geometry parallelogram rectangle search circumcircle ratio trapezoid

[b][color=DarkBlue]A triangle $\vartriangle ABC$ is given and let $ABDE,\ ACFH,\ BCKL$ be three similar rectangular parallelograms, erected outwardly to it, on sidelines $AB,\ AC,\ BC,$ respectively. We denote:
- As $A'$, the point of intersection  of the lines $EK,\ HL$.
- As $B'$, the point of intersection  of the lines $HL,\ DF$.
- As $C'$, the  point of intersection of the lines $DF,\ EK$.
Prove that the lines $AA',\ BB',\ CC',$ are concurrent at the centroid of the triangle $\vartriangle A'B'C'$[/color][/b].

Kostas Vittas.

[quote="vittasko"][color=darkred]A triangle $ABC$ is given and let $ABDE$ ,$ACFH$ , $BCKL$ be three similar rectangles
erected outwardly to it, on sides $AB$ , $AC$ , $BC$ respectively. We denote:

- As $A'$- the intersection point of the lines $EK$ , $HL\ .$
- As $B'$- the intersection point of the lines $HL$ , $DF\ .$
- As $C'$- the intersection point of the lines $DF$ , $EK\ .$

Prove that the lines $AA'$ , $BB'$ , $CC'$ are concurrently in the centroid of the triangle $A'B'C'\ .$[/color][/quote]

I have two questions : "rectangular parallelogram" means "rectangle" ? ; this problem is openly ?

[color=darkblue][b]I succeded prove [u]only two interesting properties[/u].[/b]

Denote the middlepoints $X,Y,Z,U,V,W$ of the segments $[DF]$ , $[EK]$ , $[HL]$ , $[EH]$ , $[DL]$ , $[FK]$ respectively. Then :

$P1.\blacktriangleright\{\begin{array}{ccc}XB=XC & \mathrm{\ ,\ }& XE=XH\\\\ YC=YA & \mathrm{\ ,\ }& YD=YL\\\\ ZA=ZB & \mathrm{\ ,\ }& ZF=ZK\end{array}$

$P2.\blacktriangleright$ $\{\begin{array}{ccc}UA\perp BC & \mathrm{\ ,\ }& 2\cdot UA=BL\\\\ VB\perp CA & \mathrm{\ ,\ }& 2\cdot VB=CF\\\\ WC\perp AB & \mathrm{\ ,\ }& 2\cdot WC=AE\end{array}$

[b]Remark.[/b] I"ll be continued to search another properties and maybe ... the conclusion of this problem.[/color]

Yes, ''rectangular parallelogram'' means ''rectangle''.

I don’t know if, finally, this is an open problem (as a friend of mine in Hyacinthos forum, Francois Rideau said, ''nothing is new under the geometry sun''...), but I have not found somewhere any reference about it (it is true, that I am not expert in references and I think that we need the Darij’s help).

This problem has been arose when I was studying the more general configuration when the rectangular parallelograms ABDE, ACFH, BCKL, are arbitrary ones and it is also true the result that the lines AA', BB', CC', are concurrent at one point.

A relative interesting work by Nikos Dergiades and floor van Lamoen, has already been posted in Forum Geometricorum.

You can see at http://forumgeom.fau.edu/FG2003volume3/FG200316index.html

Thank you very much for your interest about this problem and I am eager to see another (because your approach seems different than my), surely interesting proof by your self.

Kostas Vittas.

PS. Some times the schema of a configuration help (or not help) us to solve a problem. Any way in my drawing is AB = 4.7, AC = 5.3,
 BC = 3.7, BD = 3.5.

In this configuration of similar rectangular parallelograms, is also true some of the results, well known in the configuration of squares, as Vecten’s theorem states.

So, we can prove here, that the altitude of ABC through vertex A, passes through the midpoint of the segment EH (as you mentioned). Also the median through A, is perpendicular to the EH. Hence, the quadrilateral AMXU, where M is the midpoint of BC, is a parallelogram.

Kostas Vittas.

Dear all my friends.

Although I don't like to post the solution ( if I have in mind ) of a problem proposed by my self, when there was not any discussion about it, however I don't like also to stay the problems in the forum without any solution, for a long time.

In my opinion, it is better and useful to all of us, if sometime the proposers post the solutions of their problems.

So, I intend to present the solution of this interesting ( I think ) problem I have in mind, after its birthday.

Kostas Vittas.

Dear Kostas,
I will try of find some references for this problem in my own collection of the dead belgian journal Mathesis, from which I have a long run....
best regards
28121941

Thank you dear [b][size=100]28121941[/size][/b], for your interest and I am eager for your remarks and references about this problem, because the solution i have in mind, is a difficult synthetic way, rather long.

Best regards, Kostas Vittas.

Through the vertex $ A$ of the given triangle $ \bigtriangleup ABC,$ we draw the line $ (\epsilon)$ parallel to the line segment $ DF,$ which intersects the line segments $ EK,\ HL,$ at points $ N,\ N',$ respectively.

In order to be the line segment $ AA'$ as the $ A'$-median of the triangle $ \bigtriangleup A'B'C',$ it is enough to prove that $ AN \equal{} AN'\ \ \ ,(1)$

We will prove that this result is true and doesn’t depend on the rectangle $ BCKL.$

That is, if the rectangles $ ABDE,\ ACFH,$ are similar each other and $ BCKL$ an arbitrary one, the result of $ (1)$ is always true.

So, the proposed problem is equivalent to the follows one:

[b][color=DarkBlue]EQUIVALENT PROBLEM. – A triangle $ \bigtriangleup ABC$ is given and let $ ABDE,\ ACFH$ be, two similar rectangles erected outwardly to it, on its side-segments $ AB,\ AC,$ respectively. Let also be an arbitrary rectangle $ BCKL,$ erected on the side-segment $ BC$ and we denote as $ N,\ N',$ the intersection points of the line segments $ EK,\ HL$ respectively, from the line through vertex $ A$ and parallel to $ DF.$ Prove that $ AN \equal{} AN'.$[/color][/b]

[b]PROOF.[/b] – See the schema [b]t=115173(a)[/b].

Let $ (O_{1}),\ (O_{2})$ be, the circumcircles of the similar rectangles $ ABDE,\ ACFH$ respectively and it is easy to show that the point $ Z,$ as their intersection point the other than $ A,$ lies on the line segment $ DF,$ from $ \angle AZD \plus{} \angle AZF \equal{} \angle ABD \plus{} \angle ACF \equal{} 180^{o}.$

Let be the points $ K'\equiv CK\cap EZ,\ L'\equiv BL\cap HZ$ and based on the below [b][size=100]Lemma  1[/size][/b], we have that $ K'L'\parallel BC\ \ \ ,(2)$ and also we have that $ AM \equal{} AM'\ \ \ ,(3)$ where $ M\equiv (\varepsilon)\cap EK'$ and $ M'\equiv (\varepsilon)\cap HL'.$

We draw the lines through the points $ E,\ H$ and perpendicular to $ BC,$ which intersect the line $ (\epsilon),$ at points $ P,\ P'$ respectively and based on the below [b][size=100]Lemma  2[/size][/b], we have that $ AP \equal{} AP'\ \ \ ,(4)$

We denote the points $ Q\equiv BL\cap HA,\ R\equiv CK\cap EA$ and based on the below [b][size=100]Lemma  3[/size][/b], we have that $ BQ \equal{} CR$ $ \Longrightarrow$ $ QR\parallel BC\ \ \ ,(5)$

From $ (2),$ $ (5)$ $ \Longrightarrow$ $ QR\parallel BC\parallel K'L'\parallel KL\ \ \ ,(6)$

$ \bullet$ We see now, that the pencil $ E.AMNP$ is intersected from the line segment $ RK\parallel EP$ and so, based on the below [b]Lemma  4[/b], we have that $ (A,M,N,P) \equal{} (R,K',K)\ \ \ ,(7)$

Similarly, the pencil $ H.AM'N'P'$ is intersected from the line segment $ QL\parallel HP'$ and so, we have that $ (A,M',N',P') \equal{} (Q,L',L)\ \ \ ,(8)$

But, because of $ (6),$ we conclude that $ (R,K',K) \equal{} (Q,L',L)\ \ \ ,(9)$

From $ (7),$ $ (8),$ $ (9)$ $ \Longrightarrow$ $ (A,M,N,P) \equal{} (A,M',N',P')\ \ \ ,(10)$

From$ (3),\ (4),\ (10)$ $ \Longrightarrow$ $ AN \equal{} AN'\ \ \ ,(11)$ and we are done, as for the equivalent problem.

[b]REMARK.[/b] - As a consequent result in the configuration of the equivalent problem, we can say that when $ KL$ is movable such that $ KL\parallel \equal{} BC$ and $ BL\perp BC$ ( particular case when $ KL\equiv BC$ ), then, the line segment $ AA',$ bisects the segment $ B'C',$ where $ A'\equiv EK\cap HL,\ B'\equiv HL\cap DF$ and $ C'\equiv DF\cap EK.$

Return now, into the proposed problem, we conclude that the line segments $ AA',\ BB',\ CC',$ as the medians of the triangle $ \bigtriangleup A'B'C',$ are concurrent at its centroid and the proof is completed.

$ \bullet$ This proof is dedicated to [b]Nikos Dergiades[/b].

[b][color=DarkBlue]LEMMA 1. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABE,\ \bigtriangleup ACH$ be, two similar right triangles with $ \angle BAE \equal{} \angle CAH \equal{} 90^{o}$ and $ \angle ABE \equal{} \angle ACH \equal{} \angle \omega$ erected outwardly to it, on its side-segments $ AB,\ AC,$ respectively. Let $ Z$ be, the intersection point of the circumcircles $ (O_{1}),\ (O_{2})$ of $ \bigtriangleup ABE,\ \bigtriangleup ACH$ respectively, the other that $ A.$ Through vertex $ A$ we draw the line $ (\epsilon),$ parallel to the line $ O_{1}O_{2},$ connecting the midpoints of the segments $ BE,\ CH$ respectively. We draw the lines through vertices $ C,\ B$ and perpendicular to $ BC$ and we denote as $ K',\ L'$ their intersection points, from the line segments $ EZ,\ HZ,$ respectively. Prove that $ CK' \equal{} BL'$ and $ AM \equal{} AM',$ where $ M\equiv (\epsilon)\cap EZ$ and $ M'\equiv (\epsilon)\cap HZ.$[/color][/b]

[b][color=DarkBlue]LEMMA 2. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABE,\ \bigtriangleup ACH$ be, two similar right triangles with $ \angle BAE \equal{} \angle CAH \equal{} 90^{o}$ and $ \angle ABE \equal{} \angle ACH \equal{} \angle \omega$ erected outwardly to it, on its side-segments $ AB,\ AC,$ respectively. Through vertex $ A$ we draw the line $ (\epsilon),$ parallel to the line $ O_{1}O_{2},$ connecting the midpoints of the segments $ BE,\ CH$ respectively, which intersects the lines through $ E,\ H$ and perpendicular to $ BC,$ at points $ P,\ P',$ respectively. Prove that $ AP \equal{} AP'.$[/color][/b]

[b][color=DarkBlue]LEMMA 3. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABDE,\ \bigtriangleup ACFH$ be, two arbitrary rectangles erected outwardly to it, on its side-segments $ AB,\ AC,$ respectively. We draw the lines through vertices $ B,\ C$ and perpendicular to $ BC$ and we denote as $ Q,\ R$ their intersection points, from the line segments $ HA,\ EA,$ respectively. Prove that $ QR\parallel BC.$[/color][/b]

 [b][color=DarkBlue]LEMMA 4. – A line $ (\epsilon)$ intersects the rays $ P_{a},\ P_{b},\ P_{c},\ P_{d}$ of a given pencil $ P.abcd,$ at points $ A,\ B,\ C,\ D,$ respectively. We draw the line $ (\zeta)$ through the point $ A$ and parallel to $ PD$ and let be the points $ B'\equiv (\zeta)\cap PB,\ C'\equiv (\zeta)\cap PC.$ Prove that $ (A,B,C,D) \equal{} (A,B',C')$ $ ($  the double ratio of the points $ A,\ B,\ C,\ D,$ is equal to the simple ratio of the points $ A,\ B',\ C'$  $ ).$[/color][/b]

Kostas Vittas.

