because we require $ m \geq 0$


[quote="math92"][hide]
the answer is d
i used guess and check and d is the only one that works.[/hide][/quote]

Not quite.... 3 isn't bigger than 9.

oops, i got my symbols mixed up.
So the answer is[hide]1/3.[/hide]


What is the smallest integer which is larger than $ \sqrt[3]{26,991}$?

we see that 30^3=27000. we are in a countdown round, which means no time to test 29^3. we assume the answer is 30. and we are correct  :D


function
topology
complex analysis
complex analysis unsolved

Fix $ a \in D$ (open unit disc) and let $ T_{a}(z) \equal{} \frac{z\minus{}a}{1\minus{}\overline{a}z}$. Let $ f$ and $ g$ analtytic functions in an open set containing $ \overline{D}$, such that $ |f(z)| > |g(z)|$ for all $ |z| \equal{} 1$. Prove the function $ T_{a}f \plus{} g$ has at least one zero in $ D$.

This is how I find the problem, i guess $ T_{a}f$ is the composite function, first $ f$ then $ T_{a}$.

Can someone give me a hint for this one.

Thx

Suppose the contrary and conclude by [url=http://en.wikipedia.org/wiki/Rouche's_theorem]Rouché's Theorem[/url] that $ T_a f (z)$ has no zeros in $ D$, clearly a contradiction since $ T_a f(a) \equal{} 0$.

Thank you for your answer, I dont know why $ T_{a}f(a) \equal{} 0$, since

\[ T_{a}f(a)\equal{}\frac{f(a)\minus{}a}{1\minus{}\overline{a}f(a)} \equal{} 0,\] only when $ f(a)\equal{}a$ and we dont know this.
I was trying to prove that there exists some $ z_{0}$ such that $ f(z_{0}) \equal{} a$ and this is clearly a zero for $ T_{a}f$, but i think this cant be done.

I repeat i found this problem and dont know if its composite or the product, if its product then clearly
$ T_{a}f(a) \equal{} T_{a}(a)f(a) \equal{} 0$, but I dont see how $ |T_{a}f(z)| > |g(z)|$ for all $ |z|\equal{}1$ so I can use
Rouche Theorem.

My apologies, I misread your problem.

If it's composition, the statement doesn't seem to be true.  What if both $ f$ and $ g$ are constant, for example?


function
algebra
domain
real analysis
calculus
calculus computations

Hi,
While I was reading about uniform continuity of real functions, I visited this post and read the #4 on it:
[url]http://www.physicsforums.com/showthread.php?t=52795[/url]

The description is clear, but I could not figure it out, how to prove that f(x)=1/x is not uniformly continuous. Do you have recommendations please?

Continuity really means to be able to draw the graph of the function with having to take the hand of ur paper. that is it means there are no breaks in the  graph in a particular interval . For example consider the function $ f(x) = x$ in any  interval for instance in   $ ( 0, \infty )$ we say  this function is continuous throughout the interval because when we draw its graph it has no breaks .

Basically continuity needs to be checked in each and every point inorder to say its uniformly continuous in the interval.

So a function $ f(x)$ can be said as continuous  at x belonging  to  the intervalv $ [ X , Y ]$ if there exists a $ x_{0}$ in the neighborhood of of x such that $ |x - x_{0} | < \epsilon$ where $ \epsilon > 0$ is a very small real number and  $ lim_{x \rightarrow x_{0^ - } } f(x)  =    lim_{x \rightarrow x_{0^ + } } = f( x_{0} )$

Hence we see that for the function $ f(x) = \frac {1}{x}$  it is not contionuous at 0 because $ lim_{x \rightarrow{0^ - } } f(x)  \neq   lim_{x \rightarrow{0^ +  } } f(x)$and $ f( 0)$ is not defined . hence it is not uniformly continuous.

Continuity of a function is only defined at points. Continuity of a function over a certain domain means it is continuous at each point of that domain. Uniform continuity is a stronger statement, saying that in each point of the domain you can use the same distance between arguments to get the images within some distance $ \epsilon$.

vijaymenon's argument is flawed.

