geometry
circumcircle
geometric transformation
reflection
3D geometry
prism
homothety

Dear Fang-jh and Mathlinkers,
very nice problem. I give you a begining of a possible proof:
1. the circumcircle of your triangle intersect AB and AC resp. in R and Q.
2. This is only a conjecture: QR is tangent to the incircle of ABC
3. If yes, we have to prove that the A-mixtilinear incircle of ABC is also the A-mixtilinear incircle of ARQ.
Sincerely
Jean-Louis

Hello!

The proposed problem can be generalized:
In triangle $ \Delta ABC$ let $ \Pi$ be the incircle, $ \Gamma$ is a circle through the points $ B,C$ and $ \Omega$ is the circle
which tangents to $ (AB$, $ (AC$ and $ \Gamma$ such that $ \Omega\cap\Pi\neq\emptyset$. Let $ d$ be the tangent to the circle $ \Pi$
which is parallel to the side $ BC$ and $ M$ is a point on the circle $ \Gamma$. Prove that the circumcircle of the
triangle formed by the tangents from $ M$ to the circle $ \Pi$ and $ d$ is always tangent to the circle $ \Omega$.

P.S. I don't have a proof for this generalization, but I think it's true.

Sincerely, Petrisor Neagoe

Dear Jean-Louis Ayme
I think you are right, but you have a typo: we have to prove that the $ A \minus{}$mixtilinear incircle of triangle $ ABC$ is also the $ A \minus{}$mixtilinear excircle of triangle $ ARQ$.
Dear Petrisor Neagoe
By drawing, I think you are right, except for the fixed circle you refer, it  also touches the other fixed circle which is tangent to $ BA,CA$ beyond $ A$, and to $ \Gamma$ internally (when vertex $ A$ lies inside $ \Gamma.$)
.

[quote="Fang-jh"]Given a triangle $ ABC.$ Let $ M$ be a moving point on its circumcircle. Let $ l$ be a tangent of its incircle which is parallel to the side $ BC.$ Prove that the circumcircle of the triangle formed by the tangents from $ M$ to its incircle and $ l$ is always tangent to the $ A$-mixtilinear incircle of triangle $ ABC.$[/quote]
Let  $ (O), (I)$ can be 2 arbitrary circles with radii $ R, r$, even intersecting or outside of each other. $ l$ is a fixed tangent of $ (I)$ at $ D$ and $ M \in (O)$ is arbitrary, outside of $ (I)$. Circle with diameter $ IM$ cuts $ (I)$ at $ E, F$ and $ ME, MF$ are tangents of $ (I)$ from $ M.$ Let these tangents cut $ l$ at $ U, V.$ Let $ EF$ cuts $ IM$ at $ K.$ Since $ K$ is the inversion image of $ M \in (O)$ in $ (I),$ $ K$ is on the inversion image $ (P)$ of $ (O)$ in $ (I).$ Let $ L$ be reflection of $ I$ in $ EF,$ $ \overline {IL} \equal{} 2 \overline{IK}.$ This means that $ L$ is on a circle $ (Q)$ similar to $ (P)$ with similarity center $ I$ and coefficient $ 2.$ Let $ N$ be midpoint of $ DL.$ Then $ N$ is on a circle $ (S)$ similar to $ (Q)$ with similarity center $ D$ and coefficient $ \frac {_1}{^2},$ i.e. $ (S) \cong (P).$ Points $ I, D$ are external similarity centers of the circle pairs $ (P), (Q)$ and $ (Q), (S)$ $ \Longrightarrow$ external similarity center of $ (S), (P)$ is on $ ID.$ But $ (S) \cong (P)$ $ \Longrightarrow$ this external similarity center is at infinity $ \Longrightarrow$ $ PS \parallel ID.$ $ N$ is center of the 9-point circle $ (N)$ of $ \triangle DEF,$ the inversion image of the circumcircle $ \odot(MUV)$ in $ (I).$ Its radius $ \frac {_r}{^2}$ is constant, regardless of $ M.$ In addition, $ K \in (N),$ $ KN \equal{} \frac {_r}{^2}$ and $ KN \parallel ID \parallel PS,$ where $ K \in (P), N \in (S), (P) \cong (S)$ $ \Longrightarrow$ $ (S)$ is a translation of $ (P)$ by $ \overline{KN}$ and $ PS \equal{} KN \equal{} \frac {_r}{^2}.$  It follows that $ (N)$ is tangent to 2 fixed circles $ \mathcal S_1, \mathcal S_2,$ concentric with $ (S)$; the difference of their radii equal to $ r$; one of these circles can degenerate to the point $ S.$

[color=red]This part of the proof holds without change even when we replace the arbitrary circle $ \color{red} (O)$ by an arbitrary line $ \color{red}o$. When $ \color{red}M \in o$ is at infinity, the circumcircle $ \color{red}\odot (MUV)$ degenerates to the tangent $ \color{red}l$ of $ \color{red}(I)$ at $ \color{red}D$ and the circle $ \color{red}(N)$ coincides with the circle $ \color{red}\mathcal L$ with diameter $ \color{red}ID.$ It follows that the concentric circles $ \color{red}\mathcal S_1, \mathcal S_2$ with the common center $ \color{red}S$ are both tangent to $ \color{red}\mathcal L.$ But $ \color{red}\mathcal L$ is the inversion image of $ \color{red}l$ in $ \color{red}(I).$ Consequently, the inversion images of the circles $ \color{red}\mathcal S_1, \mathcal S_2$ in $ \color{red}(I)$ are tangent not only to the line $ \color{red}o,$ but also to the line $ \color{red}l$ and their centers are collinear with the inversion center $ \color{red} I.$ In addition, they are both tangent to the circumcircle $ \color{red}\odot(MUV)$ for any $ \color{red}M \in o.$ As a result, we obtained a simple inversion proof of [b]Thebault theorem[/b] ![/color]  :wow:

Let $ l$ cut $ (O)$ at $ B', C'$ and assume now that tangents to $ (I)$ from $ B', C'$ other than $ l$ intersect at $ A' \in (O),$ so that $ (O), (I)$ are circumcircle and incircle of the $ \triangle A'B'C'.$ Let $ A \in (O)$ be tangency point of the A'-mixtilinear incircle of the $ \triangle A'B'C'$ and let tangents to $ (I)$ from $ A$ cut $ (O)$ again at $ B, C,$ By Poncelet porism, $ BC$ is tangent of $ (I).$ Let the angle bisectors $ A'I, AI$ cut $ (O)$ again at $ X', X,$ then $ X'X$ is the common perpendicular bisector of $ B'C', BC$ $ \Longrightarrow$ $ BC \parallel B'C'$ and $ A' \in (O)$ is tangency point of the A-mixtilinear incircle of the $ \triangle ABC.$

Since $ (P)$ is inversion of $ (O)$ in $ (I),$ radii of $ (S) \cong (P)$ are now $ \frac {_r}{^2}.$ Since $ PS$ = $ \frac {_r}{^2}$ as well, $ P \in (S)$ and $ S \in (P).$ If $ E', F'$ are tangency points of $ (I)$ with $ C'A', A'B',$ then $ I, P,$ are circumcenter and 9-point center of $ \triangle DE'F'.$ If $ A_0 \in (P)$ is midpoint of $ E'F',$ then obviously, $ A_0P \parallel ID$ $ \Longrightarrow$ $ S, P, A_0$ are collinear. This means $ S \in (P)$ is diametrically opposite point of $ A_0 \in (P),$ i.e., the inversion image of the the tangency point $ A \in (O)$ of the A'-mixtilinear incircle of the $ \triangle A'B'C'.$ Since $ S \in (N),$ it follows that $ A \in \odot(MUV).$

[color=blue]Note that this is a natural and much nicer inversion proof of the problem [url]http://www.mathlinks.ro/viewtopic.php?t=186117[/url].[/color]

The 9-point circle $ (N)$ of the $ \triangle DEF,$ which is the inversion image of the circumcircle $ \odot(MUV)$ in $ (I),$ is tangent to 2 fixed circles concentric with $ (S).$ Since $ S \in (N),$ one of these circles degenerates to the point $ S \in (P),$ the diametriacally opposite point of $ A_0 \in (P),$ the inversion image of the tangency point $ A$ of the A'-mixtilinear incircle of the $ \triangle A'B'C'$ in $ (I).$ Since $ (N)$ has radius $ \frac {_r}{^2},$ it is tangent to a circle $ (S, r),$ which is the inversion image of the A-mixtilinear incircle of the $ \triangle ABC$ in $ (I).$

[quote="Fang-jh"]Given a triangle $ ABC.$ Let $ M$ be a moving point on its circumcircle. Let $ l$ be a tangent of its incircle which is parallel to the side $ BC.$ Prove that the circumcircle of the triangle formed by the tangents from $ M$ to its incircle and $ l$ is always tangent to the $ A$-mixtilinear incircle of triangle $ ABC.$[/quote]
First I will put some notations to the problem
[quote]Let $ \triangle ABC$ with $ (I)$ as its incircle. Denote $ l$ by the line tangents to $ (I)$ and parallel to $ BC$ and $ M$ by any point on circumcircle $ (O)$ of $ \triangle ABC$. The tangents from $ M$ to $ (I)$ intersect $ l$ respectively at $ X_1,$ $ X_2$. Prove that the circumcircle of $ \triangle MX_1X_2$ always tangents to the mixtilinear circle of $ \triangle ABC$ with respect to angle $ A$.[/quote]
Here is my proof.

Let $ D,$ $ E,$ $ F$ respectively be the tangency points of $ (I)$ with $ BC,$ $ CA,$ $ AB$. Denote $ (K_a)$ be the mixtilinear circle of $ \triangle ABC$ with respect to $ A$ and $ B'$, $ C'$, $ A'$ by the tangency points of $ (K_a)$ with $ AB$, $ AC$ and $ (O)$. Let $ S$ be second intersection of $ A'I$ with $ (O)$ and $ B''$, $ C''$ respectively be the intersections of $ l$ with $ (O)$.

Since, it is well- known that $ B',$ $ I,$ $ C'$ are collinear and $ AI$ is the perpendicular bisector of $ B'C'$, thus $ \angle AB'I$ $ \equal{} \angle AC'I$. But, we also have $ BB'IA'$ and $ CC'IA'$ are two concyclic quadrilateral, hence $ \angle AB'I$ $ \equal{} \angle SA'B$ and $ \angle AC'I$ $ \equal{} \angle SA'C$. Therefore, $ \angle SA'B$ $ \equal{} \angle SA'C$, which implies $ SB \equal{} SC$, or we can say $ S$ is the midpoint of chord $ BC$ of the circumcircle $ (O)$ of $ \triangle ABC$. On the other hand, since $ B''C''\|$ $ BC$ $ \Longrightarrow$ $ S$ is also the midpoint of chord $ B''C''$, which implies $ A'S$ is the internal bisector of $ \angle B''A'C''$. By [i]Poncelet's porism[/i], there is $ \triangle A'UV$ such that $ A'UV$ has $ (I)$ as their incircle and $ (O)$ as its circumcircle, hence $ S$ is the midpoint of chord $ UV$ that doesn't contain $ A'$. We conclude both $ B''C''$ and $ UV$ tangent to $ (I)$, and with $ B''C''\|UV$. As the consequence, $ B''C''\equiv UV$, or $ (I)$ is the incircle of $ \triangle A'B''C''$. We claim that $ A$ is the tangency point of the mixtilinear circle of $ \triangle A'B''C''$ with respect to $ A'$. $ (*)$

Indeed, let $ D'$ be the antipode of $ D$ wrt $ (I)$, then $ D$ is also the tangency point of $ (I)$ and $ B''C''$. Denote $ C_1$, $ B_1$ respectively by the tangency points of $ (I)$ with $ AB''$ and $ AC''$. Consider the inversion through pole $ I$, power $ k \equal{} r^2$, where $ r$ is the radii of $ (I)$. We have $ \mathcal {I}(I,k)$ maps $ A$, $ B$, $ C$ into the midpoints $ M_a,$ $ M_b,$ $ M_c$ of $ EF$, $ FD$, $ DE$, respectively. Hence $ (O)$ is mapped into $ (M_aM_bM_c)$ $ \equiv$ $ (\mathcal {E})$, which is the $ 9 \minus{}$ point circle of $ \triangle DEF$ as well as $ \triangle D'B_1C_1$. Now, under the inversion $ \mathcal {I}(I,k)$, from [i][1][/i], we have known that if $ A''$ is the image of $ A'$, $ A''$ will be the orthocenter of $ \triangle DM_bM_c$. Then it is well- known that $ A''$ are the reflection of the orthocenter $ H$ of $ \triangle DEF$ across to $ M_bM_c$ and $ H$ is the symmentry point of $ D'$ wrt $ M_a$. Therefore, $ IA''\|D'H$ $ \equiv D'M_a$. Note that $ A''$ is the midpoint of $ B_1C_1$ $ \Longrightarrow$ $ IA''\perp B_1C_1$, which implies $ D'M_a\perp B_1C_1$. Let $ M'_b,$ $ M'_c$ respectively be the midpoints of $ D'C_1$ and $ D'B_1$, then it is followed that $ D'M_a\perp M'_bM'_c$, $ M_a$ $ \in$ $ (\mathcal {E})$. As the result, $ M_a$ also plays the roles of the orthocenter of $ \triangle D'M'_bM'_c$. Therefore, $ A$ is the tangency point of the mixtilinear circle of $ \triangle A'B''C''$ with respect to $ A'$. We obtain result $ (*)$.

Now, let $ W_b$, $ W_c$ repsectively be the tangency points of $ MX_1,$ $ MX_2$ with $ (I)$. Then the inversion $ \mathcal {I}(I,k)$ maps $ X_1$, $ X_2$, $ M$ into the midpoints $ X'_1$, $ X'_2$, $ M'$ of $ D'W_b$, $ D'W_c$ and $ W_bW_c$. Therefore, under $ \mathcal {I}(I,k)$, $ (MX_1X_2)$ will be mapped into a circle $ (\omega)$ passes through $ X'_1,$ $ X'_2$ and $ M'$, which obviously is the $ 9 \minus{}$ point circle of $ \triangle D'W_bW_c$. But due to [i][1][/i], we have $ A\in (MX_1X_2)$ $ \Longrightarrow$ $ M_a$ also $ \in$ $ (\omega)$.  The inversion through pole $ I$, power $ k$ takes $ B'$, $ C'$ into $ X$, $ Y$ repsecively $ \Longrightarrow$ $ \mathcal {I}(I,k)$ maps $ (K_a)$ into $ (A''XY)$ $ \equiv$ $ (\omega_a)$ that tangents to $ (\mathcal {E})$ at $ A''$. We claim that $ (\omega)$ and $ (\omega_a)$ are tangent. $ (**)$

Indeed, now just taking our concern into $ \triangle DEF$ and $ (I)$. Since $ X,Y$ respectively are the inverse points of $ B',$ and $ C'$ $ \Longrightarrow$ $ X,Y$ respectively are also the projections of $ E,$ $ F$ onto $ B'C'$, which implies $ YEFX$ is rectangle. Thus, $ EF \equal{} XY$, but $ M_cM_b$ $ \|$ $ EF$, $ EF \equal{} 2.M_bM_c$. Therefore, $ M_b,$ $ M_c$ respectively are midpoints of $ A''X$ and $ A''Y$. Hence through the homothety at center $ A''$, ratio $ k' \equal{} 2$, we have $ \mathcal {H}(A'',k'):$ $ (\mathcal {E})$ $ \mapsto$ $ (\omega_a)$. Thus, $ \mathcal {H}(A'',k')$ will map $ \mathcal {E}$ into $ M_a$, or in other words, $ M_a$ is the center of $ (\omega_a)$ $ (1)$. Now, let $ H'$ be the orthocenter of $ \triangle D'W_cW_b$, then it is easy to note that $ DW_cH'W_b$ is parallelogram, which implies that $ (\omega)$ and the $ 9 \minus{}$ point circle $ (\mathcal {E'})$ of $ \triangle DW_cW_b$ are reflected each other through the midpoint $ M'$ of $ W_bW_c$. As the consequence, if we denote $ r_1$, $ r_2$, $ r_3$ by the radii of $ (\omega)$, $ (\mathcal {E'})$ and $ (\mathcal {E})$, we have $ r_1 \equal{} r_2$. But according to [i][1][/i], we have $ r_2 \equal{} r_3$, thus $ r_1 \equal{} r_3$. Also note that $ (\mathcal {E})$ intersects $ (\omega)$ at $ M'$ and $ M_a$. Therefore, $ (\mathcal {E})$ and $ (\omega)$ are reflected each across the line $ M_aM'$. Then combine with $ (1)$, we have $ (\omega)$ also tangents to $ (\omega_a)$. We obtain result $ (**)$.

To sum up, due to $ (**)$, and a note that $ (\omega)$ and $ (\omega_a)$ are the inverse circles of $ (MX_1X_2)$ and $ (K_a)$ with respect to $ \mathcal {I}(I,k)$, we obtain $ (MX_1X_2)$ always tangents to $ (K_a)$. Our proof is completed then. $ \square$
___________________________
[i]Reference[/i]
[i][1][/i]- post #3 of the topic http://www.mathlinks.ro/viewtopic.php?t=309644


Χρόνια πολλά και καλή χρονιά :)
Αφού πιάσαμε το πνεύμα των ανισοτήτων το 2010 ας βάλω και γω μια καλή:
Αν $ a,b,c,d$ είναι θετικοί πραγματικοί να δείξετε ότι $ \left(\frac{a}{a\plus{}b\plus{}c}\right)^{2}\plus{}\left(\frac{b}{b\plus{}c\plus{}d}\right)^{2}\plus{}\left(\frac{c}{c\plus{}d\plus{}a}\right)^{2}\plus{}\left(\frac{d}{d\plus{}a\plus{}b}\right)^{2}\geq \frac{4}{9}$.

Μου θύμιζε κάτι οπότε μια πιο γενική μορφή εδώ, δυστυχώς χωρίς απόδειξη...

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=322633

Νομίζω έχω βρει λύση στην απλή περίπτωση αλλά με πολλές πράξεις... Θα την ελέγξω και θα την ποστάρω...

ne kai egw exw kanei mia lush(gia thn aplh periptosh) me B-C-S kai AM-GM alla einai mia selida pra3eis :S kai dn 3erw  ama exw kanena la8os  :maybe:  [dn sou afhnoun xrono oi panellhnies akoma kai an dn diavazeis ta alla ma8hmata ]

[quote="Γιώργος"][quote="KapioPulsar"]ne kai egw exw kanei mia lush me B-C-S kai AM-GM alla einai mia selida pra3eis :S kai dn 3erw kai ama dn exw kanena la8os  :maybe:[/quote]

Prospathiste tin me Holder kai power mean :lol:[/quote]   kala odos power mean  :P

Βάζω τη λύση για να κλείσει και αυτό το post.
http://gbas2010.wordpress.com/2010/01/07/inequality-18vo-quoc-ba-can/


Let $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$.Prove that:
   $ \frac{{a^2  \plus{} 3b^2 c}}{{b \plus{} c}} \plus{} \frac{{b^2  \plus{} 3c^2 a}}{{c \plus{} a}} \plus{} \frac{{c^2  \plus{} 3a^2 b}}{{a \plus{} b}} \ge 1$ :D

[quote="leedt26"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
   $ \frac {{a^2 \plus{} 3b^2 c}}{{b \plus{} c}} \plus{} \frac {{b^2 \plus{} 3c^2 a}}{{c \plus{} a}} \plus{} \frac {{c^2 \plus{} 3a^2 b}}{{a \plus{} b}} \ge 1$ :D[/quote]
You can use : $ \sum \ \frac{a^3\plus{}3abc}{b\plus{}c} \ge 2\sum \ ab$ :)

Anyone can give a solution more clearly!

can you show your full proof,quykhtn-qa1??

[quote="leedt26"]Let $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
   $ \frac {{a^2 \plus{} 3b^2 c}}{{b \plus{} c}} \plus{} \frac {{b^2 \plus{} 3c^2 a}}{{c \plus{} a}} \plus{} \frac {{c^2 \plus{} 3a^2 b}}{{a \plus{} b}} \ge 1$ :D[/quote]
After expanding we obtain for the proof:
$ \sum_{cyc}(a^5 \plus{} a^4b \plus{} a^4c \plus{} a^3b^2 \minus{} 2a^3c^2 \minus{} 2a^2b^2c)\geq0,$ which is obviously true.

How about the stronger, Quy?  :)
http://www.mathlinks.ro/Forum/viewtopic.php?p=1526706#1526706


Here is another nice one:

there are $100$ subsets, of a line, each subste is a union of $100$ disjoint closed intervals. Prove that the intersection of these subsets is the union of at most $9900$ intervals.

Hi Pascual!

In my solution I got 9901, not 9900. It is a typo ;)

I have looked through ARO 2002 for 10 times, but I couldn't find it!
Actually, IT IS FROM ARO 2001 :)

BTW, the problems is very simple. We can easily prove that for two systems of $a$ and $b$ intervals their intersection contains at mosts $a+b-1$ intervals.


integration
trigonometry
calculus
calculus computations

Prove that $ \displaystyle 1\minus{}e^{\minus{}\frac\pi2}<\int_0^{\frac\pi2}e^{\minus{}\sin x}\textrm{d}x<\frac\pi2(1\minus{}e^{\minus{}1})$

This is your twin brothers.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=25423&search_id=161421594[/url]


function
combinatorics unsolved
combinatorics

In how many ways can $n$ identical balls be distributed to nine persons $A,B,C,D,E,F,G,H,I$ so that the number of balls recieved by $A$ is the same as the total number of balls recieved by $B,C,D,E$ together,.

these sort of problems are readily handled by the exponential generating function.
let $x_A,\ldots,x_I$ denote the distribution of the balls. Then we have $x_A=x_B+x_C+x_D+x_E$ so that we have that $2(x_B+x_C+x_D+x_E)+x_F+\cdots+x_I=n$ or in other words we need the number of ordered $8$-tuples such that $4$ terms are even. The Exponential generating function here is simply $(\sum_{k\geq 0} \dfrac{x^{2k}}{(2k)!})^4(\sum_{k\geq 0} \dfrac{x^k}{k!})^4$. what you need here is the coeff of $x^n$ which is easily obtained by simplification.

[quote="seshadri"]in other words we need the number of ordered $8$-tuples such that $4$ terms are even. The Exponential generating function here is simply $(\sum_{k\geq 0} \dfrac{x^{2k}}{(2k)!})^4(\sum_{k\geq 0} \dfrac{x^k}{k!})^4$[/quote]
I think this E.G.F. is incorrect. It is better to use regular G.F. here which is
\[ \left(\sum_{k\geq 0} x^{2k}\right)^4 \left(\sum_{k\geq 0} x^k\right)^4 = (1-x^2)^{-4} (1-x)^{-4}  \]

oh yes, you are right. i misread the 'identical balls' part and assumed that the balls are distinguishable.


Okay, here's the rules...
everyone gets to post a part of a story...
example:  person 1 posts: Bob went to the store
                 person 2 posts: Then, a killer dragon popped out of nowhere and ate everthing exept bob
                 person 3 : then a hurricane came and bob was swept into the air
                 you get the idea... so some one start us off!

[quote="mcalderbank"]math92 decided it would be a good idea to tell a story, but unfortunately......

:D[/quote]

Some mean person came and told him it was retarded

then the mean, beat up person, ran to his mom and sued math92

But the lawyer was on Mars and couldn't be contacted.

then he blew up mars and hired a new lawyer

who just happened to be on jupiter instead

and this lawyer stepped on a landmine and got his leg blown off.

And so math92 hired his pet gorilla as his lawyer

but math92 hadn't enough bananas to feed the gorilla

aAnd since math92 was wearing a banana suit, the gorilla ate him instead. Because the gorilla ate a human, the SWAT team came and took him down with a tranq-dart

And the guy from SWAT says.. Hey, isn't the idea basically the same than in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34951]this post[/url]?

Then Valentin, which was randomly passing by, says "Indeed it is ... " so he closed this topic and let the users continue the story in the other topic.
The end. :P


Two gamblers take turns rolling a fair N-sided die, with N at least 5.  The faces of the die are labeled with the numbers 1 to N.  The first player starts.  If he rolls an N or an N-1, he wins and the game is over.  Otherwise, the other player rolls the die; if she rolls a 1, 2, or 3, she wins and the game is over.  Play continues, with the players alternating rolls until one of them wins.

What is the probability that the first player will win?  Are there any values of N for which the first player has at least an even chance of winning?

[hide] The probability that the first player wins on the first move is $\frac{2}{N}$. The probability that neither player wins on their first moves, respecitively, is $\frac{(N-2)(N-3)}{N^{2}}$. This is equivalent to the probability that player one gets a second move. But we see that the probability that player one gets a $K$th move is

$\left(\frac{(N-2)(N-3)}{N^{2}}\right)^{K-1}$

and each time he has a $\frac{2}{N}$ chance of winning on that move. So his win probability is

$\frac{2}{N}+\frac{2}{N}\cdot \frac{(N-2)(N-3)}{N^{2}}+\frac{2}{N}\cdot \left(\frac{(N-2)(N-3)}{N^{2}}\right)^{2}+\cdots$

which comes out to $\frac{2N}{5N-6}$ (by infinite geometric series). This is at least $\frac{1}{2}$ for $N = 3, 4, 5, 6$.[/hide]


[quote="koredim1989"]diavasa kapou oti mia sinartisi f gia tin opoia ine f(ta,tb,tc,...)=t^kf(a,b,c,...) legete omogenis vathmou k.Mporeite na mou exigisete ti simveni me tis anisotites se afti tin periptosi?Diavasa sto vivlio tou k.stergiou oti boroume na theoroume kapoia sinthiki pou mas volevei.Giati?[/quote]

ΝΑΙ ! Μπορείς να θέτεις κάποια συνθήκη , όπως είπες , που συνήθως απλουστεύει την άσκηση. Το σκεπτικό είναι το εξής :

 Έστω ότι έχεις μεταβλητές τις $a , b$. Θέτεις $a+b= k$ και $a=kx, b=ky$. Τότε :

      $x+y = \frac{a}{k}+\frac{b}{k}= \frac{a+b}{k}=\frac{a+b}{a+b}= 1$.

 Επομένως με τις νέες μεταβλητές έχεις άθροισμα $1$ κάτι που δεν είχες με τα $a , b$ Αντί λοιπόν να κάνεις όλη αυτή τη δουλειά που σου έδειξα , θεωρείς απευθείας $a+b=1$ και προχωράς αμέσως. Κανείς δε θα σε ρωτήσει γιατί  ! :) Ανάλογα κάνεις με τα γινόμενα.Ελπίζω κάπως να σε διευκόλυνα !

  Πιο πολλές λεπτομέρειες μπορείς να δεις στο [color=red]νέο βιβλίο << Κλασικές και Νέες Ανισότητες >> , σελίδα 127 - 128.[/color]
Θα σου πρότεινα όμως - τελείως φιλικά - μέχρι τον Αρχιμήδη να το θεωρήσεις αυτό ως δεδομένο και να μελετήσεις κάποιες εφαρμογές , ώστε να μη χάνεις χρόνο.Τα υπόλοιπα τα μελετάς αργότερα.
[u]   Μπάμπης[/u]


I don't have any assigned summer reading.  Even so I would like to read more but I haven't been doing much lately.  I am going to try to read some more before the school year starts back up.

Rep, do you attend a private school or something like that?

I wasn't assigned any summer reading but I'm on my third book this summer.

I've got two books, [i]The Stranger[/i] by Albert Camus and [i]Things Fall Apart[/i] by Chinua Achebe.  I have to write detailed reportson 5 x 8 index cards about each book.

I've procrastinated so far this summer.  I feel so guilty...(sobs)

Umm...well...
Summer Reading List
Miss Grady
Grade 10 AS
Scarlett Letter - Nathaniel Hawthorne

1.Discuss Pearl as a symbol. How and why is her behavior not natural? At what point does she "transform" and why?

2. A recurring theme in literature is the "classic war between passion and responsibility". Choose either Rev. Arthur Dimmesdale or Roger Chillingworth -
Show (not tell) how this man confronts the demands of his private passion with his responsibility.

To Kill A Mockingbird - Harper Lee

1. The final scene in the trial is one of the greatest scenes in American literature.
Bring the time period of To Kill A Mockingbird into today's society and issues. Write a closing argument based on the events of the novel, yet pertaining to the current issues of today.

Farewell to Arms - Ernest Hemingway

1.Choose one of the following characters: Fredrick Henry, Catherine Barkley, or Lt. Rinaldi. Prepare five interview questions and answers for this character regarding his/her life during the novel (World War I)

You are a news correspondent assigned to the war. Using your interview questions and answers, write a news article based on your character's experiences in the war.
Cite examples, dialogue from the novel.