PS. I will post here later the proofs of the above four [b]Lemmas[/b] I have in mind.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA 1. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABE,\ \bigtriangleup ACH$ be, two similar right triangles with $ \angle BAE \equal{} \angle CAH \equal{} 90^{o}$ and $ \angle ABE \equal{} \angle ACF \equal{} \angle \omega$ erected outwardly to it, on its side-segments $ AB,\ AC,$ respectively. Let $ Z$ be, the intersection point of the circumcircles $ (O_{1}),\ (O_{2})$ of $ \bigtriangleup ABE,\ \bigtriangleup ACH$ respectively, the other than $ A.$ Through vertex $ A$ we draw the line $ (\epsilon),$ parallel to the line $ O_{1}O_{2},$ connecting the midpoints of the segments $ BE,\ CH$ respectively. We draw the lines through vertices $ C,\ B$ and perpendicular to $ BC$ and we denote as $ K',\ L'$ their intersection points, from the line segments $ EZ,\ HZ,$ respectively. Prove that $ CK' \equal{} BL'$ and $ AM \equal{} AM',$ where $ M\equiv (\epsilon)\cap EZ$ and $ M'\equiv (\epsilon)\cap HZ.$[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] – Let be the points $ D\equiv (O_{1})\cap AO_{1},\ F\equiv (O_{2})\cap AO_{2}$ as the antidiametric ( = diametric opposite ) points of $ A,$ in the circles $ (O_{1},\ (O_{2})$ respectively and it is obvious that $ AZ\perp DF,$ from $ \angle AZD \plus{} \angle AZF \equal{} 180^{o}.$

So, because of $ (\epsilon)\parallel O_{1}O_{2}\parallel DF,$ we have that $ AZ\perp (\epsilon)$ $ ,(1)$

From the cyclic quadrilaterals $ AZBE,\ AZCH$ we have that $ \angle AZE \equal{} \angle ABE \equal{} \angle \omega \equal{} \angle ACH \equal{} \angle AZH$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ AM \equal{} AM'$ $ ,(3)$

$ \bullet$ It is easy to show that the quadrilateral $ BZCK'$ is cyclic, from $ \angle BZK' \equal{} 90^{o} \equal{} \angle BCK'$ $ ,(4)$

Similarly the quadrilateral $ BZCL'$ is also cyclic, from $ \angle CZL' \equal{} 90^{o} \equal{} \angle CBL'$ $ ,(5)$

So, we have that the points $ K',\ L'$ lie on the circuncircle $ (S)$ of the triangle $ \bigtriangleup BZC$ and because of $ BL'\perp BC$ and $ CK'\perp BC,$ we conclude that the quadrilateral $ BCK'L',$ is a rectangle.

Hence, we conclude that $ CK' \equal{} BL'$ $ ,(6)$ and the proof of the [b][size=100]Lemma  1[/size][/b] is cpmpleted.

Kostas Vittas.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA 2. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABE,\ \bigtriangleup ACH$ be, two similar right triangles with $ \angle BAE \equal{} \angle CAH \equal{} 90^{o}$ and $ \angle ABE \equal{} \angle ACF \equal{} \angle \omega$ erected outwardly to it, on its side-segments $ AB,\ AC,$ respectively. Through vertex $ A$ we draw the line $ (\epsilon),$ parallel to the line $ O_{1}O_{2},$ connecting the midpoints of the segments $ BE,\ CH$ respectively, which intersects the lines through $ E,\ H$ and perpendicular to $ BC,$ at points $ P,\ P',$ respectively. Prove that $ AP \equal{} AP'.$[/color][/size][/b][/quote]
Let $ E',\ T,\ H'$ be, the orthogonal projections of the points $ E,\ A,\ H$ respectively, on the sideline $ BC,$ of the given triangle $ \bigtriangleup ABC.$

From the cyclic quadrilaterals $ EABE',\ HACH',$ we have that $ \angle AE'B \equal{} \angle AEB \equal{} 90^{o} \minus{} \angle \omega \equal{} \angle AHC \equal{} \angle AH'C$ and so, we have that the triangle $ \bigtriangleup AE'H'$ is isosceles with $ AE' \equal{} AH$ $ \Longrightarrow$ $ TE \equal{} TE'$ $ ,(1)$

From $ (1)$ and because of $ PE'\parallel AT\parallel P'H',$ in the orthogonal trapezium $ PE'H'P',$ we conclude that $ AP \equal{} AP'$ $ ,(2)$ and the proof of the [b][size=100]Lemma  2[/size][/b] is completed.

Kostas Vittas.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA 3. – A triangle $ \bigtriangleup ABC$ is given and let $ \bigtriangleup ABDE,\ \bigtriangleup ACFH$ be, two arbitrary rectangles erected outwardly to it, on its side-segments $ AB,\ AC,$ respectively. We draw the lines through vertices $ B,\ C$ and perpendicular to $ BC$ and we denote as $ Q,\ R$ their intersection points, from the line segments $ HA,\ EA,$ respectively. Prove that $ QR\parallel BC.$[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] – From $ QA\perp AC$ and $ QB\perp BC,$ we have that the quadrilateral $ AQBC$ is cyclic.

Similarly, from $ RA\perp AB$ and $ RC\perp BC,$ we have that the quadrilateral $ ARCB$ is also cyclic.

So, we conclude that he points $ Q,\ R,$ lie on the circumcircle $ (O)$ of the given triangle $ \bigtriangleup ABC.$

Hence, the quadrilateral $ BCRQ$ is a rectangle and the proof of the [b][size=100]Lemma  3[/size][/b] is completed.

Kostas  Vittas.

[quote="vittasko"][b][color=DarkBlue]LEMMA 4. – A line $ (\varepsilon)$ intersects the rays $ Pa,\ Pb,\ Pc,\ Pd$ of a given pencil $ P.abcd,$ at points $ A,\ B,\ C,\ D,$ respectively. We draw the line $ (\zeta)$ through the point $ A$ and parallel to $ PD$ and let be the points $ B'\equiv (\zeta)\cap PB,\ C'\equiv (\zeta)\cap PC.$ Prove that $ (A,B,C,D) \equal{} (A,B',C')$ $ ($  the Double ratio of the points $ A,\ B,\ C,\ D,$ is equal to the Simple ratio of the points $ A,\ B',\ C'$  $ ).$[/color][/b][/quote]
It is enough to prove that $ (A,B,C,D) \equal{} \frac {CA}{CB}\div \frac {DA}{DB} \equal{} \frac {C'A}{C'B} \equal{} (A,B',C').$

We draw the line through the point $ B$ and parallel to $ PD,$ which intersects the rays $ Pa,\ Pc$ of the given pencil $ P.abcd,$ at points $ E,\ F,$ respectively.

From $ AC'\parallel PD,$ based on the [b]Thales theorem[/b], we have that $ \frac {CA}{CD} \equal{} \frac {C'A}{PD}\ \ \ ,(1)$

By the same way, we have also that $ \frac {CB}{CD} \equal{} \frac {FB}{PD}\ \ \ ,(2)$ and $ \frac {DA}{BA} \equal{} \frac {PD}{BE}\ \ \ ,(3)$ and $ \frac {DB}{BA} \equal{} \frac {PD}{B'A}\ \ \ ,(4)$

From $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {CA}{CB} \equal{} \frac {C'A}{FB}\ \ \ ,(5)$

From $ (3),\ (4)$ $ \Longrightarrow$ $ \frac {DA}{DB} \equal{} \frac {B'A}{BE}\ \ \ ,(6)$

From $ (5),\ (6)$ $ \Longrightarrow$ $ (A,B,C,D) \equal{} \frac {CA}{CB}\div \frac {DA}{DB} \equal{} \frac {C'A}{FB}\div \frac {B'A}{BE} \equal{} \frac {C'A}{B'A}\cdot \frac {BE}{FB}\ \ \ ,(7)$

But, from $ EF\parallel AC'$ $ \Longrightarrow$ $ \frac {BE}{FB} \equal{} \frac {B'A}{C'B'}\ \ \ ,(8)$

From $ (7),\ (8)$ $ \Longrightarrow$ $ (A,B,C,D) \equal{} \frac {C'A}{C'B'} \equal{} (A,B',C')\ \ \ ,(9)$ and the proof of the [b]Lemma  4[/b] is completed.

[b]NOTE.[/b] – If $ A_{1},\ B_{1},\ C_{1},$ are the intersection points of the rays $ Pa,\ Pb,\ Pc$ respectively, from an arbitrary line $ (\eta)$ parallel to the ray $ Pd,$ because of $ (A,\ B',\ C') \equal{} (A_{1},B_{1},C_{1}),$ from $ AC'\parallel A_{1}C_{1},$ we conclude that $(P.abcd) = (A_{1},B_{1},C_{1})$.

That is, the [b]Double ratio[/b] of the given pencil $ P.abcd,$ is equal to the [b]Simple ratio[/b] of the points $ A_{1},\ B_{1},\ C_{1},$ as the points of intersection  of its three rays, from an arbitrary line parallel to its fourth one.

Kostas Vittas.


geometry
parallelogram
analytic geometry
trigonometry
parameterization
conics
parabola

An acute angle $XAY$ and a point $P$ inside the angle are given. Construct (using a ruler and a compass) a line that passes through $P$ and intersects the rays $AX$ and $AY$ at $B$ and $C$ such that the area of the triangle $ABC$ equals $AP^2$.

[i]Greece[/i]

See the solution of a bit [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61251]nicer harder[/url] problem.  Actually, 2 solutions have been posted (we have to show some respect to armpist :rotfl:). Since this problem is simpler, I will do it again. Using a parallel projection, project the rays $(AY,\ (AP$ into rays $(AX',\ (AP',$ so that the angle $\angle Y'AX = 90^\circ$ becomes right and the ray $(AP'$ bisects this right angle. In this, a square with the side AP and area $S = AP^2$ is projected into some parallelogram with area S'. The compass and straightedge constructions of the desired parallel projection and a square with an equal area S' are described under the above link. Of course, now $S' \neq AP'^2.$

Instead of choosing the the origin at A and the positive x-, y-axes along the rays $(AX,\ (AY'$ as before, we now choose the origin at A, the positive y-axis along the ray $(AP'$ and the x-axis so that the ray $(AX$ lies in the 1st quadrant. Let $b > 0,\ c > 0$ be the y-coordinates of the points B, C', respectively. Since $\angle BAP' = \angle C'AP' = 45^\circ,$ the equations of the lines AB, AC' are $y = x,\ y = -x,$ the x-coordinates of these points are $b,\ -c,$ and $AB = b \sqrt 2,\ AC = c \sqrt 2.$ The equation of the line BC' is

$y - y_B = \frac{y_C - y_B}{x_C - x_B}\ (x - x_B),\ \ \ y - b = \frac{b - c}{b + c}\ (x - b)$

Since the angles $\angle BAP = \angle C'AP = 45^\circ,$ the area S' is equal to

$S' = \frac 1 2 (AB + AC') AP' \sin 45^\circ = \frac{\sqrt 2}{4} (b \sqrt 2 + c \sqrt 2) AP' = \frac{AP'}{2}\ (b + c)$

$b + c = \frac{2S'}{AP'} = k,\ \ c = k - b$

where k is a given constant (line segment). Substituting this to the equation of the line BC',

$y - b = \frac{2b - k}{k}\ (x - b)$

This is a family of lines depending on the parameter b given by the equation

$F(x, y, b) = k(y - b) - (2b - k)(x - b) = 0$

The equation of their envelope is obtained by eliminating the parameter b from the equations $F(x, y, b) = 0$ and

$\frac{\partial F}{\partial b} = -k - 2(x - b) + (2b - k) = 0,\ \ \ b = \frac{x + k}{2}$

Substituting this b to the equation of the line family, we get the equation of their envelope:

$y - \frac{x + k}{2} = \frac{x + k - k}{k} \left(x - \frac{x + k}{2}\right) = \frac x k \cdot \frac{x - k}{2}$

$y = \frac{x^2}{2k} + \frac k 2$

This is an equation of a parabola symmetrical with respect to the y-axis (the line AP'), with the vertex V on the y-axis at the distance $AV = \frac k 2 = \frac{S'}{AP'}$ from the origin A and with the focus F also on the y-axis at the distance $VF = \frac k 2 = AV$ from the vertex. The vertex and focus are then easily constructed. The desired line BC' is a tangent to this parabola from the point P'. The line v perpendicular to the ray $(AP'$ at the vertex V is the parabola tangent at the vertex, i.e., the pedal curve of the parabola with respect to the focus F. If a circle with diameter P'F intersects the line v at 2 points $Q_1, Q_2,$, the angles $\angle FQ_1P' = \angle FQ_2P' = 90^\circ$ are both right, hence the lines $P'Q_1, P'Q_2$ are the tangents to the parabola from the point P'. These lines meet the rays $(AX,\ (AY'$ at points $B_1,\ C_1'$ and $B_2,\ C_2',$ respectively. Knowing the points $B_1, B_2 \in (AX$, we can complete the triangles $\triangle AB_1C_1,\ \triangle AB_2C_2$ by drawing the lines $B_1P,\ B_2P$ intersecting the ray $(AY$ at $C_1,\ C_2.$


One quick follow up question to the entire tournament.

W26. A student that took the Written Test of the Columbus High School Invitational Mathematics Tournament scored a 162 on the Written Portion of the exam. Given that a student's written score starts at 35; is increased by 4 for every correct answer; decreased by 1 for every incorrect answer; and not changed for blank answers; how many questions did this student miss, and how many questions did this student leave blank (the test consists of 35 questions).

[hide] 2 blank [/hide]

35 already from test, so only 127 points left to score to make a 162. Therefore, the only way is to answer 32 correctly, 1 incorrectly, and 2 blank.

hi how was the tournament? ( assuming you attended ).  :D

[quote="drunner2007"][quote="Elemennop"]Or just -- you lose 5 points for missing one and 4 points for skipping one from the 175 max, so we have 175 - 5W - 4B = 162 aka 5W+4B = 13 in positive integers yields (1,2) as the only pos. integer solution.[/quote]

Yeah. By the way, that was your written test score   :wink: . Which did you miss, and which two questions did you leave blank?[/quote]

i posted it in the first CHS tournament thread

Has anyone taken the test that was not actually at the tournament?

[quote="Xantos C. Guin"]I just did.

(Right,Blank,Wrong) = (31,2,2) = 167 points

From what I heard, Vestavia A and B got 1000 ish and 800ish points respectively, and the 3rd and 4th place teams both got under 300 points. Is this correct?