[quote]vijaymeno: ...is not defined . hence it is not uniformly continuous. [/quote]
Thank you for the fast answer.
Meanwhile, please consider as it is mentioned at #4 of the post in:
[url]http://www.physicsforums.com/showthread.php?t=52795[/url]
We are considering $ (0,\plus{}\infty)$, not [X,Y] as you mentioned.
The writer of the #4 post there said "f(x)=1/x IS contiuous, but not uniformly continuous on the interval $ (0,\plus{}\infty)$.
Hence I can't agree with the proof construction provided in your post:
"it is not uniformly continuous, because it is not continuous at all" (if I understand it right). The proof you provide, is around $ x_{0} \equal{} 0$, which is not our consideration, as "uniformly continuous" says that it is "absolutely independent of any $ x_{0}$".
The problem here is, to find an $ \epsilon>0$, for which we prove there is NO $ \delta > 0$ which breaks the uniformity on the interval. This is what I tried to do but failed. It must be possible, as in some intervals, say, between (0,1) the function itself is indefinitely big, and it may cause problem to define  $ \epsilon>0$ and $ \delta > 0$. What I am wondering about, is how to formally write down this idea! let's continue to think together around it  :wink:

For any $ \delta > 0$, consider the values of the function $ f(x)$ at $ x_0$ and $ x_0 \plus{} \delta$, where $ x_0$ is the positive root of $ x^2 \plus{} \delta x \minus{} \delta \equal{} 0$.  Now, figure out how I came up with that  :wink:

[color=green]Redundant quote deleted by moderator.[/color]

I didn't figure it out. please clarify.

First, do you understand the significance of the points $ x_0$ and $ x_0 \plus{} \delta$ that I chose?  That is, have you done what I said in my first sentence?

There's no need to quote the entire post immediately preceding yours -- you're responding to it by default.

Consider two points, $ x=\frac{1}{n}$ and $ y=\frac{1}{n+1}$, with $ n\in\mathbb{Z}_{+}$.  Then $ |x-y|=\frac{1}{n}-\frac{1}{n+1}|=|\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}|=|\frac{1}{n(n+1)}|$.  As $ n\rightarrow\infty$, this goes to 0.  However, $ |f(x)-f(y)|=|n-n-1|=1$, which does not go to 0.


quadratics
analytic geometry
calculus
conics
parabola
symmetry
special factorizations

What is the smallest value of $x^2+10x$ where $x$ is a real number?

[hide]Using the formula  to find the minimum and maximum we find it is -10/2      so the answer is -5.[/hide]

[hide] The minimum is -5, so plug that in to get:

(-5)^2 + 10(-5)

25 - 50

= -25

[/hide]

Where does this -b/2a come from?  Why does that work?

Given a quadratic in standard form $ax^2+bx+c=y$, where a>0, the lowest point on the graph lies at the vertex. The x-coordinate of the vertex point is -b/2a, and so plugging that value into the original equation for x yields the smallest value.

This can be seen by completing the square, which gives the vertex-form of the equation, $y=a(x-h)^2+k$,where (h,k) is the vertex of the function.

[quote="#include<math.h>"]-b/2a is the formula used to find the x value of the lowest, or highest, value of y depending on whether it opens up or down.  Once you get the x value you plug it into the original equation and out pops the minimum y value[/quote]

We know what the formula IS but how would you prove it?  A formula is practically useless unless you know where it comes from. nolachrymose has the right idea but hasn't really shown that -b/2a is the x-coordinate of the vertex.

$ax^2+bx+c=$

$a\left(x^2+\frac{bx}{a}\right)+c=$

$a\left(x^2+\frac{bx}{a}+\frac{b^2}{4a^2}\right)-\frac{b^2}{4a^2}+c=$

$a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+c$

We want the square to be zero to either maximize or minimize the expression, so $x=-\frac{b}{2a}$

$\frac{-b}{a}$ is the sum of the roots.$ \frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{-b}{a}$
The average of these two roots is then the x coordinate of the vertex or $\frac{-b}{2a}$

[hide="Short Answer"]
The minimum occurs at $\frac{-b}{2a}=\frac{-10}{2}=-5$ Therefore $x=-5$ and the smallest y value is -25
 :D
[/hide]

um diophantus, don't post calculus in the beginner forum.

To those who have explained their solution by simply claiming that $x=-\frac{b}{2a}$ at the min/max of a quadratic equation, how do you know that? You saw it written in a box in a textbook?