2. A true Hemingway man has self- discipline and control, avoids death at any cost, yet doesn't fear death, and accepts the joys and the turmoils in life and then moves on. Decide who the true Hemingway man is in this novel. Explain and argue your decision citing examples from the novel to support your findings.

We read To Kill a Mockingbird in class before...and I didn't do ANY of the tasks, and I'm like 10 pages into The Scarlet Letter.   Um, hopefully I do this summer reading (last year I didn't (nor did most of my class - we're horrible...come on now, it was an ADVANCE 9th grade class and she made us read Little Women, David Copperfield, and The Road To Memphis <---like four/fifth grade books!)   The tasks kind of cool this year though.

Do you students have to hand in the answers to your last year's teacher or your current (this year) teacher?

Also, do you receive any extra marks if you do answer the questions?

i have to hand it to my current teacher..and um, it is REQUIRED that you answer everything.

hmmm, i 'm going into 9th grade, and i have to read [u]the prince and the pauper[/u] for pre-ap english.  I also must highlight themes, characters, and some other stuff i forgot...i haven't even finished the book yet

I have to read the 1836 version of Mary Shelley's Frankenstein for MTU orientation; instead, I'm planning on reading the version with yellow and black stripes on the cover, and spending my free time doing things I enjoy.

One last question, do all private schools have this system in the USA?

And also, do public schools also have this system?

Or do they all differ from school to school.

Assuming that you are talking about the summer reading thing,  it differs from school to school.  Some schools give you a list of books that might be helpful to read(or I should say some teachers) other schools make you do work over the summer, and others(like mine) don't give you anything at all(which isn't a bad thing IMO :lol: ).

I didn't get any assined reading this year but if I had decided to attend a new school I would have had to read.

To Kill a Mocking Bird

and

The Diary of Ann Frank

Wow....8th...USA definetely has a harder curriculum than Canada.

I'm in Canada and i'm in grade 9 we read those books last year or we were supposed to...

I'm 10 now...hehe

The U.S. isn't really harder.  The call that I would have been taking that those books were from an advanced 9th grade english class.  I decided against the school though.

Going into 10th grade..
Loads of summer reading...

Read: [i]The Adventures of Huckleberry Finn[/i] by Mark Twain.
Take detailed, typewritten notes on plot, character, and setting. [in my case.. that usually means about 50 pages of notes.. lol *points at self and mutters dork*]
Write a well-developed, thought-provoking essay, 2-3 pages, double-spaced, 12 point Times New Roman font, one-inch margins, no cover page, page numbers in the upper right corner. This essay should adhere to the five paragraph format. [PHEW.. yes our teachers has err.. very strict layout "guidelines"]

I'm not even going to type out the whole quote and blah that our essays supposed to be about since it would take up about two pages... so yea..

Besides this, we also have to read eleven short stories:

The Necklace  Guy de Maupassant
The Story of an Hour  Kate Chopin
Young Goodman Brown  Nathaniel Hawthorne
The Cask of Amontillado  Edgar Allen Poe
Miss Brill  Katherine Mansfield
Trifles  Susan Glaspell
An Occurrence at Owl Creek Bridge  Ambrose Bierce
The Three Strangers  Thomas Hardy
The Bear: A Joke in One Act  Anton Chekhov (also called The Boor)
"Patterns  Amy Lowell
Dover Beach  Matthew Arnold

Joy ehh?

But that's not all.

We have to read The Elements of Style by Strunk and White too..
and notes for all the short stories and Elements is "suggested".. which means if you want an A.. you better do it.

Thankfully. I'm DONE! *cheers!*

[ps. sorry for that long boring entry....]

I'm surprised your school still teaches grammar.  All high school teachers expect all of us to know proper english grammar.

I think we only had one grammar class during the whole year.

I haven't even started - I'm in major trouble cuz I'm in 4 seminar classes with about 2 books for every class but math. I should really get going. However, the assignments look easy enough. How hard are your assignments?

[quote="joml88"]Is that all for the summer?  My school is so slack when it comes to english(at least the underclassmen classes).  We read about 7 books during the year, read a handful of short stories, do maybe 2 weeks worth of grammar and we're done.[/quote]

At least you guys get to actually read books in your class.  We spent our whole English class almost every single day (except for one sixth of the year where our teacher all of a sudden got really tough) reading out of our grammer books, reading A.R. books (just regular fiction books like Redwall and Harry Potter), and on the rare ocassion - write part of our esay.  We didn't even have essay /short answer questions on quizzes untill 2/3rd of the year was over.  And we only read two books as a class.

My school's already suggesting you read The Elements of Style, but we only have to read a few books over the summer, and write a short summary of them. Howevery, they have a nice incentive, you get to an hour of school for a celebration if you do it.


function
limit
real analysis
real analysis unsolved

Let $f: R \to R$ $f(x) = \displaystyle e^{2x}+2ex$. Compute the limit: $ \displaystyle \lim_{y \to \infty} \sqrt{ e^{2f^{-1}(y)}} \cdot \frac{ f^{-1}(y) - \sqrt{y}}{y- f^{-1}(y)}$.

Where $f^{-1}$ is the inverse of the function.

thank you!

$f$ is bijective increasing $\lim{_+\infty}f=+\infty$ so $\lim_{+\infty}f^{-1}=+\infty$

Do the change variable $u=f^{-1}(y)$ the given expression is

$e^{u}\frac{u-\sqrt{e^{2u}+2eu}}{e^{2u}+(2e-1)u}$ from easy to find the limit
when $u$ tends to +oo


geometry
rhombus
circumcircle
analytic geometry
parallelogram
symmetry
trigonometry

Should be quite easy

Let $ABCD$ be a rhombus of which we know the circumradius $R$ of the triangle $\triangle ABC$ and the circumradius $r$ of the triangle $\triangle BCD$. Construct the rhombus (only with ruler and compass)!

Let $a$ be the side-length of the rhombus

$(ABC) = \frac{a^2 \cdot AC}{4r} \Rightarrow 2(ABC) = \frac{a^2 \cdot AC}{2r}$

$(BCD) = \frac{a^2 \cdot BD}{4R} \Rightarrow 2(BCD) = \frac{a^2 \cdot BD}{2R}$

The total area of the rhombus is $2(ABC) =2(BCD)$, so $\frac{AC}{r} = \frac{BD}{R} \Rightarrow \frac{AC}{BD} = \frac{r}{R} \ \ (1)$

We will construct the rhombus on a coordinates system $Oxy$, so that
$A \in Ox', \ C \in Ox, \ B\in Oy , D \in Oy'$
and $O$ will be the intersection of the diagonals $AC$ and $BD$.

From $(1)$ we get $\frac{OC}{OB}=\frac{r}{R}$

We take the points $P(r,0), Q(0,R)$. Then, it's obvious that $BC \parallel PR$
$\hline$

Now, we have the direction of $BC$ but we don't have yet its length

On the ray $PO$ we take a point $S$ such that $PS=R$.
We draw the circle $(S,R)$ and let $T$ the point where this circle intersects (again) the line $PQ$ (it is $r>0$, so the line $PQ$ is not a tangent)
The triangle $STP$ is isosceles: $SP=ST=R$

We find the projection $B$ of $T$ onto $y'y$ and we construct the parallelogram $BCPT$.
After that, we can find the other two points $A,D$ of the rhombus, which are symmetric of $C,B$ w.r.t. $O$
From the construction, we have $\frac{OC}{OB}=\frac{AC}{BD}=\frac{r}{R}$
$\hline$

The quadrilateral $ABCD$ is obviously a rhombus, but we should check if the circumradiuses are correct.

The perpendicular bisector of $BC$ intersects the x-axis at $M$. It's easy to see that the isosceles triangles $MBC,STP$ are congruent, hence $MB=MC=R$. By symmetry, we have $MB=MD=R$.
So $M$ is the circumcenter of $\triangle BCD$ and the radius is $R$.

Let $r'$ be the circumradius of $\triangle ABC$. Then $2(ABC) = \frac{a^2 \cdot AC}{2r'}$ and $2(BCD) = \frac{a^2 \cdot BD}{2R} \Rightarrow$
$\Rightarrow \frac{AC}{BD} = \frac{r'}{R}$

But we constructed the rhombus such that $\frac{AC}{BD} = \frac{r}{R}$, so we have $r'=r$ and we are done

Here is another construction:

$k:$ The circumcircle of $\triangle ABC$
$m:$ The circumcircle of $\triangle BCD$

Let $2x$ be the side-length of the rhombus

From $\triangle ABC$ we get $AB=2r \sin(BCA)$
From $\triangle BCD$ we get $CD=2R \sin(CBD)$

So we have $\sin(BCA) = \frac{AB}{2r} = \frac{2x}{2r} = \frac{x}{r}$

and $\sin(CBD) = \frac{CD}{2R} = \frac{2x}{2R} = \frac{x}{R}$

Let $O$ be the intersection of the diagonals
From the right triangle $OBC$ we find $\angle BCA+\angle CBD = 90^o \Rightarrow  \sin^2(BCA)+\sin^2(CBD)=1 \Rightarrow$

$\left( \frac{x}{r} \right)^2 + \left( \frac{x}{R} \right)^2 = 1 \Rightarrow$

$x^2 = \frac{r^2R^2}{r^2+R^2} \ \ (1)$

We construct a right triangle $T$ with perpendicular sides $r,R$ and hypotenuse $z$. Its area is $E = \frac{1}{2} rR \Rightarrow  rR=2E$

$(1)$ gives $x^2 = \frac{(2E)^2}{z^2} = \left(\frac{2E}{z}\right)^2 \Rightarrow  x = \frac{2E}{z} \Rightarrow  E = \frac{1}{2}xz$

So, $x$ is the altitude of $T$ with respect to the hypotenuse

We draw the circle $k=(A,B,C)$ with radius $r$ and we take an arbitrary point $B$ on it.
Then we find the points $A,C \in w$ such that $AB=BC=2x$, and the point $D$ is on the intersection of the circles $(A,2x)$ and $(C,2x)$.

[quote="Yimin Ge"]Let $ ABCD$ be a rhombus of which we know the circumradius $ R$ of the triangle $ \triangle ABC$ and the circumradius $ r$ of the triangle $ \triangle BCD$. Construct the rhombus (only with ruler and compass)![/quote]

We construct a triangle $ MNB$, right at $ B$, with $ MB\equal{}R$ and $ NB\equal{}r$

We draw the circles $ m\equal{}(M,R)$ and $ n\equal{}(N,r)$. The circles intersect at $ B,C$.
Let $ A$ be the symmetric point of $ C$ wrt the line $ BM$. We draw the rhombus $ ABCD$
Obviously $ MA\equal{}MC\Rightarrow A\in m$

Now it's enough to prove that the circumradius of $ \triangle BCD$ is $ r$

[img]10375[/img]
We draw the rhombus $ BNCO$. It is $ OC\equal{}OB\equal{}r$
Also $ D,B$ are symmetric wrt $ OC$, so $ OD\equal{}r$.

So $ O$ is the circumcenter of $ \triangle BCD$ and the circumradius is $ r$


Midtown Taxi charges $\$1.60 plus $\$0.25 per one-eighth mile.  The distance from the hotel to the airport is 16.25 miles.  Two passengers will evenly split the fare.  How many dollars will each passenger pay?  Express your answer to the nearest hundredth.

I got $9.05, but i could be wrong i guess.

Hi MCrawford:
[hide="solution"]
If it costs 25 cents for one-eighth of a mile, then it would cost 2 dollars per mile.
so for 16.25 miles it would cost him 16.25 X 2 dollars or 32 dollars and 50 cents.
Then adding the 1 dollar and 60 cents  to 32 dollars and 50 cents, you would have
a total fare of 34 dollars and 10 cents.
[/hide]

I thought it was 25 cents per quater mile.  So the answer is[hide]34.10[/hide]


What is $ 50\%$ of the sum of the first 10 odd numbers?

It is well known that the sum of the first $ n$ odd numbers is $ n^2$, so our answer is $ 0.5\cdot100\equal{}\boxed{50}$.

The first 10 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.  Adding yields 20*5=100.  Divide by 2 to get 50.


Let a,b,c,d be four distinct real numbers such that
i)$ ac\equal{}bd$
ii)$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{d}\plus{}\frac{d}{a}\equal{}4$
Find the maximum of
 $ P\equal{}\frac{a}{c}\plus{}\frac{b}{d}\plus{}\frac{c}{a}\plus{}\frac{d}{b}\minus{}\frac{abcd}{(ab\plus{}cd)^2}$

[quote="8826"]Let a,b,c,d be four distinct real numbers such that
i)$ ac \equal{} bd$
ii)$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{d} \plus{} \frac {d}{a} \equal{} 4$
Find the maximum of
 $ P \equal{} \frac {a}{c} \plus{} \frac {b}{d} \plus{} \frac {c}{a} \plus{} \frac {d}{b} \minus{} \frac {abcd}{(ab \plus{} cd)^2}$[/quote]
It very old,in MYM :wink:


Vezi problema 1.2 din excelenta carte "Probleme de combinatorică şi teoria grafurilor" de Ioan Tomescu, Ed. Didactică şi Pedagogică, Bucureşti, 1981.

Cred ar trebui folosita relatia \[ \binom{n}{k} - \binom{n-1}{k} = \binom{n-1}{k-1} . \]

Notam membrul stang cu $c(n,k)$ si apoi demonstram ca $c(n,k) - c(n-1,k) = c(n-1,k-1)$ folosind ce-am scris eu mai sus. Apoi observam ca formula e buna pt. $n$-uri mici si apoi folosim inductie. Merge?

Dar in cartea mentionata de dl. Enescu ce solutie e? Daca e cu "dubla numarare", ai putea s-o postezi? Multumesc.

[color=blue][b]Prove:
$C_{k - 3}^{k - 3} \cdot C_{n - k + 2}^2 + C_{k - 2}^{k - 3} \cdot C_{n - k + 1}^2 + C_{k - 1}^{k - 3} \cdot C_{n - k}^2 + \cdots + C_{n - 3}^{k - 3} \cdot C_2^2 = C_n^k$
[/b][/color]
[color=red][size=75][b]
$\blacktriangleright$
Conform precizărilor D-lui enescu, problema a fost propusă şi în lucratea:
IOAN TOMESCU, "Probleme de combinatorică şi teoria mulţimilor",
Editura Didactică şi pedagogică, Bucureşti 1981. [pag. 11, prob. 1.2. a)]

[/b][/size][/color]

[b][u]Soluţie:[/u][/b]

(Voi incerca sa redactez si solutia Dlui Tomescu)

Care e titlul corect al cartii ,cel dat de metru sau cel dat de dl. Enescu ?


