Euler
geometry
circumcircle
incenter
geometric transformation
reflection
homothety

Let $ O,I$ be the circumcenter, incenter of triangle $ ABC,$ respectively. The incircle touches its sides $ BC,CA,$and $ AB$ at $ D,E,$and $ F,$ respectively. Let $ D'$ be the reflection of $ D$ wrt $ I.$ Denote by $ I_{a}$ the excenter on the side $ BC.$ Prove that the Euler line of triangle $ D'EF$ is parallel to $ OI_{a}.$

Firstly, consider $ D$, $ E$, $ F$ the intouch point of the incircle with the sides $ BC$, $ CA$, $ AB$ and $ D_{1}$, $ E_{1}$, $ F_{1}$ the points of tangency of the $ A$-excircle with $ BC$, $ CA$, respectively $ AB$.

It is known that $ A$, $ D'$ and $ D_{1}$ are collinear, since $ \frac {ID'}{I_{a}D_{1}} \equal{} \frac {AI}{AI_{a}} \equal{} \frac {p \minus{} a}{p}$. (For sake of completeness I've stressed the lines involved in the similarities in the sketch attached.)

Due to the fact above, it is clear that the homothety with pole $ A$ and ratio $ \frac {p}{p \minus{} a}$ maps the points $ D'$, $ E$, $ F$ into the points $ D_{1}$, $ E_{1}$, $ F_{1}$, respectively. Hence, the Euler line of $ D'EF$ is parallel to the Euler line of $ D_{1}E_{1}F_{1}$ and since $ I_{a}$ is the circumcenter of $ D_{1}E_{1}F_{1}$, it remains to prove that, for example, the orthocenter of $ D_{1}E_{1}F_{1}$ lies on $ OI_{a}$.

This above result reminds us of the collinearity of the orthocenter of the intouch triangle with the circumcenter and incenter. (i.e. [url]http://www.mathlinks.ro/viewtopic.php?t=6228[/url] (*))

However the equivalence of these two is clear I will detail it using homogenous barycentric coordinates:
It is known, or at least it follows from $ I_{a} \equal{} ( \minus{} a: b: c)$ and $ O \equal{} \left(a^{2}(b^{2} \plus{} c^{2} \minus{} a^{2}): b^{2}(c^{2} \plus{} a^{2} \minus{} a^{2}): c^{2}(a^{2} \plus{} b^{2} \minus{} c^{2})\right)$, that the equation of the line $ OI_{a}$ is:
\[ \minus{} \frac {(b \minus{} c)(p \minus{} a)}{a}x \plus{} \frac {(c \minus{} a)(p \minus{} b)}{b}y \plus{} \frac {(a \minus{} b)(p \minus{} c)}{c}z \equal{} 0.
\]
Now, observe that $ \angle{BF_{1}D_{1}} \equal{} \angle{BD_{1}F_{1}} \equal{} \frac {B}{2}$. Hence $ \angle{D_{1}F_{1}E_{1}} \equal{} 90^{\circ} \minus{} \frac {A}{2} \minus{} \frac {B}{2} \equal{} \frac {C}{2}$. Similarly, we deduce that $ \angle D_{1}E_{1}F_{1} \equal{} \frac {B}{2}$ and therefore $ \angle{E_{1}DF_{1}} \equal{} 90 \plus{} \frac {A}{2}$. Thus, we can now deduce that, in homogenous barycentric coordinates, the orthocenter of $ D_{1}E_{1}F_{1}$ is $ H_{1} \equal{} ( \minus{} \tan{\frac {A}{2}}: \tan{\frac {B}{2}}: \tan{\frac {C}{2}})$.

So, the condition for collinearity is:
\[ \frac {(b \minus{} c)(p \minus{} a)}{a}\tan{\frac {A}{2}} \plus{} \frac {(c \minus{} a)(p \minus{} b)}{b}\tan{\frac {A}{2}} \plus{} \frac {(a \minus{} b)(p \minus{} c)}{c}\tan{\frac {A}{2}} \equal{} 0,
\]

which obviously is equivalent with the collinearity of the point $ \left(\tan{\frac {A}{2}}: \tan{\frac {B}{2}}: \tan{\frac {C}{2}}\right)$, i.e. the orthocenter of the intouch triangle with $ O$ and $ I$, which is proved synthetically at (*).

Let $ I_{B},I_{C}$ be excenter.Use that $ I_{A}I_{B}I_{C}$ and $ DEF$ are homothety

Let $ I_{b},I_{c}$ be the excenters on the sides $ CA,AB.$we have $ D_{1}E_{1}\parallel I_{c}I,D_{1}F_{1}\parallel I_{b}I,E_{1}F_{1}\parallel I_{c}I_{b},$ hence $ \bigtriangleup D_{1}E_{1}F_{1}$ and $ \bigtriangleup II_{c}I_{b}$ are homothety. thus their Euler lines are parallel to each other. since $ I_{a},O$are the orthocenter,the nine point center of triangle $ II_{c}I_{b},$ $ I_{a}$ is the circumcenter of triangle $ D_{1}E_{1}F_{1},$ thus the orthocenter of triangle $ D_{1}E_{1}F_{1}$ lies on $ OI_{a}$.


[quote="GameBot"]In the land of Noom, all nouns are 4-letter words with consonants a
the beginning and end and a repeated vowel (a, e, i, o, or u) in the
middle. How many such words are possible?[/quote]

There are 21 letters that can be the first letter, 5 that can be the second, just 1 for the third(since the vowel has to be repeated), and 21 for the last, for a total of $ 21 \times 5 \times 1 \times 21 \equal{} 2205$.

It says "with consonants a the beginning and middle". It is supposed to say "with consonants AT the beginning and middle". Because of this typo, i thought that the first and last letters had to be "a".

there are

21 ways- 1st letter
5 ways -2nd letter ( only VOWELS!)
1ways - 3rd letter ( must repeat )
21 ways -4th letter

21*5*1*21=2205

21 ways- 1st
5 ways- 2nd
1 way- 3rd
21 ways- 4th

21x2x1x21= 2205

Answer 2205


Sarah's bowling score was 40 points more than Greg's, and the average of their two scores was 102. What was Sarah's score?

Sarah's score is $ \boxed{122}$, because the 40 points that is the difference between their scores is distributed 20 points more than 102, and 20 points less.


in ABC F is the point of tangency of incircle and the side BC.  show that incenter lies on DE, where E is the midpoint of AF and D is the midpoint of BC

Let ABCD be a quadrilateral, circuscribed to a circle of center I. If E and F are the midpoints of AC and BD then E,I,F are colinear. I think this is due to Newton.

Apply this to the degenerated quadrilateral ABFC, whose diagonals are AF and BC and whose incircle is the incircle of ABC.

Let K -- point on incircle, such that KF is diameter. AK $\cap$ BC=L. Than L -- is the point of tangence of excircle corresponding to BC with side BC (this is easy to see from homotetion). Then D--midpoint of FL. I -- midpoint of FK, E -- midpoint of AF, so E,I,D are collinear, because A,K,L are collinear.

Dear Mathlinkers,
see also
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=481938

Sincerely
Jean-Louis


We can do better:
Let $ A,B\in\mathcal{M}_m(\mathbb{C})$ where $ m$ is even. Then $ lim_{n \to \infty} \frac {\sum_{k \equal{} 1}^{n}{[det(A \plus{} kB) \plus{} det(A \minus{} kB) ]}}{n^{m\plus{}1}} \equal{} \frac {2.det(B)}{m\plus{}1}$.
Proof: $ \phi(k)\equal{}det(A\plus{}kB)\equal{}ak^m\plus{}\cdots$ is a polynomial of degree $ m$ in $ k$ with $ a\equal{}det(B)$. $ \dfrac{\sum_{k \equal{} 1}^{n}{det(A \plus{} kB) \plus{} det(A \minus{} kB) }}{n^{m\plus{}1}}\equal{}\dfrac{\sum_{k \equal{} 1}^{n}{\phi(k)\plus{}\phi(\minus{}k)}}{n^{m\plus{}1}}\equal{}\dfrac{\sum_{k \equal{} 1}^{n}{2ak^m\plus{}\cdots}}{n^{m\plus{}1}}$ converges to $ \dfrac{2a}{{m\plus{}1}}$ because $ \sum_{k \equal{} 1}^{n}{k^m}\sim\dfrac{n^{m\plus{}1}}{m\plus{}1}$. QED


geometry
incenter
circumcircle
trigonometry
geometry proposed

IN triangle $ABC$, let $I$ be the incenter and let $I_{a}$ be the excenter opposite $A$. Suppose that $II_{a}$ meets $BC$ and the circumcircle of triangle $ABC$ at $A'$ and $M$, respectively. Let $N$ be the midpoint of arc $MBA$ of the circumcircle of triangle $ABC$. Let lines $NI$ and $NI_{a}$ intersect the circumcircle of triangle $ABC$ again at $S$ and $T$, respectively. Prove that $S$,$T$, and $A'$ are collinear.

[quote="delegat"][color=purple]In a triangle $ ABC$, let $ I$ be the incenter and let $ I_{a}$ be the exincenter opposite $ A$. Suppose that $ II_{a}$ meets $ BC$ and the circumcircle of triangle $ ABC$ at $ A'$ and $ M$, respectively. Let $ N$ be the midpoint of arc $ MBA$ of the circumcircle of triangle $ ABC$. Let lines $ NI$ and $ NI_{a}$ intersect the circumcircle of triangle $ ABC$ again at $ S$ and $ T$, respectively. Prove that $ S$,$ T$, and $ A'$ are collinear.[/color][/quote]
[color=darkred][b]Lemma ([u]a remarkable property of a harmonical division[/u]).[/b] Given are four collinear points $ A,C,B,D$ in this order for which $ \frac{CA}{CB}=\frac{DA}{DB}$, i.e. the division $ (A,C,B,D)$ is harmonically. Denote the middlepoint $ M$ of the segment $ [CD]$. For a circle $ w$ which passes through the points $ A,M$ denote the its diameter $ [NP]$ which is perpendicularly on the its chord $ [AM]$ and the second intersections of the circle $ w$ with the mentioned lines $ \{\begin{array}{c}T_{1}\in ND\cap w\ ,\ T_{2}\in PD\cap w\\\ S_{1}\in NC\cap w\ ,\ S_{2}\in PC\cap w\end{array}$. Then $ B\in T_{1}S_{1}\cap T_{2}S_{2}$.[/color]

[color=darkblue][b]Proof of the proposed problem.[/b] The division $ (A,A',I,I_{a})$ is harmonically because $ \frac{AI}{AI_{a}}=\frac{p-a}{p}=\frac{r}{r_{a}}=\frac{A'I}{A'I_{a}}$. It is well-known that $ MI=MI_{a}=MB=MC$. Remain only to apply the previous lemma.[/color]

I was also thinking to this problem and I found a synthetic solution, also based on some basic harmonical division facts.

So the quadrupoint $ (AIA'I_{a})$ is harmonical, i.e $ (AIA'I_{a})=-1$, thus $ N(AIA'I_{a})=-1$ and by intersecting it with the circumcircle, we have that the quadrilateral $ ASDT$ is harmonical, where $ D \in NA' \cap w$.  $ (1)$.

On other hand, $ NN \| AM$, where $ NN$ is the tangent in $ N$ to the circumcircle, thus $ N(IMI_{a}N)=-1$, i.e the quadrilateral $ NTMS$ is harmonical. $ (2)$.

By $ (1)$, the lines $ SS$, $ TT$ and $ AD$ are concurrent, and by $ (2)$, the lines $ SS$, $ TT$, $ MN$ are concurrent, thus the all four lines $ SS$, $ TT$, $ AD$, $ MN$ are concurrent in $ X$. $ (3)$

Now because $ (ASDT)=-1$, we have also $ A'(ASDT)=-1$, which implies that the quadrilateral $ MS'NT'$ is harmonical, i.e the lines $ S'S'$, $ T'T'$ and $ MN$ are concurrent.

Similar from $ (NTMS)=-1$, we have also $ A'(NTMS)=-1$, which gives that the quadrilateral $ DS'AT'$ is harmonical, i.e the lines $ S'S'$, $ T'T'$ and $ AD$ are concurrent.

Therefore again all fourlines $ S'S'$, $ T'T'$, $ AD$ and $ MN$ are concurrent also in $ X$. $ (4)$ (because in $ (3)$ we defined $ X \in AD \cap MN$.)

Thus by $ (3)$ we have that $ ST$ is the polar of $ X$ and again by $ (4)$, $ S'T'$ is also the polar of $ X$, thus $ S' \equiv S$ and $ T' \equiv T$.

Ok Thank you
Where can I learn more about this harmonical division?

Is there any solution by using some elementary theorems like Menelaus, Ceva etc.

[quote="Virgil Nicula"]

[color=darkred][b]Lemma ([u]a remarkable property of a harmonical division[/u]).[/b] Given are four collinear points $ A,C,B,D$ in this order for which $ \frac{CA}{CB}=\frac{DA}{DB}$, i.e. the division $ (A,C,B,D)$ is harmonically. Denote the middlepoint $ M$ of the segment $ [CD]$. For a circle $ w$ which passes through the points $ A,M$ denote the its diameter $ [NP]$ which is perpendicularly on the its chord $ [AM]$ and the second intersections of the circle $ w$ with the mentioned lines $ \{\begin{array}{c}T_{1}\in ND\cap w\ ,\ T_{2}\in PD\cap w\\\ S_{1}\in NC\cap w\ ,\ S_{2}\in PC\cap w\end{array}$. Then $ B\in T_{1}S_{1}\cap T_{2}S_{2}$.[/color][/quote]

As the title of this nice problem says, "Beautiful" generalization, Virgil Nicula.

My idea as in the proof above works again, so denote $ N' \in NB \cap w$.

Because of $ N(ACBD)=-1$, we have that the quadrilateral $ AS_{1}N'T_{1}$ is harmonical, thus the lines $ S_{1}S_{1}$, $ T_{1}T_{1}$ and $ AN'$ are concurrent in a point $ Z$.

Now because $ NN \| AM$ and $ MC=MD$, we have that $ N(C,M,D, \infty)=-1$, i.e $ (NN, ND, NM, NC)=-1$, and by intersecting the pencil with $ w$, we have that the quadrilateral $ NT_{1}N'S_{1}$ is harmonical.

Thus the lines $ S_{1}S_{1}$, $ T_{1}T_{1}$ and $ NN'$ are concurrent also in $ Z$.