The algebraic solution to this problem that is not based on such assumptions is:

$(x+5)^2\ge0\Longrightarrow(x+5)^2-25\ge-25\Longrightarrow{x}^2+10x+25-25\ge-25\Longrightarrow{x}^2+10x\ge-25,$
and equality is attained for $x=-5$, so $-25$ is the minimum value.

There was a similar discussion about this in a different thread.. this is what I said:

[quote="AntonioMainenti"]... If you don't know how to use calculus or what it means, you can also do this algebraically. $y=ax^2+bx+c$ can be manipulated to $a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}$, so $\left(x+\frac{b}{2a}\right)^2\ge0\Rightarrow{a}\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}\ge\frac{4ac-b^2}{4a}$, and equality is attained for $x=\frac{-b}{2a}$, so $\frac{4ac-b^2}{4a}$ is the extreme (min or max) value in all cases. ...[/quote]

[quote="Magnara"]The arithemetic mean of the two roots will be the vertex, which is the sum of the roots divided by 2.  Hence, -b/(2a).[/quote]
Sorry, I was under the impression you were trying to determine roots from this equation rather than the vertex

[quote="Magnara"]
The arithemetic mean of the two roots will be the vertex, which is the sum of the roots divided by 2. Hence, -b/(2a).[/quote]

Yep :D This works because a parabola has a line of symmetry through the vertex.

[quote="Ravi B"]Does that argument assume that the quadratic has two real roots?[/quote]Yeah, you can't really interpret it geometrically unless it has real roots, or a double root, but still all quadratics have two roots (by the Fundamental Theorem of Something-er-rather).

I agree that a quadratic always has two roots.  But when the roots are complex, I'm not sure how Magnara was going to prove that the arithmetic mean of the roots is the x-coordinate of the vertex.  It's still true algebraically, but I'm not sure how to prove it geometrically.

:( Hi.
Are these problems solvable?  I have tried setting them up in standard form Ax^2+Bx+C=0 but not getting any where.  Thanks

Problem 1:  (4x+1)(2x-3)=3(x+4)

Problem 2: (x-11)^2=(3x+6)^2

For the 1st one. Multiply out (or distribute) both sides and rearrange terms.
[hide]You'll end up with $8x^2-13x-15=0$[/hide]

The same for the second one. Multiply them all out and rearrange terms.
[hide]You'll end up with $8x^2+58x-85=0$[/hide]

Then, you can always solve with the quadratic formula.

Refering to AntonioMainenti's post, here's the other thread discussed a few weeks ago on quadratic optimization.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26049[/url]

We can use -b/2a to get -10/2=-5 for the smallest value.
f(-5)=(-5)2+(10)*-5= -25.
I do not know how to do a rigorous proof, but It is the vertex of the parabola, so that is the minimum.


geometry
trapezoid
geometric transformation
homothety
Pythagorean Theorem

a trapezium with four lengths of sides are 1,2,3,4. Find its area?

[quote="123456789"]Letting the sides of length $ 1$ and $ 4$ be the bases, we can split the longer base into $ x$, $ 1$, and $ y$, so that $ 4 \equal{} x \plus{} y \plus{} 1 \implies x \plus{} y \equal{} 3$.
Applying the Pythagorean Theorem on two right triangles, we have
\[ 2^2 \minus{} x^2 \equal{} h^2 \equal{} 3^2 \minus{} y^2 \implies y^2 \minus{} x^2 \equal{} \left(y \plus{} x\right)\left(y \minus{} x\right) \equal{} 5.
\]
Plugging in $ x \plus{} y \equal{} 3$ and solving for $ y \minus{} x$, we have
\[ y \minus{} x \equal{} \frac {5}{3}.
\]
Then
\[ 2y \equal{} 3 \plus{} \frac {5}{3} \implies y \equal{} \frac {7}{3}.
\]
Then we find that
\[ h \equal{} \sqrt {9 \minus{} y^2} \equal{} \sqrt {9 \minus{} \frac {49}{9}} \equal{} \sqrt {\frac {32}{9}} \equal{} \frac {4\sqrt {2}}{3}.
\]
Then the area is
\[ A \equal{} \frac {1}{2}\left(b_1 \plus{} b_2\right)h \equal{} \boxed{\frac {10\sqrt {2}}{3}}.
\]
You can also use a homothety centered at the intersection of the two slants.[/quote]

would you have another idea in which the smaller base of trapezium is not 1,eg.2 or 3? Then ,is it any possible solution? :blush: