Let $ ABC$ be a triangle for which $ b\ne c$ . Consider the points  $ \{M,D\}\subset BC$ so that

$ \left\|\begin{array}{c} MB \equal{} MC \\
 \\
\widehat {DAB}\equiv\widehat {DAC} \\
 \\
m(\widehat {DAM}) \equal{} \phi\end{array}\right\|$ . Show that $ \boxed {\ \tan\phi \equal{} \frac {|b \minus{} c|}{b \plus{} c}\cdot\tan\frac A2 \equal{} \tan^2\frac A2\cdot\tan\frac {|B \minus{} C|}{2}\ }$ .

Denote the points $ \left\|\begin{array}{c} P\in AB\ ,\ PD\perp AM \\
\ R\in AD\ ,\ RP\perp AB\end{array}\right\|$ . Prove that $ \boxed {\ RM\perp BC\ \Longleftrightarrow\ B \equal{} 90^{\circ}\ \vee\ C \equal{} 90^{\circ}\ }$ .

i think the first one must be $ \tan\phi\equal{}\tan^2\dfrac{A}{2}\cdot\tan\dfrac{|B\minus{}C|}{2}$...

suppose wlog that $ B\geq C$. then the order of the points is $ B,D,M,C$, and $ DM\equal{}\dfrac{a(b\minus{}c)}{2(b\plus{}c)}$...

we have that

$ \dfrac{\sin\left(\frac{A}{2}\minus{}\phi\right)}{\sin\phi}\equal{}\dfrac{CM\cdot AD}{DM\cdot AC}\equal{}\dfrac{2c\cos\frac{A}{2}}{b\minus{}c}$

this implies that

$ \sin\frac{A}{2}\cot\phi\equal{}\cos\frac{A}{2}\plus{}\dfrac{2c\cos\frac{A}{2}}{b\minus{}c}\equal{}\dfrac{(b\plus{}c)\cos\frac{A}{2}}{b\minus{}c}$,

from where we conclude that

$ \tan \phi\equal{}\tan\frac{A}{2}\cdot\dfrac{b\minus{}c}{b\plus{}c}\equal{}\tan^2\frac{A}{2}\tan\dfrac{B\minus{}C}{2}$.

Thank you, [b]Campos[/b] ! I corrected and added ...  I am interested by the second part of it.

for the second part I assume that $ B\geq C$ still holds, i don't have checked the other case but it must follow in the same way...

note that
$ DR\equal{}\dfrac{PD\sin\left(\frac{A}{2}\plus{}\phi\right)}{\sin\left(\frac{B\plus{}C}{2}\right)}\cdot\dfrac{AD\sin\frac{A}{2}}{PD\cos\left(\frac{A}{2}\plus{}\phi\right)}\equal{}AD\tan\frac{A}{2}\tan\left(\frac{A}{2}\plus{}\phi\right)\equal{}\dfrac{2bc\sin\frac{A}{2}\tan\left(\frac{A}{2}\plus{}\phi\right)}{b\plus{}c}$

note also that $ RM\perp BC$ iff $ PBMR$ is cyclic, which holds iff $ \angle DRM\equal{}\frac{B\minus{}C}{2}$. then,

$ RM\perp BC \Leftrightarrow DR\equal{}\dfrac{DM}{\sin\left(\frac{B\minus{}C}{2}\right)}\equal{}\dfrac{a(b\minus{}c)}{2(b\plus{}c)\sin\left(\frac{B\minus{}C}{2}\right)}\equal{}\dfrac{a^2}{2(b\plus{}c)\cos\frac{A}{2}}$

then $ RM\perp BC$ iff $ \dfrac{\sin A}{2\sin B\sin C}\equal{}\tan\left(\frac{A}{2}\plus{}\phi\right)$

using the formula $ \tan(x\plus{}y)\equal{}\dfrac{\tan x\plus{}\tan y}{1\minus{}\tan x\tan y}$ it follows that the last equation holds iff

$ \dfrac{\sin A}{2\sin B\sin C}\minus{}\tan\frac{A}{2}\equal{}\tan \phi\left(1\plus{}\dfrac{\tan\frac{A}{2}\sin A}{2\sin B\sin C}\right)$

note that $ \dfrac{\sin A}{2\sin B\sin C}\minus{}\tan\frac{A}{2}\equal{}\dfrac{\sin\frac{A}{2}\sin^2\left(\frac{B\minus{}C}{2}\right)}{\cos\frac{A}{2}\sin B\sin C}$

and $ 1\plus{}\dfrac{\tan\frac{A}{2}\sin A}{2\sin B\sin C}\equal{}\dfrac{\cos^2\left(\frac{B\minus{}C}{2}\right)}{\sin B\sin C}$

therefore, $ RM\perp BC$ iff $ \tan\phi\equal{}\tan\frac{A}{2}\tan^2\frac{B\minus{}C}{2}$, which holds iff $ \tan\frac{A}{2}\equal{}\tan\frac{|B\minus{}C|}{2}$..

since we assumed that $ B\geq C$ then we must have $ \frac{A}{2}\equal{}\frac{B\minus{}C}{2}$, from where we conclude that this holds iff $ B\equal{}90^{\circ}$, and we're done.  :D


[quote="13375P34K43V312"]Triangle ABC has area 84 and is such that AB=13, AC=15. Some point D is chosen in the same plane of ABC such that BD=DC=7. E is the midpoint of AC and F is the midpoint of AB. Let G be the centroid of triangle DEF, and let H be the intersection of FD and BE. Find the area of triangle FGH.

[hide="hint"]Where/What must D be?[/hide][/quote]

[hide="Hint"]$ D$ is the midpoint of $ BC$.[/hide]
[hide="Solution"]Since $ BC \equal{} 14$, $ D$ is the midpoint of $ BC$. So, $ \triangle DEF$ is formed by connecting the midpoints of $ \triangle ABC$.

If you draw $ BE$, you'll see that it actually contains one of the medians of $ \triangle DEF$. So, $ \triangle FGH$ is just one-sixth of the area of $ \triangle DEF$, which is $ 84\cdot 1/4\cdot 1/6 \equal{}\boxed{7/2}$[/hide]

[quote="randomdragoon"]Was that whole ordeal with point D really necessary? :O

And what is it with those contest writers, they really like 13-14-15 triangles for some reason[/quote]

Thats because a 13-14-15 triangle has an altitude of [b]12[/b] xDDD

well from the side that is length 15 atleast...

Try $ 51,52,53$

By the way, memorize this triple. Contests love it :)

[hide]I mean seriously it can't be the only non-right triangle with integer side lengths AND integer area ...

.. can it?[/hide]

We had this problem in int nt...


function
calculus
derivative
real analysis
real analysis unsolved

Let $ f : [0,1] \longrightarrow \mathbb{R}$ be continuous and $ f'(x) > 0$ for all but countably many points. Prove that $ f$ is increasing. Would the result still be true if we had $ f'(x) > 0$ almost everywhere?

Here's a thought: The fact that $ f'(x)>0$ [b]at all but countably many points[/b] means that $ f'(x)\leq 0$ only at isolated points, not an interval, because any interval contains uncountably many points.

You're not going to get anywhere with the interval idea. Here's a concrete example of a continuous function with positive derivative almost everywhere, which is not increasing:
Let $ f$ be the Cantor devil's staircase function on $ [0,1]$, and let $ g(x)\equal{}\frac12x\minus{}f(x)$. With a different function with zero derivative a.e., I could make it nonicreasing on every interval.


[quote="4everwise"]A positive number $x$ satisfies the inequality $\sqrt{x} < 2x$ if and only if
$\text{(A)} \ x > \frac{1}{4} \qquad \text{(B)} \ x > 2 \qquad \text{(C)} x > 4 \qquad \text{(D)} \ x < \frac{1}{4}\qquad \text{(E)} x < 4$[/quote]
[hide]
$\sqrt{x} < 2x$
$x < 4x^2$
$0 < x(4x-1)$
$0 < 4x-1$
$1 < 4x$
$x > \frac{1}{4}$
(A)[/hide]

[hide]$\text{Answer: (A)}$

The best way to solve this seems to be by squaring and regrouping.

$\sqrt{x}<2x$
$x<4x^2$
$0<x(4x-1)$
$\frac{1}{4}<x$[/hide]

[quote="catcurio"][quote="imber.matutinis"][hide]You get two inequalities (see above messages), x>0, and x>1/4. The x>1/4 overshadows, so to say, the x>0...I think...anyone want to confirm? [/hide][/quote]
If x>1/4, then x must be greater than 0 as well. :)[/quote]

Right, that makes sense. But, what if you have an equation that yields two inequalities that don't work out that way--i.e. you would have to put the word [i]and[/i] or [i]or[/i]. Can't really think of an example...but are there specific instances/guidelines where you choose either [i]and[/i] or [i]or[/i]--how can you tell which word you're supposed to use?

[quote="ch1n353ch3s54a1l"][quote="imber.matutinis"][quote="boxedexe"][hide="Solution"]
$\\\sqrt{x}<2x\\ x<4x^2\\ \frac{1}{x}<4\\ x>\frac{1}{4}\\ \boxed{A}$
[/hide]

Masoud Zargar[/quote]

I thought dividing by the variable cancelled out a possible answer, so teachers encouraged us not to..?[/quote]

that's only if one of the solutions if it wasn't canceled is 0, and for all the answer choices, all x are greater than 0 no matter what. but i think generally you don't cancel out variables  :?  :D[/quote]

But there's an x<1/4 in the answer choices. Unless you don't count that? Why don't you count that?

But x CAN be zero! In the inequality-- x>0 and x>1/4.

[quote="imber.matutinis"][/quote][quote="ch1n353ch3s54a1l"][quote="imber.matutinis"][quote="boxedexe"][hide="Solution"]
$\\\sqrt{x}<2x\\ x<4x^2\\ \frac{1}{x}<4\\ x>\frac{1}{4}\\ \boxed{A}$
[/hide]

Masoud Zargar[/quote]

I thought dividing by the variable cancelled out a possible answer, so teachers encouraged us not to..?[/quote]

that's only if one of the solutions if it wasn't canceled is 0, and for all the answer choices, all x are greater than 0 no matter what. but i think generally you don't cancel out variables  :?  :D[/quote][quote="imber.matutinis"]

But there's an x<1/4 in the answer choices. Unless you don't count that? Why don't you count that?

But x CAN be zero! In the inequality-- x>0 and x>1/4.[/quote]

oh, i didn't see that
well according to the equation x CAN'T be zero  :P so i guess that means u can cancel out the Xs because it won't get rid of any solutions

[quote="ch1n353ch3s54a1l"][/quote][quote="imber.matutinis"][/quote][quote="ch1n353ch3s54a1l"][quote="imber.matutinis"][quote="boxedexe"][hide="Solution"]
$\\\sqrt{x}<2x\\ x<4x^2\\ \frac{1}{x}<4\\ x>\frac{1}{4}\\ \boxed{A}$
[/hide]

Masoud Zargar[/quote]

I thought dividing by the variable cancelled out a possible answer, so teachers encouraged us not to..?[/quote]

that's only if one of the solutions if it wasn't canceled is 0, and for all the answer choices, all x are greater than 0 no matter what. but i think generally you don't cancel out variables  :?  :D[/quote][quote="imber.matutinis"]

But there's an x<1/4 in the answer choices. Unless you don't count that? Why don't you count that?

But x CAN be zero! In the inequality-- x>0 and x>1/4.[/quote][quote="ch1n353ch3s54a1l"]

oh, i didn't see that
well according to the equation x CAN'T be zero  :P so i guess that means u can cancel out the Xs because it won't get rid of any solutions[/quote]

Wait, really? You can't take the square root of zero? I thought that was just for negative numbers...

[quote="catcurio"]You can take the square root of 0. It equals 0. But if we want $\sqrt{x}<2x$, then x cannot be 0. :)[/quote]

Oh...haha. That makes sense. 0 is not less than 0. Oookaay.

Does anyone know anything about the [i]and[/i] and [i]or[/i] circumstances? Not that it's really all that relevant...


function
inequalities unsolved
inequalities

Prove that for each positive integer n and each number t in the range (1/2, 1) there exist numbers a,b in the range (1999, 2000) such that (a^n)/2 + (b^n)/2 < (ta + (1-t)b)^n

We can take $b=0$ and $a$ lies in $[\frac{1}{t},t]$ with $t=\frac{2000}{1999}$

$f(a)=a^n+1-2(ta+1-t)^n$

$f'(a)=na^{n-1}-2nt(ta+1-t)^{n-1}$

because $f(1)=n(1-2t) < 0$

Then exist $u$,$1 \le u \le t$ and $f'(x) < 0$  for all $x \in [1,u]$

(Because $f'(a)$ is continuous function)

Then $f(u) \le f(1)=0$,let $u=a$ and we are done.


How do you think this year's AMC 12 A compares to 2001 AMC 12 (arguably the hardest AMC 12). Harder, easier?

2005 was much, much easier than last year (which probably explains my 34.5 point increase in score).  Of course, I started doing AoPS stuff after last year's AMC.  I took the 2001 about a week ago, and I got 125.5, and that was probably because I had done a few problems before.


Recall the identity $ \left(a^b\right)^{c}\equal{}a^{bc}.$
$ 243\equal{}3^5$ and applying our identity, we have $ 243^{3/5}\equal{}\left(3^5\right)^{3/5}\equal{}3^3\equal{}\boxed{27}.$

So basically we have to convert 243 = 3 to the 5th so we can multiply it out ans get 3 to the third. Is there any way to do so without converting 243 to a power?


calculus
integration
logarithms
calculus computations


AMC AIME probability geometry circumcircle incenter LaTeX

So iniquitus and I (Jerry and Oliver), think that math is fun.  We want to post problems for others (mostly from GA) to work on.  These should be at about an AIME level.  Who's interested?

We'll have about 35 questions.  Sadly, we probably won't give prizes.

Fun.

The message is too small. Please make the message longer before submitting.

May I participate as well, although I'm not from Georgia?  :P

@funcia- the math marathon won't be THAT hard.

@xpmath- Sure.

Sitan, you're Chinese.  Thus, you are not a ninja.

To keep this post on-topic, the last seven questions might be more proof-based than the first 28, which are straight up answers.

I'm in. :cleaning:

I've always wanted to use that smiley. -points-

Hi ! i would like to participate in the marathon too. :lol:

Wait, how are we doing this? All 35 at once or like in rounds or like the first one to put Sitan in a trash can wins?

Im confuzzled.  :huh:

I am too pro for all of you. Sitan, my name is copyrighted from now on.

Frisbeeballer- are you Heng?  I've never seen you before.

Arvind- Seriously?  Of course it's the first one to put Sitan in the trash can.

The problems will probably be given out at a rate of seven per week.  Thus, it's almost like a problem of the day.