EDIT:
(Right,Blank,Wrong) = (31,2,2) = 157 points

Obviously, I can't add/multiply correctly.[/quote]

Here's the top 10 individuals:

Ryan Muller	166
Seva Tchernov	162
Evan Zhao	161
Kevin Hu	             161
Wayne Zhu	157
Wei Yan	             147
John Reinardt	131
Liangsha Xu	125
Michael Music	120
Eric Morphis	120

Notice that the top 8 are from Vestavia.

Yeah the team results were skewed. Here are the top 3:

Team                  Written   Relay     Total     Place
Vestavia Hills A	650	350	1000	1
Vestavia Hills B	560	250	810	2
WW Academy A     323	0	323	3
(WW stands for Woodward Academy)

Rockdale was there, but they arrived late (about 20 minutes after we had already started the test), so I think that they're scores were heavily affected.

[quote="Karth"]Yea, I took it yesterday:

My score = (32, 3, 0) - (Right, Blank, Wrong) Yay!  :D

So that'd be a total of 163.[/quote]

Which 3 did you leave blank?

What about you Xantos?


A salesperson kept a mileage log of the last work week. How many miles per gallon, to the nearest tenth, did the salesperson get for the entire week?

\begin{tabular}{cccc}
Day & Miles & Gallons of & Miles per \\
 & & Gas Used & Gallon \\
Mon. & 372 & 14.8 & 25.1 \\
Tues. & 290 & 10.4 & 27.9 \\
Wed. & 542 & 17.6 & 30.8 \\
Thurs. & 601 & 21.2 & 28.3 \\
Fri. & 180 & 6.7 & 26.9
\end{tabular}

If I understood the problem correctly (which almost never happens), it's $ \frac{372\plus{}290\plus{}542\plus{}601\plus{}180}{14.8\plus{}10.4\plus{}17.6\plus{}21.2\plus{}6.7}\equal{}\frac{1985}{70.7}\approx\boxed{28.1}$.

He or she get a balance number and it s correct because is arithmetic mean equal when u go to school and ur teacher give ur avereage of all ur lessons

[quote=molocyxu]that is correct, but I don't understand why. Any help? also, why can't you just average all the miles per gallons?[/quote]

If you did it, you would need to use a weighted average, which is even more complicated


What is the simplified value of $ \frac{\sqrt{24}}{\sqrt{30}}\div\frac{\sqrt{20}}{3\sqrt{25}}$?


How many three-digit numbers satisfy the property that the middle
digit is the average of the first and the last digits?

Let the digits be $ a$, $ b$, and $ c$, where $ b$ is the middle digit and $ a$ and $ c$ are the first and last digit respectively.

$ a$ and $ c$ must have the same parity, since $ b$ needs to be an integer. There are 9 possibilities for $ a$, 5 for $ c$, and $ b$ depends on what $ a$ and $ c$ are, so just one possibility for $ b$. Thus, there are $ 9 \cdot 5 \cdot 1 \equal{} \boxed{45}$.

[quote="Python54"][quote="Alcumus"]...There are $5\cdot 5=25$ odd-odd combinations for the first and last digits. There are $4\cdot 5=20$ even-even combinations that do not use zero as the first integer... [/quote]

I don't understand how these equations 5 times 5 and 5 times 4 were created and why they work. Could I please have some help?

Thank you.[/quote]
[hide="Explanation"]For the odd-odd combination, the first number can be $1$, $3$, $5$, $7$, or $9$, and same for the second number.  That gives $5\times5$ ways.  For the even-even combination, the first number can be $2$, $4$, $6$, or $8$, but not $0$.  The second number can be $0$, $2$, $4$, $6$, or $8$, a total of $4\times5$ ways for the even-even combination.[/hide]

Solution:
[hide]Each hundred can only have 5 different combinations: take 100-199 for example. There are 5 different numbers: 111, 123, 135, 147, 159. If you do this for 200-299 you can see that there are also 5 numbers. So, since 100-999 has 9 different hundreds, 5*9 = 45.[/hide]

This is the question that was answered in the Counting&Probability 2.2 Video.


Let $ x,y,z>0$. Prove that $ \frac{xyz\left (x\plus{}y\plus{}z\plus{}\sqrt{x^2\plus{}y^2\plus{}z^2} \right )}{(x^2\plus{}y^2\plus{}z^2)\left [(x\plus{}y\plus{}z)^2\minus{}(x^2\plus{}y^2\plus{}z^2)\right ]} \le \frac{3\plus{}\sqrt{3}}{18}$

It's old inequality from Hongkong,
\[ LHS \equal{} \frac {xyz(x \plus{} y \plus{} z)}{2(x^2 \plus{} y^2 \plus{} z^2)(xy \plus{} yz \plus{} zx)} \plus{} \frac {xyz}{2\sqrt {x^2 \plus{} y^2 \plus{} z^2}(xy \plus{} yz \plus{} zx)} \le \frac{1}{6}\plus{}\frac{\sqrt {3}}{18}
\]


It is know that $ a_i$ is real, and $ a_1, ... a_n, 1$ linear dependent over rationals numbers.
Prove that $ A \equal{}\{ {(k_1, ..k_n) \plus{} m(a_1,a_2,...a_n)}\}$ isn't dense in $ R^n$, where $ k_1,..k_n, m$ is arbitrary integer numbers.

$ a_1, a_2, \ldots, 1$ being dependent over the rational numbers imply that there are fixed integers $ z_1, \ldots, z_n$, not all zero, such that $ z_1 a_1 \plus{} \ldots \plus{} z_n a_n$ is an integer.  So that $ z_1(m a_1) \plus{} \ldots z_n (m a_n)$ is also an integer for every integer $ m$.  Now pick any $ (b_1, \ldots, b_n)$ with $ z_1 b_1 \plus{} \ldots z_n b_n$ not an integer.  Then this is isolated away from $ A$.


inequalities calculus derivative function geometry trigonometry vector

Let $a,b,c \in \left[ \frac 12, 1 \right]$. Prove that \[ 2 \leq \frac{ a+b}{1+c} + \frac{ b+c}{1+a} + \frac{ c+a}{1+b} \leq 3 .  \]
[i]selected by Mircea Lascu[/i]

[hide]Let $f(a,b,c) = \frac{ a+b}{1+c} + \frac{ b+c}{1+a} + \frac{ c+a}{1+b}$

Now fix $b,c$ and note that $f(a)$ is convex, so it will be maximized at an endpoint of the given interval.  The same is true for $f(b)$ and $f(c)$.  So, $f(a,b,c)$ will be maximized for some $(a,b,c) \in \{1/2, 1\}$ (sorry for the sloppy notation there, but I think you know what I mean).  Testing these points (there are only four cases, since the expression is symmetric), we find that $f(a,b,c) \leq 3$.

Now take the derivative of the function with respect to one variable:

$f'(a) = \frac{1}{1+c} - \frac{b + c}{(1 + a)^2} + \frac{1}{1 + b} \geq 1/9 > 0$

since the expression is obviously minimized for $(a,b,c) = (1/2, 1, 1)$.  Therefore, the function is strictly increasing on the given interval, so the minimum value is given for $a = 1/2$.  Identical results hold for $f(b)$ and $f(c)$, so the minimum value of $f(a,b,c)$ is given by $(a,b,c) = (1/2, 1/2, 1/2) \Rightarrow f(a,b,c) \geq 2$.

The second part actually proves the result found in the first part as well, but since this result uses calculus, it's not preferable.[/hide]

[hide]We have:
 $2 \leq \frac{ a+b}{1+c} + \frac{ b+c}{1+a} + \frac{ c+a}{1+b} \leq 3$ $\Leftrightarrow$ $2\leq (\frac{a}{1+c}+\frac{c}{1+a})+(\frac{b}{1+a}+\frac{a}{1+b})+(\frac{c}{1+b}+\frac{b}{1+c})\leq3$.
It follows:
$\frac{a}{1+c}+\frac{c}{1+a}\geq \frac{2}{3}$ $\Leftrightarrow$ $2a^{2}+2c^2+a+c+(a-c)^2\geq2$ $\Leftrightarrow$ $2(a^2-\frac{1}{4})+2(a^2-\frac{1}{4})+(a-\frac{1}{2})+(a-\frac{1}{2})+(a-c)^2\geq 0$ and analogues.
Simultaneous:
$\frac{a}{1+c}+\frac{c}{1+a}\leq 1$ $\Leftrightarrow$ $a^{2}+c^2\leq 1+ac$ $\Leftrightarrow$ $(1-a)(1-c)+a(1-a)+c(1-c)\geq 0$ and analogues.
The conclusion is obvious.[/hide]

This nice inequality is placed very right at the grade 8 !
[b]Marius Damian[/b] - O.K. (a nice proof), but [b]Zanttrang[/b] ... no (a great proof, but it isn't for the grade 8) !
[hide="Here is a short (similar approx. 75% with Marius Damian) proof."][b]Lemma.[/b] $\{x,y\}\subset \left[\frac 12 ,1\right]\Longrightarrow$
$1.\blacktriangleright 2(1+xy)\le 3(x^2+y^2)+(x+y)\ .$
$2.\blacktriangleright x^2+y^2\le 1+xy\ .$
Indeed, if we will add the following evident inequalities $2xy\le x^2+y^2$ and $2\le2(x^2+y^2)+(x+y)$ then we obtain the first inequality and the second inequality is equivalently with the evident relation $(1-x)(1-y)+x(1-x)+y(1-y)\ge 0\ .$

[b]The proof of the proposed problem.[/b]
$S\equiv \sum \frac{a+b}{1+c}=\sum \left(\frac{a}{1+b}+\frac{b}{1+a}\right)\in [2,3]$ because :
$\frac 23\le \frac{a}{1+b}+\frac{b}{1+a}\Longleftrightarrow 2[(1+ab)+(a+b)]\le 3[(a^2+b^2)+(a+b)]\Longleftrightarrow (1).$
$\frac{a}{1+b}+\frac{b}{1+a}\le 1\Longleftrightarrow a(1+a)+b(1+b)\le (1+a)(1+b)\Longleftrightarrow (2).$[/hide]

I think that when posting problems from national Olympiads it shoud be a great idea to tell what is the level of the participants. Not just stating grade 8 (who knows what is grade 8 in Romania exept romanians :D ) but for example (no calculus)
This is very important exspecialy in geometry problems (for examle no trig, or no vectors etc.)

The seventh class (approx. 14years age).

[quote="Valentin Vornicu"]Let $a,b,c \in \left[ \frac{1}{2}, 1 \right]$. Prove that
\[2 \leq \frac{ a+b}{1+c}+\frac{ b+c}{1+a}+\frac{ c+a}{1+b}\leq 3 . \]
[i]selected by Mircea Lascu[/i][/quote]

[quote="Valentin Vornicu"]Let $ a,b,c \in \left[ \frac 12, 1 \right]$. Prove that
\[ 2 \leq \frac { a \plus{} b}{1 \plus{} c} \plus{} \frac { b \plus{} c}{1 \plus{} a} \plus{} \frac { c \plus{} a}{1 \plus{} b} \leq 3 .\]
[i]selected by Mircea Lascu[/i][/quote]
Here is my solution:
\[ \sum \frac {a \plus{} b}{1 \plus{} c} \equal{} \sum \frac {(a \plus{} b)^2}{a \plus{} b \plus{} ac \plus{} bc} \geq \frac {4(a \plus{} b \plus{} c)^2}{2(a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca)}\]

\[ \equal{} 2 \cdot \frac {(a \plus{} b \plus{} c)^2}{a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca}\]
So we only have to prove that
\[ (a \plus{} b \plus{} c)^2\geq a \plus{} b \plus{} c \plus{} ab \plus{} bc \plus{} ca \iff a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca\geq a \plus{} b \plus{} c\]

\[ \iff a(a \plus{} b \minus{} 1) \plus{} b(b \plus{} c \minus{} 1) \plus{} c(c \plus{} a \minus{} 1)\geq 0\]
Which is true since $ 1\leq a \plus{} b;b \plus{} c;c \plus{} a$.

 (EDIT: My solution to the right part was not correct so I omitted it :( :( )
[hide="PS/Edit"]Sorry for reviving this old thread, but I came to know from "Agr_94_Math" that this was Karnataka Mathematical Olympiad 2006; so I found out this solution. :) [/hide]


Find the solubility of $ \ce{PbSO4}$ ($ K_s \equal{} 1.1 \times 10^{\minus{}8}$) in a 0.1 M aqueous solution of $ \ce{KNO3}$.

[quote="Carcul"]Find the solubility of $ \ce{PbSO4}$ ($ K_s \equal{} 1.1 \times 10^{ \minus{} 8}$) in a 0.1 M aqueous solution of $ \ce{KNO3}$.[/quote]
How would $ \ce{KNO3}$ suppress or encourage solubility at $ \ce{PbSO4}$?

It is your job to explain that also. Try to think, at a molecular level, on the diferences between interactions.

[quote="kysuke"][quote="buzzer11"]But the salts have no common ions. So there will be no effect?  :?:[/quote]

Eh is there a reaction between PbSO4 and KNO3 so that changes it?[/quote]

But the double replacement happens [i]after[/i] the ions dissociate, right?  :|

The key word you are all looking for is [i]ionic strenght[/i]. Look [url=http://en.wikipedia.org/wiki/Ionic_strength]here[/url] for a start.

hmm. would this site happen to have the right ideas for the problem?

http://www.chembuddy.com/?left=pH-calculation&right=ionic-strength-activity-coefficients

1) Compute the ionic strenght (you will have to make an approximation here);

2) Find the appropriate activity coefficients;

3) Find the solubility.