[quote="matt276eagles"]Sorry to be impatient, but I was wondering when we will receive our third round scores.  Also, how soon after we get our third round scores will we find out the cut-off for the AIME?  Does anyone know what the cut-off has been in the past, or is this the first year that it's possible to qualify for the AIME through USAMTS?[/quote]

Scores should be up later this week.  We don't yet know what the cut-off will be, but I think last year it was 68.

WOW (actually, double WOW), I thought on the question sheet it says that we will get our scores in early March.  Is that a typo???

By the way, when is round 4 coming up???

Between Amc A, round 3 scores, and round 4 problems, this is a good math week :D

Yay! Scores are up!

I qualified for the AIME!  :)


integration
function
real analysis
real analysis unsolved

Let $ p\equal{}y'$. Solve the differential equation $ y \equal{} \minus{}\frac{x}{p} \plus{} ap$ where $ a$ is some real constant.

I differentiated the equation, obtained an ODE for $ p$ and found an implicit solution for $ p$ by solving it. However, this doesn't help me to find $ y$. Any suggestions?

Well if you've already got an implicit expression for $ p$, then $ y(x) \equal{} \int^x{p(t)dt}$  [i]is[/i] an implicit solution for $ y$. :)

The point is this ODE doesn't have an explicit solution for $ y$ in terms of the usual elementary functions, so an implicit solution for $ p$ is just about as good as you can get.  You can obtain other expressions more directly for $ y$ itself, but they will still be implicit.

I think it's tractable parametrically:It's of the form:$ y=xF(p)+G(p)$.  Now differentiate as Thunder suggested: $ p=F(p)+p'\bigg[xF'(p)+G'(p)\bigg]$ which can be written as:\[ \frac{dx}{dp}=x\left(\frac{F'(p)}{p-F(p)}\right)+\left(\frac{G'(p)}{p-F(p)}\right)\]
Ok, that's first order solvable via integrating factor.  The solution is in the form: $ x=\phi(p,C)$.  Now, substitute this equation for $ x$ in the differential equation and obtain the parametric solution:\begin{align*}
x&=\phi(t,C)\\
y&=\phi(t,C)F(t)+G(t)
\end{align*}Anyway, if I liked differential equations, that's what I'd be heading for.  A plot of a solution with real data wouldn't hurt neither but I don't want to push it.

See: Handbook of Differential Equations under Lagrange's Equation.


inequalities
search
inequalities proposed

$ a,b,c\,\succeq\,0$  Prove that :

$ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\,\succeq\,ab \plus{} ac \plus{} bc$
[color=red]SORRY!![/color]
with   $ a\plus{}b\plus{}c\equal{}3$

[quote="memath"]$ a,b,c\,\succeq\,0$  Prove that :

$ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\,\succeq\,ab \plus{} ac \plus{} bc$[/quote]

it is wrong $ a\equal{}1,b\equal{}4,c\equal{}9$

[quote="zaya_yc"][quote="memath"]$ a,b,c\,\succeq\,0$  Prove that :

$ \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\,\succeq\,ab \plus{} ac \plus{} bc$[/quote]

it is wrong $ a \equal{} 1,b \equal{} 4,c \equal{} 9$[/quote]
sorry.I've added a necessary condition.

note that $ ab\plus{}bc\plus{}ca\equal{}\frac{9\minus{}(a^2\plus{}b^2\plus{}c^2)}2$ so we have to show that:

$ \sqrt a\plus{}\sqrt b\plus{}\sqrt c\geq \frac{9\minus{}(a^2\plus{}b^2\plus{}c^2)}2$

$ \iff 2(\sqrt a\plus{}\sqrt b\plus{}\sqrt c)\plus{}a^2\plus{}b^2\plus{}c^2\geq 9$

now note that by AM-GM we have:

$ a^2\plus{}\sqrt a\plus{}\sqrt a\geq 3a$

summing up all these inequalities we get:

$ 2(\sqrt a\plus{}\sqrt b\plus{}\sqrt c)\plus{}a^2\plus{}b^2\plus{}c^2\geq 3(a\plus{}b\plus{}c)\equal{}9$

QED

It's Russia 2002,Serbia 2005
See here: http://www.mathlinks.ro/viewtopic.php?search_id=1719252527&t=155159


function
integration
Euler
geometry
geometric transformation
factorial
real analysis

Prove that
\[ \sum_{n\equal{}0}^{\infty}\displaystyle\frac{1}{\Gamma\left(n\plus{}3/2\right)}\equal{}\frac{2e}{\sqrt{\pi}}\int_0^1e^{\minus{}x^{2}}dx,\]
and
\[ \sum_{n\equal{}0}^{\infty}\frac{\Gamma\left(\minus{}n\plus{}1/2\right)\Gamma\left(n\plus{}1/2\right)}{\Gamma\left(n\plus{}3/2\right)}\equal{}\frac{\sqrt{\pi}}{2e}\int_0^1e^{x^{2}}dx,\]
where $ \displaystyle\Gamma(x)\equal{}\int_0^{\infty}e^{\minus{}t}t^{x\minus{}1}dt$ is the Euler Gamma function.

For the first. Using Series Expansion of Incomplete Gamma:
\[ \Gamma (a \minus{} 1) \minus{} \Gamma (a \minus{} 1,x) \equal{} x^a \left(\frac {1}{(a \minus{} 1) x} \minus{} \frac {1}{a} \plus{} \frac {x}{2 (a \plus{} 1)} \minus{} \frac {x^2}{6 (a \plus{} 2)} \plus{} \frac {x^3}{24 (a \plus{} 3)} \minus{} \frac {x^4}{120 (a \plus{} 4)} \plus{} \frac {x^5}{720 (a \plus{} 5)} \plus{} O\left(x^6\right)\right)\]
Multiplying the equation by $ x^{(1 \minus{} a)}$ to make things more polynomial.
\[ x^{1 \minus{} a} (\Gamma (a \minus{} 1) \minus{} \Gamma (a \minus{} 1,x)) \equal{} \left(\frac {1}{a \minus{} 1} \minus{} \frac {x}{a} \plus{} \frac {x^2}{2 (a \plus{} 1)} \minus{} \frac {x^3}{6 (a \plus{} 2)} \plus{} \frac {x^4}{24(a \plus{} 3)} \minus{} \frac {x^5}{120 (a \plus{} 4)} \plus{} O\left(x^6\right)\right)\]
The magic happens when you multiply by $ e^x$, It will make a translation in the $ x$ terms. Or in series form
\[ \left(\sum _{j \equal{} 0}^{\infty } \frac {x^j}{j!}\right) \left(\sum _{i \equal{} 0}^{\infty } \frac {( \minus{} 1)^i x^i}{i! (a \plus{} i \minus{} 1)}\right) \equal{} \sum _{k \equal{} 0}^{\infty } \sum _{i \equal{} 0}^k \frac {( \minus{} 1)^i x^k}{i! (a \plus{} i \minus{} 1) (k \minus{} i)!}\]

\[ e^x x^{1 \minus{} a} (\Gamma (a \minus{} 1) \minus{} \Gamma (a \minus{} 1,x)) \equal{} \frac {1}{a \minus{} 1} \plus{} \frac {x}{(a \minus{} 1) a} \plus{} \frac {x^2}{(a \minus{} 1) a (a \plus{} 1)} \plus{} \frac {x^3}{(a \minus{} 1) a (a \plus{} 1) (a \plus{} 2)} \plus{} \frac {x^4}{(a \minus{} 1) a (a \plus{} 1) (a \plus{} 2) (a \plus{} 3)} \plus{} O\left(x^5\right)\]
Finally eliminating the $ a \minus{} 1$.
\[ (a \minus{} 1) e^x x^{1 \minus{} a} (\Gamma (a \minus{} 1) \minus{} \Gamma (a \minus{} 1,x)) \equal{} 1 \plus{} \frac {x}{a} \plus{} \frac {x^2}{a (a \plus{} 1)} \plus{} \frac {x^3}{a (a \plus{} 1) (a \plus{} 2)} \plus{} \frac {x^4}{a (a \plus{} 1) (a \plus{} 2) (a \plus{} 3)} \plus{} O\left(x^5\right)\]

\[ (\Gamma ( \minus{} 1 \plus{} a) \minus{} \Gamma ( \minus{} 1 \plus{} a,x)) x^{1 \minus{} a} e^x (a \minus{} 1) \equal{} \sum _{i \equal{} 0}^{\infty } \frac {x^i (a \minus{} 1)!}{(i \plus{} a \minus{} 1)!}\]
Dividing by $ \Gamma (a)$ and using gamma function instead of factorial:
\[ \sum _{i \equal{} 0}^{\infty } \frac {x^i}{\Gamma (a \plus{} i)} \equal{} \frac {( \minus{} 1 \plus{} a) e^x x^{1 \minus{} a} (\Gamma ( \minus{} 1 \plus{} a) \minus{} \Gamma ( \minus{} 1 \plus{} a,x))}{\Gamma (a)}\]
The left and rigth side is valid for all positive real number $ a$.
Making $ x \equal{} 1$ $ a \equal{} \frac {3}{2}$ and using Erf function instead of the incomplete gamma, it's donne.

For the second just make $ x \to \minus{}x$

[quote="EulerIntegral"]For the first. Using Series Expansion of Incomplete Gamma:
\[ \Gamma (a \minus{} 1) \minus{} \Gamma (a \minus{} 1,x) \equal{} x^a \left(\frac {1}{(a \minus{} 1) x} \minus{} \frac {1}{a} \plus{} \frac {x}{2 (a \plus{} 1)} \minus{} \frac {x^2}{6 (a \plus{} 2)} \plus{} \frac {x^3}{24 (a \plus{} 3)} \minus{} \frac {x^4}{120 (a \plus{} 4)} \plus{} \frac {x^5}{720 (a \plus{} 5)} \plus{} O\left(x^6\right)\right)\]
[/quote]
this is not interesting  :( .

problem is very easy if you know just definition of $ e^{x}$ and something about beta function.
(the second one I just left you  as exercises , it easly follows from the solution of the first)
$ \frac {2e}{\sqrt {\pi}}\int_{0}^{1}e^{ \minus{} x^{2}}dx \equal{} \frac {2}{\sqrt {\pi}}\int_{0}^{1}e^{ \minus{} x^{2} \plus{} 1}dx \equal{} \frac {2}{\sqrt {\pi}}\int_{0}^{1} \sum \frac {(1 \minus{} x^{2})^{n}}{n!} \equal{} \sum \frac {1}{\Gamma(n \plus{} 3/2)}$
so you just need to note that $ \int_{0}^{1}\frac {(1 \minus{} x^{2})^{n}}{n!}dx \equal{} \frac {1}{n!2}\int_{0}^{1}(1 \minus{} t)^{n}t^{ \minus{} 1/2} \equal{} \frac {\beta (n \plus{} 1,1/2)}{n!2} \equal{} \frac {\sqrt {\pi}}{2\Gamma(n \plus{} 3/2)}$


[quote="Ignite168"]I could not find anything about brute force on the website, but if you can prove that there exist only a certain amount of cases that will work and brute force them all, would you be awarded full points if it is thorough? While I know it definitely wouldn't be elegant, I was wondering if it would net you full points at least.[/quote]

I'm pretty sure thorough brute force will get you full points because I remember I got a 5 for a problem I brute forced. It wasn't commended, but it's still a 5.

[quote="nr1337"]This begs the question:  If you have a brute force solution and a non-brute-force solution that's harder to explain, which should you submit?  (I have 2 brute force solutions, a C program, and a real solution to one of the problems and it probably took me more time to figure out how to write the real solution than it did to do the 2 brute force solutions by hand and write and debug the C program) :D

This is a serious question, by the way.[/quote]

This is also the type of question I believe they don't want people to discuss while the competition is underway (i.e. you should make that decision yourself).

Brute force solutions get full credit as long as they don't have any flaws.  These types of solutions pretty much have 0 chance of being commended though.

[quote="joml88"]
Brute force solutions get full credit as long as they don't have any flaws.  These types of solutions pretty much have 0 chance of being commended though.[/quote]
Well, if you're also the only one to get the problem right, they'd probably commend you with a brute-force solution  :D

[quote="nr1337"]This begs the question:  If you have a brute force solution and a non-brute-force solution that's harder to explain, which should you submit?  (I have 2 brute force solutions, a C program, and a real solution to one of the problems and it probably took me more time to figure out how to write the real solution than it did to do the 2 brute force solutions by hand and write and debug the C program) :D

This is a serious question, by the way.[/quote]
Joml's right that this question is getting pretty close to discussing the problems, so be careful.  If you're sure that your more elegant solution is correct, then use it, since graders will probably like it more and it could get commended.  If you're not sure, then probably for the sake of points sending one you know is right is the better thing to do (don't just send both solutions - it tends to annoy graders if they have to figure out which of your solutions you intend them to look at, and if one is right and one is wrong then it's more difficult to grade).


Suppose that given is a row: $u_{n}=n^{p}x^{n}\ (p \in \&\mathbb{N}\& , x>0)$

find the limit if $x\rightarrow \infty$

so I think to look to 3 cases first suppose that $x<1 \ and \ x>0$ then we get somthing like $\lim_{x\rightarrow \infty}\frac{n^{p}}{10^{n}}$ but then we have a very large denominator and we get 0 (zero) as the limit.

second we can get the x will bigger than 1 then we get somthing like $n^{p}$ and this will have now limit.

at last we can have that x=1 and then we have also now limit.

Does the argue hold? in orther words is this correct?

Thanks a lot.

did you mean the limit of $n$ going to infinity and $p$ is fixed right?

$x$ also is fixed, no? You wrote $x\to\infty$.

sorry of cours we have $\lim_{n\rightarrow \infty}...$

x will fixed.

Thanks.


Here are a few problems I seem completely stuck with:

1) A platform of mass $m_0$ starts moving on a frictionless horizontal surface due to a constant force $\vec F$ acting on it. We pour sand on this platform with a constant speed $\mu$ (representing the mass of sand dropped onto the platform in unit time). Find the velocity and the acceleration of the platform as a function of time. Neglect all friction.