And they should also be published REALLY soon...we're making the final edits to the first round.

Yes, it's HENG TONG (there, I said it, wut are you going to do?).  He told me.

And you can tell by the username.

Waaaaaaait.

So like 6 normal and 1 proof every week?

or one week w/ 7 proofs? that's hardcore.

[hide="hints"]1. conditional probability

2. (failed) seems bruteforce-able with coordinates + shoelace...anyone knows of a geometry solution?

3. let all three logs equal x

4. newton's sums

5. consider 200th roots of unity

6. recursion

7. split into prime power moduli and solve
[/hide]

[hide]Actually, If you used coordinates on #2, you didn't need the shoelace theorem since one side was parallel to the 14-side. So, you could use the (1/2)base*height formula[/hide]

Feel free to post solutions if you want to; if I don't see any being posted up, I will post up mine 1 by 1, in no particular order. The second round will probably start next Monday.

[hide="Solution to Question 6"]

Let the 200-gon be $ A_1 A_2...A_{200}$.

Lemma:

$ |\overline{A_1 A_2}| \times |\overline{A_1 A_3}| \times... \times |\overline{A_1 A_{200}}| = 200$

Proof:

$ x^{200}-1=(x-1)(x-w)(x-w^2)...(x-w^{199})$, where $ w=e^{\frac{2\pi i}{200}}$

Dividing $ x-1$, we get
$ x^{199}+x^{198}+...+x^0 = (x-w)(x-w^2)...(x-w^{199})$

Plug in $ x=1$ to get
$ 200 = (1-w)(1-w^2)...(x-w^{199})$

Noting that $ 1,w,w^2,...,w^{199}$ are the vertices of a regular 200-gon in the complex plane, take the magnitude of both sides to achieve the desired result.

The number we want to find, by the lemma, is $ \sqrt{200^{200}} = 200^{100}=2^{300}5^{200}$
This number has $ 60501$ factors.
[/hide]

[hide="Solution_to_Question_3"]Let $ \log_9 (a) \equal{} \log_{15} (b) \equal{} \log_{25} {(a\plus{}2b)} \equal{} x.$ Then $ 9^x\equal{}a, 15^x\equal{}b, 25^x\equal{}a\plus{}2b.$ You can see that \[ 9^x*25^x\equal{}(15^x)^2\] \[ a(a\plus{}2b)\equal{}b^2\] \[ a^2\plus{}2ab\plus{}b^2\equal{}(a\plus{}b)^2\equal{}2b^2\] \[ a\plus{}b\equal{}\sqrt 2 b\] \[ \frac{a}{b}\equal{}\boxed{\sqrt 2 \minus{} 1}\] Note that it can't be $ a\plus{}b\equal{}\minus{}\sqrt 2 b$ because then it implies the ratio is negative, which is impossible looking at the original equations.[/hide]

[hide="A MEDIOCRE SOLUTION TO QUESTION 4"]Let $ x=r_n$ for any arbitrary value $ n$ of the equation such that $ r_n$ is a root of the equation and such that $ n$ is any positive integral value between $ 1$ and $ 16$, inclusive.  Rearranging the equation to find the values of $ r^{16}_n$ and $ r^{15}_n$ gives us the values of $ 16r_n-18$ and $ 16-\frac {18} {r_n}$.  Subtracting the second value from the first gives us:

\[ \displaystyle\sum_{n=1}^{16}{r^{16}_n-r^{15}_n}=\displaystyle\sum_{n=1}^{16}{16r_n-34+\frac {18} {r_n}}\].

Separating the sum into the three parts $ 16r_n$, $ -34$, and $ \frac {18} {r_n}$ and finding these values individually, we know using Vieta's that the first value is equal to $ 0$, the second is $ -34\cdot16=-544$, and the third can be expressed as $ 18$ multiplied by the fifteenth(I think) symmetric sum of the roots and divided by the product of the roots, or $ 16$ as it turns out.

Hence, the final value is $ |16-544|=\boxed {528}$. $ \blacksquare$[/hide]

Apologies for the ugliness of language, lack of adequate explanation, and overall failure. ;\

[quote]1. Jerry has $ 3$ bags of aerodynamic marbles. Each marble is either tan or beige. Bag A has $ 40$ marbles, Bag B has $ 50$ marbles, and Bag C has $ 60$ marbles. Jerry knows that if he takes Bag A and Bag B and draws a tan marble from them, the probability that it came from Bag A is $ 25/77$. If he takes Bag B and Bag C and draws a tan marble from them, the probability it came from Bag B is $ 13/38$. If he takes Bag C and Bag A and draws a tan marble from them, the probability it came from Bag C is $ 4/5$. What is the sum of the possible total numbers of tan marbles that Jerry has in his bags?
[/quote]

Let the number of tan marbles in bag $ A$ be $ x$

Let the number of tan marbles in bag $ B$ be $ y$

Let the number of tan marbles in bag $ C$ be $ z$

Given $ (x/40)/(x/40) \plus{} (y/50)$ = $ 25/77$

Implies $ 13x/10$ = $ y/2$

Implies $ y \equal{} 13x/5$

From the second statement we have $ (y/50)/(y/50) \plus{} (z/60)$ = $ 13/38$

We have $ y/2$ = $ 13z/60$

Implies $ y \equal{} 13z/30$

The third equation i felt is extraneous.

So since the number of marbles is integral we see z is a multiple of 30.

Case 1: z is 30

then y is 13.

and x is 5.

Case 2: z is 60.

then y is 26.

and x is 10.

Summing all the possibilities the sum of  number of possible marbles is 144. :D

P.S. come on guys post solutions for some problems.

[quote="iniquitus"]Feel free to post solutions if you want to; if I don't see any being posted up, I will post up mine 1 by 1, in no particular order. The second round will probably start next Monday.

[hide="Solution to Question 6"]

Let the 200-gon be $ A_1 A_2...A_{200}$.

Lemma:

$ |\overline{A_1 A_2}| \times |\overline{A_1 A_3}| \times... \times |\overline{A_1 A_{200}}| = 200$

Proof:

$ x^{200} - 1 = (x - 1)(x - w)(x - w^2)...(x - w^{199})$, where $ w = e^{\frac {2\pi i}{200}}$

Dividing $ x - 1$, we get
$ x^{199} + x^{198} + ... + x^0 = (x - w)(x - w^2)...(x - w^{199})$

Plug in $ x = 1$ to get
$ 200 = (1 - w)(1 - w^2)...(x - w^{199})$

Noting that $ 1,w,w^2,...,w^{199}$ are the vertices of a regular 200-gon in the complex plane, take the magnitude of both sides to achieve the desired result.

The number we want to find, by the lemma, is $ \sqrt {200^{200}} = 200^{100} = 2^{300}5^{200}$
This number has $ 60501$ factors.
[/hide][/quote]

Uh, Oliver, don't you mean solution to question 5?

Yep, so here's 6:

[hide=" Question 6"]
The number of ways to put n stamps on, or $ a_n$, satisfies the recurrence relation $ a_n \equal{} a_{n\minus{}1} \plus{} 6a_{n\minus{}2}$.

By the characteristic equation $ x^2 \minus{} x \minus{} 6$, we have

$ a_n \equal{} A(3)^n \plus{} B (\minus{}2)^n$

We know that $ a_0 \equal{} 1$ and $ a_1 \equal{} 1$.
Plugging these in, we get

$ 1 \equal{} A\plus{}B$, and
$ 1 \equal{} 3A \minus{} 2B$.

This gives A = 3/5, B = 2/5

so $ a_{1337} \equal{} \frac{3}{5} 3^{1337} \plus{} \frac{2}{5} (\minus{}2)^{1337} \equal{} \frac{1}{5}(3^{1338} \minus{} 2^{1338})$

Conveniently, 223 is prime, so $ n^{222} \equiv 1 \;\; (\text{mod} 223)$

Therefore, $ a_{1337} \equiv \frac{1}{5}( 3^6 \minus{} 2^6) \equiv \frac{1}{5}( 3^3 \minus{} 2^3)(3^3\plus{}2^3) \equiv (19)(7) \equiv 133 \;\; (\text{mod} 223)$[/hide]

The solution I have to number 2 is not very "elegant," but it works.

[hide]A $ 13 \minus{} 14 \minus{} 15$ triangle can be broken into a $ 9 \minus{} 12 \minus{} 15$ and a $ 5 \minus{} 12 \minus{} 13$ triangle, which makes it very easy to use coordinates for the triangle.  Thus, place the vertexes of the triangle at $ (0,0), (5,12), (14,0)$.

The coordinates of the centroid are given as the average of the x and y values in the coordinate plane.  So the centroid is at $ (\frac {19}{3},4)$.

The incenter of the triangle can be found from the radius of the incircle.  This is given as $ A \equal{} rs$, where $ A$ is the area of the triangle, $ r$ is given as the radius, and $ s$ is the semiperimeter.  $ A \equal{} 84$, $ s \equal{} 21$, so $ r \equal{} 4$.  Now we need to find where the incircle meets the triangle in the coordinate system.  Use the two tangent theorem to discover that, WLOG, $ a \plus{} b \equal{} 13, b \plus{} c \equal{} 14, a \plus{} c \equal{} 15$.  Solving, we discover that the x coordinate of the incenter is $ 6$.  Since it is perpendicular to the triangle, the incenter is at $ (6,4)$.

Finally, to solve for the circumcenter, we will place the three vertexes of the triangle into the equation of a circle.  $ (x \minus{} h)^2 \plus{} (y \minus{} k)^2 \equal{} r^2$.  Substitute $ (0,0), (14,0)$ to get that $ x \equal{} 7$,  Then, substitute in $ (5,12)$ and solve to get that $ y \equal{} \frac {33}{8}$.

With the three coordinates, we can solve for the area.  $ A \equal{} \frac {1}{2}bh$.  $ b \equal{} \frac {1}{3}, h \equal{} \frac {1}{8}$.  $ 4*1 \plus{} 5*48 \equal{} 244$.[/hide]

[hide="Solution to Question 7"]We first split 1323 into $ 3^3$ and $ 7^2$, and then use CRT to work our way back up.

1)

$ 13x^7 \minus{}42x \plus{} 674 \equiv 0 \;\; (\text{mod} 9)$, and it's obvious that x=0 is not a solution.
$ 4x^7 \plus{} 3x \plus{}8 \equiv 0 \;\; (\text{mod} 9)$
$ 4x \plus{} 3x \plus{} 8 \equiv 0 \;\; (\text{mod} 9)$ by Euler's Theorem
$ 7x\plus{}8 \equiv 0 \;\; (\text{mod} 9)$
$ x \equiv 4 \;\; (\text{mod} 9)$.

Then, by Hansel's lemma, we get the solution

$ x \equiv 4 \plus{} 2\times 9 \equal{} 22 \;\; (\text{mod} 27)$

2)

$ 13x^7 \minus{}42x \plus{} 674 \equiv 0 \;\; (\text{mod} 7)$, and it's obvious that x=0 is not a solution.
$ \minus{}x^7 \plus{}2 \equiv 0 \;\; (\text{mod} 7)$
$ \minus{}x \plus{} 2 \equiv 0 \;\; (\text{mod} 7)$
$ x \equiv 2 \;\; (\text{mod} 7)$

By Hansel's lemma, since the derivative is a multiple of 7 and the function of 2 is a multiple of 49, the solution is

$ x \equiv 2 \;\; (\text{mod} 7)$

Combining these with CRT, we have:

$ x(7\plus{}27) \equiv (\minus{}5)(7) \plus{} (2)(27) \equiv 19 \equiv 19 \minus{} 189 \equiv \minus{}170 \;\; (\text{mod} 189)$
$ x \equiv \minus{}5 \;\; (\text{mod} 189)$, so the desired answer is $ 2 \times 189 \minus{} 5 \equal{} 373$

(In both cases, you actually only had to check 2 values, so if you didn't know Hansel's lemma, you could've used either mod manipulation or just checked 4, -5 in part (1) and 2, -5 in part (2))
[/hide]

Question for #6: How many teams are there, how many people are in each team, and do all team members have to have the same colored jersey? Or do I totally misunderstand the problem...

Sorry about the confusion.

There are only two teams.

A team does not have to have the same colored jersey (in fact, it can't, since there's only four of one color, and a team must have at least five people).

Hey guys, sorry if you've already submitted your answers.

For the team and jersey problem, it involves a lot of computation that is ugly without a calculator.

The following is a QUESTION CHANGE!

Instead of four colors of jerseys, suppose he only has two colors of jerseys.  Also suppose he only has three of each type of jersey.  THERE ARE NOW ONLY SIX JERSEYS!

AGAIN, I apologize for the poorly created question.  It's doable, but very very very ugly. :oops_sign:  :oops_sign:


The product of two consecutive even integers is 32 less than the square of the greatest integer. Find the smaller integer.


LEt $ p>3$ be a prime number. Prove that each of the divisors (bigger than $ 1$) of $ \frac{2^{p}+1}{3}$  are in the form $ 2pt+1$ where $ t$ is a positive integer. :)

It is enough to prove that for prime divisors. Let $ q$ be one of them. Then obviously $ q \not = 2$ and $ q \not = 3$ since if $ 9|2^{a}+1$ then $ 3|a$. We have $ 2^{p}\equiv-1 \mod{q}$ and $ 2^{2p}\equiv 1 \mod{q}$. Let $ k$ be the order of $ 2$ $ \mod{q}$. Then $ k|2p$ and $ k \not | p$, so $ k=2$ or $ k=2p$ since $ p$ is prime. But $ k \not = 2$ since we would have that $ 2 \equiv-1 \mod{q}$, so $ q=3$. It means that $ k=2p$ and $ 2p|q-1$, conclusion follows.


inequalities
triangle inequality
algebra unsolved
algebra

slove this system of equation: $ \begin {cases} x^2\plus{}y^2\equal{}\frac{82}{9}\\\displaystyle \left | y\plus{}\frac{1}{y}\right |\plus{}\displaystyle \left |\frac{10}{3}\minus{}x\plus{}y \right |\equal{}\frac{10}{3}\plus{}y\plus{}\frac{1}{y}\end {cases}$

(First post here!)
EDIT EDIT: I've gotten rid of the old solution. I've got to stop making careless mistakes.
[hide="Solution"]Notice that

$ \displaystyle \left | y \plus{} \frac {1}{y} \right | \ge y \plus{} \frac {1}{y}$

$ \displaystyle \left | \frac {10}{3} \minus{} x \plus{} y \right | \ge \frac {10}{3}$

Clearly $ \displaystyle \left | y \plus{} \frac {1}{y}\right | \plus{} \displaystyle \left |\frac {10}{3} \minus{} x \plus{} y \right | \equal{} \frac {10}{3} \plus{} y \plus{} \frac {1}{y}$ iff equality holds in both inequalities. So $ y \plus{} \frac {1}{y} \ge 0$ (implying $ y > 0$) and

$ \displaystyle \left | \frac {10}{3} \minus{} x \plus{} y \right | \equal{} \frac {10}{3} \implies \minus{} x \plus{} y \equal{} 0 \text{ OR } \minus{}x \plus{} y \equal{} \frac{\minus{}20}{3}\implies x \equal{} y \text{ OR } x \minus{} y \equal{} \frac{20}{3}$

The $ x \equal{} y$ case is easy.