Hey
can the activity coeffecient be given by the debye-huckel limiting law i.e.

$ \log{\gamma}{ \equal{} }{ \minus{} A}{z_ \plus{} }{z_ \minus{} }{I^2}$

$ \gamma$ represents the activity coeffcients

where Z_+ and Z_- gives us the respective charge on the ion of the salt....

then using this we can find the respective activity coeffeccients...

before this we have to fine ionic strength using the formula that carcul gave..

then write the equation of dissociation of the salt....Using the $ \delta {G}$ value of the reaction we can find the solubilities,,,....

I don think i am completely write though---
Plz point out my mistakes

plz reply

i want to know whether i am right or wrong

Carcul plz provide the answer

i don  know why i am ignored..... :(  :(  in this forum....

plz provide me the answer
i am asking this for the nth time
plz

the activity of soln is 0.1 ??? I dunno anything following carcul's instructions . I got this .using the first step only . the I didn't understand Valerium maximum's method. I dunno ,anything here would like to learn . Carcul plz reply.

To solve this problem, we must use activities in the Ksp expression instead of concentrations.
Since activity=activity coefficient x concentration, we can write Ksp as
Ksp= activity coefficient of Pb x activity coefficient of SO4 x [Pb2+][SO4 2-]

To find the activity coefficients we use the Debye-Huckel equation,

-logactivity coefficient= 0.51(charge)^2 (ionic strength of solution)^(1/2)/(1+3.3(effective diameter of hydrated ion in nanometers)(ionic strength of solution)^(1/2))

Sorry if that's a little hard to read, I still need to learn how to use Latex lol

Anyways, I had to look up the effective diameters of the hydrated ions in the table.  The table I looked at said the diamter of Pb2+ is .45 nm and SO42- is .40 nm.
The ionic strength of the solution is (0.1x1^2+0.1x1^2)/2=0.1 M
-logactivity coefficient of Pb2+ = 0.51(2)^2 x sqrt(0.10)/(1+3.3(0.45 nm) x sqrt(0.10))
                                          =.44
     activitiy coefficent of Pb2+ = 0.364
Similarly, the activity coefficient of SO4 2- is .351.

Therefore 1.1x10-8 = .364x.351 x [Pb2+][SO4 2-]
              8.6x10-8= s^2
              s=2.9 x 10-4 M PbSO4 is the molar solubility of PbSO4 in 0.1 M KNO3

Is there anyway of doing this without looking up the hydrated ion diameters?


Does anyone know the best constant $c$ such that for all $n$ the product of primes smaller than $n$ is at most $c^n$? Erdos proved it with $c=4$, I have an elementary solution for $c=2.96$, I think Rossen proved it for $c=2.83$? Anyone knows any advances?

Indeed Rosser and Schoenfeld have proved that $c = 2, 82590...$ is the best constant.There is equality if and only if $n=113$.

Could you give the reference please?

For Harazi : What is your elementary way for $c = 2,96$?

Pierre.

Pierre, the elementary approach is given as problem at ENS in PC*, it is a subject with Apery's proof of irationality of zeta(3).

Ok, thanks. I will try to find it.

Pierre.

This is maybe what you are looking for Pierre.


Let ABC be a triangle such that : pi/2>A \geq B \geq C>0
Prove that:
sinAtanB+sinBtanC+sinCtanA  \leq sinAtanC+sinBtanA+sinCtanB

Maybe i'm so stupid or st like that at this moment ,but so strange, when Ilet A=75,B=60 and C=45,then the inequality is wrong.I think it must be
sinAtanB+sinBtanC+sinCtanA  \leq  sinAtanC+sinBtanA+sinCtanB.
If we change sin,tg into the sides of a triangle.It will be:
(a-c)/(a 2 +c 2 -b 2 )+(b-a)/(a 2 +b 2 -c 2 )+(c-b)(b 2 +c 2 -a 2 ) \leq 0.(1)
But notice that a-c=a-b+b-c so
(1)<=>(a-b)(b-c)[(b+c)(b 2 +c 2 -a 2 )-(a+b)(a 2 +b 2 -c 2 )] \leq 0.
Thi is right because a \geq b \geq c. :)

[quote]I think it must be
sinAtanB+sinBtanC+sinCtanA  \leq  sinAtanC+sinBtanA+sinCtanB. [/quote]

You are right . I'm so sorry .


[quote="Dee Wells"]3x^2 +5 - -x^2 +1

I came up with 3x + x + 4

Is this right. Not sure if I have combined like terms correctly.[/quote]

If the problem is correct, the answer is $ 4x^2\plus{}6$...

[quote="Dee Wells"]3x^2 +5 - -x^2 +1

I came up with 3x + x + 4[/quote]
$ 3x^2\plus{}5\minus{}(\minus{}x^2)\plus{}1\equal{}(3x^2\plus{}x^2)\plus{}(5\plus{}1)\equal{}\boxed{4x^2\plus{}6}$
[code]$3x^2+5-(-x^2)+1=(3x^2+x^2)+(5+1)=\boxed{4x^2+6}$[/code]
Trying typing in your original equation with one dollar sign on each end. It'll help us help you. :wink:

[quote="Dee Wells"]3x^2 +5 - -x^2 +1

I came up with 3x + x + 4

Is this right. Not sure if I have combined like terms correctly.[/quote]
Yes, in simplified form, it's 4x^2+6. But plz put the parentheses around the -x^2 so that it's not so confusing.
Plus, this is a bit too elementary for high school math. This should probably belong in Classroom Math, under the Middle School forums.


Your teacher or contest manager should have a copy of the 2005 AMC 8 Solutions, it was included with the contest materials.

A simple listing of the correct answers, no solution methods, is at
http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2005-8a/2005amc8answers.html

Steve Dunbar
AMC Director

[quote="buzzer11"]Are the solutions out yet? If so, where are they?  :huh:[/quote]

If you want to see the solution to the specific problem, post the problem(s) in this forum so we can help you to get the answer.


geometry
geometric transformation
reflection
dilation
circumcircle
geometry unsolved

Let $ ABC$ be a triangle. Let $ D$ be a point inside the triangle and let $ A'$, $ B'$, and $ C'$ be the reflections of point $ D$ across sides $ BC$, $ AC$, and $ AB$ respectively. What is the locus of points $ D$ such that $ \triangle ABC\sim\triangle A'B'C'$?

I think this is right...

Dilate $ A'$ by a factor of $ \frac{1}{2}$ around $ D$ to get $ A''$, the foot of the perpendicular from $ D$ to $ BC$.  Define $ B''$, $ C''$ similarly.  Now $ AB''DC''$ is cyclic, so \[ m\angle{B''A''C''}\equal{}m\angle{A}\equal{}m\angle{DB''C''}\plus{}m\angle{DC''B''}\equal{}180^{\circ}\minus{}m\angle{B''DC''}\] The same relation holds with $ B''$ and $ C''$.  Therefore, $ D$ is the orthocenter of $ A''B''C''$ (Reason: $ D$ is the intersection of three loci which satisfy the aforementioned angle relationship.  These loci are circles which already intersect pairwise att he three vertices of the circle.  Therefore, they can have at most one concurrence point).  We therefore have $ BC\perp A''D\perp B''C''\implies BC\parallel B''C''$.  Similarly, $ AB\parallel A''B''$ and $ CA\parallel C''A''$.  The only such triangle $ A''B''C''$ which satisfies these conditions is the medial triangle of $ ABC$.  If $ D$ is the orthocenter of the medial triangle of $ ABC$, then it is the circumcenter of $ ABC$.

[quote="calc rulz"]Let $ D$ be inside $ \triangle ABC$ and define $ A'$, $ B'$, and $ C'$ as above. What point $ D$ maximizes $ [A'B'C']$?[/quote]

It seems that the previous message already gives the answer.
I saw similar, but harder for me, questions on this subject as problems 879 and 880 of Geometriagon at [url]http://www.polarprof.org/geometriagon[/url]

Mario Belluomo[/url]


I want to show you an interesting inequality composed by Petar Danchev /Plovdiv, Bulgaria/ - very good bulgarian mathematician.

1. Let x_i >0 (i = 1,...,n > 3) and  x_i^2>=x_(i-1)x_(i+1) for all integers i in range [2;n]. Prove the following inequality:
x_1...x_n>=(nx_1/(1/x_1+...+1/x_n))^(n/2).
When does equality holds?
Prove that it will be still true if in all inequalities above replace >= with <=.

Symbols:
x_i - variable x with lower index i
^ - power.
/ - division

P.S. The problem is correct. I checked it carefully. I still don't know Latex. It will looks very good if it is formated in latex. What do you think about the level of difficulty of the problem?

Is it not interesting? Are there some unclear details? Is it too easy or is too difficult? If no one cannot solve the problem, if you want I may send the solution ... but in bulgarian :-).

Write $ a_{1}\equal{}\frac{x_{1}}{x_{2}}$ up to $ a_{n}\equal{}\frac{x_{n}}{x_{1}}$. Given condition tells us that $ a_{1}\leq ...\leq a_{n}$ and of course $ a_{1}...a_{n}\equal{} 1$.
Also, $ x_{k}\equal{} a_{k}...a_{n}x_{1}$
Substitution gives that we have to prove
$ a_{1}a_{2}^{2}...a_{n}^{n}x_{1}^{n}\ge\left(\frac{nx_{1}^{2}}{1\plus{}\frac{1}{a_{n}}\plus{}\frac{1}{a_{n}a_{n\minus{}1}}\plus{}...\plus{}\frac{1}{a_{n}...a_{2}}}\right)^{n/2}$
Cancellation and squaring gives
$ \left(a_{1}a_{2}^{2}...a_{n}^{n}\right)^{2}\left(1\plus{}\frac{1}{a_{n}}\plus{}\frac{1}{a_{n}a_{n\minus{}1}}\plus{}...\plus{}\frac{1}{a_{n}...a_{2}}\right)^{n}\ge n^{n}$
By AMGM,
$ \left(1\plus{}\frac{1}{a_{n}}\plus{}\frac{1}{a_{n}a_{n\minus{}1}}\plus{}...\plus{}\frac{1}{a_{n}...a_{2}}\right)^{n}\ge\frac{n^{n}}{a_{1}a_{2}^{2}...a_{n}^{n}}$
So it suffices to show that
$ a_{n}^{n}...a_{1}\ge 1$
which is obviously because of the conditions $ a_{1}\leq ...\leq a_{n}$ and of course $ a_{1}...a_{n}\equal{} 1$.

Thank you! I think your solution is very beautiful! I'll post the solution from the magazine where it was published when I have time (in 1-2 days) I think it is a little "ugly" but you'll say. What do you think about the problem?

P.S. Please describe in more details why a_n^n...a_1>=1.

well,
$ a_{n}$ is among the largest of other singletons and their product is 1, so $ a_{n}\ge 1$.
$ a_{n}a_{n\minus{}1}$ is among the largest of other products of pair and their product is 1, so $ a_{n}a_{n\minus{}1}\ge 1$.
...
do the same thing for others.

The reason to ask for details was... if a_n is some "big number", and a_(n-1), ... a_1 are all fractions less than 1 such that
a_n>=a_(n-1)>=...>=a_1 and a_n...a_1=1. In this case I think if we may not be sure that the product a_n^n...a_1>=1.

You can be pretty certain about that actually. Think about it this way, $ a_{n}\ge ...\ge a_{1}$. If each of them carry equal weight in product, it's 1. Then of course, if i weigh the larger numbers more (by adding powers), the result would be larger than 1.

I think I have a better idea for proof for the asked question. Let a_n,...,a_(n-k) >=1. Then ... a_n^n...a_(n-k)^(n-k)>=(a_n...a_(n-k))^(n-k)=(1/(a_(n-k-1)...a_1)^(n-k). In this way: a_n^n...a_1>=a_(n-k-1)^(-1)...a_1^(1+k-n), but a_1,...a_(n-k-1) are less than 1. And required statement follows.
;-) Actually as I said, I think your proof is very beautiful. For completeness I provide the proof from the magazine. It is belongs to Veselin Nenkov - village Beli Osam, district Lovech, Bulgaria. I think he is also very good matematician. The proof is in bulgarian so if you have any questions, please ask.
I think - it is very good problem for math competition, because it doesn't require knowledge of some specific techniques like ABC-method, Karamata, Schur, Murhead, Mixing variables, Iensen, etc. but it requires math culture and experience. Also different strategies may be used. It is beautiful as statement and as solution.


algebra
polynomial
function
limit
algebra unsolved

Does there exist a real polynomial like $f$ ,that:

[b]If[/b] $x \in (0,1) \longrightarrow f(x)=\frac{1}{1+x}$

Na vojood nadaare chon ke age chanjomleie Q(x)=(x+1)f(x) ro dar nazar begiri baraaye har
0<x<1 meghdaaresh yeke.pas khodesh ham arz baa yeke.pas Q(-1) ham baayad 1 baashe dar haalike Q(-1)=(-1+1)f(-1)=0.
albate saade tar az inhaa ham mishe goftesh.

Yup ,i really love this idea of mine although its so easy to find it  :D

PS.It`s an English forum ,So kindly use English for a public post  :) .

PS.PS)Well At fist i wanted to post the $f(x)=\frac{x}{1+x}$ but its not different at all ,this time $Q(x)$ will be $x$ (cause Q(x)-x has infinite roots ),but i gave $\frac{1}{x}$
at last  :D .