2) Two bodies of masses $m_1$ and $m_2$ are connected by a massless spring and placed on a horizontal surface. The coefficient of friction between the surface and each body is $\mu$. What is the minimal horizontal force which must act on one body in order for the other body to start moving?

3) A small ball of mass $m$ is attached to a massless string of length $l$ and is left to hang vertically. At one point, the other end of the string (label it $A$) starts moving horizontally with constant speed. What is the minimal speed end $A$ must have for the ball to start moving on a circle with a center at $A$?

Any help would be appreciated :)

[quote="Djole"]
1) A platform of mass $m_0$ starts moving on a frictionless horizontal surface due to a constant force $\vec F$ acting on it. We pour sand on this platform with a constant speed $\mu$ (representing the mass of sand dropped onto the platform in unit time). Find the velocity and the acceleration of the platform as a function of time. Neglect all friction.[/quote]

[color=blue] We have the Newton's second law

$\large d(mv)=Fdt$

Consequently, $\large \int d(mv)=\int Fdt$

$\large \Longrightarrow mv=Ft+C$

When $t=0, v=0 \Longrightarrow C=0$

$\Longrightarrow v=\frac{Ft}{m}=\frac{Ft}{m_0+\mu t}$

And $\large a=\frac{dv}{dt}$[/color]

That looks right. So the platform's final speed will be $\frac{F}{\mu}$.
For the third problem, enter the reference frame of point $A$. The speed of the mass at the bottom of the circle will be the same as the speed of $A$ in the original reference frame. At the top of the circle:
$g = \frac{v_t^2}{l}$
$v_t = \sqrt{gl}$
So by conservation of energy, the speed at the bottom is $v = \sqrt{5 g l}$.

That seems to make sense. Nice solution.

Thank you for the solutions!

For problem 3, I did enter the reference frame of point $A$; my mistake was that I thought that the speed of the pendulum at the top of its path is zero...

Any candidates for solving the second problem?

Sorry if this is stupid and I'm completely missing something.

For 2, I'm assuming $\mu$ is the coefficient of kinetic friction for $m_1$ and the coefficient of static friction for $m_2$. Arbitrarily place $m_1$ on the left and $m_2$ on the right and have the horizontal force $\vec F$ be to the right on $m_1$. Let $\Delta x$ be the compression of the spring. Then, $Fnet,m_1=F-\mu m_1 g-k \Delta x=0$ (you want the minimum force so push the $m_1$ at constant velocity), and $Fnet,m_2=k \Delta x-\mu m_2 g=0$ (just before moving). Adding the two equations, you get $F-\mu m_1 g-\mu m_2 g=0$, so $F=\mu m_1 g+\mu m_2 g$.

Actually, I think this works without my assumption. The minimum $\vec F$ needed would be when $\Delta x$ is just enough to counteract $\mu m_2 g$. $\vec F$ on $m_1$ would also have to counteract $\mu m_1 g+k \Delta x$. The acceleration is 0 just before $\vec F$ is large enough to cause $m_1$ to budge forward a bit to cause the compression to be more than $\Delta x$.

[quote="Djole"]3) A small ball of mass $m$ is attached to a massless string of length $l$ and is left to hang vertically. At one point, the other end of the string (label it $A$) starts moving horizontally with constant speed. What is the minimal speed end $A$ must have for the ball to start moving on a circle with a center at $A$?[/quote]

Consider the point at rest and the ball starts moving with the speed $v$ in the opposite direction (just a different inertial system). For the ball to reach the top of the circle with zero velocity, the potential energy $2mgl$ at the top must be equal to the kinetic energy $\frac 1 2 mv^2$ at the bottom, $2mgl = \frac 1 2 mv^2,\ v = 2 \sqrt{gl}.$

@[b]yetti[/b]: Yes, that's the incorrect answer I got :) [b]durt[/b]'s solution is correct...

@[b]AnnieS1527[/b]: That's the answer I first came up with -- needless to say, it's not right and I have no idea why. The right answer is [hide]$F = \mu g(m_1 + m_2/2)$[/hide]. Can anyone get this answer?

Yes it is incorrect, the ball would fall down from the top of the loop. So the centrifugal force at the top must balance the gravitational force.

$\frac{mv_T^2}{l} = mg,\ \frac 1 2 mv_T^2 = \frac 1 2 mgl$. If $v$ is the sought after speed at the bottom, then

$\frac 1 2 mv^2 = \frac 1 2 mv_T^2 + 2mgl = \frac 5 2 mgl$

$v = \sqrt{5gl}$

[quote="Djole"]2) Two bodies of masses $m_1$ and $m_2$ are connected by a massless spring and placed on a horizontal surface. The coefficient of friction between the surface and each body is $\mu$. What is the minimal horizontal force which must act on one body in order for the other body to start moving?
[/quote]

Let k be the spring constant. Let a horizontal force $F_1 > \mu m_1 g$ act on body 1, while body 2 is at rest. The equation of motion of body 1 is

$m_1 \ddot x_1 = F_1 - \mu m_1g - k x_1$

$\ddot x_1 + \frac{k}{m_1}\ x_1 = \frac{F_1 - \mu m_1g}{m_1}$

The general solution of this differential equation is

$x_1(t) = A \sin(\omega t) + B \cos (\omega t) + \frac{F_1 - \mu m_1 g}{k}$

where $\omega = \sqrt{\frac{k}{m_1}}$. The initial conditions are $x(0) = 0,\ \dot x(0) = 0$, hence $B = -\frac{F_1 - \mu m_1 g}{k},\ A = 0$ and

$x_1(t) = \frac{F_1 - \mu m_1 g}{k}\ \left(1 - \cos (\omega t)\right)$

$\dot x_1(t) = \frac{(F_1 - \mu m_1 g) \omega}{k}\ \sin (\omega t)$

Body 1 stops again at $\tau= \frac T 2 = \frac \pi \omega$. At this time, its position is

$x_1(\tau) = \frac{F_1 - \mu m_1 g}{k}\ \left(1 - \cos \pi \right) = \frac{2(F_1 - \mu m_1 g)}{k}$

and the tension in the spring is

$T = kx_1(\tau) = 2(F_1 - \mu m_1 g)$

If body 2 is not to move, it is necessary

$T \le \mu m_2 g$

$2(F_1 - \mu m_1 g) \le \mu m_2 g$

$F_1 \le \mu g \left(m_1 + \frac{m_2}{2}\right)$

Conversely, if body 2 is to move, it must be

$F_1 > \mu g \left(m_1 + \frac{m_2}{2}\right)$


I don't understand how to make a poll, when I try it dosen't work.

go here. sigh http://www.artofproblemsolving.com/wforum/faq.php#17


find in  $ \mathbb{N}^*$
$ x\plus{}y^2\equal{}x^2\plus{}y$

[quote="mathmen"]find in  $ \mathbb{N}^*$
$ x \plus{} y^2 \equal{} x^2 \plus{} y$[/quote]

there is no solution beside $ x \equal{} y$


function
trigonometry
complex analysis
complex analysis unsolved

Prove that for $z\in\mathbb{C}\setminus\mathbb{Z}$, $f(z)=\frac{\pi^{2}}{\sin^{2}\pi{z}}$ it holds that $f(z)=\sum_{n\in\mathbb{Z}}\frac{1}{(z-n)^{2}}$


function
integration
logarithms
linear algebra
matrix
calculus
calculus computations

(a)  The function $ f$ is continuous on $ [0,1]$ such that $ f(x)>0$.  Find an equivalent of:

$ h(\epsilon)=\int_{0}^{1}\frac{f(x)}{x^{2}+\epsilon^{2}}dx$ when $ \epsilon \rightarrow 0$ and when $ \epsilon \rightarrow \infty$

(b)  Assume $ f$ is differentiable at $ 0$, that $ f(0)=0$, and $ f'(0) \neq 0$.  Find an equivalent of $ h(\epsilon)$ when $ \epsilon \rightarrow 0$.

(a) $ f$ reaches a positive minimum $ m$ and maximum $ M$ on the compact interval $ [0,1]$ .
$ h(\epsilon)\ge \int_{0}^{1}\frac{m}{x^{2}+\epsilon^{2}}=\frac{m}{\epsilon}\arctan \frac{1}{\epsilon}\to \infty$ for $ \epsilon\to 0$ and $ h(\epsilon)\le \int_{0}^{1}\frac{M}{x^{2}+\epsilon^{2}}=\frac{M}{\epsilon}\arctan \frac{1}{\epsilon}\to 0$ for $ \epsilon\to \infty$ .

(b) Is for example $ f(x)=2x$ so $ h(\epsilon)=\int_{0}^{1}\frac{2x}{x^{2}+\epsilon^{2}}dx=\log\left(1+\frac{1}{\epsilon^{2}}\right)\to \infty$ for $ \epsilon\to\infty$ .
Is for another example $ f(x)=\left\{ \begin{matrix}x^{2}& \text{for}& x\in\Bbb{Q}\\ 0 & \text{for}& x\notin\Bbb{Q}\end{matrix}\right.$ so $ h(\epsilon)=0 \; \forall \epsilon\neq 0$ and therefore $ h(\epsilon)\to 0$ for $ \epsilon\to 0$ .


Let $ a,b,c>0$ and different. Prove that

$ \sum\frac{a(b\plus{}c)(b^2\plus{}c^2)}{(b^2\minus{}c^2)}\geq\frac{7}{4}$

______________________
 :blush:  Panagiote Ligouras

[quote="Ligouras"]Let $ a,b,c > 0$ and different. Prove that

$ \sum\frac {a(b \plus{} c)(b^2 \plus{} c^2)}{(b^2 \minus{} c^2)}\geq\frac {7}{4}$

______________________
 :blush:  Panagiote Ligouras[/quote]
Hmm.. If $ f(a,b,c) \equal{} LHS$ then $ f(a,b,c) \plus{} f(a,c,b) \equal{} 0$, and hence it cannot be true. Besides $ f(ta,tb,tc) \equal{} t^2f(a,b,c)$. So it is not true because it is not homogenous. Maybe you meant this?

$ \left |\sum_{cyc} \frac {a(b \plus{} c)(b^2 \plus{} c^2)}{(b^4 \minus{} c^4)} \right | \geq\frac {7}{4}$

(Btw. this is equivalent to: $ \left | \sum_{cyc} \frac{a}{b\minus{}c} \right | \ge \frac{7}{4}$)

[quote="Ligouras"][quote="Mathias_DK"][quote="Ligouras"]Let $ a,b,c > 0$ and different. Prove that

$ \sum\frac {a(b \plus{} c)(b^2 \plus{} c^2)}{(b^2 \minus{} c^2)}\geq\frac {7}{4}$

______________________
 :blush:  Panagiote Ligouras[/quote]
Hmm.. If $ f(a,b,c) \equal{} LHS$ then $ f(a,b,c) \plus{} f(a,c,b) \equal{} 0$, and hence it cannot be true. Besides $ f(ta,tb,tc) \equal{} t^2f(a,b,c)$. So it is not true because it is not homogenous. Maybe you meant this?

$ \left |\sum_{cyc} \frac {a(b \plus{} c)(b^2 \plus{} c^2)}{(b^4 \minus{} c^4)} \right | \geq\frac {7}{4}$

(Btw. this is equivalent to: $ \left | \sum_{cyc} \frac {a}{b \minus{} c} \right | \ge \frac {7}{4}$)[/quote]

Dear mathias is correct:

$ \sum\frac {a(b \plus{} c)(b^2 \plus{} c^2)}{(b^2 \minus{} c^2)^2}\geq\frac {7}{8}$[/quote]
Let $ f(a,b,c) \equal{} \sum_{cyc} \frac {a(b^2 \plus{} c^2)}{(b \minus{} c)^2(b \plus{} c)} \equal{} LHS$. It's easy to prove $ f(a,b,c) \le f(a \plus{} k,b \plus{} k,c \plus{} k)$ when $ a,b,c \ge 0$. Wlog $ c \le a,b$. Hence $ f(a \minus{} c,b \minus{} c,0) \le f(a,b,c)$. Setting $ x \equal{} a \minus{} c, y \equal{} b \minus{} c$, we have:
$ f(x,y,0) \equal{} \frac {x}{y} \plus{} \frac {y}{x} \ge 2 > \frac {7}{4} > \frac{7}{8}$. :D

I had had a similar result:
Given $ a, b, c$ are the reals. Prove that:
$ \frac{(a\plus{}b)^2(a\plus{}c)^2}{(b^2\minus{}c^2)^2}\plus{}\frac{(b\plus{}c)^2(b\plus{}c)^2}{(c^2\minus{}a^2)^2}\plus{}\frac{(c\plus{}a)^2(c\plus{}b)^2}{(a^2\minus{}b^2)^2} \geq\ 2$
 :)

[quote="nguoivn"]I had had a similar result:
Given $ a, b, c$ are the reals. Prove that:
$ \frac {(a \plus{} b)^2(a \plus{} c)^2}{(b^2 \minus{} c^2)^2} \plus{} \frac {(b \plus{} c)^2(b \plus{} c)^2}{(c^2 \minus{} a^2)^2} \plus{} \frac {(c \plus{} a)^2(c \plus{} b)^2}{(a^2 \minus{} b^2)^2} \geq\ 2$
 :)[/quote]

it was my real problem NGUOIVN.... :wink:

according to you it exists k >2 such that

$ \sum\frac {a(b \plus{} c)(b^{2} \plus{} c^{2})}{(b^{2} \minus{} c^{2})^2}\geq\ k$

k=2 is sure

P. Ligouras

[quote="Mathias_DK"][/quote][quote="Ligouras"][quote="Mathias_DK"][quote="Ligouras"]Let $ a,b,c > 0$ and different. Prove that

$ \sum\frac {a(b + c)(b^2 + c^2)}{(b^2 - c^2)}\geq\frac {7}{4}$

______________________
 :blush:  Panagiote Ligouras[/quote]
Hmm.. If $ f(a,b,c) = LHS$ then $ f(a,b,c) + f(a,c,b) = 0$, and hence it cannot be true. Besides $ f(ta,tb,tc) = t^2f(a,b,c)$. So it is not true because it is not homogenous. Maybe you meant this?