If $ x \minus{} y \equal{} 20/3$, then note that the max for $ x$ is $ \frac{\sqrt{82}}{3}$, and the min for $ \minus{}y$ is $ \minus{}\frac{\sqrt{82}}{3}$. So the max for $ x \minus{} y$ is $ 2\sqrt{82}/3$, but this less than $ 20/3$.

We're not supposed to consider imaginary numbers, right?[/hide]

[quote="ali666"][quote="vishalarul"]

If $ | \minus{} x \plus{} y| > 0$ then this would mean that:
\[ \displaystyle \left |\frac {10}{3} \minus{} x \plus{} y \right | > \frac {10}{3} \plus{} | \minus{} x \plus{} y| > \frac {10}{3}
\]
[/quote]
I think there is a very small mistake in this step... :wink:[/quote]
? :P

vishalarul means $ \minus{} x \plus{} y > 0$.  And in that case the first inequality is an equality.  (The triangle inequality goes in the other direction.)  Unfortunately, nothing is said about the case that $ \minus{} x \plus{} y < 0$.

[quote="Amazigh"][quote="thanhnam2902"]slove this system of equation: $ \begin {cases} x^2 \plus{} y^2 \equal{} \frac {82}{9} \\
\displaystyle \left | y \plus{} \frac {1}{y}\right | \plus{} \displaystyle \left |\frac {10}{3} \minus{} x \plus{} y \right | \equal{} \frac {10}{3} \plus{} y \plus{} \frac {1}{y}\end {cases}$[/quote]
Hope it is correct.
[hide]Case 1: $ y > 0$

We have then $ |\frac {10}{3} \minus{} x \plus{} y| \equal{} \frac {10}{3}$

 1.1: suppose that $ \frac {10}{3} \minus{} x \plus{} y > 0$, then we get $ x \equal{} y$

 1.2: else we get $ x \minus{} y \equal{} 20/3$

Case 2: $ y < 0$

 Our result is then $ |\frac {10}{3} \minus{} x \plus{} y| \equal{} \frac {10}{3} \plus{} 2(y \plus{} \frac {1}{y})$, we get something absurd here because
\[ 2(y \plus{} \frac {1}{y})\leq \minus{} 4
\]
[/hide][/quote]

THANK YOU VERY MUCH!


let $f(x)=sinx+sin(x\sqrt{2})$
prove$f$ is not period  function .

We haved see that: If $f(x)=asinux+bsinvx$ is period function iff $\frac{u}{v}$ is rational number.
Hence applied the event to this problem too clearly


Let $ M_{a}$ be the $ a$ by $ a$ matrix for which every entry below the main diagonal is $ 0$ and the rest of the entries are 1's. If $ N=(M_{4})^{0}+(M_{4})^{1}+(M_{4})^{2}+...+(M_{4})^{2398}$, and the sum of the entries of $ N$ is $ A$, then find the remainder when $ A$ is divided by $ 1000$.

[hide="hint"]If an $ a \times a$ matrix $ M$ represents an a-state system in which the number of ways to get from state $ i$ to state $ j$ in one step is the element in the $ i$th row and $ j$th column, then the $ a \times a$ matrix $ M^{k}$ represents the set of numbers of ways to get from one state to another in $ k$ steps.
[/hide]

[hide="solution"]
The matrix $ M_{a}$ represents an $ a$-state system in which any step taken goes from a state to one strictly preceding it.  Then, because no paths of length greater than or equal to $ a$ can be taken, we have that $ M_{a}^{k}$ is a zero matrix for $ k \ge a$.  Therefore, all we need to calculate is $ M_{4}^{0}+M_{4}+M_{4}^{2}+M_{4}^{3}$.  The sume of the entries in the sum of these is the sum of the sum of entries in each, because there are no negative entries.  The sum of the entries of $ M_{4}^{0}$ is 4, because it is the identity matrix.  The next one gives $ 6$, the next gives $ 3$, and the last $ 1$, for a total of $ \boxed{14}$.

In general, the sum of the entries of $ M_{a}^{k}$ for $ 0<k < a$ is the $ a-k$th triangular number.[/hide]

Err...I don't think that is correct. The main diagonal has ones. Hence you can stay at the same place.

I was pretty proud of this problem. It is correct to interpret this as a graph. There is a very nice bijection.

EDIT:

[I flipped the matrix since that makes the meaning more clear; it obviously doesn't change the sum.]

1 0 0 0
1 1 0 0
1 1 1 0
1 1 1 1

If we consider this as an adjacency matrix of a directed graph, then we have the following edges.

1->1,2,3,4
2->2,3,4
3->3,4
4->4

We are creating a bijection. The first number in the sequence is where you start [a number from 1 to 4]. Each turn can be represented by a 0,1,2,3,4, representing how many spaces that you jumped. The sum of all these entries must be less than or equal to 4. Consider another dummy spot at the end that can take up the rest of the 1's. So we have:
[1 to 4] {[0 to 4] [0 to 4]....[0 to 4] } {a moves} [0 to 4]

In otherwords, we are placing 3 items in a+2 places, so this is a simple counting problem. There are $ \binom{a+2+3-1}{3}$.

We have to sum this over $ a=0,1,...,2398$. I remember that there was a slick way to circumvent this, but anyway, we can just sum them with basic binomial stuff, we get

$ \binom{2398+4}{4}=2402*2401*100*2399\equiv 200\mod 1000$.

[quote="K81o7"][quote="Altheman"]Err...I don't think that is correct. The main diagonal has ones. Hence you can stay at the same place.
[/quote]

That's a good point.  I switched the 1's and 0's (I thought it was everything below the main daigonal is a 1, and everything else is a 0).  The argument gets a little more complicated but is not that much more difficult.  You can calculate the number of paths of length $ k$ which take $ m$ backwards steps: it's just $ \binom{k}{m}$.  Then considering there are $ a-m$ places you can start, we find that the sum of the entries of $ M_{a}^{k}$ is $ \sum_{m=0}^{a-1}(a-m)\binom{k}{m}$.[/quote]

Well, right you can go through that whole mess to evaluate all the sums. I wish I could remember how to do the argument by one bijection. Maybe ask JSteinhardt, I think that he helped me check this problem when I made it.

On a slightly different note, I totally forgot about that mock test. I don't think I ever put up results because not enough people took it. It still had a lot of nice problems though. It is nice to see some of them show up on the forums again.

[quote="Altheman"]
If we consider this as an adjacency matrix of a directed graph, then we have the following edges.

1->1,2,3,4
2->2,3,4
3->3,4
4->4

We are creating a bijection. The first number in the sequence is where you start [a number from 1 to 4]. Each turn can be represented by a 0,1,2,3,4, representing how many spaces that you jumped. The sum of all these entries must be less than or equal to 4. Consider another dummy spot at the end that can take up the rest of the 1's. So we have:
[1 to 4] {[0 to 4] [0 to 4]....[0 to 4] } {a moves} [0 to 4]

[/quote]

Well, actually I don't quite get what you're doing. How come the adjacency matrix (what is it?) and the edges?  :blush:

Can you help explain it to me?

EDIT: OK, I understand now  :D

[quote="Gaku Liu"]Wait isn't the answer $ \binom{2403}{4}-1$?

[hide]
We want the number of nondecreasing sequences consisting of the numbers 1-4 and having at most 2399 terms (each sequence corresponds to the vertices of a path in the graph Altheman described). If we let a be the number of 1's, b be the number of 2's, etc., then we want the number of nonnegative integer solutions to $ 1 \le a+b+c+d \le 2399$. If we define the nonnegative integer $ e=2399-(a+b+c+d)$, we can turn this into the equality $ a+b+c+d+e=2399$. The number of solutions to this is $ \binom{2399+4}{4}$, but we have to throw out $ (0,0,0,0,2399)$ because $ a+b+c+d \ge 1$. So the final answer is $ \binom{2403}{4}-1$, which I think is 599 mod 1000.
[/hide][/quote]

I think that the M^0 term includes the 'no terms in the sequence'.

[quote="Altheman"]I think that the M^0 term includes the 'no terms in the sequence'.
[/quote]
Well, my sequence was listing the vertices encountered in a walk, not the edges. So M^n would correspond to sequences with n+1 terms. M^0 would correspond to walks without an [i]edge[/i]; but they would still have a starting vertex and therefore the sequence would have 1 term. Besides, the sum of the elements of M^0 is 4, which is the number of sequences with 1 term.


FTW MATHCOUNTS AMC AIME USA(J)MO USAMO algorithm

So, this is a tournament. In the tournament, there willl be 8 problems, 2 school/chapter difficulty, 3 state-national difficulty, 3 national and beyond. You will be each given 2 problems at the beginning. After solving a problem, pm me the answers and I will pm the next problem if you are correct. You will also gain a point. If you are incorrect, you will lose a point and will not get the next problem. (But you can try to answer the problem again, if you are correct this time you get the next problem: Otherwise you are done with the round)

When you finish all 8 problems, you will get 2 extra points.

You have a week to solve the problems. After the week, the people with the lowest score will play a tiebrake if lowest score was less than 6. (If the lowest score is 6 or above, then nobody is eliminated) The tiebraker will consist of 5 state-national level problems. Everyone with the lowest score on that will be eliminated.

When there are 4 people left, then they are matched up against each other and have an elimination tournament:

The format will be the same, except that if both people tie on the tiebraker, then they have a FTW countdown match.

The tournament will start on August 20 or when we get 20 people.

To signup, post here or PM me.

Signups:
1. shentang
2. Poincare

On second thought, I'll just post all 8 problems at once.

Signups:
1. shentang
2. Poincare
3. FantasyLover
4. qwertythecucumber
5. 5849206328x

dude when pythag says mathcounts he means IMO #1 or 4 (or since he doesn't want proofs, AIME #15)
if you can get 1 point, I'll be impressed
@pythag: I'm watching you (not really, but i know who you are)

i don't mind USAMO or AIME problems but IMO might be getting out of hand!

There will be no IMO problems except for really really easy ones. Have you seeen the first IMO problem?

Is goes like:

Prove that gcd(21n + 4, 14n +3) = 1, which is just standard euclidean algorithm on that.

gcd(21n + 4, 14n + 3) = gcd(7n +1, 14n+3) = gcd(7n+1, 1) = 1.

WOAH THAT IS PRETTY INSANELY EASY

LIKE

I CAN ACTUALLY UNDERSTAND THE PROBLEM

AND SOLVE IT

<7 HRS

THAT IS A BAD PROBLEM

BTW, IMO difficulty is about the same as USAMO difficulty.

In fact, this year's IMO is probably much easier than this year's USAMO.

Whether easy or not, proofs are beyond MathCounts level. I suggest sticking to MathCounts.  :wink:

/in, but I'm not really good at math, so I will probably fail :(

actually I sent the answer to number one to the host.

neutrus, Poincare, Eulcild_great and 5849206328x are eliminated.

Round 2! (last 3 Problems with proof)

Please answer. Deadline next Sunday September 14th midnight.

The Problems:

(Note. Next round has been purposely made pretty hard.)

1.	a + b = 5 and ab = 10. If a and b are complex numbers, find $ a^4 \plus{} b^4$.
2.	A set of positive integers has the triangle property if it has 3 numbers in it that could be the sides of a non-degenerate triangle. Consider the set {4, 5, 6, $ \ldots$ n}. If every 10-element subset of this set has the triangle property, what is the maximum possible value of n?
3.	Let’s play a game on an n x n board. You are in the center square (or one of the center squares.) Every turn, you can move to an adjacent square(but not diagonally). When it is my turn, I can signify a square to make impassable (i.e. you can never move to that square.) If you reach the edge of the board, you win. If, at some point, you cannot move, you lose. Find the largest n such that you have a winning strategy.
4.	Let S be a subset of $ {1, 2, 3, \ldots, 2000}$ such that no two members of S have a positive difference of 4 or 7. Which is maximum number of elements in S?
5.	There is an acute triangle with side lengths BC = a, AC = b, and AB = c. There are points D, E,F, and G with  D on side AC, E on side AB, and F and G on side BC such that DEFG is a square. Find the length of this square in terms of a, b, and c.
6.	We have $ 3\plus{}x^6 \plus{} y^6 \plus{} z^6 \equal{} 2x^2y \plus{} 2y^2z \plus{} 2z^2x.$ If x, y, and z are positive real numbers, find all possible values of xyz. If your answer is something like, all real numbers greater than 1/2, then use standard set-builder notation (i.e. {x| x > 1/2}. (Proof!)
7.	In a soccer tournament, every team plays every other team once. A team receives 4 points for a win, 2 points for a draw, and 1 point for a loss. After the tournament, the tournament director notes that the team who had the highest number of points at the end also won the fewest number of games out of all the teams. Find the smallest possible number of teams. (Proof!)
8.	Find all the pairs of positive integers (x, p) such that p is a prime, $ x \le 2p$ and $ x^{p\minus{}1}$ is a divisor of $ (p\minus{}1)^x \plus{} 1$. (Proof!)

PM me for clarifications.

Also: The sources for the Round 1 Problems:

5. An old tounament
6. 2007-2008 WOOT (You are allowed to use individula problems, just not a lot of problems at one time...this will be the only problem from 2007-2008 WOOT I will use.)
7. IMO Shorlist
8. Tournament of Towns.

See? IMO problems can be eaSY.

Still participating

budi713
shentang
FantasyLover
chenhsi
Textangle
Wickedstjr
xpmath
alingaling

Problems and Answers:

1.	How many two-digit integers are multiples of 4 but not 5? (18)
2.	A triangle ABC has side lengths AB = 3, AC = 4, BC = 5. Find the length of the altitude from A. Express your answer as a fraction in simplest form? (12/5)
3.	Bruce plays a game. He rolls 2 dice. If the sum of the numbers on the two dice is 7, he wins. Otherwise, he rolls again. If it is 6 or 8, he wins. Otherwise, he loses. What is the probability Bruce wins? (43/108)
4.	There are four people Albert, Bruce, Casey, and Dale who want to split up 26 apples among themselves. If Dale must have the same number of apples as Casey and Bruce combined, and Albert must get an even number of apples, how many ways are there to split up the apples if everyone must get at least one apple? (66)
5.	How many numbers less than or equal to 1000 can be expressed as the difference of two perfect squares? (749 or 750)
6.	A floor can be covered with n identical square tiles of a certain size. If a certain smaller square tile is used, then n + 76 tiles are required. Find n. (324)
7.	On a 5 x 5 board, two players alternately mark numbers on empty cells. The first player marks 1’s, the second 0’s. One number is marked per turn, until the board is filled. For each of the nine 3 x 3 squares the sum of the nine numbers on its cells is computed. Denote by A the maximum of these sums. How large can the first player make A, regardless of the responses of the second player? (6)
8.	You have n colors and a 10x10 board. You must color every square of the board, and every color must be used at least once. If a square is colored color A, then at least 2 squares adjacent to it must be colored A. What is the maximum possible value of n? (25)

Yes, I've never been in the top eight in any competition ever. I'll get working on these problems as soon as possible.