On other hand, let's denote $ S_{3}\in S_{1}B \cap w$ and $ T_{3}\in T_{1}B \cap w$, therefore the pencils $ B(AS_{1}N'T_{1})$ and $ B(NT_{1}N'S_{1})$ are harmonical,

And by intersecting them again with $ w$, we have that the quadrilaterals $ NT_{3}MS_{3}$ and $ AS_{3}N'T_{3}$ are harmonical, i.e the pairs of lines ($ MN$, $ S_{3}S_{3}$, $ T_{3}T_{3}$) and ($ AN'$, $ S_{3}S_{3}$, $ T_{3}T_{3}$) are concurrent, i.e the all four lines $ S_{3}S_{3}$, $ T_{3}T_{3}$, $ AN'$ and $ MN$ are concurrent in $ Z$.

Thus $ S_{3}T_{3}$ is the polar of $ Z$, but also $ S_{1}T_{1}$ is the polar of $ Z$, i.e $ S_{1}\equiv S_{3}$ and $ T_{1}\equiv T_{3}$.

Therefore the points $ S_{1}$, $ B$, $ T_{1}$ are collinear and similar prove that $ S_{2}$, $ B$ and $ T_{2}$ are collinear, thus $ B \in S_{1}T_{1}\cap S_{2}T_{2}$.

[quote][color=darkred][b]An equivalent enunciation.[/b] Let $ ABCD$ be a cyclic convex quadrilateral inscribed in the circle $ w$ so that $ AB=AD$. Denote : the intersection $ M\in AC\cap BD$ ; the reflection $ N$ of the point $ M$ w.r.t. the point $ D$ ; the second intersection $ E$ of the circle $ w$ with the line $ AN$ ; the point $ P\in (MN)$ for which $ \frac{PM}{PN}=\frac{BM}{BN}$. Prove that $ P\in CE$.[/color][/quote]
[color=darkblue][b][u]Proof. [/u][/b][/color]

$ \left\{\begin{array}{c}m(\widehat{ABD})=x\\\ m(\widehat{CAD})=y\\\ m(\widehat{DBE})=z\end{array}\right\|$ $ \implies$

$ \left\{\begin{array}{c}\frac{CM}{CA}=\frac{DM}{DA}\cdot\frac{\sin \widehat{CDM}}{\sin\widehat{CDA}}=\frac{DM}{DA}\cdot\frac{\sin (2x+y)}{\sin (x+y)}\\\\ \frac{EA}{EN}=\frac{DA}{DN}\cdot\frac{\sin\widehat{EDA}}{\sin\widehat{EDN}}=\frac{DA}{DN}\cdot\frac{\sin (x-z)}{\sin (2x-z)}\\\\ \frac{PN}{PM}=\frac{BN}{BM}=\frac{AN}{AM}\cdot\frac{\sin\widehat{BAN}}{\sin\widehat{BAM}}=\frac{\sin (x+y)}{\sin (x-z)}\cdot\frac{(2x-z)}{\sin (2x+y)}\end{array}\right\|$ $ \implies$

$ \frac{CM}{CA}\cdot\frac{EA}{EN}\cdot\frac{PN}{PM}=1$ $ \implies$ $ \boxed{\ P\in CE\ }\ \blacksquare$

[color=darkred][b]Lemma (prove easily !)[/b] Let $ ABC$ be a triangle inscribed in the circle $ w$. For a point $ M$ which belongs to the sideline $ BC$ denote the second intersection $ N$ of the circle $ w$ with the line $ AM$. Then exists the relation $ \boxed{\frac{MB}{MC}=\frac{AB}{AC}\cdot\frac{NB}{NC}}$. If $ N: =A$, i.e. the line $ AM$ is tangent to the circle $ w$ then $ \frac{MB}{MC}=(\frac{AB}{AC})^{2}$.[/color]
[quote][color=darkred][b][u]A very nice equivalent problem.[/u][/b]
Let $ ABCDE$ be a cyclic convex pentagon. Denote the points $ \{\begin{array}{c}M\in AC\cap BD\\\ N\in AE\cap BD\\\ P\in BD\cap CE\end{array}$.
Consider the following sentencies : $ \{\begin{array}{c}S1.\ \blacktriangleright AB=AD\\\\ S2.\ \blacktriangleright DM=DN\\\\ S3.\ \blacktriangleright \frac{BM}{BN}=\frac{PM}{PN}\end{array}$.

(The last sentence means that the division $ (B,M,P,N)$ is [u]harmonically.[/u])

Prove that if two from these sentencies are truly then and the other sentence is truly.[/color][/quote][hide="Proof."][color=darkblue]Observe that $ \boxed{\frac{BM}{BN}=\frac{AM}{AN}\cdot\frac{BC}{BE}}\ \ (1)$ and $ \{\begin{array}{c}CM=\frac{DM}{DA}\cdot BC\\\\ EN=\frac{DN}{DA}\cdot BE\end{array}$.  Apply the [u]Menelaus' theorem[/u] to the transversal $ \overline{CPE}$ in $ \triangle AMN$ : $ \frac{CM}{CA}\cdot\frac{EA}{EN}\cdot\frac{PN}{PM}=1$, i.e. $ \boxed{\frac{PM}{PN}=\frac{DM}{DN}\cdot\frac{BC}{BE}\cdot\frac{AE}{AC}}\ \ (2)$. Observe that $ S1\Longleftrightarrow\widehat{CMN}\equiv\widehat{ABC}\equiv\widehat{CEN}\Longleftrightarrow \widehat{CMN}\equiv\widehat{CEN}$, i.e. the quadrilateral $ CMEN$ is cyclically, i.e. $ AM\cdot AC=AE\cdot AN$. Thus, $ \{\begin{array}{ccc}S1 & \Longleftrightarrow & \frac{AM}{AN}=\frac{AE}{AC}\\\\ S2 & \Longleftrightarrow & \frac{PM}{PN}=\frac{BC}{BE}\cdot\frac{AE}{AC}\\\\ S3 & \Longleftrightarrow & \frac{AM}{AN}=\frac{DM}{DN}\cdot\frac{AE}{AC}\end{array}$. Therefore,

$ S1\wedge S2\implies$ $ \frac{BM}{BN}\ \begin{array}{c}(1)\\ =\end{array}\ \frac{AM}{AN}\cdot\frac{BC}{BE}\ \begin{array}{c}(S1)\\ =\end{array}\ \frac{AE}{AC}\cdot\frac{BC}{BE}\ \begin{array}{c}(S2)\\ =\end{array}\ \frac{DM}{DN}\cdot\frac{BC}{BE}\cdot\frac{AE}{AC}\ \begin{array}{c}(2)\\=\end{array}\ \frac{PM}{PN}\implies$ $ S3$.

$ S1\wedge S3\implies\frac{DM}{DN}\ \begin{array}{c}(S3)\\ =\end{array}\ \frac{AM}{AN}\cdot\frac{AC}{AE}\ \begin{array}{c}(S1)\\ =\end{array}\ 1\implies$ $ S2$.

$ S2\wedge S3\implies\frac{AM}{AN}\ \begin{array}{c}(S3)\\ =\end{array}\ \frac{DM}{DN}\cdot\frac{AE}{AC}\ \begin{array}{c}(S2)\\ =\end{array}\ \frac{AE}{AC}\implies$ $ S1$.[/color][/hide]


Ring Theory
algebra
function
domain
superior algebra
superior algebra unsolved

Show an example that shows  that the implicance [b]PID[/b] $ \implies$  [b]ED[/b] is false

[b]
PID: [/b] Principal ideal domain

[b]ED:[/b] Euclidean Domain

[quote="http://en.wikipedia.org/wiki/Principal_ideal_domain"]All Euclidean domains are principal ideal domains, but the converse is not true. An example of a principal ideal domain that is not a Euclidean domain is the ring $ \mathbb{Z}\left[\frac{1\plus{}\sqrt{\minus{}19}}{2}\right]$. [1][2].[/quote]
Apparently this is a well-known example.

[color=blue][b]t0rajir0u[/b][/color] is right! That ring is the typical counterexample to the claim that $ \mathbf{P} \mathbf{I}\mathbf{D} \Longrightarrow\mathbf{E}\mathbf{D}$. I strongly recommend that you take a minute glance at Motzkin's paper, too. He proved therein the existence of principal ideal domains that are not Euclidean domains. The rest is history.

See ya at [i][color=red]fmat.cl[/color][/i]  :wink:


Two integers are defined as "partners" if both of their prime factorizations contain all of the same prime factors. For example, 15 and 45 are partners since both are divisible by the same set of 3 and 5. How many positive integers greater than 50 and less than 100 have a partner that is greater than 50 and less than 100?

[quote="mathemagician1729"][hide]

$ 2^3 \cdot 3^2 \equal{} 72 \\
2 \cdot 3^3 \equal{} 54 \\
2^2 \cdot 3 \cdot 5 \equal{} 60 \\
2 \cdot 3^2 \cdot 5 \equal{} 90 \\
2^3 \cdot 7 \equal{} 56 \\
2^5 \cdot 3 \equal{} 96 \\
2 \cdot 7^2 \equal{} 98 \\
\implies \boxed{7}$

   [/hide][/quote]

How did you get the answers? I'm assuming there is a short cut here, besides prime factorizing all numbers between 50 and 100 and comparing?

Since you want to find number of numbers with this, lets start with this.
[hide]Suppose a number greater than 50 and less than 100 has a prime of 2. You would have to divide by a factor from your number and multiply by another or more.
Examples:54 = 2x3x3x3 and so you could try dividing by 3 and multiplying by 2x2 = 2x2x2x3x3=72 and 2x2x2x2x2x3=96.  Next is 56=2x2x2x7 which can be paired with 2x7x7=98. Then there is 60 = 2x2x3x5 and 90=2x3x3x5.
If you add the amount of numbers your answer is [b]7[/b][/hide]


Rationalize the denominators of $ \displaystyle\frac{21}{\sqrt{21}}$

$ \frac{21}{\sqrt{21}}$
obviously $ \sqrt{21}$, or sqrt21
because $ \sqrt{21}\cdot\sqrt{21}\equal{}21$


trigonometry
algebra
polynomial
Euler
modular arithmetic
pigeonhole principle
number theory

1. Prove that there are infinitely many primes.

2. Prove that cos(20 degrees) is irrational.

3. Prove that for all positive integers $ x$ and all integers $ n$ relatively prime to $ x$, there are infinitely many integers $ k$ such that $ x^k \minus{} 1$ is a multiple of $ n$.

#1 is easy. #2 is medium. #3 is hard.

[hide="classic #1"] Assume, for the sake of contradiction, that there are only a finite number of primes. Let them be $ p_1,p_2,...p_n$. Consider $ p_1p_2...p_n +1$. It is not divisible by any of our prime numbers, so it itself must be a prime. That is a contradiction, so hence there must be an infinite number of primes.[/hide]

[hide="2?"] Consider $ 8x^3 - 6x - 1 = 0$. We now prove that $ \cos{20}$ is a solution to this. Consider the cosine triple angle formula ${ \cos3x = 4\cos^3x - 3\cos}x$, where x is 20 degrees. If we let $ x = \cos20$, we have $ \frac {1}{2} = 4x^3 - 3x$ so $ 1 = 8x^3 - 6x$ and $ 8x^3 - 6x - 1 = 0$.We can use the rational root theorem to find that the aforementioned polynomial has no rational roots, so $ \cos20$ must be irrational.[/hide]

[hide="3?"]
For all $ k \equal{} c \cdot \phi(n)$ ($ c \in \mathbb{N}$), $ n$ divides $ x^k \minus{} 1$.
[/hide]

[quote="xpmath"][hide="classic #1"] Assume, for the sake of contradiction, that there are only a finite number of primes. Let them be $ p_1,p_2,...p_n$. Consider $ p_1p_2...p_n \minus{} 1$. It is not divisible by any of our prime numbers, so it itself must be a prime. That is a contradiction, so hence there must be an infinite number of primes.[/hide][/quote]
Just a minor technicality: That number you give isn't necessarily a prime, but it [i]is[/i] divisible only by primes not in the set of primes we have considered. Proof is still the same though.

[quote="xpmath"]Would it make a difference though, as it can't be divisible by any of the known primes in that scenario, so it itself must be a prime? Of course, I agree that it isn't necessarily a prime in general.

EDIT: Whoops, realized I put -1 instead of +1. Hehe...[/quote]

-1 also works

Yes, it does, but I've always seen the classic with +1  :P

[quote="xpmath"][hide=classic #1][/quote]
Also it evidently follows from the easy problem:

For $ n > 2$ there exists a prime between $ n$ and $ n!$

[quote="mathelete"][quote="aznlord1337"]3.Suppose there exists one solution $ k \equal{} a$ so $ x^a\equiv1 \mod n$. Then it's clear that all $ k \equal{} am$ for positive integers $ m$ will work. Letting $ k \equal{} \phi(n)$ will guarantee a solution by fermat-euler.[/quote] I do not get those symbols. I did it a different way. Think of digits.[/quote]

Well, i shall try to say the same by simple words.
[b]aznlord1337[/b] said. Let $ x^a \minus{}1$ is mutiple of $ n$. Then for $ k \equal{} am$ we have that $ x^k \minus{}1$ is multiple of $ n$ too. This is because of identity $ t^m\minus{}1\equal{}(t\minus{}1)(t^{m\minus{}1}\plus{}t^{m\minus{}2}\plus{}\dots \plus{} t\plus{}1)$ with $ t\equal{}x^a$.
Thus it is sufficient to show any $ a$ that works. This guarantees fermat-euler-theorem.

Maybe a more elementary solution to #3?

[hide="Solution"]Consider $ x^{k}\pmod{n}$ for $ k\equal{}0,1,\ldots n$.  There are $ n\plus{}1$ values of $ k$ and only $ n$ possible residues, so by Pigeonhole we can find $ i,j$ with $ i>j$ such that $ x^{i}\equiv x^{j}\pmod{n}$.  Then $ x^{i\minus{}j}\equiv 1\pmod{n}$ (we can divide by $ x^{j}$ because $ x$ is relatively prime to $ n$; we could also do this step by rearranging and factoring).  Then $ x^{m(i\minus{}j)}\equiv 1\pmod{n}$ for all positive integers $ m$, which is what we wanted.[/hide]

Or maybe not.  :P


quadratics
number theory proposed
number theory

Prove that there are no integers $a,b$ such that $2b^{2}+3|a^{2}-2$

$2$ is quadratic residue only for $p \equiv 1,7 \mod 8$ but $2b^{2}+3 \equiv 3,5 \mod 8$ so it has to have prime divisor $q \equiv 3,5 \mod 8$ - contradiction as $2$ has to be quadratic residue mod $q$


inequalities
geometry
circumcircle
incenter
geometric transformation
homothety
vector

Let ABC has a circumcenter O, incenter I. The incircle I meets BC,CA,AB at P,Q,R respectively, and let X be the circle of ninepoint circle of PQR. Prove that $OI \geq 4IX$.