$ \left |\sum_{cyc} \frac {a(b + c)(b^2 + c^2)}{(b^4 - c^4)} \right | \geq\frac {7}{4}$

(Btw. this is equivalent to: $ \left | \sum_{cyc} \frac {a}{b - c} \right | \ge \frac {7}{4}$)[/quote]

Dear mathias is correct:

$ \sum\frac {a(b + c)(b^2 + c^2)}{(b^2 - c^2)^2}\geq\frac {7}{8}$[/quote][quote="Mathias_DK"]
Let $ f(a,b,c) = \sum_{cyc} \frac {a(b^2 + c^2)}{(b - c)^2(b + c)} = LHS$. It's easy to prove $ f(a,b,c) \le f(a + k,b + k,c + k)$ when $ a,b,c \ge 0$. Wlog $ c \le a,b$. Hence $ f(a - c,b - c,0) \le f(a,b,c)$. Setting $ x = a - c, y = b - c$, we have:
$ f(x,y,0) = \frac {x}{y} + \frac {y}{x} \ge 2 > \frac {7}{4} > \frac {7}{8}$. :D[/quote]

Thank you Mathias very easy


[quote="solafidefarms"]Treething had posted this:

Bob the Builder had a bag of dozen bananas on sale for 3 dollars. He adds some bananas to the bag, but does not change the price tag. Now how much does a dozen bananas cost?

http://media.putfile.com/banana39

but the page got all messed up, so I've reposted it.[/quote]

  The wording is a bit vague but I'll give it a shot:

[hide="Answer"]

  Suppose bob adds $x$ bananas.

  Now the price of a dozen bananas is:

$\frac{12}{12+x} \cdot  3$          or           $\frac{36}{12+x}$ dollars. [/hide]

can the answer b just a number?, cuz it doesnt ask us the equation, it asks us the cost. if it asks us the [i]actual cost[/i], then $not$ $enough$ $information$.

oops, i was wrong, this prob is solvable. o yeah, sry 2 double post, but i forgot 2 add mi answer.
[hide="solution"]$\frac{12}{x+12}$=$\frac{1}{3}$
                       X+12=36.
                       x=24 for regular price. $\frac{24}{36}$ =$\frac{2}{3}$

 therefore 1/3 of 3 is 1, which equals a dozen bananas. a dozen bananas without a sale is therefore 1 dollar.[/hide]

He does not specify how many banana's were added, assuimg they are all sold at a uniform price.

it costs $\frac{36}{12+b}$ if b is the number of bananas he added.  If my mom met Bob, she would buy all the bananas, they are pretty cheap :D


induction
number theory solved
number theory

I found this nice-looking problem:

Prove that there are infinitely many composite numbers n s.t. n|3^(n-1)-2^(n-1). I've found that n=3^p-2^p with p prime verifies this, but I don't know if it can be proven that there are infinitely many composite numbers of the anbove form. Any other ideas? I think this is the wrong way to go...

I had an idea for this problem when I saw this one :

(*) 2^(n+2) | 3^(2^n) - 1 (easy to prove by induction)

Then if N = 3^(2^n) - 2^(2^n), let's prove that N | 3^(N-1) - 2^(N-1).

Since a-b|a^k-b^k it's enough to prove that N-1 is divisible by 2^n and that's where we use (*)  :D


The convex quadrilateral $ ABCD$ is inscribed in a circle of center $ O$ and let $ E=AC\cap BD$. Prove that if the midpoints of $ AD,BC$ and $ OE$ are collinear, then either $ AB=CD$ or $ \angle AEB = 90^{\circ}$

[hide="Hint."]In any convex quadrilateral $ ABCD$ the following theorem is true:
If $ E$ is an intersection of diagonals and $ P$ is an arbitrary point inside $ ABCD$ then midpoints of $ AD, BC, EP$ are collinear if and only if $ [ABP]=[CDP]$ (areas). With this theorem the problem becomes trivial.
[hide="Hint to the hint."]To prove this theorem, show at first that locus of points $ P$ such that $ [ABP]=[CDP]$ is a line through $ S$, where $ S$ is an intersection of lines $ AB$ and $ CD$.[/hide]
[/hide]


Prove that
\[a+b+c\geq a\cdot\frac{\sin{b}}{\sin{a}}+b\cdot\frac{\sin{c}}{\sin{b}}+c\cdot\frac{\sin{a}}{\sin{c}}\;\forall a,b,c\in (0,\frac{\pi}{2}). \]

Equiv to \[\sum a\left(\frac{\sin a-\sin b}{\sin a}\right)\geq 0 \] Now on that interval sine is increasing so it follows.
More precisely, let $S_{a}=\frac{\sin a-\sin b}{\sin a\cdot(a-b)}$ and cyclics then we must prove \[\sum a(a-b)S_{a}\geq 0\] moreover since its increasing, positive we must have $S_{a},S_{b},S_{c}\geq 0$

If $S_{a},S_{b},S_{c}\geq 0$ then $\sum \geq 0$. Why?  :maybe:

i think that
When $a\ge b \, then sina\ge sinb\, a,b\in(0;\pi/2)$
So $\sum\ge0$


trigonometry
calculus
calculus computations

i am not able to draw graph of sin[x] from sinx...where [x] represents greatest integer function....can anyone help me :(

Just split up the $ x$ axis. The curve is  equivalent to $ \sin1$ between $ 0$ and $ 1$, $ \sin2$ between $ 1$ and $ 2$ etc. You could make rough guesses as to how large $ \sin(n)$ is (for integer $ n$) by comparing $ n$ to nearby multiples of $ \pi$ but it doesn't seem worthwile, just evaluate them with a calculator.

Sorry for my reply, seriousstone89. I also find the graph of $ y=\sin [x]$ by substituting $ x=$ some numbers, such as
For $ 0\leq x<1\Longrightarrow \sin [x]=0$, $ 1\leq x<2\Longrightarrow \sin [x]=\sin1$, and so on.

kunny

one general method would be to draw the graph of $ f(x)$ and then draw vertical lines at distance$ 1,2,3 \dots$ from the origin .wherever it cuts the axis from their draw horizontal lines till the next point where the graph is cut by those vertical lines(may be above or below) of course excluding the point above or below that point.these lines are the graph of $ f([x])$
. well this is just a restatement and generalization of the fact that $ f([x])=f(n)$ for all $ n \leq x < n+1$ where $ n$ is an integer.


modular arithmetic
number theory
prime numbers
divisibility tests
number theory unsolved

Find all the triplets $ \{p, q, r\}$ of prime numbers such that $ pq\plus{}r$, $ qr\plus{}p$, and $ rp\plus{}q$  are all prime numbers also.

[quote="azjps"]If all $ p,q,r$ are odd, then the resulting numbers are even, so one number must be $ 2$. Checking $ \mod{3}$, $ \{p,q,r\} \equiv \{ \minus{} 1,1,1\},\{ \minus{} 1, \minus{} 1,1\},\{ \minus{} 1, \minus{} 1, \minus{} 1\} \pmod{3}$ all result in a resulting number being divisible $ 3$, so one of the numbers must be $ 3$. So we need all primes $ p$ such that $ p \plus{} 6, 3p \plus{} 2,2p \plus{} 3$ are primes, which can be filtered using various modulos, but there does not seem to be a small solution set (for example, $ \{2,3,1063\}$ works).[/quote]

Yes, I had found also that two of p, q, r must be 2 and 3. I suppose that there must exist infinite p such that  $ p \plus{} 6, 3p \plus{} 2,2p \plus{} 3$ are primes, but I don't know how to prove it u_u ...

If all $ p,q,r$ are odd, then the resulting numbers are even, so one number must be $ 2$. There isn't much more we can do besides restrict the other values using various modulos. For example, $ \{2,5,13\}$ and $ \{2,3,126013\}$ work.

Edit: I messed up; divisibility by $ 3$ is not necessary for $ \{p,q,r\} \equiv \{\minus{}1,\minus{}1,1\} \pmod{3}$.

What is the source of this problem?  It seems likely to be very hard unless I am missing a trick.

I dunno u_u... I thinked in the problem and I believed that it was easy u_u ... well, maybe it is, but I don't know how to prove that are infinite triplets =S (OMG! =O)

Suppose $ p,q,r>2$
then, clearly all three expression are even and greater than 2, thus not prime.

Thus wlog $ p\equal{}2$.

$ 2q\plus{}r$ is prime, so $ r>2$
$ 2r\plus{}q$ is prime, so $ q>2$

notice $ (2,3,5)$ $ (2,3,7)$ $ (2,5,7)$ $ (2,3,13)$ $ (2,3,17)$ works

Does this problem have an elementary solution? Or even any solution? What is the problem source?


AMC
AMC 10
algebra
difference of squares
special factorizations

How many ordered pairs $ (m,n)$ of positive integers, with $ m > n$, have the property that their squares differ by $ 96$?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 12$

[hide]
$(a+b)(a-b) = 96 = 2^{5}\cdot 3$, which has $(6)(2) = 12$ factors. We know that $a+b>a-b$ since $b\ne 0$. We need both factors to have the same parity so that $a = \frac{(a+b)+(a-b)}{2}$ is an integer. There would be 6 pairs, but 1 and 3 can't be paired with anything, so there are only 4. $\boxed{\text{B}}$
[/hide]

[hide]We are to find ordered pairs of positive integers $(m,\ n)$ such that $m^{2}-n^{2}=96\Longleftrightarrow (m+n)(m-n)=96.$ By $m>n>0$ and $m,\ n\in \mathbb{N},$ and $m=\frac{m+n}{2}+\frac{m-n}{2},\ n=\frac{m+n}{2}-\frac{m-n}{2},$ we have that parity of $m+n,\ m-n$ is coincides with that of $m,\ n.$ Thus $(m+n,\　m-n)=(48,\ 2),\ (24,\ 4),\ (16,\ 6),\ (12,\ 8).$ The desired answer is 4 pairs, $(B).$[/hide]

[hide="a 5th grade method"]Graph $y=\sqrt{x^{2}+94}$ on your graphing calc and count the number of integer solutions  :wink: [/hide]

[quote="Ignite168"]That's not 5th grade, more like 8th as well as basically impossible.

[hide="An 8th grade method that's actually possible"]
Factor into $(m+n)(m-n)=96$ $m+n$ must be bigger than $m-n$ and list out your pairs of factors. $(1,96), (2,48), (3,32) (4,24) (6,16) (8,12)$. So basically, you have 6 systems of equations and you want ones with integer solutions. I solved the systems of equations using augmented matrices on my calc.[/hide][/quote]
Or you could just add them...
I think he meant graph $y=\sqrt{96+x^{2}}$

[hide]
(m+n)(m-n)=96 \implies (m+n/2)(m-n/2)=24, with the first term greater, and 24 is just

1
2
3
4
6 <-- (then starts repeating factor pairs)
So 4 factors.

That was my exact logic on the test...
[/hide]

[hide]This can be done by the difference of two squares
0
1           -    1-0= 1
4            -    4-1=3
9           -     9-4=5
16         -     16-9=7
25                 .......
36                 .......
49                .......
64                ........

We see that these are consecutive odd integers 1,3,5,7,9...........
And 96 is a even integers --- o + o = e
so just try the numbers using these conditions[/hide]

[quote="King_James23"]This can be done by the difference of two squares
0
1           -    1-0= 1
4            -    4-1=3
9           -     9-4=5
16         -     16-9=7
25                 .......
36                 .......
49                .......
64                ........

We see that these are consecutive odd integers 1,3,5,7,9...........
And 96 is a even integers --- o + o = e
so just try the numbers using these conditions[/quote]
This method is vague and cumbersome... the list have to be extended up to 25 odd integers... like $25^{2}-23^{2}=96$...

[hide]m^2-n^2=96
Factor and get (m+n)(m-n)=96
96 = 2^5 x 3
6(2) = 12, there are 12 factors for 96, and 6 pairs.
Note that m-n is the lower value of the two; m+n and m-n.
m-n cannot yield 1 therefore lowering the possible amount of pairs by 1.
Now, take into consideration of the 3 in the factors of 96.
From the 3 you can deduce that m+n and m-n has to be all odd or all even.
Odd and odd yields even, even and even yields even, even and odd yields odd, odd subtracted from a higher odd yields even, even subtracted from a higher even yields even, and odd subtracted from a higher even yields odd and vice versa.
So lets say m-n denotes 3, that can only be true if a lower even or a lower odd is subtracted from a higher odd or a higher even..
Since m-n denotes 3, m+n denotes 32, an even number. However, a even added to a odd ( or vice versa ) cannot yield a even number. EX. 5+8 = 13.
Therefore the factor pairs 1,96 and 3,32 are not accounted for and lowers the total pairs by 2 so there are 4 pairs.[/hide]

Wait, you're allowed to use a calc on AMC10?

In 2007, calculators were allowed. However, starting 2008, calculators were banned. No AMC problems require a calculator.

[hide="Bashy method"]We are given that $m^2-n^2=96$.  We factor this as difference of squares, giving $(m+n)(m-n)=96$

The first term is larger than the second, so we list out pairs of numbers whose product is $96$ and where the first factor is larger than the second,  We get:

$(96,1),
(48,2),
(32,3),
(24,4),
(16,6),
(12,8)$

There are not too many values, so we just bash them out and find that $(48,2),(24,4),(16,6),$ and $(12,8)$ have positive integer values for $m$ and $n$.  This means our answer is $\boxed{\textbf{(B)}\ 4}$[/hide]

(m+n) (m-n)=96
Now factorize and do casework

[quote=ostriches88]please don't bump really old threads!

thanks[/quote]

Well, it is a C&P post, so it is okay as per the rules of the forum


A Senate committee has 8 Republicans and 6 Democrats.  In how many ways can we form a subcommittee of 5 members that has at least one member from each party?

it seems like the problem suggest that every republican is the same....then the answer would be 4

[quote="mathemonster"]it seems like the problem suggest that every republican is the same[/quote]

Then, they would be all twins.  :)

Eh Bijection's just meddling with the problem

8 ways to choose a republican, 6 for a democratic, and then (12 choose 5) to get the rest, and that multiplies out to 10560.

[quote="ernie"]I got something different...

For committees with no restrictions, you would simply have $ \binom{14}{5}$.  However, there are $ \binom{8}{5}$ ways to make the committee all Republican and $ \binom{6}{5}$ ways to make the committee all Democrat.

Thus, the answer is $ \binom{14}{5} \minus{} \binom{8}{5} \minus{} \binom{6}{5} \rightarrow 2002 \minus{} 56 \minus{} 6 \rightarrow 2002 \minus{} 62 \rightarrow \boxed{1940}$.[/quote]

I am pretty sure this is the correct way.

Just a suggestion, you might lock this.