Okay when are we going to continue this tournament. I think we shoud move on to the next round.

...you are teh only person who sent me answers  :mad:  :rotfl:

[quote="Wickedestjr"]You should send them reminders. That's what I do, and most of the time they end up answering the questions immediately after that.

OR

You could make me the champion of pythag011'stournament.  :)[/quote]

Fine. Solve problem 8 and I will.

I can't solve it, but I am obviously going to be the only active competitor in this competition. budi713 who gets on like every other day of the week hasn't been submitting answers, so therefore, he probably doesn't want to compete anymore.

Well, fine. I have a competitor who has solved all 8 problems already. So yoou are 2nd place.

/w Wickedesjr I don't think he liked your nagging....

Brut3Forc3.

I made him do it because its sad when theres only 1 person competiting...

Anywyas, he got all 7. Excpet 8. So if you solve all 8, you might get first!

BTW, #8 is an IMO level problem.

For now:
 :winner_first:  Brut3Forc3
 :winner_second: Wickedstrj
 :winner_third: spammer123

Yay!!! Even though you misspelled my name, I am glad that I got second place.


\hrulefill may be what you're looking for.  (if that doesn't work, try \hrule)


I don't know if "period" is the right name for how often something repeats itself. For example, $ 0.\overline{19}$ has a "period" of $ 2$ digits. Please let me know if that's the correct term.

Anyways, calculate how many digits are in the period of $ 0.\overline{19}\cdot0.\overline{199}$

We have that $ \overline{.19}\equal{} \frac {19}{99}$, and $ \overline{.199}\equal{}\frac{199}{999}$.  Multiplied, these fractions
equal $ \frac{3781}{98901}$, which in decimal form is equal to about
$ 0.0382301493$.  When I plugged it into my calculator, I got that
the period was $ \boxed {54}$ digits long.

Thanks [b]Z-coli[/b], but the idea is to figure out how long the period is without using calculators.

Here is an attempt I had of organizing the number, but I couldn't get anything useful from it:

[hide="Attempt"]In fractions:

$ \frac{19}{99}\cdot\frac{199}{999}\equal{}\frac{(20\minus{}1)(200\minus{}1)}{(100\minus{}1)(1000\minus{}1)}\equal{}\frac{4000\minus{}220\plus{}1}{100000\minus{}1100\plus{}1}\equal{}\frac{3781}{98901}$

Let $ y$ be our number:

$ 100000y\minus{}1000y\minus{}100y\plus{}y\equal{}3781$[/hide]

In general, to find the period of the reduced fraction $ \frac pq$, we note that factors of $ 2$ and $ 5$ in the denominator won't affect the period. If, after we take out these factors, the denominator is $ q'$, then the period is the smallest $ k$ such that $ q'|10^k \minus{} 1$. This is because if we can express $ \frac p{q'}$ in the form $ \frac {np}{\underbrace{999\ldots999}_k}$ such that $ q'n\equal{}\underbrace{999\ldots999}_k$ for some integer $ n$ and $ np \equal{} \overline{p_1p_2p_3\ldots p_k}$, then we can easily see that the decimal form of this is $ 0.\overline{p_1p_2p_3\ldots p_k}$, and the period is $ k$.

Now, for this problem, we know that $ 98901 \equal{} 3^5\cdot11\cdot37 \equal{} 3^4\cdot11\cdot111$. This means that the number of $ 9$'s in the denominator must be a multiple of $ 2$ to be divisible by $ 11$ and a multiple of $ 3$ to be divisible by $ 111$. For $ \underbrace{999\ldots999}_m$ to be divisible by $ 3^4$, it is sufficient to find the smallest $ m$ such that $ \underbrace{111\ldots111}_m$ is divisible by $ 3^2 \equal{} 9$. Since the sum of the digits of $ \underbrace{111\ldots111}_m$ is equivalent to its residue modulo $ 9$, $ m \equal{} 9$ is the smallest possible value. Finally, we see that $ k \equal{} 18i$ for some positive integer $ i$ (I don't think we can just conclude that the period is $ 2\cdot3\cdot9 \equal{} 54$, even though it gives the right answer), from which it is easier to see that $ \boxed{54}$ is the period.

I understood the idea, but there is one thing I didn't catch on.

You said that $ k \equal{} 18n$ , for some integer $ n$, and since you found $ n \equal{} 3$ , then:

$ \frac {np}{q'} \equal{} \frac {p}{q}\Leftrightarrow q' \equal{} nq$

Since $ 3\cdot 98901 \equal{} 296703$ which is very diferent from what is expected. Are the $ n$'s you used different or is there something wrong with my thinking?

Sorry, they are different. :blush: I'll edit it. The bottom paragraph is just executing the idea in the top paragraph. Again, I'm very sorry!

And for further clarification: $ q'n\equal{}10^k\minus{}1$, in case something wasn't clear about that.

Ah ok, just a small detail hehehe...

Hey, don't worry about it, no need to apologize for helping me with the problem :)

Thanks for the solution.


There will be a Putnam Competition Math Jam on Wednesday 15 September at 7:30ET / 4:30PT / 2330 GMT.

This is the first of three scheduled Math Jams for the Putnam:
Sep 15: Calculus & Analysis Problems
Oct 20: Linear Algebra & Abstract Algebra Problems
Nov 9: Miscellaneous Problems

For more info on the Putnam Competition, visit
[url]http://math.scu.edu/putnam/[/url] and
[url]http://www.unl.edu/amc/a-activities/a7-problems/putnam/[/url]

Even if you do not fit the Putnam criteria (for instance, if you're not a college student in the US or Canada), please nonetheless feel free to attend the Math Jam.  The only prerequisite is a working knowledge of basic calculus.

Also, if anyone has favorite Putnam problems that they'd like to see worked through in one of the Math Jams, please do tell.

I saw this problem in the Art and Craft of Problem Solving that looks pretty cool (but I can't solve it of course :D ):

A hole puncher is placed on a point on a plane.  When operated, the hole puncher removes every point whose distance from the hole puncher is irrational.  How many times must the hole puncher be operated to remove every point?

Sadly, I can't make it to the math jam :(

[quote="nr1337"]I saw this problem in the Art and Craft of Problem Solving that looks pretty cool (but I can't solve it of course :D ):

A hole puncher is placed on a point on a plane.  When operated, the hole puncher removes every point whose distance from the hole puncher is irrational.  How many times must the hole puncher be operated to remove every point?
[/quote]
This problem was on the 1990 Putnam (it was Problem A4).  It happens to be my all-time favorite Putnam problem!  We're going to do it in the "Miscellaneous Problem" Jam on November 9th.  Wednesday's Jam will be all calculus problems.

Oh.  I tried the problem once and got the answer 2, but then noticed a fatal error in my reasoning.  Well, I still have 7 more years before I can actually be in Putnam, so I have a while to train ;)

Doesnt that make you in 6th grade (if u mean that you are only going to take it in college)?

Seventh.  Oh wait... only six more years (13-7).

[quote="nr1337"]I saw this problem in the Art and Craft of Problem Solving that looks pretty cool (but I can't solve it of course :D ):

A hole puncher is placed on a point on a plane.  When operated, the hole puncher removes every point whose distance from the hole puncher is irrational.  How many times must the hole puncher be operated to remove every point?
[/quote]

Just take the points $(0,0),(1,0),(\sqrt 2,0)$. Now take a point $(x,y)$ in the plane. If $\sqrt{x^2+y^2}$ is irrational, then we are done. Otherwise, $x^2+y^2$ is also rational, so there are two possibilities: either $x$ is irrational, in which case $(x-1)^2+y^2$ is irrational, and then so is $\sqrt{(x-1)^2+y^2}$, or $x$ is rational, in which case $(x-\sqrt 2)^2+y^2$ is irrational, so $\sqrt{(x-\sqrt 2)^2+y^2}$ is irrational as well. On the other hand, it's easy to show that two points do not suffice, because there are two circles of rational radii, one centered at each one of the two points, which intersect. Their points of intersection will not be removed.


Do any of you play USCF chess tournaments???

If you do I'd like to know your rating.

me, im only a 1046  :blush: ...i got careless in a couple tournaments but managed a first place in one  :first:

me.... 1325... 8th grader rite now hasn't played since last year and has only played in 7 tournaments.

I went mostly because of the school team back in 5th grade. I have only gone to 2 or 3 tournaments since.

wow...all higher than me...grr! lol how long have you guys actually started chess? i'm in eighth grade and i started in first  :lol: ...

In my active playing chess heydays,1979-80,i got official
beginners rating at 1680. Though,not much higher,i appreciate
such given rating.I played couple of 'systematic blindfold games'with my brother.I found it very exercising,imaginatively.Have you ever try playing 'BLINDFOLD CHESS?'

                                                     Respectfully,
                                                    A.Eddington

yeah its fun

but i dont have a coach so i quit last summer since it got hard to improve.

math is more challenging now

uuhhh started in like 5th or 6th grade... haven't really played since like 8th or 9th grade... rating's like 1000 something :blush:

[quote="arthur eddington"]Have you ever try playing 'BLINDFOLD CHESS?'

                                                     Respectfully,
                                                    A.Eddington[/quote]

Yea except we don't use blindfolds. Instead, (like on our carpools to tournaments), we sometimes just say the notation out. Is your version of BLINDFOLD CHESS like that or do you actually use chess boards and blindfolds?

Have you ever tried to play chess on multiple boards, like with 3 to 5 boards at a time and each turn, a person can move 1 piece, but it can be on any board though. whoever wins the most, out of the number of boards you have, wins.

My chess rating is 1711. I havent played since 6th grade. When I was in 5th grade, I won the KY State chess championship. I moved to NC and I got 10th in the K-8 in 6th. Havent played an official tournament since. :(

[quote="7h3.D3m0n.117"][quote="goldendomer"]When I was in 5th grade, I won the KY State chess championship.[/quote]

WOW NICE!![/quote]

Thanks. I was tied for first with the another 5th grader but I was declared the champion because my opponents were stronger.


Aazlae yek Mosallas be toolhaye sahihe $ k,m,n$ hastand, Farz konid $ k > m > n$ va:

$ \left\{\frac{3^{k}}{10^{4}}\right\}\equal{}\left\{\frac{3^{m}}{10^{4}}\right\}\equal{}\left\{\frac{3^{n}}{10^{4}}\right\}$

matloob ast kamtarin meghdare mohite in mosallas. :)

Soale khoobie faghat ye khorde kharkari dare :) , avval az hame tavvajoh konin ke az sharte masale natije mishe :
$ 3^{m}\equiv 3^{n}\equiv 3^{k}(mod 10000)$ . hala farz konin $ Ord 3_{10000}\equal{} a$ ke ba yekam mohasebat ( hamoon kharkarishe :D) dar miad $ a \equal{} 500$ . amma az chizi ke ghabl dar ovorde boodim natije mishe $ m\equiv n\equiv k (mod 500)$
pas :
$ n \equal{} 500a_{1}\plus{}i$ , $ m \equal{} 500a_{2}\plus{}i$ , $ k \equal{} 500a_{3}\plus{}i$ va $ a_{1}< a_{2}< a_{3}$ .
hala tebghe ghazie hemar darim :
$ 500a_{1}\plus{}i\plus{}500a_{2}\plus{}i > 500a_{3}\plus{}i$ ke natije mide $ 500(a_{1}\plus{}a_{2})\plus{}i > 500a_{3}$ hala dohalat dar nazar begirin :
1)$ i \equal{} 0$ dar in halat az namosavie bala natije mishe $ a_{1}\plus{}a_{2}\geq a_{3}\plus{}1$ . hala mohite mosalas mishe
$ 500(a_{1}\plus{}a_{2}\plus{}a_{3})\geq 500(2a_{3}\plus{}1)$ amma az oonjayike $ a_{3}> a_{2}> a_{1}\geq 0$ pas agar $ a_{3}\equal{} 2$ oon moghe $ a_{2}\equal{} 1$ va $ a_{1}\equal{} 0$ ke ba namosavie $ a_{1}\plus{}a_{2}\geq a_{3}\plus{}1$ tanaghoz dare , dar gheyre in soorat
$ a_{3}\geq 3$ ke natije mide mohite mosalas hadeaghal $ 3500$ hast .
2)$ i\geq 1$ ke tebghe hamoon namosavi avvalie natije mishe $ a_{1}\plus{}a_{2}\geq a_{3}$
hala mohite mosalas mishe $ 500(a_{1}\plus{}a_{2}\plus{}a_{3})\plus{}3i\geq 1000a_{3}\plus{}3$
hala mesle bala mitoonin sabet konin ke $ a_{3}\geq 3$ ke natije mide mohite mosalas hadeaghal
$ 3003$ hastesh .
va mesal baraye $ 3003$ ine ke $ n \equal{} 501$ , $ m \equal{} 1002$ va $ k \equal{} 1501$


In the rectangle shown, the ratio of width to length is 1:4. What percent of the rectangle is shaded?

[asy]pair A,B,C,D;
A = (0,0); B = (4,0); C=(0,1); D = (1,1);
draw(A--B--B+C--C--A--D--B);
fill(A--B--D--cycle);[/asy]

The information of ratios is unnecessary. Draw a line through and you'll see parts match up, and the answer is $ 50%
$.

That is what I would do.  This is why it works though.
The height of the triangle is the short side of the rectangle, while the base is the long side of the rectangle.  If we call the long side $ b$ and the short one $ h$, we can find that the area of the rectangle is $ bh$, while the triangle is $ \frac {1}{2} bh$.  So we know that the triangle is $ \frac {1}{2}$ the area of the rectangle, so we have $ \boxed {50}$, like pinkmuskrat said.