[quote="GHAVR"][quote="Sung-yoon Kim"]Let ABC has a circumcenter O, incenter I. The incircle I meets BC,CA,AB at P,Q,R respectively, and let X be the centre of ninepoint circle of PQR. Prove that $OI \geq 4IX$.[/quote]

Let Ia, Ib,Ic  be the centres of outcircles of ABC.
Then $IaIc \parallel PR ,..$ so there exist  a homothety carrying over PQR to IaIcIb. It's centre is I. Also I is the orthocentre of IaIbIc.
O is the centre of ninepoint circle of IaIbIc. So we have to prove the coefficient of this homothety is $\geq 4$
But this homothety carries the radius of circumcircle of PQR to radius of circumcircle of IaIbIc i.o.
$r \rightarrow 2R$
And finally we have $2R \geq 4r$ so the coefficient is $\geq 4$
V :)[/quote]

Nice job! I proved using vector, but your proof is far more elegant :)

i think that I is NOT the homothety center... it is for example on angle A bisector just like Ia but P is not on it... what's wrong?


I think we didn't do great but I just want to know.

[quote="1=2"]Murph and the Magictones? You did pretty good actually. I'm fuzzy on the details but you were mentioned a few times.[/quote]
O, really, thx for that info. At least I feel better now. We did kind of bad on individual.

[quote="andersonw"]Murph and the Magictones got 5th place overall team.[/quote]

Yeah, we did okay even though we really messed up individual...

[quote="LadyKn1ght"]I got some photos of the results in A division. They're not the clearest, but are mostly readable:

Geometry: http://www.facebook.com/photo.php?pid=3023910&l=817f112b14&id=654972943
Algebra: http://www.facebook.com/photo.php?pid=3024628&l=89c871b145&id=654972943
Number Theory: http://www.facebook.com/photo.php?pid=3023911&l=3db34320aa&id=654972943
Combinatorics: http://www.facebook.com/photo.php?pid=3023913&l=77659b2846&id=654972943
Power Round: http://www.facebook.com/photo.php?pid=3023914&l=59e4b54835&id=654972943
Team Round: http://www.facebook.com/photo.php?pid=3023915&l=b08d5bc001&id=654972943
Team Overall: http://www.facebook.com/photo.php?pid=3023918&l=8d7dd024d1&id=654972943[/quote]
Thank you so much

[quote="darkdieuguerre"][quote="LadyKn1ght"]Combinatorics: http://www.facebook.com/photo.php?pid=3023913&l=77659b2846&id=654972943[/quote]

Is this public visibility? Because if it is, and people see my comment...  :blush:[/quote]

it is ok; Brian appreciates your comment so much that he made it his AIM status :O

So I've heard; I was expecting to get smote or something, but I'll go with that.


$\frac{a^3+b^3+c^3}{a+b+c} \ge \frac{a^2+b^2+c^2}{3}$ (Chebyshev)
Sum and done

Yes... It's not difficult  :lol:  :lol:  :lol:

[quote="ehku"][tex]\sum \frac{a^3+b^3+c^3}{b+c+d} \ge a^2+b^2+c^2+d^2 [/tex]
Is it true ?  :lol:[/quote]
yes because [tex]\sum \frac{a^3+b^3+c^3}{b+c+d} \ge \sum \frac{a^3+b^3+c^3}{a+b+c}[/tex]

You're right. But I don't like that solution, because.... :oops: .
 One another solution ? (not difficult)

You can't!
Using Cauchy-Schwar inequality...


The average of five numbers is 14. If three of those numbers are 15, 18, and 21, what is the average of the remaining two numbers. please explain how to solve it if you can... im really lost :?:

Well, the average of the 5 numbers is 15. Therefore, the sum of the 5 numbers is $5*15=75$

To find the average of the remaining two numbers, we subtract the given numbers and divide by 2.

$75-15-18-21=60-39=21$

The average is $\frac{21}{2}$

The average is no longer 14 because we are finding the average of the two other numbers.

The answer 21/2 makes sense because the other three terms are larger than 14 and we are not counting them. Therefore, these two terms must be smaller than 14.

The work to get it is shown in the provious post.

4everwise, I think you made a small mistake.

The average of the five numbers is [b]14[/b] (not 15), so the sum of the numbers is $14\times 5=70$

For the next step, $70-(15+18+21)=16$, so the answer is $\boxed{8}$ :)

i think i am going to move this

peoples:  please read the topic that says what types of problems to post


1. Which molecule(s) are conductive when dissolved in water?

2. Which molecule(s) are covallently bonded and has the highest boiling point?

3. Which molecule(s) have bonds that are formed by transferring electrons?

4. Which molecule(s)contain bonds that cause atoms to have partial charges?

5. Which molecule(s) have the ability to form all three intermolecular forces (hydrogen force, dipole-dipole forse, london dispersion force)?

6. Which molecule(s) are soluable in oil [note oil and water don't mix]?

There are possiblities that each question has more than one answer.

Please help me out.
Thanks.

1. All chlorides are soluble except AgCl, and a couple others.

2. Ethane has a low boiling point because it only has induced-dipole forces. (B) does not have covalent bonds. And I would assume water has a higher boiling point than (C) because on average water forms 2 H-bonds per molecule.

4. (A) and (B) are nonpolar.

5. (A) and (B) obviously can't form H-bonds.

6. Like dissolves like (i.e. nonpolar substances will dissolve nonpolar substances).

I will do some more later.

1. Molecule B,
2. Molecule D,E
3. Molecule B
4. Molecule D, E
5. Molecule C
6. Molecule A

[quote="ProblemSets"]for question #4, wouldnt A and C also cause partial charges because of the electronegativity?[/quote]
(A) would not because the difference between the electronegativies of carbon and hydrogen is not significant (less than 0.4 on the Pauling scale).

oh ok. yes, know i remember. but would C be an answer. Some is greater than .4, yet some less. so ya?

Oh ya, i also think that C can be the answer for question #4. Cl is more electronegative than C, rite? so Cl has greater tendency to attract electrons.


Let $ O$, $ Q1, Q2,.....Q20$ points in the plane. Prove that at least $ 90$ of the angles $ QiOQj$ are $ < \equal{}$ $ 120$°

I think it's $ Q_iOQ_j$, right?

if so, consider 20 points $ A_1, A_2, ... A_{20}$. Combine a segment between $ A_i\ and\ A_j$ if angle $ Q_iOQ_j$ is greater than 120 degree. Obviously, there is no three points that a segment appears between each pair of them. So the maximum segment combined is $ \frac {20^2}{4} \equal{} 100$. Since there are 190 segment in total, so at least 90 different pairs of $ A_iA_j$ doesn't have segment combined. So the conclusion follows.

90 can be achieved, assume a circle centered at O with AB one of its diameter. Let half of twenty gather at A, the other half at B. So the total number of angle less or equal to 120 is 90. :)

Hi. Maybe silly quastion but why is the maximum number of combined segments $ 20^2/4$?

It is a known lemma, that if a graph has more than $ n^2/4$ edges, then three of them form a triangle.

[quote="MathPotter"]It is a known lemma[/quote]Specifically, it's Turan's Theorem for $ k \equal{} 3$.


circumcircle
geometry
power of a point
radical axis
geometry solved

A circle k1 of the plane lies in the interior of a circle k2. Point P lies inside k1, and point Q lies outside k2 in the plane. Given the circles and the points, draw an arbitrary line e through P that does not pass through Q. Let e intersect k1 at A and B. Let the circumscribed circle of ABQ intersect k2 at C and D. Prove that all line segments CD obtained in this way are concurrent.

I don't have a solution but somehow it seems that this common point lies on the line PQ.

  Sincerely,
  Darij Grinberg

It's not that hard once you know what to look for.

First of all, we can prove that all circles QAB pass through 2 fixed points (Q and another one). To do this we observe that P is on the radical axis of the 2 circles k1 and QAB, so the powers of P with respect to the 2 circles are equal, and since the power of P with respect to k1 is constant, then so is the power of P with respect to QAB, so the circle QAB passes through a fixed point R on PQ (s.t. PQ*PR=that constant power).

The new problem is this: Let R be a point inside and Q a point outside the circle k2. A variable circle passing throuygh R and Q intersects circle k2 at C and D. Show that CD passes through a fixed pt. We can easily see that this is, in fact, the inverse problem of the one I proved in the previous paragraph (i.e. it can be obtained from that one by an inversion of pole Q).

grobber: thanks, nice solution!

[quote="grobber"]The new problem is this: Let R be a point inside and Q a point outside the circle k2. A variable circle passing throuygh R and Q intersects circle k2 at C and D. Show that CD passes through a fixed pt. We can easily see that this is, in fact, the inverse problem of the one I proved in the previous paragraph (i.e. it can be obtained from that one by an inversion of pole Q).[/quote]

Okay, this can be seen in a simpler way: All lines CD pass through a fixed point on the line QR, because if we take two different circles through Q and R, they meet k2 at C and D and at C' and D', respectively, the lines CD, C'D' and QR are the pairwise radical axes of our two circles through Q and R and of the circle k2, respectively, and hence meet at one point; i. e., the lines CD and C'D' intersect QR at the same point.

  Sincerely,
  Darij Grinberg

this is realy nice solution! :)
thanx for your regard :D


Prove that $ 2x^{2} \minus{} 6xy \plus{}5y^{2} \plus{}2x\minus{}8y\plus{}14$ is positive for all real values of $ x$ and $ y$.

This is too easy. The expression writes as $ 2\left (x\minus{} \frac {3y\minus{}1}{2}\right )^2 \plus{} \frac {1} {2}(y\minus{}5)^2 \plus{} 1$.

$ 2x^2 \minus{} 6xy \plus{} 5y^2 \plus{} 2x \minus{} 8y \plus{} 14$
$ \equal{} 2x^2 \minus{} 2(3y \minus{} 1)x \plus{} 5y^2 \minus{} 8y \plus{} 14$

$ D/4 \equal{} (3y \minus{} 1)^2\minus{}(5y^2 \minus{} 8y \plus{} 14) \equal{} \minus{} (y \minus{} 5)^2 \minus{} 2 < 0$ which completes the proof.

[i]Express the quadratic terms as the sum of two squares:[/i]
$ (x^2 \minus{} 4xy \plus{} 4y^2) \plus{} (x^2 \minus{} 2xy \plus{} y^2) \plus{} 2x \minus{} 8y \plus{} 14$
$ (x \minus{} 2y)^2 \plus{} (x \minus{} y)^2 \plus{} 2x \minus{} 8y \plus{} 14$

[i]Express the linear terms using the same binomials[/i]
$ (x \minus{} 2y)^2 \plus{} (x \minus{} y)^2 \plus{} 6(x \minus{} 2y) \minus{} 4(x \minus{} y) \plus{} 14$

[i]Change variables:[/i]
$ X^2 \plus{} Y^2 \plus{} 6X \minus{} 4Y \plus{} 14$

[i]Complete the square:[/i]
$ (X^2 \plus{} 6X \plus{} 9) \plus{} (Y^2 \minus{} 4Y \plus{} 4) \plus{} 1$
$ (X \plus{} 3)^2 \plus{} (Y \minus{} 2)^2 \plus{} 1$


LaTeX
geometry
3D geometry
algebra
factorization
sum of cubes

Let $f(x)$ with integer coefficients and a, b, and c be three integers. Suppose $f(a)=b$, $f(b)=c$, $f(c)=a$. Show that $a=b=c$.

Prove $(55+12\sqrt{21})^{\frac{1}{3}}+(55-12\sqrt{21})^{\frac{1}{3}}=5$

For #2, square the whole thing and then take the square root.

The radicands cancel out when squaring it.

#1 is already posted as a problem before.

I don't see how third roots cancel out by squaring.

Can someone show the work (For problem 2)?

It's incredibly tedious to latex out, but notice how the two terms inside the cube roots are conjugates of each other.

[quote="robinhe"][hide]For #2, all you do is cube both sides, rearrange by factoring, substitute values, and then multiply the remaining terms since they are conjugates of each other.  Then finally, you will get 125 in the LHS and of course 125 in the RHS.[/hide][/quote]

Actually, that's not [i]quite[/i] the case, but cubing helps.

[hide="1"]
Note that $b-c=f(a)-f(b)$ is divisible by $a-b$. Similarly, $c-a$ is divisible by $b-c$, and $a-b$ is divisible by $c-a$. Now suppose $a,b,c$ are not all equal. Then they must be distinct, because $a=b\Rightarrow f(a)=f(b)\Rightarrow b=c$ and similar cases for the other two pairs.

Let $x=\frac{b-c}{a-b}$, $y=\frac{c-a}{b-c}$, $z=\frac{a-b}{c-a}$. Then $xyz=1$ so $x$, $y$, and $z$ must all equal $\pm 1$. If one of $x$, $y$, or $z$ is -1, we get a contradiction because $x=-1\Rightarrow a=c$, $y=-1\Rightarrow a=b$, $z=-1\Rightarrow b=c$. So we must have $x=y=z=1$. But then $x=1\Rightarrow a+c=2b$ and $y=1\Rightarrow a+b=2c$. Subtracting the equations, $b=c$ which is a contradiction here as well. So we must have $a=b=c$.
[/hide]

[hide="2"]
Let $\alpha=\sqrt[3]{55+12\sqrt{21}}, \beta=\sqrt[3]{55-12\sqrt{21}}$, and $s=\alpha+\beta$. We have $\alpha^{3}+\beta^{3}=110$ and $\alpha \beta =55^{2}-21\cdot 12^{2}=1$.

Factoring the sum of cubes, $110=s(\alpha^{2}-\alpha \beta+\beta^{2})=s(s^{2}-3)$. So $0=s^{3}-3s-110=(s-5)(s^{2}+5s+22)$. But $s$ is clearly nonnegative, so $s^{2}+5s+22$ could not possibly equal 0. Therefore, $s=5$.
[/hide]

[quote="scorpius119"]
[hide="2"]
Let $\alpha=\sqrt[3]{55+12\sqrt{21}}, \beta=\sqrt[3]{55-12\sqrt{21}}$, and $s=\alpha+\beta$. We have $\alpha^{3}+\beta^{3}=110$ and $\alpha \beta =55^{2}-21\cdot 12^{2}=1$.

Factoring the sum of cubes, $110=s(\alpha^{2}-\alpha \beta+\beta^{2})=s(s^{2}-3)$. So $0=s^{3}-3s-110=(s-5)(s^{2}+5s+22)$. But $s$ is clearly nonnegative, so $s^{2}+5s+22$ could not possibly equal 0. Therefore, $s=5$.
[/hide][/quote]

That's not exactly the way I did it.  My approach was not as elegant.


Put $\sqrt[3] {x}=a$ . The equation is equivalent to

$a^2-2a-(a^3-4)\sqrt{a^3-7}-3a^3+28=0$

And let $f(a)=a^2-2a-(a^3-4)\sqrt{a^3-7}-3a^3+28$

We can find that  $f'(a)<0$  at  $\left[ \sqrt[3] {7},\infty\right]$

Also easily we find that  $f(2)=0$, so it is the unique solution of the equation in $R$


blogs
number theory unsolved
number theory

Any comments on my blog here?

http://mgb777.spaces.live.com/


Find integers a,b such that
|a-b|=|a|+|b| and a^3 * b^2 +a^2 * b^3 +2* a^2 *b^2 -a -b=37

In fact, |a-b|=|a|+|b| does not reduce to a=-b. Squaring both handsides of |a-b|=|a|+|b|, after some algebric manipulations we get |ab|=-ab, that means, ab<=0. And because the above manipulations are inversable, we get |a-b|=|a|+|b| if and only if ab<=0.


probability
calculus
integration
analytic geometry
inequalities
algebra
polynomial

My friend wrote another question, this time dealing with geometric probability.  When I read the question it didn't seem too bad and seemed perfectly solvable without calculus (I don't even know if that would help, but I thought integration might do something... i don't know any calculus tho, so who knows  :P) but I haven't been able to get it.   Just end up with a meeaan polynomial.  See what you can do, maybe I'm missing something.

A point P is selected at random inside square ABCD of side length 1.  What is the probability that AP+BP < DP?

Did you place the square on a coordinate system and then use x and y for P and then set up the inequality with the distance formula?  I tried this and I likewise got a very ugly polynomial which I coudn't reduce easily. Hmm...

[quote="noheadskeleton"]may we find the line which contain Q that make AQ+BQ = DQ. then find the probability that P lies in the region that contain D?[/quote]

Yeah, thats what I had tried, by putting it on a coordinate plane, but the polynomial that I got... I couldn't really work with.

I don't know, but I think it has to do with circles.

Sorry, I hate to just bump, but i would really like to see how to solve this.  If it doesn't belong in intermediate, then mods please move it to the appropriate one.  thanks

I really dont think that the points that would be above a certain line... more some curve...?

The equations come out much simpler in polar coordinates. I won't go through all the details right now, but if you let corner A be situated at the origin and point B at (1,0), you can describe the set of points that you're looking for by the polar inequality:

$|r - \frac{2}{3}(\cos{\theta} + \sin{\theta})| < \frac{2}{3}\sqrt{1 - 2\sin{(2\theta})}$, where $0 < \theta < \pi/12$.

The area of this region (and hence the probability) can be found by integrating between the inner and outer curves of this polar equation from some lower bound (which I don't know right now) to $\theta = \pi/12$, and adding the area of a small triangle underneath.

I'll fill in more details if you're [i]really[/i] interested, but I don't think this has any "elementary" solution.

I think I've got a solution...

[hide]
$A(0,0), B(1,0), C(1,1), D(0,1), P(x,y)$
$DP>AP+BP$
$\sqrt{x^{2}+(1-y)^{2}}>\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+(1-x)^{2}}$
$\sqrt{x^{2}+y^{2}-2y+1}>\sqrt{x^{2}+y^{2}}+\sqrt{y^{2}+x^{2}-2x+1}$

Switch to polar here: $r^{2}= x^{2}+y^{2}$, $x = r\cos\theta$, $y = r\sin\theta$

$\sqrt{r^{2}-2r\sin\theta+1}>r+\sqrt{r^{2}-2r\cos\theta+1}$
$r^{2}-2r\sin\theta+1>r^{2}+2r\sqrt{r^{2}-2r\cos\theta+1}+r^{2}-2r\cos\theta+1$
$-2r\sin\theta>r^{2}-2r\cos\theta+2r\sqrt{r^{2}-2r\cos\theta+1}$
$-2\sin\theta>r-2\cos\theta+2\sqrt{r^{2}-2r\cos\theta+1}$
$-r+2(\cos\theta-\sin\theta)>2\sqrt{r^{2}-2r\cos\theta+1}$
$r^{2}-4r(\cos\theta-\sin\theta)+4(\cos\theta-\sin\theta)^{2}>4r^{2}-8r\cos\theta+4$
$r^{2}-4r\cos\theta+4r\sin\theta+4(\cos\theta-\sin\theta)^{2}>4r^{2}-8r\cos\theta+4$
$r^{2}+4r\cos\theta+4r\sin\theta+4(\cos\theta-\sin\theta)^{2}>4r^{2}+4$
$r^{2}+4r(\cos\theta+\sin\theta)+4(\cos\theta-\sin\theta)^{2}>4r^{2}+4$
$r^{2}+4r(\cos\theta+\sin\theta)+4-8\cos\theta\sin\theta>4r^{2}+4$
$0>3r^{2}-4r(\cos\theta+\sin\theta)+8\cos\theta\sin\theta$

Let $z = \cos\theta+\sin\theta$
Since $4(\cos\theta+\sin\theta)^{2}= 4+8\cos\theta\sin\theta$,
$8\cos\theta\sin\theta = 4z^{2}-4$

$0>3r^{2}-4rz+4z^{2}-4$ which describes the interior of an ellipse.

Now we know that $z = \cos\theta+\sin\theta = \sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)$ and that the point $P$ lies in the first quadrant, so $\theta$ is limited to $[0,\frac{\pi}{2}]$. Then we know $1\leq z\leq\sqrt{2}$.

However, viewing $0=3r^{2}-4rz+4z^{2}-4$ as a quadratic in terms of $r$, we get $r = \frac{2z+/-2\sqrt{3-2z^{2}}}{3}$, and we know that $|z| \leq \sqrt{\frac{3}{2}}$. Since $\sqrt{\frac{3}{2}}< \sqrt{2}$, we have $1 \leq z \leq \sqrt{\frac{3}{2}}$.

If we place the ellipse on a coordinate axis with $z$-axis horizontal and the $r$-axis vertical, the ellipse will be tangent to the line $z = \sqrt{\frac{3}{2}}$ at $\left(\sqrt{\frac{3}{2}}, \sqrt{\frac{2}{3}}\right)$. When $r=0$, $z=1$ or$z=-1$ which turns out to be quite convenient. We want to find the area in the ellipse bound by the vertical lines $z=1$ and $z=\sqrt{\frac{3}{2}}$.

We then have:

$\int_{1}^{\sqrt{\frac{3}{2}}}\frac{2z+2\sqrt{3-2z^{2}}}{3}dz-\int_{1}^{\sqrt{\frac{3}{2}}}\frac{2z-2\sqrt{3-2z^{2}}}{3}dz$

which conveniently simplifies to $\frac{4}{3}\int_{1}^{\sqrt{\frac{3}{2}}}\sqrt{3-2z^{2}}dz$

Applying trig substitution $\sqrt{\frac{2}{3}}z = \sin\theta$, $dz = \sqrt{\frac{3}{2}}\cos\theta d\theta$, we get

$\frac{4}{\sqrt{2}}\int_{\arcsin\sqrt{\frac{2}{3}}}^{\frac{\pi}{2}}\cos^{2}\theta d\theta$

Applying double angle cosine and integrating, I got

$\sqrt{2}\left(\frac{\pi}{2}-\frac{\sqrt{2}}{3}-\arcsin\sqrt{\frac{2}{3}}\right)$

which is about $0.20375$.

I think that for a more elementary solution, I'd project the ellipse onto another plane to make a circle... then you'd be finding the area of a sector.

[/hide]

Let me know of any mistakes..

I haven't read your solution at all, but your answer is too large.  By construction of a very accurate diagram (using Geometer's Sketchpad), I calculate a value of about .114....

JBL, is there any chance that you could upload that very accurate diagram? Just to see what it looks like? Because I have no clue what it looks like.

Yep, here it is.  This is actually less accurate than my very accurate drawing, which divided the side into six sections instead of 4.  (The way I created it is to make P a mobile point on a line and construct a circle around D with radius $AP+BP$, then move P until it appeared to lie on the circle.)

In this particular instance, Geometer's Sketchpad calculated the area of the green hexagon to be 22.33 cm^2 and the area of the square to be 197.38 cm^2 for a ratio of about .11312... (although I doubt it's accurate beyond the second decimal point).

Here's a sketch of what I think the region of the square looks like...

The vertical line on the right is $x=1$.


limit
number theory
number theory proposed

1. Let a and b be postive integers such that a|b^2, b^2| a^3, b^3|a^4...........Prove that a=b
2. Let d be a postive integer, and let S be the set of all postive integers of the form x^2+dy^2, where x and y are non-negative integers
(a) Prove that if a belongs S and b belongs S, then ab belongs S
(b) Prove that if a belongs S and p belongs S, such that p is a prime and p|a, then a/b belongs S
(c) Asusume that the equation x^2+dy^2=p has a solution in non-negative integers x and y, where p is a given prime. Show that if d>= 2, then the solution is unique, and if d=1, then there are exactly two solution

There is sth wrong in the problem statement of (1), e.g., $ a=b^2$ fulfills the properties $ a|b^2,b^2|a^3,b^3|a^4\ldots$

(2a) follows trivially from $ x^2+dy^2=(x+i\sqrt{d}y)(x-i\sqrt{d}y)$, since complex conjugation is a (field) homomorphism.

(2b) Let $ a=(x+i\sqrt{d}y)(x-i\sqrt{d}y)$ and ${{ p=(w+i\sqrt{d}z})(w-i\sqrt{d}z})$. Since $ w\pm i\sqrt{d}z$ are prime in $ \mathbb{Z}[i\sqrt{d}]$ and $ p|a$ wlog we can assume $ \frac{x+i\sqrt{d}y}{w+i\sqrt{d}z}=A+i\sqrt{d}B\in\mathbb{Z}[i\sqrt{d}]$. This implies $ \frac{a}{p}=(A+i\sqrt{d}B)(A-i\sqrt{d}B)=A^2+dB^2\in S$, the last step by potentially multiplying $ A$ and/or $ B$ by $ -1$ if required.

(2c) The units in $ \mathbb{Z}[i\sqrt{d}]$ are $ \pm 1$ for $ d\geq 2$ and $ \pm 1,\pm i$ for $ d=1$. Decomposing $ p$ as in (2b) we know that each of the factors $ w+i\sqrt{d}z$ is prime, i.e. for any other decomposition of $ p$ wlog we can assume $ u(w+i\sqrt{d}z)=w'+i\sqrt{d}z'$ with some unit $ u$. This immediately gives the result for $ d\geq 2$ together with the non-negativeness and for $ d=1$ we get "two solutions", although there are different only with regard to ordering, i.e. $ p=x^2+dy^2=y^2+dx^2$, which for $ d=1$ are obvioulsy not distinguishable.

[quote="hjbrasch"]There is sth wrong in the problem statement of (1), e.g., $ a \equal{} b^2$ fulfills the properties $ a|b^2,b^2|a^3,b^3|a^4\ldots$[/quote]

The problem statement should read $ a | b^2, b^2 | a^3, a^3 | b^4, b^4 | a^5 \ldots$.

This is not really a number theory problem.  Let $ r \equal{} \frac{a}{b}$.  We are given $ a^{2n\minus{}1} \le b^{2n} \Leftrightarrow r^{2n\minus{}1} \le b$, but $ b^{2n} \le a^{2n\plus{}1} \Leftrightarrow r^{2n} \ge \frac{1}{a}$.

Since $ \lim_{k \to \infty} r^k \equal{} 0, 1, \infty$, we conclude that it is $ 1$ and therefore $ a \equal{} b$.


Hello
$ 7$ rectangles made of one small block
$ 8$ rectangles made of two small blocks.
$ 2$rectangles made of three small blocks.
$ 2$rectangles made of four small blocks.
Thus total=$ 19$
Thank u.

But the problem clearly states how many rectangles, not squares, and a square is a rectangle but a rectangle is not a square, so shouldn't you not count the squares?

It should be 19 after carefully counting the cases for size

[hide=Solution]We consider the separate cases by the dimensions of each type of rectangle. There are 7 of the $1 \times 1$ squares. There are 4 vertical $1 \times 2$ rectangles, and 4 horizontal $1 \times 2$ rectangles. There are also 1 each of vertical and horizontal $1 \times 3$ rectangles. And, finally, there are the two $2 \times 2$ squares. In total, there are $7 + 4 + 4 + 1 + 1 + 2 = \boxed{19}$ rectangles.[/hide]

The question should state that the sides of the rectangle must be parallel to the sides of the figure

[quote=nathanip]but can't you do a too two by root two square[/quote]

Those are not drawn, so...

[quote=themathboi101]how are there 2    1x3 rectangles?[/quote]

Middle column and middle row

@ above thanks [s] when u realize ur so dumb

[quote=jerry_li][quote=afroromanian]Each 2x2 block of four squares has $\binom{3}2 \cdot \binom{3}2 = 9$ rectangles. Multiply this by 2, take away 1 for double counting the overlapping 1x1 middle square, and add 2 for the two 3x1 rectangles that are part of both 2x2 blocks (hence were not counted above), to get $2\cdot9 - 1 + 2 = \boxed{19}$.[/quote]

I like this solution because you don't need to do heavy bashes.[/quote]

Doesn't everyone? :P

This is casework...and it's simpler to just count them, it doesn't take that much time.


A jet plane flies at constant altitude $ h$ above ground level. An observer on the ground hears a sonic
boom at time $ t$ after the plane passes directly overhead. Show that if the speed of sound $ v$ is the same at
all altitudes, the speed of the plane is given by $ v_p \equal{} \frac {hv}{\sqrt {h^2\pm v^2t^2}}$ where you will determine the appropriate sign in doing this problem.

 [geogebra]c75391c744ff2336eb984c9c4bcfea8dad8380fc[/geogebra]
$ (a,b,c)\implies a\equal{}\text{Distance},b\equal{}\text{Speed},c\equal{}\text{Time}$ so $ t\equal{}\frac{d}{v_p}\plus{}\frac{\sqrt{d^2\plus{}h^2}}v$.

I got it to $ v_p \equal{} \frac {dv}{tv\pm\sqrt {d^2 \plus{} h^2}}$ after bashing, but this doesn't look so promising. :wink:

Sorry, I don't understand the variables you are using. Maybe if you change them to greek symbols, then I will understand.

THAT? I have no idea what mewto is referring to.

I am referring to the problme obviously.


number theory
greatest common divisor
number theory proposed

Find all natural number $ n$,such that for any coprime divisors $ a,b$ of $ n$,$ a \plus{} b \minus{} 1$ is a divisor of $ n$,too.
[hide="Note:"]This is one of the most beautiful N.T problems,that i've ever seen[/hide]

It's very possible, that I've made mistakes, but nevertheless:
Assume, that $ n$, have at least two different prime divisors and then note the least of them as $ p_1$, and the next $ p_2$, so $ p_1<p_2$, and $ p_1 p_2|n$.
Then, of course $ \frac{n}{p_1}$ - is the biggest divisor of $ n$, not include the $ n$ itself.(if we find the divisor $ n>d>\frac{n}{p_1}$, then $ 1<\frac{n}{d}<p_1$, and of course $ \frac{n}{d}$ have a prime divisor, that less than $ p_1\Rightarrow$ contradiction with the choose of $ p_1$ as the least prime divisor of $ n$).But by the problem condition
\[ d'\equal{}\frac{n}{p_1}\plus{}\frac{n}{p_2}\minus{}1\ge\frac{n}{p_1}\plus{}\frac{p_1 p_2}{p_2}\minus{}1\equal{}\frac{n}{p_1}\plus{}p_1\minus{}1>\frac{n}{p_1}\]
 is also a divisor of $ n\Rightarrow$ contradiction with the fact that $ \frac{n}{p_1}$ is the biggest divisor of $ n$. So $ n$ have just one prime divisor, and have form $ n\equal{}p^k$. As we can see, this is answer.

Why $ \frac {n}{p_1}$ and $ \frac {n}{p_2}$ are coprime?Read the statement carefuly :)

[quote="silouan"]Indeed beautiful problem . :)
Consider the minimum prime divisor of $ n$ . Then $ n \equal{} p^sq$  with $ (p,q) \equal{} 1$

From the condition we have that $ p \plus{} q \minus{} 1|n$ so a prime factor of $ p \plus{} q \minus{} 1$  is also a prime factor of $ n$ .Since p+q-1 cannot have any prime factors of q therefore the only prime factor of p+q-1 is p so $ q \equal{} p^c \minus{} p \plus{} 1$ with $ c > 1$ .Now $ p^c|n$ so
$ 2p^c \minus{} p|n$  .So $ 2p^{c \minus{} 1} \minus{} 1$ must divide $ q$ . But $ \frac {p \minus{} 1}{2} < \frac {p^c \minus{} p \plus{} 1}{2p^{c \minus{} 1} \minus{} 1} < \frac {p \plus{} 1}{2}$

so a contradiction , so n cannot have two disticnt prime divisors so $ n \equal{} r$ where r is a prime .[/quote]
I think you made some mistakes.At first,the answer isn't $ n \equal{} r$,where $ r$ is a prime,the answer is $ n \equal{} r^s$,where $ r$ is prime.
Also what if $ p \equal{} 2$? Then $ p^c \minus{} p \plus{} 1$ is equal to $ 2p^{c \minus{} 1} \minus{} 1$.
I agree with your solution(it is very nice :) ),when $ n$ is odd.But you didn't consider the case when $ n$ is even(it isn't hard).So i'm waiting for your complete proof...

[quote="silouan"]Indeed beautiful problem . :)
Consider the minimum prime divisor of $ n$ . Then $ n \equal{} p^sq$  with $ (p,q) \equal{} 1$

From the condition we have that $ p \plus{} q \minus{} 1|n$ so a prime factor of $ p \plus{} q \minus{} 1$  is also a prime factor of $ n$ .Since p+q-1 cannot have any prime factors of q therefore the only prime factor of p+q-1 is p so $ q \equal{} p^c \minus{} p \plus{} 1$ with $ c > 1$ .Now $ p^c|n$ so
$ 2p^c \minus{} p|n$  .So $ 2p^{c \minus{} 1} \minus{} 1$ must divide $ q$ . But $ \frac {p \minus{} 1}{2} < \frac {p^c \minus{} p \plus{} 1}{2p^{c \minus{} 1} \minus{} 1} < \frac {p \plus{} 1}{2}$

so a contradiction , so n cannot have two disticnt prime divisors so $ n \equal{} r^f$ where r is a prime .[/quote]
Why do we know that $ p^c | n$?

[quote="Zhero"]Why do we know that $ p^c | n$?[/quote]
Because $ \text{gcd}(p,q)\equal{}1$, $ q\equal{}p^c\minus{}p\plus{}1$ and $ p^c\equal{}q\plus{}p\minus{}1 \mid n$  :D


$ S$ is a non-empty subset of the set $ \{ 1, 2, \cdots, 108 \}$, satisfying:

(1) For any two numbers $ a,b \in S$ ( may not distinct), there exists $ c \in S$, such that $ \gcd(a,c)\equal{}\gcd(b,c)\equal{}1$.

(2) For any two numbers $ a,b \in S$ ( may not distinct), there exists $ c' \in S$, $ c' \neq a$, $ c' \neq b$, such that $ \gcd(a, c') > 1$, $ \gcd(b,c') >1$.

Find the largest possible value of $ |S|$.

[hide = Solution]

We claim that the maximum value is $\boxed{79}.$

Let $T$ be the set of positive integers less than or equal to $108$ which have either $1$ or $2$ prime divisors among the set $\{2, 3, 5, 7, 11\}.$ If we consider $(T / \{55, 77\}) \cup \{30, 60, 90, 42, 84\},$ then we obtain a construction for $79.$

Let us now show that this is optimal.

[b] Lemma 1. [/b] There are at most $2$ primes in $S$ which are $>7.$

[b] Proof. [/b] Suppose that primes $p_1, p_2 > 7$ were both in $S$. Apply the second condition on them, contradiction.
$\blacksquare$

[b] Lemma 2. [/b] $1 \notin S$.

[b] Proof. [/b] Apply the second condition on $a = b = 1,$ contradiction.
$\blacksquare$

At this point, we have obtained with Lemmas $1$ and $2$ a bound of $108 - 1 - 23 = 84$.

By Lemma $1$, we may now consider only the following two cases:

[b] Case 1. [/b] There is not a prime $p > 7$ in $S.$

Then, among the pairs $(6, 35), (10, 21), (14, 15), (2, 105), (3, 70), (5, 42), (7, 30)$ at least one in each pair is not in $S$. Indeed, if both of the numbers in one of the pairs was present, then applying the first condition on them would yield that some element in $S$ has only prime divisors $>7$, contradiction. This decreases the previous upper bound of $84$ to $84-7 = 77 < 79$, so this case is resolved.

[b] Case 2. [/b] There is a prime $p > 7$ in $S.$

Let us first examine the subcases where one of $2, 3$ is not in $S.$ If $2 \notin S$, then either one of $4, 8, 16, 32, 64$ is or $|S| \le 79.$ If some power of $2$ is in $S$, then it is easily verified that we can simply add $2$ to $S$ without causing any harm. Hence, we can assume WLOG that $2 \in S.$

If $3 \notin S$, then if one of $9, 27, 81$ is, we can similarly just add $3$ to $S.$ Otherwise, let's examine the case where $3, 9, 27, 81 \notin S.$ Then, it's clear that either $|S| \le 79$, or $S \cup \{9, 27, 81\}$ is comprised of all composites, $2$, $5$, $7$, and $p$. In the former case, we win. In the latter case, applying the second condition on $7$ and $p$ yields that $7p \le 108$, and so therefore applying the first condition on $10$ and $7p$ yields a contradiction.

As a result, we may now only examine the subcases where $2, 3 \in S.$ By similar logic as before, if there exists any power of $5$ in $S$, we can assume WLOG that $5 \in S$, and similarly for $7$. We now have the following four subcases:

[b] Subcase 1. [/b] $2, 3 \in S$ but $5, 7 \notin S$

Then we have from our assumption that $5, 25, 7, 49 \notin S.$ Now, we know that either $|S| \le 79$, or $S \cup \{5, 7, 25, 49\}$ consists exactly of all composites, $2$, $3$, $5$, $7$, and $p$. In the former case, we win. Otherwise, by applying the second condition on $3$ and $p$, we know that $3p \le 108.$ Therefore, as $3p$ is composite, we have that $3p \in S.$ Thus, as $70$ is also composite, applying the first condition on $70$ and $3p$ yields a contradiction.

[b] Subcase 2. [/b] $2, 3, 5 \in S$ but $7 \notin S$

Then we have from our assumption that $7, 49 \notin S.$ By applying the second condition on $5, p$, we know that $5p \le 108.$ Now, in each of the pairs $(6, 5p), (10, 3p), (15, 2p)$ at least one number is not in $S$ (otherwise just apply the first condition to obtain a contradiction). Together with $7$ and $49$, these improve the previous bound of $84$ by $5$ to $79$, and so we're done here.

[b] Subcase 3. [/b] $2, 3, 7 \in S$ but $5 \notin S$

Then we have from our assumption that $5, 25 \notin S.$ By applying the second condition on $7, p$, we know that $7p \le 108.$ Now, in each of the pairs $(6, 7p), (14, 3p), (21, 2p)$ at least one number is not in $S$ (otherwise just apply the first condition to obtain a contradiction). Together with $5$ and $25$, these improve the previous bound of $84$ by $5$ to $79$, and so we're done here.

[b] Subcase 4. [/b] $2, 3, 5, 7 \in S$

By applying the second condition on $7, p$, we know that $7p \le 108.$ Now, in each of the pairs $(30, 7p), (42, 5p), (35, 6p), (70, 3p), (105, 2p)$ at least one number is not in $S$ (otherwise just apply the first conditionto obtain a contradiction). These five pairs improve our previous bound of $84$ by $5$ to $79$, and so we're done here.

As we've exhausted all subcases, we are done in Case $2$.

As we've exhausted both cases, we have shown that $|S| = 79$ is optimal, and so we conclude that the answer is indeed $\boxed{79}.$

$\square$

[/hide]


Let a,b,c >0
If (a+b)(b+c)(c+a)=1

Prove that

ab+bc+ca<= 3/4

[quote="conejita"]Let $a,b,c >0$
If $(a+b)(b+c)(c+a)=1$

Prove that

$ab+bc+ca\leq\frac{3}{4}$[/quote]
Let $a+b+c=3u,$ $ab+ac+bc=3v^{2}$ and $abc=w^{3},$ $v>0.$
Then $(a+b)(a+c)(b+c)=9uv^{2}-w^{3}$ and $ab+bc+ca\leq\frac{3}{4}\Leftrightarrow 3v^{2}\leq\frac{3}{4}\Leftrightarrow$
$\Leftrightarrow v^{3}\leq\frac{1}{8}\Leftrightarrow8v^{3}\leq9uv^{2}-w^{3}.$ But $w^{3}\leq\frac{v^{4}}{u}.$ Hence remain to prove that
$8v^{3}\leq9uv^{2}-\frac{v^{4}}{u}\Leftrightarrow9u^{2}-8uv-v^{2}\geq0\Leftrightarrow(u-v)(9u+v)\geq0,$ which true.

Also see here for the main page:
http://www.mathlinks.ro/Forum/viewtopic.php?p=232780#p232780

We have by Muirhead $1=(a+b)(b+c)(a+c)=\sum_{sym}+2abc\ge 8abc$ so $abc\le \frac{1}{8}$
We also have $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(\sum_{sym}+2abc) =3+a^{3}+b^{3}+c^{3}\ge 3+(3/8)=\frac{27}{8}$(by Muirhead $8(a^{3}+b^{3}+c^{3})\ge 3(\sum_{sym}+2abc)$) so $a+b+c\ge \frac{3}{2}$
Finally we have:
$(ab+bc+ac)(a+b+c)=\sum_{sym}+3abc=1+abc\le \frac{9}{8}$
$\Leftrightarrow \frac{3}{2}(ab+bc+ac)\le (ab+bc+ac)(a+b+c)\le \frac{9}{8}$
$\Leftrightarrow \frac{3}{2}(ab+bc+ac)\le \frac{9}{8}$
$\Leftrightarrow ab+bc+ac\le \frac{3}{4}$
QED


geometry
incenter
circumcircle
search
geometry unsolved

A circle which is tangent to the sides $ [AB]$ and $ [BC]$ of $ \triangle ABC$ is also tangent to its circumcircle at the point $ T$. If $ I$ is the incenter of $ \triangle ABC$ , show that $ \widehat{ATI}\equal{}\widehat{CTI}$

Dear Mathlinkers,
your circle is a mixtilinear incircle.
A proof can be seen on
http://perso.orange.fr/jl.ayme   vol. 4     A new mixtilinear incircle adventure p. 17-18  (in french)
Sincerely
Jean-Louis

[quote="sinankaral53"]A circle which is tangent to the sides $ [AB]$ and $ [BC]$ of $ \triangle ABC$ is also tangent to its circumcircle at the point $ T$. If $ I$ is the incenter of $ \triangle ABC$ , show that $ \widehat{ATI} \equal{} \widehat{CTI}$[/quote]

 It is a nice problem(probably an IMO  short list) but I can not find where we have discussed it here ! If someone has the

 link, please send it!

Mmm ! I found one :

http://www.mathlinks.ro/viewtopic.php?search_id=647921583&t=31739

 but it is also discussed and some where else.