PS.Ps.pS.another solution will be this ,$0<x<1, f(x)(x+1)=1 \longrightarrow x+1 \mid 1$  :!:  Contradiction.

Fek nemikonam in raahe halle digat doros bashe.
 :D

Can you, as lomos_lupin said, please stop that non-english posting¿ There are people that want to read what is stated there...

Just one thing: by unique continuation, every analytic function that is 'part' of a function that isn't a polynomial can't be a 'part' of a polynomial (and if the function isn't analytic when restricted to the 'part' the same holds).
So since there's a pole at $-1$, this can't be a polynomial.

[quote="lomos_lupin"]Does there exist a real polynomial like $f$,......, $x \in (0,1) \longrightarrow f(x)=\frac{1}{1+x}$[/quote]
Try with following polynomial
\[ \begin{array}{|c|} \hline \\ \displaystyle P_n(x) : = \displaystyle n\cdot \sum\limits_{k=0}^{n}{n\choose k}x^k(1-x)^{n-k} \frac{1}{n+k}\\ \\ \hline \end{array}\; .  \]
In order to justifty , I note  following :
[color=green][b]
--- $f(0)=1 ,\; f(1)=\frac{1}{2}\;$,
--- $f$  is a  decreasing function on $[0,1]$ ,
--- $(-1)^kf^{(k)}(x) >0$ on $[0,1]$ , $k\in \{0,1,...,n\} .$
[/b] [/color]
[color=red] [b]
--- $P_n(0)=1 ,\; P_n(1)=\frac{1}{2}\;$,
--- $P_n$  is a decreasing function on $[0,1]$ ,
--- $(-1)^kP_n^{(k)}(x) >0 \; , k\in \{1,2,...\}$ for  $x\in [0,1]$ and $k\in \{0,1,...,n\} .$
[/b] [/color]

[size=75] [b]
Other informations regarding  $P_n$  :
1) $P_n$ is the Bernstein polynomial (see S.N.Bernstein) corresponding to function  $f$ .
2.) $\lim\limits_{n\to \infty} P_n = f$  uniformly  on $[0,1]$ .
3) If $f$ is convex on $[0,1],$ then  $f(x)\le P_{n+1}(x) \le P_n(x)$  for  $x\in [0,1]$ .
[/b] [/size]
[color=blue] It may be said that  $P_n$ has  on $[0,1]$ the same  (graphical)  [b] shape [/b]  as $f$  ( $P_n$  [u] preserves   the shape [/u] of $f$) . [/color]

Note: There is a book devoted only to "Bernstein Polynomials" written by G.G. Lorentz (around  1950, reprinted in $\approx$  1990).
 Of interest is also the monograph "Laplace Transfor" written by D. Widder (around 1949) .


In a $n \times n$ board, the numbers from 1 to $n^{2}$ are written in the squares. A king is seated in the square labelles $n^{2}$ and from there wants to move to the square labelled 1 with the following rule: he transits between two adjacentssquares only using the shortest path (that is, he doesn't detour). From the square labelled with 1 he continues to the square labelles with a 2 in the same manner and so on until he returns to the square labelles $n^{2}$. If $N$ is the number of steps the king does, find the minimum and the maximum of $N$.

Daniel

I'm kind of confused. Wouldn't the minimum just be 2n and the maximum n^2? I don't understand the problem.

I'm not sure what the problem wants to mean...


geometry
Euler
geometric transformation
reflection
ratio
circumcircle
power of a point

[quote="Leonhard Euler"]Let the incircle of a triangle $ ABC$ touch its sides $ BC$, $ CA$, $ AB$ at $ D$, $ E$, $ F$. The line $ DE$ meets the line $ AB$ at some point $ X$. The line $ DF$ meets the line $ CA$ at some point $ Y$. Denote the Feuerbach point of triangle $ ABC$ by $ G$. Prove that the points $ G$, $ D$, $ X$, $ Y$ lie on one circle.[/quote]

 Sorry , but is the Feurbach point the so called Georgone point( In some countries some points have different names)? The problem is very challenging!

[u] Babis[/u]

No, he means the actual Feuerbach point, i. e. the point where the incircle and the nine-point circle of triangle ABC touch each other.

I don't have a solution as for now...

  Darij

[quote="Leonhard Euler"]Let the incircle of a triangle $ ABC$ touch its sides $ BC$, $ CA$, $ AB$ at $ D$, $ E$, $ F$. The line $ DE$ meets the line $ AB$ at some point $ X$. The line $ DF$ meets the line $ CA$ at some point $ Y$. Denote the Feuerbach point of triangle $ ABC$ by $ G$. Prove that the points $ G$, $ D$, $ X$, $ Y$ lie on one circle.[/quote]

Lema 1
In  triangle $ ABC$, Let $ P,Q$ be point on  $ AB,AC$ such that $ BC \equal{} BQ \equal{} CP$. and $ X,Y$ is on $ AB,AC$ such $ X,Y,P,Q$ is cyclic. then $ XY$ is perpendicular to euler line of $ ABC$.

This Lema's converse is  http://www.mathlinks.ro/viewtopic.php?t=158361.

Lema 2
In this problem Let $ I,J$ be a point that reflection of $ G$ to $ DX, DY$. Thne $ IJ$ is Euler line of trianlge $ DEF$.

Lema 3
Let $ A,B,C,D$ be a four point. Intersection of $ AB$ and $ CD$ is $ X$. then the line  joinging  midpoint of $ AC,DB$ is perpendicular to line joining orthocenter of triangle $ XAD,XBC$

Let $ P,Q$ be on $ DE,DF$ such that $ FE \equal{} FP \equal{} EQ$.  Since triangle $ DFP$ is similar to $ FXE$, $ DP$x$ XE$=$ FE$x$ FP$=$ EF^{2}$ and similarly, $ DQ$x$ YF$=$ EF^{2}$.Therefore, $ DP$x$ XE$ = $ DQ$x$ YF$. Let $ K,L$ be a point on $ DX,DY$ such that $ XE \equal{} DK$,$ YF \equal{} DL$. then $ DP$x$ DK$ = $ DQ$x$ DL$. Therefore, By Lema1, $ KL$ is perpendiuclar to Euler line of tirnalge $ DEF$. Let midpoint of $ XF,YE$ be  $ S,T$. Then we can easily show that $ [DLSK] \equal{} \frac {[FYXE]}{2} \equal{} [DLTK]$,where $ [XYZ]$ means area of $ XYZ$. Therefore  $ [KSL] \equal{} [KTL]$ . So $ KL$ is parallel to $ ST$. Let $ M,H$ be orthocenter of trinangle $ DEF,DXY$. By lema3, $ MH$ is perpendicular to $ ST$. Therefore, $ KL$ is perpendicular to $ MH$. But $ KL$ is perpendicular to Euler line of triangle $ DFE$. Therefore  $ MH$ is parallel to Euler line of triangle $ DEF$. Since Euler line of trianlge $ DEF$ pass $ M, H$ lies Euler line of triangle $ DEF$. But by lema2, Euler line of trianlge DEF is $ IJ$. therefore, $ F$ lies on $ IJ$. Then, $ G$ is Anti-Steiner point of line $ IJ$ with respect to triangle $ DXY$. So $ G,D,X,Y$ is cyclic

Oh, Anti-Steiner point? What is this?

We may just make an inversion in $ D$ and then apply Menelaus theorem for a triangle $ E'F'D$, where pry means the image under inversion. Ratios on line $ DE$ and $ DF$ are easily calculated using Menelaus on initial picture, and the placement of Feurbach point's image $ G'$ may be realized using the following observation:

if $ M$ is a midpoint of $ BC$ and $ AH$ is an altitide of $ ABC$, then an Euler circle passes through $ M$ and $ H$ and touches the incircle. So, the image of Euler's circle after inversion touches line $ E'F'$, which is parallel to $ BC$, and passes through $ M'$ and $ H'$. Hence $ G'$ projection to line $ BC$ is a midpoint of $ M'H'$.

The caclulations (really not hard) are left to the reader.

proof of lemma2 is in theorem 14,15 of http://reflections.awesomemath.org/2006_2/2006_2_balticfeuer.pdf and Anti-Steiner point is 12~15page

another solution of $ ST(S,T$ is midpoint of $ XF,YE)$ is perpendicular to Euler line of $ \triangle DEF$.
Let $ U,V,W$ be midpoint of $ EF,FD,DE$. Then $ S,U,V$ are collinear.Let $ T=AU\cap BV$ be incenter. Since $ TU\cdot TA=TF^{2}=TV\cdot TB, A,U.V.B$ are cyclic.Hence $ S$ lie on radical axis of circumcircle of $ UVW$ and circumcirle $ ABC$ and similarly $ T$ lie on radical axis of circumcircle of $ UVW$ and circumcirle $ ABC$. Hence $ ST$ is perpendicular to line that joining circumcenter of $ \triangle UVW,\triangle ABC$ and this line is Eulerline of $ \triangle DEF$ since circumcenter of $ \triangle ABC$ lie on Eulerline of $ \triangle DEF$.

Here is my solution:
Let $ G$ be the intersections of $ (DXY)$ with $ (I)$. Through $ A$, let $ d$ be the line which is parallel to BC, such that it intersects $ DX,DY$, respectively, by the point $ M,N$. We will prove that $ G$ belongs to the Euler line wrt triangle $ DMN$. Indeed, let $ H_a$ be the second intersection of $ DI$ with $ (I)$. By simple angle chasing, we get the fact that $ H_a$ is the orthocenter of triangle $ DMN$. Denote $ O_a$ the circumcenter of $ (DMN)$. Invert around $ D$, power $ \overline {DF}.\overline {DM} \equal{} \overline {DE}.\overline{DN} \equal{} k^2$. We get: $ I(D,k): X\mapsto X', Y\mapsto Y'$. Therefore, $ I(D,k): (DXY)\mapsto X'Y'$, $ G$ is mapped into $ G'$. Hence,$ G'$ belongs to $ X'Y'$. But also $ G$ belongs to $ (I)$, $ I(D,k): (I)\mapsto (MN)$. Therefore $ G'$ also belongs to $ MN$.
$ I(D,k): H_a\mapsto P$. Then, $ G'P$ is perpendicular to $ PD$. $ I(D,k): O_a\mapsto O'_a$. We will prove that $ O'_aG'$ is the polar of $ O_a$ wrt $ (D,k)$, but this is true. Therefore, $ DG'$ is perpendicular to $ G'O'_a$ which leads to the result that $ (D, P, G', O'_a)$ are concyclic. Hence, $ G$ lies on the Euler line of triangle $ DMN$.
As it has been posted before on mathlink: the Feuerbach point wrt triangle $ ABC$ lies on the Euler line of triangle $ DMN$. We will get the result of the problem.
Our proof is completed. :lol:

Do you happen to know any simple proof of the fact that the orthocenter of $ DXY$ lies on $ OI$?

$ Fe$ is the Steiner point of $ OI$ with Δ$ DEF$，and $ OI$ is the Orthocenter axis of Complete quadrilateral $ EFXY$ .
So $ Fe$ is the Miquel point of Complete quadrilateral $ EFXY$ .  From this way,we get $ Fe,D,X,Y$ are lie on one circle .

Is it true that the orthocenter of triangle $ DXY$ lies on $ OI$?? New to me!! Why do you think so [b]pohoatza[/b]? :blush:

Dear mathlinkers,

see also
http://perso.orange.fr/jl.ayme   vol. 13 Un cercle passant par Fe

Sincerely
jean-Louis


limit
integration
function
calculus
inequalities
real analysis
real analysis unsolved

$f_{k}(x) \in L^{2}[0,1]$ ,if series $\sum_{k=0}^{\infty}f_{k}(x)$ converges to $g(x)$ in $L^{2}[0,1]$ . It is
generally not true that $\sum_{k=0}^{\infty}f_{k}(x)$ converges pointwisely to $g(x)$.

But if also know $\sum_{k=0}^{\infty}\parallel f_{k}(x)\parallel _{2}<\infty$, can we say that $\sum_{k=0}^{\infty}f_{k}(x)$ converges in pointwisely to $g(x)$ a.e.?

Consider $g_{n}(x)=\sum_{k=n}^{\infty}|f_{n}(x)|$; the $g_{n}$ are in $L^{2}$ and decrease monotonically, with limit zero in $L^{2}$. If $S_{n}(x)=\sum_{k=0}^{n-1}f_{n}(x)$ and $S(x)$ is the limit of the $S_{n}$, $|S_{n}(x)-S(x)|\le g_{n}(x)$ for each $n$ and all $x$.
The problem is reduced to showing that $g_{n}\to 0$ pointwise a.e.
By the dominated convergence theorem, $0=\lim_{n}\int |g_{n}|^{2}=\int |\lim g_{n}|^{2}$. A nonnegative function with zero integral is zero a.e., and we are done.

[quote="jmerry"]Consider $g_{n}(x)=\sum_{k=n}^{\infty}|f_{n}(x)|$; the $g_{n}$ are in $L^{2}$ and decrease monotonically, with limit zero in $L^{2}$. [/quote]

Why $g_{n}$ are in $L^{2}$ ?

[quote]The problem is reduced to showing that $g_{n}\to 0$ pointwise a.e.
By the dominated convergence theorem, $0=\lim_{n}\int |g_{n}|^{2}=\int |\lim g_{n}|^{2}$. A nonnegative function with zero integral is zero a.e., and we are done.[/quote]

why $\lim_{n}\int |g_{n}|^{2}=0$?