If $a,b,c$ are positive integer satisfying

$2ab+2ac+2bc=abc$

Find the ordered triples $(a,b,c)$

 :)

[quote="socrates"]rewrite the given equation as

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$

Let WLOG $a\leq b\leq c$ then $\frac{1}{a}\geq\frac{1}{b}\geq\frac{1}{c}$.    So $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq\frac{3}{a}$

that is $a\geq6$

On the other hand $\frac{1}{2}-\frac{1}{a}=\frac{1}{b}+\frac{1}{c}>0$

that is $a>2$ or $a\geq3$. Finally $3\leq a\leq6$ so $a=3 \ \text {or}\ 4 \ \text {or}\ 5 \ \text {or}\ 6$

now substitute the values of $a$. the solution is not hard from here

What do you think?? :)[/quote]

Well , socrates , the inequality you obtained are $a\geq 3$ and $a\geq 6$ which combine and gives you $a\geq 6$ instead  ;)

[quote="Arne"][quote="shyong"]
Well , socrates , the inequality you obtained are $a\geq 3$ and $a\geq 6$ which combine and gives you $a\geq 6$ instead  ;)[/quote]

No, you definitely get $a \leq 6$.[/quote]

Oh ya , you are right Arne  :blush:  I just follows what socrates wrote but didnt notice that he has a typo there  :blush: I should have had a thorough checking next time  :lol:


geometry
circumcircle
angle bisector
geometry proposed

$M \in \gamma_1$ where $\gamma_1$ is the circumference circumscripted to OPQ which is; using chord-angles, we note that $\angle PMQ$ and $\angle QMO$ are fixed: so also $\angle QMO+0.5\angle PMQ$ is fixed: now prove the case of $M\equiv O$ m=o, and see that it's $\frac{\pi}{2}$

MP, MQ, MO and the bisector line of PMQ are an armonic fascicul. :) The problem becomes a known theorem :)

Let $MH$, with $H$ on the circumscribed circle of $POMQ$, be the angle bisector of $\measuredangle PMQ$. And $PQ \cap MH = K$. Connect $O$ with $H$. It is $OH \bot PQ$. So $\measuredangle OPM + \measuredangle MKQ = 90^{\circ}$.
$\measuredangle MKQ = 180^{\circ} - (\measuredangle KQM + \measuredangle KMQ) = 180^{\circ} - (180^{\circ} - \measuredangle POM + \measuredangle KMQ) = \measuredangle POM - \measuredangle KMQ = \\ = 180^{\circ} - \measuredangle OPM - \measuredangle PMO - \measuredangle KMQ$.
Then $90^{\circ} - \measuredangle OPM = 180^{\circ} - \measuredangle OPM - \measuredangle PMO - \measuredangle KMQ$. And finally $90^{\circ} = \measuredangle PMO + \measuredangle KMQ = \measuredangle PMO + \measuredangle PMK$.

let $\omega$ be the circumcircle of $PMOQ$,given $OP=OQ$,Let $OL$ is the bisector of $O$($L$ on circumcircle),then $OL$ is the diameter,also bisector of $\angle PMQ$ meet at $L$,hence done.

Dear Mathlinkers,
I saw this terminology:
O, L is the first, second M-perpoint of triangle MPQ....
Sincerely
Jean-Louis

That's really easy. Just assume $\angle\{POQ}=2\theta\$ and proceed.

Oil is the food of the hair; Rice is the food of the Bengalis; Maths is the food of my soul.

Let $ \omega $ denote the circumcircle of $ POMQ $, and let $ S $ be the intersection of the angle bisector of $ PMQ $ and $ \omega $. Let $ \angle SMP = \angle SMQ = \alpha $. Note that $ \angle POQ = \angle PMQ = 2\alpha $. Let $ T $ be the point diametrically opposite $ Q $. Then $ \angle POT = 180 - 2\alpha $. Since $ POT $ is isosceles, $ \angle PTQ = \alpha $, so $ \angle PQT = 90 - \angle PTQ = 90 - \alpha $. Thus $ \angle PMO = \angle PQO = \angle PQT = 90 - \alpha $, so $ \angle PMO + \angle SMP = (90 - \alpha) + \alpha = 90 $ and we are done.

[quote=k2c901_1]$P,Q$ are two fixed points on a circle centered at $O$, and $M$ is an interior point of the circle that differs from $O$. $M,P,Q,O$ are concyclic. Prove that the bisector of $\angle PMQ$ is perpendicular to line $OM$.[/quote]

I would like to present the following solution which basically trivializes the entire problem. In my opinion it is very fast and elegant :D .

[b][u][i]Solution:[/i][/u][/b] Denote by $C_1$ the circle with centre $O$ and by $C_2$ the other circle .Drop the perpendicular from point $O$ to segment $\bar{PQ}$ and extend it to the point diametrically opposite $O$, naming them $A$ and $X$ respectively. Clearly, since $OA \bot AI$, it suffices to prove that $O,A,I,M$ are concyclic. Now for the best part, consider an inversion with respect to $C_1$ ($O$ is the centre of inversion), This it suffices to prove that $A^*,I^*,M^*$ are collinear but this clearly follows from the fact that both $OX^*$ and $OM^*$ are diameters and thus $\measuredangle{OI^*X^*}= \measuredangle{M^*I^*O}= 90^o $. Thus done  :D .

Sincerely,
Aayam

Dear Mathlinkers,

[url=http://jl.ayme.pagesperso-orange.fr/Docs/Quickies%207.pdf] here [/url]

Problem 7, p. 20

Sincerely
Jean-Louis


I am helping someone but have a hard time with this calculus problem:

A person is walking due north from point P at a rate of 8m/s. After 10 min a boy walks from point M (1000 ft away from P) towards point P  at a rate of 10 m/s. After 15 mintues how far will be boy and the person? If some can make me undersatnd using calculus I would appreciate it.

You can't really solve it unless you know where m is (you don't specify a direction)

the key to related rates problems is to find the relationship between two different quantities in terms of a third. in problems like this one, i would expect this relationship to be the pythagorean theorem


trigonometry
trig identities
Law of Cosines
geometry solved
geometry

consider a quadrilateral such that all the lengths of the sides and diagonals are integers. prove that one of the six integers is divisible by 3.

[i]Hungarian TST 2003[/i]

Don't expect me to give a solution or anything :D It's from the 2'nd Hungarian TST for the 43'rd IMO. This point (a).

Point (b) (more general):

We have n>=5 points in the plane s.t. all the distances between 2 of these pts are natural. Prove that at least 1/6 of all the distances are multiples of 3.

And, if I'm not wrong, it was first given in a bulgarian competitions. I also don't have a solution except the official one.

This is one of the problems and solutions impossible to forgot. I will reproduce the official solution, so please do not ask me why it is so:
  Let ABCD the quadrilateral. Lt us suppose that none of AB, BC, CD, DA, AC, BD is a multiple of 3. We can assume that BAD=BAC+CAD (as angles). Let BAC=x and CAD=y. Let alfa=2AB*AC*cosx, beta=2AD*Ac*cosy, gamma=2Ab*AD*cos(x+y). Using law of cosines in ABC, ACD and ABD, it follows that BC^2=AB^2+C^2-alfa, CD^2=AD^2+ AC^2-beta, BD^2=AB^2+AD^2-gamma. Since the square of an integer is 1 modulo3, it follows that alfa, beta and gamma are integers congruent to 1 modulo 3. Since 2AC^2*gamma=4AC^2*AB*AD*cos(x+y)=alfa*beta-4AC^2*AB*AD*sinx*siny, it follows that 4AC^2*AB*AD*sinx*siny is an in teger congruent to 2. So, sinx*siny is a rational number that written in lowest term has the numerator not a multiple of 3.
    Now let p=2AB*AC and q=2AD*AC. Thus, cosx=alfa/p and cosy=beta/q. Since sinxsiny=sqrt((p^2-alfa^2)(q^2-beta^2))/pq ia rationla, the numerator must be an integer. This numerator is a multiple of 3 (since 3|p^2-alfa^2), but the denominator is not a multiple of 3. Thus, when written in lowest terms, the numerator of sinx*siny is a multiple of 3, false.


analytic geometry
graphing lines
slope
geometry
algebra unsolved
algebra

1.
Write the equation of a line that passes through the indicated points: O(2,3) and P(-1,1)

2.
Find the equation of a line that passes through (-2,-3) and parallel to the line whose equation is 2x - 5y = 6.

3.
Given the points P(1,-1), Q(4,2), and R(6,4),
a. Find the slopes of line segments PQ and PR.
b. From your answer to part a, what can you conclude regarding the points P, Q, and R?
c. Find the equation of the line that passes through P and R.
d. Does the point S(10,8) lie on the line PR? Why or why not?

The underlined letters are supposed to be lined on top.

i dont think this problem belongs in the olympiad section. it is trivial and routine co-ordinate geometry that you would learn at school.


function
conics
parabola
quadratics
calculus
algebra
system of equations

Let $ f(x) \equal{} ax^2 \plus{} bx \plus{} c\ \left(a\neq 0,\ x > \minus{} \frac {b}{2a}\right)$. Suppose that $ f^{ \minus{} 1}(0) \equal{} \frac {4}{3},\ f^{ \minus{} 1}(2) \equal{} 2,\ f^{ \minus{} 1}(10) \equal{} 3$.

(1) Find the values of coefficients $ a,\ b,\ c$.

(2) Use the values of coefficients $ a,\ b,$ found in (1) and let the coefficient $ c$ vary, find the values of $ c$ such that $ f(x)$ touches to $ f^{ \minus{} 1}(x)$.

[hide="1"]From the 3 equations $ f^{-1}(0)=\frac 43$, $ f^{-1}(2)=2$, and $ f^{-1}(10)=3$, we get: $ f(\frac 43)=0$, $ f(2)=2$, and $ f(3)=10$, from which we get:$  \{\begin{array}{c} (\frac 43)^2a+(\frac 43)b+c = 0\\
    4a+2b+c = 2\\
    9a+3b+c = 10 \end{array}  \;$
We can solve this to get $ (a,b,c)=\boxed{(3,-7,4)}$
[/hide]
[hide="2"]
Since $ f(x)$ and $ f^{-1}(x)$ can only intersect on the line $ y=x$, they're common point of tangency must also be on that line. Therefore, we want a $ c$ such that $ y=x$ is tangent to $ f(x)$. This occurs when $ f(x)=3x^2-7x+c=x$, and $ f'(x)=6x-7=1$. So, $ x=\frac 43$ and $ c=\boxed{\frac{16}3}$.
[/hide]

[hide]
i'm getting something different for #2

like schnoggin said, the intersection must be on the line y=x
so now solve y=x and f(x) like any other system of equations
substitute y for x in the original equation
it has only one solution now, but the equation remains quadratic
therefore it is a double root
now use the discriminant = 0
$ b^2\minus{}4\cdot a\cdot c\equal{}49\minus{}4\cdot 3\cdot c\equal{}0$
$ c\equal{}\frac{49}{12}$

my answer looks too ugly

[/hide]

once u start taking calculus, its just really hard to solve some questions without it. anyway, you're method works, there's just a slight flaw. There should only be one solution to the equation $ f(x)\equal{}x\rightarrow 3x^2\minus{}8x\plus{}c\equal{}0$. So the discriminant is $ 64\minus{}4(3)(c)\equal{}0\rightarrow c\equal{}\boxed{\frac{16}3}$.

That's right. I would add one thing: You can use $ D/4$.


Is there a set of real nos u,v,w,x,y,z such that\[ \begin{array}{l}
 u^2  \plus{} v^2  \plus{} w^2  \plus{} 3(x^2  \plus{} y^2  \plus{} z^2 ) \equal{} 6, \\
 ux \plus{} vy \plus{} wz \equal{} 2 \\
 \end{array}
\]

[quote="anorithdialga"]Is there a set of real nos u,v,w,x,y,z such that
\[ \begin{array}{l} u^2 \plus{} v^2 \plus{} w^2 \plus{} 3(x^2 \plus{} y^2 \plus{} z^2 ) \equal{} 6, \\
ux \plus{} vy \plus{} wz \equal{} 2 \\
\end{array}
\]
[/quote]

[hide="solution"]
Suppose there exist such reals,

Let $ a\equal{}u^2\plus{}v^2\plus{}w^2,b\equal{}x^2\plus{}y^2\plus{}z^2$,and we see $ a\plus{}3b\equal{}6$.

By Cauchy-schwartz inequality we have

$ 4\equal{}(ux\plus{}vy\plus{}wz)^2\le (u^2\plus{}v^2\plus{}w^2)(x^2\plus{}y^2\plus{}z^2)\equal{}ab\equal{}(6\minus{}3b)b\equal{}3\minus{}3(b\minus{}1)^2 \le 3$

:Contradiction!! So it's done.

[/hide]


linear algebra
matrix
linear algebra unsolved

Good evening.

$A \in \mathcal{M}_n(\mathbb{C})$ Demonstrate the equivalence between
i)$A\bar{A}$ has positive eigenvalues
ii)There exists $U$ unitary and $T$ triangular superior such that ${A=UT^{t}U}$

[quote="Maxbl"]
(I am sorry but I didnt find a way to write tU where t denotes "transposition"  (I dont know the english term)).[/quote]

Hi! :)

I think this is what you're looking for : ^{t}U $^{t}U$.

It is known result which you can read for example in Horn.Matrix analysis in the chapter Hermitian and symmetric matrices(I'm not sure whether it is called exactly so in the original English version-i don't have this one).
If you won't find it and nobody solve it  i can post a solution from the book.

I found the book but had only a glimpse of the proof, it appears to be recursive on n.
I'll look at it better and post it later.

For Maxbl
Wich page of this book ?

---------------------------
Eugene could you post the solution from the book ?

Just as eugene said , in the chapter Hermitian and symmetric matrix, quite far in the chapter part 4 or something.
It is in a bookshop near my school but I had not enough time when I looked to fully understand it.
I'll look back in a few days.

Here we are:

ii)=>i) is quite evident it is only calculus.
 Now suppose $A \in \mathcal{M}_n(\mathbb{C})$ verifies $A\bar{A}$ has only positive eigenvalues.

Let $\lambda$ be an eigenvalue and $v$ an nonzero eigenvector for $A\bar{A}$.
Then either $A\bar{v}$ and $v$ are linked either not.
If they are and $A\bar{v}=\mu v$ then $A\bar{A}v=A\bar{\mu v}=|\mu|^2v$

If not then let's choose $\mu$ such than $|\mu|^2=\lambda$ and consider $y=A\bar{v}+\mu v$:
$A\bar{y}=\lambda v+\bar{\mu}A\bar{v}=\bar{\mu}y$ with $y$ non zero. Hence we can always find $v$ with $||v||=1$ such than $A\bar{v}=\mu v$ with $|\mu|^2=\lambda$
 By choosing a correct $e^{i\theta}$ and multiplying $v$ with it we can even afford that $A\bar{v}=\mu v$ with $\mu \in \mathbb{R}$.
Now consider an orthonormal basis $(v,v_2,...,v_n)$ of matrix $V_i$ and look to
$V_i^{*}A\bar{V_i}$. Its first colum has only its first term non zero and you only have to recur to finish.