[quote="fmasroor"]8 ways to choose a republican, 6 for a democratic, and then (12 choose 3) to get the rest, and that multiplies out to 10560.[/quote]

Why doesn't this work?

There are a total of $\binom{14}{5}=2002$ ways of selecting a subcommittee of 5 with no restrictions on the membership. Of these committees, the only ones that will violate the given condition are the ones that consist entirely of Republicans or entirely of Democrats. There are $\binom{8}{5}=56$ possible subcommittees that have all 5 members selected from among the 8 Republicans and $\binom{6}{5}=6$ possible subcommittees that have all 5 members selected from among the 6 Democrats. Subtracting the number of subcommittees that don't work from the total number of possible subcommittees gives us our answer: $2002-56-6=\boxed{1940}$.

[hide="Why can't you do:"]
6 Democrats * 8 Republicans *12 * 11 * 10 =63360
12*11*10 represents the rest of the members. [/hide]

Can someone help?

What did I do wrong?:

You must have 2 already picked out, reserved for the D's and R's. That's $8\cdot 6$ for the choices we have to do that.

Then we have 3 spots in which we pick whatever members we'd like. 7 R's left and 5 D's left, making a total of 12 to pick from. Pick 3, and $\binom{12}{3}=220$. Multiply it all together to get $\boxed{10560}$. What's wrong here? :o

You are overcounting. It would work if you had a "head" Democrat and a "head" Republican. Do you see what I mean?


I have a problem now.
Imagine a wall which is decreasing with a relativistic velocity $\upsilon_w=\beta c$. Light ray falls on this wall at an angle $\theta$ to the horizontal. Find the shadow speed.

I'll try:

Let's introduce a cartesian system with the origin at the bottom of the wall, and the current position of the top of the wall at $(0,\ y)$. Suppose that we are in an innertial system in which the bottom of the wall doesn't move.

Let $\tau_0 = 0$ be the moment when a ray of light starts from the top of the wall. It will reach the floor at the moment $\tau_1 = y/c\sin\theta$.

Now consider another ray which started from the top of the wall at time $\tau'_0 = \Delta t$, when the height of the wall was $y - \beta c\Delta t$. This ray will reach the floor at the moment $\tau'_1 = (y-\beta c\Delta t)/c\sin\theta$, so the shadow traversed a distance $v\Delta t'$, where $v$ is the speed of the shadow and $\Delta t' = \tau'_0 - \tau_0 + \tau'_1 - \tau_1 = \Delta t \frac{\sin\theta - \beta}{\sin\theta}$. We also have the equality \[ \frac{y-\beta c \Delta t}{x - v\Delta t'} = \frac{y}{x}(= \tan\theta) \Rightarrow \beta c\Delta t = v \tan\theta \Delta t' \Rightarrow v = \frac{\cos\theta}{\sin\theta - \beta}\beta c. \]

This reduces to the expected result $v = \cot\theta \beta c$ when $\beta \rightarrow 0$.


I have some interesting inequality, let's discuss them :P  :)
1. Let $ a, b, c, x, y, z > 0$ and
$ a \plus{} x \equal{} b \plus{} y \equal{} c \plus{} z \equal{} 1$
Prove that $ (abc \plus{} xyz)(\frac {1}{ay} \plus{} \frac {1}{bz} \plus{} \frac {1}{cx}) \geq 3$
2. Let $ a,b,c,d > 0$
prove that: $ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{a \plus{} d} \plus{} \frac {d \minus{} a}{a \plus{} b}\geq0$
3. Let A=$ \frac {a^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c^2}{c^2 \plus{} (a \plus{} b)^2}$
$ a,b,c \in R$\{0}
minA=?

[quote="marsupilami"]
3. Let A=$ \frac {a^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c^2}{c^2 \plus{} (a \plus{} b)^2}$
$ a,b,c \in R$\{0}
minA=?[/quote]
I'll prove that $ \sum_{cyc}\frac {a^2}{a^2 \plus{} (b \plus{} c)^2}\geq\frac {3}{5}.$
$ \sum_{cyc}\frac {a^2}{a^2 \plus{} (b \plus{} c)^2} \equal{} \sum_{cyc}\frac {a^4}{a^4 \plus{} a^2b^2 \plus{} a^2c^2 \plus{} 2a^2bc}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^4 \plus{} 2a^2b^2 \plus{} 2a^2bc)}.$
Id est, it remains to prove that $ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^4 \plus{} 2a^2b^2 \plus{} 2a^2bc)}\geq\frac {3}{5}.$ But we obtain:
$ \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^4 \plus{} a^2b^2 \plus{} 2a^2bc)}\geq\frac {3}{5}\Leftrightarrow\sum_{cyc}(2a^4 \plus{} 4a^2b^2 \minus{} 6a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2 \minus{} b^2)^2 \plus{} 3\sum_{cyc}c^2(a \minus{} b)^2\geq0.$
$ a\equal{}b\equal{}c$ gives equality.
Thus, $ \min_{\{a,b,c\}\subset\mathbb R\setminus\{0\}}A \equal{} \frac {3}{5}.$ :)

[quote="marsupilami"]I have some interesting inequality, let's discuss them :P  :)

2. Let $ a,b,c,d > 0$
prove that: $ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{a \plus{} d} \plus{} \frac {d \minus{} a}{a \plus{} b}\geq0$

[/quote]

$ \frac {a \minus{} b}{b \plus{} c} \plus{} \frac {b \minus{} c}{c \plus{} d} \plus{} \frac {c \minus{} d}{a \plus{} d} \plus{} \frac {d \minus{} a}{a \plus{} b}\equal{}\frac{a\plus{}c}{b\plus{}c}\plus{}\frac{b\plus{}d}{c\plus{}d}\plus{}\frac{c\plus{}a}{a\plus{}d}\plus{}\frac{d\plus{}b}{a\plus{}b}\minus{}4\equal{}(a\plus{}c)(\frac{1}{b\plus{}c}\plus{}\frac{1}{a\plus{}d})\plus{}(b\plus{}d)(\frac{1}{c\plus{}d}\plus{}\frac{1}{a\plus{}b})\minus{}4\equal{}\frac{(a\plus{}c)(a\plus{}b\plus{}c\plus{}d)}{(b\plus{}c)(a\plus{}d)}\plus{}\frac{(b\plus{}d)(a\plus{}b\plus{}c\plus{}d)}{(c\plus{}d)(a\plus{}b)}\minus{}4\geq\frac{(a\plus{}c)(a\plus{}b\plus{}c\plus{}d)}{(\frac{b\plus{}c\plus{}a\plus{}d}{2})^{2}}\plus{}\frac{(b\plus{}d)(a\plus{}b\plus{}c\plus{}d)}{(\frac{c\plus{}d\plus{}a\plus{}b}{2})^{2}}\minus{}4\equal{}\frac{4(a\plus{}c)}{a\plus{}b\plus{}c\plus{}d}\plus{}\frac{4(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d}\equal{}\frac{4(a\plus{}b\plus{}c\plus{}d)}{a\plus{}b\plus{}c\plus{}d}\minus{}4\equal{}4\minus{}4\equal{}0$

$ a\equal{}b\equal{}c\equal{}d$

[quote="marsupilami"]3. Let A=$ \frac {a^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c^2}{c^2 \plus{} (a \plus{} b)^2}$
$ a,b,c \in R$\{0}
minA=?[/quote]

"Isolated fudging" also works nicely here.

WOLOG, assume that $ a \plus{} b \plus{} c \equal{} 3$.  Then
\[ \sum \frac{a^2}{a^2 \plus{} (b \plus{} c)^2} \equal{} \sum \frac{a^2}{a^2 \plus{} (3 \minus{} a)^2} \equal{} \sum \frac{a^2}{2a^2 \minus{} 6a \plus{} 9}.\]

Let
\[ f(x) \equal{} \frac{x^2}{2x^2 \minus{} 6x \plus{} 9}.\]
Then
\[ f'(x) \equal{} \frac{6x(3 \minus{} x)}{(2x^2 \minus{} 6x \plus{} 9)^2},\]
so $ f$ is increasing on the interval $ [0,3]$.

Since
\[ f(x) \minus{} \frac{7x \minus{} 2}{25} \equal{} \frac{x^2}{2x^2 \minus{} 6x \plus{} 9} \minus{} \frac{12x \minus{} 7}{25} \equal{} \frac{3(x \minus{} 1)^2 (21 \minus{} 8x)}{25(2x^2 \minus{} 6x \plus{} 9)},\]
we have that
\[ f(x) \ge \frac{7x \minus{} 2}{25}\]
for $ 0 \le x \le 21/8$.

If $ 21/8 \le a \le 3$, then $ f(a) \ge f(21/8) \equal{} 49/50$, so $ f(a) \plus{} f(b) \plus{} f(c) \ge 49/50 > 3/5$.  Similarly if $ 21/8 \le b \le 3$ and $ 21/8 \le c \le 3$.  Otherwise, $ a$, $ b$, $ c \le 21/8$, so
\[ f(a) \plus{} f(b) \plus{} f(c) \ge \frac{7(a \plus{} b \plus{} c) \minus{} 6}{25} \equal{} \frac{15}{25} \equal{} \frac{3}{5}.\]

[quote="marsupilami"]I have some interesting inequality, let's discuss them :P  :)
1. Let $ a, b, c, x, y, z > 0$ and
$ a \plus{} x \equal{} b \plus{} y \equal{} c \plus{} z \equal{} 1$
Prove that $ A\equal{}(abc \plus{} xyz)(\frac {1}{ay} \plus{} \frac {1}{bz} \plus{} \frac {1}{cx}) \geq 3$
[/quote]
$ A\equal{}(\frac{bc}{y}\plus{}\frac{yz}{c})\plus{}(\frac{ac}{z}\plus{}\frac{xz}{a})\plus{}(\frac{ab}{x}\plus{}\frac{xy}{b})\equal{}(\frac{(1\minus{}y)c}{y}\plus{}\frac{y(1\minus{}c)}{c})\plus{}(\frac{a(1\minus{}z)}{z}\plus{}\frac{z(1\minus{}a)}{a})\plus{}(\frac{(1\minus{}x)b}{x}\plus{}\frac{x(1\minus{}b)}{b})\equal{}\sum(\frac{c}{y}\plus{}\frac{y}{c})\minus{}3\geq 3$.The last inequality follows from am-gm.

Thank everybody :lol:  :)
[quote="marsupilami"]
3. Let A=$ \frac {a^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c^2}{c^2 \plus{} (a \plus{} b)^2}$
$ a,b,c \in R$\{0}
minA=?[/quote]
After I had posted the problem, I saw a problem which is similar to it.
It's is problem 3, Japan,  1997.
Let a, b, c be positive real numbers.
Prove the inequality:
$ \frac {(b\plus{}c\minus{}a)^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(c\plus{}a\minus{}b)^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(a\plus{}b\minus{}c)^2}{c^2 \plus{} (a \plus{} b)^2} \geq \frac{3}{5}$

How aout it?? :?:

[quote="marsupilami"]Thank everybody :lol:  :)
[quote="marsupilami"]
3. Let A=$ \frac {a^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c^2}{c^2 \plus{} (a \plus{} b)^2}$
$ a,b,c \in R$\{0}
minA=?[/quote]
After I had posted the problem, I saw a problem which is similar to it.
It's is problem 3, Japan,  1997.
Let a, b, c be positive real numbers.
Prove the inequality:
$ \frac {(b \plus{} c \minus{} a)^2}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(c \plus{} a \minus{} b)^2}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(a \plus{} b \minus{} c)^2}{c^2 \plus{} (a \plus{} b)^2} \geq \frac {3}{5}$

How aout it?? :?:[/quote]
See here:
http://www.mathlinks.ro/viewtopic.php?search_id=991474430&t=146


Well this is my list...
[code]
\begin{description}
	\item[a)] A line of text that is longer than two lines.
		\noindent List:
		\begin{description}
			\item[1)] Item
			\item[2)] Item
			\item[3)] Item
			\item[4)] Item
			\item[5)] Item
			\item[6)] Item
			\item[7)] Item
			\item[8)] Item
			\item[9)] Item
			\item[10)] Item
			\item[11)] Item
			\item[12)] Item
			\item[13)] Item
			\item[14)] Item
			\item[15)] Item
			\item[16)] Item
		\end{desription}
	\item[b)] Item

\end{description}
[/code]
Ok so my problem is that in the first line of \item[a)], it's normal, however, in the second line, there is an indent that lines up with the second list. Then, my second list doesn't indent, but stays the same as the text in \list[a)]. Also, my \item[16)] doesn't line up with the rest of the items, but instead lines up with the beginning in the test in \item[a)].
So is there any way I can fix this indent problem?

Use [i]itemize[/i] rather than [i]description[/i]

Doesn't itemize give you bullets?
I want numbers and such...
1)
2)
3)

It only gives bullets if you don't give a label. But that's precisely what you've done with \item[a)] and \item[1)] so you'll see a) and 1). Try it and see.
Of course the enumerate environment will put the numbers in for you instead of having to specify labels
[code]\begin{enumerate}
   \item Item
   \item Item
\end{enumerate}[/code]

Oh wait, I fixed it, using itemize does work.
 :lol:

[quote="ZhangPeijin"]yeah but enumerate gives
1.
2.
3.[/quote]
\renewcommand{\labelenumi}{\alph{enumi})}
\renewcommand{\labelenumii}{\arabic{enumii})}
makes enumerate give
a)
1)
2)
b)
c)


Competition this saturday. Got any tips or last minute preparations? I have competed before at states, and I got 40th.
Not good. Please help. I want to at least do better than last year. I am in PA if that changes any answers.

Pa is pretty competitive so u have to be good. If u think ur top 10 worthy, then make sure ur preety good at cd too(if it's official there  :wink: ). Just relax and dont get too nervous or exited because u mite blank out in the competiton.

But remember, don't get too relaxed. You have to be on your toes or else you will do terrible... I saw it happen to first place in my chapter last year (my chapter is probably #2 in the state, and this person got 16th at state). Find your balance. Look over /study some material that day, but also listen to a little music to loosen up.

Make sure your arithmetic is decent (as in, adding/subtracting/multiplying/dividing).  Cost me 2 AIME points this year, and also I would have made USAMO if I had got those 2.

1 point can make a huge difference in MATHCOUNTS.  Remember, spend enough time on the first half.  Problem 1 is worth the same as Problem 30 (of course, that doesn't count tiebreakers).