 Babis


For any natural number $ n\geq 3$ , let $ m(n)$ denote the maximum number
of points lying within or on the boundary of a regular $ n$-gon of
side length $ 1$ such that the distance between any two of the points
is greater than $ 1$. Find all $ n$ such that $ m(n) \equal{} n\minus{}1$.

m(3)=1. Because if you place a point and draw a circle of radius 1 centering the point, the triangle will be in the circle. In a regular hexagon of side length 1, the distance between the center of the hexagon and a vertex is 1. So for any regular n-gon (n>6) you can always place n-1 points on the n-gon and one point at the center. So m(n)>n-1 for all n>6. For n = 4, 5, 6 you can place n-1 points on the n-gon and no more points can be placed because the circles of radius 1 drawn centering the points will consume the wholw n-gon. So m(n)=n-1 for n=4, 5, 6.

Is it ok?

[I don't know how i made this mistake writing easy on the subject description :o  :( .actually I wanted to write "looks easy!".I must tell a moderator about it as the problem is not easy at all :mad: ]

[quote="diavolo diavol"]
Is it ok?[/quote]

The answer seems correct but the solution is not complete.as u haven't showed that we can not find more than $ n\minus{}1$ points.actually the main thing is to show that $ n\minus{}1$ is a bound for for these points.and the case for n>6 is not clear.I mean, how to place the $ n\minus{}1$ points?

The problem may be easy,but actually I could not solve it by myself :blush:

Proof of the boundary:
I’m proving it for hexagon. The rest are similar.
Suppose m(6)>5. Then there is a hexagon in that regular hexagon having each side and each diagonal greater than 1. Suppose the regular hexagon is H and the other one is abcdef. Suppose the point c is in H, not on the side. Now draw a circle of radius 1 centering c. If c is at the center of H, the circle consumes the whole H. So c is not at the center. Then let the circle intersects H at m and n. Let p is a point on the side of H and in the circle such that pm = pn. Now angle mcp = angle ncp > 90 degree. Therefore pm = pn > cm = cn = 1. So abpdef is another hexagon in H having all the sides and diagonals greater than 1. Thus we can replace all the points of abcdef which is not on the side of H with a point on the side of H and finally we get a hexagon having all the sides and diagonals greater than 1 and all the vertices on the sides of H. But this is not possible. So there can not be six points in or on H of the given criterion.

And about placing n-1 points; it’s simple. Just put one point at a vertex. Then put another on a side at an infinitesimal distance from the next vertex so that the distance between the two points is slightly greater than 1. Continue this process and you will be able to place n-1 points.


geometry vector circumcircle trigonometry

H is the orthocenter of ABC and AB = CH. Find ∠C.

[hide="vectors"]
Let the origin be the circumcenter of $ABC$ and let $A$, $B$, $C$ be the vectors to the corresponding points. It is known that $H = A+B+C$, so

$|A-B| = |A+B+C-C| = |A+B|$.

But this means $A$ and $B$ are perpendicular (you can verify this easily; one way is setting $A = (x_{1},y_{1})$ and $B = (x_{2}, y_{2})$ from which we obtain $A \cdot B = x_{1}x_{2}+y_{1}y_{2}= 0$). Then $\angle AOB = \frac{\pi}{2}$. But since $C$ is on the circumcircle, $\angle ACB = \frac{1}{2}\angle AOB = \frac{\pi}{4}$.[/hide]

project A onto BC at D, note $\angle HCD=\angle DAB=90-\angle C$, then $\triangle ABD\cong\triangle CHD$, so $HD=DB$, then $\angle CBH=45$, so $\angle C=90-45=45$

Both solutions are not full.
[hide]Missing case $\Delta ABC$ is obtuse. $\angle ACB=135^{o}$ is also possible. In general $CH=2R|\cos \gamma|$[/hide]


Let $a,b,c$ be non-negative real numbers, no two of which are zero, let $p$ and $q$ be real numbers ($p>-2$) and let
$E(a,b,c)=\sum\frac{ab+qbc+ca}{b^2+pbc+c^2}$.
Then,
(a) $E(a,b,c) \geq \frac{3(q+2)}{p+2}$,  for $q \leq p-1$;
(b) $E(a,b,c) \geq \frac{q}{p+2}+2$,  for $p-1 \leq q \leq (p+2)^2$;
(c) $E(a,b,c) \geq 2\sqrt{q} -p$,  for $q \geq (p+2)^2$.

It seems very unlikely that these problems have a nice (and not too long) solution. Vasc, do you agree/disagree?

At least case (a) can be solved by SOS theorem. I think It's not so long.


function
limit
inequalities
real analysis
real analysis unsolved

Let $g: \mathbb{R}\to\mathbb{R}$ a convex function and derivable in $0$. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that:

        1) $f(x+y)\leq f(x)+f(y)$ for any $x, y\in\mathbb{R}$;

        2) $f(x)\leq g(x)-g(0)$ for any $x\in\mathbb{R}$.

I proved that $\lim_{x\to 0}f(x)$ must exist and it is 0. Assume WLOG $g(0)=0$

Assume the contrary. Someone can find a sequence $(x_n)$ such that $0<x_n <\frac{1}{n^2},f(x_n)\to a\neq 0,\forall n\in\mathbb{N}$.
Case I: a>0. For a sufficiently large N we have that $f(x_n)>\frac a 2>0,\forall \in \mathbb{N}$. But $\frac a 2 <f(x_n)\le g(x_n) \to g(0)=0$.False.
Case II: a<0. For a sufficiently large N we have that $f(x_n)<\frac a 2<0,\forall \in \mathbb{N}$. Since $x_n>0$ and $\sum_0^{\infty}x_n<\sum_0^{\infty}\frac{1}{n^2}$ we have that $\sum_0^{\infty}x_n$ converegs to L. From hipotesys we know that $f(\sum_0^{\infty}x_n)<\sum_0^{\infty}f(x_n)<-n \frac{a}{2}$. Tending n to infinity in this inequality we get: $f(L)<-\infty$.False.
Remark: $a$ may not be $< \infty$ but if $a=\pm \infty$ the contradictions still holds.

So, $\exists \lim_{x\to 0}f(x)$ and $\lim_{x\to 0}f(x)=0$.

I actually proved that $\lim_{x\to 0,x>0}f(x)=0$. But the same method works for proving $\lim_{x\to 0,x<0}f(x)=0$.
Setting $x=y=0$ in 1) we get $f(0)\le 2f(0)\Rightarrow f(0)\ge0$. But from 2) we have that $f(0)\le 0$. So, $f(0)=0$ and $f$ continuous in 0.

Now I can prove more: $f(x)\le x \cdot g'(0)$.
Assume that there exists $f(x_0) > x_0\cdot g'(0)\Rightarrow \frac{f(x_0)} {x_0} \ge g'(0)+\epsilon$ for some $\epsilon>0$.
From 1), setting $x=y=\frac t 2$ we get: $f(\frac t 2) \ge \frac{f(t)} 2,\forall t \in \mathbb{R} \Rightarrow \frac{f(t)}{t}\le \frac{f(\frac t 2)}{\frac t 2}\Rightarrow\frac{f(t)}{t}\le \frac{f(\frac{t} {2^n})}{\frac{t }{2^n}},\forall t\in \mathbb{R},n\in \mathbb{N}$. So, we have that
$g'(0)+\epsilon \le \frac{f(x_0)} {x_0} \le\frac{f(\frac{x_0} {2^n})}{\frac{x_0} {2^n}} \le frac{g(\frac{x_0} {2^n})}{\frac{x_0} {2^n}} \to g'(0)$. False.
Since $g$ is convex we can easily prove that $g(x)\ge x\cdot g'(0)$. Now we have that  $f(x)\le x \cdot g'(0)$ is a sufficient condition for $f(x)\le g(x)$. So, the problem is equivalent to this:

Given $m\in \mathbb{R}$, find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ continuous in $0$ with $f(0)=0$ such that:
1) $f(x+y)\le f(x)+f(y)$ for any $x,y \in \mathbb{R}$
2) $f(x)\le x\cdot m \forall x\in \mathbb{R}$
This one must be easier. :D


Let triangle $\ ABC$ with $\ AB=AC$. The point $\ D$ on the side $\ BC$ so that the incircle of triangle $\ ADB$ and the $\ A$-excircle of triangle $\ ADC$ have equal radii $\ r$.
Prove that: $\ r=\frac{h_{b}}{4}$ ($\ h_{b}$ is the measures of altitude from $\ B$ of triangle $\ ABC$)


consider a "weighted" cube, where each of the edges is assign an integer range from 1 to 12.

a) we have label a cube such that the sums of the 3 edges meeting at each vertex are all equal??

b) what if we switch one of them with the number 13???

[quote="Inspired By Nature"][quote="pkerichang"]a) we have label a cube such that the sums of the 3 edges meeting at each vertex are all equal??[/quote]
Could you rephrase that?[/quote]

I think he means:

Can we label a cube such that the sums of the labels on the three edges meeting at each vertex are equal (for every vertex)?

yeah, more or less like that.  For example, maybe all eight vertices has the sum 8.

[quote="lotrgreengrapes7926"][quote="t0rajir0u"]Of course, it's not hard to calculate what the sum must be.  :)[/quote]
[hide="Is it..."] $\frac{12\cdot13}{2}\cdot\frac{3}{12}=19.5$. [/hide][/quote]

So...What can you conclude??? how about part b??


function
trigonometry
limit
algebra
domain
topology
real analysis

If real  function  $f$ is continuous on interval $D$  and the image
$f(D)$  is an open interval , prove (or disprove)  that  $D$  is also an open interval.

 [u]Babis[/u]

[b]Note[/b]

 [i]Of course D can not be closed  , but can D be an open - closed inerval ?[/i]

D can be an open interval or union of some open intervals
If f only define on D then D must be open because $D=f^{-1}f(D)$

A small remark.

If $f$ is continuous, then
1) $f^{-1}(D)$ is an open set if $D$ is an open set;
2) $f^{-1}(D)$ is a closed set if $D$ is a closed set.

D can be a closed-open interval.

D = [ 0 , $\frac\pi2$ )
f(x) =  x sin( tan(x) )
f(D) = ( - $\frac\pi2$ , $\frac\pi2$ )

Just for curiosity let's add Lipschitz condition.

Is there a Lipschitz-continuous function f : [0,1) -> R such that the image of f is open?

But we have not that
                $f^{-1}$  exists(  $f$  is not one to one)

 Second , the last post proves a different result. Why

 [u]Babis[/u]

The continuous image of a compact set is compact, so there is no counterexample with a finite closed interval. On the other hand, as cimabue pointed out, there is a counterexample on an infinite closed interval (actually, it's really easy: $\arctan$ on the whole real line works).

Oh yeah- that's an open interval, too. I was misreading the problem. Of course, we can do it for intervals that aren't open, as already shown.

[u]To cimabue[/u]

 if  $f(x) = x\sin\frac{1}{x}$ , and  $x\in D= [1 , + \infty )$ , which is $f(D)$ ?

  See that  $limf(x) = 1$ , as x tends to infinity.

 Probably the topological definition of continuity works and as posts 2 , 3  say , D must be open too !

 Otherwise  , we must find a counterexample as Cimabue does. But does it really works. I' ll see it again more carefully.

[u]Babis[/u]

there are sequences $x_n$ and $y_n$ such that $\lim f(x_n)=+\infty$ and $\lim f(y_n)=-\infty$. by the intermediate value property the image is $\mathbb R$.

I wish you are right , because then the problem will be solved. But can you find such a sequence?I can not. If the sequence has a real limit , then f(a_n) has also a real limit , since f is continuous. If a_n tends to infinity , then f(a_n) tends to 1.  What do you think? Or do I make  any mistake ?

[u]Babis[/u]

youare right. i made a mistake. the function should be $f: [1,\infty)\to\mathbb R$, $f(x)=x\sin(x)$.
if you want a finite domain, $g(x)=x\sin^3(\frac{1}{x})$ on $(0,\pi]$ for example.

Thank you ! I just thought it could be an easy question to my high school pupils. But ..... I had troubles!

   [u]Babis[/u]


LaTeX
number theory proposed
number theory

Find all positive integers $n$ such that if $a$ and $b$ are positive integers for which $n$ divides $a^{2}b+1$ , then $n$ divides $a^{2}+b$

Well, you used some strange symbols...
Is the current state the correct one¿ Otherwise edit it please again.

I think that it's okay (I have no problem with understanding and reading of question) :|

Umm...
I had edited it, what you wrote was hard to understand (especially your symbol "ˇ" instead of "^" is very unusual).
Try to use $\LaTeX$ also please.

$n|a^{2}b+1\Longrightarrow n|a{}^{b}{}^{2}+b\Longrightarrow n|a^{2}(b^{2}-1)\Longrightarrow n|b^{2}-1 .$
If n is odd $a=\frac{n-1}{2}, b=-4$ give n|15. Obviosly n=1 solution. If n=3 then from $3|a^{2}b+1\Longrightarrow 3\not |a,b=2(mod \ 3)$, it give $3|a^{2}+b$. Therefore n=3 solution too. If n=5 $5|a^{2}b+1$ give $5|a^{2}+b$. Therefore n=5 and n=15 solutions too.
If $n=2^{k}m$, were m is odd, then from $n|a^{2}b+1$ we have $m|a^{2}b+1$. It give m=1,3,5,15. Obviosly n=2,4,8 are solutions. It is easy to chek, that $n=2^{k},k>3$ is not solution. Therefore all solutions give by $n=2^{k}3^{m}5^{l}, k=0,1,2,3,m=0,1,l=0,1$ or m is divisor 120.

Yes. I don' check, that 16 works. Bur 32 don't work (a=3,b=7).

The solution is actually all positive divisors of 240.


Let x,y,z >0 such that xyz=1.Prove that
        2 \sum ( \sqrt[(x+1)(y+1)/z]) <= prod(x+1) +1/2prod(y+z).

We have that LHS=2 \sum  \sqrt x(1+y)*  \sqrt y(1+x)  \leq 2 \sum x+2 \sum xy, by AM-GM. Expanding everything we arrive at
( \sum x)( \sum xy-2) \geq 2 \sum xy-3. This one is obvious from  \sum x \geq 3 and  \sum xy>=3.