Let $ a,b,c>0$ with $ abc\equal{}1$ and the numbers $ A\equal{}\frac{1}{1\plus{}a\plus{}b}\plus{}\frac{1}{1\plus{}b\plus{}c}\plus{}\frac{1}{1\plus{}c\plus{}a}$ and
$ B\equal{}\frac{1}{1\plus{}a\plus{}bc}\plus{}\frac{1}{1\plus{}b\plus{}ca}\plus{}\frac{1}{1\plus{}c\plus{}ab}$ .

Find which of the numbers $ A,B$ is bigger  :wink:

[quote="Dimitris X"]Nice.

[i]solution[/i]

We will prove that $ A \le 1 \equal{} B$.

For $ A$ setting $ a \equal{} x^3,b \equal{} y^3,c \equal{} z^3$ we take $ xyz \equal{} 1$ and
$ A \equal{} \frac {1}{x^3 \plus{} y^3 \plus{} 1} \plus{} \frac {1}{y^3 \plus{} z^3 \plus{} 1} \plus{} \frac {1}{z^3 \plus{} x^3 \plus{} 1} \le \frac {1}{xy(x \plus{} y) \plus{} xyz} \plus{} \frac {1}{yz(y \plus{} z) \plus{} xyz} \plus{} \frac {1}{zx(z \plus{} x) \plus{} xyz} \equal{} \frac {x}{x \plus{} y \plus{} z} \plus{} \frac {y}{x \plus{} y \plus{} z} \plus{} \frac {z}{x \plus{} y \plus{} z} \equal{} 1$.

So $ A\le1$.

In $ B$ setting $ a \equal{} \frac {x}{y},b \equal{} \frac {y}{z},c \equal{} \frac {z}{x}$ we take easily enough that $ B \equal{} 1$.

Done.[/quote]
Wrong solution !!

Clearly $ B \neq 1$ ( and $ B \leq 1$ is also obvious from AM-GM's inequality )

BTW,I've already proved that $ \sum \frac{1}{a\plus{}2} \geq \sum \frac{1}{a\plus{}bc\plus{}1}$ by AM-GM,I'll try to find later a nice ( AM-GM's ) proof for the stronger ( $ A \geq B$ )  :)

Oooops sorry for my wrong solution.

I was under the impression that $ B\equal{}\sum\frac{1}{ab\plus{}a\plus{}1}$. :blush:

[url=http://www.artofproblemsolving.com/community/c6h301938p1634675]Bulgaria 1997:[/url]
Let $ a,b,c$ be positive numbers having product $ 1.$ Prove that
$ \frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \le \frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}.$


[color=darkred]In figure, two cicles $(O)$ and $(O')$ are external tangent. $OH\perp MN,\,HT=HP$ and $HT$ is a tangent of circle $(O)$

Prove that: $\angle MPN=\angle HOT$[/color]

It's a very nice problem, also $\overline{MP}\parallel\overline{O'O}$

[color=red]Please post the complete solution! [/color]

And I don't think $MP\parallel OO'$  :wink:

$[HP] \cap C(O)=S$. Let $K$ be on $[HN]$ such that $HK=OT$. Since $\triangle HTO \equiv \triangle PHK$ it follows that $\angle HOT = \angle PKH$. It's sufficient to prove that $\triangle KMP \sim \triangle PMN$. Denote $r_{O}_'$ $=r$, $r_{O}=R$, $MH=x$ and $HS=y$. $(r+x)^{2}+(y+R)^{2}=(r+R)^{2}$ $(1)$.
$\triangle KMP \sim \triangle PMN$ $\Leftrightarrow$ $MP^{2}=MK \cdot MN$ $\Leftrightarrow$ $x^{2}+y \cdot (y+2R)=(R-x) \cdot 2r$ $\Leftrightarrow$ $(1)$.

[color=darkblue][b][u]Thank you,[/color] [color=darkred]April[/color][color=darkblue], for your nice proposed problem ! Thank you,[/color] [color=darkred]SonicX[/color][color=darkblue], for your nice proof ![/u][/b]

[b][u]Remark.[/u][/b] Denote the second intersection $R$ between the line $PM$ and the circle $(O')$. Then the quadrilateral $PNRH$ is cyclically and $R\in TH$ ![/color]

[quote="April"][color=darkred]In figure, two cicles $C(O,R)$ and $C(O',r)$ are external tangent. $OH\perp MN,\,HT=HP$ and $HT$ is a tangent of circle $(O)$ Prove that $\widehat{MPN}=\widehat{HOT}$.[/color][/quote] [color=darkblue][b][u]Proof.[/u][/b] Denote $HP=HT=y$, $HM=x$. Observe that $O'O^{2}=O'H^{2}+OH^{2}=O'H^{2}+HT^{2}+OT^{2}$ $\implies$ $(R+r)^{2}=(r+x)^{2}+y^{2}+R^{2}$ $\implies$ $\boxed{2Rr=x^{2}+y^{2}+2rx}\ \ (*)$. Denote $m(\widehat{MPN})=u$, $m(\widehat{HOT})=v$, $m(\widehat{MPH})=w$. Therefore, $\tan (u+w)=\frac{HN}{HP}=\frac{2r+x}{y}$, $\tan w=\frac{MH}{HP}=\frac{x}{y}$ and $\tan v=\frac{HT}{OT}=\frac{y}{R}$. Must prove $u=v$, i.e. $\tan [(u+w)-w]=\tan v$ $\Longleftrightarrow$ $\frac{\tan (u+w)-\tan w}{1+\tan (u+w)\tan w}=\tan v$ $\Longleftrightarrow$ $\frac{\frac{2r+x}{y}-\frac{x}{y}}{1+\frac{2r+x}{y}\cdot\frac{x}{y}}=\frac{y}{R}$ $\Longleftrightarrow$ $(*)$.[/color]

[color=indigo]Thanks for all of your nice solutions and thank you for your nice [b]Remark,[/b] [b]Virgil Nicula![/b]
And now, I'll present my solution (like [b]Virgil Nicula's[/b])

[u][b]Proof.[/b][/u] We have:
\[\tan\widehat{MPN}=\tan\left(\widehat{HPN}-\widehat{HPM}\right)=\frac{\tan\widehat{HPN}-\tan\widehat{HPM}}{1+\tan\widehat{HPN}\tan\widehat{HPM}}\\ =\frac{\frac{HN}{HP}-\frac{HM}{HP}}{1+\frac{HN}{HP}\cdot\frac{HM}{HP}}=\frac{MN\cdot HP}{HP^{2}+HM\cdot HN}=\frac{2R'\cdot HP}{HO^{2}-R^{2}+HO'^{2}-R'^{2}}\\ =\frac{2R'\cdot HT}{OO'^{2}-R^{2}-R'^{2}}=\frac{2R'\cdot HT}{2RR'}=\frac{HT}{OT}=\tan\widehat{HOT}\]
Hence: $\widehat{MPN}=\widehat{HOT}$[/color]

I'm not sure if Nicula's statements are correct...?
[img]8726[/img]


A 1001-digit number begins with 6. The number formed by any two adjacent digits is divisible by 17 or 23. Write down the last six digits.

692346.

Multiples of 17: 17,34,51,68,85
Multiples of 23: 23,46,69,92

As 7 doesnt occur in tens place, so numbers with 7 can't occur => numbers with 1 => numbers with 5 => numbers with 8 cant occur
The final numbers are: 69,92,23,34,46

The number 69234 repeats itself


conics trigonometry geometry rhombus HMMT email AMC

Post your results here. I understand the Hopkins (Varsity) is giving Choate a run for its money this year...impressive.

66363
60336
36613

Apparently, the juniors this year are really crappy, we have the top junior I think and most of the sophomores are better than the juniors, and we have someone who's tied for the top senior (all perfect scores) I think the Hopkins two freshmen are the only ones in top at all.

I think the first two meets, Hopkins lost one by 2 points, and one by 3 points.

Now from what I hear, Choate brings like, people who've already been through college from other countries, and just want a high school diploma so they're going from one year which IMO is a bit unfair, considering they Choate scored a 104/105 on the most recent meet which was the highest score ever in GNHML.

[quote="Phelpedo"]Actually, I don't think that's true (the part about Choate). As far as I can see, all their best members have been there all 4 years, or  whatever year they are. I'm referring to Rosen, Cao, Watanaprakornkul, Choi, Rajagopalan, etc.[/quote]

Maybe it's just a joke I guess, I know for a fact that they do offer a program like that though.

You're probably thinking about a post-graduate program (my school has one too), which does fit your description, but most all PGs are for sports, not math...except maybe for exeter.

Well, our coach says this is probably Hopkins' 2nd best team ever.

[quote="jhredsox"]what is the GNHML?
-jorian[/quote]

Greater New Haven Mathematics League

[quote="jhredsox"]ahhhh!, i live in glastonbury

do u know of any state wides? or any hartford ones?
-jorian[/quote]

Umm.. there's one in Hartford and there's the state-wide CSAML.

[quote="jhredsox"]do most high schools participate

if so, i can't wait!
-jorian[/quote]

I don't live in Hartford so I wouldn't know. Around here I think there's about 15 that participate.

66363
60336
36614
66644

Hopkins tied with Hamden last week which is pretty sad. Do individuals count your top 4 or top 5 scores? As if it's top 5 I'm screwed (due to not being put on rounds I wanted 1st 2 meets and careless errors) but if it's top 4, I still have a chance as the coach is listening to me now after seeing the 1st two meets and the fact that I've worked on my careless errors.

For individuals, it's top 4 scores.

Choate has won the last 2 contests by 101-90 and 92-78.

66643
61463
66643
66666

I'm trying to give Choate's sophomores and freshmen a run for their money this year.

Only 3 of the numbers are counted...so what's your official score for each round?

Actually, the thing about Choate is true, so far, two members of their team have all perfect scores. The next person has a score of 59. The two members are Sarita Bunsupha and Phumpong Watanaprakornkul. The former is a 2006 IMO Bronze Medalist and the latter a 2006 IMO Gold Medalist both from Thailand.

Hm, well this makes me feel better that we are beating Choate at the neml .

Yifan, how did you do?

The two PGs from choate are in their Thai Scholars program. Two students come every year.

YAY FOR PHUMPONG WATANAKARUNAPDFGADPFOGVKADL;FVKRE

33636
60336
36614
66644
632
666

Strangely, even Choate's Thai scholars didn't do very well on the last one. Must have been hard..,

No, our coach was making fun of them. He was like IMO Medalists can't even figure out conic sections. Cause they both failed the Conic Sections round last meet and probably failed the Trig round.

Oh and they don't know what a rhombus is either.kn

Uhh, what's the point of the last post? Yeah, they sometimes struggle in rounds with a lot of english terms, maybe b/c they're not american? Just putting it out there.

I don't think they did too badly on the trig round. I mean, our A team did get 18/18 for that round, but who knows.

Btw, phelpedo, how do you get such accurate info on GNHML? I'm just curious, because it's pretty impressive, and I don't even know some of the stuff you find out.

I keep an eye out  :wink:

http://studentweb.choate.edu/mathteam/menu.htm

yifan why was everybody laughing when you won

Final scores:
66643
61463
66643
66666
36663
6-560

If I just hadn't screwed up Round 3 #1 I would have tied for third!

Oh well, first for underclassmen is enough. Hopkins did pretty well  in terms of awards. Teamwise, the things that killed us were Choate's Thai Scholars and the lack of anyone good who could do the later rounds (Genevieve didn't do so well for the most part).

New year!

5-[b]666[/b]: 18

...unless the people decide to take 2 points for me not putting a dollar sign on 3/2 (the entire problem was talking in dollars, but apparently I still needed to put it in the answer after "13"... sigh).

Edit: So apparently they announced you had to put units... but I wasn't there because I was challenging 1/3... just so stupid.

Too bad Gardner's gone.

5-[b]664[/b]: 16
3[b]66[/b]-[b]4[/b]: 16

Because of course 2 isn't an integer between 0 and 5.
Probably not gonna 72 this thing this year. Oh well. I still have two more years.

Tied Choate this meet. Looks like we might give you guys a run for your money this year, Yifan. Look out :D

I didn't particularly like these problems. 3/3 I essentially got by guesswork - I saw 30 and thought 5-12-13.

Edit: This forum is dead. Yuck.

As an interesting anecdote, look up the definition of "inchoate."

Third meet!

5-[b]664[/b]: 16
3[b]66[/b]-[b]4[/b]: 16
66[b]363[/b]: 12

Yuck. Oh well, we beat Choate (72-71, now down 2 on the year). I'm happy. It'll drop anyway. Maybe I'll be able to drop one of those 16s...

Next meet December 13 at Hillhouse. BTW, I started a [url=http://hs.facebook.com/group.php?gid=2376655891]facebook group[/url] if anyone wants to join... not that anyone ever reads this.

Lessee.. If we can get Jon, Yifan, and Phelpedo posting again... and you...

That's 5 I guess.

Anyway, back on topic, Yifan needs to post his scores.

[quote="Phelpedo"]So are any of you guys coming up for HMMT?[/quote]
Wish Hopkins was... sadly, it's hard to get the administration moving on much of anything.

Edit: Meet 5!

5-[b]664[/b]: 16
3[b]66[/b]-[b]4[/b]: 16
66[b]363[/b]: 12
--[b]615[/b]: 12
5-[b]666[/b]: 18

Yay!

Choate is 9 points ahead going into the final meet... not going to happen this year. Sadly.

Choate: 393
Hopkins: 384

SO I GOT AN EMAIL THAT SAID I GOT A PM AND I REMEMBERED THIS FORUM
uh i dunno my exact scores
BUT MY CULMULATIVE SCORE IS THE SAME AS WAWB'S
best 4 are something stupid like 1 lower or some **|* like that
I HAVE 2 13S AND 1 18 AS WELL AS A 15!!!!!!!!!!!!!!!!!!! A 13 WAS DROPPED!!!!!!!!!

so itd look something like this 16+16+12+12+18=74-13-13-18-15=15 so my scores are 13 13 15 15 18.

CHOATE CHEATS.

wawb: will you do itest comp sci with me?

PHELPEDO: HAVE YOU SEEN HOTEL RWANDA?

HEY GUYS NEW YEAR

6[b]566[/b]6

Hopkins > Choate this year I think.

did Rajopolan make USAMO?? or any of em?

[quote="academicsrule"]I recently basically got kicked off Choate A for no reason.... well, maybe it was b/c I thought 72/2=9. But still. At least there will be rotation.
Then for this meet I phailed by doing round five with no memory of absolute value inequalities.
If I'd done ANY other round I would have had 18.
instead, 12. ..... :blush:
Aren't Choate and Hopkins tied for this meet???
Did Hopkins get a perfect team round too?
By the way, I think Stephanie Choi, Andrew? Kim, and the Thai scholar made USAMO in 2008. New Thai scholars this year are awesome ppl![/quote]

Stephanie Choi = Most annoying person ever

How did she do today?