[url=http://en.wikipedia.org/wiki/Minkowski_inequality]Minkowski's inequality[/url] says that $\parallel \sum f_{j}\parallel_{p}\le \sum \parallel f_{j}\parallel_{p}$ for $1\le p\le\infty$. Hence $\parallel g_{n}\parallel_{2}\le \sum_{k=n}^{\infty}\parallel f_{k}\parallel_{2}$ which is finite and goes to $0$ as $n\to\infty$.


[hide][quote="shobber"]The right vertex of the $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $A$. If there exists a point $P$ such that $\angle{APO}=90^o$, then find the range of the eccentricity.[/quote]

Are you sure that $P$ can be any point, do you mean $P$ on the ellipse?

anyway, consider the locus of points P such that $\angle APO$ is right, this is a circle with opposite sides of a diameter at $A$ and $O$, so this locus has to intersect the ellipse

so find the intersection of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\left(x-\frac{a}{2}\right)^2+y^2=\left(\frac{a}{2}\right)^2$

and set the discriminant so you have multiple solutions $>0$[/hide]

The $\Delta$ is a square of an expression. It is always $\geq 0$. We will get $e=\frac{\sqrt{2}}{2}$ if $\Delta=0$. But there are infinity many cases that the ellipse tangent the circle. So what does $\displaystyle\frac{\sqrt{2}}{2}$ meas here?


Daca numarul $\ n$ este natural, calculati partea intreaga a numarului
$\ a=\sqrt{n^2+\frac{n}{3}}+\sqrt{n^2+\frac{2n}{3}}+\sqrt{n^2+n}$
     $\ Gazeta\ Matematica\ 2002$

Deci avem :

a= sqrt [n(n+1/3)] + sqrt [n(n+2/3)] + sqrt [n(n+1)]

desi e cam urata solutzia asta e cea care imi vine la ora asta :|...am-gm:

si obtzinem ca a > 3 * sqrt de ordin 6 din [n^3 (n+1/3)(n+2/3)(n+1)]

dar se observa ca paranteza de sub radical este mai mare decat n^ 6

deci : a > 3 * n <=> a>3n

Acum scriem n^2 + n/3 = n^2 + 2 * n * 1/6 + 1/36 - (1/6)^2
<=> n^2 + n/3 = (n+1/6)^2 - (1/6)^ 2

Deci sqrt (n^2 + n/3) < n + 1/6

Analog procedam si pt celelalte si obtine ca :
sqrt (n^2 + 2n/3) < n + 1/3
sqrt (n^2 +n) < n + 1

Prin insumare obtine cam a < 3n + 2

Deci a> 3n si a< 3n+2 => [a] = 3n + 1 ...nu prea am fost eu atent la calcule...si trebuie sa invatz LATEX :)

da, iese imediat cu medii. exact asa am facut si eu
uite pentru latex, ceva util : [url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url]

cred ca ma bag maine pe latex :) acum ma mai uit pe ce mai e pe aici si ma culc :) ms pt link  :blush:


I noticed the URL tag works differently in two ways comparing forum and blog.

1:

- Forum: clicking the URL-tag we get for "odd" clicks: [url]andfor"even"clicks:[/url]
- Blog: one click and we get [url][/url]

2:

- Forum: URL-tag put at the end of the post
- Blog: URL-tag at the current cursor position

I prefer the blog-style for both 1 and 2. And even better as previously in the forum when clicking on the URL-tag: first pop-up text box link and second pop-up text box link description.

Another thing that seems to cause problems (as some users don't know this): if one of the html-buttons was pressed an odd number of times when posting, then a closing one is automatically added at the end of the post.
I think, in most times this is more annoying than usefull, as rarely the closing tag is meant to be at the end. And if one mistakenly clicked the button, one is forced to click it again, otherwise the post is scrambled.


quadratics
Vieta
algebra
quadratic formula

There is only one real value $ a$ for which $ ax^{2} \plus{} \left( a\plus{}3 \right) x \plus{} a \minus{} 3 \equal{}0$ has two positive integer solutions.  What are those $ x$ values?

I don't know if this helps but plugging in the expressions that we know for a into the quadratic formula may help because you know that the discriminant has to be a perfect square in order for you to get integer solutions.

I tried that and could not get the two positive $ x$ values.

[hide]let the roots be p and q:
 by vieta's:
pq = (a-3)/a = 1-3/a
p+q = -(a+3)/a = -1-3/a

so we have p+q +2 = pq
pq-p-q+1=3
(p-1)(q-1) = 3
so since p and q are positive integers and 3 is prime, and wlog p>q: we get that (p,q) = (4,2)[/hide]


Find the minimum of the following expression:
$\ E=\sum {\sqrt[k]{a_i(1-a_{i+1})}}$, $\ i=1,n$, where $\ a_{n+1}=a_1,a_i\in\ (0,1)$, $\ k$ natural, such that  $\ k\geq\ 2$.

what do you mean by "the order of the root"?

Sorry, I'll correct it.I think now you understand. :)


Is there anywhere you can find these old tests for practice, especially in a free format accessible over the internet?
thanks in advance

This question has been asked repeatedly. Search for the topic in the forums.

You get the authentic old tests and solutions, which are covered by copyright, the old-fashioned way, by paying money (very little money, for the value offered) to the publisher. Links to buying information are in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12172]How to Prepare for the AMC Contests thread[/url] that I posted a while ago.

Good luck. Enjoy your reading.

If you don't want to spend money to buy them,
you can find old AHSME [url=http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm]here[/url] and old AIME [url=http://www.kalva.demon.co.uk/aime.html]here[/url]


[quote="pda"]A bag is filled with [b]m[/b] red balls and[b] n[/b] blue balls. A ball is drawn in the bag at random and returned to the bag, then [b]k [/b]balls of the same color are put into the bag. A second ball is then drawn. What isthe probability that the second ball drawn is red?[/quote]

[hide="solution"]The probability that he picks a red one is $ \frac{m}{m+n}$.  Then $ k$ reds are added so the probability that the second ball is red is $ \frac{m+k}{m+k+n}$.

The probability that he picks a blue one is $ \frac{n}{n+m}$.  The $ k$ blues are added so the probability that the second ball is red is $ \frac{m}{m+k+n}$.

Therefore $ \frac{m(m+k)+n\cdot m}{(m+n)(m+k+n)}=\boxed{\frac{m}{m+n}}$[/hide]


[quote="GameBot"]A fly starts at one vertex of a cube with edge length 1 inch and randomly walks along the edges of the cube. What is the number of inches in the longest distance that it could walk without retracing any edge?[/quote]

Since a cub is not hamiltonian, the fly can't walk 12 edges. a quick check shows 10 is impossible, so 8 is the best we can do, which works.

And here's the solution: Suppose the fly starts at the top right corner of the cube. Then the fly walks around the 4 edges of the top face. Then the fly goes down and walks around the 4 edges of the bottom face. 4+1+4=9


geometry
Asymptote
function
geometric transformation
rotation
circumcircle

Hi *,

I'm the author of the asymptote package "hyperbolic_geometry" that contains data structures and
methods for hyperbolic geometry in the poincare unit disk model.

As posted recently in the asymptote forum, there is a new version available which
is quite stable and usable. At this point the software is stable enough for many
basic and not that basic compass and straightedge constructions. So it's time to
spread the word.

Currently there are the following hyperbolic objects available:
-> Points
-> Lines including segments and rays
-> Triangles
-> Polygons
-> Circles

With methods for
-> Intersection calculation between: lines, rays segments and circles in any combination.
-> Construction of perpendiculars and common perpendiculars
-> Distance and angle calculation/measurement
-> Object properties requestment

There is little documentation for the math behind the code yet, but I'm working on
some tex files which may take it's time. Some comments are in the code, which should
be in an easy readable state.

The asymptote code for the library functions itself is about 2'000 lines including some comments.
Additionally there are over 4'000 lines of tests, demos, constructions, suitable for giving an overview
and learning how to use the code. All licensed under GPL v2 or above.

You can find the code together with parts of the mentioned demos on it's project page:
[url]http://raoul.koalatux.ch/sites/hyperbolic_geometry/hyperbolic_geometry.html[/url]

As the development is going on, there is my public git repository which contains even more features.

-- Greetings

Hi *,

After some more development, the version 0.8 is ready. New features are:

-> Mirror isometries: mirror any object at any line
-> Rotation isometries: rotate any object around any point
-> Construction methods for midpoints, angle bisectors, tangents, incircles, circumcircles, ...
-> Regular n-gons

A first version of an API documentation can be found at:
[url]http://raoul.koalatux.ch/sites/hyperbolic_geometry/API_users.txt[/url]

Download of the code at:
[url]http://raoul.koalatux.ch/sites/hyperbolic_geometry/hyperbolic_geometry.html[/url]

BTW: There are now about 2'800 lines of source code and over 6'500 lines of example code.

-- Greetings


let  $ a,b,c>0$
prove that :

$ \sqrt{\frac{2a}{a\plus{}b}}\plus{} \sqrt{\frac{2b}{b\plus{}c}}\plus{}\sqrt{\frac{2c}{c\plus{}a}} \le  3$

i think you want to say $ \sqrt{\frac{2a}{a\plus{}b}}\plus{}\sqrt{\frac{2b}{b\plus{}c}}\plus{}\sqrt{\frac{2c}{c\plus{}a}}\ge 3$

which is clearly nisbitt's inequality , you can prouve it by using chauchy or chybechev :)

[quote="crazyfehmy"]http://www.mathlinks.ro/viewtopic.php?p=1459066#1459066

[quote="R.Maths"]i think you want to say $ \sqrt {\frac {2a}{a \plus{} b}} \plus{} \sqrt {\frac {2b}{b \plus{} c}} \plus{} \sqrt {\frac {2c}{c \plus{} a}}\ge 3$

which is clearly nisbitt's inequality , you can prouve it by using chauchy or chybechev :)[/quote]

I think nisbitt's inequality is false :wink:  Try $ a \equal{} 2 \: , \: b \equal{} c \equal{} 1$[/quote]

no it's true for all $ a,b,c >0$ as $ mathmen$ said :)

[quote="Pain rinnegan"][quote="mathmen"]i say that

[quote="mathmen"]let  $ a,b,c > 0$
prove that :

$ \sqrt {\frac {2a}{a \plus{} b}} \plus{} \sqrt {\frac {2b}{b \plus{} c}} \plus{} \sqrt {\frac {2c}{c \plus{} a}} \le 3$[/quote][/quote]

Try to search. It's a famous inequality by Vasc.[/quote]

Yes indeed, it is a very popular problem by Vasc. But if you want to try it by yourself then try the substitution $ x\equal{}\sqrt{\frac{b}{a}},y\equal{}\sqrt{\frac{c}{b}},z\equal{}\sqrt{\frac{a}{c}}$. After that you will see that it reduces to a more vulnerable inequality :wink:


group theory
superior algebra
superior algebra unsolved

Let $ G$ be a group of order $ p^3$ which is not abelian.
(i) Show that the center has order $ p$

(ii) Let $ Z$ be the center. Show that $ G/Z\approx C\times C$, where $ C$ is a cyclic group of order $ p$.

This follows from: If $ G/Z(G)$ is cyclic, then $ G$ is abelian.


I found a game [url=http://www.sporcle.com/games/pi.php]here[/url] where you list the digits of pi. I used my brain and got 6. Then I used a website with the digits of pi and got 150. Post your comments here.

If you type too quickly, it says you made an error.  Bad for speed-typing, though you do have 6 minutes.

Wit my memory I got 115 :( . I kept typoing.

[quote="Thunder365"][quote="remy1140"]I got all 150 in like 20 seconds. come on...i know like 13 times more than this... why not just go up to 2000? :)[/quote]
Out of curiosity, do you spend your free time memorizing the digits of pi?

Yay I got 7 and Im proud! :lol:[/quote]

No. Well, I used to, like 20-30 minutes a day, but now I can't handle anymore :(

I got 111. The first 110 I had memorized and I guessed the last one :)

armed with google, i notice that the substring '1337' occurs 3 times in the first 10000 digits of pi
now, on topic, i know 10 digits. yay

I got 106.
I ran out of mistakes from typoing and guessing the last few  :oops:

w00t phailures. actually more like ppl you dont spend time doing that random stuff. I got, off the top of my head, 10.

[quote="Thunder365"][quote="louiethechewy123"]58 YES!!!!!  :lol:  I know somebody who recited 200 digits when they were very young.[/quote]

Lol I know someone in 6th grade who can go to 400-something. He practices for an hour a day.[/quote]

I actually know someone who knew one digit until third grade, 3 digits until 7th grade, and started memorizing more digits and got up to thousands in 8th grade. Of course, memorizing pi is really "no life". People always have something better to do, like [b]math[/b], or if that's not that awesome - hobbies or sports.

I got 93 digits--what I can normally memorize up to.  I memorized up to 107--then forgot  :(

Btw, how many digits of pi have been discovered?

13 yay
interesting stats
do you think there eventually is a pattern or end to pi?

I saw a link in google that has about 40 billion digits, but I'm sure there's lots and lots more. I doubt the computers have found more than a trillion, though.

There are much more than 1 trillion digits calculated by computers.  Also, the link in google has 40 [b]million[/b], not billion.