Wow ! You are such an expert in inequality when you give an extreme beautiful solution in such a short time. Congratulations!  :)
    In fact, I made it symmetric and here is my original inequality which is an uglier but not less interesting than the first one:
   2 \sum ( \sqrt (x+1)(y+1) /z) <=min{x^2*y+y^2*z+z^2*x, x*y^2+y*z^2+z*x^2}+prod(x+1) +1.
     I hope you will have a beautiful solution for hits one.  :)

The above solution works. All we have to prove is that x^2y+y^2z+z^2x+3>=x+y+z+xy+yz+zx. But this comes from Schur:
   \sum x^2y+3= \sum [(x^2y)^1/3]^3+3 \geq  \sum (x^2y)^1/3*(y^2z)^1/3((x^2y)^1/3+(y^2z)^1/3)= \sum x+ \sum xy

That's such beautiful solution! My solution is based on this following equality: for every xyz=1 we have
   x/(xy+x+1)+y/(yz+y+1) + z/(zx+z+1)=1.
  And then we apply Schwarzt inequality, we have
      (\sum ( \sqrt(x^2+x))^2<=(x+1)(xy+x+1) + ....
   This implies our result.


inequalities
trigonometry
function
logarithms
geometry
circumcircle
inequalities proposed

Prove the following inequality

$\sin A+\sin B+\sin C\geq \sin 2A+\sin 2B+\sin 2C$ where $A,B,C$ are angle inside a triangle .

[quote="julien_santini"]Using the concavity of sine is enough here. You get:

sin(A+B) >= 1/2*sin(2A)+1/2*sin(2B)
sin(A+C) >= 1/2*sin(2A)+1/2*sin(2C)
sin(B+C) >=  1/2*sin(2B)+1/2*sin(2C)

Adding everything gives sin(A+B)+sin(A+C)+sin(B+C) >= sin(2A)+sin(2B)+sin(2C), but sin(A+B) = sin(C), sin(A+C) = sin(B), and sin(B+C) = sin(A), which concludes the proof.

--
Julien Santini[/quote]

Excuse me Julien Santini , I agree that $f(x)=\sin x$ is concave in $[0,\pi]$ but $f(x)=\sin 2x$ doesnt nessecary concave in $[0,2\pi]$ . Please verify it .

 :)

Just make use of $\sum \sin A = 4 \prod \cos \frac{A}{2}$ and $\sum \sin 2A = 4 \prod \sin A$. It suffices to prove $\prod \sin \frac{A}{2} \le \frac{1}{8}$, but this is due to jensen and the concavity of the function $\ln(\sin(x))$ on $[0, \frac{\pi}{2}]$

[quote="pvthuan"][quote="shyong"][quote="Soarer"]Just make use of $\sum sinA = 4 \prod cos \frac{A}{2}$ and $\sum sin2A = 4 \prod sinA$. It suffices to prove $\prod \sin \frac{A}{2} \le \frac{1}{8}$, but this is due to jensen and the concavity of the function $ln(sin(x))$ on $[0, \frac{\pi}{2}]$[/quote]

Thats great soarer , I have another geometry approach here using that well-known inequality $R\geq 2r$

[hide]
We have $\sum\sin A=\frac{S}{Rr}$ $(1)$ where $S$ is the area of triangle $ABC$ .

Then we also have $\sum\sin 2A=2\sum\sin A\cos A=\frac{1}{R}\sum a\cos A$

but $a\cos A=\frac{2S_{OBC}}{R}$ where $O$ is the center of circle which $ABC$ is inscribed in it .

So we have $\sum\sin 2A=\frac{2S}{R^2}$  $(2)$

Now from $(1)$ and $(2)$ we have $\frac{(1)}{(2)}=\frac{R}{2r}\geq 1$  .Hence the inequality  is proved .

 :)

[/hide][/quote]
By sum-to-product forumula we have $\sin 2A+\sin2B=2\sin (A+B)\cos(A-B)=2\sin C\cos(A-B)$. Since $\sin C\geq0$ and $\cos(A-B)\leq 1$, we have $\sin2A+\sin2B\leq 2\sin C$. Similarly, we get two others then add them to obtain the desired one.[/quote]

That is indeed another kool method of solving it   ;) Good job [b]pvthuan[/b]

but it's in fact the way #2 mentioned, although his reasoning may not be correct.

If $a,b,c$ are the sides of a triangle $ABC$, $R$ is the radius of circumcircle, $r$ is the radius of incircle, then

$\sin 2A+\sin 2B+\sin 2C\leq\frac{3\sqrt{3}r}{R}\leq\frac{27r}{a+b+c}\leq 3r\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right)$
$\leq\sin A+\sin B+\sin C$.

[quote]If $a,b,c$ are the sides of a triangle $ABC$, $R$ is the radius of circumcircle, $r$ is the radius of incircle, then

$\frac{27r}{a+b+c}\leq 3r\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right)$
[/quote]

it's obvious

[quote]$3r\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\leq\sin A+\sin B+\sin C$ [/quote]

it's obvious, too because:
$3r\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3r}{2R}\left(\frac{1}{\sin A}+\frac{1}{\sin B}+\frac{1}{\sin C}\right)$ and $\frac{r}{R}\leq\frac{1}{2}$

We also have:
$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq\frac{3r}{R}$

Any one can prove it?

If $m_{a},m_{b},m_{c}$ are the medians of a triangle $ABC$, $R$ is the radius of circumcircle, $r$ is the radius of incircle, then

\[\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq \cos A+\cos B+\cos C\geq \frac{3r}{2}\left(\frac{1}{m_{a}}+\frac{1}{m_{b}}+\frac{1}{m_{c}}\right)\geq\frac{3r}{R}.\]

[quote="Ji Chen"]If $m_{a},m_{b},m_{c}$ are the medians of a triangle $ABC$, $R$ is the radius of circumcircle, $r$ is the radius of incircle, then
$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq \cos A+\cos B+\cos C$
[/quote]

We have: $\cos A+\cos B=2\cos\frac{A+B}{2}.\cos\frac{A-B}{2}=2\sin\frac{C}{2}.\cos\frac{A-B}{2}\leq 2\sin\frac{C}{2}$

[quote="Ji Chen"]If $m_{a},m_{b},m_{c}$ are the medians of a triangle $ABC$, $R$ is the radius of circumcircle, $r$ is the radius of incircle, then
$\cos A+\cos B+\cos C\geq \frac{3r}{2}\left(\frac{1}{m_{a}}+\frac{1}{m_{b}}+\frac{1}{m_{c}}\right)$
[/quote]

Dear Ji Chen! Could you show me hint to prove it?


$ a,b,c >0$ s.t. $ ab\plus{}bc\plus{}ca\plus{}2abc\equal{}1$.

Prove that $ 0 \leq \sqrt{a}\plus{} \sqrt{b}\plus{} \sqrt{c} \minus{} 2\sqrt{1\plus{}abc} \leq \frac{\sqrt{1\plus{} \frac{1}{abc}}}{2}$.

[quote="ttibur"]$ a,b,c > 0$ s.t. $ ab \plus{} bc \plus{} ca \plus{} 2abc \equal{} 1$.

Prove that $ 0 \leq \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \minus{} 2\sqrt {1 \plus{} abc}$.[/quote]

 Denote  $ a \equal{} \frac {x}{y \plus{} z}$ and similarly and we have

 $ 2\sqrt {1 \plus{} \frac {xyz}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}}\le \sqrt {\frac {x}{y \plus{} z}} \plus{} \sqrt {\frac {y}{z \plus{} x}} \plus{} \sqrt {\frac {z}{x \plus{} y}}$ or

 $ 2\sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x) \plus{} xyz}\le \sum\sqrt {x(x \plus{} y)(x \plus{} z)}$

 By squaring
$ 4[3xyz \plus{} \sum xy(x \plus{} y)]\le\sum x^3 \plus{} \sum xy(x \plus{} y) \plus{} 3xyz \plus{} 2\sum (x \plus{} y)\sqrt {xy(z \plus{} x)(z \plus{} y)}$ wich is true because

 $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz\ge \sum xy(x \plus{} y)$ and
 $ \sum(x \plus{} y)\sqrt {xy(z \plus{} x)(z \plus{} y)}\ge\sum (x \plus{} y)\sqrt {xy}(c \plus{} \sqrt {xy})\ge\sum xy(x \plus{} y) \plus{} \plus{} 6xyz$

For the second inequality: Define, as Marius Mainea did, $ a = \frac {x}{y + z}$, $ b = \frac {y}{z + x}$, $ c = \frac {z}{x + y}$. The inequality becomes
\[ 2\sum \sqrt {\frac {x}{y + z}} \le \sqrt {\frac {xyz + (x + y)(y + z)(z + x)}{(x + y)(y + z)(z + x)}}\left(\sqrt {\frac {(x + y)(y + z)(z + x)}{xyz}} + 4\right).\]
Note that $ \prod(x + y) + xyz = \sum x \sum xy$. Multiply the inequality by $ \sqrt{xyz(x + y)(y + z)(z + x)}$ and use AmGm to obtain the stronger inequality
\[ \sum x \sum xy + 3xyz \le \sqrt {\sum x \sum xy} \left( \sqrt {\sum x \sum xy - xyz} + 4\sqrt {xyz}\right).\]
Squaring yields
\[ 9x^2y^2z^2 \le9xyz\sum x \sum xy + 8\sum x \sum xy\prod \sqrt {x^2 + xy},\]
which is obvious.

Actually, the constraint $ 4$ may be replaced by $ 4 - 2\sqrt {2}$, which transforms the original inequality to $ \sqrt{2} \le \sum \sqrt {a} - (2 - \sqrt {2})\sqrt {1 + abc} \le \frac {1}{2}\sqrt {1 + \frac {1}{abc}}.$

[quote="Marius Mainea"][quote="ttibur"]$ a,b,c > 0$ s.t. $ ab \plus{} bc \plus{} ca \plus{} 2abc \equal{} 1$.

Prove that $ 0 \leq \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \minus{} 2\sqrt {1 \plus{} abc}$.[/quote]

 Denote  $ a \equal{} \frac {x}{y \plus{} z}$
[/quote]

I checked your substitution and it's correct, but how did you figure out to take this substitution? Cause this implies $ a\plus{}b\plus{}c\ge \frac{3}{2}$ and thats not that clear at first sight.

[quote="Rafikafi"]I checked your substitution and it's correct, but how did you figure out to take this substitution? Cause this implies $ a \plus{} b \plus{} c\ge \frac {3}{2}$ and thats not that clear at first sight.[/quote]

$ a\plus{}b\plus{}c\plus{}2\equal{}abc \Rightarrow (a\plus{}1)(b\plus{}1)\plus{}(b\plus{}1)(c\plus{}1)\plus{}(c\plus{}1)(a\plus{}1)\equal{}(a\plus{}1)(b\plus{}1)(c\plus{}1)$
$ \Rightarrow$ there exists $ x,y,z$ such that $ a\equal{}\frac{x\plus{}y}{z}$,$ b\equal{}\frac{y\plus{}z}{x}$,$ c\equal{}\frac{z\plus{}x}{y}$.

Plugging in $ \frac{1}{a}$, $ \frac{1}{b}$, $ \frac{1}{c}$ yields the result  :D


function
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

In $\mathbb{R}^{n}$ I have been given a family of balls such that the intersection of any two balls contains at most one point. Prove that the set of intersection points can't be uncountable.

It has been discussed before. The trick is to construct a function from the set of balls to $\mathbb Q$ by assigning each ball to a rational in its interior. This function is injective, so there are only countably many balls. We get at most countably many pairs of spheres that may intersect, from where we get the conclusion.


Let $ H$ be the intersection of altitudes in triangle $ ABC$. If $ \angle B\equal{}\angle C\equal{}\alpha$ and $ O$ is the center of circle passing through $ A$, $ H$ and $ C$, then find $ \angle HOC$ in terms of  $ \alpha$.

$\textbf{(A)}\  90{}^\circ \minus{}\alpha \qquad\textbf{(B)}\  90{}^\circ \plus{}\frac{\alpha }{2} \qquad\textbf{(C)}\  180{}^\circ \minus{}\alpha \\ \qquad\textbf{(D)}\  180{}^\circ \minus{}\frac{\alpha }{2} \qquad\textbf{(E)}\  180{}^\circ \minus{}2\alpha$

[hide]$ \angle HOC\equal{}2\angle HAC\equal{}2(90^\circ\minus{}\alpha)\equal{}180^\circ\minus{}2\alpha$, hence the answer is $ E$.[/hide]


$x, y$ are non-negative real numbers and $2x+y+\sqrt{2xy+3x^{2}+y^{2}}= 5$. Prove that $xy^{2}< 1$.


I don't know if this is similar to Grant's post, but I come from a school where our MATHCOUNTS team is actually pretty decent. We make State consistently, and we're really advanced. We also have a Nationals qualifier. How can we get into MOEMS?

I would assume you would join the normal way, regardless of how good your team is.

so ur allowed to just normally register?


is there an easier way to do this than multiplying this whole entire thing out?

(5x^2 - 9xy - 2y^2)/(y^2 + 4xy - 5x^2) * (y^2 + 2xy - 3x^2)/(15x^2 + 8xy + y^2) *

(15x^2 + 2xy - y^2)/(3x^2 - 5xy - 2y^2)

Am i really just suppose to multiply across and get an even bigger problem. I looked for similarities that i know of (like looking to see if i can cross out or reduce anything, but i came up with nothing.)[/code]

Have you tried factoring all those quadratics?   :wink:  :roll:

[quote="fero4u123"]i thought of that but wasnt sure since the form isnt ax^2+bx+c .... the third term in most of them are also ^2 and not like terms[/quote]
It is in that form, just treat the y's as another constant or something.
$ 5x^{2}\minus{}9xy\minus{}2y^{2}\equal{}5x^{2}\minus{}10xy\plus{}xy\minus{}2y^{2}\equal{}5x(x\minus{}2y)\plus{}y(x\minus{}2y)\equal{}(5x\plus{}y)(x\minus{}2y)$
I'll leave the rest for you to factor :)

These are just like your "standard"

$ (ax\plus{}m)(bx\plus{}n)\equal{}0$

factorizations, but with another variable. Just to show you what I mean, factor

$ 0\equal{}x^2\plus{}6x\plus{}5$

Then factor

$ 0\equal{}x^2\plus{}6xy\plus{}5y^2$

How do your solutions compare?


$ X$, $ Y$ are topological spaces. $ f: X\rightarrow Y$ is a closed, continuous onto mapping. If $ X$ is normal, how to prove $ Y$ is normal?

thx

[quote="Nukular"]Since $ f : X \to Y$ is a closed onto map, it is also an open map.

Let $ A,B \subset Y$ be disjoint closed subsets. Then if they have preimages $ A',B' \subset X$, then $ A' \cap B' \equal{} \emptyset$ and so there are open sets $ U', V'$ separating $ A'$ and $ B'$. Then $ U \equal{} f(U')$ and $ V \equal{} f(V')$ are open subsets of $ Y$ that separate $ A$ and $ B$.[/quote]

I'm sorry I don't fully understand.