A point in space $ (x,y,z)$ is randomly selected so that $ \minus{}1\le x \le 1$,$ \minus{}1\le y \le 1$,$ \minus{}1\le z \le 1$.  What is the probability that $ x^2\plus{}y^2\plus{}z^2\le 1$?

Because $ x^2\plus{}y^2\plus{}z^2 \le 1$, the point must lie in the unit sphere centered at $ (0,0,0)$.  The volume of this is $ \frac{4}{3}\pi$.  The total volume of the cube is $ 2^3\equal{}8$, so the probability is $ \frac{\frac{4}{3}\pi}{8}\equal{}\boxed{\frac{\pi}{6}}$.

It is similar to how a unit circle centered at the origin is $ x^2\plus{}y^2\equal{}1$.

[quote=cw357]Two comments:

It's nice the way they tie in some simple geometry concepts to a three dimensional graph.

But why, why, why, was this a level 25 problem?  Actually, the fact that the problem requires geometry and this is C&P might account for it, but still...[/quote]

ik its ez

because it's a little hard for some people to notice that the region we want (x,y,z) in is a sphere

Yeah, geometric probability can be a hard one sometimes.

I once got the 3 people going to meet in the restaurant from 2 to 4 for a meeting and i [size=50]almost[/size] killed myself.

Yes visualizing 3d for that problem is hard [s]until you decide to use integrals to overkill the problem[/s]

The fact that this is level 25 shows that many people taking CP don't know the area of a sphere formula.

How come I never see any prealgebra questions in these alcumus problems? All of these are like, geometry or highschool stuff.

How do integrals help except if you forgot formula for volume of sphere?

[hide=solution]The region that will satisfy $ x^2+y^2+z^2 \le 1,$ is a unit sphere centered at $(0, 0, 0),$ so the volume is $\frac{4}{3}\pi.$ The possible region is a cube with side length $2,$ so it has volume $8,$ and the desired answer is $ \frac{\frac{4}{3}\pi}{8}=\boxed{\frac{\pi}{6}}$. [/hide]

why is this #25??? bro alcumus puts AIME #15s and AMC 10 #1's in the same category

I guess it has to do with the skill level of the people doing a certain subject. AIME #15's would be more likely to be in IA or Precalc, which are the intermediate topics (and very hard ngl).

[quote=judgefan99]This is a really nice problem :D[/quote]

Really had to think for this one


calculus integration trigonometry function limit logarithms derivative

(1) Find the minimum value of $ \int_{ \minus{} \pi}^{\pi} (x \minus{} a\sin x)^2dx$

(2) Evalute $ \int_{ \minus{} 1}^{1} |e^{x} \minus{} 1|dx$

(3) Evaluate $ \int_{ \minus{} \pi}^{\frac {\pi}{2}} |\sin x|dx$

(4) Find the continuous function $ f(x)$ satisfying $ f(x) \equal{} x \plus{} \int_0^{\pi} f(t)\sin t\ dt$

(5) Find $ \frac {d}{dx}\int_0^{x} (x \minus{} t)e^{t}dt$

(6) $ \lim_{x\rightarrow 0} \int_0^{x} \frac {(1 \plus{} \sin t)^2}{x}dt$

(7) Evaluate $ \int_0^1 xe^{x}dx$

(8) Evaluate $ \int_1^{e} x\ln x\ dx$

(9) Evaluate $ \int_0^{\frac {\pi}{2}} |\cos 2x|dx$

(10) Evaluate $ \int_0^{\pi} |\sin x \plus{} \cos x|dx$

[hide="7)"]$ \int_{0}^{1}xe^{x}dx  \\ \text{Using Integration by parts,we get} \\ (x\minus{}1)e^x |_{0}^{1} \\ \equal{}1$[/hide]
[hide="8)"]$ \int_{1}^{e}x\ln xd\ x \\ \text{Using integration by parts,we get} \\ \frac{x^2}{4}(2\ln x\minus{}1)|_{1}^{e} \\ \equal{}\frac{e^2\plus{}1}{4}$[/hide]

5) $ \int_{0}^{x}xe^{t}\minus{}te^{t}\,dt\equal{}e^{t}(x\minus{}t\plus{}1)\bigg|_{0}^{x}\equal{}e^{x}\minus{}x\minus{}1$.  Taking the derivative yields $ e^{x}\minus{}1$ (assuming I didn't make a careless arithmetic mistake while doing the integral).

[hide="9)"]$ \int_{0}^{\frac{\pi}{2}}|\cos 2x|dx$

$ \cos 2x$ is positive in the interval $ [0,\frac{\pi}{4}]$ and negative in $ (\frac{\pi}{4},\frac{\pi}{2}]$

Hence the integral can be written as $ \int_{0}^{\frac{\pi}{4}}\cos 2xdx- \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos 2x dx$

$ \Longrightarrow \frac{1}{2} (\sin 2x|_0^{\frac{\pi}{4}} - \sin 2x|_{\frac{\pi}{4}}^{\frac{\pi}{2}} ) \\ =\frac{1}{2} (1+1) \\ =1$

Therefore $ \int_{0}^{\frac{\pi}{2}}|\cos 2x|dx =1$[/hide]

[hide="My doubt"]Does a definite integral denote the area under the given curve in the given interval ?(just got confused)[/hide]

I can't find so far any mistake in your solutions. :)

@picture74  A definite integral doesn't always denote the area under the given curve.

For (5), we can use Fundamental Thorem of Calculus.

For (9), we can remark the periodicity for the integral of Triangular function with absolute value, in this case set $ 2x \equal{} t$.

[hide="10)"]$ \int_{0}^{\pi}|\sin x\plus{}\cos x|dx \\ \equal{}\int_{0}^{\pi} |\sin \left(x\plus{}\frac{\pi}{4}\right)|dx$

$ \sin \left(x\plus{}\frac{\pi}{4}\right)$ is positive in $ \left[0,\frac{3\pi}{4}\right]$ and negative in $ \left(\frac{3\pi}{4},\pi\right]$

Hence,the given integral can be written as $ \int_{0}^{\frac{3\pi}{4}}\sin \left(x\plus{}\frac{\pi}{4}\right)dx \minus{} \int_{\frac{3\pi}{4}}^{\pi}\sin \left(x\plus{}\frac{\pi}{4}\right)dx$

Solving the above integral yields $ 1\plus{}\sqrt{2}$

Hence $ \int_{0}^{\pi}|\sin x\plus{}\cos x|dx\equal{}1\plus{}\sqrt{2}$[/hide]

[quote="Kunny"]@picture74 A definite integral doesn't always denote the area under the given curve.[/quote]
Other than area,what else does it represent?

[quote="picture74"]
[quote="Kunny"]@picture74 A definite integral doesn't always denote the area under the given curve.[/quote]
Other than area,what else does it represent?[/quote]

Sometimes denotes the volume, the length of arc, e.t.c.

Your answer to (10) is incorrect.

[quote="kunny"]Sometimes denotes the volume, the length of arc, e.t.c.[/quote]
Thank you  :)
[quote="kunny"]Your answer to (10) is incorrect.[/quote]
I couldn't figure out the mistake in my solution,Could you locate it?

$ a\sin x \plus{} b\cos x \equal{} \sqrt {a^2 \plus{} b^2}\sin (x \plus{} \alpha)$ for a fixed angle $ \alpha$.

@picture74
[quote]$ \int_{0}^{\pi}|\sin x\plus{}\cos x|dx \\ \equal{}\int_{0}^{\pi} |\sin \left(x\plus{}\frac{\pi}{4}\right)|dx$[/quote]
That's the trouble. Actually,
$ |\sin x \plus{}\cos x|\equal{}\sqrt{2}|\sin\left(x\plus{}\dfrac{\pi}{4}\right)|$
Further, the final step is incorrect. The integral in the final step will add up to 2.
As such the required answer is $ 2\sqrt{2}$

[quote="kaymant"]@picture74
[quote]$ \int_{0}^{\pi}|\sin x \plus{} \cos x|dx \\
\equal{} \int_{0}^{\pi} |\sin \left(x \plus{} \frac {\pi}{4}\right)|dx$[/quote]
That's the trouble. Actually,
$ |\sin x \plus{} \cos x| \equal{} \sqrt {2}|\sin\left(x \plus{} \dfrac{\pi}{4}\right)|$
Further, the final step is incorrect. The integral in the final step will add up to 2.
As such the required answer is $ 2\sqrt {2}$[/quote]

That's right.

$ \int_0^{\pi} |\sin x \plus{} \cos x|dx \equal{} \sqrt {2}\int_0^{\pi} \sqrt {2}\left|\sin \left(x \plus{} \frac {\pi}{4}\right)\right|dx$

$ \equal{} \sqrt {2}\int_{\frac {\pi}{4}}^{\pi \plus{} \frac {\pi}{4}} |\sin x|dx$

$ \equal{} \sqrt {2}\int_0^{\pi} |\sin x|dx \equal{} 2\sqrt {2}.$

Thank you kunny and kaymant,for explaining my mistake!  :)

[hide="#2"]
\[ \varphi \int_{ \minus{} 1}^{1}|e^x \minus{} 1|\text{ d}x\]
$ f(x) \equal{} e^x \minus{} 1$ has a zero at $ x \equal{} 0$ & $ f^{\prime}(0) > 0$. Therefore we compute
\[ \varphi \equal{} \int^1_0 (e^x \minus{} 1)\text{ d}x \minus{} \int^0_{ \minus{} 1} (e^x \minus{} 1)\text{ d}x\]
This has a value of
\[ \boxed{\varphi \equal{} e \plus{} \frac {1}{e} \minus{} 2}\]
[/hide]
[hide="#4"]
\[ f(x) \equal{} x \plus{} \int_{0}^{\pi}f(t)\sin{t}\text{ d}t\]
Assuming $ f(x) \equal{} x \plus{} \delta$ because $ f'(x) \equal{} 1$, I'll evaluate the definite integral
\[ \int^{\pi}_0 f(t)\sin{t}\text{ d}t \equal{} \int^{\pi}_0 (t\sin{t} \plus{} \delta \sin{t})\text{ d}t \equal{} \minus{} t\cos{t} \plus{} \sin{t} \minus{} \delta \cos{t} \plus{} C \equal{} \pi \plus{} 2\delta\]
Now,
\[ x \plus{} \delta \equal{} f(x) \equal{} x \plus{} \pi \plus{} 2\delta \Longrightarrow \delta \equal{} \minus{} \pi\]

\[ \Leftrightarrow \boxed{f(x) \equal{} x \minus{} \pi}\]
This is my favorite problem here. I don't know how to solve it without that assumption, but I'd love to see it.
[/hide]
[hide="#6"]
\[ \varphi \equal{} \lim_{x\rightarrow 0}\int_{0}^{x}\frac {(1 \plus{} \sin t)^{2}}{x}dt\]
Expand, substitute & integrate to find the indefinite representation
\[ \varphi \equal{} \frac {6t \minus{} 8\cos{t} \minus{} \sin{2t}}{4x} \plus{} C\]
before evaluation. Plug in $ t \equal{} x,0$, but hold the limit on $ x$ to obtain
\[ \varphi \equal{} \frac {8 \plus{} 6x \minus{} 8\cos{x} \minus{} \sin{2x}}{4x}\]
Apply L'Hopital:
\[ \lim_{x\to 0}\frac {3 \plus{} 4\sin{x} \minus{} \cos{x}}{2} \equal{} \boxed{1}\]
[/hide]

Those answers are correct. :lol:

Are there someone who solve the problem #6 without using l'Hopital?

6) Without L'Hospital:
Since $ x$ is constant as far as the integration is concerned, we have
$ \int_0^x (1\plus{}\sin t)^2\ \mathrm dt\equal{}\dfrac{1}{2}(4\plus{}3x\minus{}4\cos x\minus{}\cos x\sin x)$
Hence, the required limit is
$ \lim_{x\to0}\dfrac{4\plus{}3x\minus{}4\cos x\minus{}\cos x\sin x}{2x}$
$ \equal{}\lim_{x\to 0}\left(\dfrac{2(1\minus{}\cos x)}{x}\plus{}\dfrac{3}{2}\minus{}\dfrac{\cos x\sin x}{2x}\right)$
Now, note that
$ \lim_{x\to0}\dfrac{1\minus{}\cos x}{x}\equal{}\lim_{x\to0}\dfrac{2\sin^2(x/2)}{x}\equal{}2\lim_{x\to0} \dfrac{x}{4}\cdot \left(\dfrac{\sin(x/2)}{(x/2)}\right)^2$
$ \equal{}\dfrac{1}{2}\lim_{x\to0}x\cdot \lim_{x\to0}\left(\dfrac{\sin(x/2)}{(x/2)}\right)^2\equal{}\dfrac{1}{2}\cdot0\cdot1\equal{}0$
And $ \lim_{x\to0}\dfrac{\cos x\sin x}{2x}\equal{}\dfrac{1}{2}\lim_{x\to0}\cos x \cdot \lim_{x\to0}\dfrac{\sin x}{x}\equal{}\dfrac{1}{2}\cdot1\cdot1\equal{}\dfrac{1}{2}$
As such
$ \lim_{x\to 0}\left(\dfrac{2(1\minus{}\cos x)}{x}\plus{}\dfrac{3}{2}\minus{}\dfrac{\cos x\sin x}{2x}\right)$
$ \equal{}2\lim_{x\to0}\dfrac{1\minus{}\cos x}{x}\plus{}\lim_{x\to0}\dfrac{3}{2}\minus{}\lim_{x\to0}\dfrac{\cos x\sin x}{2x}$
$ \equal{}0\plus{}\dfrac{3}{2}\minus{}\dfrac{1}{2}\equal{}\boxed{1}$

Jus missed my chance , anyways thank you  sir .

hello for 3) we get $ \int_{\minus{}\pi}^{\frac{\pi}{2}}|\sin(x)|\,dx\equal{}\int_{\minus{}\pi}^{0}\minus{}\sin(x)\,dx\plus{}\int_{0}^{\frac{\pi}{2}}\sin(x)\,dx\equal{}3$.
Sonnhard.