[quote="james4l"]It's been proven irrational, in fact transcendental, so no.[/quote]

Much more interestingly, is it possible that eventually the number of occurrences of the string "1337" will be much larger than the number of occurrences on the string "1111" in pi? The answer is no if pi is normal, which is an open problem.

[quote="remy1140"]Is there a game of memorizing the euler's number? It would be so cooler than pi, because there's too many people who are memorizing digits of pi, which makes it a noob number.  :wink:  Plus, the [b]e[/b] starts out like "2.718281828..."[/quote]

And your avatar is mostly pi, and your sig is... Do you consider it now as a noob number?  :P

[quote="remy1140"]Is there a game of memorizing the euler's number? It would be so cooler than pi, because there's too many people who are memorizing digits of pi, which makes it a noob number.  :wink:  Plus, the [b]e[/b] starts out like "2.718281828..."[/quote]

The concept of "pi" is introduced much earlier in school than "e", so naturally teachers will have pi-memorizing contests before e-memorizing contests (if they have them at all!)

I'm pretty sure Japanese scientists have calculated over 1.2 trillion digits of pi! Pretty impressive. I can only get 10, through a mnemonic device.  I take the number of letters in each word in the pair of sentences "May I have a large conainer of coffee?  Thank you."  which works out to 3.141592653.......after that is just random guesses :(

Memorized a poem for English class that encodes several thousand digits of pi. Only thing is I only memorized the first 5% (about) of it.

Well, $ e$ can be confused with Euler's constant also. :P

1,241,100,000,000 digits have been discovered, according to this: http://wiki.answers.com/Q/How_many_digits_of_pi_have_been_discovered

I got 26. Great Achievement for me!!! :lol:

I got 15 using my brain the first time, then I got to 38 after playig it for like half an hour.  :P


Oπως δοθηκε....(made in  taiwan...)
Για τους μη αρνητικους πραγματικους αριθμους a,b,c,d με ab+bc+cd+da=1
ndo
a^3/(b+c+d)+b^3/(a+c+d)+c^3/(a+b+d)+d^3/(a+b+c)>=1/3
[hide]πολ/ζω με a,,b,c,d αντιστοιχα το καθε κλασμα...
Εφαρμοζω caychy-shwratz....
Χρησιμοποιω την 4(a^2+b^2+c^2+d^2)>=(a+b+c+d)^2
και προκυπτει οτι αρκει νδο 3(a^2+b^2+c^2+d^2)>=2(ab+ac+ad+bc+bd+cd)
που ισχυει....
Το προβλημα ειναι οτι σε ολο αυτο δεν χρησιμοποιησα πουθενα το ab+bc+cd+da=1.....
Ειναι αναγκαιο να δινετε??
Ο μονος λογος που καταλαβα οτι δινεται ειναι οτι θα πρεπει a+b+c diaforetiko του 0 το οποιο προκυπτει, γιατι αν a=b=c=0 tote 0=1 ,atopo
Παρολα αυτα θα μπορουσε να λεει a,b,c,d θετικοι πραγματικοι και να μην δινει καμια συνθηκη,απο το να λεει a,b,c,d μη αρνητικοι και ab+bc+cd+da=1.....Περιμενω  την γνωμη σας...[/hide]

 :D  :D  :D

Κάτι δεν μου πάει καλά στη λύση σου. :maybe:
Από Cauchy $LHS\geq\frac{(a^{2}+b^{2}+c^{2}+d^{2})^{2}}{2\sum_{sym}ab }$
Μετά αν χρησιμοποιήσεις την ανισότητα που λες αρκεί νδο
$\frac{(a+b+c+d)^{4}}{32(\sum_{sym}ab }\geq\frac{1}{3}$

an δεν κάνω και εγώ λάθος....... :maybe:


[hide="hint"]
$ \frac{\minus{}1}{2}\plus{}i\frac{\sqrt{3}}{2}\equal{}\ e^{\frac{2i\pi}{3}}$

$ \frac{\minus{}1}{2}\minus{}i\frac{\sqrt{3}}{2}\equal{}\ e^{\frac{\minus{}2i\pi}{3}}$
apply demoivre's theorem
[/hide]

Let $ \varepsilon\equal{}e^{i\frac{2\pi}{3}}$ be the third root of unity.  Clearly, $ \varepsilon\equal{}\minus{}\frac{1}{2}\plus{}i\frac{\sqrt{3}}{2}$ and $ \varepsilon^{2}\equal{}\minus{}\frac{1}{2}\minus{}i\frac{\sqrt{3}}{2}$. Since $ 1\plus{}\varepsilon\plus{}\varepsilon^{2}\equal{}0$,  $ \varepsilon^{10}\plus{}\varepsilon^{20}\equal{}\varepsilon\plus{}\varepsilon^{2}\equal{}\minus{}1$.


A plane does not pass through the vertex of a tetrahedron. How many edges can it intersect?

[hide="EDIT Thought it Said Cube"]
I found $ 6$ sides that a plane can intersect.
[/hide]

[quote="AstroPhys"]
I found $ 6$ sides that a plane can intersect.
[/quote]
[hide=" A tetrahedron has..."]
$ 4$ sides, with $ 6$ edges ; http://en.wikipedia.org/wiki/Tetrahedron
I think it's impossible to hit all $ 6$ edges with one plane. The most sides that a plane can intersect would be $ 4$ also.[/hide]

[b]EDIT:[/b] So, the question is, is it regular?


Bonjour les français et les autres francophones!

Je m'excuse si j'écris beaucoup d'erreurs, je suis Flamand et je trouve l'anglais plus facile.

Je me demandais s'il y a beacoup de gens ici qui aiment jouer le jeu d'échec.  Il y a beacoup d'années, j'ai vu un film français avec un petit homme qui était merveilleux quand à jour aux echécs.  Il était amoureux d'une jeune femme riche et il participait au concours pour le marrier. Il y avait un autre homme très bien au jeu et il devait lui vaincre pour pouvoir se marrier avec la femme.  Il employait une technique bizarre : il semblait toucher une pièce par accident, puis l'autre pense qu'il peut en profiter, mais à la fin c'est lui qui en gagne.  Après tout ça, il n'osait pas là marrier.
La film se termine le matin plus tard.  Il est tôt et il jou à l'extérieur contre soi-même.  Il joue avec des pièces très grandes.  Quand le jeu se termine il crie :"Echec!".. et il saute, donc il se suicide.

Je le trouvais très impressionant, mais je n'ai jamais pu le retrouver ou décrouvir le titre.  Il y a quelqu'un ici qui peut me aider?

Je m'excuse encore une fois pour mon français, si vous lisez une erreur inacceptable, s'il vous plaît, dites-le moi!

N'y a-t-il pas d'amateurs des films ici? :?   Je serais très content, je cherche depuis 2000!  :(

[quote="fredbel6"]Bonjour les français et les autres francophones!

Je m'excuse si j[b]e fais[/b] beaucoup d'erreurs, je suis Flamand et je trouve l'anglais plus facile.

Je me demandais s'il y a bea[b]u[/b]coup de gens ici qui aiment jouer [b]au[/b] jeu d'échec[b]s[/b].  Il y a bea[b]u[/b]coup d'années, j'ai vu un film français avec un petit homme qui était merveilleux aux echécs.  Il était amoureux d'une jeune femme riche et il participait au concours pour [b]se[/b] marier. Il y avait un autre homme très [b]bon [/b]au jeu et il devait l[b]e[/b] vaincre pour pouvoir se marier avec la femme.  Il employait une technique bizarre : il semblait toucher une pièce par accident, puis l'autre pense qu'il peut en profiter, mais à la fin c'est lui qui gagne.  Après tout ça, il n'osait pas [b]se[/b] marier [b]avec elle.[/b]
La film se termine le matin [b]suivant[/b].  Il est tôt et il jou[b]e [/b]à l'extérieur contre [b]lui[/b]-même.  Il joue avec des pièces très grandes.  Quand le jeu se termine il crie :"Echec!".. et il saute, donc il se suicide.

Je le trouvais très impressionant, mais je n'ai jamais pu le retrouver ou décrouvir le titre.  Il y a quelqu'un ici qui peut m[b]'[/b]aider?

Je m'excuse encore une fois pour mon français, si vous lisez une erreur inacceptable, s'il vous plaît, dites-le moi![/quote]

Bonjour fredbel6 !

Je viens de lire ton message et, comme j'aime aussi les échecs, je me suis intéressé sur Google et... je suis tombé sur [b]La partie d'échecs[/b], film de 1994 avec Catherine Deneuve et Pierre Richard. Je n'ai pas vu le film mais d'après ta description ça pourrait que ce soit celui que tu cherches ! :huh:

Oui, je pense que c'est ce film. :D   Je vous remercie, j'ai cherché pour plus que 5 années!

[quote="fredbel6"] j'ai cherché pour plus que 5 années![/quote]

Salut ! Je suis content de t'avoir aidé ! et...correction (comme tu le veux !):
on dit plutôt "[b]je cherche depuis plus de 5 ans[/b]"
Et...encore que je ne suis pas très sûr, vu que ma langue maternelle n'est pas le français ! mais j'habite en France depuis...15 ans !
A + ! :roll:


N is an integer greater than 5, and when 205 is divided by N, the remainder is 5. For how many distinct positive integers N is this statement true?

[hide] We must have that $ N$ is a factor of $ 200$. There are $ 4(3)\equal{}12$ factors of $ 200$ and since $ 4$ of these are less than or equal to $ 5$, the answer is $ 12\minus{}4\equal{}\boxed{8}$.[/hide]


geometry
geometric transformation
rotation
function
limit
real analysis
real analysis unsolved

Let $f: \mathbb{R}\to \mathbb{R}$ be a continuous strictly increasing function such that $f(x+1)=f(x)+1$. Prove that for every $x \in \mathbb{R}$,
$\lim_{n \to \infty}\frac{f^{(n)}(x)}{n}$
exists and is independant of $x$. (here $f^{(n)}(x)=f(f( \ldots (x) \ldots ))$)

Some other facts: what can be said if this is a rational number? (I think the discussion is much much more difficult for irrational numbers).


AMC USA(J)MO USAMO geometry probability trigonometry search

What books would you recommend to make MOSP?
I have Art of Problem Solving Volumes One and Two already.

Not sure if I entirely agree with that... "natural abilities" help, but if you do lots of problems and learn the right way from them, you don't need to be a genius.

If making MOP is your main goal, note that the qualifying test is the USAMO. Do as many past tests as you can. AOPS one and two should have most of the background you need. This site and [url=http://www.kalva.demon.co.uk/]Kalva[/url] have all of the past USAMOs, most with solutions.

I wouldn't really recommend books, except for Geometry Revisited if you have trouble with olympiad geometry problems. Once you have a modest background in facts and theorems, the USAMO is accessible, although other olympiads like the IMO emphasize knowledge more.

[quote="JRav"]Um, I think that to make MOSP, you have to have some natural abilities.  You can't just pick up a book and after a reading or two, qualify for MOSP.[/quote]

I agree completely.  If it was so easy that anyone could do it, there'd be absolutely no meaning.  Of course, I'm not saying books won't help, just that they won't suddenly implant Einstein's brain into your head.

Okay, let me be less nice about this. I think that's an unhealthy opinion to have, and you can seriously limit your potential by believing that you have to be innately good at something to succeed, and spreading that idea doesn't help. Yes, getting into MOP is rather difficult, but as I said if you work hard and learn you don't need to be a genius to make it.

Furthermore, it's not clear what purpose you think is served by telling the original poster (about whose knowledge and abilities you know nothing) that instead of answering the question that was asked.  By the fact that $ 30 < 6000000000$, not everyone can make MOP -- the O.P. surely knew that, and was asking for advice on being one of the 30.

The USAMO is usually heavy on combinatorics; so the recent AoPS book Intermediate Counting and Probability would help. The 2007 USAMO had plenty of geometry; for that I'd use the well known Geometry Revisited book. For number theory (ick) topics, this paper by Naoki Sato: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=975012#975012 might help.

I also infinitely recommend Titu's 10x Olympiad problem books.

[quote="Temperal"] The 2007 USAMO had plenty of geometry; for that I'd use the well known Geometry Revisited book.[/quote]

Geometry Revisited would not have helped a whole lot with those particular problems, #2 was basically just law of cosines, then it turned into more of an algebra problem, and reading the inversion chapter would have helped on #6, but inversion wasn't necessary. Frankly, as good of a book Geometry Revisited is, it (rigorously) presents a wide array of theorems and concepts, but its problems generally would not appear on an olympiad. Although I haven't closely read much of the book, I believe that it should probably be supplemented by problems. Then, this comes back to doing problems.

The IMO Compendium website has thousands of olympiad problems (possibly more than that AoPS Contests section?), and the IMO Compendium book is a compilation of every ILL problem ever proposed, with solutions, I believe.

[quote]The USAMO is usually heavy on combinatorics;[/quote]

Really?

This topic has been posted before...
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=659590428&t=172265
Here's one link for you; if you want more (and, believe me, there are TONS more threads about this topic) then search  :wink:

To JRav: I seriously disagree with your idea.  "Natural ability" means virtually nothing.  How much practice you get is the main factor of how well you do on the USAMO.

[quote]Really?[/quote]
Really.
The AoPSWiki contains scarcely any USAMO problems (about 3 complete ones and maybe a few others), and it has quite a few:
[[2007 USAMO Problems/Problem 3]]
[[2006 USAMO Problems/Problem 2]]
[[2004 USAMO Problems/Problem 4]]
[[1990 USAMO Problems/Problem 1]]
[[1989 USAMO Problems/Problem 2]]
[[1990 USAMO Problems#Problem 3]]
(In fact, I probably missed some. An average of 1+ per year on a 6-problem Olympiad is pretty heavy except in the most warped definition.