1. Why $ f$ is open?

2. Since $ f$ is not necessarily one-to-one, why $ f(U')$ and $ f(V')$ are disjoint?

thx again

[quote="hubanya"]

1. Why $ f$ is open?
[/quote]

let $ O\subset X$ is open then $ f(X\setminus O) \equal{} Y\setminus f(O)$

[quote="hubanya"]
2. Since $ f$ is not necessarily one-to-one, why $ f(U')$ and $ f(V')$ are disjoint?

thx again[/quote]

$ f$ is onto  , can you prove that $ f(f^{\minus{}1}(A))\equal{}A$ ?

Actually, yeah, that's a screw-up of mine... as I had it $ U'$ and $ V'$ are not necessarily disjoint.

[quote="1234567a"][quote="hubanya"]

1. Why $ f$ is open?
[/quote]

let $ O\subset X$ is open then $ f(X\setminus O) \equal{} Y\setminus f(O)$

[quote="hubanya"]
2. Since $ f$ is not necessarily one-to-one, why $ f(U')$ and $ f(V')$ are disjoint?

thx again[/quote]

$ f$ is onto  , can you prove that $ f(f^{ \minus{} 1}(A)) \equal{} A$ ?[/quote]

sorry
let $ O\subset X$ is open , then I can prove $ Y\setminus f(O)\subset f(X\setminus O)$, however, I can't prove the inverse inequality since $ f$ is not one-to-one.

[quote="Third Edition"]I am not sure whether $ f$ is necessarily open (actually I doubt it but I don't have a counterexample in stock) ;) , but the following works:

If $ A,B$ are closed subsets of $ X$, let $ U',V'$ open disjoint subsets in $ X$ with $ f^{ \minus{} 1}(A)\subset U'$ and $ f^{ \minus{} 1}(B)\subset V'$, then just check that the open sets $ U \equal{} Y\setminus f(X\setminus U')$ and $ V \equal{} Y\setminus f(X\setminus V')$ do the required job.

[hide="Hint:"]Show this equivalence first: $ y\in U$ iff $ f^{ \minus{} 1}(\{y\})\subset U'$.[/hide][/quote]

Thank you very very much!
However, I can't prove $ f^{ \minus{} 1}(\{y\})\subset U'$ implies $ y\in U$, but the other direction is enough:)

[quote="hubanya"]However, I can't prove $ f^{ \minus{} 1}(\{y\})\subset U'$ implies $ y\in U$[/quote]Assume that $ f^{ \minus{} 1}(\{y\})\subset U'$ and $ y\notin U$. Then $ y\in f(X\setminus U')$ which means there is some $ x\notin U'$ with $ f(x)\equal{}y$. This implies $ x\in f^{ \minus{} 1}(\{y\})\cap (X\setminus U')\neq\emptyset$ or $ f^{ \minus{} 1}(\{y\})\not\subset U'$. Contradiction.

BTW: There is no need to quote all of a previous posting if any at all. ;) That keeps the thread shorter and more readable.


calculus
real analysis
real analysis unsolved

Find the sum of the series
$ x\plus{}\frac{x^3}{1\cdot 3}\plus{}\frac{x^5}{1\cdot 3\cdot 5}\plus{}\cdots\plus{}\frac{x^{2n\plus{}1}}{1\cdot 3\cdot 5\cdot \cdots \cdot (2n\plus{}1)}\plus{}\cdots$

As I recall from my Calculus course, it is the Taylor expansion of the $ \tan(x)$ function.

If $ f(x) \equal{} x \plus{} \frac {x^3}{1\cdot 3} \plus{} \frac {x^5}{1\cdot 3\cdot 5} \plus{} \cdots \plus{} \frac {x^{2n \plus{} 1}}{1\cdot 3\cdot 5\cdot \cdots \cdot (2n \plus{} 1)} \plus{} \cdots$
then
$ \frac{d}{dx}f(x)\minus{}1 \equal{} xf(x)$.

This is easy to solve by standard techniques
The solution of $ \frac{d}{dx}g(x) \equal{} x*g(x)$ is $ g(x) \equal{} Ce^{\frac12 x^2}$.
Substitution of $ f(x)\equal{}C(x)e^{\frac12 x^2}$ yields
$ f(x)\equal{}\sqrt{\frac{\pi}{2}}\mbox{ erf}\left(\frac{x}{\sqrt{2}}\right)e^{\frac12 x^2}$.


integration
calculus
Functional Analysis
real analysis
real analysis unsolved

Find $ \displaystyle\min\limits_{a,b,c,d\in \mathbb{C}} \{\int\limits_{\minus{}1}^0 (x^3\plus{}ax^2\plus{}bx\plus{}c)^2dx \plus{} \int\limits_0^1 (x^3\plus{}ax^2\plus{}bx\plus{}d)^2dx\}$.

what about the scalar product on $ \mathbb{R}^5$:
if $ A\equal{}(a_4, a_3, \ldots, a_0), B\equal{}(b_4, b_3, \ldots, b_0)$ then
$ f_A(x) \equal{} a_4 x^3 \plus{} a_3 x^2 \plus{} a_2 x \plus{} a_1$
$ g_A(x) \equal{} a_4 x^3 \plus{} a_3 x^2 \plus{} a_2 x \plus{} a_0$
and the scalar product is defined by:
$ (A,B) \equal{} \int_{\minus{}1}^{0} f_A(x) f_B(x) dx \plus{} \int_{0}^{1} g_A(x) g_B(x) dx$

Are you sure the problem is $ a,b,c,d\in \mathbb{C}$?  It seems if complex coefficients are allowed a minimum shouldn't exist.  For example if $ a \equal{} b \equal{} 0$ and $ c \equal{} d \equal{} iN$, then the integral is $ \frac {2}{7} \minus{} 2N^2$.  If on the other hand $ a,b,c,d\in\mathbb{R}$ then you can find the extrema by differentiating; it's just a matter of solving a system of four linear equations.  The solution is $ a \equal{} 0,b \equal{} \minus{} \frac {9}{10},c \equal{} \minus{} \frac {1}{5},d \equal{} \frac {1}{5}$ at which point the integral has value $ \frac {9}{350}$.

Well, it is written $ a,b,c,d \in \mathbb{C}$, but it could be a mistake... I was confused with that to, because it is a functional analysis problem (Hilbert's spaces), ...
I thought it would have something to do with scalar product and norms like alekk (thanks alekk :thumbup:) has written (let suppose that is the way that problem should be resolved), and I have seen it like a $ \min_{a,b,c,d \in \mathbb{C}} \{\int_{ \minus{} 1}^0 \parallel{}f\parallel{}^2dx \plus{} \int_0^1\parallel{}g\parallel{}^2dx\}$ and the norm is a $ L^2$ norm, but it's just another representation and at the end it could be that solution is what you have posted.
Thanks for the calculations JoeBlow.  :thumbup:


a1,a2,...,an,m are reel numbers. m is the largest.

for n=2k
a1(m-a2)+a2(m-a3)+...+an(m-a1)<=(small and equal)(nm^2)/2

for n=2k+1
a1(m-a2)+a2(m-a3)+...+an(m-a1)<=[(n-1)m^2]/2

               Are they true? (Im sorry i don't know the latex. If you tell me i will be happy.)

I'm really sorry for the late reply, as the solution was rather hard to formulate. Note that you forgot to require $k\neq 0$ in the $n=2k+1$ case.

[color=blue][b]Problem.[/b] Let $n\in\mathbb N$.

[b](a)[/b] If m is a nonnegative real number, and $a_1$, $a_2$, ..., $a_n$ are n reals in the interval [0; m], then $\sum_{i=1}^n a_i\left(m-a_{i+1}\right)\leq\frac{nm^2}{2}$, where indices are modulo n (so that $a_{n+1}=a_1$).

[b](b)[/b] If, additionally, the integer n is odd and $> 1$, then the stronger inequality $\sum_{i=1}^n a_i\left(m-a_{i+1}\right)\leq\frac{\left(n-1\right)m^2}{2}$ holds.[/color]

[i]Solution.[/i] Denote $p=\frac{m}{2}$, and $k_i=a_i-p$ for every i. Then, since the number $a_i$ lies in the interval [0; m], we have $0\leq a_i\leq m$; since $p=\frac{m}{2}$ and thus $m = 2p$, this becomes $0\leq a_i\leq 2p$. Hence, $0-p\leq a_i-p\leq 2p-p$; in other words, $-p\leq k_i\leq p$. Hence, $\left|k_i\right|\leq p$.

[b](a)[/b] Now, $k_i=a_i-p$ yields $a_i=k_i+p$ and similarly $a_{i+1}=k_{i+1}+p$, so that $\underbrace{m}_{=2p}-\underbrace{a_{i+1}}_{=k_{i+1}+p}=2p-\left(k_{i+1}+p\right)=p-k_{i+1}$, so that

$\sum_{i=1}^n \underbrace{a_i}_{=k_i+p}\underbrace{\left(m-a_{i+1}\right)}_{=p-k_{i+1}}=\sum_{i=1}^n\left(k_i+p\right)\left(p-k_{i+1}\right)$
$=\sum_{i=1}^n\left(p^2-k_ik_{i+1}+k_ip-k_{i+1}p\right) =np^2-\sum_{i=1}^n k_ik_{i+1}$.

Now, $\left|k_i\right|\leq p$ and $\left|k_{i+1}\right|\leq p$ yield $k_ik_{i+1}\geq -\left|k_ik_{i+1}\right|=-\underbrace{\left|k_i\right|}_{\leq p}\cdot\underbrace{\left|k_{i+1}\right|}_{\leq p}\geq -p\cdot p=-p^2$, so that $\sum_{i=1}^n k_ik_{i+1}\geq\sum_{i=1}^n\left(-p^2\right)=-np^2$, and thus

$\sum_{i=1}^n a_i\left(m-a_{i+1}\right)=np^2-\underbrace{\sum_{i=1}^n k_ik_{i+1}}_{\geq -np^2}\leq np^2-\left(-np^2\right)$
$=2np^2=2n\left(\frac{m}{2}\right)^2=\frac{nm^2}{2}$,

and part [b](a)[/b] of the problem is solved.

[b](b)[/b] Let $n$ be an odd integer $> 1$. Then, $n\geq 3$. As in the solution of part [b](a)[/b], we note that $\sum_{i=1}^n a_i\left(m-a_{i+1}\right)=np^2-\sum_{i=1}^n k_ik_{i+1}$. Now, we are going to prove the auxiliary fact that

[b](1)[/b] if $k_1$, $k_2$, ..., $k_n$ are n reals such that $\left|k_i\right|\leq p$ for every i, then $\sum_{i=1}^n k_ik_{i+1}\geq -\left(n-2\right)p^2$.

Well, first assume that for every index i such that $1\leq i\leq n$, we have $k_ik_{i+1}<0$. Then, $\prod_{i=1}^n\left(k_ik_{i+1}\right)$ must also be < 0 (as the product of an odd number of terms which are all < 0; in fact, here we have used the condition that n is odd); but this is impossible, since $\prod_{i=1}^n\left(k_ik_{i+1}\right)=\left(\prod_{i=1}^n k_i\right)^2\geq 0$ (as squares are always $\geq 0$). Hence, our assumption that $k_ik_{i+1}<0$ for every index i was wrong, and there must exist an index u such that $k_uk_{u+1}\geq 0$. We can WLOG set u = 2 (else, just cyclically transpose the $k_i$'s); so we have $k_2k_3\geq 0$. This yields either $k_2\geq 0$ and $k_3\geq 0$, or $k_2\leq 0$ and $k_3\leq 0$. We can WLOG assume that $k_2\geq 0$ and $k_3\geq 0$ (in fact, in the case when $k_2\leq 0$ and $k_3\leq 0$, just replace each $k_i$ by $-k_i$; this obviously leaves the assertion [b](1)[/b] which we are proving invariant). Then, since $\left|k_1\right|\leq p$, we have $k_1\geq -p$, so that $k_1k_2\geq -pk_2$ (since $k_2\geq 0$). Also, since $\left|k_4\right|\leq p$, we have $k_4\geq -p$, so that $k_3k_4\geq k_3\left(-p\right)$ (since $k_3\geq 0$). Finally, for every index i, the inequalities $\left|k_i\right|\leq p$ and $\left|k_{i+1}\right|\leq p$ yield $k_ik_{i+1}\geq -\left|k_ik_{i+1}\right|=-\underbrace{\left|k_i\right|}_{\leq p}\cdot\underbrace{\left|k_{i+1}\right|}_{\leq p}\geq -p\cdot p=-p^2$. Thus,

$\sum_{i=1}^n k_ik_{i+1}=\underbrace{k_1k_2}_{\geq -pk_2}+k_2k_3+\underbrace{k_3k_4}_{\geq k_3\left(-p\right)}+\sum_{i=4}^n \underbrace{k_ik_{i+1}}_{\geq -p^2}$
$\geq \left(-pk_2\right)+k_2k_3+k_3\left(-p\right)+\sum_{i=4}^n\left(-p^2\right)=-pk_2+k_2k_3-pk_3-\left(n-3\right)p^2$
$=p^2-pk_2+k_2k_3-pk_3-\left(n-2\right)p^2=\left(p-k_2\right)\left(p-k_3\right)-\left(n-2\right)p^2$.

But $\left|k_2\right|\leq p$ yields $k_2\leq p$, so that $p-k_2\geq 0$, and similarly $p-k_3\geq 0$; thus, $\left(p-k_2\right)\left(p-k_3\right)\geq 0$, so that this yields $\sum_{i=1}^n k_ik_{i+1}\geq -\left(n-2\right)p^2$, and the assertion [b](1)[/b] is proven.

Now, using this assertion, we can finish the solution of the problem, as $\sum_{i=1}^n k_ik_{i+1}\geq -\left(n-2\right)p^2$ yields

$\sum_{i=1}^n a_i\left(m-a_{i+1}\right)=np^2-\underbrace{\sum_{i=1}^n k_ik_{i+1}}_{\geq -\left(n-2\right)p^2}\leq np^2-\left(-\left(n-2\right)p^2\right)$
$=2\left(n-1\right)p^2=2\left(n-1\right)\left(\frac{m}{2}\right)^2=\frac{\left(n-1\right)m^2}{2}$.

This solves part [b](b)[/b] of the problem.

  Darij


What is the sum of the integers that are both greater than 3 and less than 12?