My advice to you is to just relax, have fun, and pratice, practice, practice!!


geometry
3D geometry
algorithm
summer program
MathPath

How can I get better times on a Rubik's cube?

My best time is $\text{3: 19.160}$.
The algorithm I use is similar to the one described here:
http://en.wikibooks.org/wiki/How_to_solve_the_Rubik%27s_Cube

Is there a better algorithm to use?  If so, what is it?

I also saw somewhere that someone (I think on this forum) had their cube lubricated so that they could do a quarter turn by simply flicking a cubelet.  How could I do that?  What's the best way to lubricate it, and with what?

Switch to a more efficient method.  Really, a corner-edge pair method or better is necessary for speed.  Personally, I like the Petrus method, but you can explore and test some out a bit.  You definitely want to average under 60 moves.

For a lubricant, it depends on the cube, but the first thing you can do is get a god cube.  I recommend Oddzon cubes (you can get them from Toys r us, the regular rubik's cubes are oddzon, not the deluxe package).  Once you get a decent cube, making it better largely has to do with using it, stretching the springs, and spraying with silicone spray (buy at Home Depot).  I personally suggest silicone spray after you've used it a bit, sprayed under an edge from each face, but other people prefer vaseline or generally petroleum oil.  It's really up to you, but I'd say the silicone is most common.

I use silcone spray to lubricate my cube.  If you want a method with extremely little memorization (or speed  :rotfl:) try this website: http://grrroux.free.fr/begin/Begin.html
the method I use is a combination of some basic algorithims, most of which can be found on http://www.geocities.com/jasmine_ellen/RubiksCubeSolution.html
and algorithims that I discovered using the first website

Practice.

I use an algorithm similar to the Wikibooks one and the best time I've gotten is about 2 minutes.

I'm too lazy to learn any fast algorithms...

[quote="SplashD"]where do you all get your cubes? i want one badly![/quote]

I got my first one at MathPath, but I lost it a few days later. It was on the lost and found list the rest of camp. It's probably still there somewhere...

I got the second one at Mathcounts. It's really little and the stickers are about to come off; plus it's hard to turn. And the keychain thing gets in the way.

But there are a bunch of places that sell them.

Do you know where you can get one online? just a plain simple standard $3\times3\times3$ that has decent turnability and stickers.

[quote]Do you know where you can get one online? just a plain simple standard  that has decent turnability and stickers.
[/quote]
[url]http://www.rubiks.com[/url]  :wink:

Oh yes, if you get one from there, I also highly recommend you get the PVC stickers for replacements rather than the normal ones. PVC stickers don't need the extra little plastic layer, so they last longer.

I bought my cube at the national champtionships (no, I wasn't competing  :wink: )
lots of toy stores will have pretty good cubes.  u just need to lubricate them pretty well.

ive been cubing for 9 days and my best time is 1 min (with inspection). My average is 1:30 with insp.

I use this algorithm but i will start learning f2l when i get a little better.

http://peter.stillhq.com/jasmine/rubikscubesolution.html


First of all, so that it's clear what we're talking about: we're discuss the green checkmark that will appear on your "My Scores" page after we have logged your solutions as received.

If you submit via the web: the green checkmark should appear immediately.  There appears to have been a slight bug which affected about a dozen or so students for whom the checkmark didn't appear, but they should all be fixed now.

If you submit via mail: the checkmark will appear after will have logged your solution as received.  We have to do this by hand, and it may take a few days.  (Only about a dozen or so of mailed-in solutions have thus far been logged.)

[quote="andersonw"]When do we get the checkmark on the My USAMTS page after uploading our solutions?

Sorry if this question has been asked before, I searched around and couldn't find anything relating to this.

I'm also afraid that the final 5 percent of my PDF file was cut off, since there was a box that popped up (it said something like "do you want to stop the script?") when the download was at 95%, and I clicked "no" and don't know if that affected the submission.[/quote]
Go to the submit a solution page and see if it allows you to submit again or if it says that you already submitted a solution and are not allowed to submit anymore.

If you are allowed to submit again, that means the last solution wasn't uploaded and you should try again.

I've submitted my solutions online but have not received a checkmark.  Might it have to do with the fact that it says I haven't turned in the Entry form?  I faxed and mailed the entry form yesterday and today, respectively.

[quote="ProtestanT"]
Go to the submit a solution page and see if it allows you to submit again or if it says that you already submitted a solution and are not allowed to submit anymore.

If you are allowed to submit again, that means the last solution wasn't uploaded and you should try again.[/quote]

It says I'm not allowed to submit anymore.
But I just got the checkmark a few minutes ago, so it's good.

Also, a bug: From the Year 19 section, there's also a green checkmark in the round 1 row.


Obviously congruent to $ 0 \bmod 5^8$.  We have $ 5^{64} \equiv 1 \bmod 2^8$ and therefore $ 5^{1994} \equiv 5^{10} \bmod 2^8$.  The rest is computation:  we find that $ 5^{10} \equal{} \boxed{09765625}$.

[quote="ლევანი"]I calculated it by calculator (in Windows) and it wrote: 2509765625[/quote]
No.the result is $ 5^{10}$

[b]ლევანი[/b] found $ 5^{10}\times 257$, but we only need to find $ 5^{10}$, since $ 5^{1994} \equiv 5^{10} \mod 10^{8}$. Nevertheless, the last 8 digits are the same.


Let $ k\ge 8$, $ a$, $ b$ and $ c$ be positive real numbers.
Prove that
\[ \sum{\dfrac{a}{\sqrt {a + kbc}}\ge\dfrac{3}{\sqrt {k + 1}} + \dfrac{9(a + b)(b + c)(c + a) - 72abc}{2(a + b + c)^3\sqrt {k + 1}} + \dfrac{3(k - 8)\sum{(a^3 - abc)}}{2(a + b + c)^3(k + 1)\sqrt {k + 1}}
}\]
:)

[quote="ElChapin"]Let $ a$, $ b$ and $ c$ be positive real numbers.
Prove that
\[ \sum{\dfrac{a}{\sqrt {a + kbc}}\ge\dfrac{3}{\sqrt {k + 1}} + \dfrac{9(a + b)(b + c)(c + a) - 72abc}{2(a + b + c)^3\sqrt {k + 1}} + \dfrac{3(k - 8)\sum{(a^3 - abc)}}{2(a + b + c)^3(k + 1)\sqrt {k + 1}}
}\]
:)[/quote]
What is the nature of constant $ k$? $ k>0$? :huh:

I have forgotten to write that $ k\ge 8$. I've edited, thanks.

:)

I know this might appear ugly and hard but, it is not. It can be proved by using AMGM and a few operations (really few).

:)


geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

There are $2006$ points marked on the surface of a sphere. Prove that the surface can be partitioned into $2006$ congruent pieces, so that each piece contains exactly one of these points inside it.

This is also from the 1987 USA TST, I think.
[hide]Let the sphere be a globe, so that we can reference points on the sphere using latitude and longitude.  First, show that we can choose the axis of the globe such that no two points are at the same latitude.  Next, draw 1003 great circles through the endpoints of the axis so that the circle is divided into 2006 congruent wedge-shaped regions.  Next, show that for any point X that is not in some region A, we can transform each region so that the regions remain congruent, so that the point X is now in the regions A, and so that no other points have been transferred to other regions:

Draw the line of latitude $ l$ through X.  Let this latitude meet the edge of A (a line of longitude determining A) a some point $ Q$.  Take two points on this same line of longitude; one point$ Q_1$ above $ Q$, and the other point $ Q_2$ below Q.  Finally, choose one point $ Z$ on $ X$'s line of longitude such that $ X$ is contained in $ \triangle Q_1 Z Q_2$.  Then do the same for all the other regions.  Note that we can take $ Q_1 Q_2$ arbitrarily small to avoid "capturing" any other points into region A.  The diagram may clear some of this up:
[geogebra]40fdd7701901e536c0c3188a76a9a6d605f34fae[/geogebra]
[/hide]


Ring Theory
superior algebra
superior algebra unsolved

assume $ R$ is a commutative ring with unity.$ S$ is a multiplicatively closed subset of $ R$.and we have $ \forall x,y  \in R$  if $ xy \in S \longrightarrow x \in S$ and $ y \in S$.if there is no prime ideal $ P$ so that $ P \cap S \equal{} \emptyset$.then prove  $ S\equal{}R$.

HINT the complement of a saturated monoid is a union of prime ideals,
since an ideal maximal in the complement of a semigroup is prime.
This will become more clear when you study localizations, cf.
http://www.mathreference.com/id-lz,sats.html

but the proof of the total case of this quetion is incomplete.the proof is when correct when we have a prime ideal that has no intersection with S .and it will be completed when we prove the question above.

The link sketches a proof in the second paragraph in the "saturated" section. Namely, if r were an elt of R\S then, by saturation, rR < R\S so it could be enlarged to a maximal, hence, prime ideal in R\S, contradiction. Equivalently: in the localization S^-1 R, a nonunit is contained in a maximal ideal, via the usual Zorn Lemma enlargement. The problem is simply a version of this remark stripped of localization (fraction) language.


A4273B is six-digit positive number in which A and B are digits, and the total number is divisible by 72 with no remainder, decimal, or fraction. What is the value of A and the Value of B?

[hide]The answer is A=5, B=6[/hide]

Please look through forums for posting guidelines or a general level of problem difficulty.  This problem is too difficult for an ordinary middle school classroom, so I'm moving it.

[hide="Solution"]72=8*9

Using divisibility rules:

A number is divisible by 8 if it's last 3 digits are,
so $73B\equiv 0 \pmod 8 \Rightarrow B+2 \equiv 0 \pmod 8 \Rightarrow B\equiv 6 \pmod 8$

Therefore, B=6

Next, by casting out 9s, $A+6+16 \equiv 0 \pmod 9$ so A=5

 [/hide]

[hide]This number is divisible by 72, so it's also divisible by 8 and 9.

Thus, the last three digits, 73B, must be divisible by 8. B=6.

The sum of the digits, A+4+2+7+3+6, must be divisible by 9.
We have A+22 is a multiple of 9, A=5.

[/hide]


Nobody found a solution for this ? I only found out that has only one solution because $f'(x)>0$ $\rightarrow$ $f is increasing$ ,where $f: [4,\infty)\to \mathbb R$ ,$f(x)=log_{5}(x-2)+log_{8}(x-4)-log_{6}(x-1)$


PLease try the questions in the attachements:

BUMP. Why has no one attempted? Due to think there is some fault in the questions? Plz tell so cuz I think so.

hello
sorry buddy I m just going to give one answer.(having one test right now)
[hide="hint 2"]
$ f(x) \equal{} ax^2 \plus{} bx \plus{} c$
after putting the three different values the actual $ f(x) \equal{} \frac {x^2 \minus{} x \plus{} 2}{2}$
after differentiating the given equation ,find $ g(x)$
I hope u can find the suitable alternative now.

if possible i will give the answer to other two tomorrow.[/hide]
thank u

to the first one.
f''(x)>f(x), so f''(0)>f(0)=0,
so f(x) will be a concave function.
so the answer is b


There is a rubber band one end attached to the wall and the other end has a speed $v$.An ant begins motion from the left part (wall) with a speed $u$.When will the ant reach the free side of the band?

refer this
http://www.mathlinks.ro/Forum/viewtopic.php?t=131251 :)

I wasn't registered that time.Revision is mother of knowledge.(Azerbaijanian proverb) :lol:  :D


modular arithmetic
algebra
binomial theorem
number theory unsolved
number theory

Let: $ p_1 \equal{} 2, \ p_2 \equal{} 3, \ p_3 \equal{} 5$ and so on, that is $ p_i$ are following primes. Can it be true: $ p_1p_2\ldots p_n \plus{} 1 \equal{} a^k$, where $ k$ is integer, $ k>1$, and $ a$ is natural number?

[quote="Sylwek"]Let: $ p_1 \equal{} 2, \ p_2 \equal{} 3, \ p_3 \equal{} 5$ and so on, that is $ p_i$ are following primes. Can it be true: $ p_1p_2\ldots p_n \plus{} 1 \equal{} a^k$, where $ k$ is integer, $ k > 1$, and $ a$ is natural number?[/quote]
$ p_1p_2\ldots p_n \plus{} 1 \equal{} a^k$so
$ p_1p_2\ldots p_n$ is a primes or it has a prime divisor and this one$ > p_n$ so i think the ansewr is not exist

[hide]We claim no such $ a,k$ exist.

First, we prove that if $ i\le n$, then $ k\ne cp_i$:  It is sufficient to prove that there is no $ a$ such that $ p_1\cdots p_n \plus{} 1 \equal{} a^{p_i}$.  Suppose, for the sake of contradiction that there did exist such an $ a$.  Then, let $ a \equal{} qp_i \plus{} r$ where $ r < p_i$.  Then, by the Binomial Theorem, we have

$ a^{p_i} \equal{} (qp_i)^{p_i} \plus{} \binom{p_i}{1}(qp_i)^{p_i \minus{} 1}r \plus{} \cdots \plus{} \binom{p_i}{p_i \minus{} 1}(qp_i)r^{p_i \minus{} 1} \plus{} r^{p_i}$

Since $ \binom{p_i}{d}\equiv0\pmod{p_i}$ for all $ 0 < d < p_i$, we have that

$ a^{p_i}\equiv r^{p_i}\pmod{p_i^2}$

Since $ a^{p_i} \equal{} p_1\cdots p_n \plus{} 1\equiv 1\pmod{p_i}$, we must have $ r \equal{} 1$ (since $ a^{p_i}\equiv a\pmod{p_i}$).  But then we have $ a^{p_i}\equiv 1\pmod{p_i^2}$ which implies that $ p_1\cdots p_n\equiv 0\pmod{p_i^2}$, which is an obvious contradiction.

Thus, we have that if the integers $ a,k$ do satisfy the equation, then $ p_i\not|k$ for all $ i\le n$.  It is also obvious, since $ a^k \equal{} p_1\cdots p_n \plus{} 1$ that $ p_i\not|a$ for all $ i\le n$.  Thus, the minimal posible value for $ a^k$ is $ p_{n \plus{} 1}^{p_{n \plus{} 1}}$ since all numbers less than $ p_{n \plus{} 1}$ not equal to $ 1$ are multiples of a lesser prime.  But, it is obvious that

$ p_{n \plus{} 1}^{p_{n \plus{} 1}} > p_n^{p_n} \plus{} 1 > p_1p_2\cdots p_n \plus{} 1$

And so the claim is proved.[/hide]