[quote]
The IMO Compendium website has thousands of olympiad problems (possibly more than that AoPS Contests section?), and the IMO Compendium book is a compilation of every ILL problem ever proposed, with solutions, I believe. [/quote]

By the IMO Compendium, you mean this? http://www.imo.org.yu/

[quote]I don't think focusing on particular subjects is going to be too helpful. In 2006, there were pretty much 5 number theory (slash algebra) problems. 2007 was heavy on combinatorics, yes. Also, both of them were pretty light on geometry. But I don't think this tells you about 2008 or any other future year. For all we know the next test will compensate for the lack of geometry of the previous years. But just looking at these two years shows there's no consistent subject distribution.[/quote]

There's almost invariably quite a bit of number theory...

[quote="Temperal"][quote]Really?[/quote]
Really.
The AoPSWiki contains scarcely any USAMO problems (about 3 complete ones and maybe a few others), and it has quite a few:
[[2007 USAMO Problems/Problem 3]]
[[2006 USAMO Problems/Problem 2]]
[[2004 USAMO Problems/Problem 4]]
[[1990 USAMO Problems/Problem 1]]
[[1989 USAMO Problems/Problem 2]]
[[1990 USAMO Problems#Problem 3]]
(In fact, I probably missed some. An average of 1+ per year on a 6-problem Olympiad is pretty heavy except in the most warped definition.[/quote]

What are you talking about? There are 6 problems, covering 4 major areas (algebra, combinatorics, geometry, number theory), so on average there should be 1.5 of each per year. So why is it heavy on combinatorics if there is only 1+ every year? There was one in 2004, one in 2005, zero in 2006, and two in 2007. If the USAMO was consistently "heavy" on any particular area, then it wouldn't be very well written.

[quote]By the IMO Compendium, you mean this? http://www.imo.org.yu/[/quote]

That is the website, only with problems that actually made it onto the IMO (and other olympiads), without solutions. I was referring to the book which contains every longlisted problem with solutions.

[quote]There's almost invariably quite a bit of number theory...[/quote]

Again, this is not true at all.

While I agree in large part with the views already presented, I think that so much of this kind of advice has been repeated that perhaps it's time to play devil's advocate.

You do not need to be a "genius" to make MOP. But, as with pretty much anything, there is a large element of talent. I think it's naive to assume that anybody can make it with enough hard work. The fact is that a very talented person can often beat a less talented, but harder working, person. I've been in both positions. My freshman year, I walked into the AIME with pretty much no preparation and qualified for the USAMO. In contrast, last year I worked very hard studying for the USAMO: I read a bunch of books and did a bunch of problems, and I did not perform nearly as well as I had hoped I would. Let's face it: not all people are created equal, at least not in terms of abilities. Don't give up on your dreams without a serious effort, but also don't beat yourself up too much if you can't fulfill your expectations.

I also used to be one of those people that would preach, "Don't read books, don't read solutions, attempting problems is the best way to go." Now I'm not so sure. People always criticize the SAT because, thanks to all the preparation courses and books and whatnot, it has become more a measure of your preparation resources than your actual reasoning capabilities. I think that math contests, even olympiad ones, are moving in the same direction. Look at all the books, courses, and summer camps out there. It's really becoming a question of whether you've seen the right trick before: root flipping, double counting, roots of unity filter, "nikolai nikolov," etc. A problem like IMO 2005/6 is impenetrable if you don't know the simple counting in two ways trick and trivial if you do. I've remarked that at least one USAMO problems and both AIME #14's from last year appeared in Engel. The more you do contest math the more you realize that it's largely just a bag of tricks.

Now, to get onto some actually constructive advice:

Problem Solving through Problems (Larson)
The Art and Craft of Problem Solving (Zeitz)
Problem-Solving Strategies (Engel)
An Introduction to Diophantine Equations (Andreescu & Andrica)
A Path to Combinatorics for Undergraduates (Andreescu & Feng)

I've read all of these books (haven't finished Engel though) and they are quite good. Larson and Zeitz are geared towards novices to problem-solving. The Diophantine Equations has a marvelous collection of problems and techniques, but just be advised that Diophantine Equations don't come up all that often on the USAMO. The combinatorics book has great examples but not too many olympiad-caliber exercises.

[quote="CatalystOfNostalgia"]
What are you talking about? There are 6 problems, covering 4 major areas (algebra, combinatorics, geometry, number theory), so on average there should be 1.5 of each per year. So why is it heavy on combinatorics if there is only 1+ every year? There was one in 2004, one in 2005, zero in 2006, and two in 2007. If the USAMO was consistently "heavy" on any particular area, then it wouldn't be very well written.
[/quote]
I consider trigonometric manipulation to be a separate topic, and you also rather conveniently conveniently forgot probability, which reduces the expected number per year.

[quote]
Again, this is not true at all.[/quote]
Thanks for refuting my statement with no sources or any arguments to back it up. Such statements allow the thread to avoid silly things like "information", or even worse, "truth".

As it turns out, this is probably actually less than all of the number theory problems on the (as I have said before, depleted) AoPSWiki:
[[2002 USAMO Problems/Problem 5]]
[[2003 USAMO Problems/Problem 1]]
[[2006 USAMO Problems/Problem 1]]
[[2006 USAMO Problems/Problem 3]]
[[2006 USAMO Problems/Problem 4]]
[[2006 USAMO Problems/Problem 5]]
[[2007 USAMO Problems/Problem 1]]
[[2007 USAMO Problems/Problem 5]]
[[2005_USAMO_Problems#Problem_1]]
[[2005_USAMO_Problems#Problem_2]]

A more accurate look at which subjects are represented on the USAMO would be [url=http://en.wikipedia.org/wiki/USAMO#Exam_content]here[/url] (while as the subjects of the problems appearing on the wiki are more or less dependent upon the nature of their authors). Echoing MellowMelon, there really isn't anyway to surely predict the types of questions which will appear on the USAMO outside of the obvious, so subject-focusing wouldn't have any advantage over being well-rounded.

The way to get into MOP is to do math all the time. There isn't a particular book or study method. Just do lots and lots of math. You can find the best way for yourself to study. I like making problems.

I am not good at math at all (especially contest math which requires speed and time-management), and so I am not in a position to give advice; however, my method of doing mathematics may be of some help to you.  I 1) work on a problem and try to solve it; 2) think about connections to different fields of mathematics along with applicable tricks; 3) read the given (when solved) solution (if any); and 4) create related problems and find different solutions.  Hopefully, this helps.

I too may lack some of the results leading to perhaps a greater "respect", but I believe that I am just as qualified given the purpose of the format of our conversation, a forum, which is designed to accept a variety of opinions on topics. Regarding the whole mop things, it seems that their is a misconception regarding the ingredients to mop sucess. I feel that a lot of "talented" people often forget the work they point in. This is especially true when someone is an overachiever and only has math as an outlet to sucess. In other words, people will maximize their strengths by asserting that they don't put in much work. Undoubtedly, most of the people who made mop started somewhere lower. I assure you let less than 5% just walk into the competition an destroy it. Likely, they had some type of practice before, be it MATHCOUNTS, or something. I would bet 10000000000000 pesos that you simply cannot qualify for an olympiad without a lot of knowledge, point blank. If you think otherwise, look at an unknown olympiad problem and solve it without stewarts or w/e technique you use(ok well maybe not stewarts since its so easy to derive). Rarely if ever when I glance at olympiad problems do i understand the derivations of the most advanced techniques things like telescoping, and some of the crazy arbitrary lines(well not really I guess) drawn in geometry problems. Regarding the whole idea of probability, it seems almost incomprehensible to just come to multiple epiphanies in contest situations,especially for someone without experience. Coming from a background without enrichment, I find the advantages of years of posts unsurmountable, and indomitable. If you don't have it, you aren't very likely to be able to compete. Just look at all the mopers on aops, they probably all have thousands of posts; such little work and so much talent...not.
Despite the seeming negativity of the above paragraph, I hope it was clear that i meant to encourage people rather to discourage them. And i certainly believe that anyone, regardless of their level, would probably think along the same lines. For practice, I believe an individual should determine how they learn and tailor their practice to parallel their goals.  Like others have said before me, the individual should know where his or her strengths and weaknesses lie, and hence use some kind of zone of proximal development(If you're like me, you haven't read much in middle and high school,haven't done MATHCOUNTS, and haven't done my hw or paid attention much(and hence have different initial but perhaps not final goals and  subsequent means to them are obviously different).  Going with the zpd, post away on the forums, don't be afraid to fail.  At the same time, get an AOPS book and learn the material if you haven't already, while also applying it on actual contest problems such as the amc since optimal learning is created by understanding something in such as fundamental way as to be able to apply it in multiple situations.  And of course, as the individual progresses in skills and ingenuity, they should start doing AIME and USAMO problems, or whatever they like. It's a long and arduous process but shouldn't be work but rather a source of joy.


trigonometry
limit
algebra proposed
algebra

Let $ a_{0}\equal{}a$, $ f(x)\equal{}\sqrt{4\plus{}\sqrt{4\plus{}\sqrt{4\minus{}x}}}$, and $ a_{n\plus{}1}\equal{}f(a_{n})$. To show that this converges, show that $ f'(x)<0$, so $ a_{n}$ is decreasing. Clearly it is bounded below by $ 0$, so it converges.

First, lets assume that $ \lim_{n\rightarrow\infty}a_{n}\equal{}L$.
Then we are to solve:

$ L\equal{}f(L)$...

i) $ L\ge 0$.
ii) $ L^{2}\ge 4$.
iii) $ L^{4}\minus{}8L^{2}\plus{}12\ge 0$.

$ L^{8}\minus{}16L^{6}\plus{}88L^{4}\minus{}192L^{2}\plus{}L\plus{}140\equal{}0$.

I don't have a nice way to finish the problem...I presume that solution didn't come out of nowhere.

[quote="Altheman"]Let $ a_{0}\equal{} a$, $ f(x) \equal{}\sqrt{4\plus{}\sqrt{4\plus{}\sqrt{4\minus{}x}}}$, and $ a_{n\plus{}1}\equal{} f(a_{n})$. To show that this converges, show that $ f'(x) < 0$, so $ a_{n}$ is decreasing. Clearly it is bounded below by $ 0$, so it converges.
[/quote]
I think, your reasoning doesn't help for  convergation proof because if $ f$ is decreasing
then $ f\circ f$ is not necessarily. :wink:


Hi, here's another interesting problem:

"Find the minimum and maximum value of the multivariate function: y/(1+x^2+y^2)"

Using calculus, I was able to arrive at the solution -1/2 and 1/2, which occurs at (0,-1) and (0,1) respectively.

However, I was wondering if there was a more elegant non-calculus way of doing it. I like the idea of approaching it with AM-GM, but how can we seperate the "terms" as to get a constant on either side??

 :?:

The function can easily be reduced to a single-variable one: Let $ a\equal{}|y|$. Then
\[ \left| \frac{y}{1\plus{}x^2\plus{}y^2}\right| \equal{}\frac{a}{1\plus{}x^2\plus{}a^2}\leq \frac{a}{1\plus{}a^2}\]
The simplest way of using AM-GM on this one is to apply it to the denominator: $ 1\plus{}a^2\geq 2a$ so $ \frac{a}{1\plus{}a^2}\leq \frac{1}{2}$ since $ a$ nonnegative. There's equality when $ a\equal{}1,x\equal{}0$, and the original function is positive or negative depending on the sign of $ y$, so -1/2 and 1/2 are both values of the function.

$ \frac{x^2}{2(x^2\plus{}y^2\plus{}1)}\plus{}\frac{(y\minus{}1)^2}{2(x^2\plus{}y^2\plus{}1)}\ge 0 \iff \frac{y}{x^2\plus{}y^2\plus{}1}\le \frac{1}{2}$, likewise do that for the lower bound

Thanks. That was a very clever solution.  :lol:


[hide]Integers less than 5=4

Total possibilities=4*3*2=24

Outcomes that satisfy a<b<c= 1,2,3 , 2,3,4=2

1/12[/hide][/hide]

[quote="236factorial"]The elements of the ordered triple (a, b, c) are different positive integers less than 5. What is the probabilitiy that a<b<c? Express your answer as a common fraction.[/quote]
[hide]
There are 4 positive integers less than 5, $\displaystyle 4!=24$,
And the possibilities are:
1,2,3
1,2,4
2,3,4 and
1,3,4
$\displaystyle \frac{4}{24}=\boxed{\frac{1}{6}}$[/hide]


Determine the value of \[ \sup_{ 1 \leq \xi \leq 2} [\log E \xi\minus{}E \log \xi],\] where $ \xi$ is a random variable and $ E$ denotes expectation.

[i]Z. Daroczy[/i]

First, we have $ \log x \ge (x \minus{} 1)\log 2 \ \forall x \in [1,2]$. Hence
\[ \log E\xi \minus{} E \log \xi \le \log E\xi \minus{} (E\xi \minus{} 1)\log 2 \le \minus{} \log(\log 2) \minus{} 1 \plus{} \log 2
\]
The given supremum  attain when $ \xi$ takes only two value $ 2, 1$ and $ P(\xi \equal{} 1) \equal{} 2 \minus{} 1/\log 2$.