Anyway use  of L'Hospital really simplifies it.
$ \lim_{x\to0}\dfrac{\int_0^x(1\plus{}\sin t)^2\ \mathrm dt}{x}$
$ \lim_{x\to0}\left(\int_0^x(1\plus{}\sin t)^2\  \mathrm dt\right)^\prime\equal{}\lim_{x\to0}(1\plus{}\sin x)^2\equal{}1$

Let $ \int (1+\sin x)^2dx=F(x)+C$,

$ \lim_{x\rightarrow 0} \frac{\int_0^{x}(1+\sin t)^2dt}{x}$

$ =\lim_{x\rightarrow 0} \frac{[F(t)]_0^{x}}{x}$

$ =\lim_{x\rightarrow 0} \frac{F(x)-F(0)}{x}$

$ =F'(0)$ $ \because \text{The definition of differential coefficient}$

$ =1$. $ \because F'(x)=(1+\sin x)^2$


function
limit
real analysis
real analysis unsolved

if $ (X, \Sigma , \mu)$ is a finite measure space and $ f_n$ is an arbitrary sequence of real val $ \Sigma$-measurable functions on X.

show for every $ \epsilon > 0$
there exists $ E \in \Sigma$ with $ \mu (E) < \epsilon$
and a sequence $ \{a_n\}$ of positive reals with $ \lim_{n \to \infty} a_n f_n (x) \equal{} 0$ for every $ x \in E^c$

taking a random stab at this (since i have to take a test on monday)

maybe for every positive number M, and positive int n we can define

$ Q_{n,M} \equal{} \{E: |f_n|>M\}$ so we see $ Q^c_{n,M} \to X$ as $ M \to \infty$

so $ \mu(Q_{n,M}) \to 0$ as $ M \to \infty$

so fix $ \epsilon > 0$ then for each n, there exists $ M_n$ with $ \mu(Q_{n,M_n})  < \frac{\epsilon}{2^n}$

letting $ E\equal{}\cup_{n \in N} Q_{n,M_n}$ we have $ \mu (E) < \epsilon$

and hopefully... we can define $ a_n \equal{} \frac{1}{nM_n}$ so we get

$ |\lim_{n \to \infty} a_nf_n(x)| \le \lim_{n \to \infty} \frac{|f_n(x)|}{nM_n} \le \lim_{n \to \infty} n^{\minus{}1} \equal{}0$

legit?


[quote="Melissa"]If $x^2=x+3$ then $x^5$ is equal to?[/quote]
Take x= sqrt(x+3) then take it to the fifth?

[hide="Hint"]$x^5=x^2\times{x^2}\times{x}$
[hide="Further hint if you are stuck"]$x^5=(x+3)(x+3)x$[/hide][/hide]

[quote="frt"][quote="236factorial"]Melissa's second hint gave it away.

[hide]answer:

$x^3+6x^2+9x$. [/hide][/quote]
Well, you can go further.
[hide]
$x^3+6x^2+9x$
$=x^2(x+6)+9x$
$=(x+3)(x+6)+9x$
$=x^2+18x+18$
$=(x+3)+18x+18$
$=19x+21$ etc...[/hide][/quote]
I think that there's no need for an et cetera  ;) $\gcd (19,21)=1$.

IntrepidMath, please read the previous posts. We've already got the answer. Also, you forgot to multiply the expression by $x$. (Then that gives the same solution as 236factorial's, which can be simplified further.)


function
topology
advanced fields
advanced fields theorems

Just to remind myself... :

Let $ X,Y$ be topological spaces, $ A,B\subseteq X$. Is it true that if
$ f: A\rightarrow Y$, $ g: B\rightarrow Y$ are continuous functions such that $ f\equal{}g$ on $ A\cap B$ then $ h$ defined as $ h\equal{}f$ on $ A$ and $ h\equal{}g$ on $ B$ is continuous on $ A\cup B$?

Ok, and when we have $ X: \equal{} X\times [0,1]$ and $ A \equal{} X\times [0,b]$, $ B \equal{} X\times [b,1]$ then it is true ?

It's also sufficient that $ A$ and $ B$ are both closed.

The proof, because it's a bit tricky:
It suffices to show that a set $ U$ which has $ U_{A}\equal{}U\cap A$ and $ U_{B}\equal{}U\cap B$ both relatively open is itself open. Let $ A'$ be the complement of $ A$ and $ B'$ be the complement of $ B$ in $ X$; these are both open.

By hypothesis, $ U_{A}\equal{}V_{A}\cap A$ and $ U_{B}\equal{}V_{B}\cap B$ for some open sets $ V_{A},V_{B}$ in $ X$. We now write $ U$ as the union of three open sets:
$ U\equal{}(V_{A}\cap V_{B})\cup (V_{A}\cap B')\cup (V_{B}\cap A')$
The first set is contained in $ U$, and contains $ U\cap A\cap B$. The second is exactly the part of $ U$ in $ A$ but not $ B$, and the third is the part of $ U$ in $ B$ but not $ A$.

I think it's correct (assuming $ A\cup B\equal{}X$).

Yes, I did assume that. We obviously can't get any continuity information outside the union of $ A$ and $ B$, so I restricted to that union.

[quote="-oo-"]no. but it is sufficient that $ A$ and $ B$ are open.[/quote]

I agree.

To let a function to be continuous, is that it maps every open set to another open set and every open set in Y is mapped from an open set in X.

When $ A$ and $ B$ are open,
for an open set $ S\subseteq A\cup B$,
we can find two open set $ S_{A}$ and $ S_{B}$ that are open in $ A$ and $ B$ ($ S_{A}= S\cap A$ and $ S_{B}= S\cap B$), where $ S_{A}\cup S_{B}= S$
and then $ f[S] = f[S_{A}\cup S_{B}] = f[S_{A}]\cup f[S_{B}]$.
and $ f[S_{A}]\cup f[S_{B}]$ is known to be open.

the opposite is similar...

[quote="darken"]To let a function to be continuous, is that it maps every open set to another open set ...[/quote]

I don't think so  :)

[quote="darken"]It is ONE of the DEFINITION of continuous mapping...

Otherwise, how do you define "continuous" in topology (especially discreet topology.)[/quote]

$ f: X\to\mathbb R$ with $ f(x) \equal{} 1$ is continuous but $ f(X) \equal{}\{1\}$ is not open in $ \mathbb R$. Many counterexample can be found easily  :wink:

A function is continuous if the [i]preimage[/i] of every open set is open, i.e. if $ f^{\minus{}1}$ takes open sets to open sets.


inequalities
induction
inequalities unsolved

EDIT: Sorry about that [hide="Another proof"]If $ x\equal{}1\vee x\equal{}0$ the inequality is true. Assume $ 0\neq x\neq 1$ .

Denote : $ f(x) \equal{} 1 \minus{} (1 \minus{} x)^m$. The inequality is then equivalent to $ f(x)^{n}\geq f(x^n)$. I will show using induction:

$ (1 \minus{} (1 \minus{} x)^m)^2 \geq 1 \minus{} (1 \minus{} x^2)^m \Rightarrow \frac {(1 \minus{} x^2)^m \plus{} ((1 \minus{} x)^2)^m}{2} \geq (1 \minus{} x)^m \Rightarrow \frac {(1 \plus{} x)^m \plus{} (1 \minus{} x)^m}{2} \geq 1$ , true by Power mean (for every $ m \geq 1$).
Assume that it is true for some integer $ n\equal{}k>1$ , for $ n\equal{}k\plus{}1$ :

$ f(x)^{k \plus{} 1} \equal{} f(x)f(x)^k \geq f(x)f(x^{k})$

it'll be enough to prove that $ f(x)f(x^k)\geq f(x^{k \plus{} 1})$ , equivalent to:

$ (1 \minus{} x^{k \plus{} 1})^m \plus{} (1 \minus{} x \minus{} x^k \plus{} x^{k \plus{} 1})^m \geq (1 \minus{} x^k)^m \plus{} (1 \minus{} x)^m$

, true by Karamata .

So it is true for every $ m \geq 1$ , and natural $ n$.[/hide] May be is correct now .

Also probalistic interpretation [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=245389&search_id=282459874]here[/url]

[quote="alex2008"][hide="Proof"]Denote $ f(x) \equal{} (1 \minus{} x^n)^m \plus{} (1 \minus{} (1 \minus{} x)^m)^n$, $ \forall x \in [0,1]$.

Since $ f(0) \equal{} 2$ and $ f(1) \equal{} 1$ to prove $ f(x) \ge 1$ in $ [0,1]$, we just need to show that $ f$ decreases in the interval.

$ f'(x) \equal{} \minus{} n \cdot m \cdot x^{n \minus{} 1} \cdot (1 \minus{} x^n)^{m \minus{} 1} \plus{} n \cdot m \cdot (1 \minus{} x)^{m \minus{} 1} \cdot (1 \minus{} (1 \minus{} x)^m)^{n \minus{} 1} < 0$
$ \iff \left(\frac {1 \minus{} (1 \minus{} x)^m}x \right)^{n \minus{} 1} < \left(\frac {1 \minus{} x^n}{1 \minus{} x} \right)^{m \minus{} 1}$,

equivalent :
$ (1 \minus{} x \plus{} .. \plus{} m( \minus{} x)^{n \minus{} 1})^{m \minus{} 1} < (1 \plus{} x \plus{} .. \plus{} x^{m \minus{} 1})^{n \minus{} 1}$.[/hide]
f(1)=f(0)=1  !!!!!!!!!
Also probalistic interpretation [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=245389&search_id=282459874]here[/url][/quote]

[quote="alex2008"]EDIT: Sorry about that [hide="Another proof"]If $ x \equal{} 1\vee x \equal{} 0$ the inequality is true. Assume $ 0\neq x\neq 1$ .

Denote : $ f(x) \equal{} 1 \minus{} (1 \minus{} x)^m$. The inequality is then equivalent to $ f(x)^{n}\geq f(x^n)$. I will show using induction:

$ (1 \minus{} (1 \minus{} x)^m)^2 \geq 1 \minus{} (1 \minus{} x^2)^m \Rightarrow \frac {(1 \minus{} x^2)^m \plus{} ((1 \minus{} x)^2)^m}{2} \geq (1 \minus{} x)^m \Rightarrow \frac {(1 \plus{} x)^m \plus{} (1 \minus{} x)^m}{2} \geq 1$ , true by Power mean (for every $ m \geq 1$).
Assume that it is true for some integer $ n \equal{} k > 1$ , for $ n \equal{} k \plus{} 1$ :

$ f(x)^{k \plus{} 1} \equal{} f(x)f(x)^k \geq f(x)f(x^{k})$

it'll be enough to prove that $ f(x)f(x^k)\geq f(x^{k \plus{} 1})$ , equivalent to:

$ (1 \minus{} x^{k \plus{} 1})^m \plus{} (1 \minus{} x \minus{} x^k \plus{} x^{k \plus{} 1})^m \geq (1 \minus{} x^k)^m \plus{} (1 \minus{} x)^m$

, true by Karamata .

So it is true for every $ m \geq 1$ , and natural $ n$.[/hide] May be is correct now .

Also probalistic interpretation [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=245389&search_id=282459874]here[/url][/quote]

but f(1)=f(0)=1   ......


algebra
polynomial
function
induction
functional equation
algebra unsolved

Find all the functios f: Z ---> Z such that

f(x+y+ f(y)) = f(x) + 2y

Ok, but i believe is difficult to prove.
Please, iff solve this problem. posted!!

It has 2 solutions. $f(x)=x$ and $f(x)=-2x$. The problem is from St. Petersburg MO 2000. As soon as I get free time, I'll post my complete solution.

Please, posted  your solution!!!! :oops:
Im very interesting!!!

[quote="me@home"]first, let $x=y=0$ gives $f(f(0))=f(0), f(0)=0$. Now let $x=0$ to give $f(y+f(y))=2y$ so $f(1+f(1))=2$. $f(1)=a$ $f(1+a)=2$ $f(2+a)=a+2$ (let $x=y=1$ in functional equation)
Now I'm not sure about this but can't you assume that it is a polynomial, then show that it must be linear? Once you do that, we know it has exactly one zero (i. e. one $x$ such that $f(x)=0$) and we know this x is zero from above, so $f(x)=cx$, and we found a point $x=2+a$ above where $f(2+a)=2+a$ so $c(2+a)=2+a$ means $c=1$ of $a=-2$. If $c=1$ we have $f(x)=x$ if $a=-2$ we have $f(1)=-2(1)$ so $f(x)=-2x$
thats my answer, probably faulty somewhere :wink:[/quote]

you are wrong.I have solved this problem many times.I know two solutions.And also this function is injective.Now I post what came to my mind in 5 minutes(sorry I don't have enough time I want to look over other problems.)
I will show that $f(0)=0$. Putting $x=0;y=0$ and denoting $f(0)=a$ we get $f(f(0))=f(0)=a$. so $f(a)=a$.Now put $y=0;x=a$
We get $f(2a)=a$,now put $x=0;y=a$ we get $f(2a)=3a$.so we get $3a=a$ so $a=0$.Now continue yourself.....

so is my method wrong (some logical error in proof) or is my answer wrong? i'm not sure which...

No I mean your claims about the form of the function(polynomial) are wrong and (also if it is right you must prove)
in this case it is wrong.

This is my solution.
by induction,
$f(x+k(y+f(y)))=f(x)+2kx$ for all integer k.
k:$z+f(z)$ ->$f(x+(y+f(y))(z+f(z)))=f(x)+2(z+f(z))y=f(x)+2(y+f(y))z$
therefore, $\frac{f(y)}{y}=\frac{f(z)}{z}=c$ (y, z ≠0) ,
and
$f(x+f(0))=f(x)$, If f(0) isn't 0, then f is a periodic funcion, so range of f is finite set. but $f(x+y+f(y))=f(x)+2y$ y-> 1, 2, 3, 4, ... contradiction!
∴ $f(0)=0$ -> $f(0)=c\cdot 0$
∴$f(x)=cx$ for all integer x.
$c(c+1)=2$ -> $c=1,-2$.
So $f(x)=x,-2x$.

Sorry, My english is bad.

At last I post the solution :blush:
Set $x=-f(y)$ we get $f(y)=f(-f(y))+2y$
Now set $y=-f(y)$ we get $f(x-f(y)+f(-f(y)))=f(x-f(y)+f(y)-2y)=f(x)-2f(y)$.
so $f(x-2y)=f(x)-2f(y)$.
set $y=1$ we get $f(x-2)=f(x)-2f(1)$. So $f(x)=f(x-2)+2f(1)$.
Now $f(2n)=f(2n-2)+2f(1)=f(2n-4)+4f(1)=. . . . . . =f(2n-2n)+2nf(1)=2nf(1)$ because $f(0)=0$.look to my earlier post.
Now we are setting $x=2n+1$ we get $f(2n+1)=f(2n-1)+2f(1)=f(2n-3)+4f(1)=. . . . . . . =f(2n-2n+1)+2nf(1)=(2n+1)f(1)$. So for all naturals $n$ we get $f(n)=nf(1)$.
Now we go back to the equation $f(x-2y)=f(x)-2f(y)$.
Here let us take $y$ as negative integer and $x$ as natural number: We get
$xf(1)-2yf(1)=xf(1)-2f(y)$ so for all integers we get $f(z)=zf(1)$.Checking we get the answers
$f(x)=x$ and $f(x)=-2x$.